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Unformatted text preview: A HEAT TRANSFER HEAT THIRD TEXTBOOK EDITION John H. Lienhard IV / John H. Lienhard V A Heat Transfer Textbook Lienhard & Lienhard Phlogiston Press ISBN 0-9713835-0-2 PSB 01-04-0249 A Heat Transfer Textbook A Heat Transfer Textbook Third Edition by John H. Lienhard IV and John H. Lienhard V Phlogiston Press Cambridge Massachusetts Professor John H. Lienhard IV Department of Mechanical Engineering University of Houston 4800 Calhoun Road Houston TX 77204-4792 U.S.A. Professor John H. Lienhard V Department of Mechanical Engineering Massachusetts Institute of Technology 77 Massachusetts Avenue Cambridge MA 02139-4307 U.S.A. Copyright ©2008 by John H. Lienhard IV and John H. Lienhard V All rights reserved Please note that this material is copyrighted under U.S. Copyright Law. The authors grant you the right to download and print it for your personal use or for non-profit instructional use. Any other use, including copying, distributing or modifying the work for commercial purposes, is subject to the restrictions of U.S. Copyright Law. International copyright is subject to the Berne International Copyright Convention. The authors have used their best efforts to ensure the accuracy of the methods, equations, and data described in this book, but they do not guarantee them for any particular purpose. The authors and publisher offer no warranties or representations, nor do they accept any liabilities with respect to the use of this information. Please report any errata to the authors. Lienhard, John H., 1930– A heat transfer textbook / John H. Lienhard IV and John H. Lienhard V — 3rd ed. — Cambridge, MA : Phlogiston Press, c2008 Includes bibliographic references and index. 1. Heat—Transmission 2. Mass Transfer I. Lienhard, John H., V, 1961– II. Title TJ260.L445 2008 Published by Phlogiston Press Cambridge, Massachusetts, U.S.A. This book was typeset in Lucida Bright and Lucida New Math fonts (designed A by Bigelow & Holmes) using L TEX under the Y&Y TEX System. For updates and information, visit: http://web.mit.edu/lienhard/www/ahtt.html This copy is: Version 1.31 dated January 16, 2008 Preface This book is meant for students in their introductory heat transfer course — students who have learned calculus (through ordinary differential equations) and basic thermodynamics. We include the needed background in fluid mechanics, although students will be better off if they have had an introductory course in fluids. An integrated introductory course in thermofluid engineering should also be a sufficient background for the material here. Our major objectives in rewriting the 1987 edition have been to bring the material up to date and make it as clear as possible. We have substantially revised the coverage of thermal radiation, unsteady conduction, and mass transfer. We have replaced most of the old physical property data with the latest reference data. New correlations have been introduced for forced and natural convection and for convective boiling. The treatment of thermal resistance has been reorganized. Dozens of new problems have been added. And we have revised the treatment of turbulent heat transfer to include the use of the law of the wall. In a number of places we have rearranged material to make it flow better, and we have made many hundreds of small changes and corrections so that the text will be more comfortable and reliable. Lastly, we have eliminated Roger Eichhorn’s fine chapter on numerical analysis, since that topic is now most often covered in specialized courses on computation. This book reflects certain viewpoints that instructors and students alike should understand. The first is that ideas once learned should not be forgotten. We have thus taken care to use material from the earlier parts of the book in the parts that follow them. Two exceptions to this are Chapter 10 on thermal radiation, which may safely be taught at any point following Chapter 2, and Chapter 11 on mass transfer, which draws only on material through Chapter 8. v vi We believe that students must develop confidence in their own ability to invent means for solving problems. The examples in the text therefore do not provide complete patterns for solving the end-of-chapter problems. Students who study and absorb the text should have no unusual trouble in working the problems. The problems vary in the demand that they lay on the student, and we hope that each instructor will select those that best challenge their own students. The first three chapters form a minicourse in heat transfer, which is applied in all subsequent chapters. Students who have had a previous integrated course thermofluids may be familiar with this material, but to most students it will be new. This minicourse includes the study of heat exchangers, which can be understood with only the concept of the overall heat transfer coefficient and the first law of thermodynamics. We have consistently found that students new to the subject are greatly encouraged when they encounter a solid application of the material, such as heat exchangers, early in the course. The details of heat exchanger design obviously require an understanding of more advanced concepts — fins, entry lengths, and so forth. Such issues are best introduced after the fundamental purposes of heat exchangers are understood, and we develop their application to heat exchangers in later chapters. This book contains more material than most teachers can cover in three semester-hours or four quarter-hours of instruction. Typical onesemester coverage might include Chapters 1 through 8 (perhaps skipping some of the more specialized material in Chapters 5, 7, and 8), a bit of Chapter 9, and the first four sections of Chapter 10. We are grateful to the Dell Computer Corporation’s STAR Program, the Keck Foundation, and the M.D. Anderson Foundation for their partial support of this project. JHL IV, Houston, Texas JHL V, Cambridge, Massachusetts August 2003 Contents I The General Problem of Heat Exchange 1 Introduction 1.1 Heat transfer . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Relation of heat transfer to thermodynamics . 1.3 Modes of heat transfer . . . . . . . . . . . . . . . . . . 1.4 A look ahead . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 3 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Heat conduction concepts, thermal resistance, and the overall heat transfer coefficient 2.1 The heat diffusion equation . . . . . . . . . . . . . . . . . . . . . . . 2.2 Solutions of the heat diffusion equation . . . . . . . . . . . . . . 2.3 Thermal resistance and the electrical analogy . . . . . . . . . 2.4 Overall heat transfer coefficient, U . . . . . . . . . . . . . . . . . . 2.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Heat exchanger design 3.1 Function and configuration of heat exchangers . . . . . . . . 3.2 Evaluation of the mean temperature difference in a heat exchanger . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Heat exchanger effectiveness . . . . . . . . . . . . . . . . . . . . . . 3.4 Heat exchanger design . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 3 6 10 35 36 37 46 49 49 58 62 78 86 86 96 99 99 103 120 126 129 136 vii Contents viii II Analysis of Heat Conduction 4 Analysis of heat conduction and some steady one-dimensional problems 4.1 The well-posed problem . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 The general solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Dimensional analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 An illustration of dimensional analysis in a complex steady conduction problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Fin design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 III 6 Transient and multidimensional heat conduction 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Lumped-capacity solutions . . . . . . . . . . . . . . . . . . . . 5.3 Transient conduction in a one-dimensional slab . . . 5.4 Temperature-response charts . . . . . . . . . . . . . . . . . . 5.5 One-term solutions . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 Transient heat conduction to a semi-infinite region . 5.7 Steady multidimensional heat conduction . . . . . . . . 5.8 Transient multidimensional heat conduction . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Convective Heat Transfer 139 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 141 143 150 159 163 183 190 193 193 194 203 208 218 220 235 247 252 265 267 Laminar and turbulent boundary layers 269 6.1 Some introductory ideas . . . . . . . . . . . . . . . . . . . . . . . . . . 269 6.2 Laminar incompressible boundary layer on a flat surface 276 6.3 The energy equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292 6.4 The Prandtl number and the boundary layer thicknesses 296 6.5 Heat transfer coefficient for laminar, incompressible flow over a flat surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300 6.6 The Reynolds analogy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311 6.7 Turbulent boundary layers . . . . . . . . . . . . . . . . . . . . . . . . 313 6.8 Heat transfer in turbulent boundary layers . . . . . . . . . . . 322 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338 Contents 7 ix 341 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341 7.2 Heat transfer to and from laminar flows in pipes . . . . . . 342 7.3 Turbulent pipe flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355 7.4 Heat transfer surface viewed as a heat exchanger . . . . . . 367 7.5 Heat transfer coefficients for noncircular ducts . . . . . . . . 370 7.6 Heat transfer during cross flow over cylinders . . . . . . . . . 374 7.7 Other configurations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Forced convection in a variety of configurations 393 Natural convection in single-phase fluids and during film condensation 397 8.1 Scope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 397 8.2 The nature of the problems of film condensation and of natural convection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 398 Laminar natural convection on a vertical isothermal surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401 8.4 Natural convection in other situations . . . . . . . . . . . . . . . 416 8.5 Film condensation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 428 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 443 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452 8.3 9 Heat transfer in boiling and other phase-change configurations 457 9.1 Nukiyama’s experiment and the pool boiling curve . . . . . 457 9.2 Nucleate boiling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 464 9.3 Peak pool boiling heat flux . . . . . . . . . . . . . . . . . . . . . . . . 472 9.4 Film boiling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 486 9.5 Minimum heat flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 488 9.6 Transition boiling and system influences . . . . . . . . . . . . . 489 9.7 Forced convection boiling in tubes . . . . . . . . . . . . . . . . . . 496 9.8 Forced convective condensation heat transfer . . . . . . . . . 505 9.9 Dropwise condensation . . . . . . . . . . . . . . . . . . . . . . . . . . . 506 9.10 The heat pipe . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 509 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 513 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 517 Contents x IV Thermal Radiation Heat Transfer 10 Radiative heat transfer 10.1 The problem of radiative exchange . . . . . . . . . . . . . . . . . 10.2 Kirchhoff’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 Radiant heat exchange between two finite black bodies 10.4 Heat transfer among gray bodies . . . . . . . . . . . . . . . . . . 10.5 Gaseous radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6 Solar energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V Mass Transfer 523 . . . . . . . . 525 525 533 536 549 563 574 584 592 595 11 An introduction to mass transfer 597 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 597 11.2 Mixture compositions and species fluxes . . . . . . . . . . . . . 600 11.3 Diffusion fluxes and Fick’s law . . . . . . . . . . . . . . . . . . . . . 608 11.4 Transport properties of mixtures . . . . . . . . . . . . . . . . . . . 614 11.5 The equation of species conservation . . . . . . . . . . . . . . . . 627 11.6 Mass transfer at low rates . . . . . . . . . . . . . . . . . . . . . . . . . 635 11.7 Steady mass transfer with counterdiffusion . . . . . . . . . . . 648 11.8 Mass transfer coefficients at high rates of mass transfer . 654 11.9 Simultaneous heat and mass transfer . . . . . . . . . . . . . . . . 663 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 673 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 686 VI Appendices 689 A Some thermophysical properties of selected materials 691 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 694 B Units and conversion factors 721 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 722 C Nomenclature 725 Citation Index 733 Subject Index 739 Part I The General Problem of Heat Exchange 1 1. Introduction The radiation of the sun in which the planet is incessantly plunged, penetrates the air, the earth, and the waters; its elements are divided, change direction in every way, and, penetrating the mass of the globe, would raise its temperature more and more, if the heat acquired were not exactly balanced by that which escapes in rays from all points of the surface and expands through the sky. The Analytical Theory of Heat, J. Fourier 1.1 Heat transfer People have always understood that something flows from hot objects to cold ones. We call that flow heat. In the eighteenth and early nineteenth centuries, scientists imagined that all bodies contained an invisible fluid which they called caloric. Caloric was assigned a variety of properties, some of which proved to be inconsistent with nature (e.g., it had weight and it could not be created nor destroyed). But its most important feature was that it flowed from hot bodies into cold ones. It was a very useful way to think about heat. Later we shall explain the flow of heat in terms more satisfactory to the modern ear; however, it will seldom be wrong to imagine caloric flowing from a hot body to a cold one. The flow of heat is all-pervasive. It is active to some degree or another in everything. Heat flows constantly from your bloodstream to the air around you. The warmed air buoys off your body to warm the room you are in. If you leave the room, some small buoyancy-driven (or convective) motion of the air will continue because the walls can never be perfectly isothermal. Such processes go on in all plant and animal life and in the air around us. They occur throughout the earth, which is hot at its core and cooled around its surface. The only conceivable domain free from heat flow would have to be isothermal and totally isolated from any other region. It would be “dead” in the fullest sense of the word — devoid of any process of any kind. 3 Introduction 4 §1.1 The overall driving force for these heat flow processes is the cooling (or leveling) of the thermal gradients within our universe. The heat flows that result from the cooling of the sun are the primary processes that we experience naturally. The conductive cooling of Earth’s center and the radiative cooling of the other stars are processes of secondary importance in our lives. The life forms on our planet have necessarily evolved to match the magnitude of these energy flows. But while “natural man” is in balance with these heat flows, “technological man”1 has used his mind, his back, and his will to harness and control energy flows that are far more intense than those we experience naturally. To emphasize this point we suggest that the reader make an experiment. Experiment 1.1 Generate as much power as you can, in some way that permits you to measure your own work output. You might lift a weight, or run your own weight up a stairwell, against a stopwatch. Express the result in watts (W). Perhaps you might collect the results in your class. They should generally be less than 1 kW or even 1 horsepower (746 W). How much less might be surprising. Thus, when we do so small a thing as turning on a 150 W light bulb, we are manipulating a quantity of energy substantially greater than a human being could produce in sustained effort. The power consumed by an oven, toaster, or hot water heater is an order of magnitude beyond our capacity. The power consumed by an automobile can easily be three orders of magnitude greater. If all the people in the United States worked continuously like galley slaves, they could barely equal the output of even a single city power plant. Our voracious appetite for energy has steadily driven the intensity of actual heat transfer processes upward until they are far greater than those normally involved with life forms on earth. Until the middle of the thirteenth century, the energy we use was drawn indirectly from the sun 1 Some anthropologists think that the term Homo technologicus (technological man) serves to define human beings, as apart from animals, better than the older term Homo sapiens (man, the wise). We may not be as much wiser than the animals as we think we are, but only we do serious sustained tool making. §1.1 Heat transfer using comparatively gentle processes — animal power, wind and water power, and the combustion of wood. Then population growth and deforestation drove the English to using coal. By the end of the seventeenth century, England had almost completely converted to coal in place of wood. At the turn of the eighteenth century, the first commercial steam engines were developed, and that set the stage for enormously increased consumption of coal. Europe and America followed England in these developments. The development of fossil energy sources has been a bit like Jules Verne’s description in Around the World in Eighty Days in which, to win a race, a crew burns the inside of a ship to power the steam engine. The combustion of nonrenewable fossil energy sources (and, more recently, the fission of uranium) has led to remarkably intense energy releases in power-generating equipment. The energy transferred as heat in a nuclear reactor is on the order of one million watts per square meter. A complex system of heat and work transfer processes is invariably needed to bring these concentrations of energy back down to human proportions. We must understand and control the processes that divide and diffuse intense heat flows down to the level on which we can interact with them. To see how this works, consider a specific situation. Suppose we live in a town where coal is processed into fuel-gas and coke. Such power supplies used to be common, and they may return if natural gas supplies ever dwindle. Let us list a few of the process heat transfer problems that must be solved before we can drink a glass of iced tea. • A variety of high-intensity heat transfer processes are involved with combustion and chemical reaction in the gasifier unit itself. • The gas goes through various cleanup and pipe-delivery processes to get to our stoves. The heat transfer processes involved in these stages are generally less intense. • The gas is burned in the stove. Heat is transferred from the flame to the bottom of the teakettle. While this process is small, it is intense because boiling is a very efficient way to remove heat. • The coke is burned in a steam power plant. The heat transfer rates from the combustion chamber to the boiler, and from the wall of the boiler to the water inside, are very intense. 5 Introduction 6 §1.2 • The steam passes through a turbine where it is involved with many heat transfer processes, including some condensation in the last stages. The spent steam is then condensed in any of a variety of heat transfer devices. • Cooling must be provided in each stage of the electrical supply system: the winding and bearings of the generator, the transformers, the switches, the power lines, and the wiring in our houses. • The ice cubes for our tea are made in an electrical refrigerator. It involves three major heat exchange processes and several lesser ones. The major ones are the condensation of refrigerant at room temperature to reject heat, the absorption of heat from within the refrigerator by evaporating the refrigerant, and the balancing heat leakage from the room to the inside. • Let’s drink our iced tea quickly because heat transfer from the room to the water and from the water to the ice will first dilute, and then warm, our tea if we linger. A society based on power technology teems with heat transfer problems. Our aim is to learn the principles of heat transfer so we can solve these problems and design the equipment needed to transfer thermal energy from one substance to another. In a broad sense, all these problems resolve themselves into collecting and focusing large quantities of energy for the use of people, and then distributing and interfacing this energy with people in such a way that they can use it on their own puny level. We begin our study by recollecting how heat transfer was treated in the study of thermodynamics and by seeing why thermodynamics is not adequate to the task of solving heat transfer problems. 1.2 Relation of heat transfer to thermodynamics The First Law with work equal to zero The subject of thermodynamics, as taught in engineering programs, makes constant reference to the heat transfer between systems. The First Law of Thermodynamics for a closed system takes the following form on a Relation of heat transfer to thermodynamics §1.2 Figure 1.1 The First Law of Thermodynamics for a closed system. rate basis: Q positive toward the system = dU dt + Wk positive away from the system (1.1) positive when the system’s energy increases where Q is the heat transfer rate and Wk is the work transfer rate. They may be expressed in joules per second (J/s) or watts (W). The derivative dU/dt is the rate of change of internal thermal energy, U, with time, t. This interaction is sketched schematically in Fig. 1.1a. The analysis of heat transfer processes can generally be done without reference to any work processes, although heat transfer might subsequently be combined with work in the analysis of real systems. If p dV work is the only work occuring, then eqn. (1.1) is Q=p dU dV + dt dt (1.2a) This equation has two well-known special cases: Constant volume process: Constant pressure process: dU = mcv dt dH Q= = mcp dt Q= dT dt dT dt (1.2b) (1.2c) where H ≡ U + pV is the enthalpy, and cv and cp are the specific heat capacities at constant volume and constant pressure, respectively. When the substance undergoing the process is incompressible (so that V is constant for any pressure variation), the two specific heats are equal: 7 Introduction 8 §1.2 cv = cp ≡ c . The proper form of eqn. (1.2a) is then Q= dT dU = mc dt dt (1.3) Since solids and liquids can frequently be approximated as being incompressible, we shall often make use of eqn. (1.3). If the heat transfer were reversible, then eqn. (1.2a) would become2 T dS dV dU =p + dt dt dt Qrev (1.4) Wk rev That might seem to suggest that Q can be evaluated independently for inclusion in either eqn. (1.1) or (1.3). However, it cannot be evaluated using T dS , because real heat transfer processes are all irreversible and S is not defined as a function of T in an irreversible process. The reader will recall that engineering thermodynamics might better be named thermostatics, because it only describes the equilibrium states on either side of irreversible processes. Since the rate of heat transfer cannot be predicted using T dS , how can it be determined? If U (t) were known, then (when Wk = 0) eqn. (1.3) would give Q, but U (t) is seldom known a priori. The answer is that a new set of physical principles must be introduced to predict Q. The principles are transport laws, which are not a part of the subject of thermodynamics. They include Fourier’s law, Newton’s law of cooling, and the Stefan-Boltzmann law. We introduce these laws later in the chapter. The important thing to remember is that a description of heat transfer requires that additional principles be combined with the First Law of Thermodynamics. Reversible heat transfer as the temperature gradient vanishes Consider a wall connecting two thermal reservoirs as shown in Fig. 1.2. As long as T1 > T2 , heat will flow spontaneously and irreversibly from 1 to 2. In accordance with our understanding of the Second Law of Thermodynamics, we expect the entropy of the universe to increase as a consequence of this process. If T2 → T1 , the process will approach being quasistatic and reversible. But the rate of heat transfer will also approach 2 T = absolute temperature, S = entropy, V = volume, p = pressure, and “rev” denotes a reversible process. Relation of heat transfer to thermodynamics §1.2 Figure 1.2 Irreversible heat flow between two thermal reservoirs through an intervening wall. zero if there is no temperature difference to drive it. Thus all real heat transfer processes generate entropy. Now we come to a dilemma: If the irreversible process occurs at steady state, the properties of the wall do not vary with time. We know that the entropy of the wall depends on its state and must therefore be constant. How, then, does the entropy of the universe increase? We turn to this question next. Entropy production The entropy increase of the universe as the result of a process is the sum of the entropy changes of all elements that are involved in that process. ˙ The rate of entropy production of the universe, SUn , resulting from the preceding heat transfer process through a wall is ˙ ˙ SUn = Sres 1 + ˙ Swall ˙ +Sres 2 (1.5) = 0, since Swall must be constant ˙ where the dots denote time derivatives (i.e., x ≡ dx/dt ). Since the reservoir temperatures are constant, Q ˙ . Sres = Tres (1.6) Now Qres 1 is negative and equal in magnitude to Qres 2 , so eqn. (1.5) becomes 1 1 ˙ . (1.7) − SUn = Qres 1 T2 T1 9 Introduction 10 §1.3 ˙ The term in parentheses is positive, so SUn > 0. This agrees with Clausius’s statement of the Second Law of Thermodynamics. ˙ Notice an odd fact here: The rate of heat transfer, Q, and hence SUn , is determined by the wall’s resistance to heat flow. Although the wall is the agent that causes the entropy of the universe to increase, its own entropy does not change. Only the entropies of the reservoirs change. 1.3 Modes of heat transfer Figure 1.3 shows an analogy that might be useful in fixing the concepts of heat conduction, convection, and radiation as we proceed to look at each in some detail. Heat conduction Fourier’s law. Joseph Fourier (see Fig. 1.4) published his remarkable book Théorie Analytique de la Chaleur in 1822. In it he formulated a very complete exposition of the theory of heat conduction. Hebegan his treatise by stating the empirical law that bears his name: the heat flux,3 q (W/m2 ), resulting from thermal conduction is proportional to the magnitude of the temperature gradient and opposite to it in sign. If we call the constant of proportionality, k, then q = −k dT dx (1.8) The constant, k, is called the thermal conductivity. It obviously must have the dimensions W/m·K, or J/m·s·K, or Btu/h·ft·◦ F if eqn. (1.8) is to be dimensionally correct. The heat flux is a vector quantity. Equation (1.8) tells us that if temperature decreases with x , q will be positive—it will flow in the x -direction. If T increases with x , q will be negative—it will flow opposite the x direction. In either case, q will flow from higher temperatures to lower temperatures. Equation (1.8) is the one-dimensional form of Fourier’s law. We develop its three-dimensional form in Chapter 2, namely: q = −k ∇T 3 The heat flux, q, is a heat rate per unit area and can be expressed as Q/A, where A is an appropriate area. Figure 1.3 An analogy for the three modes of heat transfer. 11 Figure 1.4 Baron Jean Baptiste Joseph Fourier (1768–1830). Joseph Fourier lived a remarkable double life. He served as a high government official in Napoleonic France and he was also an applied mathematician of great importance. He was with Napoleon in Egypt between 1798 and 1801, and he was subsequently prefect of the administrative area (or “Department”) of Isère in France until Napoleon’s first fall in 1814. During the latter period he worked on the theory of heat flow and in 1807 submitted a 234-page monograph on the subject. It was given to such luminaries as Lagrange and Laplace for review. They found fault with his adaptation of a series expansion suggested by Daniel Bernoulli in the eighteenth century. Fourier’s theory of heat flow, his governing differential equation, and the nowfamous “Fourier series” solution of that equation did not emerge in print from the ensuing controversy until 1822. (Etching from Portraits et Histoire des Hommes Utiles, Collection de Cinquante Portraits, Société Montyon et Franklin 1839-1840). 12 Modes of heat transfer §1.3 13 Example 1.1 The front of a slab of lead (k = 35 W/m·K) is kept at 110◦ C and the back is kept at 50◦ C. If the area of the slab is 0.4 m2 and it is 0.03 m thick, compute the heat flux, q, and the heat transfer rate, Q. Solution. For the moment, we presume that dT /dx is a constant equal to (Tback − Tfront )/(xback − xfront ); we verify this in Chapter 2. Thus, eqn. (1.8) becomes q = −35 50 − 110 0.03 = +70, 000 W/m2 = 70 kW/m2 and Q = qA = 70(0.4) = 28 kW In one-dimensional heat conduction problems, there is never any real problem in deciding which way the heat should flow. It is therefore sometimes convenient to write Fourier’s law in simple scalar form: q=k ∆T L (1.9) where L is the thickness in the direction of heat flow and q and ∆T are both written as positive quantities. When we use eqn. (1.9), we must remember that q always flows from high to low temperatures. Thermal conductivity values. It will help if we first consider how conduction occurs in, for example, a gas. We know that the molecular velocity depends on temperature. Consider conduction from a hot wall to a cold one in a situation in which gravity can be ignored, as shown in Fig. 1.5. The molecules near the hot wall collide with it and are agitated by the molecules of the wall. They leave with generally higher speed and collide with their neighbors to the right, increasing the speed of those neighbors. This process continues until the molecules on the right pass their kinetic energy to those in the cool wall. Within solids, comparable processes occur as the molecules vibrate within their lattice structure and as the lattice vibrates as a whole. This sort of process also occurs, to some extent, in the electron “gas” that moves through the solid. The Introduction 14 §1.3 Figure 1.5 Heat conduction through gas separating two solid walls. processes are more efficient in solids than they are in gases. Notice that − q 1 ∝ k k dT = dx (1.10) since, in steady conduction, q is constant Thus solids, with generally higher thermal conductivities than gases, yield smaller temperature gradients for a given heat flux. In a gas, by the way, k is proportional to molecular speed and molar specific heat, and inversely proportional to the cross-sectional area of molecules. This book deals almost exclusively with S.I. units, or Système International d’Unités. Since much reference material will continue to be available in English units, we should have at hand a conversion factor for thermal conductivity: h ft 1.8◦ F J · · · 0.0009478 Btu 3600 s 0.3048 m K Thus the conversion factor from W/m·K to its English equivalent, Btu/h· ft·◦ F, is 1= 1 = 1.731 W/m·K Btu/h·ft·◦ F (1.11) Consider, for example, copper—the common substance with the highest conductivity at ordinary temperature: kCu at room temp = (383 W/m·K) 1.731 W/m·K = 221 Btu/h·ft·◦ F Btu/h·ft·◦ F 15 Figure 1.6 The approximate ranges of thermal conductivity of various substances.(All values are for the neighborhood of room temperature unless otherwise noted.) Introduction 16 §1.3 The range of thermal conductivities is enormous. As we see from Fig. 1.6, k varies by a factor of about 105 between gases and diamond at room temperature. This variation can be increased to about 107 if we include the effective conductivity of various cryogenic “superinsulations.” (These involve powders, fibers, or multilayered materials that have been evacuated of all air.) The reader should study and remember the order of magnitude of the thermal conductivities of different types of materials. This will be a help in avoiding mistakes in future computations, and it will be a help in making assumptions during problem solving. Actual numerical values of the thermal conductivity are given in Appendix A (which is a broad listing of many of the physical properties you might need in this course) and in Figs. 2.2 and 2.3. Example 1.2 A copper slab (k = 372 W/m·K) is 3 mm thick. It is protected from corrosion on each side by a 2-mm-thick layer of stainless steel (k = 17 W/m·K). The temperature is 400◦ C on one side of this composite wall and 100◦ C on the other. Find the temperature distribution in the copper slab and the heat conducted through the wall (see Fig. 1.7). Solution. If we recall Fig. 1.5 and eqn. (1.10), it should be clear that the temperature drop will take place almost entirely in the stainless steel, where k is less than 1/20 of k in the copper. Thus, the copper will be virtually isothermal at the average temperature of (400 + 100)/2 = 250◦ C. Furthermore, the heat conduction can be estimated in a 4 mm slab of stainless steel as though the copper were not even there. With the help of Fourier’s law in the form of eqn. (1.8), we get q = −k dT dx 17 W/m·K · 400 − 100 K/m = 1275 kW/m2 0.004 The accuracy of this rough calculation can be improved by considering the copper. To do this we first solve for ∆Ts.s. and ∆TCu (see Fig. 1.7). Conservation of energy requires that the steady heat flux through all three slabs must be the same. Therefore, q= k ∆T L s.s. =k ∆T L Cu Modes of heat transfer §1.3 17 Figure 1.7 Temperature drop through a copper wall protected by stainless steel (Example 1.2). but (400 − 100)◦ C ≡ ∆TCu + 2∆Ts.s. (k/L)Cu = ∆TCu 1 + 2 (k/L)s.s. = (30.18)∆TCu Solving this, we obtain ∆TCu = 9.94 K. So ∆Ts.s. = (300 − 9.94)/2 = 145 K. It follows that TCu, left = 255◦ C and TCu, right = 245◦ C. The heat flux can be obtained by applying Fourier’s law to any of the three layers. We consider either stainless steel layer and get q = 17 W 145 K = 1233 kW/m2 m·K 0.002 m Thus our initial approximation was accurate within a few percent. One-dimensional heat diffusion equation. In Example 1.2 we had to deal with a major problem that arises in heat conduction problems. The problem is that Fourier’s law involves two dependent variables, T and q. To eliminate q and first solve for T , we introduced the First Law of Thermodynamics implicitly: Conservation of energy required that q was the same in each metallic slab. The elimination of q from Fourier’s law must now be done in a more general way. Consider a one-dimensional element, as shown in Fig. 1.8. Introduction 18 §1.3 Figure 1.8 One-dimensional heat conduction through a differential element. From Fourier’s law applied at each side of the element, as shown, the net heat conduction out of the element during general unsteady heat flow is qnet A = Qnet = −kA ∂2T δx ∂x 2 (1.12) To eliminate the heat loss Qnet in favor of T , we use the general First Law statement for closed, nonworking systems, eqn. (1.3): −Qnet = d(T − Tref ) dT dU = ρcA δx = ρcA δx dt dt dt (1.13) where ρ is the density of the slab and c is its specific heat capacity.4 Equations (1.12) and (1.13) can be combined to give 1 ∂T ∂2T ρc ∂T ≡ = ∂x 2 k ∂t α ∂t 4 (1.14) The reader might wonder if c should be cp or cv . This is a strictly incompressible equation so cp = cv = c . The compressible equation involves additional terms, and this particular term emerges with cp in it in the conventional rearrangements of terms. Modes of heat transfer §1.3 19 Figure 1.9 The convective cooling of a heated body. This is the one-dimensional heat diffusion equation. Its importance is this: By combining the First Law with Fourier’s law, we have eliminated the unknown Q and obtained a differential equation that can be solved for the temperature distribution, T (x, t ). It is the primary equation upon which all of heat conduction theory is based. The heat diffusion equation includes a new property which is as important to transient heat conduction as k is to steady-state conduction. This is the thermal diffusivity, α: α≡ J m3 kg·K k = α m2/s (or ft2/hr). ρc m·s·K kg J The thermal diffusivity is a measure of how quickly a material can carry heat away from a hot source. Since material does not just transmit heat but must be warmed by it as well, α involves both the conductivity, k, and the volumetric heat capacity, ρc . Heat Convection The physical process. Consider a typical convective cooling situation. Cool gas flows past a warm body, as shown in Fig. 1.9. The fluid immediately adjacent to the body forms a thin slowed-down region called a boundary layer. Heat is conducted into this layer, which sweeps it away and, farther downstream, mixes it into the stream. We call such processes of carrying heat away by a moving fluid convection. In 1701, Isaac Newton considered the convective process and suggested that the cooling would be such that dTbody ∝ Tbody − T∞ dt (1.15) where T∞ is the temperature of the oncoming fluid. This statement suggests that energy is flowing from the body. But if the energy of the body Introduction 20 §1.3 is constantly replenished, the body temperature need not change. Then with the help of eqn. (1.3) we get, from eqn. (1.15) (see Problem 1.2), Q ∝ Tbody − T∞ (1.16) This equation can be rephrased in terms of q = Q/A as q = h Tbody − T∞ (1.17) This is the steady-state form of Newton’s law of cooling, as it is usually quoted, although Newton never wrote such an expression. The constant h is the film coefficient or heat transfer coefficient. The bar over h indicates that it is an average over the surface of the body. Without the bar, h denotes the “local” value of the heat transfer coefficient at a point on the surface. The units of h and h are W/m2 K or J/s·m2·K. The conversion factor for English units is: 1= 0.0009478 Btu K 3600 s (0.3048 m)2 · · · J 1.8◦ F h ft2 or 1 = 0.1761 Btu/h·ft2 ·◦ F W/m2 K (1.18) It turns out that Newton oversimplified the process of convection when he made his conjecture. Heat convection is complicated and h can depend on the temperature difference Tbody − T∞ ≡ ∆T . In Chapter 6 we find that h really is independent of ∆T in situations in which fluid is forced past a body and ∆T is not too large. This is called forced convection. When fluid buoys up from a hot body or down from a cold one, h varies as some weak power of ∆T —typically as ∆T 1/4 or ∆T 1/3 . This is called free or natural convection. If the body is hot enough to boil a liquid surrounding it, h will typically vary as ∆T 2 . For the moment, we restrict consideration to situations in which Newton’s law is either true or at least a reasonable approximation to real behavior. We should have some idea of how large h might be in a given situation. Table 1.1 provides some illustrative values of h that have been Modes of heat transfer §1.3 21 Table 1.1 Some illustrative values of convective heat transfer coefficients Situation Natural convection in gases • 0.3 m vertical wall in air, ∆T = 30◦ C Natural convection in liquids • 40 mm O.D. horizontal pipe in water, ∆T = 30◦ C • 0.25 mm diameter wire in methanol, ∆T = 50◦ C Forced convection of gases • Air at 30 m/s over a 1 m flat plate, ∆T = 70◦ C Forced convection of liquids • Water at 2 m/s over a 60 mm plate, ∆T = 15◦ C • Aniline-alcohol mixture at 3 m/s in a 25 mm I.D. tube, ∆T = 80◦ C • Liquid sodium at 5 m/s in a 13 mm I.D. tube at 370◦ C Boiling water • During film boiling at 1 atm • In a tea kettle • At a peak pool-boiling heat flux, 1 atm • At a peak flow-boiling heat flux, 1 atm • At approximate maximum convective-boiling heat flux, under optimal conditions Condensation • In a typical horizontal cold-water-tube steam condenser • Same, but condensing benzene • Dropwise condensation of water at 1 atm observed or calculated for different situations. They are only illustrative and should not be used in calculations because the situations for which they apply have not been fully described. Most of the values in the table could be changed a great deal by varying quantities (such as surface roughness or geometry) that have not been specified. The determination of h or h is a fairly complicated task and one that will receive a great deal of our attention. Notice, too, that h can change dramatically from one situation to the next. Reasonable values of h range over about six orders of magnitude. h, W/m2 K 4.33 570 4, 000 80 590 2, 600 75, 000 300 4, 000 40, 000 100, 000 106 15, 000 1, 700 160, 000 Introduction 22 §1.3 Example 1.3 The heat flux, q, is 6000 W/m2 at the surface of an electrical heater. The heater temperature is 120◦ C when it is cooled by air at 70◦ C. What is the average convective heat transfer coefficient, h? What will the heater temperature be if the power is reduced so that q is 2000 W/m2 ? Solution. h= q 6000 = = 120 W/m2 K ∆T 120 − 70 If the heat flux is reduced, h should remain unchanged during forced convection. Thus ∆T = Theater − 70◦ C = q h = 2000 W/m2 = 16.67 K 120 W/m2 K so Theater = 70 + 16.67 = 86.67◦ C Lumped-capacity solution. We now wish to deal with a very simple but extremely important, kind of convective heat transfer problem. The problem is that of predicting the transient cooling of a convectively cooled object, such as we showed in Fig. 1.9. With reference to Fig. 1.10, we apply our now-familiar First law statement, eqn. (1.3), to such a body: Q −hA(T − T∞ ) = dU dt (1.19) d [ρcV (T − Tref )] dt where A and V are the surface area and volume of the body, T is the temperature of the body, T = T (t), and Tref is the arbitrary temperature at which U is defined equal to zero. Thus5 d(T − T∞ ) hA =− (T − T∞ ) dt ρcV (1.20) 5 Is it clear why (T − Tref ) has been changed to (T − T∞ ) under the derivative? Remember that the derivative of a constant (like Tref or T∞ ) is zero. We can therefore introduce (T − T∞ ) without invalidating the equation, and get the same dependent variable on both sides of the equation. Modes of heat transfer §1.3 23 Figure 1.10 The cooling of a body for which the Biot number, hL/kb , is small. The general solution to this equation is ln(T − T∞ ) = − t (ρcV hA) +C (1.21) The group ρcV hA is the time constant, T . If the initial temperature is T (t = 0) ≡ Ti , then C = ln(Ti − T∞ ), and the cooling of the body is given by T − T∞ = e−t/T Ti − T∞ (1.22) All of the physical parameters in the problem have now been “lumped” into the time constant. It represents the time required for a body to cool to 1/e, or 37% of its initial temperature difference above (or below) T∞ . Introduction 24 §1.3 The ratio t/T can also be interpreted as t capacity for convection from surface hAt (J/◦ C) = = T ρcV (J/◦ C) heat capacity of the body (1.23) Notice that the thermal conductivity is missing from eqns. (1.22) and (1.23). The reason is that we have assumed that the temperature of the body is nearly uniform, and this means that internal conduction is not 1, the temperature of important. We see in Fig. 1.10 that, if L (kb / h) the body, Tb , is almost constant within the body at any time. Thus hL kb 1 implies that Tb (x, t) T (t) Tsurface and the thermal conductivity, kb , becomes irrelevant to the cooling process. This condition must be satisfied or the lumped-capacity solution will not be accurate. We call the group hL kb the Biot number 6 , Bi. If Bi were large, of course, the situation would be reversed, as shown in Fig. 1.11. In this 1 and the convection process offers little resistance case Bi = hL/kb to heat transfer. We could solve the heat diffusion equation 1 ∂T ∂2T = 2 ∂x α ∂t subject to the simple boundary condition T (x, t) = T∞ when x = L, to determine the temperature in the body and its rate of cooling in this case. The Biot number will therefore be the basis for determining what sort of problem we have to solve. To calculate the rate of entropy production in a lumped-capacity system, we note that the entropy change of the universe is the sum of the entropy decrease of the body and the more rapid entropy increase of the surroundings. The source of irreversibility is heat flow through the boundary layer. Accordingly, we write the time rate of change of entropy ˙ of the universe, dSUn /dt ≡ SUn , as −Qrev Qrev ˙ ˙ ˙ SUn = Sb + Ssurroundings = + Tb T∞ 6 Pronounced Bee-oh. J.B. Biot, although younger than Fourier, worked on the analysis of heat conduction even earlier—in 1802 or 1803. He grappled with the problem of including external convection in heat conduction analyses in 1804 but could not see how to do it. Fourier read Biot’s work and by 1807 had determined how to analyze the problem. (Later we encounter a similar dimensionless group called the Nusselt number, Nu = hL/kfluid . The latter relates only to the boundary layer and not to the body being cooled. We deal with it extensively in the study of convection.) Modes of heat transfer §1.3 25 Figure 1.11 The cooling of a body for which the Biot number, hL/kb , is large. or dTb ˙ SUn = −ρcV dt 1 1 − T∞ Tb . We can multiply both sides of this equation by dt and integrate the righthand side from Tb (t = 0) ≡ Tb0 to Tb at the time of interest: ∆S = −ρcV Tb Tb 0 1 1 − T∞ Tb dTb . (1.24) Equation 1.24 will give a positive ∆S whether Tb > T∞ or Tb < T∞ because the sign of dTb will always opposed the sign of the integrand. Example 1.4 A thermocouple bead is largely solder, 1 mm in diameter. It is initially at room temperature and is suddenly placed in a 200◦ C gas flow. The heat transfer coefficient h is 250 W/m2 K, and the effective values of k, ρ , and c are 45 W/m·K, 9300 kg/m3 , and c = 0.18 kJ/kg·K, respectively. Evaluate the response of the thermocouple. Introduction 26 §1.3 Solution. The time constant, T , is T ρc π D 3/6 ρcD = 2 hA h πD 6h m2·K 1000 W (9300)(0.18)(0.001) kg kJ m = 6(250) m3 kg·K W kJ/s = 1.116 s = ρcV = Therefore, eqn. (1.22) becomes T − 200◦ C = e−t/1.116 or T = 200 − 180 e−t/1.116 ◦ C (20 − 200)◦ C This result is plotted in Fig. 1.12, where we see that, for all practical purposes, this thermocouple catches up with the gas stream in less than 5 s. Indeed, it should be apparent that any such system will come within 95% of the signal in three time constants. Notice, too, that if the response could continue at its initial rate, the thermocouple would reach the signal temperature in one time constant. This calculation is based entirely on the assumption that Bi 1 for the thermocouple. We must check that assumption: Bi ≡ (250 W/m2 K)(0.001 m)/2 hL = = 0.00278 k 45 W/m·K This is very small indeed, so the assumption is valid. Experiment 1.2 Invent and carry out a simple procedure for evaluating the time constant of a fever thermometer in your mouth. Radiation Heat transfer by thermal radiation. All bodies constantly emit energy by a process of electromagnetic radiation. The intensity of such energy flux depends upon the temperature of the body and the nature of its surface. Most of the heat that reaches you when you sit in front of a fire is radiant energy. Radiant energy browns your toast in an electric toaster and it warms you when you walk in the sun. Modes of heat transfer §1.3 Figure 1.12 Thermocouple response to a hot gas flow. Objects that are cooler than the fire, the toaster, or the sun emit much less energy because the energy emission varies as the fourth power of absolute temperature. Very often, the emission of energy, or radiant heat transfer, from cooler bodies can be neglected in comparison with convection and conduction. But heat transfer processes that occur at high temperature, or with conduction or convection suppressed by evacuated insulations, usually involve a significant fraction of radiation. Experiment 1.3 Open the freezer door to your refrigerator. Put your face near it, but stay far enough away to avoid the downwash of cooled air. This way you cannot be cooled by convection and, because the air between you and the freezer is a fine insulator, you cannot be cooled by conduction. Still your face will feel cooler. The reason is that you radiate heat directly into the cold region and it radiates very little heat to you. Consequently, your face cools perceptibly. 27 28 Introduction §1.3 Table 1.2 Forms of the electromagnetic wave spectrum Characterization Wavelength, λ Cosmic rays < 0.3 pm Gamma rays 0.3–100 pm X rays 0.01–30 nm Ultraviolet light 3–400 nm Visible light 0.4–0.7 µm Near infrared radiation 0.7–30 µm Far infrared radiation 30–1000 µm Millimeter waves 1–10 mm Microwaves 10–300 mm Shortwave radio & TV 300 mm–100 m Longwave radio 100 m–30 km ⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭ Thermal Radiation 0.1–1000 µm The electromagnetic spectrum. Thermal radiation occurs in a range of the electromagnetic spectrum of energy emission. Accordingly, it exhibits the same wavelike properties as light or radio waves. Each quantum of radiant energy has a wavelength, λ, and a frequency, ν , associated with it. The full electromagnetic spectrum includes an enormous range of energy-bearing waves, of which heat is only a small part. Table 1.2 lists the various forms over a range of wavelengths that spans 17 orders of magnitude. Only the tiniest “window” exists in this spectrum through which we can see the world around us. Heat radiation, whose main component is usually the spectrum of infrared radiation, passes through the much larger window—about three orders of magnitude in λ or ν . Black bodies. The model for the perfect thermal radiator is a so-called black body. This is a body which absorbs all energy that reaches it and reflects nothing. The term can be a little confusing, since such bodies emit energy. Thus, if we possessed infrared vision, a black body would glow with “color” appropriate to its temperature. of course, perfect radiators are “black” in the sense that they absorb all visible light (and all other radiation) that reaches them. §1.3 Modes of heat transfer 29 Figure 1.13 Cross section of a spherical hohlraum. The hole has the attributes of a nearly perfect thermal black body. It is necessary to have an experimental method for making a perfectly black body. The conventional device for approaching this ideal is called by the German term hohlraum, which literally means “hollow space”. Figure 1.13 shows how a hohlraum is arranged. It is simply a device that traps all the energy that reaches the aperture. What are the important features of a thermally black body? First consider a distinction between heat and infrared radiation. Infrared radiation refers to a particular range of wavelengths, while heat refers to the whole range of radiant energy flowing from one body to another. Suppose that a radiant heat flux, q, falls upon a translucent plate that is not black, as shown in Fig. 1.14. A fraction, α, of the total incident energy, called the absorptance, is absorbed in the body; a fraction, ρ , Figure 1.14 The distribution of energy incident on a translucent slab. Introduction 30 §1.3 called the reflectance, is reflected from it; and a fraction, τ , called the transmittance, passes through. Thus 1=α+ρ+τ (1.25) This relation can also be written for the energy carried by each wavelength in the distribution of wavelengths that makes up heat from a source at any temperature: 1 = αλ + ρλ + τλ (1.26) All radiant energy incident on a black body is absorbed, so that αb or αλb = 1 and ρb = τb = 0. Furthermore, the energy emitted from a black body reaches a theoretical maximum, which is given by the StefanBoltzmann law. We look at this next. The Stefan-Boltzmann law. The flux of energy radiating from a body is commonly designated e(T ) W/m2 . The symbol eλ (λ, T ) designates the distribution function of radiative flux in λ, or the monochromatic emissive power: eλ (λ, T ) = de(λ, T ) or e(λ, T ) = dλ λ 0 eλ (λ, T ) dλ (1.27) Thus e(T ) ≡ E(∞, T ) = ∞ 0 eλ (λ, T ) dλ The dependence of e(T ) on T for a black body was established experimentally by Stefan in 1879 and explained by Boltzmann on the basis of thermodynamics arguments in 1884. The Stefan-Boltzmann law is eb (T ) = σ T 4 (1.28) where the Stefan-Boltzmann constant, σ , is 5.670400 × 10−8 W/m2 ·K4 or 1.714 × 10−9 Btu/hr·ft2 ·◦ R4 , and T is the absolute temperature. eλ vs. λ. Nature requires that, at a given temperature, a body will emit a unique distribution of energy in wavelength. Thus, when you heat a poker in the fire, it first glows a dull red—emitting most of its energy at long wavelengths and just a little bit in the visible regime. When it is §1.3 Modes of heat transfer 31 Figure 1.15 Monochromatic emissive power of a black body at several temperatures—predicted and observed. white-hot, the energy distribution has been both greatly increased and shifted toward the shorter-wavelength visible range. At each temperature, a black body yields the highest value of eλ that a body can attain. The very accurate measurements of the black-body energy spectrum by Lummer and Pringsheim (1899) are shown in Fig. 1.15. The locus of maxima of the curves is also plotted. It obeys a relation called Wien’s law: (λT )eλ=max = 2898 µm·K (1.29) About three-fourths of the radiant energy of a black body lies to the right of this line in Fig. 1.15. Notice that, while the locus of maxima leans toward the visible range at higher temperatures, only a small fraction of the radiation is visible even at the highest temperature. Predicting how the monochromatic emissive power of a black body depends on λ was an increasingly serious problem at the close of the nineteenth century. The prediction was a keystone of the most profound scientific revolution the world has seen. In 1901, Max Planck made the Introduction 32 §1.3 prediction, and his work included the initial formulation of quantum mechanics. He found that eλb = 2 2π hco 5 [exp(hc /k T λ) − 1] λ o B (1.30) where co is the speed of light, 2.99792458 × 108 m/s; h is Planck’s constant, 6.62606876×10−34 J·s; and kB is Boltzmann’s constant, 1.3806503× 10−23 J/K. Radiant heat exchange. Suppose that a heated object (1 in Fig. 1.16a) radiates only to some other object (2) and that both objects are thermally black. All heat leaving object 1 arrives at object 2, and all heat arriving at object 1 comes from object 2. Thus, the net heat transferred from object 1 to object 2, Qnet , is the difference between Q1 to 2 = A1 eb (T1 ) and Q2 to 1 = A1 eb (T2 ) 4 4 Qnet = A1 eb (T1 ) − A1 eb (T2 ) = A1 σ T1 − T2 (1.31) If the first object “sees” other objects in addition to object 2, as indicated in Fig. 1.16b, then a view factor (sometimes called a configuration factor or a shape factor ), F1–2 , must be included in eqn. (1.31): 4 4 Qnet = A1 F1–2 σ T1 − T2 (1.32) We may regard F1–2 as the fraction of energy leaving object 1 that is intercepted by object 2. Example 1.5 A black thermocouple measures the temperature in a chamber with black walls. If the air around the thermocouple is at 20◦ C, the walls are at 100◦ C, and the heat transfer coefficient between the thermocouple and the air is 75 W/m2 K, what temperature will the thermocouple read? Solution. The heat convected away from the thermocouple by the air must exactly balance that radiated to it by the hot walls if the system is in steady state. Furthermore, F1–2 = 1 since the thermocouple (1) radiates all its energy to the walls (2): 4 4 hAtc (Ttc − Tair ) = −Qnet = −Atc σ Ttc − Twall Modes of heat transfer §1.3 Figure 1.16 another. 33 The net radiant heat transfer from one object to or, with Ttc in ◦ C, 75(Ttc − 20) W/m2 = 5.6704 × 10−8 (100 + 273)4 − (Ttc + 273)4 W/m2 since T for radiation must be in kelvin. Trial-and-error solution of this equation yields Ttc = 28.4◦ C. We have seen that non-black bodies absorb less radiation than black bodies, which are perfect absorbers. Likewise, non-black bodies emit less radiation than black bodies, which also happen to be perfect emitters. We can characterize the emissive power of a non-black body using a property called emittance, ε: enon-black = εeb = εσ T 4 (1.33) where 0 < ε ≤ 1. When radiation is exchanged between two bodies that are not black, we have 4 4 Qnet = A1 F1–2 σ T1 − T2 (1.34) where the transfer factor, F1–2 , depends on the emittances of both bodies as well as the geometrical “view”. Introduction 34 §1.3 The expression for F1–2 is particularly simple in the important special case of a small object, 1, in a much larger isothermal environment, 2: F1–2 = ε1 for A1 A2 (1.35) Example 1.6 Suppose that the thermocouple in Example 1.5 was not black and had an emissivity of ε = 0.4. Further suppose that the walls were not black and had a much larger surface area than the thermocouple. What temperature would the thermocouple read? Solution. Qnet is now given by eqn. (1.34) and F1–2 can be found with eqn. (1.35): 4 4 hAtc (Ttc − Tair ) = −Atc εtc σ Ttc − Twall or 75(Ttc − 20) W/m2 = (0.4)(5.6704 × 10−8 ) (100 + 273)4 − (Ttc + 273)4 W/m2 Trial-and-error yields Ttc = 23.5◦ C. Radiation shielding. The preceding examples point out an important practical problem than can be solved with radiation shielding. The idea is as follows: If we want to measure the true air temperature, we can place a thin foil casing, or shield, around the thermocouple. The casing is shaped to obstruct the thermocouple’s “view” of the room but to permit the free flow of the air around the thermocouple. Then the shield, like the thermocouple in the two examples, will be cooler than the walls, and the thermocouple it surrounds will be influenced by this much cooler radiator. If the shield is highly reflecting on the outside, it will assume a temperature still closer to that of the air and the error will be still less. Multiple layers of shielding can further reduce the error. Radiation shielding can take many forms and serve many purposes. It is an important element in superinsulations. A glass firescreen in a fireplace serves as a radiation shield because it is largely opaque to radiation. It absorbs heat radiated by the fire and reradiates that energy (ineffectively) at a temperature much lower than that of the fire. A look ahead §1.4 Experiment 1.4 Find a small open flame that produces a fair amount of soot. A candle, kerosene lamp, or a cutting torch with a fuel-rich mixture should work well. A clean blue flame will not work well because such gases do not radiate much heat. First, place your finger in a position about 1 to 2 cm to one side of the flame, where it becomes uncomfortably hot. Now take a piece of fine mesh screen and dip it in some soapy water, which will fill up the holes. Put it between your finger and the flame. You will see that your finger is protected from the heating until the water evaporates. Water is relatively transparent to light. What does this experiment show you about the transmittance of water to infrared wavelengths? 1.4 A look ahead What we have done up to this point has been no more than to reveal the tip of the iceberg. The basic mechanisms of heat transfer have been explained and some quantitative relations have been presented. However, this information will barely get you started when you are faced with a real heat transfer problem. Three tasks, in particular, must be completed to solve actual problems: • The heat diffusion equation must be solved subject to appropriate boundary conditions if the problem involves heat conduction of any complexity. • The convective heat transfer coefficient, h, must be determined if convection is important in a problem. • The factor F1–2 or F1–2 must be determined to calculate radiative heat transfer. Any of these determinations can involve a great deal of complication, and most of the chapters that lie ahead are devoted to these three basic problems. Before becoming engrossed in these three questions, we shall first look at the archetypical applied problem of heat transfer–namely, the design of a heat exchanger. Chapter 2 sets up the elementary analytical apparatus that is needed for this, and Chapter 3 shows how to do such 35 Introduction 36 §1.5 design if h is already known. This will make it easier to see the importance of undertaking the three basic problems in subsequent parts of the book. 1.5 Problems We have noted that this book is set down almost exclusively in S.I. units. The student who has problems with dimensional conversion will find Appendix B helpful. The only use of English units appears in some of the problems at the end of each chapter. A few such problems are included to provide experience in converting back into English units, since such units will undoubtedly persist in the U.S.A. for many more years. Another matter often leads to some discussion between students and teachers in heat transfer courses. That is the question of whether a problem is “theoretical” or “practical”. Quite often the student is inclined to view as “theoretical” a problem that does not involve numbers or that requires the development of algebraic results. The problems assigned in this book are all intended to be useful in that they do one or more of five things: 1. They involve a calculation of a type that actually arises in practice (e.g., Problems 1.1, 1.3, 1.8 to 1.18, and 1.21 through 1.25). 2. They illustrate a physical principle (e.g., Problems 1.2, 1.4 to 1.7, 1.9, 1.20, 1.32, and 1.39). These are probably closest to having a “theoretical” objective. 3. They ask you to use methods developed in the text to develop other results that would be needed in certain applied problems (e.g., Problems 1.10, 1.16, 1.17, and 1.21). Such problems are usually the most difficult and the most valuable to you. 4. They anticipate development that will appear in subsequent chapters (e.g., Problems 1.16, 1.20, 1.40, and 1.41). 5. They require that you develop your ability to handle numerical and algebraic computation effectively. (This is the case with most of the problems in Chapter 1, but it is especially true of Problems 1.6 to 1.9, 1.15, and 1.17). Problems 37 Partial numerical answers to some of the problems follow them in brackets. Tables of physical property data useful in solving the problems are given in Appendix A. Actually, we wish to look at the theory, analysis, and practice of heat transfer—all three—according to Webster’s definitions: Theory: “a systematic statement of principles; a formulation of apparent relationships or underlying principles of certain observed phenomena.” Analysis: “the solving of problems by the means of equations; the breaking up of any whole into its parts so as to find out their nature, function, relationship, etc.” Practice: “the doing of something as an application of knowledge.” Problems 1.1 A composite wall consists of alternate layers of fir (5 cm thick), aluminum (1 cm thick), lead (1 cm thick), and corkboard (6 cm thick). The temperature is 60◦ C on the outside of the for and 10◦ C on the outside of the corkboard. Plot the temperature gradient through the wall. Does the temperature profile suggest any simplifying assumptions that might be made in subsequent analysis of the wall? 1.2 Verify eqn. (1.15). 1.3 q = 5000 W/m2 in a 1 cm slab and T = 140◦ C on the cold side. Tabulate the temperature drop through the slab if it is made of • Silver • Aluminum • Mild steel (0.5 % carbon) • Ice • Spruce • Insulation (85 % magnesia) • Silica aerogel Indicate which situations would be unreasonable and why. Chapter 1: Introduction 38 1.4 Explain in words why the heat diffusion equation, eqn. (1.13), shows that in transient conduction the temperature depends on the thermal diffusivity, α, but we can solve steady conduction problems using just k (as in Example 1.1). 1.5 A 1 m rod of pure copper 1 cm2 in cross section connects a 200◦ C thermal reservoir with a 0◦ C thermal reservoir. The system has already reached steady state. What are the rates of change of entropy of (a) the first reservoir, (b) the second reservoir, (c) the rod, and (d) the whole universe, as a result of the process? Explain whether or not your answer satisfies the Second Law of Thermodynamics. [(d): +0.0120 W/K.] 1.6 Two thermal energy reservoirs at temperatures of 27◦ C and −43◦ C, respectively, are separated by a slab of material 10 cm thick and 930 cm2 in cross-sectional area. The slab has a thermal conductivity of 0.14 W/m·K. The system is operating at steady-state conditions. What are the rates of change of entropy of (a) the higher temperature reservoir, (b) the lower temperature reservoir, (c) the slab, and (d) the whole universe as a result of this process? (e) Does your answer satisfy the Second Law of Thermodynamics? 1.7 (a) If the thermal energy reservoirs in Problem 1.6 are suddenly replaced with adiabatic walls, determine the final equilibrium temperature of the slab. (b) What is the entropy change for the slab for this process? (c) Does your answer satisfy the Second Law of Thermodynamics in this instance? Explain. The density of the slab is 26 lb/ft3 and the specific heat is 0.65 Btu/lb·◦ F. [(b): 30.81 J/K]. 1.8 A copper sphere 2.5 cm in diameter has a uniform temperature of 40◦ C. The sphere is suspended in a slow-moving air stream at 0◦ C. The air stream produces a convection heat transfer coefficient of 15 W/m2 K. Radiation can be neglected. Since copper is highly conductive, temperature gradients in the sphere will smooth out rapidly, and its temperature can be taken as uniform throughout the cooling process (i.e., Bi 1). Write the instantaneous energy balance between the sphere and the surrounding air. Solve this equation and plot the resulting temperatures as a function of time between 40◦ C and 0◦ C. Problems 39 1.9 Determine the total heat transfer in Problem 1.8 as the sphere cools from 40◦ C to 0◦ C. Plot the net entropy increase resulting from the cooling process above, ∆S vs. T (K). [Total heat transfer = 1123 J.] 1.10 A truncated cone 30 cm high is constructed of Portland cement. The diameter at the top is 15 cm and at the bottom is 7.5 cm. The lower surface is maintained at 6◦ C and the top at 40◦ C. The other surface is insulated. Assume one-dimensional heat transfer and calculate the rate of heat transfer in watts from top to bottom. To do this, note that the heat transfer, Q, must be the same at every cross section. Write Fourier’s law locally, and integrate it from top to bottom to get a relation between this unknown Q and the known end temperatures. [Q = −0.70 W.] 1.11 A hot water heater contains 100 kg of water at 75◦ C in a 20◦ C room. Its surface area is 1.3 m2 . Select an insulating material, and specify its thickness, to keep the water from cooling more than 3◦ C/h. (Notice that this problem will be greatly simplified if the temperature drop in the steel casing and the temperature drop in the convective boundary layers are negligible. Can you make such assumptions? Explain.) Figure 1.17 Configuration for Problem 1.12 1.12 What is the temperature at the left-hand wall shown in Fig. 1.17. Both walls are thin, very large in extent, highly conducting, and thermally black. [Tright = 42.5◦ C.] 1.13 Develop S.I. to English conversion factors for: • The thermal diffusivity, α • The heat flux, q • The density, ρ Chapter 1: Introduction 40 • The Stefan-Boltzmann constant, σ • The view factor, F1–2 • The molar entropy • The specific heat per unit mass, c In each case, begin with basic dimension J, m, kg, s, ◦ C, and check your answers against Appendix B if possible. Figure 1.18 Configuration for Problem 1.14 1.14 Three infinite, parallel, black, opaque plates transfer heat by radiation, as shown in Fig. 1.18. Find T2 . 1.15 Four infinite, parallel, black, opaque plates transfer heat by radiation, as shown in Fig. 1.19. Find T2 and T3 . [T2 = 75.53◦ C.] 1.16 Two large, black, horizontal plates are spaced a distance L from one another. The top one is warm at a controllable temperature, Th , and the bottom one is cool at a specified temperature, Tc . A gas separates them. The gas is stationary because it is warm on the top and cold on the bottom. Write the equation qrad /qcond = fn(N, Θ ≡ Th /Tc ), where N is a dimensionless group containing σ , k, L, and Tc . Plot N as a function of Θ for qrad /qcond = 1, 0.8, and 1.2 (and for other values if you wish). Now suppose that you have a system in which L = 10 cm, Tc = 100 K, and the gas is hydrogen with an average k of 0.1 W/m·K . Further suppose that you wish to operate in such a way that the conduction and radiation heat fluxes are identical. Identify the operating point on your curve and report the value of Th that you must maintain. Problems 41 Figure 1.19 Configuration for Problem 1.15 1.17 A blackened copper sphere 2 cm in diameter and uniformly at 200◦ C is introduced into an evacuated black chamber that is maintained at 20◦ C. • Write a differential equation that expresses T (t) for the sphere, assuming lumped thermal capacity. • Identify a dimensionless group, analogous to the Biot number, than can be used to tell whether or not the lumpedcapacity solution is valid. • Show that the lumped-capacity solution is valid. • Integrate your differential equation and plot the temperature response for the sphere. 1.18 As part of a space experiment, a small instrumentation package is released from a space vehicle. It can be approximated as a solid aluminum sphere, 4 cm in diameter. The sphere is initially at 30◦ C and it contains a pressurized hydrogen component that will condense and malfunction at 30 K. If we take the surrounding space to be at 0 K, how long may we expect the implementation package to function properly? Is it legitimate to use the lumped-capacity method in solving the problem? (Hint: See the directions for Problem 1.17.) [Time = 5.8 weeks.] 1.19 Consider heat conduction through the wall as shown in Fig. 1.20. Calculate q and the temperature of the right-hand side of the wall. 1.20 Throughout Chapter 1 we have assumed that the steady temperature distribution in a plane uniform wall in linear. To Chapter 1: Introduction 42 Figure 1.20 Configuration for Problem 1.19 prove this, simplify the heat diffusion equation to the form appropriate for steady flow. Then integrate it twice and eliminate the two constants using the known outside temperatures Tleft and Tright at x = 0 and x = wall thickness, L. 1.21 The thermal conductivity in a particular plane wall depends as follows on the wall temperature: k = A + BT , where A and B are constants. The temperatures are T1 and T2 on either side if the wall, and its thickness is L. Develop an expression for q. Figure 1.21 Configuration for Problem 1.22 1.22 Find k for the wall shown in Fig. 1.21. Of what might it be made? 1.23 What are Ti , Tj , and Tr in the wall shown in Fig. 1.22? [Tj = 16.44◦ C.] 1.24 An aluminum can of beer or soda pop is removed from the refrigerator and set on the table. If h is 13.5 W/m2 K, estimate Problems 43 Figure 1.22 Configuration for Problem 1.23 when the beverage will be at 15◦ C. Ignore thermal radiation. State all of your other assumptions. 1.25 One large, black wall at 27◦ C faces another whose surface is 127◦ C. The gap between the two walls is evacuated. If the second wall is 0.1 m thick and has a thermal conductivity of 17.5 W/m·K, what is its temperature on the back side? (Assume steady state.) 1.26 A 1 cm diameter, 1% carbon steel sphere, initially at 200◦ C, is cooled by natural convection, with air at 20◦ C. In this case, h is not independent of temperature. Instead, h = 3.51(∆T ◦ C)1/4 W/m2 K. Plot Tsphere as a function of t . Verify the lumpedcapacity assumption. 1.27 A 3 cm diameter, black spherical heater is kept at 1100◦ C. It radiates through an evacuated space to a surrounding spherical shell of Nichrome V. The shell has a 9 cm inside diameter and is 0.3 cm thick. It is black on the inside and is held at 25◦ C on the outside. Find (a) the temperature of the inner wall of the shell and (b) the heat transfer, Q. (Treat the shell as a plane wall.) 1.28 The sun radiates 650 W/m2 on the surface of a particular lake. At what rate (in mm/hr) would the lake evaporate away if all of this energy went to evaporating water? Discuss as many other Chapter 1: Introduction 44 ways you can think of that this energy can be distributed (hfg for water is 2,257,000 J/kg). Do you suppose much of the 650 W/m2 goes to evaporation? 1.29 It is proposed to make picnic cups, 0.005 m thick, of a new plastic for which k = ko (1 + aT 2 ), where T is expressed in ◦ C, ko = 0.15 W/m·K, and a = 10−4 ◦ C−2 . We are concerned with thermal behavior in the extreme case in which T = 100◦ C in the cup and 0◦ C outside. Plot T against position in the cup wall and find the heat loss, q. 1.30 A disc-shaped wafer of diamond 1 lb is the target of a very high intensity laser. The disc is 5 mm in diameter and 1 mm deep. The flat side is pulsed intermittently with 1010 W/m2 of energy for one microsecond. It is then cooled by natural convection from that same side until the next pulse. If h = 10 W/m2 K and T∞ =30◦ C, plot Tdisc as a function of time for pulses that are 50 s apart and 100 s apart. (Note that you must determine the temperature the disc reaches before it is pulsed each time.) 1.31 A 150 W light bulb is roughly a 0.006 m diameter sphere. Its steady surface temperature in room air is 90◦ C, and h on the outside is 7 W/m2 K. What fraction of the heat transfer from the bulb is by radiation directly from the filament through the glass? (State any additional assumptions.) 1.32 How much entropy does the light bulb in Problem 1.31 produce? 1.33 Air at 20◦ C flows over one side of a thin metal sheet (h = 10.6 W/m2 K). Methanol at 87◦ C flows over the other side (h = 141 W/m2 K). The metal functions as an electrical resistance heater, releasing 1000 W/m2 . Calculate (a) the heater temperature, (b) the heat transfer from the methanol to the heater, and (c) the heat transfer from the heater to the air. 1.34 A planar black heater is simultaneously cooled by 20◦ C air (h = 14.6 W/m2 K) and by radiation to a parallel black wall at 80◦ C. What is the temperature of the heater if it delivers 9000 W/m2 ? 1.35 An 8 oz. can of beer is taken from a 3◦ C refrigerator and placed in a 25◦ C room. The 6.3 cm diameter by 9 cm high can is placed on an insulated surface (h = 7.3 W/m2 K). How long will it take to reach 12◦ C? Ignore thermal radiation, and discuss your other assumptions. Problems 45 1.36 A resistance heater in the form of a thin sheet runs parallel with 3 cm slabs of cast iron on either side of an evacuated cavity. The heater, which releases 8000 W/m2 , and the cast iron are very nearly black. The outside surfaces of the cast iron slabs are kept at 10◦ C. Determine the heater temperature and the inside slab temperatures. 1.37 A black wall at 1200◦ C radiates to the left side of a parallel slab of type 316 stainless steel, 5 mm thick. The right side of the slab is to be cooled convectively and is not to exceed 0◦ C. Suggest a convective process that will achieve this. 1.38 A cooler keeps one side of a 2 cm layer of ice at −10◦ C. The other side is exposed to air at 15◦ C. What is h just on the edge of melting? Must h be raised or lowered if melting is to progress? 1.39 At what minimum temperature does a black heater deliver its maximum monochromatic emissive power in the visible range? Compare your result with Fig. 10.2. 1.40 The local heat transfer coefficient during the laminar flow of fluid over a flat plate of length L is equal to F /x 1/2 , where F is a function of fluid properties and the flow velocity. How does h compare with h(x = L)? (x is the distance from the leading edge of the plate.) 1.41 An object is initially at a temperature above that of its surroundings. We have seen that many kinds of convective processes will bring the object into equilibrium with its surroundings. Describe the characteristics of a process that will do so with the least net increase of the entropy of the universe. 1.42 A 250◦ C cylindrical copper billet, 4 cm in diameter and 8 cm long, is cooled in air at 25◦ C. The heat transfer coefficient is 5 W/m2 K. Can this be treated as lumped-capacity cooling? What is the temperature of the billet after 10 minutes? 1.43 The sun’s diameter is 1,392,000 km, and it emits energy as if it were a black body at 5777 K. Determine the rate at which it emits energy. Compare this with a value from the literature. What is the sun’s energy output in a year? Chapter 1: Introduction 46 Bibliography of Historical and Advanced Texts We include no specific references for the ideas introduced in Chapter 1 since these may be found in introductory thermodynamics or physics books. References 1–6 are some texts which have strongly influenced the field. The rest are relatively advanced texts or handbooks which go beyond the present textbook. References [1.1] J. Fourier. The Analytical Theory of Heat. Dover Publications, Inc., New York, 1955. [1.2] L. M. K. Boelter, V. H. Cherry, H. A. Johnson, and R. C. Martinelli. Heat Transfer Notes. McGraw-Hill Book Company, New York, 1965. Originally issued as class notes at the University of California at Berkeley between 1932 and 1941. [1.3] M. Jakob. Heat Transfer. John Wiley & Sons, New York, 1949. [1.4] W. H. McAdams. Heat Transmission. McGraw-Hill Book Company, New York, 3rd edition, 1954. [1.5] W. M. Rohsenow and H. Y. Choi. Heat, Mass and Momentum Transfer. Prentice-Hall, Inc., Englewood Cliffs, N.J., 1961. [1.6] E. R. G. Eckert and R. M. Drake, Jr. Analysis of Heat and Mass Transfer. Hemisphere Publishing Corp., Washington, D.C., 1987. [1.7] H. S. Carslaw and J. C. Jaeger. Conduction of Heat in Solids. Oxford University Press, New York, 2nd edition, 1959. Very comprehenisve, but quite dense. [1.8] D. Poulikakos. Conduction Heat Transfer. Prentice-Hall, Inc., Englewood Cliffs, NJ, 1994. This book’s approach is very accessible. Good coverage of solidification. [1.9] V. S. Arpaci. Conduction Heat Transfer. Ginn Press/Pearson Custom Publishing, Needham Heights, Mass., 1991. Abridgement of the 1966 edition, omitting numerical analysis. References [1.10] W. M. Kays and M. E. Crawford. Convective Heat and Mass Transfer. McGraw-Hill Book Company, New York, 3rd edition, 1993. Coverage is mainly of boundary layers and internal flows. [1.11] F.M. White. Viscous Fluid Flow. McGraw-Hill, Inc., New York, 2nd edition, 1991. Excellent development of fundamental results for boundary layers and internal flows. [1.12] J.A. Schetz. Foundations of Boundary Layer Theory for Momentum, Heat, and Mass Transfer. Prentice-Hall, Inc., Englewood Cliffs, NJ, 1984. This book shows many experimental results in support of the theory. [1.13] A. Bejan. Convection Heat Transfer. John Wiley & Sons, New York, 2nd edition, 1995. This book makes good use of scaling arguments. [1.14] M. Kaviany. Principles of Convective Heat Transfer. SpringerVerlag, New York, 1995. This treatise is wide-ranging and quite unique. Includes multiphase convection. [1.15] H. Schlichting and K. Gersten. Boundary-Layer Theory. SpringerVerlag, Berlin, 8th edition, 2000. Very comprehensive development of boundary layer theory. A classic. [1.16] H. C. Hottel and A. F. Sarofim. Radiative Transfer. McGraw-Hill Book Company, New York, 1967. [1.17] R. Siegel and J. R. Howell. Thermal Radiation Heat Transfer. Taylor and Francis-Hemisphere, Washington, D.C., 4th edition, 2001. [1.18] M. F. Modest. Radiative Heat Transfer. McGraw-Hill, New York, 1993. [1.19] P. B. Whalley. Boiling, Condensation, and Gas-Liquid Flow. Oxford University Press, Oxford, 1987. [1.20] J. G. Collier and J. R. Thome. Convective Boiling and Condensation. Oxford University Press, Oxford, 3rd edition, 1994. [1.21] Y. Y. Hsu and R. W. Graham. Transport Processes in Boiling and Two-Phase Systems Including Near-Critical Systems. American Nuclear Society, LaGrange Park, IL, 1986. 47 48 Chapter 1: Introduction [1.22] W. M. Kays and A. L. London. Compact Heat Exchangers. McGrawHill Book Company, New York, 3rd edition, 1984. [1.23] G. F. Hewitt, editor. Heat Exchanger Design Handbook 1998. Begell House, New York, 1998. [1.24] R. B. Bird, W. E. Stewart, and E. N. Lightfoot. Transport Phenomena. John Wiley & Sons, Inc., New York, 2nd edition, 2002. [1.25] A. F. Mills. Mass Transfer. Prentice-Hall, Inc., Upper Saddle River, 2001. Mass transfer from a mechanical engineer’s perpective with strong coverage of convective mass transfer. [1.26] D. S. Wilkinson. Mass Transfer in Solids and Fluids. Cambridge University Press, Cambridge, 2000. A systematic development of mass transfer with a materials science focus and an emphasis on modelling. [1.27] D. R. Poirier and G. H. Geiger. Transport Phenomena in Materials Processing. The Minerals, Metals & Materials Society, Warrendale, Pennsylvania, 1994. A comprehensive introduction to heat, mass, and momentum transfer from a materials science perspective. [1.28] W. M. Rohsenow, J. P. Hartnett, and Y. I. Cho, editors. Handbook of Heat Transfer. McGraw-Hill, New York, 3rd edition, 1998. 2. Heat conduction concepts, thermal resistance, and the overall heat transfer coefficient It is the fire that warms the cold, the cold that moderates the heat. . .the general coin that purchases all things. . . Don Quixote, M. de Cervantes, 1615 2.1 The heat diffusion equation Objective We must now develop some ideas that will be needed for the design of heat exchangers. The most important of these is the notion of an overall heat transfer coefficient. This is a measure of the general resistance of a heat exchanger to the flow of heat, and usually it must be built up from analyses of component resistances. In particular, we must know how to predict h and how to evaluate the conductive resistance of bodies more complicated than plane passive walls. The evaluation of h is a matter that must be deferred to Chapter 6 and 7. For the present, h values must be considered to be given information in any problem. The heat conduction component of most heat exchanger problems is more complex than the simple planar analyses done in Chapter 1. To do such analyses, we must next derive the heat conduction equation and learn to solve it. Consider the general temperature distribution in a three-dimensional body as depicted in Fig. 2.1. For some reason (heating from one side, in this case), there is a space- and time-dependent temperature field in the body. This field T = T (x, y, z, t) or T (r , t), defines instantaneous 49 50 Heat conduction, thermal resistance, and the overall heat transfer coefficient §2.1 Figure 2.1 A three-dimensional, transient temperature field. isothermal surfaces, T1 , T2 , and so on. We next consider a very important vector associated with the scalar, T . The vector that has both the magnitude and direction of the maximum increase of temperature at each point is called the temperature gradient, ∇T : ∇T ≡ i ∂T ∂T ∂T +j +k ∂x ∂y ∂z (2.1) Fourier’s law “Experience”—that is, physical observation—suggests two things about the heat flow that results from temperature nonuniformities in a body. The heat diffusion equation §2.1 51 These are: ∇T q =− |q | |∇T | This says that q and ∇T are exactly opposite one another in direction |q| ∝ |∇T | This says that the magnitude of the heat flux is directly proportional to the temperature gradient and Notice that the heat flux is now written as a quantity that has a specified direction as well as a specified magnitude. Fourier’s law summarizes this physical experience succinctly as q = −k∇T (2.2) which resolves itself into three components: qx = −k ∂T ∂x qy = −k ∂T ∂y qz = −k ∂T ∂z The coefficient k—the thermal conductivity—also depends on position and temperature in the most general case: k = k[r , T (r , t)] (2.3) Fortunately, most materials (though not all of them) are very nearly homogeneous. Thus we can usually write k = k(T ). The assumption that we really want to make is that k is constant. Whether or not that is legitimate must be determined in each case. As is apparent from Fig. 2.2 and Fig. 2.3, k almost always varies with temperature. It always rises with T in gases at low pressures, but it may rise or fall in metals or liquids. The problem is that of assessing whether or not k is approximately constant in the range of interest. We could safely take k to be a constant for iron between 0◦ and 40◦ C (see Fig. 2.2), but we would incur error between −100◦ and 800◦ C. It is easy to prove (Problem 2.1) that if k varies linearly with T , and if heat transfer is plane and steady, then q = k∆T /L, with k evaluated at the average temperature in the plane. If heat transfer is not planar or if k is not simply A + BT , it can be much more difficult to specify a single accurate effective value of k. If ∆T is not large, one can still make a reasonably accurate approximation using a constant average value of k. Figure 2.2 Variation of thermal conductivity of metallic solids with temperature 52 Figure 2.3 The temperature dependence of the thermal conductivity of liquids and gases that are either saturated or at 1 atm pressure. 53 54 Heat conduction, thermal resistance, and the overall heat transfer coefficient §2.1 Figure 2.4 Control volume in a heat-flow field. Now that we have revisited Fourier’s law in three dimensions, we see that heat conduction is more complex than it appeared to be in Chapter 1. We must now write the heat conduction equation in three dimensions. We begin, as we did in Chapter 1, with the First Law statement, eqn. (1.3): Q= dU dt (1.3) This time we apply eqn. (1.3) to a three-dimensional control volume, as shown in Fig. 2.4.1 The control volume is a finite region of a conducting body, which we set aside for analysis. The surface is denoted as S and the volume and the region as R ; both are at rest. An element of the surface, dS , is identified and two vectors are shown on dS : one is the unit normal vector, n (with |n| = 1), and the other is the heat flux vector, q = −k∇T , at that point on the surface. We also allow the possibility that a volumetric heat release equal to ˙ q(r ) W/m3 is distributed through the region. This might be the result of chemical or nuclear reaction, of electrical resistance heating, of external radiation into the region or of still other causes. With reference to Fig. 2.4, we can write the heat conducted out of dS , in watts, as (−k∇T ) · (ndS) (2.4) The heat generated (or consumed) within the region R must be added to the total heat flow into S to get the overall rate of heat addition to R : Q=− 1 S (−k∇T ) · (ndS) + ˙ q dR (2.5) R Figure 2.4 is the three-dimensional version of the control volume shown in Fig. 1.8. The heat diffusion equation §2.1 55 The rate of energy increase of the region R is dU = dt ρc R ∂T ∂t dR (2.6) where the derivative of T is in partial form because T is a function of both r and t . Finally, we combine Q, as given by eqn. (2.5), and dU /dt , as given by eqn. (2.6), into eqn. (1.3). After rearranging the terms, we obtain S k∇T · ndS = ρc R ∂T ˙ − q dR ∂t (2.7) To get the left-hand side into a convenient form, we introduce Gauss’s theorem, which converts a surface integral into a volume integral. Gauss’s theorem says that if A is any continuous function of position, then S A · ndS = R ∇ · A dR (2.8) Therefore, if we identify A with (k∇T ), eqn. (2.7) reduces to R ∇ · k∇T − ρc ∂T ˙ + q dR = 0 ∂t (2.9) Next, since the region R is arbitrary, the integrand must vanish identically.2 We therefore get the heat diffusion equation in three dimensions: ˙ ∇ · k∇T + q = ρc ∂T ∂t (2.10) The limitations on this equation are: • Incompressible medium. (This was implied when no expansion work term was included.) • No convection. (The medium cannot undergo any relative motion. However, it can be a liquid or gas as long as it sits still.) 2 Consider f (x) dx = 0. If f (x) were, say, sin x , then this could only be true over intervals of x = 2π or multiples of it. For eqn. (2.9) to be true for any range of integration one might choose, the terms in parentheses must be zero everywhere. 56 Heat conduction, thermal resistance, and the overall heat transfer coefficient §2.1 If the variation of k with T is small, k can be factored out of eqn. (2.10) to get ∇2 T + ˙ 1 ∂T q = k α ∂t (2.11) This is a more complete version of the heat conduction equation [recall eqn. (1.14)] and α is the thermal diffusivity which was discussed after eqn. (1.14). The term ∇2 T ≡ ∇ · ∇T is called the Laplacian. It arises thus in a Cartesian coordinate system: ∇ · k∇T k∇ · ∇T = k i ∂ ∂ ∂ +j +k ∂x ∂y ∂x ·i ∂T ∂T ∂T +j +k ∂x ∂y ∂z or ∇2 T = ∂2T ∂2T ∂2T + + ∂x 2 ∂y 2 ∂z2 (2.12) The Laplacian can also be expressed in cylindrical or spherical coordinates. The results are: • Cylindrical: ∇2 T ≡ 1∂ r ∂r r ∂T ∂r + 1 ∂2T ∂2T + 2 ∂θ 2 r ∂z2 (2.13) • Spherical: ∇2 T ≡ ∂T ∂ 1 ∂ 2 (r T ) 1 sin θ +2 2 r ∂r r sin θ ∂θ ∂θ + 1 ∂2T (2.14a) r 2 sin2 θ ∂φ2 or ≡ 1∂ r 2 ∂r r2 ∂T ∂r + 1 ∂T ∂ sin θ 2 sin θ ∂θ r ∂θ where the coordinates are as described in Fig. 2.5. + 1 ∂2T r 2 sin2 θ ∂φ2 (2.14b) Figure 2.5 Cylindrical and spherical coordinate schemes. 57 58 Heat conduction, thermal resistance, and the overall heat transfer coefficient 2.2 §2.2 Solutions of the heat diffusion equation We are now in position to calculate the temperature distribution and/or heat flux in bodies with the help of the heat diffusion equation. In every case, we first calculate T (r , t). Then, if we want the heat flux as well, we differentiate T to get q from Fourier’s law. The heat diffusion equation is a partial differential equation (p.d.e.) and the task of solving it may seem difficult, but we can actually do a lot with fairly elementary mathematical tools. For one thing, in onedimensional steady-state situations the heat diffusion equation becomes an ordinary differential equation (o.d.e.); for another, the equation is linear and therefore not too formidable, in any case. Our procedure can be laid out, step by step, with the help of the following example. Example 2.1 Basic Method A large, thin concrete slab of thickness L is “setting.” Setting is an ˙ exothermic process that releases q W/m3 . The outside surfaces are kept at the ambient temperature, so Tw = T∞ . What is the maximum internal temperature? Solution. Step 1. Pick the coordinate scheme that best fits the problem and identify the independent variables that determine T. In the example, T will probably vary only along the thin dimension, which we will call the x -direction. (We should want to know that the edges are insulated and that L was much smaller than the width or height. If they are, this assumption should be quite good.) Since the interior temperature will reach its maximum value when the process becomes steady, we write T = T (x only). Step 2. Write the appropriate d.e., starting with one of the forms of eqn. (2.11). ˙ ∂2T 1 ∂T ∂2T q ∂2T + + += 2 2 2 ∂x ∂y ∂z k α ∂t =0, since T ≠ T (y or z) = 0, since steady Therefore, since T = T (x only), the equation reduces to the Solutions of the heat diffusion equation §2.2 ordinary d.e. ˙ d2 T q =− dx 2 k Step 3. Obtain the general solution of the d.e. (This is usually the easiest step.) We simply integrate the d.e. twice and get T =− ˙ q2 x + C1 x + C2 2k Step 4. Write the “side conditions” on the d.e.—the initial and boundary conditions. This is always the hardest part for the beginning students; it is the part that most seriously tests their physical or “practical” understanding of problems. Normally, we have to make two specifications of temperature on each position coordinate and one on the time coordinate to get rid of the constants of integration in the general solution. (These matters are discussed at greater length in Chapter 4.) In this case there are two boundary conditions: T (x = 0) = Tw and T (x = L) = Tw Very Important Warning : Never, never introduce inaccessible information in a boundary or initial condition. Always stop and ask yourself, “Would I have access to a numerical value of the temperature (or other data) that I specify at a given position or time?” If the answer is no, then your result will be useless. Step 5. Substitute the general solution in the boundary and initial conditions and solve for the constants. This process gets very complicated in the transient and multidimensional cases. Fourier series methods are typically needed to solve the problem. However, the steady one-dimensional problems are usually easy. In the example, by evaluating at x = 0 and x = L, we get: Tw = −0 + 0 + C2 Tw = − so ˙ qL2 + C1 L + C2 2k =Tw C2 = Tw so C1 = ˙ qL 2k 59 60 Heat conduction, thermal resistance, and the overall heat transfer coefficient §2.2 Figure 2.6 Temperature distribution in the setting concrete slab Example 2.1. Step 6. Put the calculated constants back in the general solution to get the particular solution to the problem. In the example problem we obtain: T =− ˙ ˙ q q2 x+ Lx + Tw 2k 2k This should be put in neat dimensionless form: T − Tw 1 = ˙ 2 qL2 k x x − L L 2 (2.15) Step 7. Play with the solution—look it over—see what it has to tell you. Make any checks you can think of to be sure it is correct. In this case we plot eqn. (2.15) in Fig. 2.6. The resulting temperature distribution is parabolic and, as we would expect, symmetrical. It satisfies the boundary conditions at the wall and maximizes in the center. By nondimensionalizing the result, we have succeeded in representing all situations with a simple curve. That is highly desirable when the calculations are not simple, as they are here. (Notice that T actually depends on five different things, yet the solution is a single curve on a two-coordinate graph.) Solutions of the heat diffusion equation §2.2 Finally, we check to see if the heat flux at the wall is correct: qwall = −k ∂T ∂x x =0 =k ˙ ˙ qL q x− k 2k x =0 =− ˙ qL 2 Thus, half of the total energy generated in the slab comes out of the front side, as we would expect. The solution appears to be correct. Step 8. If the temperature field is now correctly established, you can, if you wish, calculate the heat flux at any point in the body by substituting T (r , t) back into Fourier’s law. We did this already, in Step 7, to check our solution. We shall run through additional examples in this section and the following one. In the process, we shall develop some important results for future use. Example 2.2 The Simple Slab A slab shown in Fig. 2.7 is at a steady state with dissimilar temperatures on either side and no internal heat generation. We want the temperature distribution and the heat flux through it. Solution. These can be found quickly by following the steps set down in Example 2.1: Figure 2.7 Heat conduction in a slab (Example 2.2). 61 62 Heat conduction, thermal resistance, and the overall heat transfer coefficient §2.3 Step 1. T = T (x) for steady x -direction heat flow Step 2. d2 T ˙ = 0, the steady 1-D heat equation with no q dx 2 Step 3. T = C1 x + C2 is the general solution of that equation Step 4. T (x = 0) = T1 and T (x = L) = T2 are the b.c.s Step 5. T1 = 0 + C2 , so C2 = T1 ; and T2 = C1 L + C2 , so C1 = Step 6. T = T1 + T2 − T1 L T − T1 T2 − T1 x x ; or = L T2 − T1 L Step 7. We note that the solution satisfies the boundary conditions and that the temperature profile is linear. Step 8. q = −k so that d T1 − T2 dT T1 − = −k x dx dx L q=k ∆T L This result, which is the simplest heat conduction solution, calls to mind Ohm’s law. Thus, if we rearrange it: Q= ∆T L/kA is like I= E R where L/kA assumes the role of a thermal resistance, to which we give the symbol Rt . Rt has the dimensions of (K/W). Figure 2.8 shows how we can represent heat flow through the slab with a diagram that is perfectly analogous to an electric circuit. 2.3 Thermal resistance and the electrical analogy Fourier’s, Fick’s, and Ohm’s laws Fourier’s law has several extremely important analogies in other kinds of physical behavior, of which the electrical analogy is only one. These analogous processes provide us with a good deal of guidance in the solution of heat transfer problems And, conversely, heat conduction analyses can often be adapted to describe those processes. §2.3 Thermal resistance and the electrical analogy Figure 2.8 Ohm’s law analogy to conduction through a slab. Let us first consider Ohm’s law in three dimensions: flux of electrical charge = I ≡ J = −γ ∇V A (2.16) I amperes is the vectorial electrical current, A is an area normal to the current vector, J is the flux of current or current density, γ is the electrical conductivity in cm/ohm·cm2 , and V is the voltage. To apply eqn. (2.16) to a one-dimensional current flow, as pictured in Fig. 2.9, we write eqn. (2.16) as J = −γ ∆V dV =γ , dx L (2.17) but ∆V is the applied voltage, E , and the resistance of the wire is R ≡ L γ A. Then, since I = J A, eqn. (2.17) becomes I= E R (2.18) which is the familiar, but restrictive, one-dimensional statement of Ohm’s law. Fick’s law is another analogous relation. It states that during mass diffusion, the flux, j1 , of a dilute component, 1, into a second fluid, 2, is 63 64 Heat conduction, thermal resistance, and the overall heat transfer coefficient §2.3 Figure 2.9 The one-dimensional flow of current. proportional to the gradient of its mass concentration, m1 . Thus j1 = −ρ D12 ∇m1 (2.19) where the constant D12 is the binary diffusion coefficient. Example 2.3 Air fills a thin tube 1 m in length. There is a small water leak at one end where the water vapor concentration builds to a mass fraction of 0.01. A desiccator maintains the concentration at zero on the other side. What is the steady flux of water from one side to the other if D12 is 2.84 × 10−5 m2/s and ρ = 1.18 kg/m3 ? Solution. jwater vapor = 1.18 kg m3 2.84 × 10−5 = 3.35 × 10−7 m2 s 0.01 kg H2 O/kg mixture 1m kg m2 ·s Contact resistance One place in which the usefulness of the electrical resistance analogy becomes immediately apparent is at the interface of two conducting media. No two solid surfaces will ever form perfect thermal contact when they are pressed together. Since some roughness is always present, a typical plane of contact will always include tiny air gaps as shown in Fig. 2.10 §2.3 Thermal resistance and the electrical analogy Figure 2.10 Heat transfer through the contact plane between two solid surfaces. (which is drawn with a highly exaggerated vertical scale). Heat transfer follows two paths through such an interface. Conduction through points of solid-to-solid contact is very effective, but conduction through the gasfilled interstices, which have low thermal conductivity, can be very poor. Thermal radiation across the gaps is also inefficient. We treat the contact surface by placing an interfacial conductance, hc , in series with the conducting materials on either side. The coefficient hc is similar to a heat transfer coefficient and has the same units, W/m2 K. If ∆T is the temperature difference across an interface of area A, then Q = Ahc ∆T . It follows that Q = ∆T /Rt for a contact resistance Rt = 1/(hc A) in K/W. The interfacial conductance, hc , depends on the following factors: • The surface finish and cleanliness of the contacting solids. • The materials that are in contact. • The pressure with which the surfaces are forced together. This may vary over the surface, for example, in the vicinity of a bolt. • The substance (or lack of it) in the interstitial spaces. Conductive shims or fillers can raise the interfacial conductance. • The temperature at the contact plane. The influence of contact pressure is usually a modest one up to around 10 atm in most metals. Beyond that, increasing plastic deformation of 65 66 Heat conduction, thermal resistance, and the overall heat transfer coefficient §2.3 Table 2.1 Some typical interfacial conductances for normal surface finishes and moderate contact pressures (about 1 to 10 atm). Air gaps not evacuated unless so indicated. hc (W/m2 K) Situation Iron/aluminum (70 atm pressure) Copper/copper Aluminum/aluminum Graphite/metals Ceramic/metals Stainless steel/stainless steel Ceramic/ceramic Stainless steel/stainless steel (evacuated interstices) Aluminum/aluminum (low pressure and evacuated interstices) 45, 000 10, 000 − 25, 000 2, 200 − 12, 000 3, 000 − 6, 000 1, 500 − 8, 500 2, 000 − 3, 700 500 − 3, 000 200 − 1, 100 100 − 400 the local contact points causes hc to increase more dramatically at high pressure. Table 2.1 gives typical values of contact resistances which bear out most of the preceding points. These values have been adapted from [2.1, Chpt. 3] and [2.2]. Theories of contact resistance are discussed in [2.3] and [2.4]. Example 2.4 Heat flows through two stainless steel slabs (k = 18 W/m·K) that are pressed together. The slab area is A = 1 m2 . How thick must the slabs be for contact resistance to be negligible? Solution. With reference to Fig. 2.11, the total or equivalent resistance is found by adding these resistances, which are in series: Rtequiv = L 1 1 L + + = kA hc A kA A L 1 L + + 18 hc 18 Since hc is about 3,000 W/m2 K, 1 2L must be = 0.00033 18 3000 Thus, L must be large compared to 18(0.00033)/2 = 0.003 m if contact resistance is to be ignored. If L = 3 cm, the error is about 10%. Thermal resistance and the electrical analogy §2.3 67 Figure 2.11 Conduction through two unit-area slabs with a contact resistance. Resistances for cylinders and for convection As we continue developing our method of solving one-dimensional heat conduction problems, we find that other avenues of heat flow may also be expressed as thermal resistances, and introduced into the solutions that we obtain. We also find that, once the heat conduction equation has been solved, the results themselves may be used as new thermal resistances. Example 2.5 Radial Heat Conduction in a Tube Find the temperature distribution and the heat flux for the long hollow cylinder shown in Fig. 2.12. Solution. Step 1. T = T (r ) Step 2. 1∂ r ∂r r ∂T ∂r + ˙ 1 ∂2T ∂2T q + + = 2 ∂φ2 2 r ∂z k =0, since T ≠ T (φ, z) Step 3. Integrate once: r =0 1 ∂T α ∂t =0, since steady ∂T = C1 ; integrate again: T = C1 ln r + C2 ∂r Step 4. T (r = ri ) = Ti and T (r = ro ) = To 68 Heat conduction, thermal resistance, and the overall heat transfer coefficient §2.3 Figure 2.12 Heat transfer through a cylinder with a fixed wall temperature (Example 2.5). Step 5. Ti = C1 ln ri + C2 To = C1 ln ro + C2 Step 6. T = Ti − ⎧ ∆T ⎪ ⎪ C = Ti − To = − ⎪1 ⎨ ln(ri /ro ) ln(ro /ri ) ⇒ ⎪ ∆T ⎪ ⎪ C2 = Ti + ln ri ⎩ ln(ro /ri ) ∆T (ln r − ln ri ) or ln(ro /ri ) ln(r /ri ) T − Ti = To − Ti ln(ro /ri ) (2.20) Step 7. The solution is plotted in Fig. 2.12. We see that the temperature profile is logarithmic and that it satisfies both boundary conditions. Furthermore, it is instructive to see what happens when the wall of the cylinder is very thin, or when ri /ro is close to 1. In this case: ln(r /ri ) r r − ri −1= ri ri Thermal resistance and the electrical analogy §2.3 and ro − ri ri ln(ro /ri ) Thus eqn. (2.20) becomes T − Ti r − ri = To − Ti ro − ri which is a simple linear profile. This is the same solution that we would get in a plane wall. Step 8. At any station, r : qradial = −k k∆T 1 ∂T =+ ∂r ln(ro /ri ) r So the heat flux falls off inversely with radius. That is reasonable, since the same heat flow must pass through an increasingly large surface as the radius increases. Let us see if this is the case for a cylinder of length l: Q (W) = (2π r l) q = 2π kl∆T ≠ f (r ) ln(ro /ri ) (2.21) Finally, we again recognize Ohm’s law in this result and write the thermal resistance for a cylinder: Rtcyl = ln(ro /ri ) 2π lk K W (2.22) This can be compared with the resistance of a plane wall: Rtwall = L kA K W Both resistances are inversely proportional to k, but each reflects a different geometry. In the preceding examples, the boundary conditions were all the same —a temperature specified at an outer edge. Next let us suppose that the temperature is specified in the environment away from a body, with a heat transfer coefficient between the environment and the body. 69 70 Heat conduction, thermal resistance, and the overall heat transfer coefficient §2.3 Figure 2.13 Heat transfer through a cylinder with a convective boundary condition (Example 2.6). Example 2.6 A Convective Boundary Condition A convective heat transfer coefficient around the outside of the cylinder in Example 2.5 provides thermal resistance between the cylinder and an environment at T = T∞ , as shown in Fig. 2.13. Find the temperature distribution and heat flux in this case. Solution. Step 1 through 3. These are the same as in Example 2.5. Step 4. The first boundary condition is T (r = ri ) = Ti . The second boundary condition must be expressed as an energy balance at the outer wall (recall Section 1.3). qconvection = qconduction at the wall or h(T − T∞ )r =ro = −k ∂T ∂r r =ro Step 5. From the first boundary condition we obtain Ti = C1 ln ri + C2 . It is easy to make mistakes when we substitute the general solution into the second boundary condition, so we will do it in Thermal resistance and the electrical analogy §2.3 detail: h (C1 ln r + C2 ) − T∞ r =ro = −k ∂ (C1 ln r + C2 ) ∂r r =ro (2.23) A common error is to substitute T = To on the lefthand side instead of substituting the entire general solution. That will do no good, because To is not an accessible piece of information. Equation (2.23) reduces to: h(T∞ − C1 ln ro − C2 ) = kC1 ro When we combine this with the result of the first boundary condition to eliminate C2 : C1 = − Ti − T∞ k (hro ) + ln(ro /ri ) = T∞ − Ti 1/Bi + ln(ro /ri ) Then C2 = Ti − T∞ − Ti ln ri 1/Bi + ln(ro /ri ) Step 6. T= T∞ − Ti ln(r /ri ) + Ti 1/Bi + ln(ro /ri ) This can be rearranged in fully dimensionless form: T − Ti ln(r /ri ) = T∞ − Ti 1/Bi + ln(ro /ri ) (2.24) Step 7. Let us fix a value of ro /ri —say, 2—and plot eqn. (2.24) for several values of the Biot number. The results are included in Fig. 2.13. Some very important things show up in this plot. When Bi 1, the solution reduces to the solution given in Example 2.5. It is as though the convective resistance to heat flow were not there. That is exactly what we anticipated in Section 1.3 for large Bi. When Bi 1, the opposite is true: (T − Ti ) (T∞ − Ti ) 71 72 Heat conduction, thermal resistance, and the overall heat transfer coefficient §2.3 Figure 2.14 Thermal circuit with two resistances. remains on the order of Bi, and internal conduction can be neglected. How big is big and how small is small? We do not really have to specify exactly. But in this case Bi < 0.1 signals constancy of temperature inside the cylinder with about ±3%. Bi > 20 means that we can neglect convection with about 5% error. Ti − T∞ 1 ∂T =k ∂r 1/Bi + ln(ro /ri ) r This can be written in terms of Q (W) = qradial (2π r l) for a cylinder of length l: Step 8. qradial = −k Q= Ti − T∞ Ti − T∞ = ln(ro /ri ) Rtconv + Rtcond + 2π kl h 2π ro l 1 (2.25) Equation (2.25) is once again analogous to Ohm’s law. But this time the denominator is the sum of two thermal resistances, as would be the case in a series circuit. We accordingly present the analogous electrical circuit in Fig. 2.14. The presence of convection on the outside surface of the cylinder causes a new thermal resistance of the form Rtconv = 1 hA (2.26) where A is the surface area over which convection occurs. Example 2.7 Critical Radius of Insulation An interesting consequence of the preceding result can be brought out with a specific example. Suppose that we insulate a 0.5 cm O.D. copper steam line with 85% magnesia to prevent the steam from condensing Thermal resistance and the electrical analogy §2.3 Figure 2.15 Thermal circuit for an insulated tube. too rapidly. The steam is under pressure and stays at 150◦ C. The copper is thin and highly conductive—obviously a tiny resistance in series with the convective and insulation resistances, as we see in Fig. 2.15. The condensation of steam inside the tube also offers very little resistance.3 But on the outside, a heat transfer coefficient of h = 20 W/m2 K offers fairly high resistance. It turns out that insulation can actually improve heat transfer in this case. The two significant resistances, for a cylinder of unit length (l = 1 m), are ln(ro /ri ) ln(ro /ri ) = K/W 2π kl 2π (0.074) 1 1 = K/W = 2π (20)ro 2π ro h Rtcond = Rtconv Figure 2.16 is a plot of these resistances and their sum. A very interesting thing occurs here. Rtconv falls off rapidly when ro is increased, because the outside area is increasing. Accordingly, the total resistance passes through a minimum in this case. Will it always do so? To find out, we differentiate eqn. (2.25), again setting l = 1 m: dQ = dro (Ti − T∞ ) ln(ro /ri ) + 2π k 2π ro h 1 2 − 1 2 2π ro h + 1 2π kro =0 When we solve this for the value of ro = rcrit at which Q is maximum and the total resistance is minimum, we obtain Bi = 1 = hrcrit k (2.27) In the present example, adding insulation will increase heat loss in3 Condensation heat transfer is discussed in Chapter 8. It turns out that h is generally enormous during condensation so that Rtcondensation is tiny. 73 Heat conduction, thermal resistance, and the overall heat transfer coefficient rcrit = 1.48 ri 4 Thermal resistance, Rt (K/W) 74 §2.3 Rtcond + Rtconv Rtconv 2 Rtcond 0 1.0 1.5 2.0 2.5 2.32 Radius ratio, ro/ri Figure 2.16 The critical radius of insulation (Example 2.7), written for a cylinder of unit length (l = 1 m). stead of reducing it, until rcrit = k h = 0.0037 m or rcrit /ri = 1.48. Indeed, insulation will not even start to do any good until ro /ri = 2.32 or ro = 0.0058 m. We call rcrit the critical radius of insulation. There is an interesting catch here. For most cylinders, rcrit < ri and the critical radius idiosyncrasy is of no concern. If our steam line had a 1 cm outside diameter, the critical radius difficulty would not have arisen. When cooling smaller diameter cylinders, such as electrical wiring, the critical radius must be considered, but one need not worry about it in the design of most large process equipment. Resistance for thermal radiation We saw in Chapter 1 that the net radiation exchanged by two objects is given by eqn. (1.34): 4 4 Qnet = A1 F1–2 σ T1 − T2 (1.34) When T1 and T2 are close, we can approximate this equation using a radiation heat transfer coefficient, hrad . Specifically, suppose that the temperature difference, ∆T = T1 − T2 , is small compared to the mean temperature, Tm = (T1 + T2 ) 2. Then we can make the following expan- Thermal resistance and the electrical analogy §2.3 sion and approximation: 4 4 Qnet = A1 F1–2 σ T1 − T2 2 2 2 2 = A1 F1–2 σ (T1 + T2 )(T1 − T2 ) = A1 F1–2 σ 2 2 (T1 + T2 ) (T1 + T2 ) (T1 − T2 ) 2 = 2Tm + (∆T )2 /2 =2Tm =∆T 3 A1 4σ Tm F1–2 ∆T (2.28) ≡hrad 2 where the last step assumes that (∆T )2 /2 2Tm or (∆T /Tm )2 /4 Thus, we have identified the radiation heat transfer coefficient ⎫ Qnet = A1 hrad ∆T ⎬ hrad = ⎭ 3 4σ Tm F1–2 for ∆T Tm 2 4 1 1. (2.29) This leads us immediately to the introduction of a radiation thermal resistance, analogous to that for convection: Rtrad = 1 A1 hrad (2.30) For the special case of a small object (1) in a much larger environment (2), the transfer factor is given by eqn. (1.35) as F1–2 = ε1 , so that 3 hrad = 4σ Tm ε1 (2.31) If the small object is black, its emittance is ε1 = 1 and hrad is maximized. For a black object radiating near room temperature, say Tm = 300 K, hrad = 4(5.67 × 10−8 )(300)3 6 W/m2 K This value is of approximately the same size as h for natural convection into a gas at such temperatures. Thus, the heat transfer by thermal radiation and natural convection into gases are similar. Both effects must be taken into account. In forced convection in gases, on the other hand, h might well be larger than hrad by an order of magnitude or more, so that thermal radiation can be neglected. 75 76 Heat conduction, thermal resistance, and the overall heat transfer coefficient §2.3 Example 2.8 An electrical resistor dissipating 0.1 W has been mounted well away from other components in an electronical cabinet. It is cylindrical with a 3.6 mm O.D. and a length of 10 mm. If the air in the cabinet is at 35◦ C and at rest, and the resistor has h = 13 W/m2 K for natural convection and ε = 0.9, what is the resistor’s temperature? Assume that the electrical leads are configured so that little heat is conducted into them. Solution. The resistor may be treated as a small object in a large isothermal environment. To compute hrad , let us estimate the resistor’s temperature as 50◦ C. Then Tm = (35 + 50)/2 43◦ C = 316 K so 3 hrad = 4σ Tm ε = 4(5.67 × 10−8 )(316)3 (0.9) = 6.44 W/m2 K Heat is lost by natural convection and thermal radiation acting in parallel. To find the equivalent thermal resistance, we combine the two parallel resistances as follows: 1 1 1 = + = Ahrad + Ah = A hrad + h Rtequiv Rtrad Rtconv Thus, Rtequiv = 1 A hrad + h A calculation shows A = 133 mm2 = 1.33 × 10−4 m2 for the resistor surface. Thus, the equivalent thermal resistance is Rtequiv = 1 = 386.8 K/W (1.33 × 10−4 )(13 + 6.44) Since Q= Tresistor − Tair Rtequiv We find Tresistor = Tair + Q · Rtequiv = 35 + (0.1)(386.8) = 73.68 ◦ C Thermal resistance and the electrical analogy §2.3 77 Tresistor Qconv Qrad 1 R t conv= – hA Qconv Tresistor Tair R t rad = 1 h rad A Figure 2.17 An electrical resistor cooled by convection and radiation. Qrad We guessed a resistor temperature of 50◦ C in finding hrad . Recomputing with this higher temperature, we have Tm = 327 K and hrad = 7.17 W/m2 K. If we repeat the rest of the calculation, we get a new value Tresistor = 72.3◦ C. Further iteration is not needed. Since the use of hrad is an approximation, we should check its applicability: 1 4 ∆T Tm 2 = 1 4 72.3 − 35.0 327 2 = 0.00325 1 In this case, the approximation is a very good one. Example 2.9 Suppose that power to the resistor in Example 2.8 is turned off. How long does it take to cool? The resistor has k 10 W/m·K, ρ 2000 kg/m3 , and cp 700 J/kg·K. Solution. The lumped capacity model, eqn. (1.22), may be applicable. To find out, we check the resistor’s Biot number, noting that the parallel convection and radiation processes have an effective heat 78 Heat conduction, thermal resistance, and the overall heat transfer coefficient §2.4 transfer coefficient heff = h + hrad = 18.44 W/m2 K. Then, Bi = (18.44)(0.0036/2) heff ro = = 0.0033 k 10 1 so eqn. (1.22) can be used to describe the cooling process. The time constant is T= ρcp V (2000)(700)π (0.010)(0.0036)2 /4 = = 58.1 s heff A (18.44)(1.33 × 10−4 ) From eqn. (1.22) with T0 = 72.3◦ C Tresistor = 35.0 + (72.3 − 35.0)e−t/58.1 ◦ C Ninety-five percent of the total temperature drop has occured when t = 3T = 174 s. 2.4 Overall heat transfer coefficient, U Definition We often want to transfer heat through composite resistances, such as the series of resistances shown in Fig. 2.18. It is very convenient to have a number, U , that works like this4 : Q = U A ∆T (2.32) This number, called the overall heat transfer coefficient, is defined largely by the system, and in many cases it proves to be insensitive to the operating conditions of the system. In Example 2.6, for instance, two resistances are in series. We can use the value Q given by eqn. (2.25) to get U= 1 Q (W) = ro ln(ro /ri ) 1 2π ro l (m2 ) ∆T (K) + k h (W/m2 K) (2.33) We have based U on the outside area, Ao = 2π ro l, in this case. We might instead have based it on inside area, Ai = 2π ri l, and obtained U= 1 ri ln(ro /ri ) + k hro ri (2.34) 4 This U must not be confused with internal energy. The two terms should always be distinct in context. Overall heat transfer coefficient, U §2.4 79 Figure 2.18 A thermal circuit with many resistances in series. The equivalent resistance is Rtequiv = i Ri . It is therefore important to remember which area an overall heat transfer coefficient is based on. It is particularly important that A and U be consistent when we write Q = U A ∆T . In general, for any composite resistance, the overall heat transfer coefficient may be obtained from the equivalent resistance. The equivalent resistance is calculated taking account of series and parallel resistors, as in Examples 2.4 and 2.8. Then, because Q = ∆T /Rtequiv = U A ∆T , it follows that U A = 1/Rtequiv . Example 2.10 Estimate the overall heat transfer coefficient for the tea kettle shown in Fig. 2.19. Note that the flame convects heat to the thin aluminum. The heat is then conducted through the aluminum and finally convected by boiling into the water. Solution. We need not worry about deciding which area to base A on because the area normal to the heat flux vector does not change. We simply write the heat flow Q= Tflame − Tboiling water ∆T = 1 1 L Rt + + hA kAl A hb A and apply the definition of U U= Q 1 = L 1 1 A∆T + + h kAl hb Let us see what typical numbers would look like in this example: h might be around 200 W/m2 K; L kAl might be 0.001 m/(160 W/m·K) or 1/160,000 W/m2 K; and hb is quite large— perhaps about 5000 W/m2 K. Thus: U 1 = 192.1 W/m2 K 1 1 1 + + 200 160, 000 5000 80 Heat conduction, thermal resistance, and the overall heat transfer coefficient §2.4 Figure 2.19 Heat transfer through the bottom of a tea kettle. It is clear that the first resistance is dominant, as is shown in Fig. 2.19. Notice that in such cases U A → 1/Rtdominant (2.35) where A is any area (inside or outside) in the thermal circuit. Experiment 2.1 Boil water in a paper cup over an open flame and explain why you can do so. [Recall eqn. (2.35) and see Problem 2.12.] Example 2.11 A wall consists of alternating layers of pine and sawdust, as shown in Fig. 2.20). The sheathes on the outside have negligible resistance and h is known on the sides. Compute Q and U for the wall. Solution. So long as the wood and the sawdust do not differ dramatically from one another in thermal conductivity, we can approximate the wall as a parallel resistance circuit, as shown in the figure.5 The 5 For this approximation to be exact, the resistances must be equal. If they differ radically, the problem must be treated as two-dimensional. Overall heat transfer coefficient, U §2.4 Figure 2.20 Heat transfer through a composite wall. equivalent thermal resistance of the circuit is 1 Rtequiv = Rtconv + 1 Rtpine + + Rtconv 1 Rtsawdust Thus Q= ∆T = Rtequiv 1 hA T∞1 − T∞r + 1 kp Ap L + ks As + 1 hA L and U= Q = A∆T 2 h 1 + 1 kp Ap LA + ks As LA The approach illustrated in this example is very widely used in calculating U values for the walls and roofs houses and buildings. The thermal resistances of each structural element — insulation, studs, siding, doors, windows, etc. — are combined to calculate U or Rtequiv , which is then used together with weather data to estimate heating and cooling loads [2.5]. 81 82 Heat conduction, thermal resistance, and the overall heat transfer coefficient §2.4 Table 2.2 Typical ranges or magnitudes of U Heat Exchange Configuration Walls and roofs dwellings with a 24 km/h outdoor wind: • Insulated roofs • Finished masonry walls • Frame walls • Uninsulated roofs Single-pane windows Air to heavy tars and oils Air to low-viscosity liquids Air to various gases Steam or water to oil Liquids in coils immersed in liquids Feedwater heaters Air condensers Steam-jacketed, agitated vessels Shell-and-tube ammonia condensers Steam condensers with 25◦ C water Condensing steam to high-pressure boiling water † U (W/m2 K) 0.3−2 0.5−6 0.3−5 1.2−4 ∼ 6† As low as 45 As high as 600 60−550 60−340 110−2, 000 110−8, 500 350−780 500−1, 900 800−1, 400 1, 500−5, 000 1, 500−10, 000 Main heat loss is by infiltration. Typical values of U In a fairly general use of the word, a heat exchanger is anything that lies between two fluid masses at different temperatures. In this sense a heat exchanger might be designed either to impede or to enhance heat exchange. Consider some typical values of U shown in Table 2.2, which were assembled from a variety of technical sources. If the exchanger is intended to improve heat exchange, U will generally be much greater than 40 W/m2 K. If it is intended to impede heat flow, it will be less than 10 W/m2 K—anywhere down to almost perfect insulation. You should have some numerical concept of relative values of U , so we recommend that you scrutinize the numbers in Table 2.2. Some things worth bearing in mind are: • The fluids with low thermal conductivities, such as tars, oils, or any of the gases, usually yield low values of h. When such fluid flows on one side of an exchanger, U will generally be pulled down. Overall heat transfer coefficient, U §2.4 • Condensing and boiling are very effective heat transfer processes. They greatly improve U but they cannot override one very small value of h on the other side of the exchange. (Recall Example 2.10.) In fact: • For a high U , all resistances in the exchanger must be low. • The highly conducting liquids, such as water and liquid metals, give high values of h and U . Fouling resistance Figure 2.21 shows one of the simplest forms of a heat exchanger—a pipe. The inside is new and clean on the left, but on the right it has built up a layer of scale. In conventional freshwater preheaters, for example, this scale is typically MgSO4 (magnesium sulfate) or CaSO4 (calcium sulfate) which precipitates onto the pipe wall after a time. To account for the resistance offered by these buildups, we must include an additional, highly empirical resistance when we calculate U . Thus, for the pipe shown in Fig. 2.21, U older pipe based on Ai = 1 1 hi + ri ln(ro /rp ) kinsul + ri ln(rp /ri ) kpipe + Figure 2.21 The fouling of a pipe. ri ro ho + Rf 83 84 Heat conduction, thermal resistance, and the overall heat transfer coefficient §2.4 Table 2.3 Some typical fouling resistances for a unit area. Fouling Resistance Rf (m2 K/W) Fluid and Situation Distilled water Seawater Treated boiler feedwater Clean river or lake water About the worst waters used in heat exchangers No. 6 fuel oil Transformer or lubricating oil Most industrial liquids Most refinery liquids Steam, non-oil-bearing Steam, oil-bearing (e.g., turbine exhaust) Most stable gases Flue gases Refrigerant vapors (oil-bearing) 0.0001 0.0001 − 0.0004 0.0001 − 0.0002 0.0002 − 0.0006 < 0.0020 0.0001 0.0002 0.0002 0.0002 − 0.0009 0.0001 0.0003 0.0002 − 0.0004 0.0010 − 0.0020 0.0040 where Rf is a fouling resistance for a unit area of pipe (in m2 K/W). And clearly Rf ≡ 1 1 − Uold Unew (2.36) Some typical values of Rf are given in Table 2.3. These values have been adapted from [2.6] and [2.7]. Notice that fouling has the effect of adding a resistance in series on the order of 10−4 m2 K/W. It is rather like another heat transfer coefficient, hf , on the order of 10,000 W/m2 K in series with the other resistances in the exchanger. The tabulated values of Rf are given to only one significant figure because they are very approximate. Clearly, exact values would have to be referred to specific heat exchanger configurations, to particular fluids, to fluid velocities, to operating temperatures, and to age [2.8, 2.9]. The resistance generally drops with increased velocity and increases with temperature and age. The values given in the table are based on reasonable Overall heat transfer coefficient, U §2.4 maintenance and the use of conventional shell-and-tube heat exchangers. With misuse, a given heat exchanger can yield much higher values of Rf . Notice too, that if U 1, 000 W/m2 K, fouling will be unimportant because it will introduce a negligibly small resistance in series. Thus, in a water-to-water heat exchanger, for which U is on the order of 2000 W/m2 K, fouling might be important; but in a finned-tube heat exchanger with hot gas in the tubes and cold gas passing across the fins on them, U might be around 200 W/m2 K, and fouling will be usually be insignificant. Example 2.12 You have unpainted aluminum siding on your house and the engineer has based a heat loss calculation on U = 5 W/m2 K. You discover that air pollution levels are such that Rf is 0.0005 m2 K/W on the siding. Should the engineer redesign the siding? Solution. From eqn. (2.36) we get 1 Ucorrected = 1 Uuncorrected + Rf = 0.2000 + 0.0005 m2 K/W Therefore, fouling is entirely irrelevant to domestic heat loads. Example 2.13 Since the engineer did not fail you in the preceding calculation, you entrust him with the installation of a heat exchanger at your plant. He installs a water-cooled steam condenser with U = 4000 W/m2 K. You discover that he used water-side fouling resistance for distilled water but that the water flowing in the tubes is not clear at all. How did he do this time? Solution. Equation (2.36) and Table 2.3 give 1 Ucorrected = 1 + (0.0006 to 0.0020) 4000 = 0.00085 to 0.00225 m2 K/W Thus, U is reduced from 4,000 to between 444 and 1,176 W/m2 K. Fouling is crucial in this case, and the engineer was in serious error. 85 86 Chapter 2: Heat conduction, thermal resistance, and the overall heat transfer coefficient 2.5 Summary Four things have been done in this chapter: • The heat diffusion equation has been established. A method has been established for solving it in simple problems, and some important results have been presented. (We say much more about solving the heat diffusion equation in Part II of this book.) • We have explored the electric analogy to steady heat flow, paying special attention to the concept of thermal resistance. We exploited the analogy to solve heat transfer problems in the same way we solve electrical circuit problems. • The overall heat transfer coefficient has been defined, and we have seen how to build it up out of component resistances. • Some practical problems encountered in the evaluation of overall heat transfer coefficients have been discussed. Three very important things have not been considered in Chapter 2: • In all evaluations of U that involve values of h, we have taken these values as given information. In any real situation, we must determine correct values of h for the specific situation. Part III deals with such determinations. • When fluids flow through heat exchangers, they give up or gain energy. Thus, the driving temperature difference varies through the exchanger. (Problem 2.14 asks you to consider this difficulty in its simplest form.) Accordingly, the design of an exchanger is complicated. We deal with this problem in Chapter 3. • The heat transfer coefficients themselves vary with position inside many types of heat exchangers, causing U to be position-dependent. Problems 2.1 Prove that if k varies linearly with T in a slab, and if heat transfer is one-dimensional and steady, then q may be evaluated precisely using k evaluated at the mean temperature in the slab. Problems 87 2.2 Invent a numerical method for calculating the steady heat flux through a plane wall when k(T ) is an arbitrary function. Use the method to predict q in an iron slab 1 cm thick if the temperature varies from −100◦ C on the left to 400◦ C on the right. How far would you have erred if you had taken kaverage = (kleft + kright )/2? 2.3 The steady heat flux at one side of a slab is a known value qo . The thermal conductivity varies with temperature in the slab, and the variation can be expressed with a power series as i=n Ai T i k= i=0 (a) Start with eqn. (2.10) and derive an equation that relates T to position in the slab, x . (b) Calculate the heat flux at any position in the wall from this expression using Fourier’s law. Is the resulting q a function of x ? 2.4 Combine Fick’s law with the principle of conservation of mass (of the dilute species) in such a way as to eliminate j1 , and obtain a second-order differential equation in m1 . Discuss the importance and the use of the result. 2.5 Solve for the temperature distribution in a thick-walled pipe if the bulk interior temperature and the exterior air temperature, T∞i , and T∞o , are known. The interior and the exterior heat transfer coefficients are hi and ho , respectively. Follow the method in Example 2.6 and put your result in the dimensionless form: T − T ∞i = fn (Bii , Bio , r /ri , ro /ri ) T∞i − T∞o 2.6 Put the boundary conditions from Problem 2.5 into dimensionless form so that the Biot numbers appear in them. Let the Biot numbers approach infinity. This should get you back to the boundary conditions for Example 2.5. Therefore, the solution that you obtain in Problem 2.5 should reduce to the solution of Example 2.5 when the Biot numbers approach infinity. Show that this is the case. 88 Chapter 2: Heat conduction, thermal resistance, and the overall heat transfer coefficient Figure 2.22 Configuration for Problem 2.8. 2.7 Write an accurate explanation of the idea of critical radius of insulation that your kid brother or sister, who is still in grade school, could understand. (If you do not have an available kid, borrow one to see if your explanation really works.) 2.8 The slab shown in Fig. 2.22 is embedded on five sides in insulating materials. The sixth side is exposed to an ambient temperature through a heat transfer coefficient. Heat is generated in the slab at the rate of 1.0 kW/m3 The thermal conductivity of the slab is 0.2 W/m·K. (a) Solve for the temperature distribution in the slab, noting any assumptions you must make. Be careful to clearly identify the boundary conditions. (b) Evaluate T at the front and back faces of the slab. (c) Show that your solution gives the expected heat fluxes at the back and front faces. 2.9 Consider the composite wall shown in Fig. 2.23. The concrete and brick sections are of equal thickness. Determine T1 , T2 , q, and the percentage of q that flows through the brick. To do this, approximate the heat flow as one-dimensional. Draw the thermal circuit for the wall and identify all four resistances before you begin. 2.10 Compute Q and U for Example 2.11 if the wall is 0.3 m thick. Five (each) pine and sawdust layers are 5 and 8 cm thick, re- Problems 89 spectively; and the heat transfer coefficients are 10 on the left and 18 on the right. T∞1 = 30◦ C and T∞r = 10◦ C. 2.11 Compute U for the slab in Example 1.2. 2.12 Consider the tea kettle in Example 2.10. Suppose that the kettle holds 1 kg of water (about 1 liter) and that the flame impinges on 0.02 m2 of the bottom. (a) Find out how fast the water temperature is increasing when it reaches its boiling point, and calculate the temperature of the bottom of the kettle immediately below the water if the gases from the flame are at 500◦ C when they touch the bottom of the kettle. Assume that the heat capacitance of the aluminum kettle is negligible. (b) There is an old parlor trick in which one puts a paper cup of water over an open flame and boils the water without burning the paper (see Experiment 2.1). Explain this using an electrical analogy. [(a): dT /dt = 0.37◦ C/s.] 2.13 Copper plates 2 mm and 3 mm in thickness are processed rather lightly together. Non-oil-bearing steam condenses under pressure at Tsat = 200◦ C on one side (h = 12, 000 W/m2 K) and methanol boils under pressure at 130◦ Con the other (h = 9000 W/m2 K). Estimate U and q initially and after extended service. List the relevant thermal resistances in order of decreasing importance and suggest whether or not any of them can be ignored. 2.14 0.5 kg/s of air at 20◦ C moves along a channel that is 1 m from wall to wall. One wall of the channel is a heat exchange surface Figure 2.23 Configuration for Problem 2.9. 90 Chapter 2: Heat conduction, thermal resistance, and the overall heat transfer coefficient (U = 300 W/m2 K) with steam condensing at 120◦ C on its back. Determine (a) q at the entrance; (b) the rate of increase of temperature of the fluid with x at the entrance; (c) the temperature and heat flux 2 m downstream. [(c): T2m = 89.7◦ C.] 2.15 An isothermal sphere 3 cm in diameter is kept at 80◦ C in a large clay region. The temperature of the clay far from the sphere is kept at 10◦ C. How much heat must be supplied to the sphere to maintain its temperature if kclay = 1.28 W/m·K? (Hint: You must solve the boundary value problem not in the sphere but in the clay surrounding it.) [Q = 16.9 W.] 2.16 Is it possible to increase the heat transfer from a convectively cooled isothermal sphere by adding insulation? Explain fully. 2.17 A wall consists of layers of metals and plastic with heat transfer coefficients on either side. U is 255 W/m2 K and the overall temperature difference is 200◦ C. One layer in the wall is stainless steel (k = 18 W/m·K) 3 mm thick. What is ∆T across the stainless steel? 2.18 A 1% carbon-steel sphere 20 cm in diameter is kept at 250◦ C on the outside. It has an 8 cm diameter cavity containing boiling water (hinside is very high) which is vented to the atmosphere. What is Q through the shell? 2.19 A slab is insulated on one side and exposed to a surrounding temperature, T∞ , through a heat transfer coefficient on the other. There is nonuniform heat generation in the slab such ˙ that q =[A (W/m4 )][x (m)], where x = 0 at the insulated wall and x = L at the cooled wall. Derive the temperature distribution in the slab. 2.20 800 W/m3 of heat is generated within a 10 cm diameter nickelsteel sphere for which k = 10 W/m·K. The environment is at 20◦ C and there is a natural convection heat transfer coefficient of 10 W/m2 K around the outside of the sphere. What is its center temperature at the steady state? [21.37◦ C.] 2.21 An outside pipe is insulated and we measure its temperature with a thermocouple. The pipe serves as an electrical resis˙ tance heater, and q is known from resistance and current mea- Problems 91 surements. The inside of the pipe is cooled by the flow of liquid with a known bulk temperature. Evaluate the heat transfer coefficient, h, in terms of known information. The pipe dimensions and properties are known. [Hint: Remember that h is not known and we cannot use a boundary condition of the third kind at the inner wall to get T (r ).] 2.22 Consider the hot water heater in Problem 1.11. Suppose that it is insulated with 2 cm of a material for which k = 0.12 W/m·K, and suppose that h = 16 W/m2 K. Find (a) the time constant T for the tank, neglecting the casing and insulation; (b) the initial rate of cooling in ◦ C/h; (c) the time required for the water to cool from its initial temperature of 75◦ C to 40◦ C; (d) the percentage of additional heat loss that would result if an outer casing for the insulation were held on by eight steel rods, 1 cm in diameter, between the inner and outer casings. 2.23 A slab of thickness L is subjected to a constant heat flux, q1 , on the left side. The right-hand side if cooled convectively by an environment at T∞ . (a) Develop a dimensionless equation for the temperature of the slab. (b) Present dimensionless equation for the left- and right-hand wall temperatures as well. (c) If the wall is firebrick, 10 cm thick, q1 is 400 W/m2 , h = 20 W/m2 K, and T∞ = 20◦ C, compute the lefthand and righthand temperatures. 2.24 Heat flows steadily through a stainless steel wall of thickness Lss = 0.06 m, with a variable thermal conductivity of kss = 1.67 + 0.0143 T(◦ C). It is partially insulated on the right side with glass wool of thickness Lgw = 0.1 m, with a thermal conductivity of kgw = 0.04. The temperature on the left-hand side of the stainless stell is 400◦ Cand on the right-hand side if the glass wool is 100◦ C. Evaluate q and Ti . 2.25 Rework Problem 1.29 with a heat transfer coefficient, ho = 40 W/m2 K on the outside (i.e., on the cold side). 2.26 A scientist proposes an experiment for the space shuttle in which he provides underwater illumination in a large tank of water at 20◦ C, using a 3 cm diameter spherical light bulb. What is the maximum wattage of the bulb in zero gravity that will not boil the water? 92 Chapter 2: Heat conduction, thermal resistance, and the overall heat transfer coefficient 2.27 A cylindrical shell is made of two layers– an inner one with inner radius = ri and outer radius = rc and an outer one with inner radius = rc and outer radius = ro . There is a contact resistance, hc , between the shells. The materials are different, and T1 (r = ri ) = Ti and T2 (r = ro ) = To . Derive an expression for the inner temperature of the outer shell (T2c ). 2.28 A 1 kW commercial electric heating rod, 8 mm in diameter and 0.3 m long, is to be used in a highly corrosive gaseous environment. Therefore, it has to be provided with a cylindrical sheath of fireclay. The gas flows by at 120◦ C, and h is 230 W/m2 K outside the sheath. The surface of the heating rod cannot exceed 800◦ C. Set the maximum sheath thickness and the outer temperature of the fireclay. [Hint: use heat flux and temperature boundary conditions to get the temperature distribution. Then use the additional convective boundary condition to obtain the sheath thickness.] 2.29 A very small diameter, electrically insulated heating wire runs down the center of a 7.5 mm diameter rod of type 304 stainless steel. The outside is cooled by natural convection (h = 6.7 W/m2 K) in room air at 22◦ C. If the wire releases 12 W/m, plot Trod vs. radial position in the rod and give the outside temperature of the rod. (Stop and consider carefully the boundary conditions for this problem.) 2.30 A contact resistance experiment involves pressing two slabs of different materials together, putting a known heat flux through them, and measuring the outside temperatures of each slab. Write the general expression for hc in terms of known quantities. Then calculate hc if the slabs are 2 cm thick copper and 1.5 cm thick aluminum, if q is 30,000 W/m2 , and if the two temperatures are 15◦ C and 22.1◦ C. 2.31 A student working heat transfer problems late at night needs a cup of hot cocoa to stay awake. She puts milk in a pan on an electric stove and seeks to heat it as rapidly as she can, without burning the milk, by turning the stove on high and stirring the milk continuously. Explain how this works using an analogous electric circuit. Is it possible to bring the entire bulk of the milk up to the burn temperature without burning part of it? Problems 93 2.32 A small, spherical hot air balloon, 10 m in diameter, weighs 130 kg with a small gondola and one passenger. How much fuel must be consumed (in kJ/h) if it is to hover at low altitude in still 27◦ C air? (houtside = 215 W/m2 K, as the result of natural convection.) 2.33 A slab of mild steel, 4 cm thick, is held at 1,000◦ C on the back side. The front side is approximately black and radiates to black surroundings at 100◦ C. What is the temperature of the front side? 2.34 With reference to Fig. 2.3, develop an empirical equation for k(T ) for ammonia vapor. Then imagine a hot surface at Tw parallel with a cool horizontal surface at a distance H below it. Develop equations for T (x) and q. Compute q if Tw = 350◦ C, Tcool = −5◦ C, and H = 0.15 m. 2.35 A type 316 stainless steel pipe has a 6 cm inside diameter and an 8 cm outside diameter with a 2 mm layer of 85% magnesia insulation around it. Liquid at 112◦ C flows inside, so hi = 346 W/m2 K. The air around the pipe is at 20◦ C, and h0 = 6 W/m2 K. Calculate U based on the inside area. Sketch the equivalent electrical circuit, showing all known temperatures. Discuss the results. 2.36 Two highly reflecting, horizontal plates are spaced 0.0005 m apart. The upper one is kept at 1000◦ C and the lower one at 200◦ C. There is air in between. Neglect radiation and compute the heat flux and the midpoint temperature in the air. Use a power-law fit of the form k = a(T ◦ C)b to represent the air data in Table A.6. 2.37 A 0.1 m thick slab with k = 3.4 W/m·K is held at 100◦ C on the left side. The right side is cooled with air at 20◦ C through a heat transfer coefficient, and h = (5.1 W/m2 (K)−5/4 )(Twall − T∞ )1/4 . Find q and Twall on the right. 2.38 Heat is generated at 54,000 W/m3 in a 0.16 m diameter sphere. The sphere is cooled by natural convection with fluid at 0◦ C, and h = [2 + 6(Tsurface − T∞ )1/4 ] W/m2 K, ksphere = 9 W/m·K. Find the surface temperature and center temperature of the sphere. 94 Chapter 2: Heat conduction, thermal resistance, and the overall heat transfer coefficient 2.39 Layers of equal thickness of spruce and pitch pine are laminated to make an insulating material. How should the laminations be oriented in a temperature gradient to achieve the best effect? 2.40 The resistances of a thick cylindrical layer of insulation must be increased. Will Q be lowered more by a small increase of the outside diameter or by the same decrease in the inside diameter? 2.41 You are in charge of energy conservation at your plant. There is a 300 m run of 6 in. O.D. pipe carrying steam at 250◦ C. The company requires that any insulation must pay for itself in one year. The thermal resistances are such that the surface of the pipe will stay close to 250◦ C in air at 25◦ C when h = 10 W/m2 K. Calculate the annual energy savings in kW·h that will result if a 1 in layer of 85% magnesia insulation is added. If energy is worth 6 cents per kW·h and insulation costs $75 per installed linear meter, will the insulation pay for itself in one year? 2.42 An exterior wall of a wood-frame house is typically composed, from outside to inside, of a layer of wooden siding, a layer glass fiber insulation, and a layer of gypsum wall board. Standard glass fiber insulation has a thickness of 3.5 inch and a conductivity of 0.038 W/m·K. Gypsum wall board is normally 0.50 inch thick with a conductivity of 0.17 W/m·K, and the siding can be assumed to be 1.0 inch thick with a conductivity of 0.10 W/m·K. a. Find the overall thermal resistance of such a wall (in K/W) if it has an area of 400 ft2 . b. Convection and radiation processes on the inside and outside of the wall introduce more thermal resistance. Assuming that the effective outside heat transfer coefficient (accounting for both convection and radiation) is ho = 20 W/m2 K and that for the inside is hi = 10 W/m2 K, determine the total thermal resistance for heat loss from the indoors to the outdoors. Also obtain an overall heat transfer coefficient, U , in W/m2 K. Problems 95 c. If the interior temperature is 20◦ C and the outdoor temperature is −5◦ C, find the heat loss through the wall in watts and the heat flux in W/m2 . d. Which of the five thermal resistances is dominant? 2.43 We found that the thermal resistance of a cylinder was Rtcyl = (1/2π kl) ln(ro /ri ). If ro = ri + δ, show that the thermal resistance of a thin-walled cylinder (δ ri ) can be approximated by that for a slab of thickness δ. Thus, Rtthin = δ/(kAi ), where Ai = 2π ri l is the inside surface area of the cylinder. How much error is introduced by this approximation if δ/ri = 0.2? [Hint: Use a Taylor series.] 2.44 A Gardon gage measures a radiation heat flux by detecting a temperature difference [2.10]. The gage consists of a circular constantan membrane of radius R , thickness t , and thermal conductivity kct which is joined to a heavy copper heat sink at its edges. When a radiant heat flux qrad is absorbed by the membrane, heat flows from the interior of the membrane to the copper heat sink at the edge, creating a radial temperature gradient. Copper leads are welded to the center of the membrane and to the copper heat sink, making two copperconstantan thermocouple junctions. These junctions measure the temperature difference ∆T between the center of the membrane, T (r = 0), and the edge of the membrane, T (r = R). The following approximations can be made: • The membrane surface has been blackened so that it absorbs all radiation that falls on it • The radiant heat flux is much larger than the heat lost from the membrane by convection or re-radiation. Thus, all absorbed radiant heat is removed from the membrane by conduction to the copper heat sink, and other loses can be ignored • The gage operates in steady state • The membrane is thin enough (t R ) that the temperature in it varies only with r , i.e., T = T (r ) only. Answer the following questions. 96 Chapter 2: Heat conduction, thermal resistance, and the overall heat transfer coefficient a. For a fixed copper heat sink temperature, T (r = R), sketch the shape of the temperature distribution in the membrane, T (r ), for two arbitrary heat radiant fluxes qrad 1 and qrad 2 , where qrad 1 > qrad 2 . b. Find the relationship between the radiant heat flux, qrad , and the temperature difference obtained from the thermocouples, ∆T . Hint: Treat the absorbed radiant heat flux as if it were a volumetric heat source of magnitude qrad /t (W/m3 ). 2.45 You have a 12 oz. (375 mL) can of soda at room temperature (70◦ F) that you would like to cool to 45◦ F before drinking. You rest the can on its side on the plastic rods of the refrigerator shelf. The can is 2.5 inches in diameter and 5 inches long. The can’s emissivity is ε = 0.4 and the natural convection heat transfer coefficient around it is a function of the temperature difference between the can and the air: h = 2 ∆T 1/4 for ∆T in kelvin. Assume that thermal interactions with the refrigerator shelf are negligible and that buoyancy currents inside the can will keep the soda well mixed. a. Estimate how long it will take to cool the can in the refrigerator compartment, which is at 40◦ F. b. Estimate how long it will take to cool the can in the freezer compartment, which is at 5◦ F. c. Are your answers for parts 1 and 2 the same? If not, what is the main reason that they are different? References [2.1] W. M. Rohsenow and J. P. Hartnett, editors. Handbook of Heat Transfer. McGraw-Hill Book Company, New York, 1973. [2.2] R. F. Wheeler. Thermal conductance of fuel element materials. USAEC Rep. HW-60343, April 1959. [2.3] M. M. Yovanovich. Recent developments in thermal contact, gap and joint conductance theories and experiment. In Proc. Eight Intl. Heat Transfer Conf., volume 1, pages 35–45. San Francisco, 1986. References [2.4] C. V. Madhusudana. Thermal Contact Conductance. SpringerVerlag, New York, 1996. [2.5] American Society of Heating, Refrigerating, and Air-Conditioning Engineers, Inc. 2001 ASHRAE Handbook—Fundamentals. Altanta, 2001. [2.6] R. K. Shah and D. P. Sekulic. Heat exchangers. In W. M. Rohsenow, J. P. Hartnett, and Y. I. Cho, editors, Handbook of Heat Transfer, chapter 17. McGraw-Hill, New York, 3rd edition, 1998. [2.7] Tubular Exchanger Manufacturer’s Association. Standards of Tubular Exchanger Manufacturer’s Association. New York, 4th and 6th edition, 1959 and 1978. [2.8] H. Müller-Steinhagen. Cooling-water fouling in heat exchangers. In T. F. Irvine, Jr., J. P. Hartnett, Y. I. Cho, and G. A. Greene, editors, Advances in Heat Transfer, volume 33, pages 415–496. Academic Press, Inc., San Diego, 1999. [2.9] W. J. Marner and J. W. Suitor. Fouling with convective heat transfer. In S. Kakaç, R. K. Shah, and W. Aung, editors, Handbook of SinglePhase Convective Heat Transfer, chapter 21. Wiley-Interscience, New York, 1987. [2.10] R. Gardon. An instrument for the direct measurement of intense thermal radiation. Rev. Sci. Instr., 24(5):366–371, 1953. Most of the ideas in Chapter 2 are also dealt with at various levels in the general references following Chapter 1. 97 3. Heat exchanger design The great object to be effected in the boilers of these engines is, to keep a small quantity of water at an excessive temperature, by means of a small amount of fuel kept in the most active state of combustion. . .No contrivance can be less adapted for the attainment of this end than one or two large tubes traversing the boiler, as in the earliest locomotive engines. The Steam Engine Familiarly Explained and Illustrated, Dionysus Lardner, 1836 3.1 Function and configuration of heat exchangers The archetypical problem that any heat exchanger solves is that of getting energy from one fluid mass to another, as we see in Fig. 3.1. A simple or composite wall of some kind divides the two flows and provides an element of thermal resistance between them. There is an exception to this configuration in the direct-contact form of heat exchanger. Figure 3.2 shows one such arrangement in which steam is bubbled into water. The steam condenses and the water is heated at the same time. In other arrangements, immiscible fluids might contact each other or noncondensible gases might be bubbled through liquids. This discussion will be restricted to heat exchangers with a dividing wall between the two fluids. There is an enormous variety of such configurations, but most commercial exchangers reduce to one of three basic types. Figure 3.3 shows these types in schematic form. They are: • The simple parallel or counterflow configuration. These arrangements are versatile. Figure 3.4 shows how the counterflow arrangement is bent around in a so-called Heliflow compact heat exchanger configuration. • The shell-and-tube configuration. Figure 3.5 shows the U-tubes of a two-tube-pass, one-shell-pass exchanger being installed in the 99 100 Heat exchanger design §3.1 Figure 3.1 Heat exchange. supporting baffles. The shell is yet to be added. Most of the really large heat exchangers are of the shell-and-tube form. • The cross-flow configuration. Figure 3.6 shows typical cross-flow units. In Fig. 3.6a and c, both flows are unmixed. Each flow must stay in a prescribed path through the exchanger and is not allowed to “mix” to the right or left. Figure 3.6b shows a typical plate-fin cross-flow element. Here the flows are also unmixed. Figure 3.7, taken from the standards of the Tubular Exchanger Manufacturer’s Association (TEMA) [3.1], shows four typical single-shell-pass heat exchangers and establishes nomenclature for such units. These pictures also show some of the complications that arise in translating simple concepts into hardware. Figure 3.7 shows an exchanger with a single tube pass. Although the shell flow is baffled so that it crisscrosses the tubes, it still proceeds from the hot to cold (or cold to hot) end of the shell. Therefore, it is like a simple parallel (or counterflow) unit. The kettle reboiler in Fig. 3.7d involves a divided shell-pass flow configuration over two tube passes (from left to right and back to the “channel header”). In this case, the isothermal shell flow could be flowing in any direction—it makes no difference to the tube flow. Therefore, this exchanger is also equivalent to either the simple parallel or counterflow configuration. Function and configuration of heat exchangers §3.1 Figure 3.2 A direct-contact heat exchanger. Notice that a salient feature of shell-and-tube exchangers is the presence of baffles. Baffles serve to direct the flow normal to the tubes. We find in Part III that heat transfer from a tube to a flowing fluid is usually better when the flow moves across the tube than when the flow moves along the tube. This augmentation of heat transfer gives the complicated shell-and-tube exchanger an advantage over the simpler single-pass parallel and counterflow exchangers. However, baffles bring with them a variety of problems. The flow patterns are very complicated and almost defy analysis. A good deal of the shell-side fluid might unpredictably leak through the baffle holes in the axial direction, or it might bypass the baffles near the wall. In certain shell-flow configurations, unanticipated vibrational modes of the tubes might be excited. Many of the cross-flow configurations also baffle the fluid so as to move it across a tube bundle. The plate-and-fin configuration (Fig. 3.6b) is such a cross-flow heat exchanger. In all of these heat exchanger arrangements, it becomes clear that a dramatic investment of human ingenuity is directed towards the task of augmenting the heat transfer from one flow to another. The variations are endless, as you will quickly see if you try Experiment 3.1. Experiment 3.1 Carry a notebook with you for a day and mark down every heat exchanger you encounter in home, university, or automobile. Classify each according to type and note any special augmentation features. The analysis of heat exchangers first becomes complicated when we account for the fact that two flow streams change one another’s temper- 101 Figure 3.3 The three basic types of heat exchangers. 102 §3.2 Evaluation of the mean temperature difference in a heat exchanger Figure 3.4 Heliflow compact counterflow heat exchanger. (Photograph coutesy of Graham Manufacturing Co., Inc., Batavia, New York.) ature. It is to the problem of predicting an appropriate mean temperature difference that we address ourselves in Section 3.2. Section 3.3 then presents a strategy to use when this mean cannot be determined initially. 3.2 Evaluation of the mean temperature difference in a heat exchanger Logarithmic mean temperature difference (LMTD) To begin with, we take U to be a constant value. This is fairly reasonable in compact single-phase heat exchangers. In larger exchangers, particularly in shell-and-tube configurations and large condensers, U is apt to vary with position in the exchanger and/or with local temperature. But in situations in which U is fairly constant, we can deal with the varying temperatures of the fluid streams by writing the overall heat transfer in terms of a mean temperature difference between the two fluid streams: Q = U A ∆Tmean (3.1) 103 Figure 3.5 Typical commercial one-shell-pass, two-tube-pass heat exchangers. 104 a. A 1980 Chevette radiator. Cross-flow exchanger with neither flow mixed. Edges of flat vertical tubes can be seen. c. The basic 1 ft. × 1 ft.× 2 ft. module for a waste heat recuperator. It is a plate-fin, gas-to-air cross-flow heat exchanger with neither flow mixed. b. A section of an automotive air conditioning condenser. The flow through the horizontal wavy fins is allowed to mix with itself while the two-pass flow through the U-tubes remains unmixed. Figure 3.6 Several commercial cross-flow heat exchangers. (Photographs courtesy of Harrison Radiator Division, General Motors Corporation.) 105 Figure 3.7 Four typical heat exchanger configurations (continued on next page). (Drawings courtesy of the Tubular Exchanger Manufacturers’ Association.) 106 §3.2 Evaluation of the mean temperature difference in a heat exchanger Figure 3.7 Continued Our problem then reduces to finding the appropriate mean temperature difference that will make this equation true. Let us do this for the simple parallel and counterflow configurations, as sketched in Fig. 3.8. The temperature of both streams is plotted in Fig. 3.8 for both singlepass arrangements—the parallel and counterflow configurations—as a function of the length of travel (or area passed over). Notice that, in the parallel-flow configuration, temperatures tend to change more rapidly with position and less length is required. But the counterflow arrangement achieves generally more complete heat exchange from one flow to the other. Figure 3.9 shows another variation on the single-pass configuration. This is a condenser in which one stream flows through with its tempera- 107 108 Heat exchanger design §3.2 Figure 3.8 The temperature variation through single-pass heat exchangers. ture changing, but the other simply condenses at uniform temperature. This arrangement has some special characteristics, which we point out shortly. The determination of ∆Tmean for such arrangements proceeds as follows: the differential heat transfer within either arrangement (see Fig. 3.8) is ˙ ˙ dQ = U ∆T dA = −(mcp )h dTh = ±(mcp )c dTc (3.2) where the subscripts h and c denote the hot and cold streams, respectively; the upper and lower signs are for the parallel and counterflow cases, respectively; and dT denotes a change from left to right in the exchanger. We give symbols to the total heat capacities of the hot and cold streams: ˙ Ch ≡ (mcp )h W/K and ˙ Cc ≡ (mcp )c W/K (3.3) Thus, for either heat exchanger, ∓Ch dTh = Cc dTc . This equation can be integrated from the lefthand side, where Th = Thin and Tc = Tcin for §3.2 Evaluation of the mean temperature difference in a heat exchanger Figure 3.9 The temperature distribution through a condenser. parallel flow or Th = Thin and Tc = Tcout for counterflow, to some arbitrary point inside the exchanger. The temperatures inside are thus: Cc Q (Tc − Tcin ) = Thin − Ch Ch Cc Q − (Tcout − Tc ) = Thin − Ch Ch parallel flow: Th = Thin − counterflow: Th = Thin (3.4) where Q is the total heat transfer from the entrance to the point of interest. Equations (3.4) can be solved for the local temperature differences: Cc Cc Tc + Tc Ch Ch in Cc Cc Tc − − 1− Tc Ch Ch out ∆Tparallel = Th − Tc = Thin − 1 + ∆Tcounter = Th − Tc = Thin (3.5) 109 Heat exchanger design 110 §3.2 Substitution of these in dQ = Cc dTc = U ∆T dA yields U dA Cc U dA Cc parallel counter = = dTc Cc Cc − 1+ Tc + Tc + Thin Ch Ch in dTc Cc Cc − 1− Tc − Tc + Thin Ch Ch out (3.6) Equations (3.6) can be integrated across the exchanger: A 0 U dA = Cc Tc out Tc in dTc [− − −] (3.7) If U and Cc can be treated as constant, this integration gives ⎡ ⎤ Cc Cc Tcout + Tc + Thin ⎥ ⎢− 1 + Cc UA Ch Ch in ⎢ ⎥ ⎥ =− parallel: ln ⎢ 1+ Cc Cc ⎣ ⎦ Cc Ch Tcin + − 1+ Tc + Thin Ch Ch in ⎤ ⎡ Cc Cc Tcout − Tcout + Thin ⎥ − 1− ⎢ Cc UA Ch Ch ⎥ ⎢ ⎥ =− 1− counter: ln ⎢ Cc Cc ⎦ ⎣ Cc Ch Tcin − − 1− Tc + Thin Ch Ch out (3.8) If U were variable, the integration leading from eqn. (3.7) to eqns. (3.8) is where its variability would have to be considered. Any such variability of U can complicate eqns. (3.8) terribly. Presuming that eqns. (3.8) are valid, we can simplify them with the help of the definitions of ∆Ta and ∆Tb , given in Fig. 3.8: parallel: ln (1 + Cc /Ch )(Tcin − Tcout ) + ∆Tb ∆Tb = −U A counter: ln ∆Ta (−1 + Cc /Ch )(Tcin − Tcout ) + ∆Ta = −U A 1 1 + Cc Ch 1 1 − Cc Ch (3.9) Conservation of energy (Qc = Qh ) requires that Th − Thin Cc = − out Ch Tcout − Tcin (3.10) §3.2 Evaluation of the mean temperature difference in a heat exchanger Then eqn. (3.9) and eqn. (3.10) give ⎡ ⎤ ∆Ta −∆Tb parallel: ⎥ ⎢ ⎢ (Tcin − Tcout ) + (Thout − Thin ) +∆Tb ⎥ ⎥ ln ⎢ ⎥ ⎢ ∆Tb ⎦ ⎣ = ln counter: ln ∆Ta ∆Tb − ∆Ta + ∆Ta ∆Ta ∆Tb = −U A 1 1 + Cc Ch = ln ∆Ta ∆Tb = −U A 1 1 − Cc Ch (3.11) Finally, we write 1/Cc = (Tcout − Tcin )/Q and 1/Ch = (Thin − Thout )/Q on the right-hand side of either of eqns. (3.11) and get for either parallel or counterflow, Q = UA ∆Ta − ∆Tb ln(∆Ta /∆Tb ) (3.12) The appropriate ∆Tmean for use in eqn. (3.11) is thus the logarithmic mean temperature difference (LMTD): ∆Tmean = LMTD ≡ ∆Ta − ∆Tb ∆Ta ln ∆Tb (3.13) Example 3.1 The idea of a logarithmic mean difference is not new to us. We have already encountered it in Chapter 2. Suppose that we had asked, “What mean radius of pipe would have allowed us to compute the conduction through the wall of a pipe as though it were a slab of thickness L = ro − ri ?” (see Fig. 3.10). To answer this, we compare Q = kA ∆T = 2π kl∆T L rmean ro − ri with eqn. (2.21): Q = 2π kl∆T 1 ln(ro /ri ) 111 112 Heat exchanger design §3.2 Figure 3.10 Calculation of the mean radius for heat conduction through a pipe. It follows that rmean = ro − ri = logarithmic mean radius ln(ro /ri ) Example 3.2 Suppose that the temperature difference on either end of a heat exchanger, ∆Ta , and ∆Tb , are equal. Clearly, the effective ∆T must equal ∆Ta and ∆Tb in this case. Does the LMTD reduce to this value? Solution. If we substitute ∆Ta = ∆Tb in eqn. (3.13), we get LMTD = ∆Tb − ∆Tb 0 = = indeterminate ln(∆Tb /∆Tb ) 0 Therefore it is necessary to use L’Hospital’s rule: limit ∆Ta →∆Tb ∆Ta − ∆Tb = ln(∆Ta /∆Tb ) = ∂ (∆Ta − ∆Tb ) ∂ ∆Ta ∂ ∆Ta ln ∂ ∆Ta ∆Tb 1 1/∆Ta ∆Ta =∆Tb ∆Ta =∆Tb = ∆Ta = ∆Tb ∆Ta =∆Tb Evaluation of the mean temperature difference in a heat exchanger §3.2 It follows that the LMTD reduces to the intuitively obvious result in the limit. Example 3.3 Water enters the tubes of a small single-pass heat exchanger at 20◦ C and leaves at 40◦ C. On the shell side, 25 kg/min of steam condenses at 60◦ C. Calculate the overall heat transfer coefficient and the required flow rate of water if the area of the exchanger is 12 m2 . (The latent heat, hfg , is 2358.7 kJ/kg at 60◦ C.) Solution. ˙ Q = mcondensate · hfg 60◦ C = 25(2358.7) = 983 kJ/s 60 and with reference to Fig. 3.9, we can calculate the LMTD without naming the exchanger “parallel” or “counterflow”, since the condensate temperature is constant. LMTD = (60 − 20) − (60 − 40) = 28.85 K 60 − 20 ln 60 − 40 Then Q A(LMTD) 983(1000) = 2839 W/m2 K = 12(28.85) U= and ˙ m H2 O = 983, 000 Q = = 11.78 kg/s cp ∆T 4174(20) Extended use of the LMTD Limitations. There are two basic limitations on the use of an LMTD. The first is that it is restricted to the single-pass parallel and counterflow configurations. This restriction can be overcome by adjusting the LMTD for other configurations—a matter that we take up in the following subsection. 113 Heat exchanger design 114 §3.2 Figure 3.11 A typical case of a heat exchanger in which U varies dramatically. The second limitation—our use of a constant value of U — is more serious. The value of U must be negligibly dependent on T to complete the integration of eqn. (3.7). Even if U ≠ fn(T ), the changing flow configuration and the variation of temperature can still give rise to serious variations of U within a given heat exchanger. Figure 3.11 shows a typical situation in which the variation of U within a heat exchanger might be great. In this case, the mechanism of heat exchange on the water side is completely altered when the liquid is finally boiled away. If U were uniform in each portion of the heat exchanger, then we could treat it as two different exchangers in series. However, the more common difficulty that we face is that of designing heat exchangers in which U varies continuously with position within it. This problem is most severe in large industrial shell-and-tube configurations1 (see, e.g., Fig. 3.5 or Fig. 3.12) and less serious in compact heat exchangers with less surface area. If U depends on the location, analyses such as we have just completed [eqn. (3.1) to eqn. (3.13)] must be done A using an average U defined as 0 U dA/A. 1 Actual heat exchangers can have areas well in excess of 10,000 m2 . Large power plant condensers and other large exchangers are often remarkably big pieces of equipment. Figure 3.12 The heat exchange surface for a steam generator. This PFT-type integral-furnace boiler, with a surface area of 4560 m2 , is not particularly large. About 88% of the area is in the furnace tubing and 12% is in the boiler (Photograph courtesy of Babcock and Wilcox Co.) 115 116 Heat exchanger design §3.2 LMTD correction factor, F . Suppose that we have a heat exchanger in which U can reasonably be taken constant, but one that involves such configurational complications as multiple passes and/or cross-flow. In such cases it is necessary to rederive the appropriate mean temperature difference in the same way as we derived the LMTD. Each configuration must be analyzed separately and the results are generally more complicated than eqn. (3.13). This task was undertaken on an ad hoc basis during the early twentieth century. In 1940, Bowman, Mueller and Nagle [3.2] organized such calculations for the common range of heat exchanger configurations. In each case they wrote ⎞ ⎛ ⎟ ⎜ ⎜ Ttout − Ttin Tsin − Tsout ⎟ ⎟ ⎜ , Q = U A(LMTD) · F ⎜ ⎟ ⎝ Tsin − Ttin Ttout − Ttin ⎠ P (3.14) R where Tt and Ts are temperatures of tube and shell flows, respectively. The factor F is an LMTD correction that varies from unity to zero, depending on conditions. The dimensionless groups P and R have the following physical significance: • P is the relative influence of the overall temperature difference (Tsin − Ttin ) on the tube flow temperature. It must obviously be less than unity. • R , according to eqn. (3.10), equals the heat capacity ratio Ct /Cs . • If one flow remains at constant temperature (as, for example, in Fig. 3.9), then either P or R will equal zero. In this case the simple LMTD will be the correct ∆Tmean and F must go to unity. The factor F is defined in such a way that the LMTD should always be calculated for the equivalent counterflow single-pass exchanger with the same hot and cold temperatures. This is explained in Fig. 3.13. Bowman et al. [3.2] summarized all the equations for F , in various configurations, that had been dervied by 1940. They presented them graphically in not-very-accurate figures that have been widely copied. The TEMA [3.1] version of these curves has been recalculated for shell-and-tube heat exchangers, and it is more accurate. We include two of these curves in Fig. 3.14(a) and Fig. 3.14(b). TEMA presents many additional curves for more complex shell-and-tube configurations. Figures 3.14(c) and 3.14(d) §3.2 Evaluation of the mean temperature difference in a heat exchanger Figure 3.13 The basis of the LMTD in a multipass exchanger, prior to correction. are the Bowman et al. curves for the simplest cross-flow configurations. Gardner and Taborek [3.3] redeveloped Fig. 3.14(c) over a different range of parameters. They also showed how Fig. 3.14(a) and Fig. 3.14(b) must be modified if the number of baffles in a tube-in-shell heat exchanger is large enough to make it behave like a series of cross-flow exchangers. We have simplified Figs. 3.14(a) through 3.14(d) by including curves only for R 1. Shamsundar [3.4] noted that for R > 1, one may obtain F using a simple reciprocal rule. He showed that so long as a heat exchanger has a uniform heat transfer coefficient and the fluid properties are constant, F (P , R) = F (P R, 1/R) (3.15) Thus, if R is greater than unity, one need only evaluate F using P R in place of P and 1/R in place of R . Example 3.4 5.795 kg/s of oil flows through the shell side of a two-shell pass, four- 117 a. F for a one-shell-pass, four, six-, . . . tube-pass exchanger. b. F for a two-shell-pass, four or more tube-pass exchanger. Figure 3.14 LMTD correction factors, F , for multipass shelland-tube heat exchangers and one-pass cross-flow exchangers. 118 c. F for a one-pass cross-flow exchanger with both passes unmixed. d. F for a one-pass cross-flow exchanger with one pass mixed. Figure 3.14 LMTD correction factors, F , for multipass shelland-tube heat exchangers and one-pass cross-flow exchangers. 119 Heat exchanger design 120 §3.3 tube-pass oil cooler. The oil enters at 181◦ C and leaves at 38◦ C. Water flows in the tubes, entering at 32◦ C and leaving at 49◦ C. In addition, cpoil = 2282 J/kg·K and U = 416 W/m2 K. Find how much area the heat exchanger must have. Solution. LMTD = (Thin − Tcout ) − (Thout − Tcin ) ln = R= Thin − Tcout Thout − Tcin (181 − 49) − (38 − 32) = 40.76 K 181 − 49 ln 38 − 32 181 − 38 = 8.412 49 − 32 P= 49 − 32 = 0.114 181 − 32 Since R > 1, we enter Fig. 3.14(b) using P = 8.412(0.114) = 0.959 and R = 1/8.412 = 0.119 and obtain F = 0.92.2 It follows that: Q = U AF (LMTD) 5.795(2282)(181 − 38) = 416(A)(0.92)(40.76) A = 121.2 m2 3.3 Heat exchanger effectiveness We are now in a position to predict the performance of an exchanger once we know its configuration and the imposed differences. Unfortunately, we do not often know that much about a system before the design is complete. Often we begin with information such as is shown in Fig. 3.15. If we sought to calculate Q in such a case, we would have to do so by guessing an exit temperature such as to make Qh = Qc = Ch ∆Th = Cc ∆Tc . Then we could calculate Q from U A(LMTD) or UAF (LMTD) and check it against Qh . The answers would differ, so we would have to guess new exit temperatures and try again. Such problems can be greatly simplified with the help of the so-called effectiveness-NTU method. This method was first developed in full detail 2 Notice that, for a 1 shell-pass exchanger, these R and P lines do not quite intersect [see Fig. 3.14(a)]. Therefore, one could not obtain these temperatures with any singleshell exchanger. Heat exchanger effectiveness §3.3 121 Figure 3.15 A design problem in which the LMTD cannot be calculated a priori. by Kays and London [3.5] in 1955, in a book titled Compact Heat Exchangers. We should take particular note of the title. It is with compact heat exchangers that the present method can reasonably be used, since the overall heat transfer coefficient is far more likely to remain fairly uniform. The heat exchanger effectiveness is defined as ε≡ Ch (Thin − Thout ) Cc (Tcout − Tcin ) = Cmin (Thin − Tcin ) Cmin (Thin − Tcin ) (3.16) where Cmin is the smaller of Cc and Ch . The effectiveness can be interpreted as ε= actual heat transferred maximum heat that could possibly be transferred from one stream to the other It follows that Q = εCmin (Thin − Tcin ) (3.17) A second definition that we will need was originally made by E.K.W. Nusselt, whom we meet again in Part III. This is the number of transfer units (NTU): NTU ≡ UA Cmin (3.18) Heat exchanger design 122 §3.3 This dimensionless group can be viewed as a comparison of the heat capacity of the heat exchanger, expressed in W/K, with the heat capacity of the flow. We can immediately reduce the parallel-flow result from eqn. (3.9) to the following equation, based on these definitions: − Cc Cmin Cmin Cmin NTU = ln − 1 + ε + +1 Cc Ch Ch Cc (3.19) We solve this for ε and, regardless of whether Cmin is associated with the hot or cold flow, obtain for the parallel single-pass heat exchanger: ε≡ 1 − exp [−(1 + Cmin /Cmax )NTU] Cmin = fn , NTU only 1 + Cmin /Cmax Cmax (3.20) The corresponding expression for the counterflow case is ε= 1 − exp [−(1 − Cmin /Cmax )NTU] 1 − (Cmin /Cmax ) exp[−(1 − Cmin /Cmax )NTU] (3.21) Equations (3.20) and (3.21) are given in graphical form in Fig. 3.16. Similar calculations give the effectiveness for the other heat exchanger configurations (see [3.5] and Problem 3.38), and we include some of the resulting effectiveness plots in Fig. 3.17. To see how the effectiveness can conveniently be used to complete a design, consider the following two examples. Example 3.5 Consider the following parallel-flow heat exchanger specification: cold flow enters at 40◦ C: ◦ hot flow enters at 150 C: 2 A = 30 m Cc = 20, 000 W/K Ch = 10, 000 W/K U = 500 W/m2 K. Determine the heat transfer and the exit temperatures. Solution. In this case we do not know the exit temperatures, so it is not possible to calculate the LMTD. Instead, we can go either to the parallel-flow effectiveness chart in Fig. 3.16 or to eqn. (3.20), using NTU = 500(30) UA = 1.5 = Cmin 10, 000 Cmin = 0.5 Cmax Heat exchanger effectiveness §3.3 Figure 3.16 The effectiveness of parallel and counterflow heat exchangers. (Data provided by A.D. Kraus.) and we obtain ε = 0.596. Now from eqn. (3.17), we find that Q = ε Cmin (Thin − Tcin ) = 0.596(10, 000)(110) = 655, 600 W = 655.6 kW Finally, from energy balances such as are expressed in eqn. (3.4), we get Q 655, 600 = 84.44◦ C = 150 − Ch 10, 000 Q 655, 600 + = 40 + = 72.78◦ C Cc 20, 000 Thout = Thin − Tcout = Tcin Example 3.6 Suppose that we had the same kind of exchanger as we considered in Example 3.5, but that the area remained unspecified as a design variable. Then calculate the area that would bring the hot flow out at 90◦ C. Solution. Once the exit cold fluid temperature is known, the problem can be solved with equal ease by either the LMTD or the effective- 123 Figure 3.17 The effectiveness of some other heat exchanger configurations. (Data provided by A.D. Kraus.) 124 Heat exchanger effectiveness §3.3 125 ness approach. Tcout = Tcin + Ch 1 (Thin − Thout ) = 40 + (150 − 90) = 70◦ C Cc 2 Then, using the effectiveness method, ε= Ch (Thin − Thout ) 10, 000(150 − 90) = = 0.5455 Cmin (Thin − Tcin ) 10, 000(150 − 40) so from Fig. 3.16 we read NTU A= 1.15 = U A/Cmin . Thus 10, 000(1.15) = 23.00 m2 500 We could also have calculated the LMTD: LMTD = (150 − 40) − (90 − 70) = 52.79 K ln(110/20) so from Q = U A(LMTD), we obtain A= 10, 000(150 − 90) = 22.73 m2 500(52.79) The answers differ by 1%, which reflects graph reading inaccuracy. When the temperature of either fluid in a heat exchanger is uniform, the problem of analyzing heat transfer is greatly simplified. We have already noted that no F -correction is needed to adjust the LMTD in this case. The reason is that when only one fluid changes in temperature, the configuration of the exchanger becomes irrelevant. Any such exchanger is equivalent to a single fluid stream flowing through an isothermal pipe.3 Since all heat exchangers are equivalent in this case, it follows that the equation for the effectiveness in any configuration must reduce to the same common expression as Cmax approaches infinity. The volumetric heat capacity rate might approach infinity because the flow rate or specific heat is very large, or it might be infinite because the flow is absorbing or giving up latent heat (as in Fig. 3.9). The limiting effectiveness expression can also be derived directly from energy-balance considerations (see Problem 3.11), but we obtain it here by letting Cmax → ∞ in either eqn. (3.20) or eqn. (3.21). The result is lim ε = 1 − e−NTU Cmax →∞ 3 (3.22) We make use of this notion in Section 7.4, when we analyze heat convection in pipes and tubes. Heat exchanger design 126 §3.4 Eqn. (3.22) defines the curve for Cmin /Cmax = 0 in all six of the effectiveness graphs in Fig. 3.16 and Fig. 3.17. 3.4 Heat exchanger design The preceding sections provided means for designing heat exchangers that generally work well in the design of smaller exchangers—typically, the kind of compact cross-flow exchanger used in transportation equipment. Larger shell-and-tube exchangers pose two kinds of difficulty in relation to U . The first is the variation of U through the exchanger, which we have already discussed. The second difficulty is that convective heat transfer coefficients are very hard to predict for the complicated flows that move through a baffled shell. We shall achieve considerable success in using analysis to predict h’s for various convective flows in Part III. The determination of h in a baffled shell remains a problem that cannot be solved analytically. Instead, it is normally computed with the help of empirical correlations or with the aid of large commercial computer programs that include relevant experimental correlations. The problem of predicting h when the flow is boiling or condensing is even more complicated. A great deal of research is at present aimed at perfecting such empirical predictions. Apart from predicting heat transfer, a host of additional considerations must be addressed in designing heat exchangers. The primary ones are the minimization of pumping power and the minimization of fixed costs. The pumping power calculation, which we do not treat here in any detail, is based on the principles discussed in a first course on fluid mechanics. It generally takes the following form for each stream of fluid through the heat exchanger: ˙ pumping power = m kg s ∆p N/m2 ρ kg/m3 ˙ m∆p N·m ρ s ˙ m∆p (W) = ρ = (3.23) ˙ where m is the mass flow rate of the stream, ∆p the pressure drop of the stream as it passes through the exchanger, and ρ the fluid density. Determining the pressure drop can be relatively straightforward in a single-pass pipe-in-tube heat exchanger or extremely difficulty in, say, a Heat exchanger design §3.4 127 shell-and-tube exchanger. The pressure drop in a straight run of pipe, for example, is given by ∆p = f L Dh ρ u2 av 2 (3.24) where L is the length of pipe, Dh is the hydraulic diameter, uav is the mean velocity of the flow in the pipe, and f is the Darcy-Weisbach friction factor (see Fig. 7.6). Optimizing the design of an exchanger is not just a matter of making ∆p as small as possible. Often, heat exchange can be augmented by employing fins or roughening elements in an exchanger. (We discuss such elements in Chapter 4; see, e.g., Fig. 4.6). Such augmentation will invariably increase the pressure drop, but it can also reduce the fixed cost of an exchanger by increasing U and reducing the required area. Furthermore, it can reduce the required flow rate of, say, coolant, by increasing the effectiveness and thus balance the increase of ∆p in eqn. (3.23). To better understand the course of the design process, faced with such an array of trade-offs of advantages and penalties, we follow Taborek’s [3.6] list of design considerations for a large shell-and-tube exchanger: • Decide which fluid should flow on the shell side and which should flow in the tubes. Normally, this decision will be made to minimize the pumping cost. If, for example, water is being used to cool oil, the more viscous oil would flow in the shell. Corrosion behavior, fouling, and the problems of cleaning fouled tubes also weigh heavily in this decision. • Early in the process, the designer should assess the cost of the calculation in comparison with: (a) The converging accuracy of computation. (b) The investment in the exchanger. (c) The cost of miscalculation. • Make a rough estimate of the size of the heat exchanger using, for example, U values from Table 2.2 and/or anything else that might be known from experience. This serves to circumscribe the subsequent trial-and-error calculations; it will help to size flow rates and to anticipate temperature variations; and it will help to avoid subsequent errors. 128 Heat exchanger design §3.4 • Evaluate the heat transfer, pressure drop, and cost of various exchanger configurations that appear reasonable for the application. This is usually done with large-scale computer programs that have been developed and are constantly being improved as new research is included in them. The computer runs suggested by this procedure are normally very complicated and might typically involve 200 successive redesigns, even when relatively efficient procedures are used. However, most students of heat transfer will not have to deal with such designs. Many, if not most, will be called upon at one time or another to design smaller exchangers in the range 0.1 to 10 m2 . The heat transfer calculation can usually be done effectively with the methods described in this chapter. Some useful sources of guidance in the pressure drop calculation are the Heat Exchanger Design Handbook [3.7], the data in Idelchik’s collection [3.8], the TEMA design book [3.1], and some of the other references at the end of this chapter. In such a calculation, we start off with one fluid to heat and one to cool. Perhaps we know the flow heat capacity rates (Cc and Ch ), certain temperatures, and/or the amount of heat that is to be transferred. The problem can be annoyingly wide open, and nothing can be done until it is somehow delimited. The normal starting point is the specification of an exchanger configuration, and to make this choice one needs experience. The descriptions in this chapter provide a kind of first level of experience. References [3.5, 3.7, 3.9, 3.10, 3.11, 3.12, 3.13] provide a second level. Manufacturer’s catalogues are an excellent source of more advanced information. Once the exchanger configuration is set, U will be approximately set and the area becomes the basic design variable. The design can then proceed along the lines of Section 3.2 or 3.3. If it is possible to begin with a complete specification of inlet and outlet temperatures, Q= C ∆T U AF (LMTD) known calculable Then A can be calculated and the design completed. Usually, a reevaluation of U and some iteration of the calculation is needed. More often, we begin without full knowledge of the outlet temperatures. In such cases, we normally have to invent an appropriate trial-anderror method to get the area and a more complicated sequence of trials if we seek to optimize pressure drop and cost by varying the configuration Problems 129 as well. If the C ’s are design variables, the U will change significantly, because h’s are generally velocity-dependent and more iteration will be needed. We conclude Part I of this book facing a variety of incomplete issues. Most notably, we face a serious need to be able to determine convective heat transfer coefficients. The prediction of h depends on a knowledge of heat conduction. We therefore turn, in Part II, to a much more thorough study of heat conduction analysis than was undertaken in Chapter 2. In addition to setting up the methodology ultimately needed to predict h’s, Part II will also deal with many other issues that have great practical importance in their own right. Problems 3.1 Can you have a cross-flow exchanger in which both flows are mixed? Discuss. 3.2 Find the appropriate mean radius, r , that will make Q = kA(r )∆T /(ro −ri ), valid for the one-dimensional heat conduction through a thick spherical shell, where A(r ) = 4π r 2 (cf. Example 3.1). 3.3 Rework Problem 2.14, using the methods of Chapter 3. 3.4 2.4 kg/s of a fluid have a specific heat of 0.81 kJ/kg·K enter a counterflow heat exchanger at 0◦ C and are heated to 400◦ C by 2 kg/s of a fluid having a specific heat of 0.96 kJ/kg·K entering the unit at 700◦ C. Show that to heat the cooler fluid to 500◦ C, all other conditions remaining unchanged, would require the surface area for a heat transfer to be increased by 87.5%. 3.5 A cross-flow heat exchanger with both fluids unmixed is used to heat water (cp = 4.18 kJ/kg·K) from 40◦ C to 80◦ C, flowing at the rate of 1.0 kg/s. What is the overall heat transfer coefficient if hot engine oil (cp = 1.9 kJ/kg·K), flowing at the rate of 2.6 kg/s, enters at 100◦ C? The heat transfer area is 20 m2 . (Note that you can use either an effectiveness or an LMTD method. It would be wise to use both as a check.) 3.6 Saturated non-oil-bearing steam at 1 atm enters the shell pass of a two-tube-pass shell condenser with thirty 20 ft tubes in Chapter 3: Heat exchanger design 130 each tube pass. They are made of schedule 160, ¾ in. steel pipe (nominal diameter). A volume flow rate of 0.01 ft3 /s of water entering at 60◦ F enters each tube. The condensing heat transfer coefficient is 2000 Btu/h·ft2 ·◦ F, and we calculate h = 1380 Btu/h·ft2 ·◦ F for the water in the tubes. Estimate the exit ˙ temperature of the water and mass rate of condensate [mc 8393 lbm /h.] 3.7 Consider a counterflow heat exchanger that must cool 3000 kg/h of mercury from 150◦ F to 128◦ F. The coolant is 100 kg/h of water, supplied at 70◦ F. If U is 300 W/m2 K, complete the design by determining reasonable value for the area and the exit-water temperature. [A = 0.147 m2 .] 3.8 An automobile air-conditioner gives up 18 kW at 65 km/h if the outside temperature is 35◦ C. The refrigerant temperature is constant at 65◦ C under these conditions, and the air rises 6◦ C in temperature as it flows across the heat exchanger tubes. The heat exchanger is of the finned-tube type shown in Fig. 3.6b, with U 200 W/m2 K. If U ∼ (air velocity)0.7 and the mass flow rate increases directly with the velocity, plot the percentage reduction of heat transfer in the condenser as a function of air velocity between 15 and 65 km/h. 3.9 Derive eqn. (3.21). 3.10 Derive the infinite NTU limit of the effectiveness of parallel and counterflow heat exchangers at several values of Cmin /Cmax . Use common sense and the First Law of Thermodynamics, and refer to eqn. (3.2) and eqn. (3.21) only to check your results. 3.11 Derive the equation ε = (NTU, Cmin /Cmax ) for the heat exchanger depicted in Fig. 3.9. 3.12 A single-pass heat exchanger condenses steam at 1 atm on the shell side and heats water from 10◦ C to 30◦ C on the tube side with U = 2500 W/m2 K. The tubing is thin-walled, 5 cm in diameter, and 2 m in length. (a) Your boss asks whether the exchanger should be counterflow or parallel-flow. How do you ˙ advise her? Evaluate: (b) the LMTD; (c) mH2 O ; (d) ε. [ε 0.222.] Problems 3.13 131 Air at 2 kg/s and 27◦ C and a stream of water at 1.5 kg/s and 60◦ C each enter a heat exchanger. Evaluate the exit temperatures if A = 12 m2 , U = 185 W/m2 K, and: a. The exchanger is parallel flow; b. The exchanger is counterflow [Thout 54.0◦ C.]; c. The exchanger is cross-flow, one stream mixed; d. The exchanger is cross-flow, neither stream mixed. [Thout = 53.62◦ C.] 3.14 Air at 0.25 kg/s and 0◦ C enters a cross-flow heat exchanger. It is to be warmed to 20◦ C by 0.14 kg/s of air at 50◦ C. The streams are unmixed. As a first step in the design process, plot U against A and identify the approximate range of area for the exchanger. 3.15 A particular two shell-pass, four tube-pass heat exchanger uses 20 kg/s of river water at 10◦ C on the shell side to cool 8 kg/s of processed water from 80◦ C to 25◦ C on the tube side. At what temperature will the coolant be returned to the river? If U is 800 W/m2 K, how large must the exchanger be? 3.16 A particular cross-flow process heat exchanger operates with the fluid mixed on one side only. When it is new, U = 2000 W/m2 K, Tcin = 25◦ C, Tcout = 80◦ C, Thin = 160◦ C, and Thout = 70◦ C. After 6 months of operation, the plant manager reports that the hot fluid is only being cooled to 90◦ C and that he is suffering a 30% reduction in total heat transfer. What is the fouling resistance after 6 months of use? (Assume no reduction of cold-side flow rate by fouling.) 3.17 Water at 15◦ C is supplied to a one-shell-pass, two-tube-pass heat exchanger to cool 10 kg/s of liquid ammonia from 120◦ C to 40◦ C. You anticipate a U on the order of 1500 W/m2 K when the water flows in the tubes. If A is to be 90 m2 , choose the correct flow rate of water. 3.18 Suppose that the heat exchanger in Example 3.5 had been a two shell-pass, four tube-pass exchanger with the hot fluid moving in the tubes. (a) What would be the exit temperature in this case? [Tcout = 75.09◦ C.] (b) What would be the area if we wanted Chapter 3: Heat exchanger design 132 the hot fluid to leave at the same temperature that it does in the example? 3.19 Plot the maximum tolerable fouling resistance as a function of Unew for a counterflow exchanger, with given inlet temperatures, if a 30% reduction in U is the maximum that can be tolerated. 3.20 Water at 0.8 kg/s enters the tubes of a two-shell-pass, fourtube-pass heat exchanger at 17◦ C and leaves at 37◦ C. It cools 0.5 kg/s of air entering the shell at 250◦ C with U = 432 W/m2 K. Determine: (a) the exit air temperature; (b) the area of the heat exchanger; and (c) the exit temperature if, after some time, the tubes become fouled with Rf = 0.0005 m2 K/W. [(c) Tairout = 140.5◦ C.] 3.21 You must cool 78 kg/min of a 60%-by-mass mixture of glycerin in water from 108◦ C to 50◦ C using cooling water available at 7◦ C. Design a one-shell-pass, two-tube-pass heat exchanger if U = 637 W/m2 K. Explain any design decision you make and report the area, TH2 Oout , and any other relevant features. 3.22 A mixture of 40%-by-weight glycerin, 60% water, enters a smooth ˙ 0.113 m I.D. tube at 30◦ C. The tube is kept at 50◦ C, and mmixture = 8 kg/s. The heat transfer coefficient inside the pipe is 1600 W/m2 K. Plot the liquid temperature as a function of position in the pipe. 3.23 Explain in physical terms why all effectiveness curves Fig. 3.16 and Fig. 3.17 have the same slope as NTU → 0. Obtain this slope from eqns. (3.20) and (3.21). 3.24 You want to cool air from 150◦ C to 60◦ C but you cannot afford a custom-built heat exchanger. You find a used cross-flow exchanger (both fluids unmixed) in storage. It was previously used to cool 136 kg/min of NH3 vapor from 200◦ C to 100◦ C using 320 kg/min of water at 7◦ C; U was previously 480 W/m2 K. How much air can you cool with this exchanger, using the same water supply, if U is approximately unchanged? (Actually, you would have to modify U using the methods of Chapters 6 and 7 once you had the new air flow rate, but that is beyond our present scope.) Problems 133 3.25 A one tube-pass, one shell-pass, parallel-flow, process heat exchanger cools 5 kg/s of gaseous ammonia entering the shell side at 250◦ C and boils 4.8 kg/s of water in the tubes. The water enters subcooled at 27◦ C and boils when it reaches 100◦ C. U = 480 W/m2 K before boiling begins and 964 W/m2 K thereafter. The area of the exchanger is 45 m2 , and hfg for water is 2.257 × 106 J/kg. Determine the quality of the water at the exit. 3.26 0.72 kg/s of superheated steam enters a crossflow heat exchanger at 240◦ C and leaves at 120◦ C. It heats 0.6 kg/s of water entering at 17◦ C. U = 612 W/m2 K. By what percentage will the area differ if a both-fluids-unmixed exchanger is used instead of a one-fluid-unmixed exchanger? [−1.8%] 3.27 Compare values of F from Fig. 3.14(c) and Fig. 3.14(d) for the same conditions of inlet and outlet temperatures. Is the one with the higher F automatically the more desirable exchanger? Discuss. 3.28 Compare values of ε for the same NTU and Cmin /Cmax in parallel and counterflow heat exchangers. Is the one with the higher ε automatically the more desirable exchanger? Discuss. 3.29 The irreversibility rate of a process is equal to the rate of entropy production times the lowest absolute sink temperature accessible to the process. Calculate the irreversibility (or lost work) for the heat exchanger in Example 3.4. What kind of configuration would reduce the irreversibility, given the same end temperatures. 3.30 Plot Toil and TH2 O as a function of position in a very long counterflow heat exchanger where water enters at 0◦ C, with CH2 O = 460 W/K, and oil enters at 90◦ C, with Coil = 920 W/K, U = 742 W/m2 K, and A = 10 m2 . Criticize the design. 3.31 Liquid ammonia at 2 kg/s is cooled from 100◦ C to 30◦ C in the shell side of a two shell-pass, four tube-pass heat exchanger by 3 kg/s of water at 10◦ C. When the exchanger is new, U = 750 W/m2 K. Plot the exit ammonia temperature as a function of the increasing tube fouling factor. Chapter 3: Heat exchanger design 134 3.32 A one shell-pass, two tube-pass heat exchanger cools 0.403 kg/s of methanol from 47◦ C to 7◦ C on the shell side. The coolant is 2.2 kg/s of Freon 12, entering the tubes at −33◦ C, with U = 538 W/m2 K. A colleague suggests that this arrangement wastes Freon. She thinks you could do almost as well if you cut the Freon flow rate all the way down to 0.8 kg/s. Calculate the new methanol outlet temperature that would result from this flow rate, and evaluate her suggestion. 3.33 The factors dictating the heat transfer coefficients in a certain two shell-pass, four tube-pass heat exchanger are such that ˙ U increases as (mshell )0.6 . The exchanger cools 2 kg/s of air ◦ C to 40◦ C using 4.4 kg/s of water at 7◦ C, and U = 312 from 200 W/m2 K under these circumstances. If we double the air flow, what will its temperature be leaving the exchanger? [Tairout = 61◦ C.] 3.34 A flow rate of 1.4 kg/s of water enters the tubes of a two-shellpass, four-tube-pass heat exchanger at 7◦ C. A flow rate of 0.6 kg/s of liquid ammonia at 100◦ C is to be cooled to 30◦ C on the shell side; U = 573 W/m2 K. (a) How large must the heat exchanger be? (b) How large must it be if, after some months, a fouling factor of 0.0015 will build up in the tubes, and we still want to deliver ammonia at 30◦ C? (c) If we make it large enough to accommodate fouling, to what temperature will it cool the ammonia when it is new? (d) At what temperature does water leave the new, enlarged exchanger? [(d) TH2 O = 49.9◦ C.] 3.35 Both C ’s in a parallel-flow heat exchanger are equal to 156 W/K, U = 327 W/m2 K and A = 2 m2 . The hot fluid enters at 140◦ C and leaves at 90◦ C. The cold fluid enters at 40◦ C. If both C ’s are halved, what will be the exit temperature of the hot fluid? 3.36 A 1.68 ft2 cross-flow heat exchanger with one fluid mixed condenses steam at atmospheric pressure (h = 2000 Btu/h·ft2 ·◦ F) and boils methanol (Tsat = 170◦ F and h = 1500 Btu/h·ft2 ·◦ F) on the other side. Evaluate U (neglecting resistance of the metal), LMTD, F , NTU, ε, and Q. 3.37 Eqn. (3.21) is troublesome when Cmin /Cmax = 1. Develop a working equation for ε in this case. Compare it with Fig. 3.16. Problems 3.38 135 The effectiveness of a cross-flow exchanger with neither fluid mixed can be calculated from the following approximate formula: ε = 1 − exp exp(−NTU0.78 r ) − 1](NTU0.22 /r ) where r ≡ Cmin /Cmax . How does this compare with correct values? 3.39 Calculate the area required in a two-tube-pass, one-shell-pass condenser that is to condense 106 kg/h of steam at 40◦ C using water at 17◦ C. Assume that U = 4700 W/m2 K, the maximum allowable temperature rise of the water is 10◦ C, and hfg = 2406 kJ/kg. 3.40 An engineer wants to divert 1 gal/min of water at 180◦ F from his car radiator through a small cross-flow heat exchanger with neither flow mixed, to heat 40◦ F water to 140◦ F for shaving when he goes camping. If he produces a pint per minute of hot water, what will be the area of the exchanger and the temperature of the returning radiator coolant if U = 720 W/m2 K? 3.41 In a process for forming lead shot, molten droplets of lead are showered into the top of a tall tower. The droplets fall through air and solidify before they reach the bottom of the tower. The solid shot is collected at the bottom. To maintain a steady state, cool air is introduced at the bottom of the tower and warm air is withdrawn at the top. For a particular tower, the droplets are 1 mm in diameter and at their melting temperature of 600 K when they are released. The latent heat of solidification is 850 kJ/kg. They fall with a mass flow rate of 200 kg/hr. There are 2430 droplets per cubic meter of air inside the tower. Air enters the bottom at 20◦ C with a mass flow rate of 1100 kg/hr. The tower has an internal diameter of 1 m with adiabatic walls. a. Sketch, qualitatively, the temperature distributions of the shot and the air along the height of the tower. b. If it is desired to remove the shot at a temperature of 60◦ C, what will be the temperature of the air leaving the top of the tower? Chapter 3: Heat exchanger design 136 c. Determine the air temperature at the point where the lead has just finished solidifying. d. Determine the height that the tower must have in order to function as desired. The heat transfer coefficient between the air and the droplets is h = 318 W/m2 K. References [3.1] Tubular Exchanger Manufacturer’s Association. Standards of Tubular Exchanger Manufacturer’s Association. New York, 4th and 6th edition, 1959 and 1978. [3.2] R. A. Bowman, A. C. Mueller, and W. M. Nagle. Mean temperature difference in design. Trans. ASME, 62:283–294, 1940. [3.3] K. Gardner and J. Taborek. Mean temperature difference: A reappraisal. AIChE J., 23(6):770–786, 1977. [3.4] N. Shamsundar. A property of the log-mean temperaturedifference correction factor. Mechanical Engineering News, 19(3): 14–15, 1982. [3.5] W. M. Kays and A. L. London. Compact Heat Exchangers. McGrawHill Book Company, New York, 3rd edition, 1984. [3.6] J. Taborek. Evolution of heat exchanger design techniques. Heat Transfer Engineering, 1(1):15–29, 1979. [3.7] G. F. Hewitt, editor. Heat Exchanger Design Handbook 1998. Begell House, New York, 1998. [3.8] E. Fried and I. E. Idelchik. Flow Resistance: A Design Guide for Engineers. Hemisphere Publishing Corp., New York, 1989. [3.9] R. H. Perry, D. W. Green, and J. Q. Maloney, editors. Perry’s Chemical Engineers’ Handbook. McGraw-Hill Book Company, New York, 7th edition, 1997. [3.10] D. M. Considine. Energy Technology Handbook. McGraw-Hill Book Company, New York, 1975. [3.11] A. P. Fraas. Heat Exchanger Design. John Wiley & Sons, Inc., New York, 2nd edition, 1989. References [3.12] R. K. Shah and D. P. Sekulic. Heat exchangers. In W. M. Rohsenow, J. P. Hartnett, and Y. I. Cho, editors, Handbook of Heat Transfer, chapter 17. McGraw-Hill, New York, 3rd edition, 1998. [3.13] R. K. Shah and D. P. Sekulic. Fundamentals of Heat Exchanger Design. John Wiley & Sons, Inc., Hoboken, NJ, 2003. 137 Part II Analysis of Heat Conduction 139 4. Analysis of heat conduction and some steady one-dimensional problems The effects of heat are subject to constant laws which cannot be discovered without the aid of mathematical analysis. The object of the theory which we are about to explain is to demonstrate these laws; it reduces all physical researches on the propagation of heat to problems of the calculus whose elements are given by experiment. The Analytical Theory of Heat, J. Fourier, 1822 4.1 The well-posed problem The heat diffusion equation was derived in Section 2.1 and some attention was given to its solution. Before we go further with heat conduction problems, we must describe how to state such problems so they can really be solved. This is particularly important in approaching the more complicated problems of transient and multidimensional heat conduction that we have avoided up to now. A well-posed heat conduction problem is one in which all the relevant information needed to obtain a unique solution is stated. A well-posed and hence solvable heat conduction problem will always read as follows: Find T (x, y, z, t) such that: 1. ˙ ∇ · (k∇T ) + q = ρc ∂T ∂t for 0 < t < T (where T can → ∞), and for (x, y, z) belonging to 141 142 Analysis of heat conduction and some steady one-dimensional problems §4.1 some region, R , which might extend to infinity.1 2. T = Ti (x, y, z) at t=0 This is called an initial condition, or i.c. (a) Condition 1 above is not imposed at t = 0. (b) Only one i.c. is required. However, (c) The i.c. is not needed: ˙ i. In the steady-state case: ∇ · (k∇T ) + q = 0. ˙ ii. For “periodic” heat transfer, where q or the boundary conditions vary periodically with time, and where we ignore the starting transient behavior. 3. T must also satisfy two boundary conditions, or b.c.’s, for each coordinate. The b.c.’s are very often of three common types. (a) Dirichlet conditions, or b.c.’s of the first kind : T is specified on the boundary of R for t > 0. We saw such b.c.’s in Examples 2.1, 2.2, and 2.5. (b) Neumann conditions, or b.c.’s of the second kind : The derivative of T normal to the boundary is specified on the boundary of R for t > 0. Such a condition arises when the heat flux, k(∂T /∂x), is specified on a boundary or when , with the help of insulation, we set ∂T /∂x equal to zero.2 (c) b.c.’s of the third kind : A derivative of T in a direction normal to a boundary is proportional to the temperature on that boundary. Such a condition most commonly arises when convection occurs at a boundary, and it is typically expressed as −k ∂T ∂x bndry = h(T − T∞ )bndry when the body lies to the left of the boundary on the x -coordinate. We have already used such a b.c. in Step 4 of Example 2.6, and we have discussed it in Section 1.3 as well. (x, y, z) might be any coordinates describing a position r : T (x, y, z, t) = T (r , t). Although we write ∂T /∂x here, we understand that this might be ∂T /∂z, ∂T /∂r , or any other derivative in a direction locally normal to the surface on which the b.c. is specified. 1 2 The general solution §4.2 Figure 4.1 The transient cooling of a body as it might occur, subject to boundary conditions of the first, second, and third kinds. This list of b.c.’s is not complete, by any means, but it includes a great number of important cases. Figure 4.1 shows the transient cooling of body from a constant initial temperature, subject to each of the three b.c.’s described above. Notice that the initial temperature distribution is not subject to the boundary condition, as pointed out previously under 2(a). The eight-point procedure that was outlined in Section 2.2 for solving the heat diffusion equation was contrived in part to assure that a problem will meet the preceding requirements and will be well posed. 4.2 The general solution Once the heat conduction problem has been posed properly, the first step in solving it is to find the general solution of the heat diffusion equation. We have remarked that this is usually the easiest part of the problem. We next consider some examples of general solutions. 143 144 Analysis of heat conduction and some steady one-dimensional problems §4.2 One-dimensional steady heat conduction Problem 4.1 emphasizes the simplicity of finding the general solutions of linear ordinary differential equations, by asking for a table of all general solutions of one-dimensional heat conduction problems. We shall work out some of those results to show what is involved. We begin the heat ˙ diffusion equation with constant k and q: ∇2 T + ˙ 1 ∂T q = k α ∂t (2.11) Cartesian coordinates: Steady conduction in the y-direction. Equation (2.11) reduces as follows: ˙ ∂2T q ∂2T ∂2T + + += 2 2 2 ∂x ∂y ∂z k =0 =0 1 ∂T α ∂t = 0, since steady Therefore, ˙ q d2 T =− 2 dy k which we integrate twice to get T =− ˙ q2 y + C1 y + C2 2k ˙ or, if q = 0, T = C1 y + C2 Cylindrical coordinates with a heat source: Tangential conduction. This time, we look at the heat flow that results in a ring when two points are held at different temperatures. We now express eqn. (2.11) in cylindrical coordinates with the help of eqn. (2.13): 1∂ r ∂r r ∂T ∂r =0 + ˙ 1 ∂2T ∂2T q + += r 2 ∂φ2 ∂z2 k r =constant =0 1 ∂T α ∂t = 0, since steady Two integrations give ˙ r 2q 2 φ + C1 φ + C2 (4.1) 2k This would describe, for example, the temperature distribution in the thin ring shown in Fig. 4.2. Here the b.c.’s might consist of temperatures specified at two angular locations, as shown. T =− The general solution §4.2 145 Figure 4.2 One-dimensional heat conduction in a ring. T = T(t only) If T is spatially uniform, it can still vary with time. In such cases ∇2 T + ˙ 1 ∂T q = k α ∂t =0 and ∂T /∂t becomes an ordinary derivative. Then, since α = k/ρc , ˙ q dT = dt ρc (4.2) This result is consistent with the lumped-capacity solution described in Section 1.3. If the Biot number is low and internal resistance is unimportant, the convective removal of heat from the boundary of a body can be prorated over the volume of the body and interpreted as ˙ qeffective = − h(Tbody − T∞ )A W/m3 volume (4.3) and the heat diffusion equation for this case, eqn. (4.2), becomes hA dT =− (T − T∞ ) dt ρcV (4.4) The general solution in this situation was given in eqn. (1.21). [A particular solution was also written in eqn. (1.22).] 146 Analysis of heat conduction and some steady one-dimensional problems §4.2 Separation of variables: A general solution of multidimensional problems Suppose that the physical situation permits us to throw out all but one of the spatial derivatives in a heat diffusion equation. Suppose, for example, that we wish to predict the transient cooling in a slab as a function of the location within it. If there is no heat generation, the heat diffusion equation is 1 ∂T ∂2T = 2 ∂x α ∂t (4.5) A common trick is to ask: “Can we find a solution in the form of a product of functions of t and x : T = T (t) · X (x)?” To find the answer, we substitute this in eqn. (4.5) and get X T= 1 TX α (4.6) where each prime denotes one differentiation of a function with respect to its argument. Thus T = dT /dt and X = d2 X/dx 2 . Rearranging eqn. (4.6), we get 1T X = X αT (4.7a) This is an interesting result in that the left-hand side depends only upon x and the right-hand side depends only upon t . Thus, we set both sides equal to the same constant, which we call −λ2 , instead of, say, λ, for reasons that will be clear in a moment: 1T X = = −λ2 X αT a constant (4.7b) It follows that the differential eqn. (4.7a) can be resolved into two ordinary differential equations: X = −λ2 X and T = −α λ2 T (4.8) The general solution of both of these equations are well known and are among the first ones dealt with in any study of differential equations. They are: X (x) = A sin λx + B cos λx X (x) = Ax + B for λ ≠ 0 for λ = 0 (4.9) The general solution §4.2 147 and T (t) = Ce−αλ T (t) = C 2t for λ ≠ 0 for λ = 0 (4.10) where we use capital letters to denote constants of integration. [In either case, these solutions can be verified by substituting them back into eqn. (4.8).] Thus the general solution of eqn. (4.5) can indeed be written in the form of a product, and that product is 2 T = XT = e−αλ t (D sin λx + E cos λx) for λ ≠ 0 T = XT = Dx + E for λ = 0 (4.11) The usefulness of this result depends on whether or not it can be fit to the b.c.’s and the i.c. In this case, we made the function X (t) take the form of sines and cosines (instead of exponential functions) by placing a minus sign in front of λ2 . The sines and cosines make it possible to fit the b.c.’s using Fourier series methods. These general methods are not developed in this book; however, a complete Fourier series solution is presented for one problem in Section 5.3. The preceding simple methods for obtaining general solutions of linear partial d.e.’s is called the method of separation of variables. It can be applied to all kinds of linear d.e.’s. Consider, for example, two-dimensional steady heat conduction without heat sources: ∂2T ∂2T + =0 ∂x 2 ∂y 2 (4.12) Set T = XY and get Y X =− = −λ2 X Y where λ can be an imaginary number. Then ⎫ X = A sin λx + B cos λx ⎬ Y = Ceλy + De−λy X = Ax + B Y = Cy + D ⎭ for λ ≠ 0 for λ = 0 The general solution is T = (E sin λx + F cos λx)(e−λy + Geλy ) for λ ≠ 0 T = (Ex + F )(y + G) for λ = 0 (4.13) 148 Analysis of heat conduction and some steady one-dimensional problems §4.2 Figure 4.3 A two-dimensional slab maintained at a constant temperature on the sides and subjected to a sinusoidal variation of temperature on one face. Example 4.1 A long slab is cooled to 0◦ C on both sides and a blowtorch is turned on the top edge, giving an approximately sinusoidal temperature distribution along the top, as shown in Fig. 4.3. Find the temperature distribution within the slab. Solution. The general solution is given by eqn. (4.13). We must therefore identify the appropriate b.c.’s and then fit the general solution to it. Those b.c.’s are: on the top surface : on the sides : as y → ∞ : T (x, 0) = A sin π x L T (0 or L, y) = 0 T (x, y → ∞) = 0 Substitute eqn. (4.13) in the third b.c.: (E sin λx + F cos λx)(0 + G · ∞) = 0 The only way that this can be true for all x is if G = 0. Substitute eqn. (4.13), with G = 0, into the second b.c.: (O + F )e−λy = 0 §4.2 The general solution so F also equals 0. Substitute eqn. (4.13) with G = F = 0, into the first b.c.: E(sin λx) = A sin π x L It follows that A = E and λ = π /L. Then eqn. (4.13) becomes the particular solution that satisfies the b.c.’s: T = A sin π x e−π y/L L Thus, the sinusoidal variation of temperature at the top of the slab is attenuated exponentially at lower positions in the slab. At a position of y = 2L below the top, T will be 0.0019 A sin π x/L. The temperature distribution in the x -direction will still be sinusoidal, but it will have less than 1/500 of the amplitude at y = 0. Consider some important features of this and other solutions: • The b.c. at y = 0 is a special one that works very well with this particular general solution. If we had tried to fit the equation to a general temperature distribution, T (x, y = 0) = fn(x), it would not have been obvious how to proceed. Actually, this is the kind of problem that Fourier solved with the help of his Fourier series method. We discuss this matter in more detail in Chapter 5. • Not all forms of general solutions lend themselves to a particular set of boundary and/or initial conditions. In this example, we made the process look simple, but more often than not, it is in fitting a general solution to a set of boundary conditions that we get stuck. • Normally, on formulating a problem, we must approximate real behavior in stating the b.c.’s. It is advisable to consider what kind of assumption will put the b.c.’s in a form compatible with the general solution. The temperature distribution imposed on the slab by the blowtorch in Example 4.1 might just as well have been approximated as a parabola. But as small as the difference between a parabola and a sine function might be, the latter b.c. was far easier to accommodate. • The twin issues of existence and uniqueness of solutions require a comment here: It has been established that solutions to all wellposed heat diffusion problems are unique. Furthermore, we know 149 150 Analysis of heat conduction and some steady one-dimensional problems §4.3 from our experience that if we describe a physical process correctly, a unique outcome exists. Therefore, we are normally safe to leave these issues to a mathematician—at least in the sort of problems we discuss here. • Given that a unique solution exists, we accept any solution as correct since we have carved it to fit the boundary conditions. In this sense, the solution of differential equations is often more of an incentive than a formal operation. The person who does it best is often the person who has done it before and so has a large assortment of tricks up his or her sleeve. 4.3 Dimensional analysis Introduction Most universities place the first course in heat transfer after an introduction to fluid mechanics: and most fluid mechanics courses include some dimensional analysis. This is normally treated using the familiar method of indices, which is seemingly straightforward to teach but is cumbersome and sometimes misleading to use. It is rather well presented in [4.1]. The method we develop here is far simpler to use than the method of indices, and it does much to protect us from the common errors we might fall into. We refer to it as the method of functional replacement. The importance of dimensional analysis to heat transfer can be made clearer by recalling Example 2.6, which (like most problems in Part I) involved several variables. Theses variables included the dependent variable of temperature, (T∞ − Ti );3 the major independent variable, which was the radius, r ; and five system parameters, ri , ro , h, k, and (T∞ − Ti ). By reorganizing the solution into dimensionless groups [eqn. (2.24)], we reduced the total number of variables to only four: ⎤ ⎡ T − Ti T∞ − Ti dependent variable ⎢ = fn⎣ r ri , ro ri , Bi ⎥ ⎦ (2.24a) indep. var. two system parameters This solution offered a number of advantages over the dimensional solution. For one thing, it permitted us to plot all conceivable solutions 3 Notice that we do not call Ti a variable. It is simply the reference temperature against which the problem is worked. If it happened to be 0◦ C, we would not notice its subtraction from the other temperatures. Dimensional analysis §4.3 for a particular shape of cylinder, (ro /ri ), in a single figure, Fig. 2.13. For another, it allowed us to study the simultaneous roles of h, k and ro in defining the character of the solution. By combining them as a Biot number, we were able to say—even before we had solved the problem— whether or not external convection really had to be considered. The nondimensionalization made it possible for us to consider, simultaneously, the behavior of all similar systems of heat conduction through cylinders. Thus a large, highly conducting cylinder might be similar in its behavior to a small cylinder with a lower thermal conductivity. Finally, we shall discover that, by nondimensionalizing a problem before we solve it, we can often greatly simplify the process of solving it. Our next aim is to map out a method for nondimensionalization problems before we have solved then, or, indeed, before we have even written the equations that must be solved. The key to the method is a result called the Buckingham pi-theorem. The Buckingham pi-theorem The attention of scientific workers was drawn very strongly toward the question of similarity at about the beginning of World War I. Buckingham first organized previous thinking and developed his famous theorem in 1914 in the Physical Review [4.2], and he expanded upon the idea in the Transactions of the ASME one year later [4.3]. Lord Rayleigh almost simultaneously discussed the problem with great clarity in 1915 [4.4]. To understand Buckingham’s theorem, we must first overcome one conceptual hurdle, which, if it is clear to the student, will make everything that follows extremely simple. Let us explain that hurdle first. Suppose that y depends on r , x, z and so on: y = y(r , x, z, . . . ) We can take any one variable—say, x —and arbitrarily multiply it (or it raised to a power) by any other variables in the equation, without altering the truth of the functional equation, like this: y y = x 2 r , x, xz x x Many people find such a rearrangement disturbing when they first see it. That is because these are not algebraic equations — they are functional equations. We have said only that if y depends upon r , x , and z that it will likewise depend upon x 2 r , x , and xz. Suppose, for example, that we gave the functional equation the following algebraic form: y = y(r , x, z) = r (sin x)e−z 151 152 Analysis of heat conduction and some steady one-dimensional problems §4.3 This need only be rearranged to put it in terms of the desired modified variables and x itself (y/x, x 2 r , x, and xz): x2r xz y = 3 (sin x) exp − x x x We can do any such multiplying or dividing of powers of any variable we wish without invalidating any functional equation that we choose to write. This simple fact is at the heart of the important example that follows. Example 4.2 Consider the heat exchanger problem described in Fig. 3.15. The “unknown,” or dependent variable, in the problem is either of the exit temperatures. Without any knowledge of heat exchanger analysis, we can write the functional equation on the basis of our physical understanding of the problem: ⎡ ⎤ ⎢ Tcout − Tcin = fn ⎢Cmax , Cmin , Thin − Tcin , ⎣ K W/K W/K U ⎥ , A⎥ ⎦ (4.14) W/m2 K m2 K where the dimensions of each term are noted under the quotation. We want to know how many dimensionless groups the variables in eqn. (4.14) should reduce to. To determine this number, we use the idea explained above—that is, that we can arbitrarily pick one variable from the equation and divide or multiply it into other variables. Then—one at a time—we select a variable that has one of the dimensions. We divide or multiply it by the other variables in the equation that have that dimension in such a way as to eliminate the dimension from them. We do this first with the variable (Thin − Tcin ), which has the dimension of K. ⎡ Tcout − Tcin ⎢ = fn ⎣Cmax (Thin − Tcin ), Cmin (Thin − Tcin ), Thin − Tcin W W ⎤ dimensionless ⎥ (Thin − Tcin ), U(Thin − Tcin ), A ⎥ ⎦ K W/m2 m2 Dimensional analysis §4.3 153 The interesting thing about the equation in this form is that the only remaining term in it with the units of K is (Thin − Tcin ). No such term can exist in the equation because it is impossible to achieve dimensional homogeneity without another term in K to balance it. Therefore, we must remove it. ⎡ ⎤ ⎢ ⎥ Tcout − Tcin = fn ⎢Cmax (Thin − Tcin ), Cmin (Thin − Tcin ), U(Thin − Tcin ), A ⎥ ⎣ ⎦ Thin − Tcin W dimensionless W W/m2 m2 Now the equation has only two dimensions in it—W and m2 . Next, we multiply U (Thin − Tcin ) by A to get rid of m2 in the second-to-last term. Accordingly, the term A (m2 ) can no longer stay in the equation, and we have ⎡ ⎤ Tcout − Tcin ⎢ ⎥ = fn ⎣Cmax (Thin − Tcin ), Cmin (Thin − Tcin ), U A(Thin − Tcin ), ⎦ Thin − Tcin W W W dimensionless Next, we divide the first and third terms on the right by the second. This leaves only Cmin (Thin − Tcin ), with the dimensions of W. That term must then be removed, and we are left with the completely dimensionless result: Tcout − Tcin Cmax U A = fn , Thin − Tcin Cmin Cmin (4.15) Equation (4.15) has exactly the same functional form as eqn. (3.21), which we obtained by direct analysis. Notice that we removed one variable from eqn. (4.14) for each dimension in which the variables are expressed. If there are n variables— including the dependent variable—expressed in m dimensions, we then expect to be able to express the equation in (n − m) dimensionless groups, or pi-groups, as Buckingham called them. This fact is expressed by the Buckingham pi-theorem, which we state formally in the following way: 154 Analysis of heat conduction and some steady one-dimensional problems §4.3 A physical relationship among n variables, which can be expressed in a minimum of m dimensions, can be rearranged into a relationship among (n − m) independent dimensionless groups of the original variables. Two important qualifications have been italicized. They will be explained in detail in subsequent examples. Buckingham called the dimensionless groups pi-groups and identified them as Π1 , Π2 , ..., Πn−m . Normally we call Π1 the dependent variable and retain Π2→(n−m) as independent variables. Thus, the dimensional functional equation reduces to a dimensionless functional equation of the form Π1 = fn (Π2 , Π3 , . . . , Πn−m ) (4.16) Applications of the pi-theorem Example 4.3 Is eqn. (2.24) consistent with the pi-theorem? Solution. To find out, we first write the dimensional functional equation for Example 2.6: T − Ti = fn K r , ri , ro , m m m h , k , (T∞ − Ti ) W/m2 K W/m·K K There are seven variables (n = 7) in three dimensions, K, m, and W (m = 3). Therefore, we look for 7 − 3 = 4 pi-groups. There are four pi-groups in eqn. (2.24): Π1 = T − Ti , T∞ − Ti Π2 = r , ri Π3 = ro , ri Π4 = hro ≡ Bi. k Consider two features of this result. First, the minimum number of dimensions was three. If we had written watts as J/s, we would have had four dimensions instead. But Joules never appear in that particular problem independently of seconds. They always appear as a ratio and should not be separated. (If we had worked in English units, this would have seemed more confusing, since there is no name for Btu/sec unless Dimensional analysis §4.3 we first convert it to horsepower.) The failure to identify dimensions that are consistently grouped together is one of the major errors that the beginner makes in using the pi-theorem. The second feature is the independence of the groups. This means that we may pick any four dimensionless arrangements of variables, so long as no group or groups can be made into any other group by mathematical manipulation. For example, suppose that someone suggested that there was a fifth pi-group in Example 4.3: Π5 = hr k It is easy to see that Π5 can be written as Π5 = hro k r ri ri = ro Bi Π2 Π3 Therefore Π5 is not independent of the existing groups, nor will we ever find a fifth grouping that is. Another matter that is frequently made much of is that of identifying the pi-groups once the variables are identified for a given problem. (The method of indices [4.1] is a cumbersome arithmetic strategy for doing this but it is perfectly correct.) We shall find the groups by using either of two methods: 1. The groups can always be obtained formally by repeating the simple elimination-of-dimensions procedure that was used to derive the pi-theorem in Example 4.2. 2. One may simply arrange the variables into the required number of independent dimensionless groups by inspection. In any method, one must make judgments in the process of combining variables and these decisions can lead to different arrangements of the pi-groups. Therefore, if the problem can be solved by inspection, there is no advantage to be gained by the use of a more formal procedure. The methods of dimensional analysis can be used to help find the solution of many physical problems. We offer the following example, not entirely with tongue in cheek: Example 4.4 Einstein might well have noted that the energy equivalent, e, of a rest 155 156 Analysis of heat conduction and some steady one-dimensional problems §4.3 mass, mo , depended on the velocity of light, co , before he developed the special relativity theory. He would then have had the following dimensional functional equation: e N ·m or e kg· m2 s2 = fn (co m/s, mo kg) The minimum number of dimensions is only two: kg and m/s, so we look for 3 − 2 = 1 pi-group. To find it formally, we eliminated the dimension of mass from e by dividing it by mo (kg). Thus, e m2 = fn co m/s, mo s2 mo kg this must be removed because it is the only term with mass in it Then we eliminate the dimension of velocity (m/s) by dividing e/mo 2 by co : e 2 = fn (co m/s) mo co This time co must be removed from the function on the right, since it is the only term with the dimensions m/s. This gives the result (which could have been written by inspection once it was known that there could only be one pi-group): Π1 = e 2 = fn (no other groups) = constant mo co or 2 e = constant · mo co Of course, it required Einstein’s relativity theory to tell us that the constant is unity. Example 4.5 What is the velocity of efflux of liquid from the tank shown in Fig. 4.4? Solution. In this case we can guess that the velocity, V , might depend on gravity, g , and the head H . We might be tempted to include Dimensional analysis §4.3 157 Figure 4.4 Efflux of liquid from a tank. the density as well until we realize that g is already a force per unit mass. To understand this, we can use English units and divide g by the conversion factor,4 gc . Thus (g ft/s2 )/(gc lbm ·ft/lbf s2 ) = g lbf /lbm . Then V = fn m/s H, g m m/s2 so there are three variables in two dimensions, and we look for 3 − 2 = 1 pi-groups. It would have to be Π1 = V = fn (no other pi-groups) = constant gH or V = constant · g H The analytical study of fluid √ mechanics tells us that this form is correct and that the constant is 2. The group V 2/gh, by the way, is called a Froude number, Fr (pronounced “Frood”). It compares inertial forces to gravitational forces. Fr is about 1000 for a pitched baseball, and it is between 1 and 10 for the water flowing over the spillway of a dam. 4 One can always divide any variable by a conversion factor without changing it. 158 Analysis of heat conduction and some steady one-dimensional problems §4.3 Example 4.6 Obtain the dimensionless functional equation for the temperature ˙ distribution during steady conduction in a slab with a heat source, q. Solution. In such a case, there might be one or two specified temperatures in the problem: T1 or T2 . Thus the dimensional functional equation is ⎡ ⎤ ⎢ ˙ T − T1 = fn ⎢(T2 − T1 ), x, L, q , ⎣ K K m k , h ⎥ ⎥ ⎦ W/m3 W/m·K W/m2 K where we presume that a convective b.c. is involved and we identify a characteristic length, L, in the x -direction. There are seven variables in three dimensions, or 7 − 3 = 4 pi-groups. Three of these groups are ones we have dealt with in the past in one form or another: T − T1 T2 − T1 x Π2 = L Π1 = Π3 = hL k dimensionless temperature, which we shall give the name Θ dimensionless length, which we call ξ which we recognize as the Biot number, Bi The fourth group is new to us: Π4 = ˙ q L2 k(T2 − T1 ) which compares the heat generation rate to the rate of heat loss; we call it Γ Thus, the solution is Θ = fn (ξ, Bi, Γ ) (4.17) In Example 2.1, we undertook such a problem, but it differed in two respects. There was no convective boundary condition and hence, no h, and only one temperature was specified in the problem. In this case, the dimensional functional equation was ˙ (T − T1 ) = fn x , L, q, k so there were only five variables in the same three dimensions. The resulting dimensionless functional equation therefore involved only two §4.4 An illustration of dimensional analysis in a complex steady conduction problem pi-groups. One was ξ = x/L and the other is a new one equal to Θ/Γ . We call it Φ: Φ≡ x T − T1 = fn ˙ L qL2 /k (4.18) And this is exactly the form of the analytical result, eqn. (2.15). Finally, we must deal with dimensions that convert into one another. For example, kg and N are defined in terms of one another through Newton’s Second Law of Motion. Therefore, they cannot be identified as separate dimensions. The same would appear to be true of J and N·m, since both are dimensions of energy. However, we must discern whether or not a mechanism exists for interchanging them. If mechanical energy remains distinct from thermal energy in a given problem, then J should not be interpreted as N·m. This issue will prove important when we do the dimensional analysis of several heat transfer problems. See, for example, the analyses of laminar convection problem at the beginning of Section 6.4, of natural convection in Section 8.3, of film condensation in Section 8.5, and of pool boiling burnout in Section 9.3. In all of these cases, heat transfer normally occurs without any conversion of heat to work or work to heat and it would be misleading to break J into N·m. Additional examples of dimensional analysis appear throughout this book. Dimensional analysis is, indeed, our court of first resort in solving most of the new problems that we undertake. 4.4 An illustration of the use of dimensional analysis in a complex steady conduction problem Heat conduction problems with convective boundary conditions can rapidly grow difficult, even if they start out simple, and so we look for ways to avoid making mistakes. For one thing, it is wise to take great care that dimensions are consistent at each stage of the solution. The best way to do this, and to eliminate a great deal of algebra at the same time, is to nondimensionalize the heat conduction equation before we apply the b.c.’s. This nondimensionalization should be consistent with the pitheorem. We illustrate this idea with a fairly complex example. 159 160 Analysis of heat conduction and some steady one-dimensional problems §4.4 Figure 4.5 Heat conduction through a heat-generating slab with asymmetric boundary conditions. Example 4.7 A slab shown in Fig. 4.5 has different temperatures and different heat transfer coefficients on either side and the heat is generated within it. Calculate the temperature distribution in the slab. Solution. The differential equation is ˙ q d2 T =− 2 dx k and the general solution is T =− ˙ qx 2 + C1 x + C2 2k (4.19) §4.4 An illustration of dimensional analysis in a complex steady conduction problem with b.c.’s h1 (T1 − T )x =0 = −k dT dx x =0 , h2 (T − T2 )x =L = −k dT dx x =L . (4.20) There are eight variables involved in the problem: (T − T2 ), (T1 − T2 ), ˙ x , L, k, h1 , h2 , and q; and there are three dimensions: K, W, and m. This results in 8 − 3 = 5 pi-groups. For these we choose Π1 ≡ Θ = T − T2 , T1 − T2 Π4 ≡ Bi2 = h2 L , k Π2 ≡ ξ = and x , L Π3 ≡ Bi1 = Π5 ≡ Γ = h1 L , k ˙ qL2 , 2k(T1 − T2 ) where Γ can be interpreted as a comparison of the heat generated in the slab to that which could flow through it. Under this nondimensionalization, eqn. (4.19) becomes5 Θ = −Γ ξ 2 + C3 ξ + C4 (4.21) and b.c.’s become Bi1 (1 − Θξ =0 ) = −Θξ =0 , Bi2 Θξ =1 = −Θξ =1 (4.22) where the primes denote differentiation with respect to ξ . Substituting eqn. (4.21) in eqn. (4.22), we obtain Bi1 (1 − C4 ) = −C3 , Bi2 (−Γ + C3 + C4 ) = 2Γ − C3 . (4.23) Substituting the first of eqns. (4.23) in the second we get C4 = 1 + −Bi1 + 2(Bi1 /Bi2 )Γ + Bi1 Γ Bi1 + Bi2 Bi2 + Bi2 1 1 C3 = Bi1 (C4 − 1) Thus, eqn. (4.21) becomes Θ=1+Γ 5 2(Bi1 Bi2 ) + Bi1 2(Bi1 Bi2 ) + Bi1 ξ − ξ2 + 1 + Bi1 Bi2 + Bi1 Bi1 + Bi2 Bi2 + Bi2 1 1 Bi1 Bi1 ξ− − 1 + Bi1 Bi2 + Bi1 Bi1 + Bi2 Bi2 + Bi2 1 1 (4.24) The rearrangement of the dimensional equations into dimensionless form is straightforward algebra. If the results shown here are not immediately obvious to you, sketch the calculation on a piece of paper. 161 162 Analysis of heat conduction and some steady one-dimensional problems §4.4 This is a complicated result and one that would have required enormous patience and accuracy to obtain without first simplifying the problem statement as we did. If the heat transfer coefficients were the same on either side of the wall, then Bi1 = Bi2 ≡ Bi, and eqn. (4.24) would reduce to Θ = 1 + Γ ξ − ξ 2 + 1/Bi − ξ + 1/Bi 1 + 2/Bi (4.25) which is a very great simplification. Equation (4.25) is plotted on the left-hand side of Fig. 4.5 for Bi equal to 0, 1, and ∞ and for Γ equal to 0, 0.1, and 1. The following features should be noted: • When Γ 0.1, the heat generation can be ignored. • When Γ 1, Θ → Γ /Bi + Γ (ξ − ξ 2 ). This is a simple parabolic temperature distribution displaced upward an amount that depends on the relative external resistance, as reflected in the Biot number. • If both Γ and 1/Bi become large, Θ → Γ /Bi. This means that when internal resistance is low and the heat generation is great, the slab temperature is constant and quite high. If T2 were equal to T1 in this problem, Γ would go to infinity. In such a situation, we should redo the dimensional analysis of the problem. The dimensional functional equation now shows (T − T1 ) to be a function of ˙ x , L, k, h, and q. There are six variables in three dimensions, so there are three pi-groups T − T1 = fn (ξ, Bi) ˙ qL/h where the dependent variable is like Φ [recall eqn. (4.18)] multiplied by Bi. We can put eqn. (4.25) in this form by multiplying both sides of it by ˙ h(T1 − T2 )/qδ. The result is 1 h(T − T1 ) 1 = Bi ξ − ξ 2 + ˙ 2 2 qL (4.26) The result is plotted on the right-hand side of Fig. 4.5. The following features of the graph are of interest: • Heat generation is the only “force” giving rise to temperature nonuniformity. Since it is symmetric, the graph is also symmetric. Fin design §4.5 • When Bi 1, the slab temperature approaches a uniform value ˙ equal to T1 + qL/2h. (In this case, we would have solved the problem with far greater ease by using a simple lumped-capacity heat balance, since it is no longer a heat conduction problem.) • When Bi > 100, the temperature distribution is a very large parabola with ½ added to it. In this case, the problem could have been solved using boundary conditions of the first kind because the surface temperature stays very close to T∞ (recall Fig. 1.11). 4.5 Fin design The purpose of fins The convective removal of heat from a surface can be substantially improved if we put extensions on that surface to increase its area. These extensions can take a variety of forms. Figure 4.6, for example, shows many different ways in which the surface of commercial heat exchanger tubing can be extended with protrusions of a kind we call fins. Figure 4.7 shows another very interesting application of fins in a heat exchanger design. This picture is taken from an issue of Science magazine [4.5], which presents an intriguing argument by Farlow, Thompson, and Rosner. They offered evidence suggesting that the strange rows of fins on the back of the Stegosaurus were used to shed excess body heat after strenuous activity, which is consistent with recent suspicions that Stegosaurus was warm-blooded. These examples involve some rather complicated fins. But the analysis of a straight fin protruding from a wall displays the essential features of all fin behavior. This analysis has direct application to a host of problems. Analysis of a one-dimensional fin The equations. Figure 4.8 shows a one-dimensional fin protruding from a wall. The wall—and the roots of the fin—are at a temperature T0 , which is either greater or less than the ambient temperature, T∞ . The length of the fin is cooled or heated through a heat transfer coefficient, h, by the ambient fluid. The heat transfer coefficient will be assumed uniform, although (as we see in Part III) that can introduce serious error in boil- 163 a. Eight examples of externally finned tubing: 1) and 2) typical commercial circular fins of constant thickness; 3) and 4) serrated circular fins and dimpled spirally-wound circular fins, both intended to improve convection; 5) spirally-wound copper coils outside and inside; 6) and 8) bristle fins, spirally wound and machined from base metal; 7) a spirally indented tube to improve convection and increase surface area. b. An array of commercial internally finned tubing (photo courtesy of Noranda Metal Industries, Inc.) Figure 4.6 Some of the many varieties of finned tubes. 164 Fin design §4.5 165 Figure 4.7 The Stegosaurus with what might have been cooling fins (etching by Daniel Rosner). ing, condensing, or other natural convection situations, and will not be strictly accurate even in forced convection. The tip may or may not exchange heat with the surroundings through a heat transfer coefficient, hL , which would generally differ from h. The length of the fin is L, its uniform cross-sectional area is A, and its circumferential perimeter is P . The characteristic dimension of the fin in the transverse direction (normal to the x -axis) is taken to be A/P . Thus, for a circular cylindrical fin, A/P = π (radius)2 /(2π radius) = (radius/2). We define a Biot number for conduction in the transverse direction, based on this dimension, and require that it be small: Bifin = h(A/P ) k 1 (4.27) 166 Analysis of heat conduction and some steady one-dimensional problems §4.5 Figure 4.8 The analysis of a one-dimensional fin. This condition means that the transverse variation of T at any axial position, x , is much less than (Tsurface − T∞ ). Thus, T T (x only) and the heat flow can be treated as one-dimensional. An energy balance on the thin slice of the fin shown in Fig. 4.8 gives −kA dT dx x +δx + kA dT dx x + h(P δx)(T − T∞ )x = 0 (4.28) but dT /dx |x +δx − dT /dx |x d2 T d2 (T − T∞ ) → = δx dx 2 dx 2 (4.29) Fin design §4.5 167 so hP d2 (T − T∞ ) (T − T∞ ) = dx 2 kA (4.30) The b.c.’s for this equation are (T − T∞ )x =0 = T0 − T∞ −kA d(T − T∞ ) dx x =L = hL A(T − T∞ )x =L (4.31a) Alternatively, if the tip is insulated, or if we can guess that hL is small enough to be unimportant, the b.c.’s are (T − T∞ )x =0 = T0 − T∞ and d(T − T∞ ) dx x =L =0 (4.31b) Before we solve this problem, it will pay to do a dimensional analysis of it. The dimensional functional equation is T − T∞ = fn (T0 − T∞ ) , x, L, kA, hP , hL A (4.32) Notice that we have written kA, hP , and hL A as single variables. The reason for doing so is subtle but important. Setting h(A/P )/k 1, erases any geometric detail of the cross section from the problem. The only place where P and A enter the problem is as product of k, h, orhL . If they showed up elsewhere, they would have to do so in a physically incorrect way. Thus, we have just seven variables in W, K, and m. This gives four pi-groups if the tip is uninsulated: ⎛ ⎞ ⎜ ⎜x T − T∞ ⎜ = fn ⎜ , ⎜L T0 − T∞ ⎝ ⎟ hP 2 hL AL ⎟ ⎟ ⎟ L, kA kA ⎟ ⎠ =hL L k or if we rename the groups, Θ = fn (ξ, mL, Biaxial ) (4.33a) where we call hP L2 /kA ≡ mL because that terminology is common in the literature on fins. If the tip of the fin is insulated, hL will not appear in eqn. (4.32). There is one less variable but the same number of dimensions; hence, there will 168 Analysis of heat conduction and some steady one-dimensional problems §4.5 be only three pi-groups. The one that is removed is Biaxial , which involves hL . Thus, for the insulated fin, Θ = fn(ξ, mL) (4.33b) We put eqn. (4.30) in these terms by multiplying it by L2 /(T0 − T∞ ). The result is d2 Θ = (mL)2 Θ dξ 2 (4.34) This equation is satisfied by Θ = Ce±(mL)ξ . The sum of these two solutions forms the general solution of eqn. (4.34): Θ = C1 emLξ + C2 e−mLξ (4.35) Temperature distribution in a one-dimensional fin with the tip insulated The b.c.’s [eqn. (4.31b)] can be written as Θξ =0 = 1 dΘ dξ and =0 (4.36) ξ =1 Substituting eqn. (4.35) into both eqns. (4.36), we get C1 + C2 = 1 and C1 emL − C2 e−mL = 0 (4.37) Mathematical Digression 4.1 To put the solution of eqn. (4.37) for C1 and C2 in the simplest form, we need to recall a few properties of hyperbolic functions. The four basic functions that we need are defined as ex − e−x 2 x + e−x e cosh x ≡ 2 sinh x tanh x ≡ cosh x ex + e−x coth x ≡ x e − e−x sinh x ≡ = ex − e−x ex + e−x (4.38) Fin design §4.5 169 where x is the independent variable. Additional functions are defined by analogy to the trigonometric counterparts. The differential relations can be written out formally, and they also resemble their trigonometric counterparts. d sinh x = dx d cosh x = dx 1x e − (−e−x ) = cosh x 2 1x e + (−e−x ) = sinh x 2 (4.39) These are analogous to the familiar results, d sin x/dx = cos x and d cos x/dx = − sin x , but without the latter minus sign. The solution of eqns. (4.37) is then C1 = e−mL 2 cosh mL and C2 = 1 − e−mL 2 cosh mL (4.40) Therefore, eqn. (4.35) becomes Θ= e−mL(1−ξ) + (2 cosh mL)e−mLξ − e−mL(1+ξ) 2 cosh mL which simplifies to Θ= cosh mL(1 − ξ) cosh mL (4.41) for a one-dimensional fin with its tip insulated. One of the most important design variables for a fin is the rate at which it removes (or delivers) heat the wall. To calculate this, we write Fourier’s law for the heat flow into the base of the fin:6 Q = −kA d(T − T∞ ) dx x =0 (4.42) We multiply eqn. (4.42) by L/kA(T0 − T∞ ) and obtain, after substituting eqn. (4.41) on the right-hand side, QL sinh mL = mL = mL tanh mL kA(T0 − T∞ ) cosh mL (4.43) We could also integrate h(T − T∞ ) over the outside area of the fin to get Q. The answer would be the same, but the calculation would be a little more complicated. 6 170 Analysis of heat conduction and some steady one-dimensional problems §4.5 Figure 4.9 The temperature distribution, tip temperature, and heat flux in a straight one-dimensional fin with the tip insulated. which can be written Q kAhP (T0 − T∞ ) = tanh mL (4.44) Figure 4.9 includes two graphs showing the behavior of one-dimensional fin with an insulated tip. The top graph shows how the heat removal increases with mL to a virtual maximum at mL 3. This means that no such fin should have a length in excess of 2/m or 3/m if it is being used to cool (or heat) a wall. Additional length would simply increase the cost without doing any good. Also shown in the top graph is the temperature of the tip of such a fin. Setting ξ = 1 in eqn. (4.41), we discover that Θtip = 1 cosh mL (4.45) Fin design §4.5 171 This dimensionless temperature drops to about 0.014 at the tip when mL reaches 5. This means that the end is 0.014(T0 − T∞ ) K above T∞ at the end. Thus, if the fin is actually functioning as a holder for a thermometer or a thermocouple that is intended to read T∞ , the reading will be in error if mL is not significantly greater than five. The lower graph in Fig. 4.9 hows how the temperature is distributed in insulated-tip fins for various values of mL. Experiment 4.1 Clamp a 20 cm or so length of copper rod by one end in a horizontal position. Put a candle flame very near the other end and let the arrangement come to a steady state. Run your finger along the rod. How does what you feel correspond to Fig. 4.9? (The diameter for the rod should not exceed about 3 mm. A larger rod of metal with a lower conductivity will also work.) Exact temperature distribution in a fin with an uninsulated tip. The approximation of an insulated tip may be avoided using the b.c’s given in eqn. (4.31a), which take the following dimensionless form: Θξ =0 = 1 and − dΘ dξ ξ =1 = Biax Θξ =1 (4.46) Substitution of the general solution, eqn. (4.35), in these b.c.’s yields C1 + C2 −mL(C1 emL − C2 e−mL ) =1 = Biax (C1 emL + C2 e−mL ) (4.47) It requires some manipulation to solve eqn. (4.47) for C1 and C2 and to substitute the results in eqn. (4.35). We leave this as an exercise (Problem 4.11). The result is Θ= cosh mL(1 − ξ) + (Biax /mL) sinh mL(1 − ξ) cosh mL + (Biax /mL) sinh mL (4.48) which is the form of eqn. (4.33a), as we anticipated. The corresponding heat flux equation is Q (kA)(hP ) (T0 − T∞ ) = (Biax /mL) + tanh mL 1 + (Biax /mL) tanh mL (4.49) 172 Analysis of heat conduction and some steady one-dimensional problems §4.5 We have seen that mL is not too much greater than unity in a welldesigned fin with an insulated tip. Furthermore, when hL is small (as it might be in natural convection), Biax is normally much less than unity. Therefore, in such cases, we expect to be justified in neglecting terms multiplied by Biax . Then eqn. (4.48) reduces to Θ= cosh mL(1 − ξ) cosh mL (4.41) which we obtained by analyzing an insulated fin. It is worth pointing out that we are in serious difficulty if hL is so large that we cannot assume the tip to be insulated. The reason is that hL is nearly impossible to predict in most practical cases. Example 4.8 A 2 cm diameter aluminum rod with k = 205 W/m·K, 8 cm in length, protrudes from a 150◦ C wall. Air at 26◦ C flows by it, and h = 120 W/m2 K. Determine whether or not tip conduction is important in this hL . problem. To do this, make the very crude assumption that h Then compare the tip temperatures as calculated with and without considering heat transfer from the tip. Solution. mL = hP L2 = kA Biax = 120(0.08)2 = 0.8656 205(0.01/2) 120(0.08) hL = = 0.0468 k 205 Therefore, eqn. (4.48) becomes cosh 0 + (0.0468/0.8656) sinh 0 cosh(0.8656) + (0.0468/0.8656) sinh(0.8656) 1 = = 0.6886 1.3986 + 0.0529 Θ (ξ = 1) = Θtip = so the exact tip temperature is Ttip = T∞ + 0.6886(T0 − T∞ ) = 26 + 0.6886(150 − 26) = 111.43◦ C Fin design §4.5 173 Equation (4.41) or Fig. 4.9, on the other hand, gives Θtip = 1 = 0.7150 1.3986 so the approximate tip temperature is Ttip = 26 + 0.715(150 − 26) = 114.66◦ C Thus the insulated-tip approximation is adequate for the computation in this case. Very long fin. If a fin is so long that mL 1, then eqn. (4.41) becomes emL(1−ξ) emL(1−ξ) + e−mL(1−ξ) = mL→∞ emL + e−mL emL limit Θ = limit mL→∞ or limit Θ = e−mLξ mL→large (4.50) Substituting this result in eqn. (4.42), we obtain [cf. eqn. (4.44)] Q = (kAhP ) (T0 − T∞ ) (4.51) A heating or cooling fin would have to be terribly overdesigned for these results to apply—that is, mL would have been made much larger than necessary. Very long fins are common, however, in a variety of situations related to undesired heat losses. In practice, a fin may be regarded as “infinitely long” in computing its temperature if mL 5; in computing Q, mL 3 is sufficient for the infinite fin approximation. Physical significance of mL. The group mL has thus far proved to be extremely useful in the analysis and design of fins. We should therefore say a brief word about its physical significance. Notice that (mL)2 = L/kA 1/h(P L) = internal resistance in x -direction gross external resistance Thus (mL)2 is a hybrid Biot number. When it is big, Θ|ξ =1 → 0 and we can neglect tip convection. When it is small, the temperature drop along the axis of the fin becomes small (see the lower graph in Fig. 4.9). 174 Analysis of heat conduction and some steady one-dimensional problems §4.5 The group (mL)2 also has a peculiar similarity to the NTU (Chapter 3) and the dimensionless time, t/T , that appears in the lumped-capacity solution (Chapter 1). Thus, h(P L) kA/L is like UA Cmin is like hA ρcV /t In each case a convective heat rate is compared with a heat rate that characterizes the capacity of a system; and in each case the system temperature asymptotically approaches its limit as the numerator becomes large. This was true in eqn. (1.22), eqn. (3.21), eqn. (3.22), and eqn. (4.50). The problem of specifying the root temperature Thus far, we have assmed the root temperature of a fin to be given information. There really are many circumstances in which it might be known; however, if a fin protrudes from a wall of the same material, as sketched in Fig. 4.10a, it is clear that for heat to flow, there must be a temperature gradient in the neighborhood of the root. Consider the situation in which the surface of a wall is kept at a temperature Ts . Then a fin is placed on the wall as shown in the figure. If T∞ < Ts , the wall temperature will be depressed in the neighborhood of the root as heat flows into the fin. The fin’s performance should then be predicted using the lowered root temperature, Troot . This heat conduction problem has been analyzed for several fin arrangements by Sparrow and co-workers. Fig. 4.10b is the result of Sparrow and Hennecke’s [4.6] analysis for a single circular cylinder. They give 1− hr Ts − Troot Qactual , (mr ) tanh(mL) = = fn Qno temp. depression Ts − T∞ k (4.52) where r is the radius of the fin. From the figure we see that the actual heat flux into the fin, Qactual , and the actual root temperature are both reduced when the Biot number, hr /k, is large and the fin constant, m, is small. Example 4.9 Neglect the tip convection from the fin in Example 4.8 and suppose that it is embedded in a wall of the same material. Calculate the error in Q and the actual temperature of the root if the wall is kept at 150◦ C. Figure 4.10 The influence of heat flow into the root of circular cylindrical fins [4.6]. 175 176 Analysis of heat conduction and some steady one-dimensional problems §4.5 Solution. From Example 4.8 we have mL = 0.8656 and hr /k = 120(0.010)/205 = 0.00586. Then, with mr = mL(r /L), we have (mr ) tanh(mL) = 0.8656(0.010/0.080) tanh(0.8656) = 0.0756. The lower portion of Fig. 4.10b then gives 1− Ts − Troot Qactual = = 0.05 Qno temp. depression Ts − T∞ so the heat flow is reduced by 5% and the actual root temperature is Troot = 150 − (150 − 26)0.05 = 143.8◦ C The correction is modest in this case. Fin design Two basic measures of fin performance are particularly useful in a fin design. The first is called the efficiency, ηf . ηf ≡ actual heat transferred by a fin heat that would be transferred if the entire fin were at T = T0 (4.53) To see how this works, we evaluate ηf for a one-dimensional fin with an insulated tip: ηf = (hP )(kA)(T0 − T∞ ) tanh mL h(P L)(T0 − T∞ ) = tanh mL mL (4.54) This says that, under the definition of efficiency, a very long fin will give tanh(mL)/mL → 1/large number, so the fin will be inefficient. On the other hand, the efficiency goes up to 100% as the length is reduced to zero, because tanh(mL) → mL as mL → 0. While a fin of zero length would accomplish little, a fin of small m might be designed in order to keep the tip temperature near the root temperature; this, for example, is desirable if the fin is the tip of a soldering iron. It is therefore clear that, while ηf provides some useful information as to how well a fin is contrived, it is not generally advisable to design toward a particular value of ηf . A second measure of fin performance is called the effectiveness, εf : εf ≡ heat flux from the wall with the fin heat flux from the wall without the fin (4.55) Fin design §4.5 177 This can easily be computed from the efficiency: εf = ηf surface area of the fin cross-sectional area of the fin (4.56) Normally, we want the effectiveness to be as high as possible, But this can always be done by extending the length of the fin, and that—as we have seen—rapidly becomes a losing proposition. The measures ηf and εf probably attract the interest of designers not because their absolute values guide the designs, but because they are useful in characterizing fins with more complex shapes. In such cases the solutions are often so complex that ηf and εf plots serve as laborsaving graphical solutions. We deal with some of these curves later in this section. The design of a fin thus becomes an open-ended matter of optimizing, subject to many factors. Some of the factors that have to be considered include: • The weight of material added by the fin. This might be a cost factor or it might be an important consideration in its own right. • The possible dependence of h on (T − T∞ ), flow velocity past the fin, or other influences. • The influence of the fin (or fins) on the heat transfer coefficient, h, as the fluid moves around it (or them). • The geometric configuration of the channel that the fin lies in. • The cost and complexity of manufacturing fins. • The pressure drop introduced by the fins. Fin thermal resistance When fins occur in combination with other thermal elements, it can simplify calculations to treat them as a thermal resistance between the root and the surrounding fluid. Specifically, for a straight fin with an insulated tip, we can rearrange eqn. (4.44) as Q= (T0 − T∞ ) kAhP tanh mL −1 ≡ (T0 − T∞ ) Rtfin (4.57) 178 Analysis of heat conduction and some steady one-dimensional problems §4.5 where 1 Rtfin = for a straight fin kAhP tanh mL (4.58) In general, for a fin of any shape, fin thermal resistance can be written in terms of fin efficiency and fin effectiveness. From eqns. (4.53) and (4.55), we obtain Rtfin = 1 ηf Asurface h = 1 (4.59) εf Aroot h Example 4.10 Consider again the resistor described in Examples 2.8 and 2.9, starting on page 76. Suppose that the two electrical leads are long straight wires 0.62 mm in diameter with k = 16 W/m·K and heff = 23 W/m2 K. Recalculate the resistor’s temperature taking account of heat conducted into the leads. Solution. The wires act as very long fins connected to the resistor, so that tanh mL 1 (see Prob. 4.44). Each has a fin resistance of Rtfin = 1 kAhP = 1 (16)(23)(π )2 (0.00062)3 /4 = 2, 150 K/W These two thermal resistances are in parallel to the thermal resistances for natural convection and thermal radiation from the resistor surface found in Example 2.8. The equivalent thermal resistance is now Rtequiv = = 1 1 1 1 + + + Rtfin Rtfin Rtrad Rtconv −1 2 + (1.33 × 10−4 )(7.17) + (1.33 × 10−4 )(13) 2, 150 −1 = 276.8 K/W The leads reduce the equivalent resistance by about 30% from the value found before. The resistor temperature becomes Tresistor = Tair + Q · Rtequiv = 35 + (0.1)(276.8) = 62.68 ◦ C or about 10◦ C lower than before. Fin design §4.5 179 Figure 4.11 A general fin of variable cross section. Fins of variable cross section Let us consider what is involved is the design of a fin for which A and P are functions of x . Such a fin is shown in Fig. 4.11. We restrict our attention to fins for which h(A/P ) k 1 and d(a/P ) d(x) 1 so the heat flow will be approximately one-dimensional in x . We begin the analysis, as always, with the First Law statement: Qnet = Qcond − Qconv = dU dt or7 kA(x + δx) = dT dx x =δx − kA(x) dT dx x −hP δx (T − T∞ ) dT d kA(x) δx dx dx = ρcA(x)δx dT dt =0, since steady 7 Note that we approximate the external area of the fin as horizontal when we write it as P δx . The actual area is negligibly larger than this in most cases. An exception would be the tip of the fin in Fig. 4.11. 180 Analysis of heat conduction and some steady one-dimensional problems §4.5 Figure 4.12 A two-dimensional wedge-shaped fin. Therefore, d(T − T∞ ) d hP A(x) − (T − T∞ ) = 0 dx dx k (4.60) If A(x) = constant, this reduces to Θ − (mL)2 Θ = 0, which is the straight fin equation. To see how eqn. (4.60) works, consider the triangular fin shown in Fig. 4.12. In this case eqn. (4.60) becomes d(T − T∞ ) 2hb x d 2δ b − (T − T∞ ) = 0 dx L dx k or ξ d2 Θ dΘ + − dξ 2 dξ hL2 kδ Θ=0 (4.61) a kind of (mL)2 This second-order linear differential equation is difficult to solve because it has a variable coefficient. Its solution is expressible in Bessel functions: Θ= Io 2 hLx/kδ (4.62) Io 2 hL2 /kδ Fin design 181 where the modified Bessel function of the first kind, Io , can be looked up in appropriate tables. Rather than explore the mathematics of solving eqn. (4.60), we simply show the result for several geometries in terms of the fin efficiency, ηf , in Fig. 4.13. These curves were given by Schneider [4.7]. Kraus, Aziz, and Welty [4.8] provide a very complete discussion of fins and show a great many additional efficiency curves. Example 4.11 A thin brass pipe, 3 cm in outside diameter, carries hot water at 85◦ C. It is proposed to place 0.8 mm thick straight circular fins on the pipe to cool it. The fins are 8 cm in diameter and are spaced 2 cm apart. It is determined that h will equal 20 W/m2 K on the pipe and 15 W/m2 K on the fins, when they have been added. If T∞ = 22◦ C, compute the heat loss per meter of pipe before and after the fins are added. Solution. Before the fins are added, Q = π (0.03 m)(20 W/m2 K)[(85 − 22) K] = 199 W/m where we set Twall = Twater since the pipe is thin. Notice that, since the wall is constantly heated by the water, we should not have a roottemperature depression problem after the fins are added. Then we can enter Fig. 4.13a with r2 = 2.67 r1 and mL L = P hL3 = kA 15(0.04 − 0.15)3 = 0.306 125(0.025)(0.0008) and we obtain ηf = 89%. Thus, the actual heat transfer given by Qwithout fin 0.02 − 0.0008 0.02 119 W/m fraction of unfinned area + 0.89 [2π (0.042 − 0.0152 )] 50 fins m 15 W [(85 − 22) K] m2 K area per fin (both sides), m2 so Qnet = 478 W/m = 4.02 Qwithout fins Figure 4.13 The efficiency of several fins with variable cross section. 182 Problems 183 Problems 4.1 Make a table listing the general solutions of all steady, unidimensional constant-properties heat conduction problemns in Cartesian, cylindrical and spherical coordinates, with and without uniform heat generation. This table should prove to be a very useful tool in future problem solving. It should include a total of 18 solutions. State any restrictions on your solutions. Do not include calculations. 4.2 The left side of a slab of thickness L is kept at 0◦ C. The right side is cooled by air at T∞ ◦ C blowing on it. hRHS is known. An exothermic reaction takes place in the slab such that heat is generated at A(T − T∞ ) W/m3 , where A is a constant. Find a fully dimensionless expression for the temperature distribution in the wall. 4.3 A long, wide plate of known size, material, and thickness L is connected across the terminals of a power supply and serves as a resistance heater. The voltage, current and T∞ are known. The plate is insulated on the bottom and transfers heat out the top by convection. The temperature, Ttc , of the botton is measured with a thermocouple. Obtain expressions for (a) temperature distribution in the plate; (b) h at the top; (c) temperature at the top. (Note that your answers must depend on known information only.) [Ttop = Ttc − EIL2 /(2k · volume)] 4.4 The heat tansfer coefficient, h, resulting from a forced flow over a flat plate depends on the fluid velocity, viscosity, density, specific heat, and thermal conductivity, as well as on the length of the plate. Develop the dimensionless functional equation for the heat transfer coefficient (cf. Section 6.5). 4.5 Water vapor condenses on a cold pipe and drips off the bottom in regularly spaced nodes as sketched in Fig. 3.9. The wavelength of these nodes, λ, depends on the liquid-vapor density difference, ρf − ρg , the surface tension, σ , and the gravity, g . Find how λ varies with its dependent variables. 4.6 A thick film flows down a vertical wall. The local film velocity at any distance from the wall depends on that distance, gravity, the liquid kinematic viscosity, and the film thickness. Obtain 184 Chapter 4: Analysis of heat conduction and some steady one-dimensional problems the dimensionless functional equation for the local velocity (cf. Section 8.5). 4.7 A steam preheater consists of a thick, electrically conducting, cylindrical shell insulated on the outside, with wet stream flowing down the middle. The inside heat transfer coefficient is highly variable, depending on the velocity, quality, and so on, but the flow temperature is constant. Heat is released at ˙ q J/m3 s within the cylinder wall. Evaluate the temperature within the cylinder as a function of position. Plot Θ against ρ , where Θ is an appropriate dimensionless temperature and ρ = r /ro . Use ρi = 2/3 and note that Bi will be the parameter of a family of solutions. On the basis of this plot, recommend criteria (in terms of Bi) for (a) replacing the convective boundary condition on the inside with a constant temperature condition; (b) neglecting temperature variations within the cylinder. 4.8 Steam condenses on the inside of a small pipe, keeping it at a specified temperature, Ti . The pipe is heated by electrical ˙ resistance at a rate q W/m3 . The outside temperature is T∞ and there is a natural convection heat transfer coefficient, h around the outside. (a) Derive an expression for the dimensionless expression temperature distribution, Θ = (T − T∞ )/(Ti − T∞ ), as a function of the radius ratios, ρ = r /ro and ρi = ri /ro ; ˙2 a heat generation number, Γ = qro /k(Ti − T∞ ); and the Biot number. (b) Plot this result for the case ρi = 2/3, Bi = 1, and for several values of Γ . (c) Discuss any interesting aspects of your result. 4.9 Solve Problem 2.5 if you have not already done so, putting it in dimensionless form before you begin. Then let the Biot numbers approach infinity in the solution. You should get the same solution we got in Example 2.5, using b.c.’s of the first kind. Do you? 4.10 Complete the algebra that is missing between eqns. (4.30) and eqn. (4.31b) and eqn. (4.41). 4.11 Complete the algebra that is missing between eqns. (4.30) and eqn. (4.31a) and eqn. (4.48). Problems 185 4.12 Obtain eqn. (4.50) from the general solution for a fin [eqn. (4.35)], using the b.c.’s T (x = 0) = T0 and T (x = L) = T∞ . Comment on the significance of the computation. 4.13 What is the minimum length, l, of a thermometer well necessary to ensure an error less than 0.5% of the difference between the pipe wall temperature and the temperature of fluid flowing in a pipe? The well consists of a tube with the end closed. It has a 2 cm O.D. and a 1.88 cm I.D. The material is type 304 stainless steel. Assume that the fluid is steam at 260◦ C and that the heat transfer coefficient between the steam and the tube wall is 300 W/m2 K. [3.44 cm.] 4.14 Thin fins with a 0.002 m by 0.02 m rectangular cross section and a thermal conductivity of 50 W/m·K protrude from a wall and have h 600 W/m2 K and T0 = 170◦ C. What is the heat flow rate into each fin and what is the effectiveness? T∞ = 20◦ C. 4.15 A thin rod is anchored at a wall at T = T0 on one end and is insulated at the other end. Plot the dimensionless temperature distribution in the rod as a function of dimensionless length: (a) if the rod is exposed to an environment at T∞ through a heat transfer coefficient; (b) if the rod is insulated but heat is ˙ removed from the fin material at the unform rate −q = hP (T0 − T∞ )/A. Comment on the implications of the comparison. 4.16 A tube of outside diameter do and inside diameter di carries fluid at T = T1 from one wall at temperature T1 to another wall a distance L away, at Tr . Outside the tube ho is negligible, and inside the tube hi is substantial. Treat the tube as a fin and plot the dimensionless temperature distribution in it as a function of dimensionless length. 4.17 (If you have had some applied mathematics beyond the usual two years of calculus, this problem will not be difficult.) The shape of the fin in Fig. 4.12 is changed so that A(x) = 2δ(x/L)2 b instead of 2δ(x/L)b. Calculate the temperature distribution and the heat flux at the base. Plot the temperature distribution and fin thickness against x/L. Derive an expression for ηf . 186 Chapter 4: Analysis of heat conduction and some steady one-dimensional problems 4.18 Work Problem 2.21, if you have not already done so, nondimensionalizing the problem before you attempt to solve it. It should now be much simpler. 4.19 One end of a copper rod 30 cm long is held at 200◦ C, and the other end is held at 93◦ C. The heat transfer coefficient in between is 17 W/m2 K (including both convection and radiation). If T∞ = 38◦ C and the diameter of the rod is 1.25 cm, what is the net heat removed by the air around the rod? [19.13 W.] 4.20 How much error will the insulated-tip assumption give rise to in the calculation of the heat flow into the fin in Example 4.8? 4.21 A straight cylindrical fin 0.6 cm in diameter and 6 cm long protrudes from a magnesium block at 300◦ C. Air at 35◦ C is forced past the fin so that h is 130 W/m2 K. Calculate the heat removed by the fin, considering the temperature depression of the root. 4.22 Work Problem 4.19 considering the temperature depression in both roots. To do this, find mL for the two fins with insulated tips that would give the same temperature gradient at each wall. Base the correction on these values of mL. 4.23 A fin of triangular axial section (cf. Fig. 4.12) 0.1 m in length and 0.02 m wide at its base is used to extend the surface area of a 0.5% carbon steel wall. If the wall is at 40◦ C and heated gas flows past at 200◦ C (h = 230 W/m2 K), compute the heat removed by the fin per meter of breadth, b, of the fin. Neglect temperature distortion at the root. 4.24 Consider the concrete slab in Example 2.1. Suppose that the heat generation were to cease abruptly at time t = 0 and the slab were to start cooling back toward Tw . Predict T = Tw as a function of time, noting that the initial parabolic temperature profile can be nicely approximated as a sine function. (Without the sine approximation, this problem would require the series methods of Chapter 5.) 4.25 Steam condenses in a 2 cm I.D. thin-walled tube of 99% aluminum at 10 atm pressure. There are circular fins of constant thickness, 3.5 cm in diameter, every 0.5 cm on the outside. The Problems 187 fins are 0.8 mm thick and the heat transfer coefficient from them h = 6 W/m2 K (including both convection and radiation). What is the mass rate of condensation if the pipe is 1.5 m in length, the ambient temperature is 18◦ C, and h for condensa˙ tion is very large? [mcond = 0.802 kg/hr.] 4.26 How long must a copper fin, 0.4 cm in diameter, be if the temperature of its insulated tip is to exceed the surrounding air temperature by 20% of (T0 − T∞ )? Tair = 20◦ C and h = 28 W/m2 K (including both convection and radiation). 4.27 A 2 cm ice cube sits on a shelf of widely spaced aluminum rods, 3 mm in diameter, in a refrigerator at 10◦ C. How rapidly, in mm/min, do the rods melt their way through the ice cube if h at the surface of the rods is 10 W/m2 K (including both convection and radiation). Be sure that you understand the physical mechanism before you make the calculation. Check your result experimentally. hsf = 333, 300 J/kg. 4.28 The highest heat flux that can be achieved in nucleate boiling (called qmax —see the qualitative discussion in Section 9.1) depends upon ρg , the saturated vapor density; hfg , the latent heat vaporization; σ , the surface tension; a characteristic length, l; and the gravity force per unit volume, g(ρf − ρg ), where ρf is the saturated liquid density. Develop the dimensionless functional equation for qmax in terms of dimensionless length. 4.29 You want to rig a handle for a door in the wall of a furnace. The door is at 160◦ C. You consider bending a 40 cm length of 6.35 mm diam. 0.5% carbon steel rod into a U-shape and welding the ends to the door. Surrounding air at 24◦ C will cool the handle (h = 12 W/m2 K including both convection and radiation). What is the coolest temperature of the handle? How close to the door can you grasp the handle without getting burned if Tburn = 65◦ C? How might you improve the design? 4.30 A 14 cm long by 1 cm square brass rod is supplied with 25 W at its base. The other end is insulated. It is cooled by air at 20◦ C, with h = 68 W/m2 K. Develop a dimensionless expression for Θ as a function of εf and other known information. Calculate the base temperature. 188 Chapter 4: Analysis of heat conduction and some steady one-dimensional problems 4.31 A cylindrical fin has a constant imposed heat flux of q1 at one end and q2 at the other end, and it is cooled convectively along its length. Develop the dimensionless temperature distribution in the fin. Specialize this result for q2 = 0 and L → ∞, and compare it with eqn. (4.50). 4.32 A thin metal cylinder of radius ro serves as an electrical resistance heater. The temperature along an axial line in one side is kept at T1 . Another line, θ2 radians away, is kept at T2 . Develop dimensionless expressions for the temperature distributions in the two sections. 4.33 Heat transfer is augmented, in a particular heat exchanger, with a field of 0.007 m diameter fins protruding 0.02 m into a flow. The fins are arranged in a hexagonal array, with a minimum spacing of 1.8 cm. The fins are bronze, and hf around the fins is 168 W/m2 K. On the wall itself, hw is only 54 W/m2 K. Calculate heff for the wall with its fins. (heff = Qwall divided by Awall and [Twall − T∞ ].) 4.34 Evaluate d(tanh x)/dx . 4.35 An engineer seeks to study the effect of temperature on the curing of concrete by controlling the temperature of curing in the following way. A sample slab of thickness L is subjected to a heat flux, qw , on one side, and it is cooled to temperature T1 on the other. Derive a dimensionless expression for the steady temperature in the slab. Plot the expression and offer a criterion for neglecting the internal heat generation in the slab. 4.36 Develop the dimensionless temperature distribution in a spherical shell with the inside wall kept at one temperature and the outside wall at a second temperature. Reduce your solution rinside and in which to the limiting cases in which routside routside is very close to rinside . Discuss these limits. 4.37 Does the temperature distribution during steady heat transfer in an object with b.c.’s of only the first kind depend on k? Explain. 4.38 A long, 0.005 m diameter duralumin rod is wrapped with an electrical resistor over 3 cm of its length. The resistor imparts Problems 189 a surface flux of 40 kW/m2 . Evaluate the temperature of the rod in either side of the heated section if h = 150 W/m2 K around the unheated rod, and Tambient = 27◦ C. 4.39 The heat transfer coefficient between a cool surface and a saturated vapor, when the vapor condenses in a film on the surface, depends on the liquid density and specific heat, the temperature difference, the buoyant force per unit volume (g[ρf − ρg ]), the latent heat, the liquid conductivity and the kinematic viscosity, and the position (x ) on the cooler. Develop the dimensionless functional equation for h. 4.40 A duralumin pipe through a cold room has a 4 cm I.D. and a 5 cm O.D. It carries water that sometimes sits stationary. It is proposed to put electric heating rings around the pipe to protect it against freezing during cold periods of −7◦ C. The heat transfer coefficient outside the pipe is 9 W/m2 K (including both convection and radiation). Neglect the presence of the water in the conduction calculation, and determine how far apart the heaters would have to be if they brought the pipe temperature to 40◦ C locally. How much heat do they require? 4.41 The specific entropy of an ideal gas depends on its specific heat at constant pressure, its temperature and pressure, the ideal gas constant and reference values of the temperature and pressure. Obtain the dimensionless functional equation for the specific entropy and compare it with the known equation. 4.42 A large freezer’s door has a 2.5 cm thick layer of insulation (kin = 0.04 W/m·K) covered on the inside, outside, and edges with a continuous aluminum skin 3.2 mm thick (kAl = 165 W/m·K). The door closes against a nonconducting seal 1 cm wide. Heat gain through the door can result from conduction straight through the insulation and skins (normal to the plane of the door) and from conduction in the aluminum skin only, going from the skin outside, around the edge skin, and to the inside skin. The heat transfer coefficients to the inside, hi , and outside, ho , are each 12 W/m2 K, accounting for both convection and radiation. The temperature outside the freezer is 25◦ C, and the temperature inside is −15◦ C. a. If the door is 1 m wide, estimate the one-dimensional heat gain through the door, neglecting any conduction around 190 Chapter 4: Analysis of heat conduction and some steady one-dimensional problems the edges of the skin. Your answer will be in watts per meter of door height. b. Now estimate the heat gain by conduction around the edges of the door, assuming that the insulation is perfectly adiabatic so that all heat flows through the skin. This answer will also be per meter of door height. 4.43 A thermocouple epoxied onto a high conductivity surface is intended to measure the surface temperature. The thermocouple consists of two each bare, 0.51 mm diameter wires. One wire is made of Chromel (Ni-10% Cr with kcr = 17 W/m·K) and the other of constantan (Ni-45% Cu with kcn = 23 W/m·K). The ends of the wires are welded together to create a measuring junction having has dimensions of Dw by 2Dw . The wires extend perpendicularly away from the surface and do not touch one another. A layer of epoxy (kep = 0.5 W/m·K separates the thermocouple junction from the surface by 0.2 mm. Air at 20◦ C surrounds the wires. The heat transfer coefficient between each wire and the surroundings is h = 28 W/m2 K, including both convection and radiation. If the thermocouple reads Ttc = 40◦ C, estimate the actual temperature Ts of the surface and suggest a better arrangement of the wires. 4.44 The resistor leads in Example 4.10 were assumed to be “infinitely long” fins. What is the minimum length they each must have if they are to be modelled this way? What are the effectiveness, εf , and efficiency, ηf , of the wires? References [4.1] V. L. Streeter and E. B. Wylie. Fluid Mechanics. McGraw-Hill Book Company, New York, 7th edition, 1979. Chapter 4. [4.2] E. Buckingham. Phy. Rev., 4:345, 1914. [4.3] E. Buckingham. Model experiments and the forms of empirical equations. Trans. ASME, 37:263–296, 1915. [4.4] Lord Rayleigh, John Wm. Strutt. The principle of similitude. Nature, 95:66–68, 1915. References [4.5] J. O. Farlow, C. V. Thompson, and D. E. Rosner. Plates of the dinosaur stegosaurus: Forced convection heat loss fins? Science, 192(4244): 1123–1125 and cover, 1976. [4.6] D. K. Hennecke and E. M. Sparrow. Local heat sink on a convectively cooled surface—application to temperature measurement error. Int. J. Heat Mass Transfer, 13:287–304, 1970. [4.7] P. J. Schneider. Conduction Heat Transfer. Addison-Wesley Publishing Co., Inc., Reading, Mass., 1955. [4.8] A. D. Kraus, A. Aziz, and J.R. Welty. Extended Surface Heat Transfer. John Wiley & Sons, Inc., New York, 2001. 191 5. Transient and multidimensional heat conduction When I was a lad, winter was really cold. It would get so cold that if you went outside with a cup of hot coffee it would freeze. I mean it would freeze fast. That cup of hot coffee would freeze so fast that it would still be hot after it froze. Now that’s cold! Old North-woods tall-tale 5.1 Introduction James Watt, of course, did not invent the steam engine. What he did do was to eliminate a destructive transient heating and cooling process that wasted a great amount of energy. By 1763, the great puffing engines of Savery and Newcomen had been used for over half a century to pump the water out of Cornish mines and to do other tasks. In that year the young instrument maker, Watt, was called upon to renovate the Newcomen engine model at the University of Glasgow. The Glasgow engine was then being used as a demonstration in the course on natural philosophy. Watt did much more than just renovate the machine—he first recognized, and eventually eliminated, its major shortcoming. The cylinder of Newcomen’s engine was cold when steam entered it and nudged the piston outward. A great deal of steam was wastefully condensed on the cylinder walls until they were warm enough to accommodate it. When the cylinder was filled, the steam valve was closed and jets of water were activated inside the cylinder to cool it again and condense the steam. This created a powerful vacuum, which sucked the piston back in on its working stroke. First, Watt tried to eliminate the wasteful initial condensation of steam by insulating the cylinder. But that simply reduced the vacuum and cut the power of the working stroke. 193 194 Transient and multidimensional heat conduction §5.2 Then he realized that, if he led the steam outside to a separate condenser, the cylinder could stay hot while the vacuum was created. The separate condenser was the main issue in Watt’s first patent (1769), and it immediately doubled the thermal efficiency of steam engines from a maximum of 1.1% to 2.2%. By the time Watt died in 1819, his invention had led to efficiencies of 5.7%, and his engine had altered the face of the world by powering the Industrial Revolution. And from 1769 until today, the steam power cycles that engineers study in their thermodynamics courses are accurately represented as steady flow—rather than transient—processes. The repeated transient heating and cooling that occurred in Newcomen’s engine was the kind of process that today’s design engineer might still carelessly ignore, but the lesson that we learn from history is that transient heat transfer can be of overwhelming importance. Today, for example, designers of food storage enclosures know that such systems need relatively little energy to keep food cold at steady conditions. The real cost of operating them results from the consumption of energy needed to bring the food down to a low temperature and the losses resulting from people entering and leaving the system with food. The transient heat transfer processes are a dominant concern in the design of food storage units. We therefore turn our attention, first, to an analysis of unsteady heat transfer, beginning with a more detailed consideration of the lumpedcapacity system that we looked at in Section 1.3. 5.2 Lumped-capacity solutions We begin by looking briefly at the dimensional analysis of transient conduction in general and of lumped-capacity systems in particular. Dimensional analysis of transient heat conduction We first consider a fairly representative problem of one-dimensional transient heat conduction: ⎧ ⎪ i.c.: T (t = 0) = Ti ⎪ ⎪ ⎪ ⎨ 2T 1 ∂T ∂ b.c.: T (t > 0, x = 0) = T1 = with ⎪ ∂x 2 α ∂t ⎪ ⎪ ⎪ b.c.: − k ∂T ⎩ = h (T − T1 )x =L ∂x x =L Lumped-capacity solutions §5.2 195 The solution of this problem must take the form of the following dimensional functional equation: T − T1 = fn (Ti − T1 ), x, L, t, α, h, k There are eight variables in four dimensions (K, s, m, W), so we look for 8 − 4 = 4 pi-groups. We anticipate, from Section 4.3, that they will include Θ≡ (T − T1 ) , (Ti − T1 ) ξ≡ x , L and Bi ≡ hL , k and we write Θ = fn (ξ, Bi, Π4 ) (5.1) One possible candidate for Π4 , which is independent of the other three, is Π4 ≡ Fo = αt/L2 (5.2) where Fo is the Fourier number. Another candidate that we use later is x Π4 ≡ ζ = √ αt ξ this is exactly √ Fo (5.3) If the problem involved only b.c.’s of the first kind, the heat transfer coefficient, h—and hence the Biot number—would go out of the problem. Then the dimensionless function eqn. (5.1) is Θ = fn (ξ, Fo) (5.4) By the same token, if the b.c.’s had introduced different values of h at x = 0 and x = L, two Biot numbers would appear in the solution. The lumped-capacity problem is particularly interesting from the standpoint of dimensional analysis [see eqns. (1.19)–(1.22)]. In this case, neither k nor x enters the problem because we do not retain any features of the internal conduction problem. Therefore, we have ρc rather than α. Furthermore, we do not have to separate ρ and c because they only appear as a product. Finally, we use the volume-to-external-area ratio, V /A, as a characteristic length. Thus, for the transient lumped-capacity problem, the dimensional equation is T − T∞ = fn (Ti − T∞ ) , ρc, V /A, h, t (5.5) 196 Transient and multidimensional heat conduction §5.2 Figure 5.1 A simple resistance-capacitance circuit. With six variables in the dimensions J, K, m, and s, only two pi-groups will appear in the dimensionless function equation. Θ = fn hAt ρcV = fn t T (5.6) This is exactly the form of the simple lumped-capacity solution, eqn. (1.22). Notice, too, that the group t/T can be viewed as h(V /A) hk(V /A)t αt t = = · = Bi Fo 2k T ρc(V /A) k (V /A)2 (5.7) Electrical and mechanical analogies to the lumped-thermal-capacity problem The term capacitance is adapted from electrical circuit theory to the heat transfer problem. Therefore, we sketch a simple resistance-capacitance circuit in Fig. 5.1. The capacitor is initially charged to a voltage, Eo . When the switch is suddenly opened, the capacitor discharges through the resistor and the voltage drops according to the relation E dE + =0 dt RC (5.8) The solution of eqn. (5.8) with the i.c. E(t = 0) = Eo is E = Eo e−t/RC (5.9) and the current can be computed from Ohm’s law, once E(t) is known. I= E R (5.10) Normally, in a heat conduction problem the thermal capacitance, ρcV , is distributed in space. But when the Biot number is small, T (t) Lumped-capacity solutions §5.2 197 is uniform in the body and we can lump the capacitance into a single circuit element. The thermal resistance is 1/hA, and the temperature difference (T − T∞ ) is analogous to E(t). Thus, the thermal response, analogous to eqn. (5.9), is [see eqn. (1.22)] T − T∞ = (Ti − T∞ ) exp − hAt ρcV Notice that the electrical time constant, analogous to ρcV /hA, is RC . Now consider a slightly more complex system. Figure 5.2 shows a spring-mass-damper system. The well-known response equation (actually, a force balance) for this system is m d2 x dx +c + k x = F (t) 2 dt dt (5.11) where k is analogous to 1/C or to hA the damping coefficient is analogous to R or to ρcV What is the mass analogous to? A term analogous to mass would arise from electrical inductance, but we Figure 5.2 A spring-mass-damper system with a forcing function. did not include it in the electrical circuit. Mass has the effect of carrying the system beyond its final equilibrium point. Thus, in an underdamped mechanical system, we might obtain the sort of response shown in Fig. 5.3 if we specified the velocity at x = 0 and provided no forcing function. Electrical inductance provides a similar effect. But the Second Law of Thermodynamics does not permit temperatures to overshoot their equilibrium values spontaneously. There are no physical elements analogous to mass or inductance in thermal systems. 198 Transient and multidimensional heat conduction §5.2 Figure 5.3 Response of an unforced spring-mass-damper system with an initial velocity. Next, consider another mechanical element that does have a thermal analogy—namely, the forcing function, F . We consider a (massless) spring-damper system with a forcing function F that probably is timedependent, and we ask: “What might a thermal forcing function look like?” Lumped-capacity solution with a variable ambient temperature To answer the preceding question, let us suddenly immerse an object at 1, into a cool bath whose temperature is a temperature T = Ti , with Bi rising as T∞ (t) = Ti + bt , where Ti and b are constants. Then eqn. (1.20) becomes T − T∞ T − Ti − bt d(T − Ti ) =− =− dt T T where we have arbitrarily subtracted Ti under the differential. Then bt d(T − Ti ) T − Ti + = dt T T (5.12) To solve eqn. (5.12) we must first recall that the general solution of a linear ordinary differential equation with constant coefficients is equal to the sum of any particular integral of the complete equation and the general solution of the homogeneous equation. We know the latter; it is T − Ti = (constant) exp(−t/T ). A particular integral of the complete equation can often be formed by guessing solutions and trying them in the complete equation. Here we discover that T − Ti = bt − bT Lumped-capacity solutions §5.2 199 satisfies eqn. (5.12). Thus, the general solution of eqn. (5.12) is T − Ti = C1 e−t/T + b(t − T ) (5.13) The solution for arbitrary variations of T∞ (t) is given in Problem 5.52 (see also Problems 5.3, 5.53, and 5.54). Example 5.1 The flow rates of hot and cold water are regulated into a mixing chamber. We measure the temperature of the water as it leaves, using a thermometer with a time constant, T . On a particular day, the system started with cold water at T = Ti in the mixing chamber. Then hot water is added in such a way that the outflow temperature rises linearly, as shown in Fig. 5.4, with Texit flow = Ti + bt . How will the thermometer report the temperature variation? Solution. The initial condition in eqn. (5.13), which describes this process, is T − Ti = 0 at t = 0. Substituting eqn. (5.13) in the i.c., we get 0 = C1 − bT so C1 = bT and the response equation is T − (Ti + bt) = bT e−t/T − 1 (5.14) This result is graphically shown in Fig. 5.4. Notice that the thermometer reading reflects a transient portion, bT e−t/T , which decays for a few time constants and then can be neglected, and a steady portion, Ti + b(t − T ), which persists thereafter. When the steady response is established, the thermometer follows the bath with a temperature lag of bT . This constant error is reduced when either T or the rate of temperature increase, b, is reduced. Second-order lumped-capacity systems Now we look at situations in which two lumped-thermal-capacity systems are connected in series. Such an arrangement is shown in Fig. 5.5. Heat is transferred through two slabs with an interfacial resistance, h−1 between c them. We shall require that hc L1 /k1 , hc L2 /k2 , and hL2 /k2 are all much 200 Transient and multidimensional heat conduction §5.2 Figure 5.4 Response of a thermometer to a linearly increasing ambient temperature. less than unity so that it will be legitimate to lump the thermal capacitance of each slab. The differential equations dictating the temperature response of each slab are then dT1 = hc A(T1 − T2 ) dt dT2 = hA(T2 − T∞ ) − hc A(T1 − T2 ) slab 2 : −(ρcV )2 dt slab 1 : −(ρcV )1 (5.15) (5.16) and the initial conditions on the temperatures T1 and T2 are T1 (t = 0) = T2 (t = 0) = Ti (5.17) We next identify two time constants for this problem:1 T1 ≡ (ρcV )1 hc A and T2 ≡ (ρcV )2 hA Then eqn. (5.15) becomes T2 = T1 1 dT1 + T1 dt (5.18) Notice that we could also have used (ρcV )2 /hc A for T2 since both hc and h act on slab 2. The choice is arbitrary. Lumped-capacity solutions §5.2 201 Figure 5.5 Two slabs conducting in series through an interfacial resistance. which we substitute in eqn. (5.16) to get T1 hc dT1 dT1 d2 T1 dT1 + T1 − T∞ + = T1 T2 − T2 T1 2 dt dt dt dt h or d2 T1 + dt 2 1 1 hc + + T1 T2 hT2 ≡b T1 − T∞ dT1 =0 + dt T1 T2 (5.19a) c (T1 − T∞ ) if we call T1 − T∞ ≡ θ , then eqn. (5.19a) can be written as d2 θ dθ + cθ = 0 +b 2 dt dt (5.19b) Thus we have reduced the pair of first-order equations, eqn. (5.15) and eqn. (5.16), to a single second-order equation, eqn. (5.19b). The general solution of eqn. (5.19b) is obtained by guessing a solution of the form θ = C1 eDt . Substitution of this guess into eqn. (5.19b) gives D 2 + bD + c = 0 (5.20) from which we find that D = −(b/2) ± (b/2)2 − c . This gives us two values of D , from which we can get two exponential solutions. By adding 202 Transient and multidimensional heat conduction §5.2 them together, we form a general solution: ⎡ ⎤ ⎡ b b b2 − c ⎦ t + C2 exp ⎣− − θ = C1 exp ⎣− + 2 2 2 b 2 2 ⎤ − c ⎦t (5.21) To solve for the two constants we first substitute eqn. (5.21) in the first of i.c.’s (5.17) and get Ti − T∞ = θi = C1 + C2 (5.22) The second i.c. can be put into terms of T1 with the help of eqn. (5.15): − dT1 dt t =0 = hc A (T1 − T2 )t =0 = 0 (ρcV )1 We substitute eqn. (5.21) in this and obtain ⎤ ⎡ ⎡ b b2 b b − c ⎦ C1 + ⎣− − 0 = ⎣− + 2 2 2 2 2 ⎤ −c⎦ C2 = θi − C 1 so C1 = −θi −b/2 − (b/2)2 − c 2 (b/2)2 − c and C2 = θi −b/2 + (b/2)2 − c 2 (b/2)2 − c So we obtain at last: ⎡ T1 − T∞ −c θ b/2 + b ≡ = exp ⎣− + Ti − T∞ θi 2 2 (b/2)2 − c ⎡ −b/2 + (b/2)2 − c b + exp ⎣− − 2−c 2 2 (b/2) (b/2)2 b 2 2 b 2 2 ⎤ − c⎦ t ⎤ (5.23) − c⎦ t This is a pretty complicated result—all the more complicated when we remember that b involves three algebraic terms [recall eqn. (5.19a)]. Yet there is nothing very sophisticated about it; it is easy to understand. A system involving three capacitances in series would similarly yield a third-order equation of correspondingly higher complexity, and so forth. §5.3 Transient conduction in a one-dimensional slab 203 Figure 5.6 The transient cooling of a slab; ξ = (x/L) + 1. 5.3 Transient conduction in a one-dimensional slab We next extend consideration to heat flow in bodies whose internal resistance is significant—to situations in which the lumped capacitance assumption is no longer appropriate. When the temperature within, say, a one-dimensional body varies with position as well as time, we must solve the heat diffusion equation for T (x, t). We shall do this somewhat complicated task for the simplest case and then look at the results of such calculations in other situations. A simple slab, shown in Fig. 5.6, is initially at a temperature Ti . The temperature of the surface of the slab is suddenly changed to Ti , and we wish to calculate the interior temperature profile as a function of time. The heat conduction equation is 1 ∂T ∂2T = 2 ∂x α ∂t (5.24) with the following b.c.’s and i.c.: T (−L, t > 0) = T (L, t > 0) = T1 and T (x, t = 0) = Ti (5.25) In fully dimensionless form, eqn. (5.24) and eqn. (5.25) are ∂Θ ∂2Θ = 2 ∂ξ ∂ Fo (5.26) 204 Transient and multidimensional heat conduction §5.3 and Θ(0, Fo) = Θ(2, Fo) = 0 and Θ(ξ, 0) = 1 (5.27) where we have nondimensionalized the problem in accordance with eqn. (5.4), using Θ ≡ (T − T1 )/(Ti − T1 ) and Fo ≡ αt/L2 ; but for convenience in solving the equation, we have set ξ equal to (x/L) + 1 instead of x/L. The general solution of eqn. (5.26) may be found using the separation of variables technique described in Sect. 4.2, leading to the dimensionless form of eqn. (4.11): ˆ2 Fo Θ = e−λ ˆ ˆ G sin(λξ) + E cos(λξ) (5.28) ˆ Direct nondimensionalization of eqn. (4.11) would show that λ ≡ λL, −1 . The solution therefore appears to have since λ had units of (length) ˆ introduced a fourth dimensionless group, λ. This needs explanation. The number λ, which was introduced in the separation-of-variables process, ˆ is called an eigenvalue.2 In the present problem, λ = λL will turn out to be a number—or rather a sequence of numbers—that is independent of system parameters. Substituting the general solution, eqn. (5.28), in the first b.c. gives ˆ2 Fo 0 = e−λ (0 + E) so E=0 and substituting it in the second yields ˆ2 Fo 0 = e−λ ˆ G sin 2λ so either G=0 or ˆ ˆ 2λ = 2λn = nπ , n = 0, 1, 2, . . . In the second case, we are presented with two choices. The first, G = 0, would give Θ ≡ 0 in all situations, so that the initial condition could never be accommodated. (This is what mathematicians call a trivial ˆ solution.) The second choice, λn = nπ /2, actually yields a string of solutions, each of the form Θ = Gn e−n 2 2 π 2 Fo/4 sin nπ ξ 2 (5.29) The word eigenvalue is a curious hybrid of the German term eigenwert and its English translation, characteristic value. Transient conduction in a one-dimensional slab §5.3 where Gn is the constant appropriate to the nth one of these solutions. We still face the problem that none of eqns. (5.29) will fit the initial condition, Θ(ξ, 0) = 1. To get around this, we remember that the sum of any number of solutions of a linear differential equation is also a solution. Then we write ∞ Gn e−n Θ= 2 π 2 Fo/4 sin n n=1 π ξ 2 (5.30) where we drop n = 0 since it gives zero contribution to the series. And we arrive, at last, at the problem of choosing the Gn ’s so that eqn. (5.30) will fit the initial condition. ∞ Θ (ξ, 0) = Gn sin n n=1 π ξ =1 2 (5.31) The problem of picking the values of Gn that will make this equation true is called “making a Fourier series expansion” of the function f (ξ) = 1. We shall not pursue strategies for making Fourier series expansions in any general way. Instead, we merely show how to accomplish the task for the particular problem at hand. We begin with a mathematical trick. We multiply eqn. (5.31) by sin(mπ /2), where m may or may not equal n, and we integrate the result between ξ = 0 and 2. 2 sin 0 mπ ξ 2 ∞ dξ = 2 Gn n=1 sin 0 nπ mπ ξ sin ξ 2 2 dξ (5.32) (The interchange of summation and integration turns out to be legitimate, although we have not proved, here, that it is.3 ) With the help of a table of integrals, we find that 2 sin 0 nπ mπ ξ sin ξ 2 2 0 for n ≠ m 1 dξ = for n = m Thus, when we complete the integration of eqn. (5.32), we get − 3 2 mπ cos ξ mπ 2 ∞ 2 = 0 Gn × n=1 0 for n ≠ m 1 for n = m What is normally required is that the series in eqn. (5.31) be uniformly convergent. 205 206 Transient and multidimensional heat conduction §5.3 This reduces to − 2 (−1)n − 1 = Gn mπ so Gn = 4 nπ where n is an odd number Substituting this result into eqn. (5.30), we finally obtain the solution to the problem: 4 Θ (ξ, Fo) = π ∞ 1 −(nπ /2)2 Fo nπ e ξ sin n 2 n=odd (5.33) Equation (5.33) admits a very nice simplification for large time (or at large Fo). Suppose that we wish to evaluate Θ at the outer center of the slab—at x = 0 or ξ = 1. Then 4 × Θ (0, Fo) = π ⎧ ⎪ ⎨ 1 π2 3π exp − Fo − exp − ⎪ 2 3 2 ⎩ = 0.085 at Fo = 1 = 0.781 at Fo = 0.1 = 0.976 at Fo = 0.01 2 Fo + 10−10 at Fo = 1 = 0.036 at Fo = 0.1 = 0.267 at Fo = 0.01 1 5π exp − 5 2 2 ⎫ ⎪ ⎬ Fo + · · · ⎪ ⎭ 10−27 at Fo = 1 = 0.0004 at Fo = 0.1 = 0.108 at Fo = 0.01 Thus for values of Fo somewhat greater than 0.1, only the first term in the series need be used in the solution (except at points very close to the boundaries). We discuss these one-term solutions in Sect. 5.5. Before we move to this matter, let us see what happens to the preceding problem if the slab is subjected to b.c.’s of the third kind. Suppose that the walls of the slab had been cooled by symmetrical convection such that the b.c.’s were h(T∞ − T )x =−L = −k ∂T ∂x x =−L and h(T − T∞ )x =L = −k ∂T ∂x x =L or in dimensionless form, using Θ ≡ (T − T∞ )/(Ti − T∞ ) and ξ = (x/L) + 1, −Θ ξ =0 =− 1 ∂Θ Bi ∂ξ and ξ =0 ∂Θ ∂ξ =0 ξ =1 Transient conduction in a one-dimensional slab §5.3 207 Table 5.1 Terms of series solutions for slabs, cylinders, and spheres. J0 and J1 are Bessel functions of the first kind. An ˆ Equation for λn fn ˆ 2 sin λn ˆ ˆ ˆ λn + sin λn cos λn Slab Cylinder ˆx cos λn L ˆ 2 J1(λn ) ˆ ˆ J 2(λn ) + J 2(λn ) ˆr J0 λn ro ˆ λn 2 Sphere 0 1 ˆ ˆ ˆ sin λn − λn cos λn ˆ ˆ ˆn − sin λn cos λn λ ro ˆn r λ sin ˆ cot λn = ˆ ˆ ˆ λn J1(λn ) = Biro J0(λn ) ˆ λn r ro ˆ ˆ λn cot λn = 1 − Biro The solution is somewhat harder to find than eqn. (5.33) was, but the result is4 ∞ ˆ exp −λ2 Fo n Θ= n=1 ˆ ˆ 2 sin λn cos[λn (ξ − 1)] ˆ ˆ ˆ λn + sin λn cos λn (5.34) ˆ where the values of λn are given as a function of n and Bi = hL/k by the transcendental equation ˆ cot λn = ˆ λn Bi (5.35) ˆˆ ˆ The successive positive roots of this equation, which are λn = λ1 , λ2 , ˆ3 , . . . , depend upon Bi. Thus, Θ = fn(ξ, Fo, Bi), as we would expect. This λ result, although more complicated than the result for b.c.’s of the first kind, still reduces to a single term for Fo 0.2. Similar series solutions can be constructed for cylinders and spheres that are convectively cooled at their outer surface, r = ro . The solutions for slab, cylinders, and spheres all have the form ∞ Θ= T − T∞ ˆ = An exp −λ2 Fo fn n Ti − T∞ n=1 (5.36) where the coefficients An , the functions fn , and the equations for the ˆ dimensionless eigenvalues λn are given in Table 5.1. 4 ˆ λn BiL See, for example, [5.1, §2.3.4] or [5.2, §3.4.3] for details of this calculation. 208 Transient and multidimensional heat conduction 5.4 §5.4 Temperature-response charts Figure 5.7 is a graphical presentation of eqn. (5.34) for 0 Fo 1.5 and for six x -planes in the slab. (Remember that the x -coordinate goes from zero in the center to L on the boundary, while ξ goes from 0 up to 2 in the preceding solution.) Notice that, with the exception of points for which 1/Bi < 0.25 on the outside boundary, the curves are all straight lines when Fo 0.2. Since the coordinates are semilogarithmic, this portion of the graph corresponds to the lead term—the only term that retains any importance— in eqn. (5.34). When we take the logarithm of the one-term version of eqn. (5.34), the result is ln Θ ln ˆ ˆ 2 sin λ1 cos[λ1 (ξ − 1)] ˆ ˆ ˆ λ1 + sin λ1 cos λ1 Θ-intercept at Fo = 0 of the straight portion of the curve ˆ λ2 Fo 1 − slope of the straight portion of the curve If Fo is greater than 1.5, the following options are then available to us for solving the problem: • Extrapolate the given curves using a straightedge. • Evaluate Θ using the first term of eqn. (5.34), as discussed in Sect. 5.5. • If Bi is small, use a lumped-capacity result. Figure 5.8 and Fig. 5.9 are similar graphs for cylinders and spheres. Everything that we have said in general about Fig. 5.7 is also true for these graphs. They were simply calculated from different solutions, and the numerical values on them are somewhat different. These charts are from [5.3, Chap. 5], although such charts are often called Heisler charts, after a collection of related charts subsequently published by Heisler [5.4]. Another useful kind of chart derivable from eqn. (5.34) is one that gives heat removal from a body up to a time of interest: ⌠t t ∂T ⎮ Q dt = −⌡ kA dt ∂x surface 0 0 ⌠ Fo Ti − T∞ ∂ Θ ⎮ = −⌡ kA L ∂ξ 0 surface L2 α dFo 209 Figure 5.7 The transient temperature distribution in a slab at six positions: x/L = 0 is the center, x/L = 1 is one outside boundary. 210 Figure 5.8 The transient temperature distribution in a long cylinder of radius ro at six positions: r /ro = 0 is the centerline; r /ro = 1 is the outside boundary. 211 Figure 5.9 The transient temperature distribution in a sphere of radius ro at six positions: r /ro = 0 is the center; r /ro = 1 is the outside boundary. 212 Transient and multidimensional heat conduction §5.4 Dividing this by the total energy of the body above T∞ , we get a quantity, Φ, which approaches unity as t → ∞ and the energy is all transferred to the surroundings: t ⌠ Fo ∂Θ ⎮ = −⌡ Φ≡ ρcV (Ti − T∞ ) ∂ξ 0 Q dt 0 dFo (5.37) surface where the volume, V = AL. Substituting the appropriate temperature distribution [e.g., eqn. (5.34) for a slab] in eqn. (5.37), we obtain Φ(Fo, Bi) in the form of an infinite series ∞ Φ (Fo, Bi) = 1 − ˆ Dn exp −λ2 Fo n (5.38) n=1 ˆ The coefficients Dn are different functions of λn — and thus of Bi — for ˆˆ slabs, cylinders, and spheres (e.g., for a slab Dn = An sin λn λn ). These functions can be used to plot Φ(Fo, Bi) once and for all. Such curves are given in Fig. 5.10. The quantity Φ has a close relationship to the mean temperature of a body at any time, T (t). Specifically, the energy lost as heat by time t determines the difference between the initial temperature and the mean temperature at time t t 0 Q dt = U (0) − U (t) = ρcV Ti − T (t) . (5.39) Thus, if we define Θ as follows, we find the relationship of T (t) to Φ t Q(t) dt T (t) − T∞ 0 Θ≡ = 1 − Φ. =1− Ti − T∞ ρcV (Ti − T∞ ) (5.40) Example 5.2 A dozen approximately spherical apples, 10 cm in diameter are taken from a 30◦ C environment and laid out on a rack in a refrigerator at 5◦ C. They have approximately the same physical properties as water, and h is approximately 6 W/m2 K as the result of natural convection. What will be the temperature of the centers of the apples after 1 hr? How long will it take to bring the centers to 10◦ C? How much heat will the refrigerator have to carry away to get the centers to 10◦ C? Figure 5.10 The heat removal from suddenly-cooled bodies as a function of h and time. 213 214 Transient and multidimensional heat conduction §5.4 Solution. After 1 hr, or 3600 s: Fo = αt 2= ro k ρc = 20◦ C 3600 s (0.05 m)2 (0.603 J/m·s·K)(3600 s) (997.6 kg/m3 )(4180 J/kg·K)(0.0025 m2 ) = 0.208 Furthermore, Bi−1 = (hro /k)−1 = [6(0.05)/0.603]−1 = 2.01. Therefore, we read from Fig. 5.9 in the upper left-hand corner: Θ = 0.85 After 1 hr: Tcenter = 0.85(30 − 5)◦ C + 5◦ C = 26.3◦ C To find the time required to bring the center to 10◦ C, we first calculate Θ= 10 − 5 = 0.2 30 − 5 and Bi−1 is still 2.01. Then from Fig. 5.9 we read Fo = 1.29 = αt 2 ro so t= 1.29(997.6)(4180)(0.0025) = 22, 300 s = 6 hr 12 min 0.603 Finally, we look up Φ at Bi = 1/2.01 and Fo = 1.29 in Fig. 5.10, for spheres: t Q dt Φ = 0.80 = 0 ρc 4 3 3 π r0 (Ti − T∞ ) so t 0 Q dt = 997.6(4180) 4 π (0.05)3 (25)(0.80) = 43, 668 J/apple 3 Therefore, for the 12 apples, total energy removal = 12(43.67) = 524 kJ §5.4 Temperature-response charts 215 The temperature-response charts in Fig. 5.7 through Fig. 5.10 are without doubt among the most useful available since they can be adapted to a host of physical situations. Nevertheless, hundreds of such charts have been formed for other situations, a number of which have been cataloged by Schneider [5.5]. Analytical solutions are available for hundreds more problems, and any reader who is faced with a complex heat conduction calculation should consult the literature before trying to solve it. An excellent place to begin is Carslaw and Jaeger’s comprehensive treatise on heat conduction [5.6]. Example 5.3 A 1 mm diameter Nichrome (20% Ni, 80% Cr) wire is simultaneously being used as an electric resistance heater and as a resistance thermometer in a liquid flow. The laboratory workers who operate it are attempting to measure the boiling heat transfer coefficient, h, by supplying an alternating current and measuring the difference between the average temperature of the heater, Tav , and the liquid temperature, T∞ . They get h = 30, 000 W/m2 K at a wire temperature of 100◦ C and are delighted with such a high value. Then a colleague suggests that h is so high because the surface temperature is rapidly oscillating as a result of the alternating current. Is this hypothesis correct? Solution. Heat is being generated in proportion to the product of voltage and current, or as sin2 ωt , where ω is the frequency of the current in rad/s. If the boiling action removes heat rapidly enough in comparison with the heat capacity of the wire, the surface temperature may well vary significantly. This transient conduction problem was first solved by Jeglic in 1962 [5.7]. It was redone in a different form two years later by Switzer and Lienhard (see, e.g. [5.8]), who gave response curves in the form Tmax − Tav = fn (Bi, ψ) Tav − T∞ (5.41) where the left-hand side is the dimensionless range of the temperature oscillation, and ψ = ωδ2 /α, where δ is a characteristic length [see Problem 5.56]. Because this problem is common and the solution is not widely available, we include the curves for flat plates and cylinders in Fig. 5.11 and Fig. 5.12 respectively. 216 Figure 5.11 Temperature deviation at the surface of a flat plate heated with alternating current. 217 Figure 5.12 Temperature deviation at the surface of a cylinder heated with alternating current. 218 Transient and multidimensional heat conduction §5.5 In the present case: 30, 000(0.0005) h radius = = 1.09 k 13.8 [2π (60)](0.0005)2 ωr 2 = = 27.5 α 0.00000343 Bi = and from the chart for cylinders, Fig. 5.12, we find that Tmax − Tav Tav − T∞ 0.04 A temperature fluctuation of only 4% is probably not serious. It therefore appears that the experiment was valid. 5.5 One-term solutions As we have noted previously, when the Fourier number is greater than 0.2 or so, the series solutions from eqn. (5.36) may be approximated using only their first term: ˆ Θ ≈ A1 · f1 · exp −λ2 Fo . 1 (5.42) Likewise, the fractional heat loss, Φ, or the mean temperature Θ from eqn. (5.40), can be approximated using just the first term of eqn. (5.38): ˆ Θ = 1 − Φ ≈ D1 exp −λ2 Fo . 1 (5.43) ˆ Table 5.2 lists the values of λ1 , A1 , and D1 for slabs, cylinders, and spheres as a function of the Biot number. The one-term solution’s error in Θ is less than 0.1% for a sphere with Fo ≥ 0.28 and for a slab with Fo ≥ 0.43. These errors are largest for Biot numbers near unity. If high accuracy is not required, these one-term approximations may generally be used whenever Fo ≥ 0.2 Table 5.2 One-term coefficients for convective cooling [5.1]. Bi Plate Cylinder D1 ˆ λ1 A1 1.0017 1.0033 1.0082 1.0000 1.0000 0.9999 0.14124 0.19950 0.31426 0.31105 0.37788 0.43284 0.52179 0.59324 0.65327 0.70507 0.75056 0.79103 0.82740 1.0161 1.0237 1.0311 1.0450 1.0580 1.0701 1.0814 1.0918 1.1016 1.1107 0.9998 0.9995 0.9992 0.9983 0.9971 0.9956 0.9940 0.9922 0.9903 0.9882 1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.80 2.00 2.20 2.40 0.86033 0.89035 0.91785 0.94316 0.96655 0.98824 1.00842 1.04486 1.07687 1.10524 1.13056 1.1191 1.1270 1.1344 1.1412 1.1477 1.1537 1.1593 1.1695 1.1785 1.1864 1.1934 3.00 4.00 5.00 6.00 8.00 10.00 20.00 50.00 100.00 ∞ 1.19246 1.26459 1.31384 1.34955 1.39782 1.42887 1.49613 1.54001 1.55525 1.57080 1.2102 1.2287 1.2402 1.2479 1.2570 1.2620 1.2699 1.2727 1.2731 1.2732 ˆ λ1 A1 0.01 0.02 0.05 0.09983 0.14095 0.22176 0.10 0.15 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 Sphere D1 ˆ λ1 A1 D1 1.0025 1.0050 1.0124 1.0000 1.0000 0.9999 0.17303 0.24446 0.38537 1.0030 1.0060 1.0150 1.0000 1.0000 1.0000 0.44168 0.53761 0.61697 0.74646 0.85158 0.94077 1.01844 1.08725 1.14897 1.20484 1.0246 1.0365 1.0483 1.0712 1.0931 1.1143 1.1345 1.1539 1.1724 1.1902 0.9998 0.9995 0.9992 0.9983 0.9970 0.9954 0.9936 0.9916 0.9893 0.9869 0.54228 0.66086 0.75931 0.92079 1.05279 1.16556 1.26440 1.35252 1.43203 1.50442 1.0298 1.0445 1.0592 1.0880 1.1164 1.1441 1.1713 1.1978 1.2236 1.2488 0.9998 0.9996 0.9993 0.9985 0.9974 0.9960 0.9944 0.9925 0.9904 0.9880 0.9861 0.9839 0.9817 0.9794 0.9771 0.9748 0.9726 0.9680 0.9635 0.9592 0.9549 1.25578 1.30251 1.34558 1.38543 1.42246 1.45695 1.48917 1.54769 1.59945 1.64557 1.68691 1.2071 1.2232 1.2387 1.2533 1.2673 1.2807 1.2934 1.3170 1.3384 1.3578 1.3754 0.9843 0.9815 0.9787 0.9757 0.9727 0.9696 0.9665 0.9601 0.9537 0.9472 0.9408 1.57080 1.63199 1.68868 1.74140 1.79058 1.83660 1.87976 1.95857 2.02876 2.09166 2.14834 1.2732 1.2970 1.3201 1.3424 1.3640 1.3850 1.4052 1.4436 1.4793 1.5125 1.5433 0.9855 0.9828 0.9800 0.9770 0.9739 0.9707 0.9674 0.9605 0.9534 0.9462 0.9389 0.9431 0.9264 0.9130 0.9021 0.8858 0.8743 0.8464 0.8260 0.8185 0.8106 1.78866 1.90808 1.98981 2.04901 2.12864 2.17950 2.28805 2.35724 2.38090 2.40483 1.4191 1.4698 1.5029 1.5253 1.5526 1.5677 1.5919 1.6002 1.6015 1.6020 0.9224 0.8950 0.8721 0.8532 0.8244 0.8039 0.7542 0.7183 0.7052 0.6917 2.28893 2.45564 2.57043 2.65366 2.76536 2.83630 2.98572 3.07884 3.11019 3.14159 1.6227 1.7202 1.7870 1.8338 1.8920 1.9249 1.9781 1.9962 1.9990 2.0000 0.9171 0.8830 0.8533 0.8281 0.7889 0.7607 0.6922 0.6434 0.6259 0.6079 219 220 Transient and multidimensional heat conduction 5.6 §5.6 Transient heat conduction to a semi-infinite region Introduction Bronowksi’s classic television series, The Ascent of Man [5.9], included a brilliant reenactment of the ancient ceremonial procedure by which the Japanese forged Samurai swords (see Fig. 5.13). The metal is heated, folded, beaten, and formed, over and over, to create a blade of remarkable toughness and flexibility. When the blade is formed to its final configuration, a tapered sheath of clay is baked on the outside of it, so the cross section is as shown in Fig. 5.13. The red-hot blade with the clay sheath is then subjected to a rapid quenching, which cools the uninsulated cutting edge quickly and the back part of the blade very slowly. The result is a layer of case-hardening that is hardest at the edge and less hard at points farther from the edge. Figure 5.13 The ceremonial case-hardening of a Samurai sword. §5.6 Transient heat conduction to a semi-infinite region 221 Figure 5.14 The initial cooling of a thin sword blade. Prior to t = t4 , the blade might as well be infinitely thick insofar as cooling is concerned. The blade is then tough and ductile, so it will not break, but has a fine hard outer shell that can be honed to sharpness. We need only look a little way up the side of the clay sheath to find a cross section that was thick enough to prevent the blade from experiencing the sudden effects of the cooling quench. The success of the process actually relies on the failure of the cooling to penetrate the clay very deeply in a short time. Now we wish to ask: “How can we say whether or not the influence of a heating or cooling process is restricted to the surface of a body?” Or if we turn the question around: “Under what conditions can we view the depth of a body as infinite with respect to the thickness of the region that has felt the heat transfer process?” Consider next the cooling process within the blade in the absence of the clay retardant and when h is very large. Actually, our considerations will apply initially to any finite body whose boundary suddenly changes temperature. The temperature distribution, in this case, is sketched in Fig. 5.14 for four sequential times. Only the fourth curve—that for which t = t4 —is noticeably influenced by the opposite wall. Up to that time, the wall might as well have infinite depth. Since any body subjected to a sudden change of temperature is infinitely large in comparison with the initial region of temperature change, we must learn how to treat heat transfer in this period. Solution aided by dimensional analysis The calculation of the temperature distribution in a semi-infinite region poses a difficulty in that we can impose a definite b.c. at only one position— the exposed boundary. We shall be able to get around that difficulty in a nice way with the help of dimensional analysis. 222 Transient and multidimensional heat conduction §5.6 When the one boundary of a semi-infinite region, initially at T = Ti , is suddenly cooled (or heated) to a new temperature, T∞ , as in Fig. 5.14, the dimensional function equation is T − T∞ = fn [t, x, α, (Ti − T∞ )] where there is no characteristic length or time. Since there are five variables in ◦ C, s, and m, we should look for two dimensional groups. T − T∞ = fn Ti − T∞ x √ αt Θ (5.44) ζ The very important thing that we learn from this exercise in dimensional analysis is that position and time collapse into one independent variable. This means that the heat conduction equation and its b.c.s must transform from a partial differential equation into a simpler ordinary dif√ ferential equation in the single variable, ζ = x αt . Thus, we transform each side of ∂2T 1 ∂T = 2 ∂x α ∂t as follows, where we call Ti − T∞ ≡ ∆T : ∂T ∂ Θ ∂ζ x ∂Θ = (Ti − T∞ ) = ∆T = ∆T − √ ∂t ∂ζ ∂t ∂t 2t αt ∂Θ ; ∂ζ ∂ Θ ∂ζ ∆T ∂ Θ ∂T = ∆T =√ ; ∂x ∂ζ ∂x αt ∂ζ and ∂2T ∆T ∂ 2 Θ ∆T ∂ 2 Θ ∂ζ = =√ . ∂x 2 αt ∂ζ 2 αt ∂ζ 2 ∂x Substituting the first and last of these derivatives in the heat conduction equation, we get d2 Θ ζ dΘ =− 2 dζ 2 dζ (5.45) Notice that we changed from partial to total derivative notation, since Θ now depends solely on ζ . The i.c. for eqn. (5.45) is T (t = 0) = Ti or Θ (ζ → ∞) = 1 (5.46) Transient heat conduction to a semi-infinite region §5.6 and the one known b.c. is T (x = 0) = T∞ or Θ (ζ = 0) = 0 (5.47) If we call dΘ/dζ ≡ χ , then eqn. (5.45) becomes the first-order equation ζ dχ =− χ dζ 2 which can be integrated once to get χ≡ dΘ 2 = C1 e−ζ /4 dζ (5.48) and we integrate this a second time to get ζ Θ = C1 e−ζ 2 /4 0 dζ + Θ(0) (5.49) = 0 according to the b.c. The b.c. is now satisfied, and we need only substitute eqn. (5.49) in the i.c., eqn. (5.46), to solve for C1 : 1 = C1 ∞ e−ζ 2 /4 dζ 0 The definite integral is given by integral tables as √ π , so 1 C1 = √ π Thus the solution to the problem of conduction in a semi-infinite region, subject to a b.c. of the first kind is 1 Θ= √ π ζ 0 e−ζ 2 /4 2 dζ = √ π ζ /2 0 2 e−s ds ≡ erf (ζ/2) (5.50) The second integral in eqn. (5.50), obtained by a change of variables, is called the error function (erf). Its name arises from its relationship to certain statistical problems related to the Gaussian distribution, which describes random errors. In Table 5.3, we list values of the error function and the complementary error function, erfc(x) ≡ 1 − erf (x). Equation (5.50) is also plotted in Fig. 5.15. 223 Transient and multidimensional heat conduction 224 §5.6 Table 5.3 Error function and complementary error function. ζ2 erf (ζ/2) erfc(ζ/2) 0.00 0.05 0.10 0.15 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 0.00000 0.05637 0.11246 0.16800 0.22270 0.32863 0.42839 0.52050 0.60386 0.67780 0.74210 0.79691 0.84270 1.00000 0.94363 0.88754 0.83200 0.77730 0.67137 0.57161 0.47950 0.39614 0.32220 0.25790 0.20309 0.15730 ζ2 erf (ζ/2) erfc(ζ/2) 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.8214 1.90 2.00 2.50 3.00 0.88021 0.91031 0.93401 0.95229 0.96611 0.97635 0.98379 0.98909 0.99000 0.99279 0.99532 0.99959 0.99998 0.11980 0.08969 0.06599 0.04771 0.03389 0.02365 0.01621 0.01091 0.01000 0.00721 0.00468 0.00041 0.00002 In Fig. 5.15 we see the early-time curves shown in Fig. 5.14 have collapsed into a single curve. This was accomplished by the similarity trans√ formation, as we call it5 : ζ/2 = x/2 αt . From the figure or from Table 5.3, we see that Θ ≥ 0.99 when x ζ ≥ 1.8214 =√ 2 2 αt or x ≥ δ99 ≡ 3.64 αt (5.51) In other words, the local value of (T − T∞ ) is more than 99% of (Ti − T∞ ) for positions in the slab beyond farther from the surface than δ99 = √ 3.64 αt . Example 5.4 For what maximum time can a samurai sword be analyzed as a semiinfinite region after it is quenched, if it has no clay coating and hexternal ∞? Solution. First, we must guess the half-thickness of the sword (say, 3 mm) and its material (probably wrought iron with an average α 5 The transformation is based upon the “similarity” of spatial an temporal changes in this problem. Transient heat conduction to a semi-infinite region §5.6 225 Figure 5.15 Temperature distribution in a semi-infinite region. around 1.5 × 10−5 m2 /s). The sword will be semi-infinite until δ99 equals the half-thickness. Inverting eqn. (5.51), we find t δ2 (0.003 m)2 99 = 0.045 s = 2α 3.64 13.3(1.5)(10)−5 m2 /s Thus the quench would be felt at the centerline of the sword within only 1/20 s. The thermal diffusivity of clay is smaller than that of steel by a factor of about 30, so the quench time of the coated steel must continue for over 1 s before the temperature of the steel is affected at all, if the clay and the sword thicknesses are comparable. Equation (5.51) provides an interesting foretaste of the notion of a fluid boundary layer. In the context of Fig. 1.9 and Fig. 1.10, we observe that free stream flow around an object is disturbed in a thick layer near the object because the fluid adheres to it. It turns out that the thickness of this boundary layer of altered flow velocity increases in the downstream direction. For flow over a flat plate, this thickness is ap√ proximately 4.92 νt , where t is the time required for an element of the stream fluid to move from the leading edge of the plate to a point of interest. This is quite similar to eqn. (5.51), except that the thermal diffusivity, α, has been replaced by its counterpart, the kinematic viscosity, ν , and the constant is a bit larger. The velocity profile will resemble Fig. 5.15. If we repeated the problem with a boundary condition of the third kind, we would expect to get Θ = Θ(Bi, ζ), except that there is no length, L, upon which to build a Biot number. Therefore, we must replace L with √ αt , which has the dimension of length, so √ h αt ≡ Θ(ζ, β) (5.52) Θ = Θ ζ, k 226 Transient and multidimensional heat conduction §5.6 √ √ The term β ≡ h αt k is like the product: Bi Fo. The solution of this problem (see, e.g., [5.6], §2.7) can be conveniently written in terms of the complementary error function, erfc(x) ≡ 1 − erf (x): Θ = erf ζ + exp βζ + β2 2 erfc ζ +β 2 (5.53) This result is plotted in Fig. 5.16. Example 5.5 Most of us have passed our finger through an 800◦ C candle flame and know that if we limit exposure to about 1/4 s we will not be burned. Why not? Solution. The short exposure to the flame causes only a very superficial heating, so we consider the finger to be a semi-infinite region and go to eqn. (5.53) to calculate (Tburn − Tflame )/(Ti − Tflame ). It turns out that the burn threshold of human skin, Tburn , is about 65◦ C. (That is why 140◦ F or 60◦ C tap water is considered to be “scalding.”) Therefore, we shall calculate how long it will take for the surface temperature of the finger to rise from body temperature (37◦ C) to 65◦ C, when it is protected by an assumed h 100 W/m2 K. We shall assume that the thermal conductivity of human flesh equals that of its major component—water—and that the thermal diffusivity is equal to the known value for beef. Then Θ= βζ = hx =0 k 65 − 800 = 0.963 37 − 800 since x = 0 at the surface 2 1002 (0.135 × 10−6 )t h αt β= = = 0.0034(t s) k2 0.632 2 The situation is quite far into the corner of Fig. 5.16. We read β2 0.001, which corresponds with t 0.3 s. For greater accuracy, we must go to eqn. (5.53): 0.963 = erf 0 +e0.0034t erfc 0 + 0.0034 t =0 Figure 5.16 The cooling of a semi-infinite region by an environment at T∞ , through a heat transfer coefficient, h. 227 228 Transient and multidimensional heat conduction By trial and error, we get t §5.6 0.33 s. In fact, it can be shown that Θ(ζ = 0, β) 2β 1− √ π for β 1 √ which can be solved directly for β = (1 − 0.963) π /2 = 0.03279, leading to the same answer. Thus, it would require about 1/3 s to bring the skin to the burn point. Experiment 5.1 Immerse your hand in the subfreezing air in the freezer compartment of your refrigerator. Next immerse your finger in a mixture of ice cubes and water, but do not move it. Then, immerse your finger in a mixture of ice cubes and water , swirling it around as you do so. Describe your initial sensation in each case, and explain the differences in terms of Fig. 5.16. What variable has changed from one case to another? Heat transfer Heat will be removed from the exposed surface of a semi-infinite region, with a b.c. of either the first or the third kind, in accordance with Fourier’s law: q = −k ∂T ∂x x =0 = k(T∞ − Ti ) dΘ √ dζ αt ζ =0 Differentiating Θ as given by eqn. (5.50), we obtain, for the b.c. of the first kind, q= k(T∞ − Ti ) √ αt 1 2 √ e−ζ /4 π ζ =0 = k(T∞ − Ti ) √ π αt (5.54) Thus, q decreases with increasing time, as t −1/2 . When the temperature of the surface is first changed, the heat removal rate is enormous. Then it drops off rapidly. It often occurs that we suddenly apply a specified input heat flux, qw , at the boundary of a semi-infinite region. In such a case, we can Transient heat conduction to a semi-infinite region §5.6 differentiate the heat diffusion equation with respect to x , so α ∂2T ∂3T = ∂x 3 ∂t∂x When we substitute q = −k ∂T /∂x in this, we obtain α ∂q ∂2q = 2 ∂x ∂t with the b.c.’s: q(x = 0, t > 0) = qw q(x qw − q qw or 0, t = 0) = 0 qw − q qw or =0 x =0 =1 t =0 What we have done here is quite elegant. We have made the problem of predicting the local heat flux q into exactly the same form as that of predicting the local temperature in a semi-infinite region subjected to a step change of wall temperature. Therefore, the solution must be the same: x qw − q √ . = erf qw 2 αt (5.55) The temperature distribution is obtained by integrating Fourier’s law. At the wall, for example: Tw Ti 0 dT = − ∞ q dx k where Ti = T (x → ∞) and Tw = T (x = 0). Then Tw = Ti + qw k Tw = Ti + qw k ∞ erfc(x/2 αt) dx 0 This becomes ∞ αt erfc(ζ/2) dζ 0 √ =2/ π so Tw (t) = Ti + 2 qw k αt π (5.56) 229 230 Transient and multidimensional heat conduction §5.6 Figure 5.17 A bubble growing in a superheated liquid. Example 5.6 Predicting the Growth Rate of a Vapor Bubble in an Infinite Superheated Liquid This prediction is relevant to a large variety of processes, ranging from nuclear thermodynamics to the direct-contact heat exchange. It was originally presented by Max Jakob and others in the early 1930s (see, e.g., [5.10, Chap. I]). Jakob (pronounced Yah -kob) was an important figure in heat transfer during the 1920s and 1930s. He left Nazi Germany in 1936 to come to the United States. We encounter his name again later. Figure 5.17 shows how growth occurs. When a liquid is superheated to a temperature somewhat above its boiling point, a small gas or vapor cavity in that liquid will grow. (That is what happens in the superheated water at the bottom of a teakettle.) This bubble grows into the surrounding liquid because its boundary is kept at the saturation temperature, Tsat , by the near-equilibrium coexistence of liquid and vapor. Therefore, heat must flow from the superheated surroundings to the interface, where evaporation occurs. So long as the layer of cooled liquid is thin, we should not suffer too much error by using the one-dimensional semi-infinite region solution to predict the heat flow. Transient heat conduction to a semi-infinite region §5.6 Thus, we can write the energy balance at the bubble interface: −q W m2 4π R 2 m2 = ρg hfg Q into bubble J m3 dV m3 dt s rate of energy increase of the bubble and then substitute eqn. (5.54) for q and 4π R 3 /3 for the volume, V . This gives k(Tsup − Tsat ) dR √ = ρg hfg dt απ t (5.57) Integrating eqn. (5.57) from R = 0 at t = 0 up to R at t , we obtain Jakob’s prediction: k∆T 2 √ t R=√ π ρg hfg α (5.58) This analysis was done without assuming the curved bubble interface to be plane, 24 years after Jakob’s work, by Plesset and Zwick [5.11]. It was verified in a more exact way after another 5 years by Scriven [5.12]. These calculations are more complicated, but they lead to a very similar result: √ √ 2 3 k∆T √ t = 3 RJakob . (5.59) R= √ π ρg hfg α Both predictions are compared with some of the data of Dergarabedian [5.13] in Fig. 5.18. The data and the exact theory match almost perfectly. The simple theory of Jakob et al. shows the correct dependence on R on√ its variables, but it shows growth rates that are low all by a factor of 3. This is because the expansion of the spherical bubble causes a relative motion of liquid toward the bubble surface, which helps to thin the region of thermal influence in the radial direction. Consequently, the temperature gradient and heat transfer rate are higher than in Jakob’s model, which neglected the liquid motion. Therefore, the temperature profile flattens out more slowly than Jakob predicts, and the bubble grows more rapidly. Experiment 5.2 Touch various objects in the room around you: glass, wood, corkboard, paper, steel, and gold or diamond, if available. Rank them in 231 232 Transient and multidimensional heat conduction §5.6 Figure 5.18 The growth of a vapor bubble—predictions and measurements. order of which feels coldest at the first instant of contact (see Problem 5.29). The more advanced theory of heat conduction (see, e.g., [5.6]) shows that if two semi-infinite regions at uniform temperatures T1 and T2 are placed together suddenly, their interface temperature, Ts , is given by6 Ts − T2 = T1 − T2 (kρcp )1 (kρcp )1 + (kρcp )2 If we identify one region with your body (T1 37◦ C) and the other with the object being touched (T2 20◦ C), we can determine the temperature, Ts , that the surface of your finger will reach upon contact. Compare the ranking you obtain experimentally with the ranking given by this equation. 6 For semi-infinite regions, initially at uniform temperatures, Ts does not vary with time. For finite bodies, Ts will eventually change. A constant value of Ts means that each of the two bodies independently behaves as a semi-infinite body whose surface temperature has been changed to Ts at time zero. Consequently, our previous results— eqns. (5.50), (5.51), and (5.54)—apply to each of these bodies while they may be treated as semi-infinite. We need only replace T∞ by Ts in those equations. Transient heat conduction to a semi-infinite region §5.6 Notice that your bloodstream and capillary system provide a heat source in your finger, so the equation is valid only for a moment. Then you start replacing heat lost to the objects. If you included a diamond among the objects that you touched, you will notice that it warmed up almost instantly. Most diamonds are quite small but are possessed of the highest known value of α. Therefore, they can behave as a semi-infinite region only for an instant, and they usually feel warm to the touch. Conduction to a semi-infinite region with a harmonically oscillating temperature at the boundary Suppose that we approximate the annual variation of the ambient temperature as sinusoidal and then ask what the influence of this variation will be beneath the ground. We want to calculate T − T (where T is the time-average surface temperature) as a function of: depth, x ; thermal diffusivity, α; frequency of oscillation, ω; amplitude of oscillation, ∆T ; and time, t . There are six variables in K, m, and s, so the problem can be represented in three dimensionless variables: Θ≡ T −T ; ∆T Ω ≡ ωt ; ξ≡x ω . 2α We pose the problem as follows in these variables. The heat conduction equation is ∂Θ 1 ∂2Θ = 2 ∂ξ 2 ∂Ω (5.60) and the b.c.’s are Θ ξ =0 = cos ωt and Θ ξ >0 = finite (5.61) No i.c. is needed because, after the initial transient decays, the remaining steady oscillation must be periodic. The solution is given by Carslaw and Jaeger (see [5.6, §2.6] or work Problem 5.16). It is Θ (ξ, Ω) = e−ξ cos (Ω − ξ) (5.62) This result is plotted in Fig. 5.19. It shows that the surface temperature variation decays exponentially into the region and suffers a phase shift as it does so. 233 234 Transient and multidimensional heat conduction §5.6 Figure 5.19 The temperature variation within a semi-infinite region whose temperature varies harmonically at the boundary. Example 5.7 How deep in the earth must we dig to find the temperature wave that was launched by the coldest part of the last winter if it is now high summer? Solution. ω = 2π rad/yr, and Ω = ωt = 0 at the present. First, we must find the depths at which the Ω = 0 curve reaches its local extrema. (We pick the Ω = 0 curve because it gives the highest temperature at t = 0.) dΘ dξ = −e−ξ cos(0 − ξ) + e−ξ sin(0 − ξ) = 0 Ω =0 This gives tan(0 − ξ) = 1 so ξ= 3π 7π , ,... 4 4 and the first minimum occurs where ξ = 3π /4 = 2.356, as we can see in Fig. 5.19. Thus, ξ = x ω/2α = 2.356 Steady multidimensional heat conduction §5.7 or, if we take α = 0.139×10−6 m2 /s (given in [5.14] for coarse, gravelly earth), x = 2.356 1 2π = 2.783 m −6 365(24)(3600) 2 0.139 × 10 If we dug in the earth, we would find it growing older and colder until it reached a maximum coldness at a depth of about 2.8 m. Farther down, it would begin to warm up again, but not much. In midwinter (Ω = π ), the reverse would be true. 5.7 Steady multidimensional heat conduction Introduction The general equation for T (r ) during steady conduction in a region of constant thermal conductivity, without heat sources, is called Laplace’s equation: ∇2 T = 0 (5.63) It looks easier to solve than it is, since [recall eqn. (2.12) and eqn. (2.14)] the Laplacian, ∇2 T , is a sum of several second partial derivatives. We solved one two-dimensional heat conduction problem in Example 4.1, but this was not difficult because the boundary conditions were made to order. Depending upon your mathematical background and the specific problem, the analytical solution of multidimensional problems can be anything from straightforward calculation to a considerable challenge. The reader who wishes to study such analyses in depth should refer to [5.6] or [5.15], where such calculations are discussed in detail. Faced with a steady multidimensional problem, three routes are open to us: • Find out whether or not the analytical solution is already available in a heat conduction text or in other published literature. • Solve the problem. (a) Analytically. (b) Numerically. • Obtain the solution graphically if the problem is two-dimensional. It is to the last of these options that we give our attention next. 235 236 Transient and multidimensional heat conduction §5.7 Figure 5.20 The two-dimensional flow of heat between two isothermal walls. The flux plot The method of flux plotting will solve all steady planar problems in which all boundaries are held at either of two temperatures or are insulated. With a little skill, it will provide accuracies of a few percent. This accuracy is almost always greater than the accuracy with which the b.c.’s and k can be specified; and it displays the physical sense of the problem very clearly. Figure 5.20 shows heat flowing from one isothermal wall to another in a regime that does not conform to any convenient coordinate scheme. We identify a series of channels, each which carries the same heat flow, δQ W/m. We also include a set of equally spaced isotherms, δT apart, between the walls. Since the heat fluxes in all channels are the same, δQ = k δT δs δn (5.64) Notice that if we arrange things so that δQ, δT , and k are the same for flow through each rectangle in the flow field, then δs/δn must be the same for each rectangle. We therefore arbitrarily set the ratio equal to unity, so all the elements appear as distorted squares. The objective then is to sketch the isothermal lines and the adiabatic,7 7 These are lines in the direction of heat flow. It immediately follows that there can §5.7 Steady multidimensional heat conduction or heat flow, lines which run perpendicular to them. This sketch is to be done subject to two constraints • Isothermal and adiabatic lines must intersect at right angles. • They must subdivide the flow field into elements that are nearly square—“nearly” because they have slightly curved sides. Once the grid has been sketched, the temperature anywhere in the field can be read directly from the sketch. And the heat flow per unit depth into the paper is Q W/m = Nk δT N δs = k∆T δn I (5.65) where N is the number of heat flow channels and I is the number of temperature increments, ∆T /δT . The first step in constructing a flux plot is to draw the boundaries of the region accurately in ink, using either drafting software or a straightedge. The next is to obtain a soft pencil (such as a no. 2 grade) and a soft eraser. We begin with an example that was executed nicely in the influential Heat Transfer Notes [5.3] of the mid-twentieth century. This example is shown in Fig. 5.21. The particular example happens to have an axis of symmetry in it. We immediately interpret this as an adiabatic boundary because heat cannot cross it. The problem therefore reduces to the simpler one of sketching lines in only one half of the area. We illustrate this process in four steps. Notice the following steps and features in this plot: • Begin by dividing the region, by sketching in either a single isothermal or adiabatic line. • Fill in the lines perpendicular to the original line so as to make squares. Allow the original line to move in such a way as to accommodate squares. This will always require some erasing. Therefore: • Never make the original lines dark and firm. • By successive subdividing of the squares, make the final grid. Do not make the grid very fine. If you do, you will lose accuracy because the lack of perpendicularity and squareness will be less evident to the eye. Step IV in Fig. 5.21 is as fine a grid as should ever be made. be no component of heat flow normal to them; they must be adiabatic. 237 Figure 5.21 The evolution of a flux plot. 238 §5.7 Steady multidimensional heat conduction • If you have doubts about whether any large, ill-shaped regions are correct, fill them in with an extra isotherm and adiabatic line to be sure that they resolve into appropriate squares (see the dashed lines in Fig. 5.21). • Fill in the final grid, when you are sure of it, either in hard pencil or pen, and erase any lingering background sketch lines. • Your flow channels need not come out even. Notice that there is an extra 1/7 of a channel in Fig. 5.21. This is simply counted as 1/7 of a square in eqn. (5.65). • Never allow isotherms or adiabatic lines to intersect themselves. When the sketch is complete, we can return to eqn. (5.65) to compute the heat flux. In this case 2(6.14) N k∆T = k∆T = 3.07 k∆T Q= I 4 When the authors of [5.3] did this problem, they obtained N/I = 3.00—a value only 2% below ours. This kind of agreement is typical when flux plotting is done with care. Figure 5.22 A flux plot with no axis of symmetry to guide construction. 239 240 Transient and multidimensional heat conduction §5.7 One must be careful not to grasp at a false axis of symmetry. Figure 5.22 shows a shape similar to the one that we just treated, but with unequal legs. In this case, no lines must enter (or leave) the corners A and B . The reason is that since there is no symmetry, we have no guidance as to the direction of the lines at these corners. In particular, we know that a line leaving A will no longer arrive at B . Example 5.8 A structure consists of metal walls, 8 cm apart, with insulating material (k = 0.12 W/m·K) between. Ribs 4 cm long protrude from one wall every 14 cm. They can be assumed to stay at the temperature of that wall. Find the heat flux through the wall if the first wall is at 40◦ C and the one with ribs is at 0◦ C. Find the temperature in the middle of the wall, 2 cm from a rib, as well. Figure 5.23 Heat transfer through a wall with isothermal ribs. Steady multidimensional heat conduction §5.7 Solution. The flux plot for this configuration is shown in Fig. 5.23. For a typical section, there are approximately 5.6 isothermal increments and 6.15 heat flow channels, so Q= 2(6.15) N k∆T = (0.12)(40 − 0) = 10.54 W/m I 5.6 where the factor of 2 accounts for the fact that there are two halves in the section. We deduce the temperature for the point of interest, A, by a simple proportionality: Tpoint A = 2.1 (40 − 0) = 15◦ C 5.6 The shape factor A heat conduction shape factor S may be defined for steady problems involving two isothermal surfaces as follows: Q ≡ S k∆T . (5.66) Thus far, every steady heat conduction problem we have done has taken this form. For these situations, the heat flow always equals a function of the geometric shape of the body multiplied by k∆T . The shape factor can be obtained analytically, numerically, or through flux plotting. For example, let us compare eqn. (5.65) and eqn. (5.66): Q W W = (S dimensionless) k∆T m m = N k∆T I (5.67) This shows S to be dimensionless in a two-dimensional problem, but in three dimensions S has units of meters: Q W = (S m) k∆T W . m (5.68) It also follows that the thermal resistance of a two-dimensional body is Rt = 1 kS where Q= ∆T Rt (5.69) For a three-dimensional body, eqn. (5.69) is unchanged except that the dimensions of Q and Rt differ.8 8 Recall that we noted after eqn. (2.22) that the dimensions of Rt changed, depending on whether or not Q was expressed in a unit-length basis. 241 242 Transient and multidimensional heat conduction §5.7 Figure 5.24 The shape factor for two similar bodies of different size. The virtue of the shape factor is that it summarizes a heat conduction solution in a given configuration. Once S is known, it can be used again and again. That S is nondimensional in two-dimensional configurations means that Q is independent of the size of the body. Thus, in Fig. 5.21, S is always 3.07—regardless of the size of the figure—and in Example 5.8, S is 2(6.15)/5.6 = 2.196, whether or not the wall is made larger or smaller. When a body’s breadth is increased so as to increase Q, its thickness in the direction of heat flow is also increased so as to decrease Q by the same factor. Example 5.9 Calculate the shape factor for a one-quarter section of a thick cylinder. Solution. We already know Rt for a thick cylinder. It is given by eqn. (2.22). From it we compute Scyl = 1 2π = kRt ln(ro /ri ) so on the case of a quarter-cylinder, S= π 2 ln(ro /ri ) The quarter-cylinder is pictured in Fig. 5.24 for a radius ratio, ro /ri = 3, but for two different sizes. In both cases S = 1.43. (Note that the same S is also given by the flux plot shown.) Steady multidimensional heat conduction §5.7 Figure 5.25 Heat transfer through a thick, hollow sphere. Example 5.10 Calculate S for a thick hollow sphere, as shown in Fig. 5.25. Solution. The general solution of the heat diffusion equation in spherical coordinates for purely radial heat flow is: C1 + C2 r when T = fn(r only). The b.c.’s are T= T (r = ri ) = Ti and T (r = ro ) = To substituting the general solution in the b.c.’s we get C1 + C2 = Ti ri and C1 + C1 = To ro Therefore, C1 = Ti − To ri ro ro − ri and C2 = Ti − Ti − To ro ro − ri Putting C1 and C2 in the general solution, and calling Ti − To ≡ ∆T , we get T = Ti + ∆T ro ri ro − r (ro − ri ) ro − ri Then 4π (ri ro ) dT = k∆T dr ro − ri 4π (ri ro ) m S= ro − ri Q = −kA where S now has the dimensions of m. 243 244 Transient and multidimensional heat conduction §5.7 Table 5.4 includes a number of analytically derived shape factors for use in calculating the heat flux in different configurations. Notice that these results will not give local temperatures. To obtain that information, one must solve the Laplace equation, ∇2 T = 0, by one of the methods listed at the beginning of this section. Notice, too, that this table is restricted to bodies with isothermal and insulated boundaries. In the two-dimensional cases, both a hot and a cold surface must be present in order to have a steady-state solution; if only a single hot (or cold) body is present, steady state is never reached. For example, a hot isothermal cylinder in a cooler, infinite medium never reaches steady state with that medium. Likewise, in situations 5, 6, and 7 in the table, the medium far from the isothermal plane must also be at temperature T2 in order for steady state to occur; otherwise the isothermal plane and the medium below it would behave as an unsteady, semi-infinite body. Of course, since no real medium is truly infinite, what this means in practice is that steady state only occurs after the medium “at infinity” comes to a temperature T2 . Conversely, in three-dimensional situations (such as 4, 8, 12, and 13), a body can come to steady state with a surrounding infinite or semi-infinite medium at a different temperature. Example 5.11 A spherical heat source of 6 cm in diameter is buried 30 cm below the surface of a very large box of soil and kept at 35◦ C. The surface of the soil is kept at 21◦ C. If the steady heat transfer rate is 14 W, what is the thermal conductivity of this sample of soil? Solution. Q = S k∆T = 4π R k∆T 1 − R/2h where S is that for situation 7 in Table 5.4. Then k= 1 − (0.06/2) 2(0.3) 14 W = 2.545 W/m·K (35 − 21)K 4π (0.06/2) m Readers who desire a broader catalogue of shape factors should refer to [5.16], [5.18], or [5.19]. Table 5.4 Conduction shape factors: Q = S k∆T . Situation Shape factor, S 1. Conduction through a slab A/L Dimensions meter Source Example 2.2 2. Conduction through wall of a long thick cylinder 2π ln (ro /ri ) none Example 5.9 3. Conduction through a thick-walled hollow sphere 4π (ro ri ) ro − ri meter Example 5.10 4π R meter Problems 5.19 and 2.15 meter [5.16] none [5.16] meter [5.16, 5.17] 4. The boundary of a spherical hole of radius R conducting into an infinite medium 5. Cylinder of radius R and length L, transferring heat to a parallel isothermal plane; h L 2π L cosh−1 (h/R) 6. Same as item 5, but with L → ∞ (two-dimensional conduction) 2π cosh −1 (h/R) 7. An isothermal sphere of radius R transfers heat to an isothermal plane; R/h < 0.8 (see item 4) 4π R 1 − R/2h 245 Table 5.4 Conduction shape factors: Q = S k∆T (con’t). Situation Shape factor, S 8. An isothermal sphere of radius R , near an insulated plane, transfers heat to a semi-infinite medium at T∞ (see items 4 and 7) Dimensions 4π R 1 + R/2h meter Source [5.18] 9. Parallel cylinders exchange heat in an infinite conducting medium −1 cosh 10. Same as 9, but with cylinders widely spaced; L R1 and R2 11. Cylinder of radius Ri surrounded by eccentric cylinder of radius Ro > Ri ; centerlines a distance L apart (see item 2) 246 none 2π cosh−1 L 2R1 cosh−1 12. Isothermal disc of radius R on an otherwise insulated plane conducts heat into a semi-infinite medium at T∞ below it 13. Isothermal ellipsoid of semimajor axis b and semiminor axes a conducts heat into an infinite medium at T∞ ; b > a (see 4) 2π 2 2 L2 − R1 − R2 2R1 R2 + cosh−1 2π 2 2 Ro + Ri − L2 2Ro Ri 4R 4π b 1 − a2 b2 tanh−1 1 − a2 b2 L 2R2 [5.6] none [5.16] none [5.6] meter [5.6] meter [5.16] §5.8 Transient multidimensional heat conduction 247 Figure 5.26 Resistance vanishes where two isothermal boundaries intersect. The problem of locally vanishing resistance Suppose that two different temperatures are specified on adjacent sides of a square, as shown in Fig. 5.26. The shape factor in this case is S= ∞ N = =∞ I 4 (It is futile to try and count channels beyond N 10, but it is clear that they multiply without limit in the lower left corner.) The problem is that we have violated our rule that isotherms cannot intersect and have created a 1/r singularity. If we actually tried to sustain such a situation, the figure would be correct at some distance from the corner. However, where the isotherms are close to one another, they will necessarily influence and distort one another in such a way as to avoid intersecting. And S will never really be infinite, as it appears to be in the figure. 5.8 Transient multidimensional heat conduction— The tactic of superposition Consider the cooling of a stubby cylinder, such as the one shown in Fig. 5.27a. The cylinder is initially at T = Ti , and it is suddenly subjected to a common b.c. on all sides. It has a length 2L and a radius ro . Finding the temperature field in this situation is inherently complicated. 248 Transient and multidimensional heat conduction §5.8 It requires solving the heat conduction equation for T = fn(r , z, t) with b.c.’s of the first, second, or third kind. However, Fig. 5.27a suggests that this can somehow be viewed as a combination of an infinite cylinder and an infinite slab. It turns out that the problem can be analyzed from that point of view. If the body is subject to uniform b.c.’s of the first, second, or third kind, and if it has a uniform initial temperature, then its temperature response is simply the product of an infinite slab solution and an infinite cylinder solution each having the same boundary and initial conditions. For the case shown in Fig. 5.27a, if the cylinder begins convective cooling into a medium at temperature T∞ at time t = 0, the dimensional temperature response is T (r , z, t) − T∞ = Tslab (z, t) − T∞ × Tcyl (r , t) − T∞ (5.70a) Observe that the slab has as a characteristic length L, its half thickness, while the cylinder has as its characteristic length R , its radius. In dimensionless form, we may write eqn. (5.70a) as Θ≡ T (r , z, t) − T∞ = Θinf slab (ξ, Fos , Bis ) Ti − T∞ Θinf cyl (ρ, Foc , Bic ) (5.70b) For the cylindrical component of the solution, ρ= r , ro Foc = αt 2, ro and Bic = hro , k while for the slab component of the solution ξ= z + 1, L Fos = αt , L2 and Bis = hL . k The component solutions are none other than those discussed in Sections 5.3–5.5. The proof of the legitimacy of such product solutions is given by Carlsaw and Jaeger [5.6, §1.15]. Figure 5.27b shows a point inside a one-eighth-infinite region, near the corner. This case may be regarded as the product of three semi-infinite bodies. To find the temperature at this point we write Θ≡ T (x1 , x2 , x3 , t) − T∞ = [Θsemi (ζ1 , β)] [Θsemi (ζ2 , β)] [Θsemi (ζ3 , β)] Ti − T∞ (5.71) Figure 5.27 Various solid bodies whose transient cooling can be treated as the product of one-dimensional solutions. 249 250 Transient and multidimensional heat conduction §5.8 in which Θsemi is either the semi-infinite body solution given by eqn. (5.53) when convection is present at the boundary or the solution given by eqn. (5.50) when the boundary temperature itself is changed at time zero. Several other geometries can also be represented by product solutions. Note that for of these solutions, the value of Θ at t = 0 is one for each factor in the product. Example 5.12 A very long 4 cm square iron rod at Ti = 100◦ C is suddenly immersed in a coolant at T∞ = 20◦ C with h = 800 W/m2 K. What is the temperature on a line 1 cm from one side and 2 cm from the adjoining side, after 10 s? Solution. With reference to Fig. 5.27c, see that the bar may be treated as the product of two slabs, each 4 cm thick. We first evaluate Fo1 = Fo2 = αt/L2 = (0.0000226 m2 /s)(10 s) (0.04 m/2)2 = 0.565, and Bi1 = Bi2 = hL k = 800(0.04/2)/76 = 0.2105, and we then write Θ x L 1 x L = 0, = Θ1 2 x L = 1 1 , Fo1 , Fo2 , Bi−1 , Bi−1 1 2 2 = 0, Fo1 = 0.565, Bi−1 = 4.75 1 = 0.93 from upper left-hand side of Fig. 5.7 × Θ2 x L 2 = 1 , Fo2 = 0.565, Bi−1 = 4.75 2 2 = 0.91 from interpolation between lower lefthand side and upper righthand side of Fig. 5.7 Thus, at the axial line of interest, Θ = (0.93)(0.91) = 0.846 so T − 20 = 0.846 100 − 20 or T = 87.7◦ C Transient multidimensional heat conduction 251 Product solutions can also be used to determine the mean temperature, Θ, and the total heat removal, Φ, from a multidimensional object. For example, when two or three solutions (Θ1 , Θ2 , and perhaps Θ3 ) are multiplied to obtain Θ, the corresponding mean temperature of the multidimensional object is simply the product of the one-dimensional mean temperatures from eqn. (5.40) Θ = Θ1 (Fo1 , Bi1 ) × Θ2 (Fo2 , Bi2 ) for two factors Θ = Θ1 (Fo1 , Bi1 ) × Θ2 (Fo2 , Bi2 ) × Θ3 (Fo3 , Bi3 ) (5.72a) for three factors. (5.72b) Since Φ = 1 − Θ, a simple calculation shows that Φ can found from Φ1 , Φ2 , and Φ3 as follows: Φ = Φ1 + Φ2 (1 − Φ1 ) for two factors Φ = Φ1 + Φ2 (1 − Φ1 ) + Φ3 (1 − Φ2 ) (1 − Φ1 ) (5.73a) for three factors. (5.73b) Example 5.13 For the bar described in Example 5.12, what is the mean temperature after 10 s and how much heat has been lost at that time? Solution. For the Biot and Fourier numbers given in Example 5.12, we find from Fig. 5.10a Φ1 (Fo1 = 0.565, Bi1 = 0.2105) = 0.10 Φ2 (Fo2 = 0.565, Bi2 = 0.2105) = 0.10 and, with eqn. (5.73a), Φ = Φ1 + Φ2 (1 − Φ1 ) = 0.19 The mean temperature is Θ= T − 20 = 1 − Φ = 0.81 100 − 20 so T = 20 + 80(0.81) = 84.8◦ C Chapter 5: Transient and multidimensional heat conduction 252 Problems 5.1 Rework Example 5.1, and replot the solution, with one change. This time, insert the thermometer at zero time, at an initial temperature < (Ti − bT ). 5.2 A body of known volume and surface area and temperature Ti is suddenly immersed in a bath whose temperature is rising as Tbath = Ti + (T0 − Ti )et/τ . Let us suppose that h is known, that τ = 10ρcV /hA, and that t is measured from the time of immersion. The Biot number of the body is small. Find the temperature response of the body. Plot the response and the bath temperature as a function of time up to t = 2τ . (Do not use Laplace transform methods except, perhaps, as a check.) 5.3 A body of known volume and surface area is immersed in a bath whose temperature is varying sinusoidally with a frequency ω about an average value. The heat transfer coefficient is known and the Biot number is small. Find the temperature variation of the body after a long time has passed, and plot it along with the bath temperature. Comment on any interesting aspects of the solution. A suggested program for solving this problem: • Write the differential equation of response. • To get the particular integral of the complete equation, guess that T − Tmean = C1 cos ωt + C2 sin ωt . Substitute this in the differential equation and find C1 and C2 values that will make the resulting equation valid. • Write the general solution of the complete equation. It will have one unknown constant in it. • Write any initial condition you wish—the simplest one you can think of—and use it to get rid of the constant. • Let the time be large and note which terms vanish from the solution. Throw them away. • Combine two trigonometric terms in the solution into a term involving sin(ωt − β), where β = fn(ωT ) is the phase lag of the body temperature. 5.4 A block of copper floats within a large region of well-stirred mercury. The system is initially at a uniform temperature, Ti . Problems 253 There is a heat transfer coefficient, hm , on the inside of the thin metal container of the mercury and another one, hc , between the copper block and the mercury. The container is then suddenly subjected to a change in ambient temperature from Ti to Ts < Ti . Predict the temperature response of the copper block, neglecting the internal resistance of both the copper and the mercury. Check your result by seeing that it fits both initial conditions and that it gives the expected behavior at t → ∞. 5.5 Sketch the electrical circuit that is analogous to the secondorder lumped capacity system treated in the context of Fig. 5.5 and explain it fully. 5.6 A one-inch diameter copper sphere with a thermocouple in its center is mounted as shown in Fig. 5.28 and immersed in water that is saturated at 211◦ F. The figure shows the thermocouple reading as a function of time during the quenching process. If the Biot number is small, the center temperature can be interpreted as the uniform temperature of the sphere during the quench. First draw tangents to the curve, and graphically differentiate it. Then use the resulting values of dT /dt to construct a graph of the heat transfer coefficient as a function of (Tsphere − Tsat ). The result will give actual values of h during boiling over the range of temperature differences. Check to see whether or not the largest value of the Biot number is too great to permit the use of lumped-capacity methods. 5.7 A butt-welded 36-gage thermocouple is placed in a gas flow whose temperature rises at the rate 20◦ C/s. The thermocouple steadily records a temperature 2.4◦ C below the known gas flow temperature. If ρc is 3800 kJ/m3 K for the thermocouple material, what is h on the thermocouple? [h = 1006 W/m2 K.] 5.8 Check the point on Fig. 5.7 at Fo = 0.2, Bi = 10, and x/L = 0 analytically. 5.9 Prove that when Bi is large, eqn. (5.34) reduces to eqn. (5.33). 5.10 Check the point at Bi = 0.1 and Fo = 2.5 on the slab curve in Fig. 5.10 analytically. Chapter 5: Transient and multidimensional heat conduction 254 Figure 5.28 Problem 5.6 5.11 Configuration and temperature response for Sketch one of the curves in Fig. 5.7, 5.8, or 5.9 and identify: • The region in which b.c.’s of the third kind can be replaced with b.c.’s of the first kind. • The region in which a lumped-capacity response can be assumed. • The region in which the solid can be viewed as a semiinfinite region. 5.12 Water flows over a flat slab of Nichrome, 0.05 mm thick, which serves as a resistance heater using AC power. The apparent value of h is 2000 W/m2 K. How much surface temperature fluctuation will there be? Problems 255 5.13 Put Jakob’s bubble growth formula in dimensionless form, identifying a “Jakob number”, Ja ≡ cp (Tsup − Tsat )/hfg as one of the groups. (Ja is the ratio of sensible heat to latent heat.) Be certain that your nondimensionalization is consistent with the Buckingham pi-theorem. 5.14 A 7 cm long vertical glass tube is filled with water that is uniformly at a temperature of T = 102◦ C. The top is suddenly opened to the air at 1 atm pressure. Plot the decrease of the height of water in the tube by evaporation as a function of time until the bottom of the tube has cooled by 0.05◦ C. 5.15 A slab is cooled convectively on both sides from a known initial temperature. Compare the variation of surface temperature with time as given in Fig. 5.7 with that given by eqn. (5.53) if Bi = 2. Discuss the meaning of your comparisons. 5.16 To obtain eqn. (5.62), assume a complex solution of the type √ Θ = fn(ξ)exp(iΩ), where i ≡ −1. This will assure that the real part of your solution has the required periodicity and, when you substitute it in eqn. (5.60), you will get an easy-tosolve ordinary d.e. in fn(ξ). 5.17 A certain steel cylinder wall is subjected to a temperature oscillation that we approximate at T = 650◦ C + (300◦ C) cos ωt , where the piston fires eight times per second. For stress design purposes, plot the amplitude of the temperature variation in the steel as a function of depth. If the cylinder is 1 cm thick, can we view it as having infinite depth? 5.18 A 40 cm diameter pipe at 75◦ C is buried in a large block of Portland cement. It runs parallel with a 15◦ C isothermal surface at a depth of 1 m. Plot the temperature distribution along the line normal to the 15◦ C surface that passes through the center of the pipe. Compute the heat loss from the pipe both graphically and analytically. 5.19 Derive shape factor 4 in Table 5.4. 5.20 Verify shape factor 9 in Table 5.4 with a flux plot. Use R1 /R2 = 2 and R1 /L = ½. (Be sure to start out with enough blank paper surrounding the cylinders.) Chapter 5: Transient and multidimensional heat conduction 256 5.21 5.22 Obtain the shape factor for any or all of the situations pictured in Fig. 5.29a through j on pages 258–259. In each case, 1.03, Sc Sd , Sg = present a well-drawn flux plot. [Sb 1.] 5.23 Eggs cook as their proteins denature and coagulate. The time to cook depends on whether a soft or hard cooked egg desired. Eggs may be cooked by placing them (cold or warm) into cold water before heating starts or by placing warm eggs directly into simmering water [5.20]. A copper block 1 in. thick and 3 in. square is held at 100◦ F on one 1 in. by 3 in. surface. The opposing 1 in. by 3 in. surface is adiabatic for 2 in. and 90◦ F for 1 inch. The remaining surfaces are adiabatic. Find the rate of heat transfer. [Q = 36.8 W.] Two copper slabs, 3 cm thick and insulated on the outside, are suddenly slapped tightly together. The one on the left side is initially at 100◦ C and the one on the right side at 0◦ C. Determine the left-hand adiabatic boundary’s temperature after 2.3 s have elapsed. [Twall 80.5◦ C] 5.24 Estimate the time required to hard-cook an egg if: • The minor diameter is 45 mm. • k for the entire egg is about the same as for egg white. No significant heat release or change of properties occurs during cooking. • h between the egg and the water is 1000 W/m2 K. • The egg has a uniform temperature of 20◦ C when it is put into simmering water at 85◦ C. • The egg is done when the center reaches 75◦ C. 5.25 Prove that T1 in Fig. 5.5 cannot oscillate. 5.26 Show that when isothermal and adiabatic lines are interchanged in a two-dimenisonal body, the new shape factor is the inverse of the original one. 5.27 A 0.5 cm diameter cylinder at 300◦ C is suddenly immersed in saturated water at 1 atm. If h = 10, 000 W/m2 K, find the centerline and surface temperatures after 0.2 s: a. If the cylinder is copper. b. If the cylinder is Nichrome V. [Tsfc 200◦ C.] c. If the cylinder is Nichrome V, obtain the most accurate value of the temperatures after 0.04 s that you can. Problems 257 5.28 A large, flat electrical resistance strip heater is fastened to a firebrick wall, unformly at 15◦ C. When it is suddenly turned on, it releases heat at the uniform rate of 4000 W/m2 . Plot the temperature of the brick immediately under the heater as a function of time if the other side of the heater is insulated. What is the heat flux at a depth of 1 cm when the surface reaches 200◦ C. 5.29 Do Experiment 5.2 and submit a report on the results. 5.30 An approximately spherical container, 2 cm in diameter, containing electronic equipment is placed in wet mineral soil with its center 2 m below the surface. The soil surface is kept at 0◦ C. What is the maximum rate at which energy can be released by the equipment if the surface of the sphere is not to exceed 30◦ C? 5.31 A semi-infinite slab of ice at −10◦ C is exposed to air at 15◦ C through a heat transfer coefficient of 10 W/m2 K. What is the initial rate of melting of ice in kg/m2 s? What is the asymptotic rate of melting? Describe the melting process in physical terms. (The latent heat of fusion of ice, hsf = 333, 300 J/kg.) 5.32 One side of an insulating firebrick wall, 10 cm thick, initially at 20◦ C is exposed to 1000◦ C flame through a heat transfer coefficient of 230 W/m2 K. How long will it be before the other side is too hot to touch, say at 65◦ C? (Estimate properties at 500◦ C, and assume that h is quite low on the cool side.) 5.33 A particular lead bullet travels for 0.5 sec within a shock wave that heats the air near the bullet to 300◦ C. Approximate the bullet as a cylinder 0.8 cm in diameter. What is its surface temperature at impact if h = 600 W/m2 K and if the bullet was initially at 20◦ C? What is its center temperature? 5.34 A loaf of bread is removed from an oven at 125◦ C and set on the (insulating) counter to cool in a kitchen at 25◦ C. The loaf is 30 cm long, 15 cm high, and 12 cm wide. If k = 0.05 W/m·K and α = 5 × 10−7 m2 /s for bread, and h = 10 W/m2 K, when will the hottest part of the loaf have cooled to 60◦ C? [About 1 h 5 min.] Figure 5.29 Configurations for Problem 5.22 258 Figure 5.29 Configurations for Problem 5.22 (con’t) 259 Chapter 5: Transient and multidimensional heat conduction 260 5.35 A lead cube, 50 cm on each side, is initially at 20◦ C. The surroundings are suddenly raised to 200◦ C and h around the cube is 272 W/m2 K. Plot the cube temperature along a line from the center to the middle of one face after 20 minutes have elapsed. 5.36 A jet of clean water superheated to 150◦ C issues from a 1/16 inch diameter sharp-edged orifice into air at 1 atm, moving at 27 m/s. The coefficient of contraction of the jet is 0.611. Evaporation at T = Tsat begins immediately on the outside of the jet. Plot the centerline temperature of the jet and T (r /ro = 0.6) as functions of distance from the orifice up to about 5 m. Neglect any axial conduction and any dynamic interactions between the jet and the air. 5.37 A 3 cm thick slab of aluminum (initially at 50◦ C) is slapped tightly against a 5 cm slab of copper (initially at 20◦ C). The outsides are both insulated and the contact resistance is neglible. What is the initial interfacial temperature? Estimate how long the interface will keep its initial temperature. 5.38 A cylindrical underground gasoline tank, 2 m in diameter and 4 m long, is embedded in 10◦ C soil with k = 0.8 W/m2 K and α = 1.3 × 10−6 m2 /s. water at 27◦ C is injected into the tank to test it for leaks. It is well-stirred with a submerged ½ kW pump. We observe the water level in a 10 cm I.D. transparent standpipe and measure its rate of rise and fall. What rate of change of height will occur after one hour if there is no leakage? Will the level rise or fall? Neglect thermal expansion and deformation of the tank, which should be complete by the time the tank is filled. 5.39 A 47◦ C copper cylinder, 3 cm in diameter, is suddenly immersed horizontally in water at 27◦ C in a reduced gravity environment. Plot Tcyl as a function of time if g = 0.76 m/s2 and if h = [2.733 + 10.448(∆T ◦ C)1/6 ]2 W/m2 K. (Do it numerically if you cannot integrate the resulting equation analytically.) 5.40 The mechanical engineers at the University of Utah end spring semester by roasting a pig and having a picnic. The pig is roughly cylindrical and about 26 cm in diameter. It is roasted Problems 261 over a propane flame, whose products have properties similar to those of air, at 280◦ C. The hot gas flows across the pig at about 2 m/s. If the meat is cooked when it reaches 95◦ C, and if it is to be served at 2:00 pm, what time should cooking commence? Assume Bi to be large, but note Problem 7.40. The pig is initially at 25◦ C. 5.41 People from cold northern climates know not to grasp metal with their bare hands in subzero weather. A very slightly frosted peice of, say, cast iron will stick to your hand like glue in, say, −20◦ C weather and might tear off patches of skin. Explain this quantitatively. 5.42 A 4 cm diameter rod of type 304 stainless steel has a very small hole down its center. The hole is clogged with wax that has a melting point of 60◦ C. The rod is at 20◦ C. In an attempt to free the hole, a workman swirls the end of the rod—and about a meter of its length—in a tank of water at 80◦ C. If h is 688 W/m2 K on both the end and the sides of the rod, plot the depth of the melt front as a function of time up to say, 4 cm. 5.43 A cylindrical insulator contains a single, very thin electrical resistor wire that runs along a line halfway between the center and the outside. The wire liberates 480 W/m. The thermal conductivity of the insulation is 3 W/m2 K, and the outside perimeter is held at 20◦ C. Develop a flux plot for the cross section, considering carefully how the field should look in the neighborhood of the point through which the wire passes. Evaluate the temperature at the center of the insulation. 5.44 A long, 10 cm square copper bar is bounded by 260◦ C gas flows on two opposing sides. These flows impose heat transfer coefficients of 46 W/m2 K. The two intervening sides are cooled by natural convection to water at 15◦ C, with a heat transfer coefficient of 30 W/m2 K. What is the heat flow through the block and the temperature at the center of the block? (This could be a pretty complicated problem, but take the trouble to think about Biot numbers before you begin.) 5.45 Lord Kelvin made an interesting estimate of the age of the earth in 1864. He assumed that the earth originated as a mass of Chapter 5: Transient and multidimensional heat conduction 262 molten rock at 4144 K (7000◦ F) and that it had been cooled by outer space at 0 K ever since. To do this, he assumed that Bi for the earth is very large and that cooling had thus far penetrated through only a relatively thin (one-dimensional) layer. Using αrock = 1.18 × 10−6 m2 /s and the measured sur1 face temperature gradient of the earth, 27 ◦ C/m, Find Kelvin’s value of Earth’s age. (Kelvin’s result turns out to be much less than the accepted value of 4 billion years. His calculation fails because internal heat generation by radioactive decay of the material in the surface layer causes the surface temperature gradient to be higher than it would otherwise be.) 5.46 A pure aluminum cylinder, 4 cm diam. by 8 cm long, is initially at 300◦ C. It is plunged into a liquid bath at 40◦ C with h = 500 W/m2 K. Calculate the hottest and coldest temperatures in the cylinder after one minute. Compare these results with the lumped capacity calculation, and discuss the comparison. 5.47 When Ivan cleaned his freezer, he accidentally put a large can of frozen juice into the refrigerator. The juice can is 17.8 cm tall and has an 8.9 cm I.D. The can was at −15◦ C in the freezer, but the refrigerator is at 4◦ C. The can now lies on a shelf of widely-spaced plastic rods, and air circulates freely over it. Thermal interactions with the rods can be ignored. The effective heat transfer coefficient to the can (for simultaneous convection and thermal radiation) is 8 W/m2 K. The can has a 1.0 mm thick cardboard skin with k = 0.2 W/m·K. The frozen juice has approximately the same physical properties as ice. a. How important is the cardboard skin to the thermal response of the juice? Justify your answer quantitatively. b. If Ivan finds the can in the refrigerator 30 minutes after putting it in, will the juice have begun to melt? 5.48 A cleaning crew accidentally switches off the heating system in a warehouse one Friday night during the winter, just ahead of the holidays. When the staff return two weeks later, the warehouse is quite cold. In some sections, moisture that con- Problems 263 densed has formed a layer of ice 1 to 2 mm thick on the concrete floor. The concrete floor is 25 cm thick and sits on compacted earth. Both the slab and the ground below it are now at 20◦ F. The building operator turns on the heating system, quickly warming the air to 60◦ F. If the heat transfer coefficient between the air and the floor is 15 W/m2 K, how long will it take for the ice to start melting? Take αconcr = 7.0 × 10−7 m2 /s and kconcr = 1.4 W/m·K, and make justifiable approximations as appropriate. 5.49 A thick wooden wall, initially at 25◦ C, is made of fir. It is suddenly exposed to flames at 800◦ C. If the effective heat transfer coefficient for convection and radiation between the wall and the flames is 80 W/m2 K, how long will it take the wooden wall to reach its ignition temperature of 430◦ C? 5.50 Cold butter does not spread as well as warm butter. A small tub of whipped butter bears a label suggesting that, before use, it be allowed to warm up in room air for 30 minutes after being removed from the refrigerator. The tub has a diameter of 9.1 cm with a height of 5.6 cm, and the properties of whipped butter are: k = 0.125 W/m·K, cp = 2520 J/kg·K, and ρ = 620 kg/m3 . Assume that the tub’s cardboard walls offer negligible thermal resistance, that h = 10 W/m2 K outside the tub. Negligible heat is gained through the low conductivity lip around the bottom of the tub. If the refrigerator temperature was 5◦ C and the tub has warmed for 30 minutes in a room at 20◦ C, find: the temperature in the center of the butter tub, the temperature around the edge of the top surface of the butter, and the total energy (in J) absorbed by the butter tub. 5.51 A two-dimensional, 90◦ annular sector has an adiabatic inner arc, r = ri , and an adiabatic outer arc, r = ro . The flat surface along θ = 0 is isothermal at T1 , and the flat surface along θ = π /2 is isothermal at T2 . Show that the shape factor is S = (2/π ) ln(ro /ri ). 5.52 Suppose that T∞ (t) is the time-dependent environmental temperature surrounding a convectively-cooled, lumped object. Chapter 5: Transient and multidimensional heat conduction 264 a. Show that eqn. (1.20) leads to d dT∞ (T − T∞ ) =− (T − T∞ ) + dt T dt where the time constant T is defined as usual. b. If the initial temperature of the object is Ti , use either an integrating factor or a Laplace transform to show that T (t) is T (t) = T∞ (t)+[Ti − T∞ (0)] e−t/τ −e−t/τ t es/τ 0 d T∞ (s) ds. ds 5.53 Use the result of Problem 5.52 to verify eqn. (5.13). 5.54 Suppose that a thermocouple with an initial temperature Ti is placed into an airflow for which its Bi 1 and its time constant is T . Suppose also that the temperature of the airflow varies harmonically as T∞ (t) = Ti + ∆T cos (ωt). a. Use the result of Problem 5.52 to find the temperature of the thermocouple, Ttc (t), for t > 0. (If you wish, note that the real part of eiωt is Re eiωt = cos ωt and use complex variables to do the integration.) b. Approximate your result for t T . Then determine the 1 and for ωT 1. Explain value of Ttc (t) for ωT in physical terms the relevance of these limits to the frequency response of the thermocouple. c. If the thermocouple has a time constant of T = 0.1 sec, estimate the highest frequency temperature variation that it will measure accurately. 5.55 A particular tungsten lamp filament has a diameter of 100 µm and sits inside a glass bulb filled with inert gas. The effective heat transfer coefficient for conduction and radiation is 750 W/m·K and the electrical current is at 60 Hz. How much does the filament’s surface temperature fluctuate if the gas temperature is 200◦ C and the average wire temperature is 2900◦ C? 5.56 The consider the parameter ψ in eqn. (5.41). a. If the timescale for heat to diffuse a distance δ is δ2 /α, explain the physical significance of ψ and the consequence of large or small values of ψ. References 265 b. Show that the timescale for the thermal response of a wire with Bi 1 is ρcp δ/(2h). Then explain the meaning of the new parameter φ = ρcp ωδ/(4π h). c. When Bi 1, is φ or ψ a more relevant parameter? References [5.1] H. D. Baehr and K. Stephan. Heat and Mass Transfer. SpringerVerlag, Berlin, 1998. [5.2] A. F. Mills. Basic Heat and Mass Transfer. Prentice-Hall, Inc., Upper Saddle River, NJ, 2nd edition, 1999. [5.3] L. M. K. Boelter, V. H. Cherry, H. A. Johnson, and R. C. Martinelli. Heat Transfer Notes. McGraw-Hill Book Company, New York, 1965. [5.4] M. P. Heisler. Temperature charts for induction and constant temperature heating. Trans. ASME, 69:227–236, 1947. [5.5] P. J. Schneider. Temperature Response Charts. John Wiley & Sons, Inc., New York, 1963. [5.6] H. S. Carslaw and J. C. Jaeger. Conduction of Heat in Solids. Oxford University Press, New York, 2nd edition, 1959. [5.7] F. A. Jeglic. An analytical determination of temperature oscillations in wall heated by alternating current. NASA TN D-1286, July 1962. [5.8] F. A. Jeglic, K. A. Switzer, and J. H. Lienhard. Surface temperature oscillations of electric resistance heaters supplied with alternating current. J. Heat Transfer, 102(2):392–393, 1980. [5.9] J. Bronowski. The Ascent of Man. Chapter 4. Little, Brown and Company, Boston, 1973. [5.10] N. Zuber. Hydrodynamic aspects of boiling heat transfer. AEC Report AECU-4439, Physics and Mathematics, June 1959. [5.11] M. S. Plesset and S. A. Zwick. The growth of vapor bubbles in superheated liquids. J. Appl. Phys., 25:493–500, 1954. 266 Chapter 5: Transient and multidimensional heat conduction [5.12] L. E. Scriven. On the dynamics of phase growth. Chem. Eng. Sci., 10:1–13, 1959. [5.13] P. Dergarabedian. The rate of growth of bubbles in superheated water. J. Appl. Mech., Trans. ASME, 75:537, 1953. [5.14] E. R. G. Eckert and R. M. Drake, Jr. Analysis of Heat and Mass Transfer. Hemisphere Publishing Corp., Washington, D.C., 1987. [5.15] V. S. Arpaci. Conduction Heat Transfer. Ginn Press/Pearson Custom Publishing, Needham Heights, Mass., 1991. [5.16] E. Hahne and U. Grigull. Formfactor and formwiderstand der stationären mehrdimensionalen wärmeleitung. Int. J. Heat Mass Transfer, 18:751–767, 1975. [5.17] P. M. Morse and H. Feshbach. Methods of Theoretical Physics. McGraw-Hill Book Company, New York, 1953. [5.18] R. Rüdenberg. Die ausbreitung der luft—und erdfelder um hochspannungsleitungen besonders bei erd—und kurzschlüssen. Electrotech. Z., 36:1342–1346, 1925. [5.19] M. M. Yovanovich. Conduction and thermal contact resistances (conductances). In W. M. Rohsenow, J. P. Hartnett, and Y. I. Cho, editors, Handbook of Heat Transfer, chapter 3. McGraw-Hill, New York, 3rd edition, 1998. [5.20] S. H. Corriher. Cookwise: the hows and whys of successful cooking. Wm. Morrow and Company, New York, 1997. Includes excellent desciptions of the physical and chemical processes of cooking. The cookbook for those who enjoyed freshman chemistry. Part III Convective Heat Transfer 267 6. Laminar and turbulent boundary layers In cold weather, if the air is calm, we are not so much chilled as when there is wind along with the cold; for in calm weather, our clothes and the air entangled in them receive heat from our bodies; this heat. . .brings them nearer than the surrounding air to the temperature of our skin. But in windy weather, this heat is prevented. . .from accumulating; the cold air, by its impulse. . .both cools our clothes faster and carries away the warm air that was entangled in them. notes on “The General Effects of Heat”, Joseph Black, c. 1790s 6.1 Some introductory ideas Joseph Black’s perception about forced convection (above) represents a very correct understanding of the way forced convective cooling works. When cold air moves past a warm body, it constantly sweeps away warm air that has become, as Black put it, “entangled” with the body and replaces it with cold air. In this chapter we learn to form analytical descriptions of these convective heating (or cooling) processes. Our aim is to predict h and h, and it is clear that such predictions must begin in the motion of fluid around the bodies that they heat or cool. It is by predicting such motion that we will be able to find out how much heat is removed during the replacement of hot fluid with cold, and vice versa. Flow boundary layer Fluids flowing past solid bodies adhere to them, so a region of variable velocity must be built up between the body and the free fluid stream, as 269 Laminar and turbulent boundary layers 270 §6.1 Figure 6.1 A boundary layer of thickness δ. indicated in Fig. 6.1. This region is called a boundary layer, which we will often abbreviate as b.l. The b.l. has a thickness, δ. The boundary layer thickness is arbitrarily defined as the distance from the wall at which the flow velocity approaches to within 1% of u∞ . The boundary layer is normally very thin in comparison with the dimensions of the body immersed in the flow.1 The first step that has to be taken before h can be predicted is the mathematical description of the boundary layer. This description was first made by Prandtl2 (see Fig. 6.2) and his students, starting in 1904, and it depended upon simplifications that followed after he recognized how thin the layer must be. The dimensional functional equation for the boundary layer thickness on a flat surface is δ = fn(u∞ , ρ, µ, x) where x is the length along the surface and ρ and µ are the fluid density in kg/m3 and the dynamic viscosity in kg/m·s. We have five variables in 1 We qualify this remark when we treat the b.l. quantitatively. Prandtl was educated at the Technical University in Munich and finished his doctorate there in 1900. He was given a chair in a new fluid mechanics institute at Göttingen University in 1904—the same year that he presented his historic paper explaining the boundary layer. His work at Göttingen, during the period up to Hitler’s regime, set the course of modern fluid mechanics and aerodynamics and laid the foundations for the analysis of heat convection. 2 Some introductory ideas §6.1 271 Figure 6.2 Ludwig Prandtl (1875–1953). (Courtesy of Appl. Mech. Rev. [6.1]) kg, m, and s, so we anticipate two pi-groups: δ = fn(Rex ) x Rex ≡ u∞ x ρu∞ x = µ ν (6.1) where ν is the kinematic viscosity µ/ρ and Rex is called the Reynolds number. It characterizes the relative influences of inertial and viscous forces in a fluid problem. The subscript on Re—x in this case—tells what length it is based upon. We discover shortly that the actual form of eqn. (6.1) for a flat surface, where u∞ remains constant, is 4.92 δ = x Rex (6.2) which means that if the velocity is great or the viscosity is low, δ/x will be relatively small. Heat transfer will be relatively high in such cases. If the velocity is low, the b.l. will be relatively thick. A good deal of nearly 272 Laminar and turbulent boundary layers §6.1 Osborne Reynolds (1842 to 1912) Reynolds was born in Ireland but he taught at the University of Manchester. He was a significant contributor to the subject of fluid mechanics in the late 19th C. His original laminar-toturbulent flow transition experiment, pictured below, was still being used as a student experiment at the University of Manchester in the 1970s. Figure 6.3 Osborne Reynolds and his laminar–turbulent flow transition experiment. (Detail from a portrait at the University of Manchester.) stagnant fluid will accumulate near the surface and be “entangled” with the body, although in a different way than Black envisioned it to be. The Reynolds number is named after Osborne Reynolds (see Fig. 6.3), who discovered the laminar–turbulent transition during fluid flow in a tube. He injected ink into a steady and undisturbed flow of water and found that, beyond a certain average velocity, uav , the liquid streamline marked with ink would become wobbly and then break up into increasingly disorderly eddies, and it would finally be completely mixed into the Some introductory ideas §6.1 273 Figure 6.4 Boundary layer on a long, flat surface with a sharp leading edge. water, as is suggested in the sketch. To define the transition, we first note that (uav )crit , the transitional value of the average velocity, must depend on the pipe diameter, D , on µ , and on ρ —four variables in kg, m, and s. There is therefore only one pi-group: Recritical ≡ ρD(uav )crit µ (6.3) The maximum Reynolds number for which fully developed laminar flow in a pipe will always be stable, regardless of the level of background noise, is 2100. In a reasonably careful experiment, laminar flow can be made to persist up to Re = 10, 000. With enormous care it can be increased still another order of magnitude. But the value below which the flow will always be laminar—the critical value of Re—is 2100. Much the same sort of thing happens in a boundary layer. Figure 6.4 shows fluid flowing over a plate with a sharp leading edge. The flow is laminar up to a transitional Reynolds number based on x : Rexcritical = u∞ xcrit ν (6.4) At larger values of x the b.l. exhibits sporadic vortexlike instabilities over a fairly long range, and it finally settles into a fully turbulent b.l. 274 Laminar and turbulent boundary layers §6.1 For the boundary layer shown, Rexcritical = 3.5 × 105 , but in general the critical Reynolds number depends strongly on the amount of turbulence in the freestream flow over the plate, the precise shape of the leading edge, the roughness of the wall, and the presence of acoustic or structural vibrations [6.2, §5.5]. On a flat plate, a boundary layer will remain laminar even when such disturbances are very large if Rex ≤ 6 × 104 . With relatively undisturbed conditions, transition occurs for Rex in the range of 3 × 105 to 5 × 105 , and in very careful laboratory experiments, turbulent transition can be delayed until Rex ≈ 3 × 106 or so. Turbulent transition is essentially always complete before Rex = 4 × 106 and usually much earlier. These specifications of the critical Re are restricted to flat surfaces. If the surface is curved away from the flow, as shown in Fig. 6.1, turbulence might be triggered at much lower values of Rex . Thermal boundary layer If the wall is at a temperature Tw , different from that of the free stream, T∞ , there is a thermal boundary layer thickness, δt —different from the flow b.l. thickness, δ. A thermal b.l. is pictured in Fig. 6.5. Now, with reference to this picture, we equate the heat conducted away from the wall by the fluid to the same heat transfer expressed in terms of a convective heat transfer coefficient: −k f ∂T ∂y y =0 = h(Tw − T∞ ) (6.5) conduction into the fluid where kf is the conductivity of the fluid. Notice two things about this result. In the first place, it is correct to express heat removal at the wall using Fourier’s law of conduction, because there is no fluid motion in the direction of q. The other point is that while eqn. (6.5) looks like a b.c. of the third kind, it is not. This condition defines h within the fluid instead of specifying it as known information on the boundary. Equation (6.5) can be arranged in the form ∂ Tw − T Tw − T∞ ∂ (y/L) = y /L=0 hL = NuL , the Nusselt number kf (6.5a) §6.1 Some introductory ideas 275 Figure 6.5 The thermal boundary layer during the flow of cool fluid over a warm plate. where L is a characteristic dimension of the body under consideration— the length of a plate, the diameter of a cylinder, or [if we write eqn. (6.5) at a point of interest along a flat surface] Nux ≡ hx/kf . From Fig. 6.5 we see immediately that the physical significance of Nu is given by NuL = L δt (6.6) In other words, the Nusselt number is inversely proportional to the thickness of the thermal b.l. The Nusselt number is named after Wilhelm Nusselt,3 whose work on convective heat transfer was as fundamental as Prandtl’s was in analyzing the related fluid dynamics (see Fig. 6.6). We now turn to the detailed evaluation of h. And, as the preceding remarks make very clear, this evaluation will have to start with a development of the flow field in the boundary layer. 3 Nusselt finished his doctorate in mechanical engineering at the Technical University in Munich in 1907. During an indefinite teaching appointment at Dresden (1913 to 1917) he made two of his most important contributions: He did the dimensional analysis of heat convection before he had access to Buckingham and Rayleigh’s work. In so doing, he showed how to generalize limited data, and he set the pattern of subsequent analysis. He also showed how to predict convective heat transfer during film condensation. After moving about Germany and Switzerland from 1907 until 1925, he was named to the important Chair of Theoretical Mechanics at Munich. During his early years in this post, he made seminal contributions to heat exchanger design methodology. He held this position until 1952, during which time his, and Germany’s, great influence in heat transfer and fluid mechanics waned. He was succeeded in the chair by another of Germany’s heat transfer luminaries, Ernst Schmidt. 276 Laminar and turbulent boundary layers §6.2 Figure 6.6 Ernst Kraft Wilhelm Nusselt (1882–1957). This photograph, provided by his student, G. Lück, shows Nusselt at the Kesselberg waterfall in 1912. He was an avid mountain climber. 6.2 Laminar incompressible boundary layer on a flat surface We predict the boundary layer flow field by solving the equations that express conservation of mass and momentum in the b.l. Thus, the first order of business is to develop these equations. Conservation of mass—The continuity equation A two- or three-dimensional velocity field can be expressed in vectorial form: u = iu + jv + kw where u, v , and w are the x , y , and z components of velocity. Figure 6.7 shows a two-dimensional velocity flow field. If the flow is steady, the paths of individual particles appear as steady streamlines. The streamlines can be expressed in terms of a stream function, ψ(x, y) = constant, where each value of the constant identifies a separate streamline, as shown in the figure. The velocity, u, is directed along the streamlines so that no flow can cross them. Any pair of adjacent streamlines thus resembles a heat flow §6.2 Laminar incompressible boundary layer on a flat surface Figure 6.7 A steady, incompressible, two-dimensional flow field represented by streamlines, or lines of constant ψ. channel in a flux plot (Section 5.7); such channels are adiabatic—no heat flow can cross them. Therefore, we write the equation for the conservation of mass by summing the inflow and outflow of mass on two faces of a triangular element of unit depth, as shown in Fig. 6.7: ρv dx − ρu dy = 0 (6.7) If the fluid is incompressible, so that ρ = constant along each streamline, then −v dx + u dy = 0 (6.8) But we can also differentiate the stream function along any streamline, ψ(x, y) = constant, in Fig. 6.7: dψ = ∂ψ ∂x y dx + ∂ψ ∂y dy = 0 (6.9) x If we compare eqns. (6.8) and (6.9), we immediately see that the coefficients of dx and dy must be the same, so v=− ∂ψ ∂x y and u = ∂ψ ∂y (6.10) x 277 278 Laminar and turbulent boundary layers §6.2 Furthermore, ∂2ψ ∂2ψ = ∂y∂x ∂x∂y so it follows that ∂v ∂u + =0 ∂x ∂y (6.11) This is called the two-dimensional continuity equation for incompressible flow, because it expresses mathematically the fact that the flow is continuous ; it has no breaks in it. In three dimensions, the continuity equation for an incompressible fluid is ∇·u= ∂v ∂w ∂u + + =0 ∂x ∂y ∂z Example 6.1 Fluid moves with a uniform velocity, u∞ , in the x -direction. Find the stream function and see if it gives plausible behavior (see Fig. 6.8). Solution. u = u∞ and v = 0. Therefore, from eqns. (6.10) u∞ = ∂ψ ∂y and x 0= ∂ψ ∂x y Integrating these equations, we get ψ = u∞ y + fn(x) and ψ = 0 + fn(y) Comparing these equations, we get fn(x) = constant and fn(y) = u∞ y + constant, so ψ = u∞ y + constant This gives a series of equally spaced, horizontal streamlines, as we would expect (see Fig. 6.8). We set the arbitrary constant equal to zero in the figure. §6.2 Laminar incompressible boundary layer on a flat surface Figure 6.8 Streamlines in a uniform horizontal flow field, ψ = u∞ y . Conservation of momentum The momentum equation in a viscous flow is a complicated vectorial expression called the Navier-Stokes equation. Its derivation is carried out in any advanced fluid mechanics text (see, e.g., [6.3, Chap. III]). We shall offer a very restrictive derivation of the equation—one that applies only to a two-dimensional incompressible b.l. flow, as shown in Fig. 6.9. Here we see that shear stresses act upon any element such as to continuously distort and rotate it. In the lower part of the figure, one such element is enlarged, so we can see the horizontal shear stresses4 and the pressure forces that act upon it. They are shown as heavy arrows. We also display, as lighter arrows, the momentum fluxes entering and leaving the element. Notice that both x - and y -directed momentum enters and leaves the element. To understand this, one can envision a boxcar moving down the railroad track with a man standing, facing its open door. A child standing at a crossing throws him a baseball as the car passes. When he catches the ball, its momentum will push him back, but a component of momentum will also jar him toward the rear of the train, because of the relative motion. Particles of fluid entering element A will likewise influence its motion, with their x components of momentum carried into the element by both components of flow. The velocities must adjust themselves to satisfy the principle of conservation of linear momentum. Thus, we require that the sum of the external forces in the x -direction, which act on the control volume, A, must be balanced by the rate at which the control volume, A, forces x 4 The stress, τ , is often given two subscripts. The first one identifies the direction normal to the plane on which it acts, and the second one identifies the line along which it acts. Thus, if both subscripts are the same, the stress must act normal to a surface—it must be a pressure or tension instead of a shear stress. 279 280 Laminar and turbulent boundary layers §6.2 Figure 6.9 Forces acting in a two-dimensional incompressible boundary layer. directed momentum out. The external forces, shown in Fig. 6.9, are τyx + ∂τyx ∂p dy dx − τyx dx + p dy − p + dx ∂y ∂x = dy ∂ τyx ∂p − ∂y ∂x dx dy The rate at which A loses x -directed momentum to its surroundings is ρ u2 + ∂ρu2 dx ∂x dy − ρu2 dy + u(ρv) + − ρuv dx = ∂ρuv dy ∂y dx ∂ ρu2 ∂ρuv + ∂x ∂y dx dy Laminar incompressible boundary layer on a flat surface §6.2 We equate these results and obtain the basic statement of conservation of x -directed momentum for the b.l.: ∂τyx dp dy dx − dx dy = ∂y dx ∂ ρu2 ∂ρuv + ∂x ∂y dx dy The shear stress in this result can be eliminated with the help of Newton’s law of viscous shear: τyx = µ ∂u ∂y so the momentum equation becomes ∂ ∂y µ ∂u ∂y − dp = dx ∂ ρu2 ∂ρuv + ∂x ∂y Finally, we remember that the analysis is limited to ρ constant, and we limit use of the equation to temperature ranges in which µ constant. Then ∂uv 1 dp ∂2u ∂u2 + =− +ν ∂x ∂y ρ dx ∂y 2 (6.12) This is one form of the steady, two-dimensional, incompressible boundary layer momentum equation. Although we have taken ρ constant, a more complete derivation reveals that the result is valid for compressible flow as well. If we multiply eqn. (6.11) by u and subtract the result from the left-hand side of eqn. (6.12), we obtain a second form of the momentum equation: u ∂u ∂u 1 dp ∂2u +v =− +ν ∂x ∂y ρ dx ∂y 2 (6.13) Equation (6.13) has a number of so-called boundary layer approximations built into it: • ∂ u/∂x is generally • v is generally • p ≠ fn(y) u. ∂ u/∂y . 281 Laminar and turbulent boundary layers 282 §6.2 The Bernoulli equation for the free stream flow just above the boundary layer where there is no viscous shear, u2 p + ∞ = constant ρ 2 can be differentiated and used to eliminate the pressure gradient, 1 dp du∞ = −u∞ ρ dx dx so from eqn. (6.12): ∂(uv) ∂2u du∞ ∂u2 + = u∞ +ν ∂x ∂y dx ∂y 2 (6.14) And if there is no pressure gradient in the flow—if p and u∞ are constant as they would be for flow past a flat plate—then eqns. (6.12), (6.13), and (6.14) become ∂(uv) ∂u ∂u ∂2u ∂u2 + =u +v =ν ∂x ∂y ∂x ∂y ∂y 2 (6.15) Predicting the velocity profile in the laminar boundary layer without a pressure gradient Exact solution. Two strategies for solving eqn. (6.15) for the velocity profile have long been widely used. The first was developed by Prandtl’s student, H. Blasius,5 before World War I. It is exact, and we shall sketch it only briefly. First we introduce the stream function, ψ, into eqn. (6.15). This reduces the number of dependent variables from two (u and v ) to just one—namely, ψ. We do this by substituting eqns. (6.10) in eqn. (6.15): ∂ψ ∂ 2 ψ ∂3ψ ∂ψ ∂ 2 ψ − =ν ∂y ∂y∂x ∂x ∂y 2 ∂y 3 (6.16) It turns out that eqn. (6.16) can be converted into an ordinary d.e. with the following change of variables: ψ(x, y) ≡ 5 √ u∞ νx f (η) where η ≡ u∞ y νx (6.17) Blasius achieved great fame for many accomplishments in fluid mechanics and then gave it up. He is quoted as saying: “I decided that I had no gift for it; all of my ideas came from Prandtl.” §6.2 Laminar incompressible boundary layer on a flat surface where f (η) is an as-yet-undertermined function. [This transformation is rather similar to the one that we used to make an ordinary d.e. of the heat conduction equation, between eqns. (5.44) and (5.45).] After some manipulation of partial derivatives, this substitution gives (Problem 6.2) f d3 f d2 f +2 =0 dη2 dη3 (6.18) and 1 v = 2 u∞ ν/x df u = u∞ dη η df −f dη (6.19) The boundary conditions for this flow are u(y = 0) = 0 or u(y = ∞) = u∞ v(y = 0) = 0 df dη η=0 df dη ⎫ ⎪ ⎪ =0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ =1 ⎪ ⎪ ⎪ ⎪ ⎪ η=∞ ⎪ ⎪ ⎪ or f (η = 0) = 0 ⎭ or (6.20) The solution of eqn. (6.18) subject to these b.c.’s must be done numerically. (See Problem 6.3.) The solution of the Blasius problem is listed in Table 6.1, and the dimensionless velocity components are plotted in Fig. 6.10. The u component increases from zero at the wall (η = 0) to 99% of u∞ at η = 4.92. Thus, the b.l. thickness is given by 4.92 = δ ν x/u∞ or, as we anticipated earlier [eqn. (6.2)], δ = x 4.92 4.92 = Rex u∞ x/ν Concept of similarity. The exact solution for u(x, y) reveals a most useful fact—namely, that u can be expressed as a function of a single variable, η: u = f (η) = f u∞ y u∞ νx 283 284 Laminar and turbulent boundary layers §6.2 Table 6.1 Exact velocity profile in the boundary layer on a flat surface with no pressure gradient y u∞ /νx η 0.00 0.20 0.40 0.60 0.80 1.00 2.00 3.00 4.00 4.918 6.00 8.00 u u∞ f (η) f (η) 0.00000 0.00664 0.02656 0.05974 0.10611 0.16557 0.65003 1.39682 2.30576 3.20169 4.27964 6.27923 v x /νu∞ (ηf − f ) 2 f (η) 0.00000 0.06641 0.13277 0.19894 0.26471 0.32979 0.62977 0.84605 0.95552 0 .99000 0.99898 1.00000− 0.00000 0.00332 0.01322 0.02981 0.05283 0.08211 0.30476 0.57067 0.75816 0.83344 0.85712 0.86039 0.33206 0.33199 0.33147 0.33008 0.32739 0.32301 0.26675 0.16136 0.06424 0.01837 0.00240 0.00001 This is called a similarity solution. To see why, we solve eqn. (6.2) for 4.92 u∞ = νx δ(x) and substitute this in f (y u∞ /νx). The result is f= u y = fn u∞ δ(x) (6.21) The velocity profile thus has the same shape with respect to the b.l. thickness at each x -station. We say, in other words, that the profile is similar at each station. This is what we found to be true for conduction √ into a semi-infinite region. In that case [recall eqn. (5.51)], x/ t always had the same value at the outer limit of the thermally disturbed region. Boundary layer similarity makes it especially easy to use a simple approximate method for solving other b.l. problems. This method, called the momentum integral method, is the subject of the next subsection. Example 6.2 Air at 27◦ C blows over a flat surface with a sharp leading edge at 1 1.5 m/s. Find the b.l. thickness 2 m from the leading edge. Check the b.l. assumption that u v at the trailing edge. Laminar incompressible boundary layer on a flat surface §6.2 Figure 6.10 The dimensionless velocity components in a laminar boundary layer. Solution. The dynamic and kinematic viscosities are µ = 1.853 × 10−5 kg/m·s and ν = 1.566 × 10−5 m2 /s. Then Rex = 1.5(0.5) u∞ x = = 47, 893 ν 1.566 × 10−5 The Reynolds number is low enough to permit the use of a laminar flow analysis. Then δ= 4.92x 4.92(0.5) = 0.01124 = 1.124 cm = Rex 47, 893 (Remember that the b.l. analysis is only valid if δ/x 1. In this case, δ/x = 1.124/50 = 0.0225.) From Fig. 6.10 or Table 6.1, we observe that v/u is greatest beyond the outside edge of the b.l, at large η. Using data from Table 6.1 at η = 8, v at x = 0.5 m is v= 0.8604 = 0.8604 x /νu∞ (1.566)(10−5 )(1.5) (0.5) = 0.00590 m/s 285 Laminar and turbulent boundary layers 286 §6.2 or, since u/u∞ → 1 at large η v 0.00590 v = = 0.00393 = u u∞ 1.5 Since v grows larger as x grows smaller, the condition v u is not satisfied very near the leading edge. There, the b.l. approximations themselves break down. We say more about this breakdown after eqn. (6.34). Momentum integral method.6 A second method for solving the b.l. momentum equation is approximate and much easier to apply to a wide range of problems than is any exact method of solution. The idea is this: We are not really interested in the details of the velocity or temperature profiles in the b.l., beyond learning their slopes at the wall. [These slopes give us the shear stress at the wall, τw = µ(∂u/∂y)y =0 , and the heat flux at the wall, qw = −k(∂T /∂y)y =0 .] Therefore, we integrate the b.l. equations from the wall, y = 0, to the b.l. thickness, y = δ, to make ordinary d.e.’s of them. It turns out that while these much simpler equations do not reveal anything new about the temperature and velocity profiles, they do give quite accurate explicit equations for τw and qw . Let us see how this procedure works with the b.l. momentum equation. We integrate eqn. (6.15), as follows, for the case in which there is no pressure gradient (dp/dx = 0): δ 0 ∂u2 dy + ∂x δ 0 ∂(uv) dy = ν ∂y δ 0 ∂2u dy ∂y 2 At y = δ, u can be approximated as the free stream value, u∞ , and other quantities can also be evaluated at y = δ just as though y were infinite: ⎡ ⎤ δ 0 ∂u2 ⎢ dy + (uv)y =δ − (uv)y =0 = ν ⎣ ∂x =u∞ v∞ ∂u ∂y − y =δ =0 ∂u ∂y ⎥ ⎦ y =0 0 (6.22) The continuity equation (6.11) can be integrated thus: δ v∞ − vy =0 = − 0 ∂u dy ∂x (6.23) =0 6 This method was developed by Pohlhausen, von Kármán, and others. See the discussion in [6.3, Chap. XII]. Laminar incompressible boundary layer on a flat surface §6.2 Multiplying this by u∞ gives δ u∞ v∞ = − 0 ∂uu∞ dy ∂x Using this result in eqn. (6.22), we obtain δ 0 ∂ ∂u [u(u − u∞ )] dy = −ν ∂x ∂y y =0 Finally, we note that µ(∂u/∂y)y =0 is the shear stress on the wall, τw = τw (x only), so this becomes7 d dx δ(x) 0 u(u − u∞ ) dy = − τw ρ (6.24) Equation (6.24) expresses the conservation of linear momentum in integrated form. It shows that the rate of momentum loss caused by the b.l. is balanced by the shear force on the wall. When we use it in place of eqn. (6.15), we are said to be using an integral method. To make use of eqn. (6.24), we first nondimensionalize it as follows: d dx 1 δ 0 u u∞ y u −1 d u∞ δ =− ν ∂(u/u∞ ) u∞ δ ∂(y/δ) y =0 τw (x) 1 =− 2 ≡ − 2 Cf (x) ρu∞ (6.25) where τw /(ρu2 /2) is defined as the skin friction coefficient, Cf . ∞ Equation (6.25) will be satisfied precisely by the exact solution (Problem 6.4) for u/u∞ . However, the point is to use eqn. (6.25) to determine u/u∞ when we do not already have an exact solution. To do this, we recall that the exact solution exhibits similarity. First, we guess the solution in the form of eqn. (6.21): u/u∞ = fn(y/δ). This guess is made in such a way that it will fit the following four things that are true of the velocity profile: ⎫ ⎪ • u/u∞ = 0 at y/δ = 0 ⎪ ⎪ ⎪ ⎬ • u/u∞ 1 at y/δ = 1 (6.26) ⎪ ⎪ u y ⎪ •d 0 at y/δ = 1 ⎪ d ⎭ u∞ δ 7 The interchange of integration and differentiation is consistent with Leibnitz’s rule for differentiation of an integral (Problem 6.14). 287 Laminar and turbulent boundary layers 288 • §6.2 and from eqn. (6.15), we know that at y/δ = 0: u ∂u ∂u ∂2u =ν +v ∂x ∂y ∂y 2 =0 y =0 =0 so ∂ 2 (u/u∞ ) ∂(y/δ)2 =0 (6.27) y /δ=0 If fn(y/δ) is written as a polynomial with four constants—a, b, c , and d—in it, y y u =a+b +c u∞ δ δ 2 +d y δ 3 (6.28) the four things that are known about the profile give • 0 = a, which eliminates a immediately • 1=0+b+c+d • 0 = b + 2c + 3d • 0 = 2c , which eliminates c as well Solving the middle two equations (above) for b and d, we obtain d = − 1 2 3 and b = + 2 , so 3y 1 u = − u∞ 2δ 2 y δ 3 (6.29) This approximate velocity profile is compared with the exact Blasius profile in Fig. 6.11, and they prove to be equal within a maximum error of 8%. The only remaining problem is then that of calculating δ(x). To do this, we substitute eqn. (6.29) in eqn. (6.25) and get, after integration (see Problem 6.5): − d 39 δ dx 280 =− ν u∞ δ 3 2 or − 39 280 2 3 1 2 dδ2 ν =− dx u∞ (6.30) §6.2 Laminar incompressible boundary layer on a flat surface Figure 6.11 Comparison of the third-degree polynomial fit with the exact b.l. velocity profile. (Notice that the approximate result has been forced to u/u∞ = 1 instead of 0.99 at y = δ.) We integrate this using the b.c. δ2 = 0 at x = 0: δ2 = 280 νx 13 u∞ (6.31a) or δ 4.64 = x Rex (6.31b) This b.l. thickness is of the correct functional form, and the constant is low by only 5.6%. The skin friction coefficient The fact that the function f (η) gives all information about flow in the b.l. must be stressed. For example, the shear stress can be obtained from it 289 Laminar and turbulent boundary layers 290 §6.2 by using Newton’s law of viscous shear: τw =µ ∂u ∂y =µ y =0 ∂ u∞ f ∂y √ u∞ d2 f =µu∞ √ νx dη2 y =0 = µu∞ df ∂η dη ∂y y =0 η=0 But from Fig. 6.10 and Table 6.1, we see that (d2 f /dη2 )η=0 = 0.33206, so µu∞ Rex (6.32) τw = 0.332 x The integral method that we just outlined would have given 0.323 for the constant in eqn. (6.32) instead of 0.332 (Problem 6.6). The local skin friction coefficient, or local skin drag coefficient, is defined as Cf ≡ 0.664 τw = 2 Rex ρu∞ /2 (6.33) The overall skin friction coefficient, C f , is based on the average of the shear stress, τw , over the length, L, of the plate ⌠L ⌠L u2 ρu2 ⎮ 1⎮ 0.664 ν ∞ ∞ ⌡ dx = 1.328 τ w = ⌡ τw dx = L0 2L 0 u∞ x/ν 2 u∞ L so Cf = 1.328 ReL (6.34) As a matter of interest, we note that Cf (x) approaches infinity at the leading edge of the flat surface. This means that to stop the fluid that first touches the front of the plate—dead in its tracks—would require infinite shear stress right at that point. Nature, of course, will not allow such a thing to happen; and it turns out that the boundary layer analysis is not really valid right at the leading edge. In fact, the range x 5δ is too close to the edge to use this analysis with accuracy because the b.l. is relatively thick and v is no longer u. With eqn. (6.2), this converts to x > 600 ν/u∞ for a boundary layer to exist Laminar incompressible boundary layer on a flat surface §6.2 or simply Rex 600. In Example 6.2, this condition is satisfied for all x ’s greater than about 6 mm. This region is usually very small. Example 6.3 Calculate the average shear stress and the overall friction coefficient for the surface in Example 6.2 if its total length is L = 0.5 m. Compare τ w with τw at the trailing edge. At what point on the surface does τw = τ w ? Finally, estimate what fraction of the surface can legitimately be analyzed using boundary layer theory. Solution. Cf = 1.328 = Re0.5 1.328 = 0.00607 47, 893 and τw = ρu2 1.183(1.5)2 ∞ 0.00607 = 0.00808 kg/m·s2 Cf = 2 2 N/m2 (This is very little drag. It amounts only to about 1/50 ounce/m2 .) At x = L, τw (x) τw x =L = ρu2 /2 ∞ ρu2 /2 ∞ 0.664 1.328 ReL ReL = 1 2 and τw (x) = τ w where 0.664 1.328 √ =√ x 0.5 so the local shear stress equals the average value, where x= 1 8 m or 1 x = L 4 Thus, the shear stress, which is initially infinite, plummets to τ w onefourth of the way from the leading edge and drops only to one-half of τ w in the remaining 75% of the plate. The boundary layer assumptions fail when x < 600 1.566 × 10−5 ν = 600 = 0.0063 m u∞ 1.5 Thus, the preceding analysis should be good over almost 99% of the 0.5 m length of the surface. 291 292 Laminar and turbulent boundary layers 6.3 §6.3 The energy equation Derivation We now know how fluid moves in the b.l. Next, we must extend the heat conduction equation to allow for the motion of the fluid. This equation can be solved for the temperature field in the b.l., and its solution can be used to calculate h, using Fourier’s law: h= q k ∂T =− Tw − T∞ Tw − T∞ ∂y (6.35) y =0 To predict T , we extend the analysis done in Section 2.1. Figure 2.4 shows a volume containing a solid subjected to a temperature field. We now allow this volume to contain fluid with a velocity field u(x, y, z) in it, as shown in Fig. 6.12. We make the following restrictive approximations: • Pressure variations in the flow are not large enough to affect thermodynamic properties. From thermodynamics, we know that the ˆ specific internal energy, u, is related to the specific enthalpy as ˆ ˆ ˆ ˆ h = u + p/ρ , and that dh = cp dT + (∂ h/∂p)T dp . We shall neglect the effect of dp on enthalpy, internal energy, and density. This approximation is reasonable for most liquid flows and for gas flows moving at speeds less than about 1/3 the speed of sound. • Under these conditions, density changes result only from temperature changes and will also be small; and the flow will behave as if incompressible. For such flows, ∇ · u = 0 (Sect. 6.2). • Temperature variations in the flow are not large enough to change k significantly. When we consider the flow field, we will also presume µ to be unaffected by temperature change. • Potential and kinetic energy changes are negligible in comparison to thermal energy changes. Since the kinetic energy of a fluid can change owing to pressure gradients, this again means that pressure variations may not be too large. • The viscous stresses do not dissipate enough energy to warm the fluid significantly. The energy equation §6.3 293 Figure 6.12 Control volume in a heat-flow and fluid-flow field. Just as we wrote eqn. (2.7) in Section 2.1, we now write conservation of energy in the form d dt R ˆ ρ u dR = − rate of internal energy increase in R S ˆ (ρ h) u · n dS rate of internal energy and flow work out of R − S (−k∇T ) · n dS + net heat conduction rate out of R ˙ q dR (6.36) R rate of heat generation in R In the third integral, u · n dS represents the volume flow rate through an element dS of the control surface. The position of R is not changing in time, so we can bring the time derivative inside the first integral. If we then we call in Gauss’s theorem [eqn. (2.8)] to make volume integrals of the surface integrals, eqn. (6.36) becomes R ˆ ∂ (ρ u) ˆ ˙ + ∇ · ρ uh − ∇ · k∇T − q dR = 0 ∂t Because the integrand must vanish identically (recall the footnote on pg. 55 in Chap. 2) and because k depends weakly on T , ˆ ∂(ρ u) ˆ ˙ + ∇ · ρ uh − k∇2 T − q = 0 ∂t ˆˆ = ρ u · ∇h + h∇ · (ρ u) 294 Laminar and turbulent boundary layers §6.3 Since we are neglecting pressure effects, we may introduce the following approximation: ˆ ˆ ˆˆ ˆ d(ρ u) = d(ρ h) − dp ≈ d(ρ h) = ρdh + h dρ Thus, collecting and rearranging terms ρ ˆ ∂h ˆ ˆ ∂ ρ + ∇ · ρu + u · ∇h + h ∂t ∂t ˙ = k∇2 T + q neglect The term involving density derivatives may be neglected on the basis that density changes are small and the flow is nearly incompressible (but see Problem 6.36 for a more general result). ˆ Upon substituting dh ≈ cp dT , we obtain our final result: ρcp ∂T ∂t + u · ∇T energy storage = enthalpy convection k∇2 T + heat conduction ˙ q (6.37) heat generation This is the energy equation for a constant pressure flow field. It is the same as the corresponding equation (2.11) for a solid body, except for the enthalpy transport, or convection, term, ρcp u · ∇T . Consider the term in parentheses in eqn. (6.37): ∂T ∂T ∂T ∂T ∂T DT + u · ∇T = +u +v +w ≡ ∂t ∂t ∂x ∂y ∂z Dt (6.38) DT /Dt is exactly the so-called material derivative, which is treated in some detail in every fluid mechanics course. DT /Dt is the rate of change of the temperature of a fluid particle as it moves in a flow field. In a steady two-dimensional flow field without heat sources, eqn. (6.37) takes the form u ∂T ∂T +v =α ∂x ∂y Furthermore, in a b.l., ∂ 2 T /∂x 2 u ∂2T ∂2T + 2 ∂x ∂y 2 (6.39) ∂ 2 T /∂y 2 , so the b.l. form is ∂T ∂2T ∂T +v =α ∂x ∂y ∂y 2 (6.40) The energy equation §6.3 295 Heat and momentum transfer analogy Consider a b.l. in a fluid of bulk temperature T∞ , flowing over a flat surface at temperature Tw . The momentum equation and its b.c.’s can be written as ⎧u ⎪ ⎪ =0 ⎪ ⎪ ⎪ u∞ y =0 ⎪ ⎪ ⎪ ⎨u ∂ ∂2 u u u ∂ =1 +v =ν u ⎪ u∞ y =∞ ∂x u∞ ∂y u∞ ∂y 2 u∞ ⎪ ⎪ ⎪ ⎪∂ ⎪ u ⎪ ⎪ =0 ⎩ ∂y u∞ y =∞ (6.41) And the energy equation (6.40) can be written in terms of a dimensionless temperature, Θ = (T − Tw )/(T∞ − Tw ), as ⎧ ⎪ Θ(y = 0) = 0 ⎪ ⎪ ⎪ ⎪ ⎨ 2Θ ∂Θ ∂ ∂Θ Θ(y = ∞) = 1 (6.42) +v =α u 2 ⎪ ∂Θ ∂x ∂y ∂y ⎪ ⎪ ⎪ ⎪ =0 ⎩ ∂y y =∞ Notice that the problems of predicting u/u∞ and Θ are identical, with one exception: eqn. (6.41) has ν in it whereas eqn. (6.42) has α. If ν and α should happen to be equal, the temperature distribution in the b.l. is for ν = α : T − Tw = f (η) T∞ − Tw derivative of the Blasius function since the two problems must have the same solution. In this case, we can immediately calculate the heat transfer coefficient using eqn. (6.5): h= k ∂(T − Tw ) T∞ − Tw ∂y =k y =0 ∂ f ∂η ∂ η ∂y η=0 but (∂ 2 f /∂η2 )η=0 = 0.33206 (see Fig. 6.10) and ∂η/∂y = u∞ /νx , so hx = Nux = 0.33206 Rex k for ν = α (6.43) Normally, in using eqn. (6.43) or any other forced convection equation, properties should be evaluated at the film temperature, Tf = (Tw +T∞ )/2. 296 Laminar and turbulent boundary layers §6.4 Example 6.4 Water flows over a flat heater, 0.06 m in length, at 15 atm pressure and 440 K. The free stream velocity is 2 m/s and the heater is held at 460 K. What is the average heat flux? Solution. At Tf = (460 + 440)/2 = 450 K: ν = 1.725 × 10−7 m2 /s α = 1.724 × 10−7 m2 /s Therefore, ν α, and we can use eqn. (6.43). First, we must calculate the average heat flux, q. To do this, we set ∆T ≡ Tw − T∞ and write q= 1 L L 0 (h∆T ) dx = ∆T L L 0 k∆T k Nux dx = 0.332 x L L 0 =2 u∞ dx νx √ u∞ L/ν so k ReL ∆T = 2qx =L L Note that the average heat flux is twice that at the trailing edge, x = L. Using k = 0.674 W/m·K for water at the film temperature, q = 2 0.332 q = 2(0.332) 0.674 0.06 2(0.06) (460 − 440) 1.72 × 10−7 = 124, 604 W/m2 = 125 kW/m2 Equation (6.43) is clearly a very restrictive heat transfer solution. We now want to find how to evaluate q when ν does not equal α. 6.4 The Prandtl number and the boundary layer thicknesses Dimensional analysis We must now look more closely at the implications of the similarity between the velocity and thermal boundary layers. We first ask what dimensional analysis reveals about heat transfer in the laminar b.l. We know by now that the dimensional functional equation for the heat transfer coefficient, h, should be h = fn(k, x, ρ, cp , µ, u∞ ) The Prandtl number and the boundary layer thicknesses §6.4 We have excluded Tw − T∞ on the basis of Newton’s original hypothesis, borne out in eqn. (6.43), that h ≠ fn(∆T ) during forced convection. This gives seven variables in J/K, m, kg, and s, or 7 − 4 = 3 pi-groups. Note that, as we indicated at the end of Section 4.3, there is no conversion between heat and work so it we should not regard J as N·m, but rather as a separate unit. The dimensionless groups are then: Π1 = hx ≡ Nux k Π2 = ρu∞ x ≡ Rex µ and a new group: Π3 = µcp ν ≡ ≡ Pr, Prandtl number k α Thus, Nux = fn(Rex , Pr) (6.44) in forced convection flow situations. Equation (6.43) was developed for the case in which ν = α or Pr = 1; therefore, it is of the same form as eqn. (6.44), although it does not display the Pr dependence of Nux . To better understand the physical meaning of the Prandtl number, let us briefly consider how to predict its value in a gas. Kinetic theory of µ and k Figure 6.13 shows a small neighborhood of a point of interest in a gas in which there exists a velocity or temperature gradient. We identify the mean free path of molecules between collisions as and indicate planes at y ± /2 which bracket the average travel of those molecules found at plane y . (Actually, these planes should be located closer to y ± for a variety of subtle reasons. This and other fine points of these arguments are explained in detail in [6.4].) The shear stress, τyx , can be expressed as the change of momentum of all molecules that pass through the y -plane of interest, per unit area: τyx = change in fluid mass flux of molecules · velocity from y − /2 to y + /2 The mass flux from top to bottom is proportional to ρ C , where C , the mean molecular speed of the stationary fluid, is u or v in incompressible flow. Thus, τyx = C1 ρ C du dy N du and this also equals µ 2 m dy (6.45) 297 Laminar and turbulent boundary layers 298 §6.4 Figure 6.13 Momentum and energy transfer in a gas with a velocity or temperature gradient. By the same token, qy = C2 ρ cv C dT dy and this also equals − k dT dy where cv is the specific heat at constant volume. The constants, C1 and C2 , are on the order of unity. It follows immediately that µ = C1 ρ C so ν = C1 C k = C2 ρ cv C so α = C2 and C γ where γ ≡ cp /cv is approximately a constant on the order of unity for a given gas. Thus, for a gas, Pr ≡ ν = a constant on the order of unity α More detailed use of the kinetic theory of gases reveals more specific information as to the value of the Prandtl number, and these points are borne out reasonably well experimentally, as you can determine from Appendix A: 2 • For simple monatomic gases, Pr = 3 . §6.4 The Prandtl number and the boundary layer thicknesses • For diatomic gases in which vibration is unexcited (such as N2 and 5 O2 at room temperature), Pr = 7 . • As the complexity of gas molecules increases, Pr approaches an upper value of unity. • Pr is most insensitive to temperature in gases made up of the simplest molecules because their structure is least responsive to temperature changes. In a liquid, the physical mechanisms of molecular momentum and energy transport are much more complicated and Pr can be far from unity. For example (cf. Table A.3): • For liquids composed of fairly simple molecules, excluding metals, Pr is of the order of magnitude of 1 to 10. • For liquid metals, Pr is of the order of magnitude of 10−2 or less. • If the molecular structure of a liquid is very complex, Pr might reach values on the order of 105 . This is true of oils made of long-chain hydrocarbons, for example. Thus, while Pr can vary over almost eight orders of magnitude in common fluids, it is still the result of analogous mechanisms of heat and momentum transfer. The numerical values of Pr, as well as the analogy itself, have their origins in the same basic process of molecular transport. Boundary layer thicknesses, δ and δt , and the Prandtl number We have seen that the exact solution of the b.l. equations gives δ = δt for Pr = 1, and it gives dimensionless velocity and temperature profiles that are identical on a flat surface. Two other things should be easy to see: • When Pr > 1, δ > δt , and when Pr < 1, δ < δt . This is true because high viscosity leads to a thick velocity b.l., and a high thermal diffusivity should give a thick thermal b.l. • Since the exact governing equations (6.41) and (6.42) are identical for either b.l., except for the appearance of α in one and ν in the other, we expect that ν δt = fn only δ α 299 Laminar and turbulent boundary layers 300 §6.5 Therefore, we can combine these two observations, defining δt /δ ≡ φ, and get φ = monotonically decreasing function of Pr only (6.46) The exact solution of the thermal b.l. equations proves this to be precisely true. The fact that φ is independent of x will greatly simplify the use of the integral method. We shall establish the correct form of eqn. (6.46) in the following section. 6.5 Heat transfer coefficient for laminar, incompressible flow over a flat surface The integral method for solving the energy equation Integrating the b.l. energy equation in the same way as the momentum equation gives δt u 0 ∂T dy + ∂x δt v 0 ∂T dy = α ∂y δt 0 ∂2T dy ∂y 2 And the chain rule of differentiation in the form xdy ≡ dxy − ydx , reduces this to δt 0 ∂uT dy − ∂x δt 0 δt ∂u dy + T ∂x 0 ∂vT dy − ∂y δt 0 δt ∂T ∂v dy = α T ∂y ∂y 0 or δt 0 ∂uT dy + ∂x δt vT 0 − δt ∂v ∂u + ∂x ∂y T 0 =T∞ v |y =δt −0 = 0, eqn. (6.11) dy ⎡ = α⎣ ⎤ ∂T ∂y − δt ∂T ∂y ⎦ 0 =0 We evaluate v at y = δt , using the continuity equation in the form of eqn. (6.23), in the preceeding expression: δt 0 1 ∂ u(T − T∞ ) dy = ∂x ρcp −k ∂T ∂y = fn(x only) 0 §6.5 Heat transfer coefficient for laminar, incompressible flow over a flat surface or d dx δt 0 u(T − T∞ ) dy = qw ρcp (6.47) Equation (6.47) expresses the conservation of thermal energy in integrated form. It shows that the rate thermal energy is carried away by the b.l. flow is matched by the rate heat is transferred in at the wall. Predicting the temperature distribution in the laminar thermal boundary layer We can continue to paraphrase the development of the velocity profile in the laminar b.l., from the preceding section. We previously guessed the velocity profile in such a way as to make it match what we know to be true. We also know certain things to be true of the temperature profile. The temperatures at the wall and at the outer edge of the b.l. are known. Furthermore, the temperature distribution should be smooth as it blends into T∞ for y > δt . This condition is imposed by setting dT /dy equal to zero at y = δt . A fourth condition is obtained by writing eqn. (6.40) at the wall, where u = v = 0. This gives (∂ 2 T /∂y 2 )y =0 = 0. These four conditions take the following dimensionless form: ⎫ T − T∞ ⎪ ⎪ = 1 at y/δt = 0⎪ ⎪ ⎪ Tw − T∞ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ T − T∞ ⎪ = 0 at y/δt = 1⎪ ⎪ ⎪ ⎬ Tw − T∞ (6.48) ⎪ d[(T − T∞ )/(Tw − T∞ )] ⎪ ⎪ = 0 at y/δt = 1⎪ ⎪ ⎪ d(y/δt ) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2 [(T − T )/(T − T )] ⎪ ⎪ ∂ ∞ w ∞ ⎪ = 0 at y/δt = 0⎪ ⎭ 2 ∂(y/δt ) Equations (6.48) provide enough information to approximate the temperature profile with a cubic function. T − T∞ y y =a+b +c Tw − T∞ δt δt 2 +d y δt 3 (6.49) Substituting eqn. (6.49) into eqns. (6.48), we get a=1 −1=b+c+d 0 = b + 2c + 3d 0 = 2c 301 302 Laminar and turbulent boundary layers §6.5 which gives 3 a=1 c=0 b = −2 d= 1 2 so the temperature profile is 3y T − T∞ 1 =1− + Tw − T∞ 2 δt 2 y δt 3 (6.50) Predicting the heat flux in the laminar boundary layer Equation (6.47) contains an as-yet-unknown quantity—the thermal b.l. thickness, δt . To calculate δt , we substitute the temperature profile, eqn. (6.50), and the velocity profile, eqn. (6.29), in the integral form of the energy equation, (6.47), which we first express as u∞ (Tw − T∞ ) d dx 1 δt 0 u u∞ T − T∞ y d Tw − T∞ δt α(Tw − T∞ ) =− δt d T − T∞ Tw − T∞ d(y/δt ) (6.51) y /δt =0 There is no problem in completing this integration if δt < δ. However, if δt > δ, there will be a problem because the equation u/u∞ = 1, instead of eqn. (6.29), defines the velocity beyond y = δ. Let us proceed for the δ will be satisfied. moment in the hope that the requirement that δt Introducing φ ≡ δt /δ in eqn. (6.51) and calling y/δt ≡ η, we get ⎡ d⎢ δt ⎣δt dx ⎤ 1 0 3 1 3α ⎥ 1 − η + η3 dη ⎦ = 2 2 2u∞ 3 1 ηφ − η3 φ3 2 2 (6.52) 3 3 = 20 φ− 280 φ3 Since φ is a constant for any Pr [recall eqn. (6.46)], we separate variables: 2δt dδ2 dδt t = = dx dx 3α/u∞ 3 3 φ− φ3 20 280 §6.5 Heat transfer coefficient for laminar, incompressible flow over a flat surface Figure 6.14 The exact and approximate Prandtl number influence on the ratio of b.l. thicknesses. Integrating this result with respect to x and taking δt = 0 at x = 0, we get δt = 3αx u∞ 3 3 φ− φ3 20 280 (6.53) But δ = 4.64x/ Rex in the integral formulation [eqn. (6.31b)]. We divide by this value of δ to be consistent and obtain δt ≡ φ = 0.9638 δ Pr φ 1 − φ2 /14 Rearranging this gives 1 δt = δ 1.025 Pr1/3 1 − (δ2 /14δ2 ) t 1/3 1 1.025 Pr1/3 (6.54) The unapproximated result above is shown in Fig. 6.14, along with the results of Pohlhausen’s precise calculation (see Schlichting [6.3, Chap. 14]). It turns out that the exact ratio, δ/δt , is represented with great accuracy 303 Laminar and turbulent boundary layers 304 §6.5 by δt = Pr−1/3 δ 0.6 Pr 50 (6.55) So the integral method is accurate within 2.5% in the Prandtl number range indicated. Notice that Fig. 6.14 is terminated for Pr less than 0.6. The reason for doing this is that the lowest Pr for pure gases is 0.67, and the next lower values of Pr are on the order of 10−2 for liquid metals. For Pr = 0.67, δ, but only by a δt /δ = 1.143, which violates the assumption that δt small margin. For, say, mercury at 100◦ C, Pr = 0.0162 and δt /δ = 3.952, which violates the condition by an intolerable margin. We therefore have a theory that is acceptable for gases and all liquids except the metallic ones. The final step in predicting the heat flux is to write Fourier’s law: ∂T q = −k ∂y y =0 Tw − T∞ = −k δt ∂ T − T∞ Tw − T∞ ∂ (y/δt ) (6.56) y /δt =0 Using the dimensionless temperature distribution given by eqn. (6.50), we get q = +k Tw − T∞ 3 δt 2 or h≡ 3k 3k δ q = = ∆T 2δt 2 δ δt (6.57) and substituting eqns. (6.54) and (6.31b) for δ/δt and δ, we obtain Nux ≡ hx 3 Rex 1/2 = 1.025 Pr1/3 = 0.3314 Rex Pr1/3 k 2 4.64 Considering the various approximations, this is very close to the result of the exact calculation, which turns out to be 1/2 Nux = 0.332 Rex Pr1/3 0.6 Pr 50 (6.58) This expression gives very accurate results under the assumptions on which it is based: a laminar two-dimensional b.l. on a flat surface, with Tw = constant and 0.6 Pr 50. Heat transfer coefficient for laminar, incompressible flow over a flat surface §6.5 Figure 6.15 A laminar b.l. in a low-Pr liquid. The velocity b.l. is so thin that u u∞ in the thermal b.l. Some other laminar boundary layer heat transfer equations High Pr. is At high Pr, eqn. (6.58) is still close to correct. The exact solution 1/2 Nux → 0.339 Rex Pr1/3 , Pr → ∞ (6.59) Low Pr. Figure 6.15 shows a low-Pr liquid flowing over a flat plate. In δ, and for all practical purposes u = u∞ everywhere within this case δt the thermal b.l. It is as though the no-slip condition [u(y = 0) = 0] and the influence of viscosity were removed from the problem. Thus, the dimensional functional equation for h becomes h = fn x , k, ρcp , u∞ (6.60) There are five variables in J/K, m, and s, so there are only two pi-groups. They are Nux = hx k and Π2 ≡ Rex Pr = u∞ x α The new group, Π2 , is called a Péclét number, Pex , where the subscript identifies the length upon which it is based. It can be interpreted as follows: Pex ≡ ρcp u∞ ∆T heat capacity rate of fluid in the b.l. u∞ x = (6.61) = α axial heat conductance of the b.l. k∆T 305 306 Laminar and turbulent boundary layers §6.5 So long as Pex is large, the b.l. assumption that ∂ 2 T /∂x 2 ∂ 2 T /∂y 2 will be valid; but for small Pex (i.e., Pex 100), it will be violated and a boundary layer solution cannot be used. The exact solution of the b.l. equations gives, in this case: ⎧ ⎪ Pex ≥ 100 and ⎪ ⎪ ⎨ 1/2 1 or Pr 100 (6.62) Nux = 0.565 Pex ⎪ ⎪ ⎪ ⎩ Re ≥ 104 x General relationship. Churchill and Ozoe [6.5] recommend the following empirical correlation for laminar flow on a constant-temperature flat surface for the entire range of Pr: 1/2 Nux = 0.3387 Rex Pr1/3 1 + (0.0468/Pr)2/3 1/4 Pex > 100 (6.63) This relationship proves to be quite accurate, and it approximates eqns. (6.59) and (6.62), respectively, in the high- and low-Pr limits. The calculations of an average Nusselt number for the general case is left as an exercise (Problem 6.10). Boundary layer with an unheated starting length Figure 6.16 shows a b.l. with a heated region that starts at a distance x0 from the leading edge. The heat transfer in this instance is easily obtained using integral methods (see Prob. 6.41). 1/2 Nux = 0.332 Rex Pr1/3 1 − (x0 /x)3/4 1/3 , x > x0 (6.64) Average heat transfer coefficient, h. The heat transfer coefficient h, is the ratio of two quantities, q and ∆T , either of which might vary with x . So far, we have only dealt with the uniform wall temperature problem. Equations (6.58), (6.59), (6.62), and (6.63), for example, can all be used to calculate q(x) when (Tw − T∞ ) ≡ ∆T is a specified constant. In the next subsection, we discuss the problem of predicting [T (x) − T∞ ] when q is a specified constant. That is called the uniform wall heat flux problem. Heat transfer coefficient for laminar, incompressible flow over a flat surface §6.5 Figure 6.16 A b.l. with an unheated region at the leading edge. The term h is used to designate either q/∆T in the uniform wall temperature problem or q/∆T in the uniform wall heat flux problem. Thus, uniform wall temp.: h ≡ 1 q = ∆T ∆T L 1 L = q dx 0 1 L L h(x) dx 0 (6.65) uniform heat flux: h≡ q = ∆T 1 L q (6.66) L ∆T (x) dx 0 The Nusselt number based on h and a characteristic length, L, is designated NuL . This is not to be construed as an average of Nux , which would be meaningless in either of these cases. Thus, for a flat surface (with x0 = 0), we use eqn. (6.58) in eqn. (6.65) to get 1 h= L L 0 0.332 k Pr1/3 h(x) dx = L k x u∞ ν L 0 √ x dx x Nux 1/2 = 0.664 ReL Pr1/3 k L (6.67) Thus, h = 2h(x = L) in a laminar flow, and NuL = hL 1/2 = 0.664 ReL Pr1/3 k (6.68) Likewise for liquid metal flows: 1/2 NuL = 1.13 PeL (6.69) 307 308 Laminar and turbulent boundary layers §6.5 Some final observations. The preceding results are restricted to the two-dimensional, incompressible, laminar b.l. on a flat isothermal wall at velocities that are not too high. These conditions are usually met if: • Rex or ReL is not above the turbulent transition value, which is typically a few hundred thousand. • The Mach number of the flow, Ma ≡ u∞ /(sound speed), is less than about 0.3. (Even gaseous flows behave incompressibly at velocities well below sonic.) A related condition is: • The Eckert number, Ec ≡ u2 /cp (Tw − T∞ ), is substantially less than ∞ unity. (This means that heating by viscous dissipation—which we have neglected—does not play any role in the problem. This assumption was included implicitly when we treated J as an independent unit in the dimensional analysis of this problem.) It is worthwhile to notice how h and Nu depend on their independent variables: 1 1 h or h ∝ √ or √ , x L Nux or NuL ∝ x or L, √ u∞ , ν −1/6 , (ρcp )1/3 , k2/3 √ u∞ , ν −1/6 , (ρcp )1/3 , k−1/3 (6.70) Thus, h → ∞ and Nux vanishes at the leading edge, x = 0. Of course, an infinite value of h, like infinite shear stress, will not really occur at the leading edge because the b.l. description will actually break down in a small neighborhood of x = 0. In all of the preceding considerations, the fluid properties have been assumed constant. Actually, k, ρcp , and especially µ might all vary noticeably with T within the b.l. It turns out that if properties are all evaluated at the average temperature of the b.l. or film temperature Tf = (Tw + T∞ )/2, the results will normally be quite accurate. It is also worth noting that, although properties are given only at one pressure in Appendix A, µ , k, and cp change very little with pressure, especially in liquids. Example 6.5 Air at 20◦ C and moving at 15 m/s is warmed by an isothermal steamheated plate at 110◦ C, ½ m in length and ½ m in width. Find the average heat transfer coefficient and the total heat transferred. What are h, δt , and δ at the trailing edge? Heat transfer coefficient for laminar, incompressible flow over a flat surface §6.5 Solution. We evaluate properties at Tf = (110 + 20)/2 = 65◦ C. Then Pr = 0.707 and ReL = 15(0.5) u∞ L = = 386, 600 ν 0.0000194 so the flow ought to be laminar up to the trailing edge. The Nusselt number is then 1/2 NuL = 0.664 ReL Pr1/3 = 367.8 and h = 367.8 367.8(0.02885) k = = 21.2 W/m2 K L 0.5 The value is quite low because of the low conductivity of air. The total heat flux is then Q = hA ∆T = 21.2(0.5)2 (110 − 20) = 477 W By comparing eqns. (6.58) and (6.68), we see that h(x = L) = ½ h, so 1 h(trailing edge) = 2 (21.2) = 10.6 W/m2 K And finally, δ(x = L) = 4.92L ReL = 4.92(0.5) = 0.00396 m 386, 600 = 3.96 mm and 3.96 δ = 4.44 mm δt = √ = √ 3 3 Pr 0.707 The problem of uniform wall heat flux When the heat flux at the heater wall, qw , is specified instead of the temperature, it is Tw that we need to know. We leave the problem of finding Nux for qw = constant as an exercise (Problem 6.11). The exact result is 1/2 Nux = 0.453 Rex Pr1/3 for Pr 0.6 (6.71) 309 Laminar and turbulent boundary layers 310 §6.5 where Nux = hx/k = qw x/k(Tw − T∞ ). The integral method gives the same result with a slightly lower constant (0.417). We must be very careful in discussing average results in the constant heat flux case. The problem now might be that of finding an average temperature difference (cf. (6.66)): Tw − T∞ = 1 L L 0 (Tw − T∞ ) dx = 1 L L 0 dx qw x √ 1/3 k(0.453 u∞ /ν Pr ) x or Tw − T∞ = qw L/k 1/2 0.6795 ReL Pr (6.72) 1/3 1/2 1/3 (although the which can be put into the form NuL = 0.6795 ReL Pr Nusselt number yields an awkward nondimensionalization for Tw − T∞ ). Churchill and Ozoe [6.5] have pointed out that their eqn. (6.63) will describe (Tw − T∞ ) with high accuracy over the full range of Pr if the constants are changed as follows: 1/2 Nux = 0.4637 Rex Pr1/3 1 + (0.02052/Pr)2/3 1/4 Pex > 100 (6.73) Example 6.6 Air at 15◦ C flows at 1.8 m/s over a 0.6 m-long heating panel. The panel is intended to supply 420 W/m2 to the air, but the surface can sustain only about 105◦ C without being damaged. Is it safe? What is the average temperature of the plate? Solution. In accordance with eqn. (6.71), ∆Tmax = ∆Tx =L = qL qL/k = 1/2 k Nux =L 0.453 Rex Pr1/3 or if we evaluate properties at (85 + 15)/2 = 50◦ C, for the moment, ∆Tmax = 420(0.6)/0.0278 0.453 1/2 (0.709)1/3 0.6(1.8)/1.794 × 10−5 = 91.5◦ C This will give Twmax = 15 + 91.5 = 106.5◦ C. This is very close to 105◦ C. If 105◦ C is at all conservative, q = 420 W/m2 should be safe— particularly since it only occurs over a very small distance at the end of the plate. The Reynolds analogy §6.6 311 From eqn. (6.72) we find that ∆T = 0.453 ∆Tmax = 61.0◦ C 0.6795 so Tw = 15 + 61.0 = 76.0◦ C 6.6 The Reynolds analogy The analogy between heat and momentum transfer can now be generalized to provide a very useful result. We begin by recalling eqn. (6.25), which is restricted to a flat surface with no pressure gradient: 1 d dx δ 0 u u∞ u y −1 d u∞ δ =− Cf 2 (6.25) and by rewriting eqns. (6.47) and (6.51), we obtain for the constant wall temperature case: d dx 1 φδ 0 u u∞ T − T∞ y d Tw − T∞ δt = qw ρcp u∞ (Tw − T∞ ) (6.74) But the similarity of temperature and flow boundary layers to one another [see, e.g., eqns. (6.29) and (6.50)], suggests the following approximation, which becomes exact only when Pr = 1: u T − T∞ δt δ= 1− Tw − T∞ u∞ Substituting this result in eqn. (6.74) and comparing it to eqn. (6.25), we get − d dx 1 δ 0 u u∞ u y −1 d u∞ δ =− Cf 2 =− qw ρcp u∞ (Tw − T∞ )φ2 (6.75) Finally, we substitute eqn. (6.55) to eliminate φ from eqn. (6.75). The result is one instance of the Reynolds-Colburn analogy :8 Cf h Pr2/3 = ρcp u∞ 2 8 (6.76) Reynolds [6.6] developed the analogy in 1874. Colburn made important use of it in this century. The form given is for flat plates with 0.6 ≤ Pr ≤ 50. The Prandtl number factor is usually a little different for other flows or other ranges of Pr. 312 Laminar and turbulent boundary layers §6.6 For use in Reynolds’ analogy, Cf must be a pure skin friction coefficient. The profile drag that results from the variation of pressure around the body is unrelated to heat transfer. The analogy does not apply when profile drag is included in Cf . The dimensionless group h/ρcp u∞ is called the Stanton number. It is defined as follows: St, Stanton number ≡ Nux h = ρcp u∞ Rex Pr The physical significance of the Stanton number is St = h∆T actual heat flux to the fluid = ρcp u∞ ∆T heat flux capacity of the fluid flow (6.77) The group St Pr2/3 was dealt with by the chemical engineer Colburn, who gave it a special symbol: j ≡ Colburn j -factor = St Pr2/3 = Nux Rex Pr1/3 (6.78) Example 6.7 Does the equation for the Nusselt number on an isothermal flat surface in laminar flow satisfy the Reynolds analogy? Solution. If we rewrite eqn. (6.58), we obtain Nux 0.332 2/3 = 1/3 = St Pr Rex Rex Pr (6.79) But comparison with eqn. (6.33) reveals that the left-hand side of eqn. (6.79) is precisely Cf /2, so the analogy is satisfied perfectly. Likewise, from eqns. (6.68) and (6.34), we get Cf 0.664 NuL 2/3 = = 1/3 ≡ St Pr 2 ReL ReL Pr (6.80) The Reynolds-Colburn analogy can be used directly to infer heat transfer data from measurements of the shear stress, or vice versa. It can also be extended to turbulent flow, which is much harder to predict analytically. We shall undertake that problem in Sect. 6.8. Turbulent boundary layers §6.7 Example 6.8 How much drag force does the air flow in Example 6.5 exert on the heat transfer surface? Solution. From eqn. (6.80) in Example 6.7, we obtain Cf = 2 NuL ReL Pr1/3 From Example 6.5 we obtain NuL , ReL , and Pr1/3 : Cf = 2(367.8) = 0.002135 (386, 600)(0.707)1/3 so τyx = (0.002135) 1 (0.002135)(1.05)(15)2 ρu2 = ∞ 2 2 = 0.2522 kg/m·s2 and the force is τyx A = 0.2522(0.5)2 = 0.06305 kg·m/s2 = 0.06305 N = 0.23 oz 6.7 Turbulent boundary layers Turbulence Big whirls have little whirls, That feed on their velocity. Little whirls have littler whirls, And so on, to viscosity. This bit of doggerel by the English fluid mechanic, L. F. Richardson, tells us a great deal about the nature of turbulence. Turbulence in a fluid can be viewed as a spectrum of coexisting vortices in which kinetic energy from the larger ones is dissipated to successively smaller ones until the very smallest of these vortices (or “whirls”) are damped out by viscous shear stresses. The next time the weatherman shows a satellite photograph of North America on the 10:00 p.m. news, notice the cloud patterns. There will be 313 Laminar and turbulent boundary layers 314 §6.7 one or two enormous vortices of continental proportions. These huge vortices, in turn, feed smaller “weather-making” vortices on the order of hundreds of miles in diameter. These further dissipate into vortices of cyclone and tornado proportions—sometimes with that level of violence but more often not. These dissipate into still smaller whirls as they interact with the ground and its various protrusions. The next time the wind blows, stand behind any tree and feel the vortices. In the great plains, where there are not many ground vortex generators (such as trees), you will see small cyclonic eddies called “dust devils.” The process continues right on down to millimeter or even micrometer scales. There, momentum exchange is no longer identifiable as turbulence but appears simply as viscous stretching of the fluid. The same kind of process exists within, say, a turbulent pipe flow at high Reynolds number. Such a flow is shown in Fig. 6.17. Turbulence in such a case consists of coexisting vortices which vary in size from a substantial fraction of the pipe radius down to micrometer dimensions. The spectrum of sizes varies with location in the pipe. The size and intensity of vortices at the wall must clearly approach zero, since the fluid velocity goes to zero at the wall. Figure 6.17 shows the fluctuation of a typical flow variable—namely, velocity—both with location in the pipe and with time. This fluctuation arises because of the turbulent motions that are superposed on the average local flow. Other flow variables, such as T or ρ , can vary in the same manner. For any variable we can write a local time-average value as u≡ 1 T T u dt (6.81) 0 where T is a time that is much longer than the period of typical fluctuations.9 Equation (6.81) is most useful for so-called stationary processes — ones for which u is nearly time-independent. If we substitute u = u + u in eqn. (6.81), where u is the actual local velocity and u is the instantaneous magnitude of the fluctuation, we obtain u= 1 T T 0 u dt + =u 9 1 T T u dt (6.82) 0 =u Take care not to interpret this T as the thermal time constant that we introduced in Chapter 1; we denote time constants are as T . §6.7 Turbulent boundary layers 315 Figure 6.17 Fluctuation of u and other quantities in a turbulent pipe flow. This is consistent with the fact that u or any other average fluctuation = 0 (6.83) since the fluctuations are defined as deviations from the average. We now want to create a measure of the size, or lengthscale, of turbulent vortices. This might be done experimentally by placing two velocitymeasuring devices very close to one another in a turbulent flow field. When the probes are close, their measurements will be very highly correlated with one one another. Then, suppose that the two velocity probes are moved apart until the measurements first become unrelated to one another. That spacing gives an indication of the average size of the turbulent motions. Prandtl invented a slightly different (although related) measure of the lengthscale of turbulence, called the mixing length, . He saw as an average distance that a parcel of fluid moves between interactions. It has a physical significance similar to that of the molecular mean free path. It is harder to devise a clean experimental measure of than of the 316 Laminar and turbulent boundary layers §6.7 correlation lengthscale of turbulence. But we can still use the concept of to examine the notion of a turbulent shear stress. The shear stresses of turbulence arise from the same kind of momentum exchange process that gives rise to the molecular viscosity. Recall that, in the latter case, a kinetic calculation gave eqn. (6.45) for the laminar shear stress ∂u ∂y τyx = (constant) ρ C (6.45) =u where was the molecular mean free path and u was the velocity difference for a molecule that had travelled a distance in the mean velocity gradient. In the turbulent flow case, pictured in Fig. 6.18, we can think of Prandtl’s parcels of fluid (rather than individual molecules) as carrying the x -momentum. Let us rewrite eqn. (6.45) in the following way: • The shear stress τyx becomes a fluctuation in shear stress, τy x , resulting from the turbulent movement of a parcel of fluid • changes from the mean free path to the mixing length • C is replaced by v = v + v , the instantaneous vertical speed of the fluid parcel • The velocity fluctuation, u , is for a fluid parcel that moves a distance through the mean velocity gradient, ∂ u/∂y . It is given by (∂ u/∂y). Then τy x = (constant) ρ v + v u (6.84) Equation (6.84) can also be derived formally and precisely with the help of the Navier-Stokes equation. When this is done, the constant comes out equal to −1. The average of the fluctuating shear stress is τy x = − ρ T T 0 vu + v u dt = −ρ v u −ρ v u =0 (6.85) Turbulent boundary layers §6.7 317 Figure 6.18 The shear stress, τyx , in a laminar or turbulent flow. Notice that, while u = v = 0, averages of cross products of fluctuations (such as u v or u 2 ) do not generally vanish. Thus, the time average of the fluctuating component of shear stress is τy x = −ρ v u (6.86) In addition to the fluctuating shear stress, the flow will have a mean shear stress associated with the mean velocity gradient, ∂ u/∂y . That stress is µ(∂ u/∂y), just as in Newton’s law of viscous shear. It is not obvious how to calculate v u (although it can be measured), so we shall not make direct use of eqn. (6.86). Instead, we can try to model v u . From the preceding discussion, we see that v u should go to zero when the velocity gradient, (∂ u/∂y), is zero, and that it should increase when the velocity gradient increases. We might therefore assume it to be proportional to (∂ u/∂y). Then the total time-average shear stress, τyx , can be expressed as a sum of the mean flow and turbulent contributions that are each proportional to the mean velocity gradient. Specifically, τyx = µ ∂u − ρv u ∂y τyx = µ some other factor, which ∂u + reflects turbulent mixing ∂y (6.87a) ∂u ∂y (6.87b) ≡ ρ · εm or τyx = ρ (ν + εm ) ∂u ∂y (6.87c) 318 Laminar and turbulent boundary layers §6.7 where εm is called the eddy diffusivity for momentum. We shall use this characterization in examining the flow field and the heat transfer. The eddy diffusivity itself may be expressed in terms of the mixing length. Suppose that u increases in the y -direction (i.e., ∂ u/∂y > 0). Then, when a fluid parcel moves downward into slower moving fluid, (∂ u/∂y). If that parcel moves upward into faster fluid, it has u the sign changes. The vertical velocity fluctation, v , is positive for an upward moving parcel and negative for a downward motion. On average, u and v for the eddies should be about the same size. Hence, we expect that ρεm ∂u = −ρ v u = −ρ (constant) ∂y = ρ (constant) 2 ± ∂u ∂y ∂u ∂y ∂u ∂y ∓ ∂u ∂y (6.88a) (6.88b) where the absolute value is needed to get the right sign when ∂ u/∂y < 0. Both ∂ u/∂y and v u can be measured, so we may arbitrarily set the constant in eqn. (6.88) to unity to obtain a measurable definition of the mixing length. We also obtain an expression for the eddy diffusivity: εm = 2 ∂u . ∂y (6.89) Turbulence near walls The most important convective heat transfer issue is how flowing fluids cool solid surfaces. Thus, we are principally interested in turbulence near walls. In a turbulent boundary layer, the gradients are very steep near the wall and weaker farther from the wall where the eddies are larger and turbulent mixing is more efficient. This is in contrast to the gradual variation of velocity and temperature in a laminar boundary layer, where heat and momentum are transferred by molecular diffusion rather than the vertical motion of vortices. In fact,the most important processes in turbulent convection occur very close to walls, perhaps within only a fraction of a millimeter. The outer part of the b.l. is less significant. Let us consider the turbulent flow close to a wall. When the boundary layer momentum equation is time-averaged for turbulent flow, the result Turbulent boundary layers §6.7 319 is ∂u ∂u +v ∂x ∂y ∂u − ρv u ∂y = ∂ ∂y = ∂ τyx ∂y = ρu ∂ ∂y µ (6.90a) neglect very near wall (6.90b) ρ (ν + εm ) ∂u ∂y (6.90c) In the innermost region of a turbulent boundary layer — y/δ 0.2, where δ is the b.l. thickness — the mean velocities are small enough that the convective terms in eqn. (6.90a) can be neglected. As a result, ∂τyx /∂y 0. The total shear stress is thus essentially constant in y and must equal the wall shear stress: τw τyx = ρ (ν + εm ) ∂u ∂y (6.91) Equation (6.91) shows that the near-wall velocity profile does not depend directly upon x . In functional form u = fn τw , ρ, ν, y (6.92) (Note that εm does not appear because it is defined by the velocity field.) The effect of the streamwise position is carried in τw , which varies slowly with x . As a result, the flow field near the wall is not very sensitive to upstream conditions, except through their effect on τw . When the velocity profile is scaled in terms of the local value τw , essentially the same velocity profile is obtained in every turbulent boundary layer. Equation (6.92) involves five variables in three dimensions (kg, m, s), so just two dimensionless groups are needed to describe the velocity profile: u u∗ y = fn u∗ ν (6.93) where the velocity scale u∗ ≡ τw /ρ is called the friction velocity. The friction velocity is a speed characteristic of the turbulent fluctuations in the boundary layer. 320 Laminar and turbulent boundary layers §6.7 Equation (6.91) can be integrated to find the near wall velocity profile: u 0 du = τw ρ y 0 dy ν + εm (6.94) =u(y) To complete the integration, an equation for εm (y) is needed. Measurements show that the mixing length varies linearly with the distance from the wall for small y = κy for y/δ 0.2 (6.95) where κ = 0.41 is called the von Kármán constant. Physically, this says that the turbulent eddies at a location y must be no bigger that the distance to wall. That makes sense, since eddies cannot cross into the wall. The viscous sublayer. Very near the wall, the eddies must become tiny; εm . In other words, in and thus εm will tend to zero, so that ν this region turbulent shear stress is negligible compared to viscous shear stress. If we integrate eqn. (6.94) in that range, we find u(y) = τw ρ y 0 dy τw y = ρν ν (u∗ )2 y = ν (6.96) Experimentally, eqn. (6.96) is found to apply for (u∗ y/ν) 7, a thin region called the viscous sublayer. Depending upon the fluid and the shear stress, the sublayer is on the order of tens to hundreds of micrometers thick. Because turbulent mixing is ineffective in the sublayer, the sublayer is responsible for a major fraction of the thermal resistance of a turbulent boundary layer. Even a small wall roughness can disrupt this thin sublayer, causing a large decrease in the thermal resistance (but also a large increase in the wall shear stress). The log layer. Farther away from the wall, is larger and turbulent ν . Then, from eqns. (6.91) and (6.89) shear stress is dominant: εm τw ρ εm ∂u =ρ ∂y 2 ∂u ∂y ∂u ∂y (6.97) Turbulent boundary layers §6.7 321 Assuming the velocity gradient to be positive, we may take the square root of eqn. (6.97), rearrange, and integrate it: du = u(y) = u∗ = dy τw ρ (6.98a) dy + constant κy u∗ ln y + constant κ (6.98b) (6.98c) Experimental data may be used to fix the constant, with the result that 1 u(y) u∗ y = ln u∗ κ ν +B (6.99) for B 5.5. Equation (6.99) is sometimes called the log law. Experimentally, it is found to apply for (u∗ y/ν) 30 and y/δ 0.2. Other regions of the turbulent b.l. For the range 7 < (u∗ y/ν) < 30, the so-called buffer layer, more complicated equations for , εm , or u are used to connect the viscous sublayer to the log layer [6.7, 6.8]. Here, actually decreases a little faster than shown by eqn. (6.95), as y 3/2 [6.9]. In contrast, for the outer part of the turbulent boundary layer (y/δ 0.2), the mixing length is approximately constant: 0.09δ. Gradients in this part of the boundary layer are weak and do not directly affect transport at the wall. This part of the b.l. is nevertheless essential to the streamwise momentum balance that determines how τw and δ vary along the wall. Analysis of that momentum balance [6.2] leads to the following expressions for the boundary thickness and the skin friction coefficient as a function of x : 0.16 δ(x) = 1/7 x Rex Cf (x) = 0.027 1/7 Rex (6.100) (6.101) To write these expressions, we assume that the turbulent b.l. begins at x = 0, neglecting the initial laminar region. They are reasonably accurate for Reynolds numbers ranging from about 106 to 109 . A more accurate 322 Laminar and turbulent boundary layers §6.8 formula for Cf , valid for all turbulent Rex , was given by White [6.10]: Cf (x) = 6.8 0.455 ln(0.06 Rex ) (6.102) 2 Heat transfer in turbulent boundary layers Like the turbulent momentum boundary layer, the turbulent thermal boundary layer is characterized by inner and outer regions. In the inner part of the thermal boundary layer, turbulent mixing is increasingly weak; there, heat transport is controlled by heat conduction in the sublayer. Farther from the wall, a logarithmic temperature profile is found, and in the outermost parts of the boundary layer, turbulent mixing is the dominant mode of transport. The boundary layer ends where turbulence dies out and uniform freestream conditions prevail, with the result that the thermal and momentum boundary layer thicknesses are the same. At first, this might seem to suggest that an absence of any Prandtl number effect on turbulent heat transfer, but that is not the case. The effect of Prandtl number is now found in the sublayers near the wall, where molecular viscosity and thermal conductivity still control the transport of heat and momentum. The Reynolds-Colburn analogy for turbulent flow The eddy diffusivity for momentum was introduced by Boussinesq [6.11] in 1877. It was subsequently proposed that Fourier’s law might likewise be modified for turbulent flow as follows: q = −k another constant, which ∂T − reflects turbulent mixing ∂y ∂T ∂y ≡ ρcp · εh where T is the local time-average value of the temperature. Therefore, q = −ρcp (α + εh ) ∂T ∂y (6.103) where εh is called the eddy diffusivity of heat. This immediately suggests yet another definition: εm (6.104) turbulent Prandtl number, Prt ≡ εh Heat transfer in turbulent boundary layers §6.8 Equation (6.103) can be written in terms of ν and εm by introducing Pr and Prt into it. Thus, q = −ρcp εm ν + Pr Prt ∂T ∂y (6.105) Before trying to build a form of the Reynolds analogy for turbulent flow, we must note the behavior of Pr and Prt : • Pr is a physical property of the fluid. It is both theoretically and actually near unity for ideal gases, but for liquids it may differ from unity by orders of magnitude. • Prt is a property of the flow field more than of the fluid. The numerical value of Prt is normally well within a factor of 2 of unity. It varies with location in the b.l., but, for nonmetallic fluids, it is often near 0.85. The time-average boundary-layer energy equation is similar to the time-average momentum equation [eqn. (6.90a)] ρcp u ∂T ∂T +v ∂x ∂y =− ∂ ∂ q= ∂y ∂y ρ cp ν εm + Pr Prt ∂T ∂y (6.106) neglect very near wall and in the near wall region the convective terms are again negligible. This means that ∂q/∂y 0 near the wall, so that the heat flux is constant in y and equal to the wall heat flux: q = qw = −ρcp εm ν + Pr Prt ∂T ∂y (6.107) We may integrate this equation as we did eqn. (6.91), with the result that ⎧ ∗ ⎪ ⎪Pr u y thermal sublayer ⎪ ⎨ ν Tw − T (y) (6.108) = ∗ qw /(ρcp u∗ ) ⎪ 1 ⎪ ⎪ ln u y + A(Pr) thermal log layer ⎩ κ ν The constant A depends upon the Prandtl number. It reflects the thermal resistance of the sublayer near the wall. As was done for the constant B in the velocity profile, experimental data or numerical simulation may be used to determine A(Pr) [6.12, 6.13]. For Pr ≥ 0.5, A(Pr) = 12.8 Pr0.68 − 7.3 (6.109) 323 324 Laminar and turbulent boundary layers §6.8 To obtain the Reynolds analogy, we can subtract the dimensionless log-law, eqn. (6.99), from its thermal counterpart, eqn. (6.108): u(y) Tw − T (y) − = A(Pr) − B qw /(ρcp u∗ ) u∗ In the outer part of the boundary layer, T (y) T∞ and u(y) u∞ Tw − T∞ − ∗ = A(Pr) − B ∗) qw /(ρcp u u (6.110a) u∞ , so (6.110b) We can eliminate the friction velocity in favor of the skin friction coefficient by using the definitions of each: u∗ = u∞ τw = ρu2 ∞ Cf (6.110c) 2 Hence, Tw − T∞ qw /(ρcp u∞ ) Cf 2 2 = A(Pr) − B Cf − (6.110d) Rearrangment of the last equation gives Cf 2 qw = (ρcp u∞ )(Tw − T∞ ) 1 + [A(Pr) − B] Cf 2 (6.110e) The lefthand side is simply the Stanton number, St = h (ρcp u∞ ). Upon substituting B = 5.5 and eqn. (6.109) for A(Pr), we obtain the ReynoldsColburn analogy for turbulent flow: Stx = Cf 2 1 + 12.8 Pr0.68 − 1 Cf 2 Pr ≥ 0.5 (6.111) This result can be used with eqn. (6.102) for Cf , or with data for Cf , to calculate the local heat transfer coefficient in a turbulent boundary layer. The equation works for either uniform Tw or uniform qw . This is because the thin, near-wall part of the boundary layer controls most of the thermal resistance and that thin layer is not strongly dependent on upstream history of the flow. Equation (6.111) is valid for smooth walls with a mild or a zero pressure gradient. The factor 12.8 (Pr0.68 − 1) in the denominator accounts for the thermal resistance of the sublayer. If the walls are rough, the sublayer will be disrupted and that term must be replaced by one that takes account of the roughness (see Sect. 7.3). Heat transfer in turbulent boundary layers §6.8 Other equations for heat transfer in the turbulent b.l. Although eqn. (6.111) gives an excellent prediction of the local value of h in a turbulent boundary layer, a number of simplified approximations to it have been suggested in the literature. For example, for Prandtl numbers not too far from unity and Reynolds numbers not too far above transition, the laminar flow Reynolds-Colburn analogy can be used Stx = Cf Pr−2/3 2 for Pr near 1 (6.76) The best exponent for the Prandtl number in such an equation actually depends upon the Reynolds and Prandtl numbers. For gases, an exponent of −0.4 gives somewhat better results. A more wide-ranging approximation can be obtained after introducing a simplifed expression for Cf . For example, Schlichting [6.3, Chap. XXI] shows that, for turbulent flow over a smooth flat plate in the low-Re range, Cf 0.0592 1/5 Rex , 5 × 105 Rex 107 (6.112) With this Reynolds number dependence, Žukauskas and coworkers [6.14, 6.15] found that Stx = Cf 2 Pr−0.57 , 0.7 ≤ Pr ≤ 380 (6.113) so that when eqn. (6.112) is used to eliminate Cf Nux = 0.0296 Re0.8 Pr0.43 x Somewhat better agreement with data, for 2 × 105 obtained by adjusting the constant [6.15]: (6.114) Rex Nux = 0.032 Re0.8 Pr0.43 x 5 × 106 , is (6.115) The average Nusselt number for uniform Tw is obtained from eqn. (6.114) as follows: NuL = L 0.0296 Pr0.43 L h= k k k L L 0 1 Re0.8 dx x x 325 326 Laminar and turbulent boundary layers §6.8 where we ignore the fact that there is a laminar region at the front of the plate. Thus, NuL = 0.0370 Re0.8 Pr0.43 L (6.116) This equation may be used for either uniform Tw or uniform qw , and for ReL up to about 3 × 107 [6.14, 6.15]. A flat heater with a turbulent b.l. on it actually has a laminar b.l. between x = 0 and x = xtrans , as is indicated in Fig. 6.4. The obvious way to calculate h in this case is to write h= L 1 L∆T 1 = L q dx 0 xtrans 0 (6.117) L hlaminar dx + xtrans hturbulent dx where xtrans = (ν/u∞ )Retrans . Thus, we substitute eqns. (6.58) and (6.114) in eqn. (6.117) and obtain, for 0.6 Pr 50, NuL = 0.037 Pr0.43 Re0.8 − Re0.8 − 17.95 Pr−0.097 (Retrans )1/2 trans L (6.118) If ReL Retrans , this result reduces to eqn. (6.116). Whitaker [6.16] suggested setting Pr−0.097 ≈ 1 and Retrans ≈ 200, 000 in eqn. (6.118): 0.43 NuL = 0.037 Pr Re0.8 L − 9200 µ∞ µw 1/4 0.6 ≤ Pr ≤ 380 (6.119) This expression has been corrected to account for the variability of liquid viscosity with the factor (µ∞ /µw )1/4 , where µ∞ is the viscosity at the freestream temperature, T∞ , and µw is that at the wall temperature, Tw ; other physical properties should be evaluated at T∞ . If eqn. (6.119) is used to predict heat transfer to a gaseous flow, the viscosity-ratio correction term should not be used and properties should be evaluated at the film temperature. This is because the viscosity of a gas rises with temperature instead of dropping, and the correction will be incorrect. Finally, it is important to remember that eqns. (6.118) and (6.119) should be used only when ReL is substantially above the transitional value. §6.8 Heat transfer in turbulent boundary layers A correlation for laminar, transitional, and turbulent flow A problem with the two preceding relations is that they do not really deal with the question of heat transfer in the rather lengthy transition region. Both eqns. (6.118) and (6.119) are based on the assumption that flow abruptly passes from laminar to turbulent at a critical value of x , and we have noted in the context of Fig. 6.4 that this is not what occurs. The location of the transition depends upon such variables as surface roughness and the turbulence, or lack of it, in the stream approaching the heater. Churchill [6.17] suggests correlating any particular set of data with Nux = 0.45 + 0.3387 φ1/2 ⎧ ⎪ ⎨ (φ/2, 600) ⎪ ⎩ 1 + (φu /φ)7/2 ⎫1/2 ⎪ ⎬ 3/5 1+ 2/5 ⎪ ⎭ (6.120a) where φ ≡ Rex Pr 0.0468 1+ Pr 2/3 2/3 −1/2 (6.120b) and φu is a number between about 105 and 107 . The actual value of φu must be fit to the particular set of data. In a very “clean” system, φu will be larger; in a very “noisy” one, it will be smaller. If the Reynolds number at the end of the turbulent transition region is Reu , an estimate is φu ≈ φ(Rex = Reu ). The equation is for uniform Tw , but it may be used for uniform qw if the constants 0.3387 and 0.0468 are replaced by 0.4637 and 0.02052, respectively. Churchill also gave an expression for the average Nusselt number: NuL = 0.45 + 0.6774 φ1/2 ⎧ ⎪ ⎨ ⎪ ⎩ 1+ (φ/12, 500) 3/5 1 + (φum /φ)7/2 ⎫1/2 ⎪ ⎬ 2/5 ⎪ ⎭ (6.120c) where φ is defined as in eqn. (6.120b), using ReL in place of Rex , and φum ≈ 1.875 φ(ReL = Reu ). This equation may be used for either uniform Tw or uniform qw . The advantage of eqns. (6.120a) or (6.120c) is that, once φu or φum is known, they will predict heat transfer from the laminar region, through the transition regime, and into the turbulent regime. 327 328 Laminar and turbulent boundary layers §6.8 Example 6.9 After loading its passengers, a ship sails out of the mouth of a river, where the water temperature is 24◦ C, into 10◦ C ocean water. The forward end of the ship’s hull is sharp and relatively flat. If the ship travels at 5 knots, find Cf and h at a distance of 1 m from the forward edge of the hull. Solution. If we assume that the hull’s heat capacity holds it at the river temperature for a time, we can take the properties of water at Tf = (10 + 24)/2 = 17◦ C: ν = 1.085 × 10−6 m2 /s, k = 0.5927 W/m·K, ρ = 998.8 kg/m3 , cp = 4187 J/kg·K, and Pr = 7.66. One knot equals 0.5144 m/s, so u∞ = 5(0.5144) = 2.572 m/s. Then, Rex = (2.572)(1)/(1.085 × 10−6 ) = 2.371 × 106 , indicating that the flow is turbulent at this location. We have given several different equations for Cf in a turbulent boundary layer, but the most accurate of these is eqn. (6.102): Cf (x) = = 0.455 ln(0.06 Rex ) 2 0.455 ln[0.06(2.371 × 106 )] 2 = 0.003232 For the heat transfer coefficient, we can use either eqn. (6.115) h(x) = k · 0.032 Re0.8 Pr0.43 x x (0.5927)(0.032)(2.371 × 106 )0.8 (7.66)0.43 (1.0) = 5, 729 W/m2 K = or its more complex counterpart, eqn. (6.111): h(x) = ρcp u∞ · = Cf 2 1 + 12.8 Pr0.68 − 1 Cf 2 998.8(4187)(2.572)(0.003232/2) 1 + 12.8 (7.66)0.68 − 1 0.003232/2 = 6, 843 W/m2 K The two values of h differ by about 18%, which is within the uncertainty of eqn. (6.115). Heat transfer in turbulent boundary layers §6.8 Example 6.10 In a wind tunnel experiment, an aluminum plate 2.0 m in length is electrically heated at a power density of 1 kW/m2 and is cooled on one surface by air flowing at 10 m/s. The air in the wind tunnel has a temperature of 290 K and is at 1 atm pressure, and the Reynolds number at the end of turbulent transition regime is observed to be 400,000. Estimate the average temperature of the plate. Solution. For this low heat flux, we expect the plate temperature to be near the air temperature, so we evaluate properties at 300 K: ν = 1.578 × 10−5 m2 /s, k = 0.02623 W/m·K, and Pr = 0.713. At 10 m/s, the plate Reynolds number is ReL = (10)(2)/(1.578 × 10−5 ) = 1.267 × 106 . From eqn. (6.118), we get NuL = 0.037(0.713)0.43 (1.267 × 106 )0.8 − (400, 000)0.8 − 17.95(0.713)−0.097 (400, 000)1/2 = 1, 845 so h= 1845(0.02623) 1845 k = = 24.19 W/m2 K L 2.0 It follows that the average plate temperature is T w = 290 K + 103 W/m2 = 331 K. 24.19 W/m2 K The film temperature is (331 + 290)/2 = 311 K; if we recalculate using properties at 311 K, the h changes by less than 4%, and T w by 1.3 K. To take better account of the transition regime, we can use Churchill’s equation, (6.120c). First, we evaluate φ: φ= (1.267 × 106 )(0.713)2/3 1/2 1 + (0.0468/0.713)2/3 = 9.38 × 105 We then estimate φum = 1.875 · φ(ReL = 400, 000) = (1.875)(400, 000)(0.713)2/3 1 + (0.0468/0.713)2/3 1/2 = 5.55 × 105 329 Chapter 6: Laminar and turbulent boundary layers 330 Finally, 1/2 NuL = 0.45 + (0.6774) 9.38 × 105 ⎧ ⎪ 3/5 ⎨ 9.38 × 105 /12, 500 × 1+ ⎪ ⎩ 1 + (5.55 × 105 /9.38 × 105 )7/2 ⎫1/2 ⎪ ⎬ 2/5 ⎪ ⎭ = 2, 418 which leads to 2418(0.02623) 2418 k = = 31.71 W/m2 K h= L 2.0 and T w = 290 K + 103 W/m2 = 322 K. 31.71 W/m2 K Thus, in this case, the average heat transfer coefficient is 33% higher when the transition regime is included. A word about the analysis of turbulent boundary layers The preceding discussion has circumvented serious analysis of heat transfer in turbulent boundary layers. In the past, boundary layer heat transfer has been analyzed in many flows (with and without pressure gradients, dp/dx ) using sophisticated integral methods. In recent decades, however, computational techniques have largely replaced integral analyses. Various computational schemes, particularly those based on turbulent kinetic energy and viscous dissipation (so-called k-ε methods), are widely-used and have been implemented in a variety of commercial fluiddynamics codes. These methods are described in the technical literature and in monographs on turbulence [6.18, 6.19]. We have found our way around analysis by presenting some correlations for the simple plane surface. In the next chapter, we deal with more complicated configurations. A few of these configurations will be amenable to elementary analyses, but for others we shall only be able to present the best data correlations available. Problems 6.1 Verify that eqn. (6.13) follows from eqns. (6.11) and (6.12). Problems 331 6.2 The student with some analytical ability (or some assistance from the instructor) should complete the algebra between eqns. (6.16) and (6.20). 6.3 Use a computer to solve eqn. (6.18) subject to b.c.’s (6.20). To do this you need all three b.c.’s at η = 0, but one is presently at η = ∞. There are three ways to get around this: • Start out by guessing a value of ∂f /∂η at η = 0—say, ∂f /∂η = 1. When η is large—say, 6 or 10—∂f /∂η will asymptotically approach a constant. If the constant > 1, go back and guess a lower value of ∂f /∂η, or vice versa, until the constant converges on unity. (There are many ways to automate the successive guesses.) • The correct value of df /dη is approximately 0.33206 at η = 0. You might cheat and begin with it. • There exists a clever way to map df /dη = 1 at η = ∞ back into the origin. (Consult your instructor.) 6.4 Verify that the Blasius solution (Table 6.1) satisfies eqn. (6.25). To do this, carry out the required integration. 6.5 Verify eqn. (6.30). 6.6 Obtain the counterpart of eqn. (6.32) based on the velocity profile given by the integral method. 6.7 Assume a laminar b.l. velocity profile of the simple form u/u∞ = y/δ and calculate δ and Cf on the basis of this very rough estimate, using the momentum integral method. How accurate is each? [Cf is about 13% low.] 6.8 √ In a certain flow of water at 40◦ C over a flat plate δ = 0.005 x , for δ and x measured in meters. Plot to scale on a common graph (with an appropriately expanded y -scale): • δ and δt for the water. • δ and δt for air at the same temperature and velocity. 6.9 A thin film of liquid with a constant thickness, δ0 , falls down a vertical plate. It has reached its terminal velocity so that viscous shear and weight are in balance and the flow is steady. Chapter 6: Laminar and turbulent boundary layers 332 The b.l. equation for such a flow is the same as eqn. (6.13), except that it has a gravity force in it. Thus, u ∂u 1 dp ∂2u ∂u +v =− +g+ν ∂x ∂y ρ dx ∂y 2 where x increases in the downward direction and y is normal to the wall. Assume that the surrounding air density 0, so there is no hydrostatic pressure gradient in the surrounding air. Then: • Simplify the equation to describe this situation. • Write the b.c.’s for the equation, neglecting any air drag on the film. • Solve for the velocity distribution in the film, assuming that you know δ0 (cf. Chap. 8). (This solution is the starting point in the study of many process heat and mass transfer problems.) 6.10 Develop an equation for NuL that is valid over the entire range of Pr for a laminar b.l. over a flat, isothermal surface. 6.11 Use an integral method to develop a prediction of Nux for a laminar b.l. over a uniform heat flux surface. Compare your result with eqn. (6.71). What is the temperature difference at the leading edge of the surface? 6.12 Verify eqn. (6.118). 6.13 It is known from flow measurements that the transition to turbulence occurs when the Reynolds number based on mean velocity and diameter exceeds 4000 in a certain pipe. Use the fact that the laminar boundary layer on a flat plate grows according to the relation δ = 4.92 x ν umax x to find an equivalent value for the Reynolds number of transition based on distance from the leading edge of the plate and umax . (Note that umax = 2uav during laminar flow in a pipe.) Problems 333 6.14 Execute the differentiation in eqn. (6.24) with the help of Leibnitz’s rule for the differentiation of an integral and show that the equation preceding it results. 6.15 Liquid at 23◦ C flows at 2 m/s over a smooth, sharp-edged, flat surface 12 cm in length which is kept at 57◦ C. Calculate h at the trailing edge (a) if the fluid is water; (b) if the fluid is glycerin (h = 346 W/m2 K). (c) Compare the drag forces in the two cases. [There is 23.4 times as much drag in the glycerin.] 6.16 Air at −10◦ C flows over a smooth, sharp-edged, almost-flat, aerodynamic surface at 240 km/hr. The surface is at 10◦ C. Find (a) the approximate location of the laminar turbulent transition; (b) the overall h for a 2 m chord; (c) h at the trailing edge for a 2 m chord; (d) δ and h at the beginning of the transition region. [δxt = 0.54 mm.] 6.17 Find h in Example 6.10 using eqn. (6.120c) with Reu = 105 and 2 × 105 . Discuss the results. 6.18 For system described in Example 6.10, plot the local value of h over the whole length of the plate using eqn. (6.120c). On the same graph, plot h from eqn. (6.71) for Rex < 400, 000 and from eqn. (6.115) for Rex > 200, 000. Discuss the results. 6.19 Mercury at 25◦ C flows at 0.7 m/s over a 4 cm-long flat heater at 60◦ C. Find h, τ w , h(x = 0.04 m), and δ(x = 0.04 m). 6.20 A large plate is at rest in water at 15◦ C. The plate is suddenly translated parallel to itself, at 1.5 m/s. The resulting fluid movement is not exactly like that in a b.l. because the velocity profile builds up uniformly, all over, instead of from an edge. The governing transient momentum equation, Du/Dt = ν(∂ 2 u/∂y 2 ), takes the form ∂2u 1 ∂u = ν ∂t ∂y 2 Determine u at 0.015 m from the plate for t = 1, 10, and 1000 s. Do this by first posing the problem fully and then comparing it with the solution in Section 5.6. [u 0.003 m/s after 10 s.] Chapter 6: Laminar and turbulent boundary layers 334 6.21 Notice that, when Pr is large, the velocity b.l. on an isothermal, flat heater is much larger than δt . The small part of the velocity b.l. inside the thermal b.l. is approximately u/u∞ = 3 3 2 y/δ = 2 φ(y/δt ). Derive Nux for this case based on this velocity profile. 6.22 Plot the ratio of h(x)laminar to h(x)turbulent against Rex in the range of Rex that might be either laminar or turbulent. What does the plot suggest about heat transfer design? 6.23 Water at 7◦ C flows at 0.38 m/s across the top of a 0.207 m-long, thin copper plate. Methanol at 87◦ C flows across the bottom of the same plate, at the same speed but in the opposite direction. Make the obvious first guess as to the temperature at which to evaluate physical properties. Then plot the plate temperature as a function of position. (Do not bother to correct the physical properties in this problem, but note Problem 6.24.) 6.24 Work Problem 6.23 taking full account of property variations. 6.25 If the wall temperature in Example 6.6 (with a uniform qw = 420 W/m2 ) were instead fixed at its average value of 76◦ C, what would the average wall heat flux be? 6.26 A cold, 20 mph westerly wind at 20◦ F cools a rectangular building, 35 ft by 35 ft by 22 ft high, with a flat roof. The outer walls are at 27◦ F. Find the heat loss, conservatively assuming that the east and west faces have the same h as the north, south, and top faces. Estimate U for the walls. 6.27 A 2 ft-square slab of mild steel leaves a forging operation 0.25 in. thick at 1000◦ C. It is laid flat on an insulating bed and 27◦ C air is blown over it at 30 m/s. How long will it take to cool to 200◦ C. (State your assumptions about property evaluation.) 6.28 Do Problem 6.27 numerically, recalculating properties at successive points. If you did Problem 6.27, compare results. 6.29 Plot Tw against x for the situation described in Example 6.10. 6.30 Consider the plate in Example 6.10. Suppose that instead of specifying qw = 1000 W/m2 , we specified Tw = 200◦ C. Plot qw against x for this case. Problems 335 6.31 A thin metal sheet separates air at 44◦ C, flowing at 48 m/s, from water at 4◦ C, flowing at 0.2 m/s. Both fluids start at a leading edge and move in the same direction. Plot Tplate and q as a function of x up to x = 0.1 m. 6.32 A mixture of 60% glycerin and 40% water flows over a 1-mlong flat plate. The glycerin is at 20◦ C and the plate is at 40◦ . A thermocouple 1 mm above the trailing edge records 35◦ C. What is u∞ , and what is u at the thermocouple? 6.33 What is the maximum h that can be achieved in laminar flow over a 5 m plate, based on data from Table A.3? What physical circumstances give this result? 6.34 A 17◦ C sheet of water, ∆1 m thick and moving at a constant speed u∞ m/s, impacts a horizontal plate at 45◦ , turns, and flows along it. Develop a dimensionless equation for the thickness ∆2 at a distance L from the point of impact. Assume that δ ∆2 . Evaluate the result for u∞ = 1 m/s, ∆1 = 0.01 m, and L = 0.1 m, in water at 27◦ C. 6.35 A good approximation to the temperature dependence of µ in gases is given by the Sutherland formula: µ = µref T Tref 1.5 Tref + S , T +S where the reference state can be chosen anywhere. Use data for air at two points to evaluate S for air. Use this value to predict a third point. (T and Tref are expressed in kelvin.) 6.36 We have derived a steady-state continuity equation in Section 6.3. Now derive the time-dependent, compressible, three-dimensional version of the equation: ∂ρ + ∇ · (ρ u) = 0 ∂t To do this, paraphrase the development of equation (2.10), requiring that mass be conserved instead of energy. 6.37 Various considerations show that the smallest-scale motions in a turbulent flow have no preferred spatial orientation at Chapter 6: Laminar and turbulent boundary layers 336 large enough values of Re. Moreover, these small eddies are responsible for most of the viscous dissipation of kinetic energy. The dissipation rate, ε (W/kg), may be regarded as given information about the small-scale motion, since it is set by the larger-scale motion. Both ε and ν are governing parameters of the small-scale motion. a. Find the characteristic length and velocity scales of the small-scale motion. These are called the Kolmogorov scales of the flow. b. Compute Re for the small-scale motion and interpret the result. c. The Kolmogorov length scale characterizes the smallest motions found in a turbulent flow. If ε is 10 W/kg and the mean free path is 7 × 10−8 m, show that turbulent motion is a continuum phenomenon and thus is properly governed by the equations of this chapter. 6.38 The temperature outside is 35◦ F, but with the wind chill it’s −15◦ F. And you forgot your hat. If you go outdoors for long, are you in danger of freezing your ears? 6.39 To heat the airflow in a wind tunnel, an experimenter uses an array of electrically heated, horizontal Nichrome V strips. The strips are perpendicular to the flow. They are 20 cm long, very thin, 2.54 cm wide (in the flow direction), with the flat sides parallel to the flow. They are spaced vertically, each 1 cm above the next. Air at 1 atm and 20◦ C passes over them at 10 m/s. a. How much power must each strip deliver to raise the mean temperature of the airstream to 30◦ C? b. What is the heat flux if the electrical heating in the strips is uniformly distributed? c. What are the average and maximum temperatures of the strips? 6.40 An airflow sensor consists of a 5 cm long, heated copper slug that is smoothly embedded 10 cm from the leading edge of a flat plate. The overall length of the plate is 15 cm, and the width of the plate and the slug are both 10 cm. The slug is electrically heated by an internal heating element, but, owing Problems 337 to its high thermal conductivity, the slug has an essentially uniform temperature along its airside surface. The heater’s controller adjusts its power to keep the slug surface at a fixed temperature. The air velocity is found from measurements of the slug temperature, the air temperature, and the heating power needed to hold the slug at the set temperature. a. If the air is at 280 K, the slug is at 300 K, and the heater power is 5.0 W, find the airspeed assuming the flow is laminar. Hint: For x1 /x0 = 1.5 x1 x0 x −1/2 1 − (x0 /x)3/4 −1/3 √ dx = 1.0035 x0 b. Suppose that a disturbance trips the boundary layer near the leading edge, causing it to become turbulent over the whole plate. The air speed, air temperature, and the slug’s set-point temperature remain the same. Make a very rough estimate of the heater power that the controller now delivers, without doing a lot of analysis. 6.41 Equation (6.64) gives Nux for a flat plate with an unheated starting length. This equation may be derived using the integral energy equation [eqn. (6.47)], modelling the velocity and temperature profiles with eqns. (6.29) and (6.50), respectively, and taking δ(x) from eqn. (6.31a). Equation (6.52) is again obtained; however, in this case, φ = δt /δ is a function of x for x > x0 . Derive eqn. (6.64) by starting with eqn. (6.52), neglecting the term 3φ3 /280, and replacing δt by φδ. After some manipulation, you will obtain x 13 4d 3 φ + φ3 = 3 dx 14 Pr Show that its solution is φ3 = Cx −3/4 + 13 14 Pr for an unknown constant C . Then apply an appropriate initial condition and the definition of qw and Nux to obtain eqn. (6.64). Chapter 6: Laminar and turbulent boundary layers 338 References [6.1] S. Juhasz. Notes on Applied Mechanics Reviews – Referativnyi Zhurnal Mekhanika exhibit at XIII IUTAM, Moscow 1972. Appl. Mech. Rev., 26(2):145–160, 1973. [6.2] F.M. White. Viscous Fluid Flow. McGraw-Hill, Inc., New York, 2nd edition, 1991. [6.3] H. Schlichting. Boundary-Layer Theory. (trans. J. Kestin). McGrawHill Book Company, New York, 6th edition, 1968. [6.4] C. L. Tien and J. H. Lienhard. Statistical Thermodynamics. Hemisphere Publishing Corp., Washington, D.C., rev. edition, 1978. [6.5] S. W. Churchill and H. Ozoe. Correlations for laminar forced convection in flow over an isothermal flat plate and in developing and fully developed flow in an isothermal tube. J. Heat Trans., Trans. ASME, Ser. C, 95:78, 1973. [6.6] O. Reynolds. On the extent and action of the heating surface for steam boilers. Proc. Manchester Lit. Phil. Soc., 14:7–12, 1874. [6.7] J.A. Schetz. Foundations of Boundary Layer Theory for Momentum, Heat, and Mass Transfer. Prentice-Hall, Inc., Englewood Cliffs, NJ, 1984. [6.8] P. S. Granville. A modified Van Driest formula for the mixing length of turbulent boundary layers in pressure gradients. J. Fluids Engr., 111(1):94–97, 1989. [6.9] P. S. Granville. A near-wall eddy viscosity formula for turbulent boundary layers in pressure gradients suitable for momentum, heat, or mass transfer. J. Fluids Engr., 112(2):240–243, 1990. [6.10] F. M. White. A new integral method for analyzing the turbulent boundary layer with arbitrary pressure gradient. J. Basic Engr., 91: 371–378, 1969. [6.11] J. Boussinesq. Théorie de l’écoulement tourbillant. Mem. Pres. Acad. Sci., (Paris), 23:46, 1877. [6.12] F. M. White. Viscous Fluid Flow. McGraw-Hill Book Company, New York, 1974. References [6.13] B. S. Petukhov. Heat transfer and friction in turbulent pipe flow with variable physical properties. In T.F. Irvine, Jr. and J. P. Hartnett, editors, Advances in Heat Transfer, volume 6, pages 504–564. Academic Press, Inc., New York, 1970. [6.14] A. A. Žukauskas and A. B. Ambrazyavichyus. Heat transfer from a plate in a liquid flow. Int. J. Heat Mass Transfer, 3(4):305–309, 1961. [6.15] A. Žukauskas and A. Šlanciauskas. Heat Transfer in Turbulent Fluid Flows. Hemisphere Publishing Corp., Washington, 1987. [6.16] S. Whitaker. Forced convection heat transfer correlation for flow in pipes past flat plates, single cylinders, single spheres, and for flow in packed beds and tube bundles. AIChE J., 18:361, 1972. [6.17] S. W. Churchill. A comprehensive correlating equation for forced convection from flat plates. AIChE J., 22:264–268, 1976. [6.18] S. B. Pope. Turbulent Flows. Cambridge University Press, Cambridge, 2000. [6.19] P. A. Libby. Introduction to Turbulence. Taylor & Francis, Washington D.C., 1996. 339 7. Forced convection in a variety of configurations The bed was soft enough to suit me. . .But I soon found that there came such a draught of cold air over me from the sill of the window that this plan would never do at all, especially as another current from the rickety door met the one from the window and both together formed a series of small whirlwinds in the immediate vicinity of the spot where I had thought to spend the night. Moby Dick, H. Melville, 1851 7.1 Introduction Consider for a moment the fluid flow pattern within a shell-and-tube heat exchanger, such as that shown in Fig. 3.5. The shell-pass flow moves up and down across the tube bundle from one baffle to the next. The flow around each pipe is determined by the complexities of the one before it, and the direction of the mean flow relative to each pipe can vary. Yet the problem of determining the heat transfer in this situation, however difficult it appears to be, is a task that must be undertaken. The flow within the tubes of the exchanger is somewhat more tractable, but it, too, brings with it several problems that do not arise in the flow of fluids over a flat surface. Heat exchangers thus present a kind of microcosm of internal and external forced convection problems. Other such problems arise everywhere that energy is delivered, controlled, utilized, or produced. They arise in the complex flow of water through nuclear heating elements or in the liquid heating tubes of a solar collector—in the flow of a cryogenic liquid coolant in certain digital computers or in the circulation of refrigerant in the spacesuit of a lunar astronaut. We dealt with the simple configuration of flow over a flat surface in 341 342 Forced convection in a variety of configurations §7.2 Chapter 6. This situation has considerable importance in its own right, and it also reveals a number of analytical methods that apply to other configurations. Now we wish to undertake a sequence of progressively harder problems of forced convection heat transfer in more complicated flow configurations. Incompressible forced convection heat transfer problems normally admit an extremely important simplification: the fluid flow problem can be solved without reference to the temperature distribution in the fluid. Thus, we can first find the velocity distribution and then put it in the energy equation as known information and solve for the temperature distribution. Two things can impede this procedure, however: • If the fluid properties (especially µ and ρ ) vary significantly with temperature, we cannot predict the velocity without knowing the temperature, and vice versa. The problems of predicting velocity and temperature become intertwined and harder to solve. We encounter such a situation later in the study of natural convection, where the fluid is driven by thermally induced density changes. • Either the fluid flow solution or the temperature solution can, itself, become prohibitively hard to find. When that happens, we resort to the correlation of experimental data with the help of dimensional analysis. Our aim in this chapter is to present the analysis of a few simple problems and to show the progression toward increasingly empirical solutions as the problems become progressively more unwieldy. We begin this undertaking with one of the simplest problems: that of predicting laminar convection in a pipe. 7.2 Heat transfer to and from laminar flows in pipes Not many industrial pipe flows are laminar, but laminar heating and cooling does occur in an increasing variety of modern instruments and equipment: micro-electro-mechanical systems (MEMS), laser coolant lines, and many compact heat exchangers, for example. As in any forced convection problem, we first describe the flow field. This description will include a number of ideas that apply to turbulent as well as laminar flow. §7.2 Heat transfer to and from laminar flows in pipes Figure 7.1 The development of a laminar velocity profile in a pipe. Development of a laminar flow Figure 7.1 shows the evolution of a laminar velocity profile from the entrance of a pipe. Throughout the length of the pipe, the mass flow rate, ˙ m (kg/s), is constant, of course, and the average, or bulk, velocity uav is also constant: ˙ m= Ac ρu dAc = ρuav Ac (7.1) where Ac is the cross-sectional area of the pipe. The velocity profile, on the other hand, changes greatly near the inlet to the pipe. A b.l. builds up from the front, generally accelerating the otherwise undisturbed core. The b.l. eventually occupies the entire flow area and defines a velocity profile that changes very little thereafter. We call such a flow fully developed. A flow is fully developed from the hydrodynamic standpoint when ∂u =0 ∂x or v=0 (7.2) at each radial location in the cross section. An attribute of a dynamically fully developed flow is that the streamlines are all parallel to one another. The concept of a fully developed flow, from the thermal standpoint, is a little more complicated. We must first understand the notion of the ˆ mixing-cup, or bulk, enthalpy and temperature, hb and Tb . The enthalpy is of interest because we use it in writing the First Law of Thermodynamics when calculating the inflow of thermal energy and flow work to open control volumes. The bulk enthalpy is an average enthalpy for the fluid 343 344 Forced convection in a variety of configurations §7.2 flowing through a cross section of the pipe: ˙ˆ m hb ≡ Ac ˆ ρuh dAc (7.3) If we assume that fluid pressure variations in the pipe are too small to affect the thermodynamic state much (see Sect. 6.3) and if we assume a ˆ constant value of cp , then h = cp (T − Tref ) and ˙ m cp (Tb − Tref ) = Ac ρcp u (T − Tref ) dAc (7.4) or simply Ac Tb = ρcp uT dAc (7.5) ˙ mcp In words, then, Tb ≡ rate of flow of enthalpy through a cross section rate of flow of heat capacity through a cross section Thus, if the pipe were broken at any x -station and allowed to discharge into a mixing cup, the enthalpy of the mixed fluid in the cup would equal the average enthalpy of the fluid flowing through the cross section, and the temperature of the fluid in the cup would be Tb . This definition of Tb is perfectly general and applies to either laminar or turbulent flow. For a circular pipe, with dAc = 2π r dr , eqn. (7.5) becomes R Tb = ρcp uT 2π r dr 0 (7.6) R 0 ρcp u 2π r dr A fully developed flow, from the thermal standpoint, is one for which the relative shape of the temperature profile does not change with x . We state this mathematically as ∂ ∂x Tw − T Tw − Tb =0 (7.7) where T generally depends on x and r . This means that the profile can be scaled up or down with Tw − Tb . Of course, a flow must be hydrodynamically developed if it is to be thermally developed. §7.2 Heat transfer to and from laminar flows in pipes Figure 7.2 The thermal development of flows in tubes with a uniform wall heat flux and with a uniform wall temperature (the entrance region ). Figures 7.2 and 7.3 show the development of two flows and their subsequent behavior. The two flows are subjected to either a uniform wall heat flux or a uniform wall temperature. In Fig. 7.2 we see each flow develop until its temperature profile achieves a shape which, except for a linear stretching, it will retain thereafter. If we consider a small length of pipe, dx long with perimeter P , then its surface area is P dx (e.g., 2π R dx for a circular pipe) and an energy balance on it is1 ˙ˆ dQ = qw P dx = mdhb ˙ = mcp dTb (7.8) (7.9) so that qw P dTb = ˙ mcp dx 1 (7.10) Here we make the same approximations as were made in deriving the energy equation in Sect. 6.3. 345 346 Forced convection in a variety of configurations §7.2 Figure 7.3 The thermal behavior of flows in tubes with a uniform wall heat flux and with a uniform temperature (the thermally developed region ). This result is also valid for the bulk temperature in a turbulent flow. In Fig. 7.3 we see the fully developed variation of the temperature profile. If the flow is fully developed, the boundary layers are no longer growing thicker, and we expect that h will become constant. When qw is constant, then Tw − Tb will be constant in fully developed flow, so that the temperature profile will retain the same shape while the temperature rises at a constant rate at all values of r . Thus, at any radial position, dTb qw P ∂T = = = constant ˙ ∂x dx mcp (7.11) In the uniform wall temperature case, the temperature profile keeps the same shape, but its amplitude decreases with x , as does qw . The lower right-hand corner of Fig. 7.3 has been drawn to conform with this requirement, as expressed in eqn. (7.7). Heat transfer to and from laminar flows in pipes §7.2 The velocity profile in laminar tube flows The Buckingham pi-theorem tells us that if the hydrodynamic entry length, xe , required to establish a fully developed velocity profile depends on uav , µ , ρ , and D in three dimensions (kg, m, and s), then we expect to find two pi-groups: xe = fn (ReD ) D where ReD ≡ uav D/ν . The matter of entry length is discussed by White [7.1, Chap. 4], who quotes xe D 0.03 ReD (7.12) The constant, 0.03, guarantees that the laminar shear stress on the pipe wall will be within 5% of the value for fully developed flow when x > xe . The number 0.05 can be used, instead, if a deviation of just 1.4% is desired. The thermal entry length, xet , turns out to be different from xe . We deal with it shortly. The hydrodynamic entry length for a pipe carrying fluid at speeds near the transitional Reynolds number (2100) will extend beyond 100 diameters. Since heat transfer in pipes shorter than this is very often important, we will eventually have to deal with the entry region. The velocity profile for a fully developed laminar incompressible pipe flow can be derived from the momentum equation for an axisymmetric flow. It turns out that the b.l. assumptions all happen to be valid for a fully developed pipe flow: • The pressure is constant across any section. • ∂ 2 u ∂ x 2 is exactly zero. • The radial velocity is not just small, but it is zero. • The term ∂u ∂ x is not just small, but it is zero. The boundary layer equation for cylindrically symmetrical flows is quite similar to that for a flat surface, eqn. (6.13): u ∂u 1 dp ν∂ ∂u +v =− + ∂x ∂r ρ dx r ∂r r ∂u ∂r (7.13) 347 Forced convection in a variety of configurations §7.2 For fully developed flows, we go beyond the b.l. assumptions and set v and ∂u/∂x equal to zero as well, so eqn. (7.13) becomes 1d r dr r du dr = 1 dp µ dx We integrate this twice and get 1 dp 4µ dx u= r 2 + C1 ln r + C2 The two b.c.’s on u express the no-slip (or zero-velocity) condition at the wall and the fact that u must be symmetrical in r : u(r = R) = 0 and du dr r =0 =0 They give C1 = 0 and C2 = (−dp/dx)R 2 /4µ , so u= dp R2 − 4µ dx 1− r R 2 (7.14) This is the familiar Hagen-Poiseuille2 parabolic velocity profile. We can identify the lead constant (−dp/dx)R 2 4µ as the maximum centerline velocity, umax . In accordance with the conservation of mass (see Problem 7.1), 2uav = umax , so r u =2 1− uav R 2 (7.15) Thermal behavior of a flow with a uniform heat flux at the wall The b.l. energy equation for a fully developed laminar incompressible flow, eqn. (6.40), takes the following simple form in a pipe flow where the radial velocity is equal to zero: u 1∂ ∂T =α ∂x r ∂r r ∂T ∂r (7.16) 2 The German scientist G. Hagen showed experimentally how u varied with r , dp/dx , µ , and R , in 1839. J. Poiseuille (pronounced Pwa-zói or, more precisely, Pwä-z´¯) did e the same thing, almost simultaneously (1840), in France. Poiseuille was a physician interested in blood flow, and we find today that if medical students know nothing else about fluid flow, they know “Poiseuille’s law.” e 348 Heat transfer to and from laminar flows in pipes §7.2 For a fully developed flow with qw = constant, Tw and Tb increase linearly with x . In particular, by integrating eqn. (7.10), we find x Tb (x) − Tbin = 0 qw P x qw P dx = ˙ ˙ mcp mcp (7.17) Then, from eqns. (7.11) and (7.1), we get ∂T dTb qw P qw (2π R) 2qw α = = = = 2) ˙p mc ∂x dx ρcp uav (π R uav Rk Using this result and eqn. (7.15) in eqn. (7.16), we obtain r R 4 1− 2 1d qw = Rk r dr r dT dr (7.18) This ordinary d.e. in r can be integrated twice to obtain T= r2 r4 − 4 16R 2 4qw Rk + C1 ln r + C2 (7.19) The first b.c. on this equation is the symmetry condition, ∂T /∂r = 0 at r = 0, and it gives C1 = 0. The second b.c. is the definition of the mixing-cup temperature, eqn. (7.6). Substituting eqn. (7.19) with C1 = 0 into eqn. (7.6) and carrying out the indicated integrations, we get C2 = Tb − 7 qw R 24 k so T − Tb = qw R k r R 2 − 1 4 r R 4 − 7 24 (7.20) and at r = R , eqn. (7.20) gives Tw − Tb = 11 qw D 11 qw R = 24 k 48 k (7.21) so the local NuD for fully developed flow, based on h(x) = qw [Tw (x) − Tb (x)], is NuD ≡ qw D 48 = 4.364 = (Tw − Tb )k 11 (7.22) 349 350 Forced convection in a variety of configurations §7.2 Equation (7.22) is surprisingly simple. Indeed, the fact that there is only one dimensionless group in it is predictable by dimensional analysis. In this case the dimensional functional equation is merely h = fn (D, k) We exclude ∆T , because h should be independent of ∆T in forced convection; µ , because the flow is parallel regardless of the viscosity; and ρu2 , av because there is no influence of momentum in a laminar incompressible flow that never changes direction. This gives three variables, effectively in only two dimensions, W/K and m, resulting in just one dimensionless group, NuD , which must therefore be a constant. Example 7.1 Water at 20◦ C flows through a small-bore tube 1 mm in diameter at a uniform speed of 0.2 m/s. The flow is fully developed at a point beyond which a constant heat flux of 6000 W/m2 is imposed. How much farther down the tube will the water reach 74◦ C at its hottest point? Solution. As a fairly rough approximation, we evaluate properties at (74 + 20)/2 = 47◦ C: k = 0.6367 W/m·K, α = 1.541 × 10−7 , and ν = 0.556 × 10−6 m2 /s. Therefore, ReD = (0.001 m)(0.2 m/s)/0.556 × 10−6 m2 /s = 360, and the flow is laminar. Then, noting that T is greatest at the wall and setting x = L at the point where Twall = 74◦ C, eqn. (7.17) gives: Tb (x = L) = 20 + qw P 4qw α L L = 20 + ˙ mcp uav Dk And eqn. (7.21) gives 74 = Tb (x = L) + 4qw α 11 qw D 11 qw D = 20 + L+ 48 k uav Dk 48 k so 11 qw D L = 54 − D 48 k uav k 4qw α or 11 6000(0.001) L = 54 − D 48 0.6367 0.2(0.6367) = 1785 4(6000)1.541(10)−7 Heat transfer to and from laminar flows in pipes §7.2 so the wall temperature reaches the limiting temperature of 74◦ C at L = 1785(0.001 m) = 1.785 m While we did not evaluate the thermal entry length here, it may be shown to be much, much less than 1785 diameters. In the preceding example, the heat transfer coefficient is actually rather large h = NuD 0.6367 k = 4.364 = 2, 778 W/m2 K D 0.001 The high h is a direct result of the small tube diameter, which limits the thermal boundary layer to a small thickness and keeps the thermal resistance low. This trend leads directly to the notion of a microchannel heat exchanger. Using small scale fabrication technologies, such as have been developed in the semiconductor industry, it is possible to create channels whose characteristic diameter is in the range of 100 µm, resulting in heat transfer coefficients in the range of 104 W/m2 K for water [7.2]. If, instead, liquid sodium (k ≈ 80 W/m·K) is used as the working fluid, the laminar flow heat transfer coefficient is on the order of 106 W/m2 K — a range that is usually associated with boiling processes! Thermal behavior of the flow in an isothermal pipe The dimensional analysis that showed NuD = constant for flow with a uniform heat flux at the wall is unchanged when the pipe wall is isothermal. Thus, NuD should still be constant. But this time (see, e.g., [7.3, Chap. 8]) the constant changes to NuD = 3.657, Tw = constant (7.23) for fully developed flow. The behavior of the bulk temperature is discussed in Sect. 7.4. The thermal entrance region The thermal entrance region is of great importance in laminar flow because the thermally undeveloped region becomes extremely long for higherPr fluids. The entry-length equation (7.12) takes the following form for 351 352 Forced convection in a variety of configurations §7.2 the thermal entry region3 , where the velocity profile is assumed to be fully developed before heat transfer starts at x = 0: xet D 0.034 ReD Pr (7.24) Thus, the thermal entry length for the flow of cold water (Pr 10) can be over 600 diameters in length near the transitional Reynolds number, and oil flows (Pr on the order of 104 ) practically never achieve fully developed temperature profiles. A complete analysis of the heat transfer rate in the thermal entry region becomes quite complicated. The reader interested in details should look at [7.3, Chap. 8]. Dimensional analysis of the entry problem shows that the local value of h depends on uav , µ , ρ , D , cp , k, and x —eight variables in m, s, kg, and J K. This means that we should anticipate four pi-groups: NuD = fn (ReD , Pr, x/D) (7.25) In other words, to the already familiar NuD , ReD , and Pr, we add a new length parameter, x/D . The solution of the constant wall temperature problem, originally formulated by Graetz in 1885 [7.6] and solved in convenient form by Sellars, Tribus, and Klein in 1956 [7.7], includes an arrangement of these dimensionless groups, called the Graetz number: Graetz number, Gz ≡ ReD Pr D x (7.26) Figure 7.4 shows values of NuD ≡ hD/k for both the uniform wall temperature and uniform wall heat flux cases. The independent variable in the figure is a dimensionless length equal to 2/Gz. The figure also presents an average Nusselt number, NuD for the isothermal wall case: NuD ≡ 3 D hD = k k 1 L L h dx 0 = 1 L L 0 NuD dx (7.27) The Nusselt number will be within 5% of the fully developed value if xet 0.034 ReD PrD for Tw = constant. The error decreases to 1.4% if the coefficient is raised from 0.034 to 0.05 [Compare this with eqn. (7.12) and its context.]. For other situations, the coefficient changes. With qw = constant, it is 0.043 at a 5% error level; when the velocity and temperature profiles develop simultaneously, the coefficient ranges between about 0.028 and 0.053 depending upon the Prandtl number and the wall boundary condition [7.4, 7.5]. §7.2 Heat transfer to and from laminar flows in pipes Figure 7.4 Local and average Nusselt numbers for the thermal entry region in a hydrodynamically developed laminar pipe flow. where, since h = q(x) [Tw − Tb (x)], it is not possible to average just q or ∆T . We show how to find the change in Tb using h for an isothermal wall in Sect. 7.4. For a fixed heat flux, the change in Tb is given by eqn. (7.17), and a value of h is not needed. For an isothermal wall, the following curve fits are available for the Nusselt number in thermally developing flow [7.4]: NuD = 3.657 + NuD = 3.657 + 0.0018 Gz1/3 0.04 + Gz−2/3 0.0668 Gz1/3 0.04 + Gz−2/3 2 (7.28) (7.29) The error is less than 14% for Gz > 1000 and less than 7% for Gz < 1000. For fixed qw , a more complicated formula reproduces the exact result for local Nusselt number to within 1%: ⎧ ⎪1.302 Gz1/3 − 1 for 2 × 104 ≤ Gz ⎪ ⎨ NuD = 1.302 Gz1/3 − 0.5 for 667 ≤ Gz ≤ 2 × 104 (7.30) ⎪ ⎪ ⎩ 4.364 + 0.263 Gz0.506 e−41/Gz for 0 ≤ Gz ≤ 667 353 354 Forced convection in a variety of configurations §7.2 Example 7.2 A fully developed flow of air at 27◦ C moves at 2 m/s in a 1 cm I.D. pipe. An electric resistance heater surrounds the last 20 cm of the pipe and supplies a constant heat flux to bring the air out at Tb = 40◦ C. What power input is needed to do this? What will be the wall temperature at the exit? Solution. This is a case in which the wall heat flux is uniform along the pipe. We first must compute Gz20 cm , evaluating properties at (27 + 40) 2 34◦ C. Gz20 cm ReD Pr D x (2 m/s)(0.01 m) (0.711)(0.01 m) 16.4 × 10−6 m2 /s = 43.38 = 0.2 m = From eqn. 7.30, we compute NuD = 5.05, so Twexit − Tb = qw D 5.05 k Notice that we still have two unknowns, qw and Tw . The bulk temperature is specified as 40◦ C, and qw is obtained from this number by a simple energy balance: qw (2π Rx) = ρcp uav (Tb − Tentry )π R 2 so qw = 1.159 m kg J R · 2 · (40 − 27)◦ C · · 1004 = 378 W/m2 m3 kg·K s 2x 1/80 Then Twexit = 40◦ C + (378 W/m2 )(0.01 m) = 68.1◦ C 5.05(0.0266 W/m·K) Turbulent pipe flow §7.3 7.3 Turbulent pipe flow Turbulent entry length The entry lengths xe and xet are generally shorter in turbulent flow than in laminar flow. Table 7.1 gives the thermal entry length for various values of Pr and ReD , based on NuD lying within 5% of its fully developed value. These results are for a uniform wall heat flux imposed on a hydrodynamically fully developed flow. Similar results are obtained for a uniform wall temperature. For Prandtl numbers typical of gases and nonmetallic liquids, the entry length is not strongly sensitive to the Reynolds number. For Pr > 1 in particular, the entry length is just a few diameters. This is because the heat transfer rate is controlled by the thin thermal sublayer on the wall, which develops very quickly. Only liquid metals give fairly long thermal entrance lengths, and, for these fluids, xet depends on both Re and Pr in a complicated way. Since liquid metals have very high thermal conductivities, the heat transfer rate is also more strongly affected by the temperature distribution in the center of the pipe. We discusss liquid metals in more detail at the end of this section. When heat transfer begins at the inlet to a pipe, the velocity and temperature profiles develop simultaneously. The entry length is then very strongly affected by the shape of the inlet. For example, an inlet that induces vortices in the pipe, such as a sharp bend or contraction, can create Table 7.1 Thermal entry lengths, xet /D , for which NuD will be no more than 5% above its fully developed value in turbulent flow Pr 0.01 0.7 3.0 ReD 20,000 7 10 4 100,000 22 12 3 500,000 32 14 3 355 356 Forced convection in a variety of configurations §7.3 Table 7.2 Constants for the gas-flow simultaneous entry length correlation, eqn. (7.31), for various inlet configurations Inlet configuration C n Long, straight pipe Square-edged inlet 180◦ circular bend 90◦ circular bend 90◦ sharp elbow 0.9756 2.4254 0.9759 1.0517 2.0152 0.760 0.676 0.700 0.629 0.614 a much longer entry length than occurs for a thermally developing flow. These vortices may require 20 to 40 diameters to die out. For various types of inlets, Bhatti and Shah [7.8] provide the following correlation for NuD with L/D > 3 for air (or other fluids with Pr ≈ 0.7) C NuD =1+ Nu∞ (L/D)n for Pr = 0.7 (7.31) where Nu∞ is the fully developed value of the Nusselt number, and C and n depend on the inlet configuration as shown in Table 7.2. Whereas the entry effect on the local Nusselt number is confined to a few ten’s of diameters, the effect on the average Nusselt number may persist for a hundred diameters. This is because much additional length is needed to average out the higher heat transfer rates near the entry. The discussion that follows deals almost entirely with fully developed turbulent pipe flows. Illustrative experiment Figure 7.5 shows average heat transfer data given by Kreith [7.9, Chap. 8] for air flowing in a 1 in. I.D. isothermal pipe 60 in. in length. Let us see how these data compare with what we know about pipe flows thus far. The data are plotted for a single Prandtl number on NuD vs. ReD coordinates. This format is consistent with eqn. (7.25) in the fully developed range, but the actual pipe incorporates a significant entry region. Therefore, the data will reflect entry behavior. For laminar flow, NuD 3.66 at ReD = 750. This is the correct value for an isothermal pipe. However, the pipe is too short for flow to be fully developed over much, if any, of its length. Therefore NuD is not constant Turbulent pipe flow §7.3 357 Figure 7.5 Heat transfer to air flowing in a 1 in. I.D., 60 in. long pipe (after Kreith [7.9]). in the laminar range. The rate of rise of NuD with ReD becomes very great in the transitional range, which lies between ReD = 2100 and about 5000 5000, the flow is turbulent and it turns out in this case. Above ReD that NuD Re0.8 . D The Reynolds analogy and heat transfer A form of the Reynolds analogy appropriate to fully developed turbulent pipe flow can be derived from eqn. (6.111) Stx = Cf (x) 2 h = ρcp u∞ 1 + 12.8 Pr0.68 − 1 (6.111) Cf (x) 2 where h, in a pipe flow, is defined as qw /(Tw − Tb ). We merely replace u∞ with uav and Cf (x) with the friction coefficient for fully developed pipe flow, Cf (which is constant), to get St = Cf 2 h = ρcp uav 1 + 12.8 Pr0.68 − 1 (7.32) Cf 2 This should not be used at very low Pr’s, but it can be used in either uniform qw or uniform Tw situations. It applies only to smooth walls. 358 Forced convection in a variety of configurations §7.3 The frictional resistance to flow in a pipe is normally expressed in terms of the Darcy-Weisbach friction factor, f [recall eqn. (3.24)]: f≡ head loss u2 av pipe length D 2 = ∆p L ρu2 av D2 (7.33) where ∆p is the pressure drop in a pipe of length L. However, τw = ∆p (π /4)D 2 ∆pD frictional force on liquid = = surface area of pipe π DL 4L so f= τw = 4Cf ρu2 /8 av (7.34) Substituting eqn. (7.34) in eqn. (7.32) and rearranging the result, we obtain, for fully developed flow, NuD = f 8 ReD Pr 1 + 12.8 Pr0.68 − 1 (7.35) f8 The friction factor is given graphically in Fig. 7.6 as a function of ReD and the relative roughness, ε/D , where ε is the root-mean-square roughness of the pipe wall. Equation (7.35) can be used directly along with Fig. 7.6 to calculate the Nusselt number for smooth-walled pipes. Historical formulations. A number of the earliest equations for the Nusselt number in turbulent pipe flow were based on Reynolds analogy in the form of eqn. (6.76), which for a pipe flow becomes St = Cf 2 Pr−2/3 = f Pr−2/3 8 (7.36) or NuD = ReD Pr1/3 f /8 (7.37) For smooth pipes, the curve ε/D = 0 in Fig. 7.6 is approximately given by this equation: f 0.046 = Cf = 4 Re0.2 D (7.38) 359 Figure 7.6 Pipe friction factors. 360 Forced convection in a variety of configurations §7.3 in the range 20, 000 < ReD < 300, 000, so eqn. (7.37) becomes NuD = 0.023 Pr1/3 Re0.8 D for smooth pipes. This result was given by Colburn [7.10] in 1933. Actually, it is quite similar to an earlier result developed by Dittus and Boelter in 1930 (see [7.11, pg. 552]) for smooth pipes. NuD = 0.0243 Pr0.4 Re0.8 D (7.39) These equations are intended for reasonably low temperature differences under which properties can be evaluated at a mean temperature (Tb + Tw )/2. In 1936, a study by Sieder and Tate [7.12] showed that when |Tw − Tb | is large enough to cause serious changes of µ , the Colburn equation can be modified in the following way for liquids: NuD = 0.023 Re0.8 Pr1/3 D µb µw 0.14 (7.40) where all properties are evaluated at the local bulk temperature except µw , which is the viscosity evaluated at the wall temperature. These early relations proved to be reasonably accurate. They gave maximum errors of +25% and −40% in the range 0.67 Pr < 100 and usually were considerably more accurate than this. However, subsequent research has provided far more data, and better theoretical and physical understanding of how to represent them accurately. Modern formulations. During the 1950s and 1960s, B. S. Petukhov and his co-workers at the Moscow Institute for High Temperature developed a vastly improved description of forced convection heat transfer in pipes. Much of this work is described in a 1970 survey article by Petukhov [7.13]. Petukhov recommends the following equation, which is built from eqn. (7.35), for the local Nusselt number in fully developed flow in smooth pipes where all properties are evaluated at Tb . NuD = (f /8) ReD Pr 1.07 + 12.7 f /8 Pr2/3 − 1 where 104 < ReD < 5 × 106 0.5 < Pr < 200 200 Pr < 2000 for 6% accuracy for 10% accuracy (7.41) Turbulent pipe flow §7.3 361 and where the friction factor for smooth pipes is given by f= 1 1.82 log10 ReD − 1.64 2 (7.42) Gnielinski [7.14] later showed that the range of validity could be extended down to the transition Reynolds number by making a small adjustment to eqn. (7.41): NuD = (f /8) (ReD − 1000) Pr 1 + 12.7 f /8 Pr2/3 − 1 (7.43) for 2300 ≤ ReD ≤ 5 × 106 . Variations in physical properties. Sieder and Tate’s work on property variations was also refined in later years [7.13]. The effect of variable physical properties is dealt with differently for liquids and gases. In both cases, the Nusselt number is first calculated with all properties evaluated at Tb using eqn. (7.41) or (7.43). For liquids, one then corrects by multiplying with a viscosity ratio. Over the interval 0.025 ≤ (µb /µw ) ≤ 12.5, ⎧ n ⎨0.11 for Tw > T µb b NuD = NuD where n = (7.44) ⎩0.25 for Tw < Tb Tb µw For gases a ratio of temperatures in kelvins is used, with 0.27 ≤ (Tb /Tw ) ≤ 2.7, ⎧ ⎨0.47 for Tw > T n Tb b where n = (7.45) NuD = NuD ⎩0 Tb Tw for Tw < Tb After eqn. (7.42) is used to calculate NuD , it should also be corrected for the effect of variable viscosity. For liquids, with 0.5 ≤ (µb /µw ) ≤ 3 ⎧ ⎪(7 − µ /µ )/6 for T > T ⎨ w w b b ×K where K = (7.46) f =f ⎪ Tb ⎩(µb /µw )−0.24 for Tw < Tb For gases, the data are much weaker [7.15, 7.16]. For 0.14 ≤ (Tb /Tw ) ≤ 3.3 ⎧ ⎨0.23 for Tw > T m Tb b f =f where m ≈ (7.47) ⎩0.23 for Tw < Tb Tb Tw 362 Forced convection in a variety of configurations §7.3 Example 7.3 A 21.5 kg/s flow of water is dynamically and thermally developed in a 12 cm I.D. pipe. The pipe is held at 90◦ C and ε/D = 0. Find h and f where the bulk temperature of the fluid has reached 50◦ C. Solution. uav = ˙ m 21.5 = = 1.946 m/s ρAc 977π (0.06)2 ReD = 1.946(0.12) uav D = = 573, 700 ν 4.07 × 10−7 so and Pr = 2.47, 5.38 × 10−4 µb = = 1.74 µw 3.10 × 10−4 From eqn. (7.42), f = 0.0128 at Tb , and since Tw > Tb , n = 0.11 in eqn. (7.44). Thus, with eqn. (7.41) we have NuD = (0.0128/8)(5.74 × 105 )(2.47) (1.74)0.11 = 1617 1.07 + 12.7 0.0128/8 2.472/3 − 1 or h = NuD 0.661 k = 1617 = 8, 907 W/m2 K D 0.12 The corrected friction factor, with eqn. (7.46), is f = (0.0128) (7 − 1.74)/6 = 0.0122 Rough-walled pipes. Roughness on a pipe wall can disrupt the viscous and thermal sublayers if it is sufficiently large. Figure 7.6 shows the effect of increasing root-mean-square roughness height ε on the friction factor, f . As the Reynolds number increases, the viscous sublayer becomes thinner and smaller levels of roughness influence f . Some typical pipe roughnesses are given in Table 7.3. The importance of a given level of roughness on friction and heat transfer can determined by comparing ε to the sublayer thickness. We saw in Sect. 6.7 that the thickness of the sublayer is around 30 times Turbulent pipe flow §7.3 363 Table 7.3 Typical wall roughness of commercially available pipes when new. Pipe ε (µm) Glass Drawn tubing Steel or wrought iron Pipe 0.31 1.5 46. Asphalted cast iron Galvanized iron Cast iron ε (µm) 120. 150. 260. ν/u∗ , where u∗ = w /ρ was the friction velocity. We can define the ratio of ε and ν/u∗ as the roughness Reynolds number, Reε Reε ≡ u∗ ε ε = ReD ν D f 8 (7.48) where the second equality follows from the definitions of u∗ and f (and a little algebra). Experimental data then show that the smooth, transitional, and fully rough regions seen in Fig. 7.6 correspond to the following ranges of Reε : Reε < 5 hydraulically smooth 5 ≤ Reε ≤ 70 70 < Reε transitionally rough fully rough In the fully rough regime, Bhatti and Shah [7.8] provide the following correlation for the local Nusselt number NuD = (f /8) ReD Pr (7.49) 1 + f /8 4.5 Re0.2 Pr0.5 − 8.48 ε which applies for the ranges 104 ReD , 0.5 Pr 10, and 0.002 ε D 0.05 The corresponding friction factor may be computed from Haaland’s equation [7.17]: 1 f= 1.8 log10 6.9 ε/D + ReD 3.7 1.11 2 (7.50) 364 Forced convection in a variety of configurations §7.3 The heat transfer coefficient on a rough wall can be several times that for a smooth wall at the same Reynolds number. The friction factor, and thus the pressure drop and pumping power, will also be higher. Nevertheless, designers sometimes deliberately roughen tube walls so as to raise h and reduce the surface area needed for heat transfer. Several manufacturers offer tubing that has had some pattern of roughness impressed upon its interior surface. Periodic ribs are one common configuration. Specialized correlations have been developed for a number of such configurations [7.18, 7.19]. Example 7.4 Repeat Example 7.3, now assuming the pipe to be cast iron with a wall roughness of ε = 260 µm. Solution. The Reynolds number and physical properties are unchanged. From eqn. (7.50) ⎧ ⎡ ⎤⎫ 1.11 ⎬−2 ⎨ 260 × 10−6 0.12 6.9 ⎦ + f = 1.8 log10 ⎣ ⎭ ⎩ 573, 700 3.7 =0.02424 The roughness Reynolds number is then Reε = (573, 700) 260 × 10−6 0.12 0.02424 = 68.4 8 This corresponds to fully rough flow. With eqn. (7.49) we have NuD = (0.02424/8)(5.74 × 105 )(2.47) 1 + 0.02424/8 4.5(68.4)0.2 (2.47)0.5 − 8.48 = 2, 985 so h = 2985 0.661 = 16.4 kW/m2 K 0.12 In this case, wall roughness causes a factor of 1.8 increase in h and a factor of 2.0 increase in f and the pumping power. We have omitted the variable properties corrections here because they were developed for smooth-walled pipes. §7.3 Turbulent pipe flow 365 Figure 7.7 Velocity and temperature profiles during fully developed turbulent flow in a pipe. Heat transfer to fully developed liquid-metal flows in tubes A dimensional analysis of the forced convection flow of a liquid metal over a flat surface [recall eqn. (6.60) et seq.] showed that Nu = fn(Pe) (7.51) because viscous influences were confined to a region very close to the wall. Thus, the thermal b.l., which extends far beyond δ, is hardly influenced by the dynamic b.l. or by viscosity. During heat transfer to liquid metals in pipes, the same thing occurs as is illustrated in Fig. 7.7. The region of thermal influence extends far beyond the laminar sublayer, when Pr 1, and the temperature profile is not influenced by the sublayer. Conversely, if Pr 1, the temperature profile is largely shaped within the laminar sublayer. At high or even moderate Pr’s, ν is therefore very important, but at low Pr’s it vanishes from the functional equation. Equation (7.51) thus applies to pipe flows as well as to flow over a flat surface. Numerous measured values of NuD for liquid metals flowing in pipes with a constant wall heat flux, qw , were assembled by Lubarsky and Kaufman [7.20]. They are included in Fig. 7.8. It is clear that while most of the data correlate fairly well on NuD vs. Pe coordinates, certain sets of data are badly scattered. This occurs in part because liquid metal experiments are hard to carry out. Temperature differences are small and must often be measured at high temperatures. Some of the very low data might possibly result from a failure of the metals to wet the inner surface of the pipe. Another problem that besets liquid metal heat transfer measurements is the very great difficulty involved in keeping such liquids pure. Most 366 Forced convection in a variety of configurations §7.3 Figure 7.8 Comparison of measured and predicted Nusselt numbers for liquid metals heated in long tubes with uniform wall heat flux, qw . (See NACA TN 336, 1955, for details and data source references.) impurities tend to result in lower values of h. Thus, most of the Nusselt numbers in Fig. 7.8 have probably been lowered by impurities in the liquids; the few high values are probably the more correct ones for pure liquids. There is a body of theory for turbulent liquid metal heat transfer that yields a prediction of the form NuD = C1 + C2 Pe0.8 D (7.52) where the Péclét number is defined as PeD = uav D/α. The constants are 7 and 0.0185 C2 0.386 according normally in the ranges 2 C1 to the test circumstances. Using the few reliable data sets available for uniform wall temperature conditions, Reed [7.21] recommends NuD = 3.3 + 0.02 Pe0.8 D (7.53) (Earlier work by Seban and Shimazaki [7.22] had suggested C1 = 4.8 and C2 = 0.025.) For uniform wall heat flux, many more data are available, Heat transfer surface viewed as a heat exchanger §7.4 and Lyon [7.23] recommends the following equation, shown in Fig. 7.8: NuD = 7 + 0.025 Pe0.8 D (7.54) In both these equations, properties should be evaluated at the average of the inlet and outlet bulk temperatures and the pipe flow should have L/D > 60 and PeD > 100. For lower PeD , axial heat conduction in the liquid metal may become significant. Although eqns. (7.53) and (7.54) are probably correct for pure liquids, we cannot overlook the fact that the liquid metals in actual use are seldom pure. Lubarsky and Kaufman [7.20] put the following line through the bulk of the data in Fig. 7.8: NuD = 0.625 Pe0.4 D (7.55) The use of eqn. (7.55) for qw = constant is far less optimistic than the use of eqn. (7.54). It should probably be used if it is safer to err on the low side. 7.4 Heat transfer surface viewed as a heat exchanger Let us reconsider the problem of a fluid flowing through a pipe with a uniform wall temperature. By now we can predict h for a pretty wide range of conditions. Suppose that we need to know the net heat transfer to a pipe of known length once h is known. This problem is complicated by the fact that the bulk temperature, Tb , is varying along its length. However, we need only recognize that such a section of pipe is a heat exchanger whose overall heat transfer coefficient, U (between the wall and the bulk), is just h. Thus, if we wish to know how much pipe surface area is needed to raise the bulk temperature from Tbin to Tbout , we can calculate it as follows: ˙ Q = (mcp)b Tbout − Tbin = hA(LMTD) or A= ˙ (mcp)b Tbout − Tbin h ln Tbout − Tw Tbin − Tw Tbout − Tw − Tbin − Tw (7.56) By the same token, heat transfer in a duct can be analyzed with the effectiveness method (Sect. 3.3) if the exiting fluid temperature is unknown. 367 368 Forced convection in a variety of configurations §7.4 Suppose that we do not know Tbout in the example above. Then we can write an energy balance at any cross section, as we did in eqn. (7.8): ˙ dQ = qw P dx = hP (Tw − Tb ) dx = mcP dTb Integration can be done from Tb (x = 0) = Tbin to Tb (x = L) = Tbout L 0 P ˙ mcp hP dx = − ˙ mcp Tbout Tbin L 0 h dx = − ln d(Tw − Tb ) (Tw − Tb ) Tw − Tbout Tw − Tbin We recognize in this the definition of h from eqn. (7.27). Hence, hP L = − ln ˙ mcp Tw − Tbout Tw − Tbin which can be rearranged as Tbout − Tbin hP L = 1 − exp − ˙ mcp Tw − Tbin (7.57) This equation can be used in either laminar or turbulent flow to compute the variation of bulk temperature if Tbout is replaced by Tb (x), L is replaced by x , and h is adjusted accordingly. The left-hand side of eqn. (7.57) is the heat exchanger effectiveness. On the right-hand side we replace U with h; we note that P L = A, the ˙ exchanger surface area; and we write Cmin = mcp . Since Tw is uniform, the stream that it represents must have a very large capacity rate, so that Cmin /Cmax = 0. Under these substitutions, we identify the argument of the exponential as NTU = U A/Cmin , and eqn. (7.57) becomes ε = 1 − exp (−NTU) (7.58) which we could have obtained directly, from either eqn. (3.20) or (3.21), by setting Cmin /Cmax = 0. A heat exchanger for which one stream is isothermal, so that Cmin /Cmax = 0, is sometimes called a single-stream heat exchanger. Equation (7.57) applies to ducts of any cross-sectional shape. We can cast it in terms of the hydraulic diameter, Dh = 4Ac /P , by substituting Heat transfer surface viewed as a heat exchanger §7.4 ˙ m = ρuav Ac : Tbout − Tbin hP L = 1 − exp − Tw − Tbin ρuav cp Ac = 1 − exp − h 4L ρuav cp Dh (7.59a) (7.59b) For a circular tube, with Ac = π D 2 /4 and P = π D , Dh = 4(π D 2 /4) (π D) = D . To use eqn. (7.59b) for a noncircular duct, of course, we will need the value of h for its more complex geometry. We consider this issue in the next section. Example 7.5 Air at 20◦ C is hydrodynamically fully developed as it flows in a 1 cm I.D. pipe. The average velocity is 0.7 m/s. If it enters a section where the pipe wall is at 60◦ C, what is the temperature 0.25 m farther downstream? Solution. ReD = (0.7)(0.01) uav D = = 422 ν 1.66 × 10−5 The flow is therefore laminar. To account for the thermal entry region, we compute the Graetz number from eqn. (7.26) Gz = ReD Pr D (422)(0.709)(0.01) = = 12.0 x 0.25 Substituting this value into eqn. (7.29), we find NuD = 4.32. Thus, h= 3.657(0.0268) = 11.6 W/m2 K 0.01 Then, using eqn. (7.59b), Tbout − Tbin 11.6 4(0.25) = 1 − exp − Tw − Tbin 1.14(1007)(0.7) 0.01 so that Tb − 20 = 0.764 60 − 20 or Tb = 50.6◦ C 369 370 Forced convection in a variety of configurations 7.5 §7.5 Heat transfer coefficients for noncircular ducts So far, we have focused on flows within circular tubes, which are by far the most common configuration. Nevertheless, other cross-sectional shapes often occur. For example, the fins of a heat exchanger may form a rectangular passage through which air flows. Sometimes, the passage crosssection is very irregular, as might happen when fluid passes through a clearance between other objects. In situations like these, all the qualitative ideas that we developed in Sections 7.1–7.3 still apply, but the Nusselt numbers for circular tubes cannot be used in calculating heat transfer rates. The hydraulic diameter, which was introduced in connection with eqn. (7.59b), provides a basis for approximating heat transfer coefficients in noncircular ducts. Recall that the hydraulic diameter is defined as Dh ≡ 4 Ac P (7.60) where Ac is the cross-sectional area and P is the passage’s wetted perimeter (Fig. 7.9). The hydraulic diameter measures the fluid area per unit length of wall. In turbulent flow, where most of the convection resistance is in the sublayer on the wall, this ratio determines the heat transfer coefficient to within about ±20% across a broad range of duct shapes. In fully-developed laminar flow, where the thermal resistance extends into the core of the duct, the heat transfer coefficient depends on the details of the duct shape, and Dh alone cannot define the heat transfer coefficient. Nevertheless, the hydraulic diameter provides an appropriate characteristic length for cataloging laminar Nusselt numbers. Figure 7.9 Flow in a noncircular duct. Heat transfer coefficients for noncircular ducts §7.5 The factor of four in the definition of Dh ensures that it gives the actual diameter of a circular tube. We noted in the preceding section that, for a circular tube of diameter D , Dh = D . Some other important cases include: a rectangular duct of width a and height b Dh = an annular duct of inner diameter Di and outer diameter Do Dh = 2ab 4 ab = 2a + 2b a+b 2 2 4 π Do 4 − π Di 4 π (Do + Di ) = (Do − Di ) and, for very wide parallel plates, eqn. (7.61a) with a two parallel plates a distance b apart (7.61a) Dh = 2b (7.61b) b gives (7.61c) Turbulent flow in noncircular ducts With some caution, we may use Dh directly in place of the circular tube diameter when calculating turbulent heat transfer coefficients and bulk temperature changes. Specifically, Dh replaces D in the Reynolds number, which is then used to calculate f and NuDh from the circular tube formulas. The mass flow rate and the bulk velocity must be based on 2 the true cross-sectional area, which does not usually equal π Dh /4 (see Problem 7.46). The following example illustrates the procedure. Example 7.6 An air duct carries chilled air at an inlet bulk temperature of Tbin = 17◦ C and a speed of 1 m/s. The duct is made of thin galvanized steel, has a square cross-section of 0.3 m by 0.3 m, and is not insulated. A length of the duct 15 m long runs outdoors through warm air at T∞ = 37◦ C. The heat transfer coefficient on the outside surface, due to natural convection and thermal radiation, is 5 W/m2 K. Find the bulk temperature change of the air over this length. Solution. The hydraulic diameter, from eqn. (7.61a) with a = b, is simply Dh = a = 0.3 m 371 372 Forced convection in a variety of configurations §7.5 Using properties of air at the inlet temperature (290 K), the Reynolds number is ReDh = (1)(0.3) uav Dh = = 19, 011 ν (1.578 × 10−5 ) The Reynolds number for turbulent transition in a noncircular duct is typically approximated by the circular tube value of about 2300, so this flow is turbulent. The friction factor is obtained from eqn. (7.42) f = 1.82 log10 (19, 011) − 1.64 −2 = 0.02646 and the Nusselt number is found with Gnielinski’s equation, (7.43) NuDh = (0.02646/8)(19, 011 − 1, 000)(0.713) = 49.82 1 + 12.7 0.02646/8 (0.713)2/3 − 1 The heat transfer coefficient is h = NuDh (49.82)(0.02623) k = 4.371 W/m2 K = Dh 0.3 The remaining problem is to find the bulk temperature change. The thin metal duct wall offers little thermal resistance, but convection resistance outside the duct must be considered. Heat travels first from the air at T∞ through the outside heat transfer coefficient to the duct wall, through the duct wall, and then through the inside heat transfer coefficient to the flowing air — effectively through three resistances in series from the fixed temperature T∞ to the rising temperature Tb . We have seen in Section 2.4 that an overall heat transfer coefficient may be used to describe such series resistances. Here, with Ainside Aoutside , we find U based on inside area to be U= 1 Ainside 1 1 + Rt wall + (hA)inside (hA)outside −1 neglect = 1 1 + 4.371 5 −1 = 2.332 W/m2 K We then adapt eqn. (7.59b) by replacing h by U and Tw by T∞ : Tbout − Tbin U 4L = 1 − exp − T∞ − Tbin ρuav cp Dh = 1 − exp − 4(15) 2.332 (1.217)(1)(1007) 0.3 = 0.3165 The outlet bulk temperature is therefore Tbout = [17 + (37 − 17)(0.3165)] ◦ C = 23.3 ◦ C §7.5 Heat transfer coefficients for noncircular ducts The results obtained by substituting Dh for D in turbulent circular tube formulæ are generally accurate to within ±20% and are often within ±10%. Worse results are obtained for duct cross-sections having sharp corners, such as an acute triangle. Specialized equations for “effective” hydraulic diameters have been developed for specific geometries and can improve the accuracy to 5 or 10% [7.8]. When only a portion of the duct cross-section is heated — one wall of a rectangle, for example — the procedure for finding h is the same. The hydraulic diameter is based upon the entire wetted perimeter, not simply the heated part. However, in eqn. (7.59a) P is the heated perimeter: eqn. (7.59b) does not apply for nonuniform heating. One situation in which one-sided or unequal heating often occurs is an annular duct, for which the inner tube might be a heating element. The hydraulic diameter procedure will typically predict the heat transfer coefficient on the outer tube to within ±10%, irrespective of the heating configuration. The heat transfer coefficient on the inner surface, however, is sensitive to both the diameter ratio and the heating configuration. For that surface, the hydraulic diameter approach is not very accurate, Do ; other methods have been developed to accurately especially if Di predict heat transfer in annular ducts (see [7.3] or [7.8]). Laminar flow in noncircular ducts Laminar velocity profiles in noncircular ducts develop in essentially the same way as for circular tubes, and the fully developed velocity profiles are generally paraboloidal in shape. For example, for fully developed flow between parallel plates located at y = b/2 and y = −b/2, y 3 u 1−4 = uav 2 b 2 (7.62) for uav the bulk velocity. This should be compared to eqn. (7.15) for a circular tube. The constants and coordinates differ, but the equations are otherwise identical. Likewise, an analysis of the temperature profiles between parallel plates leads to constant Nusselt numbers, which may be expressed in terms of the hydraulic diameter for various boundary conditions: ⎧ ⎪7.541 for fixed plate temperatures ⎪ ⎨ hDh = 8.235 for fixed flux at both plates (7.63) NuDh = ⎪ k ⎪ ⎩ 5.385 one plate fixed flux, one adiabatic Some other cases are summarized in Table 7.4. Many more have been considered in the literature (see, especially, [7.5]). The latter include 373 374 Forced convection in a variety of configurations §7.6 Table 7.4 Laminar, fully developed Nusselt numbers based on hydraulic diameters given in eqn. (7.61) Cross-section Tw fixed qw fixed Circular Square Rectangular a = 2b a = 4b a = 8b Parallel plates 3.657 2.976 4.364 3.608 3.391 4.439 5.597 7.541 4.123 5.331 6.490 8.235 different wall boundary conditions and a wide variety cross-sectional shapes, both practical and ridiculous: triangles, circular sectors, trapezoids, rhomboids, hexagons, limaçons, and even crescent moons! The boundary conditions, in particular, should be considered when the duct is small (so that h will be large): if the conduction resistance of the tube wall is comparable to the convective resistance within the duct, then temperature or flux variations around the tube perimeter must be expected. This will significantly affect the laminar Nusselt number. The rectangular duct values in Table 7.4 for fixed wall flux, for example, assume a uniform temperature around the perimeter of the tube, as if the wall has no conduction resistance around its perimeter. This might be true for a copper duct heated at a fixed rate in watts per meter of duct length. Laminar entry length formulæ for noncircular ducts are also given by Shah and London [7.5]. 7.6 Heat transfer during cross flow over cylinders Fluid flow pattern It will help us to understand the complexity of heat transfer from bodies in a cross flow if we first look in detail at the fluid flow patterns that occur in one cross-flow configuration—a cylinder with fluid flowing normal to it. Figure 7.10 shows how the flow develops as Re ≡ u∞ D/ν is increased from below 5 to near 107 . An interesting feature of this evolving flow pattern is the fairly continuous way in which one flow transition follows §7.6 Heat transfer during cross flow over cylinders Figure 7.10 Regimes of fluid flow across circular cylinders [7.24]. 375 376 Forced convection in a variety of configurations §7.6 Figure 7.11 The Strouhal–Reynolds number relationship for circular cylinders, as defined by existing data [7.24]. another. The flow field degenerates to greater and greater degrees of disorder with each successive transition until, rather strangely, it regains order at the highest values of ReD . An important reflection of the complexity of the flow field is the vortex-shedding frequency, fv . Dimensional analysis shows that a dimensionless frequency called the Strouhal number, Str, depends on the Reynolds number of the flow: Str ≡ fv D = fn (ReD ) u∞ (7.64) Figure 7.11 defines this relationship experimentally on the basis of about 550 of the best data available (see [7.24]). The Strouhal numbers stay a little over 0.2 over most of the range of ReD . This means that behind a given object, the vortex-shedding frequency rises almost linearly with velocity. Experiment 7.1 When there is a gentle breeze blowing outdoors, go out and locate a large tree with a straight trunk or the shaft of a water tower. Wet your §7.6 Heat transfer during cross flow over cylinders 377 Figure 7.12 Giedt’s local measurements of heat transfer around a cylinder in a normal cross flow of air. finger and place it in the wake a couple of diameters downstream and about one radius off center. Estimate the vortex-shedding frequency and use Str 0.21 to estimate u∞ . Is your value of u∞ reasonable? Heat transfer The action of vortex shedding greatly complicates the heat removal process. Giedt’s data [7.25] in Fig. 7.12 show how the heat removal changes as the constantly fluctuating motion of the fluid to the rear of the cylin- 378 Forced convection in a variety of configurations §7.6 der changes with ReD . Notice, for example, that NuD is near its minimum at 110◦ when ReD = 71, 000, but it maximizes at the same place when ReD = 140, 000. Direct prediction by the sort of b.l. methods that we discussed in Chapter 6 is out of the question. However, a great deal can be done with the data using relations of the form NuD = fn (ReD , Pr) The broad study of Churchill and Bernstein [7.26] probably brings the correlation of heat transfer data from cylinders about as far as it is possible. For the entire range of the available data, they offer 1/2 NuD = 0.3 + 0.62 ReD Pr1/3 1 + (0.4/Pr)2/3 1/4 1+ ReD 282, 000 5/8 4/5 (7.65) This expression underpredicts most of the data by about 20% in the range 20, 000 < ReD < 400, 000 but is quite good at other Reynolds numbers above PeD ≡ ReD Pr = 0.2. This is evident in Fig. 7.13, where eqn. (7.65) is compared with data. Greater accuracy and, in most cases, greater convenience results from breaking the correlation into component equations: • Below ReD = 4000, the bracketed term [1 + (ReD /282, 000)5/8 ]4/5 is 1, so 1/2 NuD = 0.3 + 0.62 ReD Pr1/3 1 + (0.4/Pr)2/3 (7.66) 1/4 • Below Pe = 0.2, the Nakai-Okazaki [7.27] relation NuD = 1 0.8237 − ln Pe1/2 (7.67) should be used. • In the range 20, 000 < ReD < 400, 000, somewhat better results are given by 1/2 NuD = 0.3 + 0.62 ReD Pr1/3 1 + (0.4/Pr)2/3 than by eqn. (7.65). 1/4 1+ ReD 282, 000 1/2 (7.68) Heat transfer during cross flow over cylinders §7.6 Figure 7.13 Comparison of Churchill and Bernstein’s correlation with data by many workers from several countries for heat transfer during cross flow over a cylinder. (See [7.26] for data sources.) Fluids include air, water, and sodium, with both qw and Tw constant. All properties in eqns. (7.65) to (7.68) are to be evaluated at a film temperature Tf = (Tw + T∞ ) 2. Example 7.7 An electric resistance wire heater 0.0001 m in diameter is placed perpendicular to an air flow. It holds a temperature of 40◦ C in a 20◦ C air flow while it dissipates 17.8 W/m of heat to the flow. How fast is the air flowing? Solution. h = (17.8 W/m) [π (0.0001 m)(40 − 20) K] = 2833 W/m2 K. Therefore, NuD = 2833(0.0001)/0.0264 = 10.75, where we have evaluated k = 0.0264 at T = 30◦ C. We now want to find the ReD for which NuD is 10.75. From Fig. 7.13 we see that ReD is around 300 379 380 Forced convection in a variety of configurations §7.6 when the ordinate is on the order of 10. This means that we can solve eqn. (7.66) to get an accurate value of ReD : ReD = ⎧ ⎨ ⎡ (NuD − 0.3) ⎣1 + ⎩ 0.4 Pr 2/3 1/4 0.62 Pr1/3 ⎫2 ⎬ ⎭ but Pr = 0.71, so ReD = ⎧ ⎨ ⎡ (10.75 − 0.3) ⎣1 + ⎩ 0.40 0.71 2/3 1/4 1/3 0.62(0.71) ⎫2 ⎬ ⎭ = 463 Then u∞ = ν ReD = D 1.596 × 10−5 10−4 463 = 73.9 m/s The data scatter in ReD is quite small—less than 10%, it would appear—in Fig. 7.13. Therefore, this method can be used to measure local velocities with good accuracy. If the device is calibrated, its accuracy is improved further. Such an air speed indicator is called a hot-wire anemometer, as discussed further in Problem 7.45. Heat transfer during flow across tube bundles A rod or tube bundle is an arrangement of parallel cylinders that heat, or are being heated by, a fluid that might flow normal to them, parallel with them, or at some angle in between. The flow of coolant through the fuel elements of all nuclear reactors being used in this country is parallel to the heating rods. The flow on the shell side of most shell-and-tube heat exchangers is generally normal to the tube bundles. Figure 7.14 shows the two basic configurations of a tube bundle in a cross flow. In one, the tubes are in a line with the flow; in the other, the tubes are staggered in alternating rows. For either of these configurations, heat transfer data can be correlated reasonably well with power-law relations of the form NuD = C Ren Pr1/3 D (7.69) but in which the Reynolds number is based on the maximum velocity, umax = uav in the narrowest transverse area of the passage §7.6 Heat transfer during cross flow over cylinders Figure 7.14 Aligned and staggered tube rows in tube bundles. Thus, the Nusselt number based on the average heat transfer coefficient over any particular isothermal tube is NuD = hD k and ReD = umax D ν Žukauskas at the Lithuanian Academy of Sciences Institute in Vilnius has written two comprehensive review articles on tube-bundle heat trans- 381 382 Forced convection in a variety of configurations §7.6 fer [7.28, 7.29]. In these he summarizes his work and that of other Soviet workers, together with earlier work from the West. He was able to correlate data over very large ranges of Pr, ReD , ST /D , and SL /D (see Fig. 7.14) with an expression of the form ⎧ ⎨0 for gases (7.70) NuD = Pr0.36 (Pr/Prw )n fn (ReD ) with n = 1 ⎩ for liquids 4 where properties are to be evaluated at the local fluid bulk temperature, except for Prw , which is evaluated at the uniform tube wall temperature, Tw . The function fn(ReD ) takes the following form for the various circumstances of flow and tube configuration: 100 ReD 103 : fn (ReD ) = 0.52 Re0.5 D (7.71a) staggered rows: fn (ReD ) = 0.71 Re0.5 D (7.71b) aligned rows: 103 ReD 2 × 105 : aligned rows: fn (ReD ) = 0.27 Re0.63 , ST /SL D 0.7 (7.71c) For ST /SL < 0.7, heat exchange is much less effective. Therefore, aligned tube bundles are not designed in this range and no correlation is given. staggered rows: fn (ReD ) = 0.35 (ST /SL )0.2 Re0.6 , D ST /SL 2 (7.71d) fn (ReD ) = 0.40 Re0.6 , ST /SL > 2 D (7.71e) fn (ReD ) = 0.033 Re0.8 D (7.71f) ReD > 2 × 105 : aligned rows: staggered rows: fn (ReD ) = 0.031 (ST /SL )0.2 Re0.8 , D Pr > 1 (7.71g) NuD = 0.027 (ST /SL )0.2 Re0.8 , D Pr = 0.7 (7.71h) All of the preceding relations apply to the inner rows of tube bundles. The heat transfer coefficient is smaller in the rows at the front of a bundle, §7.6 Heat transfer during cross flow over cylinders 383 Figure 7.15 Correction for the heat transfer coefficients in the front rows of a tube bundle [7.28]. facing the oncoming flow. The heat transfer coefficient can be corrected so that it will apply to any of the front rows using Fig. 7.15. Early in this chapter we alluded to the problem of predicting the heat transfer coefficient during the flow of a fluid at an angle other than 90◦ to the axes of the tubes in a bundle. Žukauskas provides the empirical corrections in Fig. 7.16 to account for this problem. The work of Žukauskas does not extend to liquid metals. However, Kalish and Dwyer [7.30] present the results of an experimental study of heat transfer to the liquid eutectic mixture of 77.2% potassium and 22.8% sodium (called NaK). NaK is a fairly popular low-melting-point metallic coolant which has received a good deal of attention for its potential use in certain kinds of nuclear reactors. For isothermal tubes in an equilateral triangular array, as shown in Fig. 7.17, Kalish and Dwyer give NuD = 5.44 + 0.228 Pe0.614 C P −D P sin φ + sin2 φ 1 + sin2 φ (7.72) Figure 7.16 Correction for the heat transfer coefficient in flows that are not perfectly perpendicular to heat exchanger tubes [7.28]. 384 Forced convection in a variety of configurations §7.7 Figure 7.17 Geometric correction for the Kalish-Dwyer equation (7.72). where • φ is the angle between the flow direction and the rod axis. • P is the “pitch” of the tube array, as shown in Fig. 7.17, and D is the tube diameter. • C is the constant given in Fig. 7.17. • PeD is the Péclét number based on the mean flow velocity through the narrowest opening between the tubes. • For the same uniform heat flux around each tube, the constants in eqn. (7.72) change as follows: 5.44 becomes 4.60; 0.228 becomes 0.193. 7.7 Other configurations At the outset, we noted that this chapter would move further and further beyond the reach of analysis in the heat convection problems that it dealt with. However, we must not forget that even the most completely empirical relations in Section 7.6 were devised by people who were keenly aware of the theoretical framework into which these relations had to fit. Notice, for example, that eqn. (7.66) reduces to NuD ∝ PeD as Pr becomes small. That sort of theoretical requirement did not just pop out of a data plot. Instead, it was a consideration that led the authors to select an empirical equation that agreed with theory at low Pr. Thus, the theoretical considerations in Chapter 6 guide us in correlating limited data in situations that cannot be analyzed. Such correlations Other configurations §7.7 can be found for all kinds of situations, but all must be viewed critically. Many are based on limited data, and many incorporate systematic errors of one kind or another. In the face of a heat transfer situation that has to be predicted, one can often find a correlation of data from similar systems. This might involve flow in or across noncircular ducts; axial flow through tube or rod bundles; flow over such bluff bodies as spheres, cubes, or cones; or flow in circular and noncircular annuli. The Handbook of Heat Transfer [7.31], the shelf of heat transfer texts in your library, or the journals referred to by the Engineering Index are among the first places to look for a correlation curve or equation. When you find a correlation, there are many questions that you should ask yourself: • Is my case included within the range of dimensionless parameters upon which the correlation is based, or must I extrapolate to reach my case? • What geometric differences exist between the situation represented in the correlation and the one I am dealing with? (Such elements as these might differ: (a) inlet flow conditions; (b) small but important differences in hardware, mounting brackets, and so on; (c) minor aspect ratio or other geometric nonsimilarities • Does the form of the correlating equation that represents the data, if there is one, have any basis in theory? (If it is only a curve fit to the existing data, one might be unjustified in using it for more than interpolation of those data.) • What nuisance variables might make our systems different? For example: (a) surface roughness; (b) fluid purity; (c) problems of surface wetting • To what extend do the data scatter around the correlation line? Are error limits reported? Can I actually see the data points? (In this regard, you must notice whether you are looking at a correlation 385 Chapter 7: Forced convection in a variety of configurations 386 on linear or logarithmic coordinates. Errors usually appear smaller than they really are on logarithmic coordinates. Compare, for example, the data of Figs. 8.3 and 8.10.) • Are the ranges of physical variables large enough to guarantee that I can rely on the correlation for the full range of dimensionless groups that it purports to embrace? • Am I looking at a primary or secondary source (i.e., is this the author’s original presentation or someone’s report of the original)? If it is a secondary source, have I been given enough information to question it? • Has the correlation been signed by the persons who formulated it? (If not, why haven’t the authors taken responsibility for the work?) Has it been subjected to critical review by independent experts in the field? Problems 7.1 Prove that in fully developed laminar pipe flow, (−dp/dx)R 2 4µ is twice the average velocity in the pipe. To do this, set the mass flow rate through the pipe equal to (ρuav )(area). 7.2 A flow of air at 27◦ C and 1 atm is hydrodynamically fully developed in a 1 cm I.D. pipe with uav = 2 m/s. Plot (to scale) Tw , qw , and Tb as a function of the distance x after Tw is changed or qw is imposed: a. In the case for which Tw = 68.4◦ C = constant. b. In the case for which qw = 378 W/m2 = constant. Indicate xet on your graphs. 7.3 Prove that Cf is 16/ReD in fully developed laminar pipe flow. 7.4 Air at 200◦ C flows at 4 m/s over a 3 cm O.D. pipe that is kept at 240◦ C. (a) Find h. (b) If the flow were pressurized water at 200◦ C, what velocities would give the same h, the same NuD , and the same ReD ? (c) If someone asked if you could model the water flow with an air experiment, how would you answer? [u∞ = 0.0156 m/s for same NuD .] Problems 387 7.5 Compare the h value calculated in Example 7.3 with those calculated from the Dittus-Boelter, Colburn, and Sieder-Tate equations. Comment on the comparison. 7.6 Water at Tblocal = 10◦ C flows in a 3 cm I.D. pipe at 1 m/s. The pipe walls are kept at 70◦ C and the flow is fully developed. Evaluate h and the local value of dTb /dx at the point of interest. The relative roughness is 0.001. 7.7 Water at 10◦ C flows over a 3 cm O.D. cylinder at 70◦ C. The velocity is 1 m/s. Evaluate h. 7.8 Consider the hot wire anemometer in Example 7.7. Suppose that 17.8 W/m is the constant heat input, and plot u∞ vs. Twire over a reasonable range of variables. Must you deal with any changes in the flow regime over the range of interest? 7.9 Water at 20◦ C flows at 2 m/s over a 2 m length of pipe, 10 cm in diameter, at 60◦ C. Compare h for flow normal to the pipe with that for flow parallel to the pipe. What does the comparison suggest about baffling in a heat exchanger? 7.10 A thermally fully developed flow of NaK in a 5 cm I.D. pipe moves at uav = 8 m/s. If Tb = 395◦ C and Tw is constant at 403◦ C, what is the local heat transfer coefficient? Is the flow laminar or turbulent? 7.11 Water enters a 7 cm I.D. pipe at 5◦ C and moves through it at an average speed of 0.86 m/s. The pipe wall is kept at 73◦ C. Plot Tb against the position in the pipe until (Tw − Tb )/68 = 0.01. Neglect the entry problem and consider property variations. 7.12 Air at 20◦ C flows over a very large bank of 2 cm O.D. tubes that are kept at 100◦ C. The air approaches at an angle 15◦ off normal to the tubes. The tube array is staggered, with SL = 3.5 cm and ST = 2.8 cm. Find h on the first tubes and on the tubes deep in the array if the air velocity is 4.3 m/s before it enters the array. [hdeep = 118 W/m2 K.] 7.13 Rework Problem 7.11 using a single value of h evaluated at 3(73 − 5)/4 = 51◦ C and treating the pipe as a heat exchanger. At what length would you judge that the pipe is no longer efficient as an exchanger? Explain. Chapter 7: Forced convection in a variety of configurations 388 7.14 Go to the periodical engineering literature in your library. Find a correlation of heat transfer data. Evaluate the applicability of the correlation according to the criteria outlined in Section 7.7. 7.15 Water at 24◦ C flows at 0.8 m/s in a smooth, 1.5 cm I.D. tube that is kept at 27◦ C. The system is extremely clean and quiet, and the flow stays laminar until a noisy air compressor is turned on in the laboratory. Then it suddenly goes turbulent. Calculate the ratio of the turbulent h to the laminar h. [hturb = 4429 W/m2 K.] 7.16 Laboratory observations of heat transfer during the forced flow of air at 27◦ C over a bluff body, 12 cm wide, kept at 77◦ C yield q = 646 W/m2 when the air moves 2 m/s and q = 3590 W/m2 when it moves 18 m/s. In another test, everything else is the same, but now 17◦ C water flowing 0.4 m/s yields 131,000 W/m2 . The correlations in Chapter 7 suggest that, with such limited data, we can probably create a fairly good correlation in the form: NuL = C Rea Prb . Estimate the constants C , a, and b by cross-plotting the data on log-log paper. 7.17 Air at 200 psia flows at 12 m/s in an 11 cm I.D. duct. Its bulk temperature is 40◦ C and the pipe wall is at 268◦ C. Evaluate h if ε/D = 0.00006. 7.18 How does h during cross flow over a cylindrical heater vary with the diameter when ReD is very large? 7.19 Air enters a 0.8 cm I.D. tube at 20◦ C with an average velocity of 0.8 m/s. The tube wall is kept at 40◦ C. Plot Tb (x) until it reaches 39◦ C. Use properties evaluated at [(20 + 40)/2]◦ C for the whole problem, but report the local error in h at the end to get a sense of the error incurred by the simplification. 7.20 ˙ Write ReD in terms of m in pipe flow and explain why this representation could be particularly useful in dealing with compressible pipe flows. 7.21 NaK at 394◦ C flows at 0.57 m/s across a 1.82 m length of 0.036 m O.D. tube. The tube is kept at 404◦ C. Find h and the heat removal rate from the tube. 7.22 Verify the value of h specified in Problem 3.22. Problems 389 7.23 Check the value of h given in Example 7.3 by using Reynolds’s analogy directly to calculate it. Which h do you deem to be in error, and by what percent? 7.24 A homemade heat exchanger consists of a copper plate, 0.5 m square, with twenty 1.5 cm I.D. copper tubes soldered to it. The ten tubes on top are evenly spaced across the top and parallel with two sides. The ten on the bottom are also evenly spaced, but they run at 90◦ to the top tubes. The exchanger is used to cool methanol flowing at 0.48 m/s in the tubes from an initial temperature of 73◦ C, using water flowing at 0.91 m/s and entering at 7◦ C. What is the temperature of the methanol when it is mixed in a header on the outlet side? Make a judgement of the heat exchanger. 7.25 Given that NuD = 12.7 at (2/Gz) = 0.004, evaluate NuD at (2/Gz) = 0.02 numerically, using Fig. 7.4. Compare the result with the value you read from the figure. 7.26 Report the maximum percent scatter of data in Fig. 7.13. What is happening in the fluid flow when the scatter is worst? 7.27 Water at 27◦ C flows at 2.2 m/s in a 0.04 m I.D. thin-walled pipe. Air at 227◦ C flows across it at 7.6 m/s. Find the pipe wall temperature. 7.28 Freshly painted aluminum rods, 0.02 m in diameter, are withdrawn from a drying oven at 150◦ C and cooled in a 3 m/s cross flow of air at 23◦ C. How long will it take to cool them to 50◦ C so that they can be handled? 7.29 At what speed, u∞ , must 20◦ C air flow across an insulated tube before the insulation on it will do any good? The tube is at 60◦ C and is 6 mm in diameter. The insulation is 12 mm in diameter, with k = 0.08 W/m·K. (Notice that we do not ask for the u∞ for which the insulation will do the most harm.) 7.30 Water at 37◦ C flows at 3 m/s across at 6 cm O.D. tube that is held at 97◦ C. In a second configuration, 37◦ C water flows at an average velocity of 3 m/s through a bundle of 6 cm O.D. tubes that are held at 97◦ C. The bundle is staggered, with ST /SL = 2. Compare the heat transfer coefficients for the two situations. Chapter 7: Forced convection in a variety of configurations 390 7.31 It is proposed to cool 64◦ C air as it flows, fully developed, in a 1 m length of 8 cm I.D. smooth, thin-walled tubing. The coolant is Freon 12 flowing, fully developed, in the opposite direction, in eight smooth 1 cm I.D. tubes equally spaced around the periphery of the large tube. The Freon enters at −15◦ C and is fully developed over almost the entire length. The average speeds are 30 m/s for the air and 0.5 m/s for the Freon. Determine the exiting air temperature, assuming that soldering provides perfect thermal contact between the entire surface of the small tubes and the surface of the large tube. Criticize the heat exchanger design and propose some design improvement. 7.32 Evaluate NuD using Giedt’s data for air flowing over a cylinder at ReD = 140, 000. Compare your result with the appropriate correlation and with Fig. 7.13. 7.33 A 25 mph wind blows across a 0.25 in. telephone line. What is the pitch of the hum that it emits? 7.34 A large Nichrome V slab, 0.2 m thick, has two parallel 1 cm I.D. holes drilled through it. Their centers are 8 cm apart. One carries liquid CO2 at 1.2 m/s from a −13◦ C reservoir below. The other carries methanol at 1.9 m/s from a 47◦ C reservoir above. Take account of the intervening Nichrome and compute the heat transfer. Need we worry about the CO2 being warmed up by the methanol? 7.35 Consider the situation described in Problem 4.38 but suppose that you do not know h. Suppose, instead, that you know there is a 10 m/s cross flow of 27◦ C air over the rod. Then rework the problem. 7.36 A liquid whose properties are not known flows across a 40 cm O.D. tube at 20 m/s. The measured heat transfer coefficient is 8000 W/m2 K. We can be fairly confident that ReD is very large indeed. What would h be if D were 53 cm? What would h be if u∞ were 28 m/s? 7.37 Water flows at 4 m/s, at a temperature of 100◦ C, in a 6 cm I.D. thin-walled tube with a 2 cm layer of 85% magnesia insulation on it. The outside heat transfer coefficient is 6 W/m2 K, and the outside temperature is 20◦ C. Find: (a) U based on the inside Problems 391 area, (b) Q W/m, and (c) the temperature on either side of the insulation. 7.38 Glycerin is added to water in a mixing tank at 20◦ C. The mixture discharges through a 4 m length of 0.04 m I.D. tubing under a constant 3 m head. Plot the discharge rate in m3 /hr as a function of composition. 7.39 Plot h as a function of composition for the discharge pipe in Problem 7.38. Assume a small temperature difference. 7.40 Rework Problem 5.40 without assuming the Bi number to be very large. 7.41 Water enters a 0.5 cm I.D. pipe at 24◦ C. The pipe walls are held at 30◦ C. Plot Tb against distance from entry if uav is 0.27 m/s, neglecting entry behavior in your calculation. (Indicate the entry region on your graph, however.) 7.42 Devise a numerical method to find the velocity distribution and friction factor for laminar flow in a square duct of side length a. Set up a square grid of size N by N and solve the difference equations by hand for N = 2, 3, and 4. Hint : First show that the velocity distribution is given by the solution to the equation ∂2u ∂2u + =1 ∂x2 ∂y 2 where u = 0 on the sides of the square and we define u = u [(a2 /µ)(dp/dz)], x = (x/a), and y = (y/a). Then show that the friction factor, f [eqn. (7.34)], is given by f= −2 ρuav a µ u dxdy Note that the area integral can be evaluated as 7.43 u/N 2 . Chilled air at 15◦ C enters a horizontal duct at a speed of 1 m/s. The duct is made of thin galvanized steel and is not insulated. A 30 m section of the duct runs outdoors through humid air at 30◦ C. Condensation of moisture on the outside of the duct is undesirable, but it will occur if the duct wall is at or below Chapter 7: Forced convection in a variety of configurations 392 the dew point temperature of 20◦ C. For this problem, assume that condensation rates are so low that their thermal effects can be ignored. a. Suppose that the duct’s square cross-section is 0.3 m by 0.3 m and the effective outside heat transfer coefficient is 5 W/m2 K in still air. Determine whether condensation occurs. b. The single duct is replaced by four circular horizontal ducts, each 0.17 m in diameter. The ducts are parallel to one another in a vertical plane with a center-to-center separation of 0.5 m. Each duct is wrapped with a layer of fiberglass insulation 6 cm thick (ki = 0.04 W/m·K) and carries air at the same inlet temperature and speed as before. If a 15 m/s wind blows perpendicular to the plane of the circular ducts, find the bulk temperature of the air exiting the ducts. 7.44 An x-ray “monochrometer” is a mirror that reflects only a single wavelength from a broadband beam of x-rays. Over 99% of the beam’s energy arrives on other wavelengths and is absorbed creating a high heat flux on part of the surface of the monochrometer. Consider a monochrometer made from a silicon block 10 mm long and 3 mm by 3 mm in cross-section which absorbs a flux of 12.5 W/mm2 over an area of 6 mm2 on one face (a heat load of 75 W). To control the temperature, it is proposed to pump liquid nitrogen through a circular channel bored down the center of the silicon block. The channel is 10 mm long and 1 mm in diameter. LN2 enters the channel at 80 K and a pressure of 1.6 MPa (Tsat = 111.5 K). The entry to this channel is a long, straight, unheated passage of the same diameter. a. For what range of mass flow rates will the LN2 have a bulk temperature rise of less than 1.5 K over the length of the channel? b. At your minimum flow rate, estimate the maximum wall temperature in the channel. As a first approximation, assume that the silicon conducts heat well enough to distribute the 75 W heat load uniformly over the channel References 393 surface. Could boiling occur in the channel? Discuss the influence of entry length and variable property effects. 7.45 Turbulent fluid velocities are sometimes measured with a constant temperature hot-wire anemometer, which consists of a long, fine wire (typically platinum, 4µm in diameter and 1.25 mm long) supported between two much larger needles. The needles are connected to an electronic bridge circuit which electrically heats the wire while adjusting the heating voltage, Vw , so that the wire’s temperature — and thus its resistance, Rw — stays constant. The electrical power dissipated in the 2 wire, Vw /Rw , is convected away at the surface of the wire. Analyze the heat loss from the wire to show 2 Vw = (Twire − Tflow ) A + Bu1/2 where u is the instantaneous flow speed perpendicular to the wire. Assume that u is between 2 and 100 m/s and that the fluid is an isothermal gas. The constants A and B depend on properties, dimensions, and resistance; they are usually found by calibration of the anemometer. This result is called King’s law. 7.46 (a) Show that the Reynolds number for a circular tube may be ˙ written in terms of the mass flow rate as ReD = 4m π µD . (b) Show that this result does not apply to a noncircular tube, ˙ specifically ReDh ≠ 4m π µDh . References [7.1] F. M. White. Viscous Fluid Flow. McGraw-Hill Book Company, New York, 1974. [7.2] S. S. Mehendale, A. M. Jacobi, and R. K. Shah. Fluid flow and heat transfer at micro- and meso-scales with application to heat exchanger design. Appl. Mech. Revs., 53(7):175–193, 2000. [7.3] W. M. Kays and M. E. Crawford. Convective Heat and Mass Transfer. McGraw-Hill Book Company, New York, 3rd edition, 1993. 394 Chapter 7: Forced convection in a variety of configurations [7.4] R. K. Shah and M. S. Bhatti. Laminar convective heat transfer in ducts. In S. Kakaç, R. K. Shah, and W. Aung, editors, Handbook of Single-Phase Convective Heat Transfer, chapter 3. WileyInterscience, New York, 1987. [7.5] R. K. Shah and A. L. London. Laminar Flow Forced Convection in Ducts. Academic Press, Inc., New York, 1978. Supplement 1 to the series Advances in Heat Transfer. [7.6] L. Graetz. Über die wärmeleitfähigkeit von flüssigkeiten. Ann. Phys., 25:337, 1885. [7.7] S. R. Sellars, M. Tribus, and J. S. Klein. Heat transfer to laminar flow in a round tube or a flat plate—the Graetz problem extended. Trans. ASME, 78:441–448, 1956. [7.8] M. S. Bhatti and R. K. Shah. Turbulent and transition flow convective heat transfer in ducts. In S. Kakaç, R. K. Shah, and W. Aung, editors, Handbook of Single-Phase Convective Heat Transfer, chapter 4. Wiley-Interscience, New York, 1987. [7.9] F. Kreith. Principles of Heat Transfer. Intext Press, Inc., New York, 3rd edition, 1973. [7.10] A. P. Colburn. A method of correlating forced convection heat transfer data and a comparison with fluid friction. Trans. AIChE, 29:174, 1933. [7.11] L. M. K. Boelter, V. H. Cherry, H. A. Johnson, and R. C. Martinelli. Heat Transfer Notes. McGraw-Hill Book Company, New York, 1965. [7.12] E. N. Sieder and G. E. Tate. Heat transfer and pressure drop of liquids in tubes. Ind. Eng. Chem., 28:1429, 1936. [7.13] B. S. Petukhov. Heat transfer and friction in turbulent pipe flow with variable physical properties. In T.F. Irvine, Jr. and J. P. Hartnett, editors, Advances in Heat Transfer, volume 6, pages 504–564. Academic Press, Inc., New York, 1970. [7.14] V. Gnielinski. New equations for heat and mass transfer in turbulent pipe and channel flow. Int. Chemical Engineering, 16:359–368, 1976. References [7.15] D. M. McEligot. Convective heat transfer in internal gas flows with temperature-dependent properties. In A. S. Majumdar and R. A. Mashelkar, editors, Advances in Transport Processes, volume IV, pages 113–200. Wiley, New York, 1986. [7.16] M. F. Taylor. Prediction of friction and heat-transfer coefficients with large variations in fluid properties. NASA TM X-2145, December 1970. [7.17] S. E. Haaland. Simple and explicit formulas for the friction factor in turbulent pipe flow. J. Fluids Engr., 105:89–90, 1983. [7.18] T. S. Ravigururajan and A. E. Bergles. Development and verification of general correlations for pressure drop and heat transfer in single-phase turbulent flow in enhanced tubes. Exptl. Thermal Fluid Sci., 13:55–70, 1996. [7.19] R. L. Webb. Enhancement of single-phase heat transfer. In S. Kakaç, R. K. Shah, and W. Aung, editors, Handbook of Single-Phase Convective Heat Transfer, chapter 17. Wiley-Interscience, New York, 1987. [7.20] B. Lubarsky and S. J. Kaufman. Review of experimental investigations of liquid-metal heat transfer. NACA Tech. Note 3336, 1955. [7.21] C. B. Reed. Convective heat transfer in liquid metals. In S. Kakaç, R. K. Shah, and W. Aung, editors, Handbook of Single-Phase Convective Heat Transfer, chapter 8. Wiley-Interscience, New York, 1987. [7.22] R. A. Seban and T. T. Shimazaki. Heat transfer to a fluid flowing turbulently in a smooth pipe with walls at a constant temperature. Trans. ASME, 73:803, 1951. [7.23] R. N. Lyon, editor. Liquid Metals Handbook. A.E.C. and Dept. of the Navy, Washington, D.C., 3rd edition, 1952. [7.24] J. H. Lienhard. Synopsis of lift, drag, and vortex frequency data for rigid circular cylinders. Bull. 300. Wash. State Univ., Pullman, 1966. May be downloaded as a 2.3 MB pdf file from http://www.uh.edu/engines/vortexcylinders.pdf. [7.25] W. H. Giedt. Investigation of variation of point unit-heat-transfer coefficient around a cylinder normal to an air stream. Trans. ASME, 71:375–381, 1949. 395 396 Chapter 7: Forced convection in a variety of configurations [7.26] S. W. Churchill and M. Bernstein. A correlating equation for forced convection from gases and liquids to a circular cylinder in crossflow. J. Heat Transfer, Trans. ASME, Ser. C, 99:300–306, 1977. [7.27] S. Nakai and T. Okazaki. Heat transfer from a horizontal circular wire at small Reynolds and Grashof numbers—1 pure convection. Int. J. Heat Mass Transfer, 18:387–396, 1975. [7.28] A. Žukauskas. Heat transfer from tubes in crossflow. In T.F. Irvine, Jr. and J. P. Hartnett, editors, Advances in Heat Transfer, volume 8, pages 93–160. Academic Press, Inc., New York, 1972. [7.29] A. Žukauskas. Heat transfer from tubes in crossflow. In T. F. Irvine, Jr. and J. P. Hartnett, editors, Advances in Heat Transfer, volume 18, pages 87–159. Academic Press, Inc., New York, 1987. [7.30] S. Kalish and O. E. Dwyer. Heat transfer to NaK flowing through unbaffled rod bundles. Int. J. Heat Mass Transfer, 10:1533–1558, 1967. [7.31] W. M. Rohsenow, J. P. Hartnett, and Y. I. Cho, editors. Handbook of Heat Transfer. McGraw-Hill, New York, 3rd edition, 1998. 8. Natural convection in singlephase fluids and during film condensation There is a natural place for everything to seek, as: Heavy things go downward, fire upward, and rivers to the sea. The Anatomy of Melancholy, R. Burton, 1621 8.1 Scope The remaining convection mechanisms that we deal with are to a large degree gravity-driven. Unlike forced convection, in which the driving force is external to the fluid, these so-called natural convection processes are driven by body forces exerted directly within the fluid as the result of heating or cooling. Two such mechanisms that are rather alike are: • Natural convection. When we speak of natural convection without any qualifying words, we mean natural convection in a single-phase fluid. • Film condensation. This natural convection process has much in common with single-phase natural convection. We therefore deal with both mechanisms in this chapter. The governing equations are developed side by side in two brief opening sections. Then each mechanism is developed independently in Sections 8.3 and 8.4 and in Section 8.5, respectively. Chapter 9 deals with other natural convection heat transfer processes that involve phase change—for example: 397 398 Natural convection in single-phase fluids and during film condensation §8.2 • Nucleate boiling. This heat transfer process is highly disordered as opposed to the processes described in Chapter 8. • Film boiling. This is so similar to film condensation that it is usually treated by simply modifying film condensation predictions. • Dropwise condensation. This bears some similarity to nucleate boiling. 8.2 The nature of the problems of film condensation and of natural convection Description The natural convection problem is sketched in its simplest form on the left-hand side of Fig. 8.1. Here we see a vertical isothermal plate that cools the fluid adjacent to it. The cooled fluid sinks downward to form a b.l. The figure would be inverted if the plate were warmer than the fluid next to it. Then the fluid would buoy upward. On the right-hand side of Fig. 8.1 is the corresponding film condensation problem in its simplest form. An isothermal vertical plate cools an adjacent vapor, which condenses and forms a liquid film on the wall.1 The film is normally very thin and it flows off, rather like a b.l., as the figure suggests. While natural convection can carry fluid either upward or downward, a condensate film can only move downward. The temperature in the film rises from Tw at the cool wall to Tsat at the outer edge of the film. In both problems, but particularly in film condensation, the b.l. and the film are normally thin enough to accommodate the b.l. assumptions [recall the discussion following eqn. (6.13)]. A second idiosyncrasy of both problems is that δ and δt are closely related. In the condensing film they are equal, since the edge of the condensate film forms the edge of both b.l.’s. In natural convection, δ and δt are approximately equal when Pr is on the order of unity or less, because all cooled (or heated) fluid must buoy downward (or upward). When Pr is large, the cooled (or heated) fluid will fall (or rise) and, although it is all very close to the wall, this fluid, with its high viscosity, will also drag unheated liquid with it. 1 It might instead condense into individual droplets, which roll of without forming into a film. This process, called dropwise condensation, is dealt with in Section 9.9. §8.2 The nature of the problems of film condensation and of natural convection Figure 8.1 The convective boundary layers for natural convection and film condensation. In both sketches, but particularly in that for film condensation, the y -coordinate has been stretched. In this case, δ can exceed δt . We deal with cases for which δ subsequent analysis. δt in the Governing equations To describe laminar film condensation and laminar natural convection, we must add a gravity term to the momentum equation. The dimensions of the terms in the momentum equation should be examined before we do this. Equation (6.13) can be written as u ∂u ∂u +v ∂x ∂y m s2 = kg·m kg·s2 =− = N kg N 1 dp m3 ∂ 2 u m2 m +ν ρ dx kg m2 · m y 2 s s · m2 = N kg = m s2 = N kg where ∂p/∂x dp/dx in the b.l. and where µ constant. Thus, every term in the equation has units of acceleration or (equivalently) force per unit mass. The component of gravity in the x -direction therefore enters 399 400 Natural convection in single-phase fluids and during film condensation §8.2 the momentum balance as (+g). This is because x and g point in the same direction. Gravity would enter as −g if it acted opposite the x direction. u ∂u 1 dp ∂2u ∂u +v =− +g+ν ∂x ∂y ρ dx ∂y 2 (8.1) In the two problems at hand, the pressure gradient is the hydrostatic gradient outside the b.l. Thus, dp = ρ∞ g dx dp = ρg g dx natural convection film condensation (8.2) where ρ∞ is the density of the undisturbed fluid and ρg (and ρf below) are the saturated vapor and liquid densities. Equation (8.1) then becomes u ∂u ∂u +v = ∂x ∂y 1− ρ∞ ρ g+ν ∂2u ∂y 2 for natural convection (8.3) u ∂u ∂u +v = ∂x ∂y 1− ρg ρf g+ν ∂2u ∂y 2 for film condensation (8.4) Two boundary conditions, which apply to both problems, are u y =0 =0 the no-slip condition v y =0 =0 no flow into the wall (8.5a) The third b.c. is different for the film condensation and natural convection problems: ⎫ ⎪ ∂u condensation: ⎪ ⎪ ⎪ =0 ⎬ no shear at the edge of the film ∂y y =δ (8.5b) ⎪ ⎪ ⎪ natural convection: ⎪ ⎭ u y =δ =0 undisturbed fluid outside the b.l. The energy equation for either of the two cases is eqn. (6.40): u ∂T ∂2T ∂T +v =α ∂x ∂y ∂y 2 We leave the identification of the b.c.’s for temperature until later. The crucial thing we must recognize about the momentum equation at the moment is that it is coupled to the energy equation. Let us consider how that occurs: Laminar natural convection on a vertical isothermal surface §8.3 In natural convection : The velocity, u, is driven by buoyancy, which is reflected in the term (1 − ρ∞ /ρ)g in the momentum equation. The density, ρ = ρ(T ), varies with T , so it is impossible to solve the momentum and energy equations independently of one another. In film condensation : The third boundary condition (8.5b) for the momentum equation involves the film thickness, δ. But to calculate δ we must make an energy balance on the film to find out how much latent heat—and thus how much condensate—it has absorbed. This will bring (Tsat − Tw ) into the solution of the momentum equation. Recall that the boundary layer on a flat surface, during forced convection, was easy to analyze because the momentum equation could be solved completely before any consideration of the energy equation was attempted. We do not have that advantage in predicting natural convection or film condensation. 8.3 Laminar natural convection on a vertical isothermal surface Dimensional analysis and experimental data Before we attempt a dimensional analysis of the natural convection problem, let us simplify the buoyancy term, (ρ − ρ∞ )g ρ , in the momentum equation (8.3). The equation was derived for incompressible flow, but we modified it by admitting a small variation of density with temperature in this term only. Now we wish to eliminate (ρ − ρ∞ ) in favor of (T − T∞ ) with the help of the coefficient of thermal expansion, β: β≡ 1 ∂v v ∂T p =− 1 ∂ρ ρ ∂T p − 1 − ρ∞ ρ 1 ρ − ρ∞ =− ρ T − T∞ T − T∞ (8.6) where v designates the specific volume here, not a velocity component. Figure 8.2 shows natural convection from a vertical surface that is hotter than its surroundings. In either this case or on the cold plate shown in Fig. 8.1, we replace (1 − ρ∞ /ρ)g with −gβ(T − T∞ ). The sign (see Fig. 8.2) is the same in either case. Then u ∂u ∂2u ∂u +v = −gβ(T − T∞ ) + ν ∂x ∂y ∂y 2 (8.7) 401 402 Natural convection in single-phase fluids and during film condensation §8.3 Figure 8.2 Natural convection from a vertical heated plate. where the minus sign corresponds to plate orientation in Fig. 8.1a. This conveniently removes ρ from the equation and makes the coupling of the momentum and energy equations very clear. The functional equation for the heat transfer coefficient, h, in natural convection is therefore (cf. Section 6.4) h or h = fn k, |Tw − T∞ | , x or L, ν, α, g, β where L is a length that must be specified for a given problem. Notice that while h was assumed to be independent of ∆T in the forced convection problem (Section 6.4), the explicit appearance of (T − T∞ ) in eqn. (8.7) suggests that we cannot make that assumption here. There are thus eight variables in W, m, s, and ◦ C (where we again regard J as a unit independent of N and m); so we look for 8 − 4 = 4 pi-groups. For h and a characteristic length, L, the groups may be chosen as NuL ≡ hL , k Pr ≡ ν , α Π3 ≡ L3 ν2 g, Π4 ≡ β |Tw − T∞ | = β ∆T where we set ∆T ≡ |Tw − T∞ |. Two of these groups are new to us: • Π3 ≡ gL3 /ν 2 : This characterizes the importance of buoyant forces relative to viscous forces.2 Note that gL is dimensionally the same as a velocity squared—say, u2 . Then Π3 can be interpreted as a Reynolds number: uL/ν . In a laminar b.l. we recall that Nu ∝ 1/4 Re1/2 ; so here we expect that Nu ∝ Π3 . 2 §8.3 Laminar natural convection on a vertical isothermal surface • Π4 ≡ β∆T : This characterizes the thermal expansion of the fluid. For an ideal gas, β= 1∂ v ∂T RT p = p 1 T∞ where R is the gas constant. Therefore, for ideal gases β ∆T = ∆T T∞ (8.8) It turns out that Π3 and Π4 (which do not bear the names of famous people) usually appear as a product. This product is called the Grashof (pronounced Gráhs-hoff) number,3 GrL , where the subscript designates the length on which it is based: Π3 Π4 ≡ GrL = gβ∆T L3 ν2 (8.9) Two exceptions in which Π3 and Π4 appear independently are rotating systems (where Coriolis forces are part of the body force) and situations in which β∆T is no longer 1 but instead approaches unity. We therefore expect to correlate data in most other situations with functional equations of the form Nu = fn(Gr, Pr) (8.10) Another attribute of the dimensionless functional equation is that the primary independent variable is usually the product of Gr and Pr. This is called the Rayleigh number, RaL , where the subscript designates the length on which it is based: RaL ≡ GrL Pr = gβ∆T L3 αν (8.11) 3 Nu, Pr, Π3 , Π4 , and Gr were all suggested by Nusselt in his pioneering paper on convective heat transfer [8.1]. Grashof was a notable nineteenth-century mechanical engineering professor who was simply given the honor of having a dimensionless group named after him posthumously (see, e.g., [8.2]). He did not work with natural convection. 403 404 Natural convection in single-phase fluids and during film condensation §8.3 Figure 8.3 The correlation of h data for vertical isothermal surfaces by Churchill and Chu [8.3], using NuL = fn(RaL , Pr). (Applies to full range of Pr.) Thus, most (but not all) analyses and correlations of natural convection yield Nu = fn Ra , Pr (8.12) secondary parameter primary (or most important) independent variable Figure 8.3 is a careful selection of the best data available for natural convection from vertical isothermal surfaces. These data were organized by Churchill and Chu [8.3] and they span 13 orders of magnitude of the Rayleigh number. The correlation of these data in the coordinates of Fig. 8.2 is exactly in the form of eqn. (8.12), and it brings to light the dominant influence of RaL , while any influence of Pr is small. The data correlate on these coordinates within a few percent up to RaL [1 + (0.492/Pr9/16 )]16/9 108 . That is about where the b.l. starts exhibiting turbulent behavior. Beyond that point, the overall Nusselt number, NuL , rises more sharply, and the data scatter increases somewhat because the heat transfer mechanisms change. Laminar natural convection on a vertical isothermal surface §8.3 Prediction of h in natural convection on a vertical surface The analysis of natural convection using an integral method was done independently by Squire (see [8.4]) and by Eckert [8.5] in the 1930s. We shall refer to this important development as the Squire-Eckert formulation. The analysis begins with the integrated momentum and energy equations. We assume δ = δt and integrate both equations to the same value of δ: d dx δ 0 u2 − uu∞ = 0, since u∞ = 0 dy = −ν ∂u ∂y δ + gβ y =0 0 (T − T∞ ) dy (8.13) and [eqn. (6.47)] d dx δ 0 u (T − T∞ ) dy = ∂T qw = −α ρcp ∂y y =0 The integrated momentum equation is the same as eqn. (6.24) except that it includes the buoyancy term, which was added to the differential momentum equation in eqn. (8.7). We now must estimate the temperature and velocity profiles for use in eqns. (8.13) and (6.47). This is done here in much the same way as it was done in Sections 6.2 and 6.3 for forced convection. We write down a set of known facts about the profiles and then use these things to evaluate the constants in power-series expressions for u and T . Since the temperature profile has a fairly simple shape, a simple quadratic expression can be used: y T − T∞ =a+b Tw − T∞ δ +c y δ 2 (8.14) Notice that the thermal boundary layer thickness, δt , is assumed equal to δ in eqn. (8.14). This would seemingly limit the results to Prandtl numbers not too much larger than unity. Actually, the analysis will also prove useful for large Pr’s because the velocity profile exerts diminishing influence on the temperature profile as Pr increases. We require the following 405 406 Natural convection in single-phase fluids and during film condensation §8.3 things to be true of this profile: T y = 0 = Tw or T − T∞ Tw − T∞ y /δ=0 • T y = δ = T∞ or T − T∞ Tw − T∞ y /δ=1 T − T∞ Tw − T∞ y /δ=1 • • ∂T ∂y =0 or y =δ d d(y/δ) =1=a =0=1+b+c = 0 = b + 2c so a = 1, b = −2, and c = 1. This gives the following dimensionless temperature profile: T − T∞ y =1−2 Tw − T∞ δ + 2 y δ = 1− y δ 2 (8.15) We anticipate a somewhat complicated velocity profile (recall Fig. 8.1) and seek to represent it with a cubic function: u = uc (x) y δ +c y δ 2 +d y δ 3 (8.16) where, since there is no obvious characteristic velocity in the problem, we write uc as an as-yet-unknown function. (uc will have to increase with x , since u must increase with x .) We know three things about u: • • u(y = δ) = 0 • we have already satisfied this condition by writing eqn. (8.16) with no lead constant u(y = 0) = 0 ∂u ∂y =0 or u = 0 = (1 + c + d) uc or ∂u ∂(y/δ) y =δ = 0 = (1 + 2c + 3d) uc y /δ=1 These give c = −2 and d = 1, so u y = uc (x) δ 1− y δ 2 (8.17) We could also have written the momentum equation (8.7) at the wall, where u = v = 0, and created a fourth condition: ∂2u ∂y 2 =− y =0 gβ (Tw − T∞ ) ν §8.3 Laminar natural convection on a vertical isothermal surface Figure 8.4 The temperature and velocity profiles for air (Pr = 0.7) in a laminar convection b.l. and then we could have evaluated uc (x) as βg |Tw − T∞ |δ2 4ν . A correct expression for uc will eventually depend upon these variables, but we will not attempt to make uc fit this particular condition. Doing so would yield two equations, (8.13) and (6.47), in a single unknown, δ(x). It would be impossible to satisfy both of them. Instead, we shall allow the velocity profile to violate this condition slightly and write uc (x) = C1 βg |Tw − T∞ | 2 δ (x) ν (8.18) Then we shall solve the two integrated conservation equations for the two unknowns, C1 (which should ¼) and δ(x). The dimensionless temperature and velocity profiles are plotted in Fig. 8.4. With them are included Schmidt and Beckmann’s exact calculation for air (Pr = 0.7), as presented in [8.4]. Notice that the integral approximation to the temperature profile is better than the approximation to the velocity profile. That is fortunate, since the temperature profile exerts the major influence in the heat transfer solution. When we substitute eqns. (8.15) and (8.17) in the momentum equa- 407 408 Natural convection in single-phase fluids and during film condensation §8.3 tion (8.13), using eqn. (8.18) for uc (x), we get 2 C1 g β |Tw − T∞ | ν 2 d δ5 dx 1 2 y δ 0 1− 4 y δ y δ d 1 = 105 1 = gβ |Tw − T∞ | δ 1− 0 2 y δ y δ d 1 =3 − C1 gβ |Tw − T∞ | δ(x) ∂ ∂yδ y δ 1− y δ 2 (8.19) y δ =0 =1 where we change the sign of the terms on the left by replacing (Tw − T∞ ) with its absolute value. Equation (8.19) then becomes 1 dδ 1 2 gβ |Tw − T∞ | C = − C1 δ3 21 1 dx ν2 3 or dδ4 = dx 1 − C1 3 2 gβ |Tw − T∞ | C1 ν2 84 Integrating this with the b.c., δ(x = 0) = 0, gives 1 − C1 3 δ4 = 2 gβ |Tw − T∞ | C1 x ν2 84 (8.20) Substituting eqns. (8.15), (8.17), and (8.18) in eqn. (6.47) likewise gives (Tw − T∞ ) C1 gβ |Tw − T∞ | d δ3 ν dx 1 0 y δ 1− y δ 4 d y δ 1 = 30 = −α d Tw − T∞ δ d(y/δ) 1− =−2 y δ 2 y /δ=0 Laminar natural convection on a vertical isothermal surface §8.3 or 3 C1 dδ4 C1 3 dδ δ = = 30 dx 40 dx 2 gβ |Tw − T∞ | Pr ν2 Integrating this with the b.c., δ(x = 0) = 0, we get δ4 = 80 x gβ|Tw − T∞ | C1 Pr ν2 (8.21) Equating eqns. (8.20) and (8.21) for δ4 , we then obtain 21 20 1 − C1 1 3 x= x gβ |Tw − T∞ | gβ |Tw − T∞ | C1 Pr ν2 ν2 or C1 = Pr 20 + Pr 3 21 (8.22) Then, from eqn. (8.21): 20 + Pr 21 δ4 = x gβ |Tw − T∞ | Pr2 ν2 240 or 0.952 + Pr δ = 3.936 x Pr2 1/4 1 (8.23) 1/4 Grx Equation (8.23) can be combined with the known temperature profile, eqn. (8.15), and substituted in Fourier’s law to find q: ∂T q = −k ∂y y =0 k(Tw − T∞ ) =− δ d T − T∞ Tw − T∞ y d δ =−2 =2 y /δ=0 k∆T δ (8.24) 409 410 Natural convection in single-phase fluids and during film condensation §8.3 so, writing h = q |Tw − T∞ | ≡ q/∆T , we obtain4 Nux ≡ x 2 qx Pr =2 = (PrGrx )1/4 ∆T k δ 3.936 0.952 + Pr 1/4 or 1/4 Nux = 0.508 Rax Pr 0.952 + Pr 1/4 (8.25) This is the Squire-Eckert result for the local heat transfer from a vertical isothermal wall during laminar natural convection. It applies for either Tw > T∞ or Tw < T∞ . The average heat transfer coefficient can be obtained from L L q(x) dx h= 0 h(x) dx = L∆T 0 L Thus, NuL = 1 hL = k k L 0 4 k Nux dx = Nux x 3 x =L or 1/4 NuL = 0.678 RaL Pr 0.952 + Pr 1/4 (8.26) All properties in eqn. (8.26) and the preceding equations should be evaluated at T = (Tw + T∞ ) 2 except in gases, where β should be evaluated at T∞ . Example 8.1 A thin-walled metal tank containing fluid at 40◦ C cools in air at 14◦ C; h is very large inside the tank. If the sides are 0.4 m high, compute h, q, and δ at the top. Are the b.l. assumptions reasonable? Solution. βair = 1 T∞ = 1 (273 + 14) = 0.00348 K−1 . RaL = gβ∆T L3 να = Then 9.8(0.00348)(40 − 14)(0.4)3 = 1.645 × 108 1.566 × 10−5 2.203 × 10−5 Recall that, in footnote 2, we anticipated that Nu would vary as Gr1/4 . We now see that this is the case. 4 Laminar natural convection on a vertical isothermal surface §8.3 and Pr = 0.711, where the properties are evaluated at 300 K = 27◦ C. Then, from eqn. (8.26), NuL = 0.678 1.645 × 108 1/4 0.711 0.952 + 0.711 1/4 = 62.1 so h= 62.1(0.02614) 62.1k = = 4.06 W/m2 K L 0.4 and q = h ∆T = 4.06(40 − 14) = 105.5 W/m2 The b.l. thickness at the top of the tank is given by eqn. (8.23) at x = L: 0.952 + 0.711 δ = 3.936 L 0.7112 1/4 1 RaL Pr 1/4 = 0.0430 Thus, the b.l. thickness at the end of the plate is only 4% of the height, or 1.72 cm thick. This is thicker than typical forced convection b.l.’s, but it is still reasonably thin. Example 8.2 Large thin metal sheets of length L are dipped in an electroplating bath in the vertical position. Their average temperature is initially cooler than the liquid in the bath. How rapidly will they come up to bath temperature, Tb ? Solution. We can probably take Bi 1 and use the lumped-capacity response equation (1.20). We obtain h for use in eqn. (1.20) from eqn. (8.26): h = 0.678 k L Pr 0.952 + Pr 1/4 g βL3 αν 1/4 ∆T 1/4 call this B Since h ∝ ∆T 1/4 , with ∆T = Tb − T , eqn. (1.20) becomes d(Tb − T ) BA =− (Tb − T )5/4 dt ρcV 411 412 Natural convection in single-phase fluids and during film condensation §8.3 where V /A = the half-thickness of the plate, w . Integrating this between the initial temperature of the plate, Ti , and the temperature at time t , we get T Ti d(Tb − T ) (Tb − T ) 5/4 t =− 0 B dt ρcw so Tb − T = 1 (Tb − Ti )1/4 B t + 4ρcw −4 (Before we use this result, we should check Bi = Bw ∆T 1/4 k to be certain that it is, in fact, less than unity.) The temperature can be put in dimensionless form as B (Tb − Ti )1/4 Tb − T t = 1+ Tb − Ti 4ρcw −4 where the coefficient of t is a kind of inverse time constant of the response. Thus, the temperature dependence of h in natural convection leads to a solution quite different from the exponential response that resulted from a constant h [eqn. (1.22)]. Comparison of analysis and correlations with experimental data Churchill and Chu [8.3] have proposed two equations for the data correlated in Fig. 8.3. The simpler of the two is shown in the figure. It is 1/4 NuL = 0.68 + 0.67 RaL 1+ 0.492 Pr 9/16 −4/9 (8.27) which applies for all Pr and for the range of Ra shown in the figure. The Squire–Eckert prediction is within 1.2% of this correlation for high Pr and high RaL , and it differs by only 5.5% if the fluid is a gas and RaL > 105 . Typical Rayleigh numbers usually exceed 105 , so we conclude that the Squire–Eckert prediction is remarkably accurate in the range of practical interest, despite the approximations upon which it is built. The additive constant of 0.68 in eqn. (8.27) is a correction for low RaL , where the b.l. 1/4 assumptions are inaccurate and NuL is no longer proportional to RaL . At low Prandtl numbers, the Squire-Eckert prediction fails and eqn. (8.27) has to be used. In the turbulent regime, Gr 109 [8.6], eqn. (8.27) Laminar natural convection on a vertical isothermal surface §8.3 predicts a lower bound on the data (see Fig. 8.3), although it is really intended only for laminar boundary layers. In this correlation, as in eqn. (8.26), the thermal properties should all be evaluated at a film temperature, Tf = (T∞ + Tw )/2, except for β, which is to be evaluated at T∞ if the fluid is a gas. Example 8.3 Verify the first heat transfer coefficient in Table 1.1. It is for air at 20◦ C next to a 0.3 m high wall at 50◦ C. Solution. At T = 35◦ C = 308 K, we find Pr = 0.71, ν = 16.45 × 10−6 m2 /s, α = 2.318×10−5 m2 /s, and β = 1 (273+20) = 0.00341 K−1 . Then RaL = 9.8(0.00341)(30)(0.3)3 gβ∆T L3 = = 7.10 × 107 αν (16.45)(0.2318)10−10 The Squire-Eckert prediction gives NuL = 0.678 7.10 × 107 1/4 0.71 0.952 + 0.71 1/4 = 50.3 so h = 50.3 0.0267 k = 50.3 L 0.3 = 4.48 W/m2 K. And the Churchill-Chu correlation gives NuL = 0.68 + 0.67 7.10 × 107 1/4 1 + (0.492/0.71)9/16 4/9 = 47.88 so h = 47.88 0.0267 0.3 = 4.26 W/m2 K The prediction is therefore within 5% of the correlation. We should use the latter result in preference to the theoretical one, although the difference is slight. 413 414 Natural convection in single-phase fluids and during film condensation §8.3 Variable-properties problem Sparrow and Gregg [8.7] provide an extended discussion of the influence of physical property variations on predicted values of Nu. They found that while β for gases should be evaluated at T∞ , all other properties should be evaluated at Tr , where Tr = Tw − C (Tw − T∞ ) (8.28) and where C = 0.38 for gases. Most books recommend that a simple mean between Tw and T∞ (or C = 0.50) be used. A simple mean seldom differs much from the more precise result above, of course. It has also been shown by Barrow and Sitharamarao [8.8] that when β∆T is no longer 1, the Squire-Eckert formula should be corrected as follows: 3 Nu = Nusq−Ek 1 + 5 β∆T + O(β∆T )2 1/4 (8.29) This same correction can be applied to the Churchill-Chu correlation or to other expressions for Nu. Since β = 1 T∞ for an ideal gas, eqn. (8.29) gives only about a 1.5% correction for a 330 K plate heating 300 K air. Note on the validity of the boundary layer approximations The boundary layer approximations are sometimes put to a rather severe test in natural convection problems. Thermal b.l. thicknesses are often fairly large, and the usual analyses that take the b.l. to be thin can be significantly in error. This is particularly true as Gr becomes small. Figure 8.5 includes three pictures that illustrate this. These pictures are interferograms (or in the case of Fig. 8.5c, data deduced from interferograms). An interferogram is a photograph made in a kind of lighting that causes regions of uniform density to appear as alternating light and dark bands. Figure 8.5a was made at the University of Kentucky by G.S. Wang and R. Eichhorn. The Grashof number based on the radius of the leading edge is 2250 in this case. This is low enough to result in a b.l. that is larger than the radius near the leading edge. Figure 8.5b and c are from Kraus’s classic study of natural convection visualization methods [8.9]. Figure 8.5c shows that, at Gr = 585, the b.l. assumptions are quite unreasonable since the cylinder is small in comparison with the large region of thermal disturbance. a. A 1.34 cm wide flat plate with a rounded leading edge in air. Tw = 46.5◦ C, ∆T = 17.0◦ C, Grradius 2250 b. A square cylinder with a fairly low value of Gr. (Rendering of an interferogram shown in [8.9].) c. Measured isotherms around a cylinder in airwhen GrD ≈ 585 (from [8.9]). Figure 8.5 The thickening of the b.l. during natural convection at low Gr, as illustrated by interferograms made on two-dimensional bodies. (The dark lines in the pictures are isotherms.) 415 416 Natural convection in single-phase fluids and during film condensation §8.4 The analysis of free convection becomes a far more complicated problem at low Gr’s, since the b.l. equations can no longer be used. We shall not discuss any of the numerical solutions of the full Navier-Stokes equations that have been carried out in this regime. We shall instead note that correlations of data using functional equations of the form Nu = fn(Ra, Pr) will be the first thing that we resort to in such cases. Indeed, Fig. 8.3 reveals that Churchill and Chu’s equation (8.27) already serves this purpose in the case of the vertical isothermal plate, at low values of Ra ≡ Gr Pr. 8.4 Natural convection in other situations Natural convection from horizontal isothermal cylinders Churchill and Chu [8.10] provide yet another comprehensive correlation of existing data. For horizontal isothermal cylinders, they find that an equation with the same form as eqn. (8.27) correlates the data for horizontal cylinders as well. Horizontal cylinder data from a variety of sources, over about 24 orders of magnitude of the Rayleigh number based on the diameter, RaD , are shown in Fig. 8.6. The equation that correlates them is 1/4 NuD = 0.36 + 0.518 RaD 1 + (0.559/Pr)9/16 4/9 (8.30) They recommend that eqn. (8.30) be used in the range 10−6 RaD 109 . When RaD is greater than 109 , the flow becomes turbulent. The following equation is a little more complex, but it gives comparable accuracy over a larger range: NuD = ⎧ ⎪ ⎨ ⎡ 0.60 + 0.387⎣ ⎪ ⎩ RaD 1 + (0.559/Pr)9/16 ⎤1/6 ⎫2 ⎪ ⎬ ⎦ 16/9 ⎪ ⎭ The recommended range of applicability of eqn. (8.31) is 10−6 RaD (8.31) Natural convection in other situations §8.4 417 Figure 8.6 The data of many investigators for heat transfer from isothermal horizontal cylinders during natural convection, as correlated by Churchill and Chu [8.10]. Example 8.4 Space vehicles are subject to a “g -jitter,” or background variation of acceleration, on the order of 10−6 or 10−5 earth gravities. Brief periods of gravity up to 10−4 or 10−2 earth gravities can be exerted by accelerating the whole vehicle. A certain line carrying hot oil is ½ cm in diameter and it is at 127◦ C. How does Q vary with g -level if T∞ = 27◦ C in the air around the tube? Solution. The average b.l. temperature is 350 K. We evaluate properties at this temperature and write g as ge × (g -level), where ge is g at the earth’s surface and the g -level is the fraction of ge in the space vehicle. With β = 1/T∞ for an ideal gas 400 − 300 9.8 (0.005)3 gβ∆T D 3 300 = RaD = g -level να 2.062(10)−5 2.92(10)−5 = (678.2) g -level From eqn. (8.31), with Pr = 0.706, we compute NuD = ⎧ ⎨ ⎩ 0.6 + 0.387 678.2 16/9 1 + (0.559/0.706)9/16 =0.952 so 1/6 (g -level)1/6 ⎫2 ⎬ ⎭ 418 Natural convection in single-phase fluids and during film condensation g -level NuD 10−6 10−5 10−4 10−2 0.483 0.547 0.648 1.086 0.0297 0.005 h = NuD 2.87 3.25 3.85 6.45 W/m2 K W/m2 K W/m2 K W/m2 K §8.4 Q = π D h∆T 4.51 5.10 6.05 10.1 W/m W/m W/m W/m of of of of tube tube tube tube The numbers in the rightmost column are quite low. Cooling is clearly inefficient at these low gravities. Natural convection from vertical cylinders The heat transfer from the wall of a cylinder with its axis running vertically is the same as that from a vertical plate, so long as the thermal b.l. is thin. However, if the b.l. is thick, as is indicated in Fig. 8.7, heat transfer will be enhanced by the curvature of the thermal b.l. This correction was first considered some years ago by Sparrow and Gregg, and the analysis was subsequently extended with the help of more powerful numerical methods by Cebeci [8.11]. Figure 8.7 includes the corrections to the vertical plate results that were calculated for many Pr’s by Cebeci. The left-hand graph gives a correction that must be multiplied by the local flat-plate Nusselt number to get the vertical cylinder result. Notice that the correction increases when the Grashof number decreases. The right-hand curve gives a similar correction for the overall Nusselt number on a cylinder of height L. Notice that in either situation, the correction for all but liquid metals is less than 1/4 10% if (x or L)/R < 0.08 Grx or L . Heat transfer from general submerged bodies Spheres. The sphere is an interesting case because it has a clearly specifiable value of NuD as RaD → 0. We look first at this limit. When the buoyancy forces approach zero by virtue of: • low gravity, • very high viscosity, • small diameter, • a very small value of β, then heated fluid will no longer be buoyed away convectively. In that case, only conduction will serve to remove heat. Using shape factor number 4 Natural convection in other situations §8.4 Figure 8.7 Corrections for h and h on vertical isothermal plates to make them apply to vertical isothermal cylinders [8.11]. in Table 5.4, we compute in this case lim NuD = RaD →0 QD k∆T (S)D 4π (D/2) = = =2 2 ∆T k A∆T k 4π (D/2) 4π (D/4) (8.32) Every proper correlation of data for heat transfer from spheres therefore has the lead constant, 2, in it.5 A typical example is that of Yuge [8.12] for spheres immersed in gases: 1/4 NuD = 2 + 0.43 RaD , RaD < 105 (8.33) A more complex expression [8.13] encompasses other Prandtl numbers: 1/4 NuD = 2 + 0.589 RaD 4/9 1 + (0.492/Pr)9/16 RaD < 1012 (8.34) This result has an estimated uncertainty of 5% in air and an rms error of about 10% at higher Prandtl numbers. 5 It is important to note that while NuD for spheres approaches a limiting value at small RaD , no such limit exists for cylinders or vertical surfaces. The constants in eqns. (8.27) and (8.30) are not valid at extremely low values of RaD . 419 420 Natural convection in single-phase fluids and during film condensation §8.4 Rough estimate of Nu for other bodies. In 1973 Lienhard [8.14] noted that, for laminar convection in which the b.l. does not separate, the expression Nuτ 1/4 0.52 Raτ (8.35) would predict heat transfer from any submerged body within about 10% if Pr is not 1. The characteristic dimension in eqn. (8.35) is the length of travel, τ , of fluid in the unseparated b.l. In the case of spheres without separation, for example, τ = π D/2, the distance from the bottom to the top around the circumference. Thus, for spheres, eqn. (8.35) becomes g β∆T (π D/2)3 hπ D = 0.52 2k να 1/4 or 2 hD = 0.52 k π π 2 3/4 g β∆T D 3 να 1/4 or 1/4 NuD = 0.465 RaD This is within 8% of Yuge’s correlation if RaD remains fairly large. Laminar heat transfer from inclined and horizontal plates In 1953, Rich [8.15] showed that heat transfer from inclined plates could be predicted by vertical plate formulas if the component of the gravity vector along the surface of the plate was used in the calculation of the Grashof number. Thus, g is replaced by g cos θ , where θ is the angle of inclination measured from the vertical, as shown in Fig. 8.8. The heat transfer rate decreases as (cos θ)1/4 . Subsequent studies have shown that Rich’s result is substantially correct for the lower surface of a heated plate or the upper surface of a cooled plate. For the upper surface of a heated plate or the lower surface of a cooled plate, the boundary layer becomes unstable and separates at a relatively low value of Gr. Experimental observations of such instability have been reported by Fujii and Imura [8.16], Vliet [8.17], Pera and Gebhart [8.18], and Al-Arabi and El-Riedy [8.19], among others. §8.4 Natural convection in other situations Figure 8.8 Natural convection b.l.’s on some inclined and horizontal surfaces. The b.l. separation, shown here for the unstable cases in (a) and (b), occurs only at sufficiently large values of Gr. In the limit θ = 90◦ — a horizontal plate — the fluid flow above a hot plate or below a cold plate must form one or more plumes, as shown in Fig. 8.8c and d. In such cases, the b.l. is unstable for all but small Rayleigh numbers, and even then a plume must leave the center of the plate. The unstable cases can only be represented with empirical correlations. Theoretical considerations, and experiments, show that the Nusselt number for laminar b.l.s on horizontal and slightly inclined plates varies as Ra1/5 [8.20, 8.21]. For the unstable cases, when the Rayleigh number exceeds 104 or so, the experimental variation is as Ra1/4 , and once the flow is fully turbulent, for Rayleigh numbers above about 107 , experi- 421 422 Natural convection in single-phase fluids and during film condensation §8.4 ments show a Ra1/3 variation of the Nusselt number [8.22, 8.23]. In the 1/3 latter case, both NuL and RaL are proportional to L, so that the heat transfer coefficient is independent of L. Moreover, the flow field in these situations is driven mainly by the component of gravity normal to the plate. Unstable Cases: For the lower side of cold plates and the upper side of hot plates, the boundary layer becomes increasingly unstable as Ra is increased. 45◦ and 105 • For inclinations θ in eqn. (8.27). 109 , replace g with g cos θ RaL • For horizontal plates with Rayleigh numbers above 107 , nearly identical results have been obtained by many investigators. From these results, Raithby and Hollands propose [8.13]: 1/3 NuL = 0.14 RaL 1 + 0.0107 Pr , 1 + 0.01 Pr 0.024 Pr 2000 (8.36) This formula is consistent with available data up to RaL = 2 × 1011 , and probably goes higher. As noted before, the choice of lengthscale L is immaterial. Fujii and Imura’s results support using the above for 60◦ θ 90◦ with g in the Rayleigh number. For high Ra in gases, temperature differences and variable properties effects can be large. From experiments on upward facing plates, Clausing and Berton [8.23] suggest evaluating all gas properties at a reference temperature, in kelvin, of Tref = Tw − 0.83 (Tw − T∞ ) for 1 Tw /T∞ 3. • For horizontal plates of area A and perimeter P at lower Rayleigh numbers, Raithby and Hollands suggest [8.13] 1/4 NuL∗ = 0.560 RaL∗ 1 + (0.492/Pr)9/16 4/9 (8.37a) where, following Lloyd and Moran [8.22], a characteristic lengthscale L∗ = A/P , is used in the Rayleigh and Nusselt numbers. If Natural convection in other situations §8.4 NuL∗ 10, the b.l.s will be thick, and they suggest correcting the result to Nucorrected = 1.4 ln 1 + 1.4 NuL∗ (8.37b) These equations are recommended6 for 1 < RaL∗ < 107 . • In general, for inclined plates in the unstable cases, Raithby and Hollands [8.13] recommend that the heat flow be computed first using the formula for a vertical plate with g cos θ and then using the formula for a horizontal plate with g sin θ (i.e., the component of gravity normal to the plate) and that the larger value of the heat flow be taken. Stable Cases: For the upper side of cold plates and the lower side of hot plates, the flow is generally stable. The following results assume that the flow is not obstructed at the edges of the plate; a surrounding adiabatic surface, for example, will lower h [8.24, 8.25]. RaL 1011 , eqn. (8.27) is still valid for the • For θ < 88◦ and 105 upper side of cold plates and the lower side of hot plates when g is replaced with g cos θ in the Rayleigh number [8.16]. • For downward-facing hot plates and upward-facing cold plates of width L with very slight inclinations, Fujii and Imura give: 1/5 NuL = 0.58 RaL (8.38) θ 90◦ and for 109 This is valid for 106 < RaL < 109 if 87◦ θ 90◦ . RaL is based on g (not g cos θ ). RaL < 1011 if 89◦ Fujii and Imura’s results are for two-dimensional plates—ones in which infinite breadth has been approximated by suppression of end effects. For circular plates of diameter D in the stable horizontal configurations, the data of Kadambi and Drake [8.26] suggest that 1/5 NuD = 0.82 RaD Pr0.034 6 (8.39) Raithby and Hollands also suggest using a blending formula for 1 < RaL∗ < 1010 Nublended,L∗ = Nucorrected 10 + Nuturb 10 1/10 (8.37c) in which Nuturb is calculated from eqn. (8.36) using L∗ . The formula is useful for numerical progamming, but its effect on h is usually small. 423 424 Natural convection in single-phase fluids and during film condensation §8.4 Natural convection with uniform heat flux When qw is specified instead of ∆T ≡ (Tw − T∞ ), ∆T becomes the unknown dependent variable. Because h ≡ qw /∆T , the dependent variable appears in the Nusselt number; however, for natural convection, it also appears in the Rayleigh number. Thus, the situation is more complicated than in forced convection. Since Nu often varies as Ra1/4 , we may write Nux = qw x 1/4 ∝ Rax ∝ ∆T 1/4 x 3/4 ∆T k The relationship between x and ∆T is then ∆T = C x 1/5 (8.40) where the constant of proportionality C involves qw and the relevant physical properties. The average of ∆T over a heater of length L is ∆T = 1 L L 0 C x 1/5 dx = 5 C 6 (8.41) We plot ∆T /C against x/L in Fig. 8.9. Here, ∆T and ∆T (x/L = ½) are within 4% of each other. This suggests that, if we are interested in average values of ∆T , we can use ∆T evaluated at the midpoint of the plate in both the Rayleigh number, RaL , and the average Nusselt number, NuL = qw L/k∆T . Churchill and Chu, for example, show that their vertical plate correlation, eqn. (8.27), represents qw = constant data exceptionally well in the range RaL > 1 when RaL is based on ∆T at the middle of the plate. This approach eliminates the variation of ∆T with x from the calculation, but the temperature difference at the middle of the plate must still be found by iteration. To avoid iterating, we need to eliminate ∆T from the Rayleigh number. We can do this by introducing a modified Rayleigh number, Ra∗ , defined x as Ra∗ ≡ Rax Nux ≡ x gβqw x 4 gβ∆T x 3 qw x = να ∆T k kνα (8.42) For example, in eqn. (8.27), we replace RaL with Ra∗ NuL . The result is L NuL = 0.68 + 0.67 1/4 Ra∗ L 1/4 NuL 0.492 1+ Pr 9/16 4/9 Natural convection in other situations §8.4 Figure 8.9 The mean value of ∆T ≡ Tw − T∞ during natural convection. which may be rearranged as 1/4 NuL When NuL NuL − 0.68 = 1/4 0.67 Ra∗ L 1 + (0.492/Pr)9/16 4/9 (8.43a) 5, the term 0.68 may be neglected, with the result NuL = 0.73 Ra∗ L 1/5 1 + (0.492/Pr)9/16 16/45 (8.43b) Raithby and Hollands [8.13] give the following, somewhat simpler correlations for laminar natural convection from vertical plates with a uniform wall heat flux: Nux = 0.630 NuL = 6 5 Ra∗ Pr √x 4 + 9 Pr + 10 Pr Ra∗ Pr √L 4 + 9 Pr + 10 Pr These equations apply for all Pr and for Nu or Ra∗ are given in [8.13]). 1/5 (8.44a) 1/5 (8.44b) 5 (equations for lower Nu 425 426 Natural convection in single-phase fluids and during film condensation §8.4 Example 8.5 A horizontal circular disk heater of diameter 0.17 m faces downward in air at 27◦ C. If it delivers 15 W, estimate its average surface temperature. Solution. We have no formula for this situation, so the problem calls for some judicious guesswork. Following the lead of Churchill and Chu, we replace RaD with Ra∗ /NuD in eqn. (8.39): D NuD 6/5 = qw D ∆T k 6/5 = 0.82 Ra∗ D 1/5 Pr0.034 so ∆T = 1.18 = 1.18 qw D k g βqw D 4 kνα 1/6 Pr0.028 15 π (0.085)2 0.17 0.02614 9.8[15/π (0.085)2 ]0.174 300(0.02164)(1.566)(2.203)10−10 1/6 (0.711)0.028 = 140 K In the preceding computation, all properties were evaluated at T∞ . Now we must return the calculation, reevaluating all properties except β at 27 + (140/2) = 97◦ C: ∆T corrected = 1.18 661(0.17)/0.03104 9.8[15/π (0.085)2 ]0.174 300(0.03104)(3.231)(2.277)10−10 1/6 (0.99) = 142 K so the surface temperature is 27 + 142 = 169◦ C. That is rather hot. Obviously, the cooling process is quite ineffective in this case. Some other natural convection problems There are many natural convection situations that are beyond the scope of this book but which arise in practice. Two examples follow. §8.4 Natural convection in other situations Natural convection in enclosures. When a natural convection process occurs within a confined space, the heated fluid buoys up and then follows the contours of the container, releasing heat and in some way returning to the heater. This recirculation process normally enhances heat transfer beyond that which would occur by conduction through the stationary fluid. These processes are of importance to energy conservation processes in buildings (as in multiply glazed windows, uninsulated walls, and attics), to crystal growth and solidification processes, to hot or cold liquid storage systems, and to countless other configurations. Survey articles on natural convection in enclosures have been written by Yang [8.27], Raithby and Hollands [8.13], and Catton [8.28]. Combined natural and forced convection. When forced convection along, say, a vertical wall occurs at a relatively low velocity but at a relatively high heating rate, the resulting density changes can give rise to a superimposed natural convection process. We saw in footnote 2 on page 402 1/2 that GrL plays the role of of a natural convection Reynolds number, it follows that we can estimate of the relative importance of natural and forced convection can be gained by considering the ratio strength of natural convection flow GrL 2 = strength of forced convection flow ReL (8.45) where ReL is for the forced convection along the wall. If this ratio is small compared to one, the flow is essentially that due to forced convection, whereas if it is large compared to one, we have natural convection. When GrL Re2 is on the order of one, we have a mixed convection process. L It should be clear that the relative orientation of the forced flow and the natural convection flow matters. For example, compare cool air flowing downward past a hot wall to cool air flowing upward along a hot wall. The former situation is called opposing flow and the latter is called assisting flow. Opposing flow may lead to boundary layer separation and degraded heat transfer. Churchill [8.29] has provided an extensive discussion of both the conditions that give rise to mixed convection and the prediction of heat transfer for it. Review articles on the subject have been written by Chen and Armaly [8.30] and by Aung [8.31]. 427 428 Natural convection in single-phase fluids and during film condensation 8.5 §8.5 Film condensation Dimensional analysis and experimental data The dimensional functional equation for h (or h) during film condensation is7 h or h = fn cp , ρf , hfg , g ρf − ρg , k, µ, (Tsat − Tw ) , L or x where hfg is the latent heat of vaporization. It does not appear in the differential equations (8.4) and (6.40); however, it is used in the calculation of δ [which enters in the b.c.’s (8.5)]. The film thickness, δ, depends heavily on the latent heat and slightly on the sensible heat, cp ∆T , which the film must absorb to condense. Notice, too, that g(ρf − ρg ) is included as a product, because gravity only enters the problem as it acts upon the density difference [cf. eqn. (8.4)]. The problem is therefore expressed nine variables in J, kg, m, s, and ◦ C (where we once more avoid resolving J into N · m since heat is not being converted into work in this situation). It follows that we look for 9 − 5 = 4 pi-groups. The ones we choose are Π1 = NuL ≡ Π3 = Ja ≡ hL k cp (Tsat − Tw ) hfg Π2 = Pr ≡ Π4 ≡ ν α ρf (ρf − ρg )ghfg L3 µk(Tsat − Tw ) Two of these groups are new to us. The group Π3 is called the Jakob number, Ja, to honor Max Jakob’s important pioneering work during the 1930s on problems of phase change. It compares the maximum sensible heat absorbed by the liquid to the latent heat absorbed. The group Π4 does not normally bear anyone’s name, but, if it is multiplied by Ja, it may be regarded as a Rayleigh number for the condensate film. Notice that if we condensed water at 1 atm on a wall 10◦ C below Tsat , then Ja would equal 4.174(10/2257) = 0.0185. Although 10◦ C is a fairly large temperature difference in a condensation process, it gives a maximum sensible heat that is less than 2% of the latent heat. The Jakob number is accordingly small in most cases of practical interest, and it turns out that sensible heat can often be neglected. (There are important 7 Note that, throughout this section, k, µ , cp , and Pr refer to properties of the liquid, rather than the vapor. Film condensation §8.5 429 exceptions to this.) The same is true of the role of the Prandtl number. Therefore, during film condensation ⎛ ⎞ ρf (ρf − ρg )ghfg L3 NuL = fn ⎝ , Pr, Ja ⎠ (8.46) µk(Tsat − Tw ) secondary independent variables primary independent variable, Π4 Equation (8.46) is not restricted to any geometrical configuration, since the same variables govern h during film condensation on any body. Figure 8.10, for example, shows laminar film condensation data given for spheres by Dhir8 [8.32]. They have been correlated according to eqn. (8.12). The data are for only one value of Pr but for a range of Π4 and Ja. They generally correlate well within ±10%, despite a broad variation of the not-very-influential variable, Ja. A predictive curve [8.32] is included in Fig. 8.10 for future reference. Laminar film condensation on a vertical plate Consider the following feature of film condensation. The latent heat of a liquid is normally a very large number. Therefore, even a high rate of heat transfer will typically result in only very thin films. These films move relatively slowly, so it is safe to ignore the inertia terms in the momentum equation (8.4): u ∂u ∂u = +v ∂x ∂y 1− ρg ρf g+ν ∂2u ∂y 2 d2 u dy 2 0 This result will give u = u(y, δ) (where δ is the local b.l. thickness) when it is integrated. We recognize that δ = δ(x), so that u is not strictly dependent on y alone. However, the y -dependence is predominant, and it is reasonable to use the approximate momentum equation ρf − ρg g d2 u =− 2 dy ν ρf 8 (8.47) Professor Dhir very kindly recalculated his data into the form shown in Fig. 8.10 for use here. 430 Natural convection in single-phase fluids and during film condensation §8.5 Figure 8.10 Correlation of the data of Dhir [8.32] for laminar film condensation on spheres at one value of Pr and for a range of Π4 and Ja, with properties evaluated at (Tsat + Tw )/2. Analytical prediction from [8.33]. This simplification was made by Nusselt in 1916 when he set down the original analysis of film condensation [8.34]. He also eliminated the convective terms from the energy equation (6.40): u ∂T ∂2T ∂T +v =α ∂x ∂y ∂y 2 0 Film condensation §8.5 431 on the same basis. The integration of eqn. (8.47) subject to the b.c.’s u y =0 =0 ∂u ∂y and =0 y =δ gives the parabolic velocity profile: u= (ρf − ρg )gδ2 2 2µ y δ y δ − 2 (8.48) And integration of the energy equation subject to the b.c.’s T y = 0 = Tw and T y = δ = Tsat gives the linear temperature profile: T = Tw + (Tsat − Tw ) y δ (8.49) To complete the analysis, we must calculate δ. This can be done in ˙ two steps. First, we express the mass flow rate per unit width of film, m, in terms of δ, with the help of eqn. (8.48): δ ˙ m= 0 ρf u dy = ρf (ρf − ρg ) 3µ gδ3 (8.50) Second, we neglect the sensible heat absorbed by that part of the film cooled below Tsat and express the local heat flux in terms of the rate of ˙ change of m (see Fig. 8.11): q =k ∂T ∂y =k y =0 ˙ Tsat − Tw dm = hfg δ dx (8.51) Substituting eqn. (8.50) in eqn. (8.51), we obtain a first-order differential equation for δ: k hfg ρf (ρf − ρg ) dδ Tsat − Tw = gδ2 δ µ dx (8.52) This can be integrated directly, subject to the b.c., δ(x = 0) = 0. The result is δ= 4k(Tsat − Tw )µx ρf (ρf − ρg )ghfg 1/4 (8.53) 432 Natural convection in single-phase fluids and during film condensation §8.5 Figure 8.11 Heat and mass flow in an element of a condensing film. Both Nusselt and, subsequently, Rohsenow [8.35] showed how to correct the film thickness calculation for the sensible heat that is needed to cool the inner parts of the film below Tsat . Rohsenow’s calculation was, in part, an assessment of Nusselt’s linear-temperature-profile assumption, and it led to a corrected latent heat—designated hfg —which accounted for subcooling in the liquid film when Pr is large. Rohsenow’s result, which we show below to be strictly true only for large Pr, was ⎡ ⎤ cp (Tsat − Tw ) ⎦ (8.54) hfg = hfg ⎣ 1 + 0.68 hfg ≡ Ja, Jakob number Thus, we simply replace hfg with hfg wherever it appears explicitly in the analysis, beginning with eqn. (8.51). Finally, the heat transfer coefficient is obtained from h≡ 1 q = Tsat − Tw Tsat − Tw k(Tsat − Tw ) δ = k δ (8.55) so Nux = x hx = k δ (8.56) Thus, with the help of eqn. (8.54), we substitute eqn. (8.53) in eqn. (8.56) Film condensation §8.5 433 and get ⎡ Nux = 0.707 ⎣ ρf (ρf − ρg )ghfg x 3 µk(Tsat − Tw ) ⎤1/4 ⎦ (8.57) This equation carries out the functional dependence that we anticipated in eqn. (8.46): Nux = fn Π4 , Ja , Pr eliminated in so far as we neglected convective terms in the energy equation this is carried implicitly in hfg this is clearly the dominant variable The liquid properties in Π4 , Ja, and Pr (with the exception of hfg ) are to be evaluated at the mean film temperature. However, if Tsat − Tw is small—and it often is—one might approximate them at Tsat . At this point we should ask just how great the missing influence of Pr is and what degree of approximation is involved in representing the influence of Ja with the use of hfg . Sparrow and Gregg [8.36] answered these questions with a complete b.l. analysis of film condensation. They did not introduce Ja in a corrected latent heat but instead showed its influence directly. Figure 8.12 displays two figures from the Sparrow and Gregg paper. The first shows heat transfer results plotted in the form Nux = fn (Ja, Pr) → constant as Ja → 0 4 Π4 (8.58) Notice that the calculation approaches Nusselt’s simple result for all Pr as Ja → 0. It also approaches Nusselt’s result, even for fairly large values of Ja, if Pr is not small. The second figure shows how the temperature deviates from the linear profile that we assumed to exist in the film in developing eqn. (8.49). If we remember that a Jakob number of 0.02 is about as large as we normally find in laminar condensation, it is clear that the linear temperature profile is a very sound assumption for nonmetallic liquids. 434 Natural convection in single-phase fluids and during film condensation §8.5 Figure 8.12 Results of the exact b.l. analysis of laminar film condensation on a vertical plate [8.36]. Sadasivan and Lienhard [8.37] have shown that the Sparrow-Gregg formulation can be expressed with high accuracy, for Pr 0.6, by including Pr in the latent heat correction. Thus they wrote hfg = hfg 1 + 0.683 − 0.228 Pr Ja which includes eqn. (8.54) for Pr → ∞ as we anticipated. (8.59) Film condensation §8.5 435 The Sparrow and Gregg analysis proves that Nusselt’s analysis is quite accurate for all Prandtl numbers above the liquid-metal range. The very high Ja flows, for which Nusselt’s theory requires some correction, usually result in thicker films, which become turbulent so the exact analysis no longer applies. The average heat transfer coefficient is calculated in the usual way for Twall = constant: h= 1 L L h(x) dx = 0 4 3 h(L) so ⎡ NuL = 0.9428 ⎣ ⎤1/4 ρf (ρf − ρg )ghfg L3 ⎦ µk(Tsat − Tw ) (8.60) Example 8.6 Water at atmospheric pressure condenses on a strip 30 cm in height that is held at 90◦ C. Calculate the overall heat transfer per meter, the film thickness at the bottom, and the mass rate of condensation per meter. Solution. ⎤1/4 4k(Tsat − Tw )µx ⎦ δ=⎣ ρf (ρf − ρg )ghfg ⎡ where we have replaced hfg with hfg : hfg = 2257 1 + 0.683 − 0.228 1.86 4.211(10) 2257 = 2281 kJ/kg so δ= 4(0.677)(10)(2.99 × 10−4 ) x 961.9(961.9 − 0.6)(9.806)(2281 × 103 ) 1/4 Then δ(L) = 0.000104 m = 0.104 mm = 0.000141 x 1/4 436 Natural convection in single-phase fluids and during film condensation §8.5 Notice how thin the film is. Finally, we use eqns. (8.56) and (8.59) to compute NuL = 4(0.3) 4L = = 3846 3δ 3(0.000104) so q= NuL k∆T 3846(0.677)(10) = = 8.68 × 104 W/m2 L 0.3 (This would correspond to a heat flow of 86.8 kW on an area about half the size of a desk top. That is very high for such a small temperature difference.) Then Q = (8.68 × 104 )(0.3) = 26, 040 W/m = 26.0 kW/m ˙ The rate of condensate flow, m is ˙ m= 26.0 Q = 0.0114 kg/m·s = hfg 2281 Condensation on other bodies Nusselt himself extended his prediction to certain other bodies but was restricted by the lack of a digital computer from evaluating as many cases as he might have. In 1971 Dhir and Lienhard [8.33] showed how Nusselt’s method could be readily extended to a large class of problems. They showed that one need only to replace the gravity, g , with an effective gravity, geff : geff ≡ x gR x g 1/3 R 4/3 (8.61) 4/3 dx 0 in eqns. (8.53) and (8.57), to predict δ and Nux for a variety of bodies. The terms in eqn. (8.61) are (see Fig. 8.13): • x is the distance along the liquid film measured from the upper stagnation point. • g = g(x), the component of gravity (or other body force) along x ; g can vary from point to point as it does in Fig. 8.13b and c. Figure 8.13 Condensation on various bodies. g(x) is the component of gravity or other body force in the x -direction. 437 438 Natural convection in single-phase fluids and during film condensation §8.5 • R(x) is a radius of curvature about the vertical axis. In Fig. 8.13a, it is a constant that factors out of eqn. (8.61). In Fig. 8.13c, R is infinite. Since it appears to the same power in both the numerator and the denominator, it again can be factored out of eqn. (8.61). Only in axisymmetric bodies, where R varies with x , need it be included. When it can be factored out, xg 4/3 geff reduces to x g 1/3 (8.62) dx 0 • ge is earth-normal gravity. We introduce ge at this point to distinguish it from g(x). Example 8.7 Find Nux for laminar film condensation on the top of a flat surface sloping at θ ◦ from the vertical plane. Solution. In this case g = ge cos θ and R = ∞. Therefore, eqn. (8.61) or (8.62) reduces to 4/3 geff = xge 1/3 ge (cos θ)4/3 (cos θ)1/3 x = ge cos θ dx 0 as we might expect. Then, for a slanting plate, ⎤1/4 ⎡ ρf (ρf − ρg )(ge cos θ)hfg x 3 ⎦ Nux = 0.707 ⎣ µk(Tsat − Tw ) (8.63) Example 8.8 Find the overall Nusselt number for a horizontal cylinder. Solution. There is an important conceptual hurdle here. The radius R(x) is infinity, as shown in Fig. 8.13c—it is not the radius of the cylinder. It is also very easy to show that g(x) is equal to ge sin(2x/D), where D is the diameter of the cylinder. Then 4/3 geff = xge 1/3 ge x 0 (sin 2x/D)4/3 (sin 2x/D)1/3 dx Film condensation §8.5 439 and, with h(x) from eqn. (8.57), ⌠ π D/2 2⎮ 1k ⎮ √ ⎮ h= πD ⌡ 2x ⎤1/4 ⎡ ⎢ ρf ρf − ρg h x 3 fg ⎢ ⎢ ⎣ µk (Tsat − Tw ) 0 xge (sin 2x/D)4/3 ⎥ ⎥ ⎥ x ⎦ 1/3 dx (sin 2x/D) dx 0 This integral can be evaulated in terms of gamma functions. The result, when it is put back in the form of a Nusselt number, is ⎡ NuD = 0.728 ⎣ ρf ρf − ρg ge hfg D 3 µk (Tsat − Tw ) ⎤1/4 ⎦ (8.64) for a horizontal cylinder. (Nusselt got 0.725 for the lead constant, but he had to approximate the integral with a hand calculation.) Some other results of this calculation include the following cases. Sphere of diameter D : ⎡ NuD = 0.828 ⎣ ρf ρf − ρg ge hfg D 3 µk (Tsat − Tw ) ⎤1/4 ⎦ (8.65) This result9 has already been compared with the experimental data in Fig. 8.10. Vertical cone with the apex on top, the bottom insulated, and a cone angle of α◦ : ⎡ Nux = 0.874 [cos(α/2)]1/4 ⎣ ρf ρf − ρg ge hfg x 3 µk (Tsat − Tw ) ⎤1/4 ⎦ (8.66) Rotating horizontal disk 10 : In this case, g = ω2 x , where x is the distance from the center and ω is the speed of rotation. The Nusselt number, based on L = (µ/ρf ω)1/2 , is ⎡ Nu = 0.9034 ⎣ 9 µ ρf − ρg hfg ρf k (Tsat − Tw ) ⎤1/4 ⎦ = constant (8.67) There is an error in [8.33]: the constant given there is 0.785. The value of 0.828 given here is correct. 10 This problem was originally solved by Sparrow and Gregg [8.38]. 440 Natural convection in single-phase fluids and during film condensation §8.5 This result might seem strange at first glance. It says that Nu ≠ fn(x or ω). The reason is that δ just happens to be independent of x in this configuration. The Nusselt solution can thus be bent to fit many complicated geometric figures. One of the most complicated ones that have been dealt with is the reflux condenser shown in Fig. 8.14. In such a configuration, cooling water flows through a helically wound tube and vapor condenses on the outside, running downward along the tube. As the condensate flows, centripetal forces sling the liquid outward at a downward angle. This complicated flow was analyzed by Karimi [8.39], who found that ⎡ ⎤ 3 1/4 d hd cos α ⎣ ρf − ρg ρf hfg g(d cos α) ⎦ = ,B Nu ≡ fn k µk∆T D (8.68) where B is a centripetal parameter: B≡ ρf − ρg cp ∆T tan2 α ρf hfg Pr and α is the helix angle (see Fig. 8.14). The function on the righthand side of eqn. (8.68) was a complicated one that must be evaluated numerically. Karimi’s result is plotted in Fig. 8.14. Laminar–turbulent transition ˙ The mass flow rate of condensate per unit width of film, m, is more commonly designated as Γc (kg/m · s). Its calculation in eqn. (8.50) involved substituting eqn. (8.48) in δ ˙ m or Γc = ρf u dy 0 Equation (8.48) gives u(y) independently of any geometric features. [The geometry is characterized by δ(x).] Thus, the resulting equation for the mass flow rate is still Γc = ρf ρf − ρg g δ3 3µ (8.50a) This expression is valid for any location along any film, regardless of the geometry of the body. The configuration will lead to variations of g(x) and δ(x), but eqn. (8.50a) still applies. Film condensation §8.5 441 Figure 8.14 Fully developed film condensation heat transfer on a helical reflux condenser [8.39]. It is useful to define a Reynolds number in terms of Γc . This is easy to do, because Γc is equal to ρuav δ. Rec = ρf (ρf − ρg )gδ3 Γc = µ 3µ 2 (8.69) It turns out that the Reynolds number dictates the onset of film instability, just as it dictates the instability of a b.l. or of a pipe flow.11 When 7, scallop-shaped ripples become visible on the condensate film. Rec When Rec reaches about 400, a full-scale laminar-to-turbulent transition occurs. Gregorig, Kern, and Turek [8.40] reviewed many data for the film condensation of water and added their own measurements. Figure 8.15 shows these data in comparison with Nusselt’s theory, eqn. (8.60). The comparison is almost perfect up to Rec 7. Then the data start yielding somewhat higher heat transfer rates than the prediction. This is because Two Reynolds numbers are defined for film condensation: Γc /µ and 4Γc /µ . The latter one, which is simply four times as large as the one we use, is more common in the American literature. 11 442 Natural convection in single-phase fluids and during film condensation §8.5 Figure 8.15 Film condensation on vertical plates. Data are for water [8.40]. the ripples improve heat transfer—just a little at first and by about 20% when the full laminar-to-turbulent transition occurs at Rec = 400. Above Rec = 400, NuL begins to rise with Rec . The Nusselt number begins to exhibit an increasingly strong dependence on the Prandtl number in this turbulent regime. Therefore, one can use Fig. 8.15, directly as a data correlation, to predict the heat transfer coefficient for steam condensating at 1 atm. But for other fluids with different Prandtl numbers, one should consult [8.41] or [8.42]. Two final issues in natural convection film condensation • Condensation in tube bundles. Nusselt showed that if n horizontal tubes are arrayed over one another, and if the condensate leaves each one and flows directly onto the one below it without splashing, then NuDfor n tubes = NuD1 tube n1/4 (8.70) This is a fairly optimistic extension of the theory, of course. In addition, the effects of vapor shear stress on the condensate and of pressure losses on the saturation temperature are often important in tube bundles. These effects are discussed by Rose et al. [8.42] and Marto [8.41]. Problems 443 • Condensation in the presence of noncondensable gases. When the condensing vapor is mixed with noncondensable air, uncondensed air must constantly diffuse away from the condensing film and vapor must diffuse inward toward the film. This coupled diffusion process can considerably slow condensation. The resulting h can easily be cut by a factor of five if there is as little as 5% by mass of air mixed into the steam. This effect was first analyzed in detail by Sparrow and Lin [8.43]. More recent studies of this problem are reviewed in [8.41, 8.42]. Problems 8.1 Show that Π4 in the film condensation problem can properly be interpreted as Pr Re2 Ja. 8.2 A 20 cm high vertical plate is kept at 34◦ C in a 20◦ C room. Plot (to scale) δ and h vs. height and the actual temperature and velocity vs. y at the top. 8.3 Redo the Squire-Eckert analysis, neglecting inertia, to get a high-Pr approximation to Nux . Compare your result with the Squire-Eckert formula. 8.4 Assume a linear temperature profile and a simple triangular velocity profile, as shown in Fig. 8.16, for natural convection on a vertical isothermal plate. Derive Nux = fn(Pr, Grx ), compare your result with the Squire-Eckert result, and discuss the comparison. 8.5 A horizontal cylindrical duct of diamond-shaped cross section (Fig. 8.17) carries air at 35◦ C. Since almost all thermal resistance is in the natural convection b.l. on the outside, take Tw to be approximately 35◦ C. T∞ = 25◦ C. Estimate the heat loss per meter of duct if the duct is uninsulated. [Q = 24.0 W/m.] 8.6 The heat flux from a 3 m high electrically heated panel in a wall is 75 W/m2 in an 18◦ C room. What is the average temperature of the panel? What is the temperature at the top? at the bottom? 444 Chapter 8: Natural convection in single-phase fluids and during film condensation Figure 8.16 Configuration for Problem 8.4. Figure 8.17 Configuration for Problem 8.5. 8.7 Find pipe diameters and wall temperatures for which the film condensation heat transfer coefficients given in Table 1.1 are valid. 8.8 Consider Example 8.6. What value of wall temperature (if any), or what height of the plate, would result in a laminar-to-turbulent transition at the bottom in this example? 8.9 A plate spins, as shown in Fig. 8.18, in a vapor that rotates synchronously with it. Neglect earth-normal gravity and calculate NuL as a result of film condensation. 8.10 A laminar liquid film of temperature Tsat flows down a vertical wall that is also at Tsat . Flow is fully developed and the film thickness is δo . Along a particular horizontal line, the wall temperature has a lower value, Tw , and it is kept at that temperature everywhere below that position. Call the line where the wall temperature changes x = 0. If the whole system is Problems 445 Figure 8.18 Configuration for Problem 8.9. immersed in saturated vapor of the flowing liquid, calculate δ(x), Nux , and NuL , where x = L is the bottom edge of the wall. (Neglect any transition behavior in the neighborhood of x = 0.) 8.11 Prepare a table of formulas of the form h (W/m2 K) = C [∆T ◦ C/L m]1/4 for natural convection at normal gravity in air and in water at T∞ = 27◦ C. Assume that Tw is close to 27◦ C. Your table should include results for vertical plates, horizontal cylinders, spheres, and possibly additional geometries. Do not include your calculations. 8.12 For what value of Pr is the condition ∂2u ∂y 2 = y =0 gβ(Tw − T∞ ) ν satisfied exactly in the Squire-Eckert b.l. solution? [Pr = 2.86.] 8.13 The overall heat transfer coefficient on the side of a particular house 10 m in height is 2.5 W/m2 K, excluding exterior convection. It is a cold, still winter night with Toutside = −30◦ C and Tinside air = 25◦ C. What is h on the outside of the house? Is external convection laminar or turbulent? 8.14 Consider Example 8.2. The sheets are mild steel, 2 m long and 6 mm thick. The bath is basically water at 60◦ C, and the sheets 446 Chapter 8: Natural convection in single-phase fluids and during film condensation are put in it at 18◦ C. (a) Plot the sheet temperature as a function of time. (b) Approximate h at ∆T = [(60 + 18)/2 − 18]◦ C and plot the conventional exponential response on the same graph. 8.15 A vertical heater 0.15 m in height is immersed in water at 7◦ C. Plot h against (Tw − T∞ )1/4 , where Tw is the heater temperature, in the range 0 < (Tw − T∞ ) < 100◦ C. Comment on the result. should the line be straight? 8.16 A 77◦ C vertical wall heats 27◦ C air. Evaluate δtop /L, RaL , and L where the line in Fig. 8.3 ceases to be straight. Comment on the implications of your results. [δtop /L 0.6.] 8.17 A horizontal 8 cm O.D. pipe carries steam at 150◦ C through a room at 17◦ C. The pipe has a 1.5 cm layer of 85% magnesia insulation on it. Evaluate the heat loss per meter of pipe. [Q = 97.3 W/m.] 8.18 What heat rate (in W/m) must be supplied to a 0.01 mm horizontal wire to keep it 30◦ C above the 10◦ C water around it? 8.19 A vertical run of copper tubing, 5 mm in diameter and 20 cm long, carries condensation vapor at 60◦ C through 27◦ C air. What is the total heat loss? 8.20 A body consists of two cones joined at their bases. The diameter is 10 cm and the overall length of the joined cones is 25 cm. The axis of the body is vertical, and the body is kept at 27◦ C in 7◦ C air. What is the rate of heat removal from the body? [Q = 3.38 W.] 8.21 Consider the plate dealt with in Example 8.3. Plot h as a function of the angle of inclination of the plate as the hot side is tilted both upward and downward. Note that you must make do with discontinuous formulas in different ranges of θ . 8.22 You have been asked to design a vertical wall panel heater, 1.5 m high, for a dwelling. What should the heat flux be if no part of the wall should exceed 33◦ C? How much heat will be added to the room if the panel is 7 m in width? 8.23 A 14 cm high vertical surface is heated by condensing steam at 1 atm. If the wall is kept at 30◦ C, how would the average Problems 447 heat transfer coefficient change if ammonia, R22, methanol, or acetone were used instead of steam to heat it? How would the heat flux change? (Data for methanol and acetone must be obtained from sources outside this book.) 8.24 A 1 cm diameter tube extends 27 cm horizontally through a region of saturated steam at 1 atm. The outside of the tube can be maintained at any temperature between 50◦ C and 150◦ C. Plot the total heat transfer as a function of tube temperature. 8.25 A 2 m high vertical plate condenses steam at 1 atm. Below what temperature will Nusselt’s prediction of h be in error? Below what temperature will the condensing film be turbulent? 8.26 A reflux condenser is made of copper tubing 0.8 cm in diameter with a wall temperature of 30◦ C. It condenses steam at 1 atm. Find h if α = 18◦ and the coil diameter is 7 cm. 8.27 The coil diameter of a helical condenser is 5 cm and the tube diameter is 5 mm. The condenser carries water at 15◦ C and is in a bath of saturated steam at 1 atm. Specify the number of coils and a reasonable helix angle if 6 kg/hr of steam is to be condensed. hinside = 600 W/m2 K. 8.28 A schedule 40 type 304 stainless steam pipe with a 4 in. nominal diameter carries saturated steam at 150 psia in a processing plant. Calculate the heat loss per unit length of pipe if it is bare and the surrounding air is still at 68◦ F. How much would this heat loss be reduced if the pipe were insulated with a 1 in. layer of 85% magnesia insulation? [Qsaved 127 W/m.] 8.29 What is the maximum speed of air in the natural convection b.l. in Example 8.1? 8.30 All of the uniform-Tw , natural convection formulas for Nu take the same form, within a constant, at high Pr and Ra. What is that form? (Exclude any equation that includes turbulence.) 8.31 A large industrial process requires that water be heated by a large horizontal cylinder using natural convection. The water is at 27◦ C. The diameter of the cylinder is 5 m, and it is kept at 67◦ C. First, find h. Then suppose that D is increased to 10 m. 448 Chapter 8: Natural convection in single-phase fluids and during film condensation What is the new h? Explain the similarity of these answers in the turbulent natural convection regime. 8.32 A vertical jet of liquid of diameter d and moving at velocity u∞ impinges on a horizontal disk rotating ω rad/s. There is no heat transfer in the system. Develop an expression for δ(r ), where r is the radial coordinate on the disk. Contrast the r dependence of δ with that of a condensing film on a rotating disk and explain the difference qualitatively. 8.33 We have seen that if properties are constant, h ∝ ∆T 1/4 in natural convection. If we consider the variation of properties as Tw is increased over T∞ , will h depend more or less strongly on ∆T in air? in water? 8.34 A film of liquid falls along a vertical plate. It is initially saturated and it is surrounded by saturated vapor. The film thickness is δo . If the wall temperature below a certain point on the wall (call it x = 0) is raised to a value of Tw , slightly above Tsat , derive expressions for δ(x), Nux , and xf —the distance at which the plate becomes dry. Calculate xf if the fluid is water at 1 atm, if Tw = 105◦ C and δo = 0.1 mm. 8.35 In a particular solar collector, dyed water runs down a vertical plate in a laminar film with thickness δo at the top. The sun’s rays pass through parallel glass plates (see Section 10.6) and deposit qs W/m2 in the film. Assume the water to be saturated at the inlet and the plate behind it to be insulated. Develop an expression for δ(x) as the water evaporates. Develop an expression for the maximum length of wetted plate, and provide a criterion for the laminar solution to be valid. 8.36 What heat removal flux can be achieved at the surface of a horizontal 0.01 mm diameter electrical resistance wire in still 27◦ C air if its melting point is 927◦ C? Neglect radiation. 8.37 A 0.03 m O.D. vertical pipe, 3 m in length, carries refrigerant through a 24◦ C room. How much heat does it absorb from the room if the pipe wall is at 10◦ C? 8.38 A 1 cm O.D. tube at 50◦ C runs horizontally in 20◦ C air. What is the critical radius of 85% magnesium insulation on the tube? Problems 449 8.39 A 1 in. cube of ice is suspended in 20◦ C air. Estimate the drip rate in gm/min. (Neglect ∆T through the departing water film. hsf = 333, 300 J/kg.) 8.40 A horizontal electrical resistance heater, 1 mm in diameter, releases 100 W/m in water at 17◦ C. What is the wire temperature? 8.41 Solve Problem 5.39 using the correct formula for the heat transfer coefficient. 8.42 A red-hot vertical rod, 0.02 m in length and 0.005 m in diameter, is used to shunt an electrical current in air at room temperature. How much power can it dissipate if it melts at 1200◦ C? Note all assumptions and corrections. Include radiation using Frod-room = 0.064. 8.43 A 0.25 mm diameter platinum wire, 0.2 m long, is to be held horizontally at 1035◦ C. It is black. How much electric power is needed? Is it legitimate to treat it as a constant-wall-temperature heater in calculating the convective part of the heat transfer? The surroundings are at 20◦ C and the surrounding room is virtually black. 8.44 A vertical plate, 11.6 m long, condenses saturated steam at 1 atm. We want to be sure that the film stays laminar. What is the lowest allowable plate temperature, and what is q at this temperature? 8.45 A straight horizontal fin exchanges heat by laminar natural convection with the surrounding air. a. Show that d2 θ = m2 L2 θ 5/4 dξ 2 where m is based on ho ≡ h(T = To ). b. Develop an iterative numerical method to solve this equation for T (x = 0) = To and an insulated tip. (Hint : linearize the right side by writing it as (m2 L2 θ 1/4 )θ , and evaluate the term in parenthesis at the previous iteration step.) 450 Chapter 8: Natural convection in single-phase fluids and during film condensation c. Solve the resulting difference equations for m2 L2 values ranging from 10−3 to 103 . Use Gauss elimination or the tridiagonal algorithm. Express the results as η/ηo where η is the fin efficiency and ηo is the efficiency that would result if ho were the uniform heat transfer coefficient over the entire fin. 8.46 A 2.5 cm black sphere (F = 1) is in radiation-convection equilibrium with air at 20◦ C. The surroundings are at 1000 K. What is the temperature of the sphere? 8.47 Develop expressions for h(D) and NuD during condensation on a vertical circular plate. 8.48 A cold copper plate is surrounded by a 5 mm high ridge which forms a shallow container. It is surrounded by saturated water vapor at 100◦ C. Estimate the steady heat flux and the rate of condensation. a. When the plate is perfectly horizontal and filled to overflowing with condensate. b. When the plate is in the vertical position. c. Did you have to make any idealizations? Would they result in under- or over-estimation of the condensation? 8.49 A proposed design for a nuclear power plant uses molten lead to remove heat from the reactor core. The heated lead is then used to boil water that drives a steam turbine. Water at 5 atm pressure (Tsat = 152◦ C) enters a heated section of a pipe at ˙ 60◦ C with a mass flow rate of m = 2 kg/s. The pipe is stainless steel (ks = 15 W/m·K) with a wall thickness of 12 mm and an outside diameter of 6.2 cm. The outside surface of the pipe is surrounded by an almost-stationary pool of molten lead at 477◦ C. a. At point where the liquid water has a bulk temperature of Tb = 80◦ C, estimate the inside and outside wall temperatures of the pipe, Twi and Two , to within about 5◦ C. Neglect entry length and variable properties effects and take β ≈ 0.000118 K−1 for lead. Hint: Guess an outside wall temperature above 370◦ C when computing h for the lead. Problems 451 b. At what distance from the inlet will the inside wall of the pipe reach Tsat ? What redesign may be needed? 8.50 A flat plate 10 cm long and 40 cm wide is inclined at 30◦ from the vertical. It is held at a uniform temperature of 250 K. Saturated HCFC-22 vapor at 260 K condenses onto the plate. Determine the following: a. The ratio hfg /hfg . b. The average heat transfer coefficient, h, and the rate at which the plate must be cooled, Q (watts). c. The film thickness, δ (µm), at the bottom of the plate, and the plate’s rate of condensation in g/s. 8.51 One component in a particular automotive air-conditioning system is a “receiver”, a small vertical cylindrical tank that contains a pool of liquid refrigerant, HFC-134a, with vapor above it. The receiver stores extra refrigerant for the system and helps to regulate the pressure. The receiver is at equilibrium with surroundings at 330 K. A 5 mm diameter, spherical thermistor inside the receiver monitors the liquid level. The thermistor is a temperature-sensing resistor driven by a small electric current; it dissipates a power of 0.1 W. When the system is fully charged with refrigerant, the thermistor sits below the liquid surface. When refrigerant leaks from the system, the liquid level drops and the thermistor eventually sits in vapor. The thermistor is small compared to the receiver, and its power is too low to affect the bulk temperature in the receiver. a. If the system is fully charged, determine the temperature of the thermistor. b. If enough refrigerant has leaked that the thermistor sits in vapor, find the thermistor’s temperature. Neglect thermal radiation. 8.52 Ammonia vapor at 300 K and 1.062 MPa pressure condenses onto the outside of a horizontal tube. The tube has an O.D. of 1.91 cm. a. Suppose that the outside of the tube has a uniform temperature of 290 K. Determine the average condensation 452 Chapter 8: Natural convection in single-phase fluids and during film condensation heat transfer cofficient of the tube. b. The tube is cooled by cold water flowing through it and the thin wall of the copper tube offers negligible thermal resistance. If the bulk temperature of the water is 275 K at a location where the outside surface of the tube is at 290 K, what is the heat transfer coefficient inside the tube? c. Using the heat transfer coefficients you just found, estimate the largest wall thickness for which the thermal resistance of the tube could be neglected. Discuss the variation the tube wall temperature around the circumference and along the length of the tube. 8.53 An inclined plate in a piece of process equipment is tilted 30◦ above the horizontal and is 20 cm long and 25 cm wide (in the horizontal direction). The plate is held at 280 K by a stream of liquid flowing past its bottom side; the liquid in turn is cooled by a refrigeration system capable of removing 12 watts from it. If the heat transfer from the plate to the stream exceeds 12 watts, the temperature of both the liquid and the plate will begin to rise. The upper surface of the plate is in contact with gaseous ammonia vapor at 300 K and a varying pressure. An engineer suggests that any rise in the bulk temperature of the liquid will signal that the pressure has exceeded a level of about pcrit = 551 kPa. a. Explain why the gas’s pressure will affect the heat transfer to the coolant. b. Suppose that the pressure is 255.3 kPa. What is the heat transfer (in watts) from gas to the plate, if the plate temperature is Tw = 280 K? Will the coolant temperature rise? Data for ammonia are given in App. A. c. Suppose that the pressure rises to 1062 kPa. What is the heat transfer to the plate if the plate is still at Tw = 280 K? Will the coolant temperature rise? References [8.1] W. Nusselt. Das grundgesetz des wärmeüberganges. Gesund. Ing., 38:872, 1915. References [8.2] C. J. Sanders and J. P. Holman. Franz Grashof and the Grashof Number. Int. J. Heat Mass Transfer, 15:562–563, 1972. [8.3] S. W. Churchill and H. H. S. Chu. Correlating equations for laminar and turbulent free convection from a vertical plate. Int. J. Heat Mass Transfer, 18:1323–1329, 1975. [8.4] S. Goldstein, editor. Modern Developments in Fluid Mechanics, volume 2, chapter 14. Oxford University Press, New York, 1938. [8.5] E. R. G. Eckert and R. M. Drake, Jr. Analysis of Heat and Mass Transfer. Hemisphere Publishing Corp., Washington, D.C., 1987. [8.6] A. Bejan and J. L. Lage. The Prandtl number effect on the transition in natural convection along a vertical surface. J. Heat Transfer, Trans. ASME, 112:787–790, 1990. [8.7] E. M. Sparrow and J. L. Gregg. The variable fluid-property problem in free convection. In J. P. Hartnett, editor, Recent Advances in Heat and Mass Transfer, pages 353–371. McGraw-Hill Book Company, New York, 1961. [8.8] H. Barrow and T. L. Sitharamarao. The effect of variable β on free convection. Brit. Chem. Eng., 16(8):704, 1971. [8.9] W. Kraus. Messungen des Temperatur- und Geschwindigskeitsfeldes bei freier Konvection. Verlag G. Braun, Karlsruhe, 1955. Chapter F. [8.10] S. W. Churchill and H. H. S. Chu. Correlating equations for laminar and turbulent free convection from a horizontal cylinder. Int. J. Heat Mass Transfer, 18:1049–1053, 1975. [8.11] T. Cebeci. Laminar-free-convective-heat transfer from the outer surface of a vertical slender circular cylinder. In Proc. Fifth Intl. Heat Transfer Conf., volume 3, pages 15–19. Tokyo, September 3–7 1974. [8.12] T. Yuge. Experiments on heat transfer from spheres including combined forced and natural convection. J. Heat Transfer, Trans. ASME, Ser. C, 82(1):214, 1960. [8.13] G. D. Raithby and K. G. T. Hollands. Natural convection. In W. M. Rohsenow, J. P. Hartnett, and Y. I. Cho, editors, Handbook of Heat Transfer, chapter 4. McGraw-Hill, New York, 3rd edition, 1998. 453 454 Chapter 8: Natural convection in single-phase fluids and during film condensation [8.14] J. H. Lienhard. On the commonality of equations for natural convection from immersed bodies. Int. J. Heat Mass Transfer, 16:2121, 1973. [8.15] B. R. Rich. An investigation of heat transfer from an inclined flat plate in free convection. Trans. ASME, 75:489–499, 1953. [8.16] T. Fujii and H. Imura. Natural convection heat transfer from a plate with arbitrary inclination. Int. J. Heat Mass Transfer, 15(4): 755–767, 1972. [8.17] G. C. Vliet. Natural convection local heat transfer on constant heat transfer inclined surface. J. Heat Transfer, Trans. ASME, Ser. C, 91:511–516, 1969. [8.18] L. Pera and B. Gebhart. On the stability of natural convection boundary layer flow over horizontal and slightly inclined surfaces. Int. J. Heat Mass Transfer, 16(6):1147–1163, 1973. [8.19] M. Al-Arabi and M. K. El-Riedy. Natural convection heat transfer from isothermal horizontal plates of different shapes. Int. J. Heat Mass Transfer, 19:1399–1404, 1976. [8.20] L. Pera and B. Gebhart. Natural convection boundary layer flow over horizontal and slightly inclined surfaces. Int. J. Heat Mass Transfer, 16(6):1131–1147, 1973. [8.21] B. Gebhart, Y. Jaluria, R. L. Mahajan, and B. Sammakia. BuoyancyInduced Flows and Transport. Hemisphere Publishing Corp., Washington, 1988. [8.22] J. R. Lloyd and W. R. Moran. Natural convection adjacent to horizontal surface of various planforms. J. Heat Transfer, Trans. ASME, Ser. C, 96(4):443–447, 1974. [8.23] A. M. Clausing and J. J. Berton. An experimental investigation of natural convection from an isothermal horizontal plate. J. Heat Transfer, Trans. ASME, 111(4):904–908, 1989. [8.24] F. Restrepo and L. R. Glicksman. The effect of edge conditions on natural convection heat transfer from a horizontal plates. Int. J. Heat Mass Transfer, 17(1):135–142, 1974. References [8.25] D. W. Hatfield and D. K. Edwards. Edge and aspect ratio effects on natural convection from the horizontal heated plate facing downwards. Int. J. Heat Mass Transfer, 24(6):1019–1024, 1981. [8.26] V. Kadambi and R. M. Drake, Jr. Free convection heat transfer from horizontal surfaces for prescribed variations in surface temperature and mass flow through the surface. Tech. Rept. Mech. Eng. HT-1, Princeton Univ., June 30 1959. [8.27] K. T. Yang. Natural convection in enclosures. In S. Kakaç, R. K. Shah, and W. Aung, editors, Handbook of Single-Phase Convective Heat Transfer, chapter 13. Wiley-Interscience, New York, 1987. [8.28] I. Catton. Natural convection in enclosures. In Proc. Sixth Intl. Heat Transfer Conf., volume 6, pages 13–31. Toronto, Aug. 7–11 1978. [8.29] S. W. Churchill. A comprehensive correlating equation for laminar, assisting, forced and free convection. AIChE J., 23(1):10–16, 1977. [8.30] T. S. Chen and B. F. Armaly. Mixed convection in external flow. In S. Kakaç, R. K. Shah, and W. Aung, editors, Handbook of SinglePhase Convective Heat Transfer, chapter 14. Wiley-Interscience, New York, 1987. [8.31] W. Aung. Mixed convection in internal flow. In S. Kakaç, R. K. Shah, and W. Aung, editors, Handbook of Single-Phase Convective Heat Transfer, chapter 15. Wiley-Interscience, New York, 1987. [8.32] V. K. Dhir. Quasi-steady laminar film condensation of steam on copper spheres. J. Heat Transfer, Trans. ASME, Ser. C, 97(3):347– 351, 1975. [8.33] V. K. Dhir and J. H. Lienhard. Laminar film condensation on plane and axi-symmetric bodies in non-uniform gravity. J. Heat Transfer, Trans. ASME, Ser. C, 93(1):97–100, 1971. [8.34] W. Nusselt. Die oberflächenkondensation des wasserdampfes. Z. Ver. Dtsch. Ing., 60:541 and 569, 1916. [8.35] W. M. Rohsenow. Heat transfer and temperature distribution in laminar-film condensation. Trans. ASME, 78:1645–1648, 1956. [8.36] E. M. Sparrow and J. L. Gregg. A boundary-layer treatment of laminar-film condensation. J. Heat Transfer, Trans. ASME, Ser. C, 81:13–18, 1959. 455 456 Chapter 8: Natural convection in single-phase fluids and during film condensation [8.37] P. Sadasivan and J. H. Lienhard. Sensible heat correction in laminar film boiling and condensation. J. Heat Transfer, Trans. ASME, 109: 545–547, 1987. [8.38] E. M. Sparrow and J. L. Gregg. A theory of rotating condensation. J. Heat Transfer, Trans. ASME, Ser. C, 81:113–120, 1959. [8.39] A. Karimi. Laminar film condensation on helical reflux condensers and related configurations. Int. J. Heat Mass Transfer, 20:1137– 1144, 1977. [8.40] R. Gregorig, J. Kern, and K. Turek. Improved correlation of film condensation data based on a more rigorous application of similarity parameters. Wärme- und Stoffübertragung, 7:1–13, 1974. [8.41] P. J. Marto. Condensation. In W. M. Rohsenow, J. P. Hartnett, and Y. I. Cho, editors, Handbook of Heat Transfer, chapter 14. McGrawHill, New York, 3rd edition, 1998. [8.42] J. Rose, H. Uehara, S. Koyama, and T. Fujii. Film condensation. In S. G. Kandlikar, M. Shoji, and V. K. Dhir, editors, Handbook of Phase Change: Boiling and Condensation, chapter 19. Taylor & Francis, Philadelphia, 1999. [8.43] E. M. Sparrow and S. H. Lin. Condensation in the presence of a non-condensible gas. J. Heat Transfer, Trans. ASME, Ser. C, 86: 430, 1963. 9. Heat transfer in boiling and other phase-change configurations For a charm of powerful trouble, like a Hell-broth boil and bubble.. . . . . .Cool it with a baboon’s blood, then the charm is firm and good. Macbeth, Wm. Shakespeare “A watched pot never boils”—the water in a teakettle takes a long time to get hot enough to boil because natural convection initially warms it rather slowly. Once boiling begins, the water is heated the rest of the way to the saturation point very quickly. Boiling is of interest to us because it is remarkably effective in carrying heat from a heater into a liquid. The heater in question might be a red-hot horseshoe quenched in a bucket or the core of a nuclear reactor with coolant flowing through it. Our aim is to learn enough about the boiling process to design systems that use boiling for cooling. We begin by considering pool boiling—the boiling that occurs when a stationary heater transfers heat to an otherwise stationary liquid. 9.1 Nukiyama’s experiment and the pool boiling curve Hysteresis in the q vs. ∆T relation for pool boiling In 1934, Nukiyama [9.1] did the experiment described in Fig. 9.1. He boiled saturated water on a horizontal wire that functioned both as an electric resistance heater and as a resistance thermometer. By calibrating the resistance of a Nichrome wire as a function of temperature before the 457 458 Heat transfer in boiling and other phase-change configurations §9.1 Figure 9.1 Nukiyama’s boiling hysteresis loop. experiment, he was able to obtain both the heat flux and the temperature using the observed current and voltage. He found that, as he increased the power input to the wire, the heat flux rose sharply but the temperature of the wire increased relatively little. Suddenly, at a particular high value of the heat flux, the wire abruptly melted. Nukiyama then obtained a platinum wire and tried again. This time the wire reached the same limiting heat flux, but then it turned almost white-hot without melting. §9.1 Nukiyama’s experiment and the pool boiling curve As he reduced the power input to the white-hot wire, the temperature dropped in a continuous way, as shown in Fig. 9.1, until the heat flux was far below the value where the first temperature jump occurred. Then the temperature dropped abruptly to the original q vs. ∆T = (Twire − Tsat ) curve, as shown. Nukiyama suspected that the hysteresis would not occur if ∆T could be specified as the independent controlled variable. He conjectured that such an experiment would result in the connecting line shown between the points where the temperatures jumped. In 1937, Drew and Mueller [9.2] succeeded in making ∆T the independent variable by boiling organic liquids outside a tube. Steam was allowed to condense inside the tube at an elevated pressure. The steam saturation temperature—and hence the tube-wall temperature—was varied by controlling the steam pressure. This permitted them to obtain a few scattered data that seemed to bear out Nukiyama’s conjecture. Measurements of this kind are inherently hard to make accurately. For the next forty years, the relatively few nucleate boiling data that people obtained were usually—and sometimes imaginatively—interpreted as verifying Nukiyama’s suggestion that this part of the boiling curve is continuous. Figure 9.2 is a completed boiling curve for saturated water at atmospheric pressure on a particular flat horizontal heater. It displays the behavior shown in Fig. 9.1, but it has been rotated to place the independent variable, ∆T , on the abscissa. (We represent Nukiyama’s connecting region as two unconnected extensions of the neighboring regions for reasons that we explain subsequently.) Modes of pool boiling The boiling curve in Fig. 9.2 has been divided into five regimes of behavior. These regimes, and the transitions that divide them, are discussed next. Natural convection. Water that is not in contact with its own vapor does not boil at the so-called normal boiling point,1 Tsat . Instead, it continues to rise in temperature until bubbles finally to begin to form. On conventional machined metal surfaces, this occurs when the surface is a few degrees above Tsat . Below the bubble inception point, heat is removed by natural convection, and it can be predicted by the methods laid out in Chapter 8. 1 This notion might be new to some readers. It is explained in Section 9.2. 459 460 Heat transfer in boiling and other phase-change configurations §9.1 Figure 9.2 Typical boiling curve and regimes of boiling for an unspecified heater surface. Nucleate boiling. The nucleate boiling regime embraces the two distinct regimes that lie between bubble inception and Nukiyama’s first transition point: 1. The region of isolated bubbles. In this range, bubbles rise from isolated nucleation sites, more or less as they are sketched in Fig. 9.1. As q and ∆T increase, more and more sites are activated. Figure 9.3a is a photograph of this regime as it appears on a horizontal plate. 2. The region of slugs and columns. When the active sites become very numerous, the bubbles start to merge into one another, and an entirely different kind of vapor escape path comes into play. Vapor formed at the surface merges immediately into jets that feed into large overhead bubbles or “slugs” of vapor. This process is shown as it occurs on a horizontal cylinder in Fig. 9.3b. 461 d. Film boiling of acetone on a 22 gage wire at earth-normal gravity. The true width of this image is 3.48 cm. b. Two views of transitional boiling in acetone on a 0.32 cm diam. tube. Figure 9.3 Typical photographs of boiling in the four regimes identified in Fig. 9.2. c. Two views of the regime of slugs and columns. 3.75 cm length of 0.164 cm diam. wire in benzene at earth-normal gravity. q=0.35×106 W/m2 3.45 cm length of 0.0322 cm diam. wire in methanol at 10 earth-normal gravities. q=1.04×106 W/m2 a. Isolated bubble regime—water. 462 Heat transfer in boiling and other phase-change configurations §9.1 Peak heat flux. Clearly, it is very desirable to be able to operate heat exchange equipment at the upper end of the region of slugs and columns. Here the temperature difference is low while the heat flux is very high. Heat transfer coefficients in this range are enormous. However, it is very dangerous to run equipment near qmax in systems for which q is the independent variable (as in nuclear reactors). If q is raised beyond the upper limit of the nucleate boiling regime, such a system will suffer a sudden and damaging increase of temperature. This transition2 is known by a variety of names: the burnout point (although a complete burning up or melting away does not always accompany it); the peak heat flux (a modest descriptive term); the boiling crisis (a Russian term); the DNB, or departure from nucleate boiling, and the CHF, or critical heat flux (terms more often used in flow boiling); and the first boiling transition (which term ignores previous transitions). We designate the peak heat flux as qmax . Transitional boiling regime. It is a curious fact that the heat flux actually diminishes with ∆T after qmax is reached. In this regime the effectiveness of the vapor escape process becomes worse and worse. Furthermore, the hot surface becomes completely blanketed in vapor and q reaches a minimum heat flux which we call qmin . Figure 9.3c shows two typical instances of transitional boiling just beyond the peak heat flux. Film boiling. Once a stable vapor blanket is established, q again increases with increasing ∆T . The mechanics of the heat removal process during film boiling, and the regular removal of bubbles, has a great deal in common with film condensation, but the heat transfer coefficients are much lower because heat must be conducted through a vapor film instead of through a liquid film. We see an instance of film boiling in Fig. 9.3d. Experiment 9.1 Set an open pan of cold tap water on your stove to boil. Observe the following stages as you watch: • At first nothing appears to happen; then you notice that numerous small, stationary bubbles have formed over the bottom of the pan. 2 We defer a proper physical explanation of the transition to Section 9.3. Nukiyama’s experiment and the pool boiling curve §9.1 These bubbles have nothing to do with boiling—they contain air that was driven out of solution as the temperature rose. • Suddenly the pan will begin to “sing.” There will be a somewhat high-pitched buzzing-humming sound as the first vapor bubbles are triggered. They grow at the heated surface and condense very suddenly when their tops encounter the still-cold water above them. This cavitation collapse is accompanied by a small “ping” or “click,” over and over, as the process is repeated at a fairly high frequency. • As the temperature of the liquid bulk rises, the singing is increasingly muted. You may then look in the pan and see a number of points on the bottom where a feathery blur appears to be affixed. These blurred images are bubble columns emanating scores of bubbles per second. The bubbles in these columns condense completely at some distance above the surface. Notice that the air bubbles are all gradually being swept away. • The “singing” finally gives way to a full rolling boil, accompanied by a gentle burbling sound. Bubbles no longer condense but now reach the surface, where they break. • A full rolling-boil process, in which the liquid bulk is saturated, is a kind of isolated-bubble process, as plotted in Fig. 9.2. No kitchen stove supplies energy fast enough to boil water in the slugs-andcolumns regime. You might, therefore, reflect on the relative intensity of the slugs-and-columns process. Experiment 9.2 Repeat Experiment 9.1 with a glass beaker instead of a kitchen pan. Place a strobe light, blinking about 6 to 10 times per second, behind the beaker with a piece of frosted glass or tissue paper between it and the beaker. You can now see the evolution of bubble columns from the first singing mode up to the rolling boil. You will also be able to see natural convection in the refraction of the light before boiling begins. 463 464 Heat transfer in boiling and other phase-change configurations §9.2 Figure 9.4 Enlarged sketch of a typical metal surface. 9.2 Nucleate boiling Inception of boiling Figure 9.4 shows a highly enlarged sketch of a heater surface. Most metalfinishing operations score tiny grooves on the surface, but they also typically involve some chattering or bouncing action, which hammers small holes into the surface. When a surface is wetted, liquid is prevented by surface tension from entering these holes, so small gas or vapor pockets are formed. These little pockets are the sites at which bubble nucleation occurs. To see why vapor pockets serve as nucleation sites, consider Fig. 9.5. Here we see the problem in highly idealized form. Suppose that a spherical bubble of pure saturated steam is at equilibrium with an infinite superheated liquid. To determine the size of such a bubble, we impose the conditions of mechanical and thermal equilibrium. The bubble will be in mechanical equilibrium when the pressure difference between the inside and the outside of the bubble is balanced by the forces of surface tension, σ , as indicated in the cutaway sketch in Fig. 9.5. Since thermal equilibrium requires that the temperature must be the same inside and outside the bubble, and since the vapor inside must be saturated at Tsup because it is in contact with its liquid, the force balance takes the form Rb = 2σ psat at Tsup − pambient (9.1) The p –v diagram in Fig. 9.5 shows the state points of the internal vapor and external liquid for a bubble at equilibrium. Notice that the external liquid is superheated to (Tsup − Tsat ) K above its boiling point at the ambient pressure; but the vapor inside, being held at just the right elevated pressure by surface tension, is just saturated. Nucleate boiling §9.2 465 Figure 9.5 The conditions required for simultaneous mechanical and thermal equilibrium of a vapor bubble. Physical Digression 9.1 The surface tension of water in contact with its vapor is given with great accuracy by [9.3]: σwater = 235.8 1 − Tsat Tc 1.256 1 − 0.625 1 − Tsat Tc mN m (9.2a) where both Tsat and the thermodynamical critical temperature, Tc = 647.096 K, are expressed in K. The units of σ are millinewtons (mN) per meter. Table 9.1 gives additional values of σ for several substances. Equation 9.2a is a specialized refinement of a simple, but quite accurate and widely-used, semi-empirical equation for correlating surface Table 9.1 Surface tension of various substances from the collection of Jasper [9.4]a and other sources. Substance Acetone Ammonia Aniline Benzene Butyl alcohol Carbon tetrachloride Cyclohexanol Ethyl alcohol Ethylene glycol Hydrogen Isopropyl alcohol Mercury Methane Methyl alcohol Naphthalene Nicotine Nitrogen Octane Oxygen Pentane Toluene Water Temperature Range (◦ C) 25 to 50 −70 −60 −50 −40 15 to 90 10 30 50 70 10 to 100 15 to 105 20 to 100 10 to 100 20 to 140 −258 −255 −253 10 to 100 5 to 200 90 100 115 10 to 60 100 to 200 −40 to 90 −195 to −183 10 to 120 −202 to −184 10 to 30 10 to 100 10 to 100 σ (mN/m) σ = a − bT (◦ C) a (mN/m) b (mN/m·◦ C) 26.26 0.112 44.83 0.1085 27.18 29.49 35.33 24.05 50.21 0.08983 0.1224 0.0966 0.0832 0.089 42.39 40.25 37.91 35.38 30.21 27.56 24.96 22.40 2.80 2.29 1.95 22.90 490.6 0.0789 0.2049 24.00 42.84 41.07 26.42 23.52 −33.72 18.25 30.90 75.83 0.0773 0.1107 0.1112 0.2265 0.09509 −0.2561 0.11021 0.1189 0.1477 18.877 16.328 12.371 n Substance Carbon dioxide CFC-12 (R12) [9.5] HCFC-22 (R22) [9.5] Temperature Range (◦ C) −56 to 31 σ = σo [1 − T (K)/Tc ] σo (mN/m) 75.00 Tc (K) n 304.26 1.25 −148 to 112 56.52 385.01 1.27 −158 to 96 61.23 369.32 1.23 HFC-134a (R134a) [9.6] −30 to 101 59.69 374.18 1.266 Propane [9.7] −173 to 96 53.13 369.85 1.242 a The function σ = σ (T ) is not really linear, but Jasper was able to linearize it over modest ranges of temperature [e.g., compare the water equation above with eqn. (9.2a)]. 466 Nucleate boiling §9.2 467 tension: σ = σo 1 − Tsat Tc 11/9 (9.2b) We include correlating equations of this form for CO2 , propane, and some refrigerants at the bottom of Table 9.1. Equations of this general form are discussed in Reference [9.8]. It is easy to see that the equilibrium bubble, whose radius is described by eqn. (9.1), is unstable. If its radius is less than this value, surface tension will overbalance [psat (Tsup ) − pambient ]. Thus, vapor inside will condense at this higher pressure and the bubble will collapse. If the bubble radius is slightly larger than the equation specifies, liquid at the interface will evaporate and the bubble will begin to grow. Thus, as the heater surface temperature is increased, higher and higher values of [psat (Tsup ) − pambient ] will result and the equilibrium radius, Rb , will decrease in accordance with eqn. (9.1). It follows that smaller and smaller vapor pockets will be triggered into active bubble growth as the temperature is increased. As an approximation, we can use eqn. (9.1) to specify the radius of those vapor pockets that become active nucleation sites. More accurate estimates can be made using Hsu’s [9.9] bubble inception theory, the subsequent work by Rohsenow and others (see, e.g., [9.10]), or the still more recent technical literature. Example 9.1 Estimate the approximate size of active nucleation sites in water at 1 atm on a wall superheated by 8 K and by 16 K. This is roughly in the regime of isolated bubbles indicated in Fig. 9.2. Solution. psat = 1.203 × 105 N/m2 at 108◦ C and 1.769 × 105 N/m2 at 116◦ C, and σ is given as 57.36 mN/m at Tsat = 108◦ C and as 55.78 mN/m at Tsat = 116◦ C by eqn. (9.2a). Then, at 108◦ C, Rb from eqn. (9.1) is Rb = 2(57.36 × 10−3 ) N/m 1.203 × 105 − 1.013 × 105 N/m2 and similarly for 116◦ C, so the radius of active nucleation sites is on the order of Rb = 0.0060 mm at T = 108◦ C or 0.0015 mm at 116◦ C 468 Heat transfer in boiling and other phase-change configurations §9.2 This means that active nucleation sites would be holes with diameters very roughly on the order of magnitude of 0.005 mm or 5µm—at least on the heater represented by Fig. 9.2. That is within the range of roughness of commercially finished surfaces. Region of isolated bubbles The mechanism of heat transfer enhancement in the isolated bubble regime was hotly argued in the years following World War II. A few conclusions have emerged from that debate, and we shall attempt to identify them. There is little doubt that bubbles act in some way as small pumps that keep replacing liquid heated at the wall with cool liquid. The question is that of specifying the correct mechanism. Figure 9.6 shows the way bubbles probably act to remove hot liquid from the wall and introduce cold liquid to be heated. It is apparent that the number of active nucleation sites generating bubbles will strongly influence q. On the basis of his experiments, Yamagata showed in 1955 (see, e.g., [9.11]) that q ∝ ∆T a nb (9.3) where ∆T ≡ Tw − Tsat and n is the site density or number of active sites per square meter. A great deal of subsequent work has been done to fix the constant of proportionality and the constant exponents, a and b. 1 The exponents turn out to be approximately a = 1.2 and b = 3 . The problem with eqn. (9.3) is that it introduces what engineers call a nuisance variable. A nuisance variable is one that varies from system to system and cannot easily be evaluated—the site density, n, in this case. Normally, n increases with ∆T in some way, but how? If all sites were identical in size, all sites would be activated simultaneously, and q would be a discontinuous function of ∆T . When the sites have a typical distribution of sizes, n (and hence q) can increase very strongly with ∆T . It is a lucky fact that for a large class of factory-finished materials, n varies approximately as ∆T 5 or 6 , so q varies roughly as ∆T 3 . This has made it possible for various authors to correlate q approximately for a large variety of materials. One of the first and most useful correlations for nucleate boiling was that of Rohsenow [9.12] in 1952. It is cp (Tw − Tsat ) = Csf hfg Prs q µhfg σ g ρf − ρg 0.33 (9.4) §9.2 Nucleate boiling A bubble growing and departing in saturated liquid. The bubble grows, absorbing heat from the superheated liquid on its periphery. As it leaves, it entrains cold liquid onto the plate which then warms up until nucleation occurs and the cycle repeats. 469 A bubble growing in subcooled liquid. When the bubble protrudes into cold liquid, steam can condense on the top while evaporation continues on the bottom. This provides a short-circuit for cooling the wall. Then, when the bubble caves in, cold liquid is brought to the wall. Figure 9.6 Heat removal by bubble action during boiling. Dark regions denote locally superheated liquid. where all properties, unless otherwise noted, are for liquid at Tsat . The constant Csf is an empirical correction for typical surface conditions. Table 9.2 includes a set of values of Csf for common surfaces (taken from [9.12]) as well as the Prandtl number exponent, s . A more extensive compilation of these constants was published by Pioro in 1999 [9.13]. We noted, initially, that there are two nucleate boiling regimes, and the Yamagata equation (9.3) applies only to the first of them. Rohsenow’s equation is frankly empirical and does not depend on the rational analysis of either nucleate boiling process. It turns out that it represents q(∆T ) in both regimes, but it is not terribly accurate in either one. Figure 9.7 shows Rohsenow’s original comparison of eqn. (9.4) with data for water over a large range of conditions. It shows typical errors in heat flux of 100% and typical errors in ∆T of about 25%. Thus, our ability to predict the nucleate pool boiling heat flux is poor. Our ability to predict ∆T is better because, with q ∝ ∆T 3 , a large error in q gives a much smaller error in ∆T . It appears that any substantial improvement in this situation will have to wait until someone has managed to deal realistically with the nuisance variable, n. Current research efforts are dealing with this matter, and we can simply hope that such work will eventually produce a method for achieving reliable heat transfer design relationships for nucleate boiling. 470 Heat transfer in boiling and other phase-change configurations §9.2 Table 9.2 Selected values of the surface correction factor for use with eqn. (9.4) [9.12] Surface–Fluid Combination Water–nickel Water–platinum Water–copper Water–brass CCl4 –copper Benzene–chromium n-Pentane–chromium Ethyl alcohol–chromium Isopropyl alcohol–copper 35% K2 CO3 –copper 50% K2 CO3 –copper n-Butyl alcohol–copper Csf s 0.006 0.013 0.013 0.006 0.013 0.010 0.015 0.0027 0.0025 0.0054 0.0027 0.0030 1.0 1.0 1.0 1.0 1.7 1.7 1.7 1.7 1.7 1.7 1.7 1.7 It is indeed fortunate that we do not often have to calculate q, given ∆T , in the nucleate boiling regime. More often, the major problem is to avoid exceeding qmax . We turn our attention in the next section to predicting this limit. Example 9.2 What is Csf for the heater surface in Fig. 9.2? Solution. From eqn. (9.4) we obtain 3 µcp q 3 Csf = 2 ∆T 3 hfg Pr3 g ρf − ρg σ where, since the liquid is water, we take s to be 1.0. Then, for water at Tsat = 100◦ C: cp = 4.22 kJ/kg·K, Pr = 1.75, (ρf − ρg ) = 958 kg/m3 , σ = 0.0589 N/m or kg/s2 , hfg = 2257 kJ/kg, µ = 0.000282 kg/m·s. Nucleate boiling §9.2 471 Figure 9.7 Illustration of Rohsenow’s [9.12] correlation applied to data for water boiling on 0.61 mm diameter platinum wire. Thus, kW q C 3 = 3.10 × 10−7 2 3 ∆T 3 sf mK At q = 800 kW/m2 , we read ∆T = 22 K from Fig. 9.2. This gives Csf = 3.10 × 10−7 (22)3 800 1/3 = 0.016 This value compares favorably with Csf for a platinum or copper surface under water. 472 Heat transfer in boiling and other phase-change configurations 9.3 §9.3 Peak pool boiling heat flux Transitional boiling regime and Taylor instability It will help us to understand the peak heat flux if we first consider the process that connects the peak and the minimum heat fluxes. During high heat flux transitional boiling, a large amount of vapor is glutted about the heater. It wants to buoy upward, but it has no clearly defined escape route. The jets that carry vapor away from the heater in the region of slugs and columns are unstable and cannot serve that function in this regime. Therefore, vapor buoys up in big slugs—then liquid falls in, touches the surface briefly, and a new slug begins to form. Figure 9.3c shows part of this process. The high and low heat flux transitional boiling regimes are different in character. The low heat flux region does not look like Fig. 9.2c but is almost indistinguishable from the film boiling shown in Fig. 9.2d. However, both processes display a common conceptual key: In both, the heater is almost completely blanketed with vapor. In both, we must contend with the unstable configuration of a liquid on top of a vapor. Figure 9.8 shows two commonplace examples of such behavior. In either an inverted honey jar or the water condensing from a cold water pipe, we have seen how a heavy fluid falls into a light one (water or honey, in this case, collapses into air). The heavy phase falls down at one node of a wave and the light fluid rises into the other node. The collapse process is called Taylor instability after G. I. Taylor, who first predicted it. The so-called Taylor wavelength, λd , is the length of the wave that grows fastest and therefore predominates during the collapse of an infinite plane horizontal interface. It can be predicted using dimensional analysis. The dimensional functional equation for λd is λd = fn σ , g ρf − ρg (9.5) since the wave is formed as a result of the balancing forces of surface tension against inertia and gravity. There are three variables involving m and kg/s2 , so we look for just one dimensionless group: λd g ρf − ρg σ = constant This relationship was derived analytically by Bellman and Pennington [9.14] for one-dimensional waves and by Sernas [9.15] for the two-dimensional Peak pool boiling heat flux §9.3 473 a. Taylor instability in the surface of the honey in an inverted honey jar b. Taylor instability in the interface of the water condensing on the underside of a small cold water pipe. Figure 9.8 Two examples of Taylor instabilities that one might commonly experience. waves that actually occur in a plane horizontal interface. The results were λd g ρf − ρg σ = √ 2π √3 for one-dimensional waves 2π 6 for two-dimensional waves (9.6) 474 Heat transfer in boiling and other phase-change configurations §9.3 Experiment 9.3 Hang a metal rod in the horizontal position by threads at both ends. The rod should be about 30 cm in length and perhaps 1 to 2 cm in diameter. Pour motor oil or glycerin in a narrow cake pan and lift the pan up under the rod until it is submerged. Then lower the pan and watch the liquid drain into it. Take note of the wave action on the underside of the rod. The same thing can be done in an even more satisfactory way by running cold water through a horizontal copper tube above a beaker of boiling water. The condensing liquid will also come off in a Taylor wave such as is shown in Fig. 9.8. In either case, the waves will approximate λd1 (the length of a one-dimensional wave, since they are arrayed on a line), but the wavelength will be influenced by the curvature of the rod. Throughout the transitional boiling regime, vapor rises into liquid on the nodes of Taylor waves, and at qmax this rising vapor forms into jets. These jets arrange themselves on a staggered square grid, as shown in Fig. 9.9. The basic spacing of the grid is λd2 (the two-dimensional Taylor wavelength). Since √ λd2 = 2 λd1 (9.7) [recall eqn. (9.6)], the spacing of the most basic module of jets is actually λd1 , as shown in Fig. 9.9. Next we must consider how the jets become unstable at the peak, to bring about burnout. Helmholtz instability of vapor jets Figure 9.10 shows a commonplace example of what is called Helmholtz instability. This is the phenomenon that causes the vapor jets to cave in when the vapor velocity in them reaches a critical value. Any flag in a breeze will constantly be in a state of collapse as the result of relatively high pressures where the velocity is low and relatively low pressures where the velocity is high, as is indicated in the top view. This same instability is shown as it occurs in a vapor jet wall in Fig. 9.11. This situation differs from the flag in one important particular. There is surface tension in the jet walls, which tends to balance the flow-induced pressure forces that bring about collapse. Thus, while the flag is unstable in any breeze, the vapor velocity in the jet must reach a limiting value, ug , before the jet becomes unstable. a. Plan view of bubbles rising from surface b. Waveform underneath the bubbles shown in a. Figure 9.9 The array of vapor jets as seen on an infinite horizontal heater surface. 475 476 Heat transfer in boiling and other phase-change configurations §9.3 Figure 9.10 The flapping of a flag due to Helmholtz instability. Lamb [9.16] gives the following relation between the vapor flow ug , shown in Fig. 9.11, and the wavelength of a disturbance in the jet wall, λH : ug = 2π σ ρg λH (9.8) [This result, like eqn. (9.6), can be predicted within a constant using dimensional analysis. See Problem 9.19.] A real liquid–vapor interface will usually be irregular, and therefore it can be viewed as containing all possible sinusoidal wavelengths superposed on one another. One problem we face is that of guessing whether or not one of those wavelengths Peak pool boiling heat flux §9.3 Figure 9.11 Helmholtz instability of vapor jets. will be better developed than the others and therefore more liable to collapse. Example 9.3 Saturated water at 1 atm flows down the periphery of the inside of a 10 cm I.D. vertical tube. Steam flows upward in the center. The wall of the pipe has circumferential corrugations in it, with a 4 cm wavelength in the axial direction. Neglect problems raised by curvature and the finite thickness of the liquid, and estimate the steam velocity required to destabilize the liquid flow over these corrugations, assuming that the liquid moves slowly. Solution. The flow will be Helmholtz-stable until the steam velocity reaches the value given by eqn. (9.8): ug = 2π (0.0589) 0.577(0.04 m) Thus, the maximum stable steam velocity would be ug = 4 m/s. Beyond that, the liquid will form whitecaps and be blown back upward. 477 478 Heat transfer in boiling and other phase-change configurations §9.3 Example 9.4 Capillary forces hold mercury in place between two parallel steel plates with a lid across the top. The plates are slowly pulled apart until the mercury interface collapses. Approximately what is the maximum spacing? Solution. The mercury is most susceptible to Taylor instability when the spacing reaches the wavelength given by eqn. (9.6): √ λd1 = 2π 3 √ σ = 2π 3 g(ρf − ρg ) 0.487 = 0.021 m = 2.1 cm 9.8(13600) (Actually, this spacing would give the maximum rate of collapse. It √ can be shown that collapse would begin at 1 3 times this value, or at 1.2 cm.) Prediction of qmax General expression for qmax The heat flux must be balanced by the latent heat carried away in the jets when the liquid is saturated. Thus, we can write immediately qmax = ρg hfg ug Aj Ah (9.9) where Aj is the cross-sectional area of a jet and Ah is the heater area that supplies each jet. For any heater configuration, two things must be determined. One is the length of the particular disturbance in the jet wall, λH , which will trigger Helmholtz instability and fix ug in eqn. (9.8) for use in eqn. (9.9). The other is the ratio Aj Ah . The prediction of qmax in any pool boiling configuration always comes down to these two problems. qmax on an infinite horizontal plate. The original analysis of this type was done by Zuber in his doctoral dissertation at UCLA in 1958 (see [9.17]). He first guessed that the jet radius was λd1 4. This guess has received corroboration by subsequent investigators, and (with reference to Fig. 9.9) Peak pool boiling heat flux §9.3 479 it gives Aj cross-sectional area of circular jet Ah area of the square portion of the heater that feeds the jet π (λd1 /4)2 π (9.10) = = (λd1 )2 16 = Lienhard and Dhir ([9.18, 9.19, 9.20]) guessed that the Helmholtz-unstable wavelength might be equal to λd1 , so eqn. (9.9) became g (ρf − ρg ) 2π σ 1 √ ρg 2π 3 qmax = ρg hfg σ × π 16 or3 1/2 qmax = 0.149 ρg hfg 4 g (ρf − ρg )σ (9.11) Equation (9.11) is compared with available data for large flat heaters, with vertical sidewalls to prevent any liquid sideflow, in Fig. 9.12. So long as the diameter or width of the heater is more than about 3λd1 , the prediction is quite accurate. When the width or diameter is less than this, there is a small integral number of jets on a plate which may be larger or smaller in area than 16/π per jet. When this is the case, the actual qmax may be larger or smaller than that predicted by eqn. (9.11) (see Problem 9.13). The form of the preceding prediction is usually credited to Kutateladze [9.21] and Zuber [9.17]. Kutateladze (then working in Leningrad and later director of the Heat Transfer Laboratory near Novosibirsk, Siberia) recognized that burnout resembled the flooding of a distillation column. At any level in a distillation column, alcohol-rich vapor (for example) rises while water-rich liquid flows downward in counterflow. If the process is driven too far, the flows become Helmholtz-unstable and the process collapses. The liquid then cannot move downward and the column is said to “flood.” Kutateladze did the dimensional analysis of qmax based on the flooding mechanism and obtained the following relationship, which, lacking a characteristic length and being of the same form as eqn. (9.11), is really valid only for an infinite horizontal plate: 1/2 qmax = C ρg hfg 3 Readers are reminded that √ n x ≡ x 1/n . 4 g ρf − ρg σ 480 Heat transfer in boiling and other phase-change configurations §9.3 Figure 9.12 Comparison of the qmax prediction for infinite horizontal heaters with data reported in [9.18]. He then suggested that C was equal to 0.131 on the basis of data from configurations other than infinite flat plates (horizontal cylinders, for example). Zuber’s analysis yielded C = π /24 = 0.1309, which was quite close to Kutateladze’s value but lower by 14% than eqn. (9.11). We therefore designate the Zuber-Kutateladze prediction as qmaxz . However, we shall not use it directly, since it does not predict any actual physical configuration. 1/2 qmaxz ≡ 0.131 ρg hfg 4 g ρf − ρg σ (9.12) It is very interesting that C. F. Bonilla, whose qmax experiments in the early 1940s are included in Fig. 9.12, also suggested that qmax should be compared with the column-flooding mechanism. He presented these ideas in a paper, but A. P. Colburn wrote to him: “A correlation [of the flooding velocity plots with] boiling data would not serve any great purpose and would perhaps be very misleading.” And T. H. Chilton—another eminent chemical engineer of that period—wrote to him: “I venture to suggest that you delete from the manuscript…the relationship between boiling rates and loading velocities in packed towers.” Thus, the technical conservativism of the period prevented the idea from gaining acceptance for another decade. Peak pool boiling heat flux §9.3 Example 9.5 Predict the peak heat flux for Fig. 9.2. Solution. We use eqn. (9.11) to evaluate qmax for water at 100◦ C on an infinite flat plate: 1/2 qmax = 0.149 ρg hfg 4 g (ρf − ρg )σ = 0.149(0.597)1/2 (2, 257, 000) 9.8(958.2 − 0.6)(0.0589) 4 = 1.260 × 106 W/m2 = 1.260 MW/m2 Figure 9.2 shows qmax 8%. 1.160 MW/m2 , which is less by only about Example 9.6 What is qmax in mercury on a large flat plate at 1 atm? Solution. The normal boiling point of mercury is 355◦ C. At this temperature, hfg = 292, 500 J/kg, ρf = 13, 400 kg/m3 , ρg = 4.0 kg/m3 , and σ 0.418 kg/s2 , so qmax = 0.149(4.0)1/2 (292, 500) 9.8(13, 400 − 4)(0.418) 4 = 1.334 MW/m2 The result is very close to that for water. The increases in density and surface tension have been compensated by a much lower latent heat. Peak heat flux in other pool boiling configurations The prediction of qmax in configurations other than an infinite flat heater will involve a characteristic length, L. Thus, the dimensional functional equation for qmax becomes qmax = fn ρg , hfg , σ , g ρf − ρg , L which involves six variables and four dimensions: J, m, s, and kg, where, once more in accordance with Section 4.3, we note that no significant conversion from work to heat is occurring so that J must be retained as a separate unit. There are thus two pi-groups. The first group can 481 482 Heat transfer in boiling and other phase-change configurations §9.3 arbitrarily be multiplied by 24/π to give qmax qmax Π1 = = 1/2 qmaxz (π /24) ρg hfg 4 σ g(ρf − ρg ) (9.13) Notice that the factor of 24/π has served to make the denominator equal to qmaxz (Zuber’s expression for qmax ). Thus, for qmax on a flat plate, Π1 equals 0.149/0.131, or 1.14. The second pi-group is √L L Π2 = = 2π 3 ≡L (9.14) λd1 σ g (ρf − ρg ) The latter group, Π2 , is the square root of the Bond number, Bo, which is used to compare buoyant force with capillary forces. Predictions and correlations of qmax have been made for several finite geometries in the form qmax = fn L (9.15) qmaxz The dimensionless characteristic length in eqn. (9.15) might be a dimensionless radius (R ), a dimensionless diameter (D ), or a dimensionless height (H ). The graphs in Fig. 9.13 are comparisons of several of the existing predictions and correlations with experimental data. These predictions and others are listed in Table 9.3. Notice that the last three items in Table 9.3 (10, 11, and 12) are general expressions from which several of the preceding expressions in the table can be obtained. The equations in Table 9.3 are all valid within ±15% or 20%, which is very little more than the inherent scatter of qmax data. However, they are subject to the following conditions: • The bulk liquid is saturated. • There are no pathological surface imperfections. • There is no forced convection. Another limitation on all the equations in Table 9.3 is that neither the size of the heater nor the relative force of gravity can be too small. When L < 0.15 in most configurations, the Bond number is Bo ≡ L 2 = g(ρf − ρg )L3 σL = buoyant force capillary force < 1 44 In this case, the process becomes completely dominated by surface tension and the Taylor-Helmholtz wave mechanisms no longer operate. As L is reduced, the peak and minimum heat fluxes cease to occur and the Figure 9.13 The peak pool boiling heat flux on several heaters. 483 484 Small flat heater Horizontal cylinder Large horizontal cylinder Small horizontal cylinder Large sphere Small sphere 2. 3. 4. 5. 6. 7. √ −3.44 R Characteristic length, L Transverse perimeter, P Characteristic length, L 1.4/(P )1/4 Constant (L )1/2 11. Small slender cylinder of any cross section 12. Small bluff body Height of side, H Height of side, H Sphere radius, R Sphere radius, R Cylinder radius, R Cylinder radius, R Cylinder radius, R Heater width or diameter Heater width or diameter Basis for L ∼ 0.90 1.4/(H ) 1/4 1.18/(H )1/4 1.734/(R ) 0.84 1/2 1/4 0.94/(R ) 0.90 0.89 + 2.27e 1.14(λd1 /Aheater ) 1.14 qmax /qmaxz 10. Any large finite body 9. 1 side insulated 8. plain Small horizontal ribbon oriented vertically Infinite flat heater 1. Situation Source [9.22] [9.23] [9.23] 0.15 ≤ R ≤ 1.2 R ≥ 4.26 0.15 ≤ R ≤ 4.26 [9.20] [9.20] R ≥ 1.2 [9.20] 0.15 ≤ H ≤ 5.86 [9.20] 0.15 ≤ P ≤ 5.86 cannot specify generally; L 4 [9.20] cannot specify generally; L 4 [9.20] 0.15 ≤ H ≤ 2.96 [9.20] R ≥ 0.15 [9.19] L ≥ 27 9 < L < 20 [9.19] Range of L Table 9.3 Predictions of the peak pool boiling heat flux (9.27) (9.26) (9.25) (9.24) (9.23) (9.22) (9.21) (9.20) (9.19) (9.18) (9.17) (9.16) Eqn. No. Peak pool boiling heat flux §9.3 boiling curve becomes monotonic. When nucleation occurs on a very small wire, the wire is immediately enveloped in vapor and the mechanism of heat removal passes directly from natural convection to film boiling. Example 9.7 A spheroidal metallic body of surface area 400 cm2 and volume 600 cm3 is quenched in saturated water at 1 atm. What is the most rapid rate of heat removal during the quench? Solution. As the cooling process progresses, it goes through the boiling curve from film boiling, through qmin , up the transitional boiling regime, through qmax , and down the nucleate boiling curve. Cooling is finally completed by natural convection. One who has watched the quenching of a red-hot horseshoe will recall the great gush of bubbling that occurs as qmax is reached. We therefore calculate the required heat flow as Q = qmax Aspheroid , where qmax is given by eqn. (9.25) in Table 9.3: 1/2 qmax = 0.9 qmaxz = 0.9(0.131)ρg hfg 4 g σ (ρf − ρg ) so Q = 0.9(0.131)(0.597)1/2 (2, 257, 000) 9.8(0.0589)(958) W/m2 4 × 400 × 10−4 m2 or Q = 39, 900 W or 39.9 kW This is a startingly large rate of energy removal for such a small object. To complete the calculation, it is necessary to check whether or not R is large enough to justify the use of eqn. (9.25): R= V /A σ /g(ρf − ρg ) = 0.0006 0.04 9.8(958) = 6.0 0.0589 This is larger than the specified lower bound of about 4. 485 486 Heat transfer in boiling and other phase-change configurations 9.4 §9.4 Film boiling Film boiling bears an uncanny similarity to film condensation. The similarity is so great that in 1950, Bromley [9.24] was able to use the eqn. (8.64) for condensation on cylinders—almost directly—to predict film boiling from cylinders. He observed that the boundary condition (∂u/∂y)y =δ = 0 at the liquid–vapor interface in film condensation would have to change to something in between (∂u/∂y)y =δ = 0 and u(y = δ) = 0 during film boiling. The reason is that the external liquid is not so easily set into motion. He then redid the film condensation analysis, merely changing k and ν from liquid to vapor properties. The change of boundary conditions gave eqn. (8.64) with the constant changed from 0.729 to 0.512 and with k and ν changed to vapor values. By comparing the equation with experimental data, he fixed the constant at the intermediate value of 0.62. Thus, NuD based on kg became ⎡ NuD = 0.62 ⎣ (ρf − ρg )ghfg D 3 νg kg (Tw − Tsat ) ⎤1/4 ⎦ (9.28) where vapor and liquid properties should be evaluated at Tsat + ∆T /2 and at Tsat , respectively. The latent heat correction in this case is similar in form to that for film condensation, but with different constants in it. Sadasivan and Lienhard [9.25] have shown it to be hfg = hfg 1 + 0.968 − 0.163 Prg Jag (9.29) for Prg ≥ 0.6, where Jag = cpg (Tw − Tsat ) hfg . Dhir and Lienhard [9.26] did the same thing for spheres, as Bromley did for cylinders, 20 years later. Their result [cf. eqn. (8.65)] was ⎡ NuD = 0.67 ⎣ (ρf − ρg )ghfg D 3 νg kg (Tw − Tsat ) ⎤1/4 ⎦ (9.30) The preceding expressions are based on heat transfer by convection through the vapor film, alone. However, when film boiling occurs much beyond qmin in water, the heater glows dull cherry-red to white-hot. Radiation in such cases can be enormous. One’s first temptation might Film boiling §9.4 487 be simply to add a radiation heat transfer coefficient, hrad to hboiling as obtained from eqn. (9.28) or (9.30), where hrad 4 4 εσ Tw − Tsat qrad = = Tw − Tsat Tw − Tsat and where ε is a surface radiation property of the heater called the emittance (see Section 10.1). Unfortunately, such addition is not correct, because the additional radiative heat transfer will increase the vapor blanket thickness, reducing the convective contribution. Bromley [9.24] suggested for cylinders the approximate relation htotal = hboiling + 3 4 hrad , hrad < hboiling (9.31) More accurate corrections that have subsequently been offered are considerably more complex than this [9.10]. One of the most comprehensive is that of Pitschmann and Grigull [9.27]. Their correlation, which is fairly intricate, brings together an enormous range of heat transfer data for cylinders, within 20%. It is worth noting that radiation is seldom important when the heater temperature is less than 300◦ C. The use of the analogy between film condensation and film boiling is somewhat questionable during film boiling on a vertical surface. In this case, the liquid–vapor interface becomes Helmholtz-unstable at a short distance from the leading edge. However, Leonard, Sun, and Dix [9.28] √ have shown that by using λd1 3 in place of D in eqn. (9.28), one obtains a very satisfactory prediction of h for rather tall vertical plates. The analogy between film condensation and film boiling also deteriorates when it is applied to small curved bodies. The reason is that the thickness of the vapor film in boiling is far greater than the liquid film during condensation. Consequently, a curvature correction, which could be ignored in film condensation, must be included during film boiling from small cylinders, spheres, and other curved bodies. The first curvature correction to be made was an empirical one given by Westwater and Breen [9.29] in 1962. They showed that the equation NuD = 0.715 + 0.263 R R 1/4 NuDBromley (9.32) applies when R < 1.86. Otherwise, Bromley’s equation should be used directly. 488 Heat transfer in boiling and other phase-change configurations 9.5 §9.5 Minimum heat flux Zuber [9.17] also provided a prediction of the minimum heat flux, qmin , along with his prediction of qmax . He assumed that as Tw − Tsat is reduced in the film boiling regime, the rate of vapor generation eventually becomes too small to sustain the Taylor wave action that characterizes film boiling. Zuber’s qmin prediction, based on this assumption, has to include an arbitrary constant. The result for flat horizontal heaters is qmin = C ρg hfg 4 σ g(ρf − ρg ) (ρf + ρg )2 (9.33) Zuber guessed a value of C which Berenson [9.30] subsequently corrected on the basis of experimental data. Berenson used measured values of qmin on horizontal heaters to get qminBerenson = 0.09 ρg hfg 4 σ g(ρf − ρg ) (ρf + ρg )2 (9.34) Lienhard and Wong [9.31] did the parallel prediction for horizontal wires and found that qmin = 0.515 18 2 (2R 2 + 1) R 1/4 qmin Berenson (9.35) The problem with all of these expressions is that some contact frequently occurs between the liquid and the heater wall at film boiling heat fluxes higher than the minimum. When this happens, the boiling curve deviates above the film boiling curve and finds a higher minimum than those reported above. The values of the constants shown above should therefore be viewed as practical lower limits of qmin . We return to this matter subsequently. Example 9.8 Check the value of qmin shown in Fig. 9.2. Solution. The heater is a flat surface, so we use eqn. (9.34) and the physical properties given in Example 9.5. qmin = 0.09(0.597)(2, 257, 000) 4 9.8(0.0589)(958) (959)2 §9.6 Transition boiling and system influences or qmin = 18, 990 W/m2 From Fig. 9.2 we read 20,000 W/m2 , which is the same, within the accuracy of the graph. 9.6 Transition boiling and system influences Many system features influence the pool boiling behavior we have discussed thus far. These include forced convection, subcooling, gravity, surface roughness and surface chemistry, and the heater configuration, among others. To understand one of the most serious of these—the influence of surface roughness and surface chemistry—we begin by thinking about transition boiling, which is extremely sensitive to both. Surface condition and transition boiling Less is known about transition boiling than about any other mode of boiling. Data are limited, and there is no comprehensive body of theory. The first systematic sets of accurate measurements of transition boiling were reported by Berenson [9.30] in 1960. Figure 9.14 shows two sets of his data. The upper set of curves shows the typical influence of surface chemistry on transition boiling. It makes it clear that a change in the surface chemistry has little effect on the boiling curve except in the transition boiling region and the low heat flux film boiling region. The oxidation of the surface has the effect of changing the contact angle dramatically— making it far easier for the liquid to wet the surface when it touches it. Transition boiling is more susceptible than any other mode to such a change. The bottom set of curves shows the influence of surface roughness on boiling. In this case, nucleate boiling is far more susceptible to roughness than any other mode of boiling except, perhaps, the very lowest end of the film boiling range. That is because as roughness increases the number of active nucleation sites, the heat transfer rises in accordance with the Yamagata relation, eqn. (9.3). It is important to recognize that neither roughness nor surface chemistry affects film boiling, because the liquid does not touch the heater. 489 Figure 9.14 Typical data from Berenson’s [9.30] study of the influence of surface condition on the boiling curve. 490 §9.6 Transition boiling and system influences Figure 9.15 The transition boiling regime. The fact that both effects appear to influence the lower film boiling range means that they actually cause film boiling to break down by initiating liquid–solid contact at low heat fluxes. Figure 9.15 shows what an actual boiling curve looks like under the influence of a wetting (or even slightly wetting) contact angle. This figure is based on the work of Witte and Lienhard ([9.32] and [9.33]). On it are identified a nucleate-transition and a film-transition boiling region. These are continuations of nucleate boiling behavior with decreasing liquid– solid contact (as shown in Fig. 9.3c) and of film boiling behavior with increasing liquid–solid contact, respectively. These two regions of transition boiling are often connected by abrupt jumps. However, no one has yet seen how to predict where such jumps take place. Reference [9.33] is a full discussion of the hydrodynamic theory of boiling, which includes an extended discussion of the transition boiling problem and a correlation for the transition-film boiling heat flux by Ramilison and Lienhard [9.34]. 491 492 Heat transfer in boiling and other phase-change configurations §9.6 Figure 9.14 also indicates fairly accurately the influence of roughness and surface chemistry on qmax . It suggests that these influences normally can cause significant variations in qmax that are not predicted in the hydrodynamic theory. Ramilison et al. [9.35] correlated these effects for large flat-plate heaters using the rms surface roughness, r in µm, and the receding contact angle for the liquid on the heater material, βr in radians: qmax = 0.0336 (π − βr )3.0 r 0.0125 qmaxZ (9.36) This correlation collapses the data to ±6%. Uncorrected, variations from the predictions of hydrodynamic theory reached 40% as a result of roughness and finish. Equivalent results are needed for other geometries. Subcooling A stationary pool will normally not remain below its saturation temperature over an extended period of time. When heat is transferred to the pool, the liquid soon becomes saturated—as it does in a teakettle (recall Experiment 9.1). However, before a liquid comes up to temperature, or if a very small rate of forced convection continuously replaces warm liquid with cool liquid, we can justly ask what the effect of a cool liquid bulk might be. Figure 9.16 shows how a typical boiling curve might be changed if Tbulk < Tsat : We know, for example, that in laminar natural convection, q will increase as (Tw − Tbulk )5/4 or as [(Tw − Tsat ) + ∆Tsub ]5/4 , where ∆Tsub ≡ Tsat − Tbulk . During nucleate boiling, the influence of subcooling on q is known to be small. The peak and minimum heat fluxes are known to increase linearly with ∆Tsub . These increases are quite significant. The film boiling heat flux increases rather strongly, especially at lower heat fluxes. The influence of ∆Tsub on transitional boiling is not well documented. Gravity The influence of gravity (or any other such body force) is of concern because boiling processes frequently take place in rotating or accelerating systems. The reduction of gravity has a significant impact on boiling processes aboard space vehicles. Since g appears explicitly in the equations for qmax , qmin , and qfilm boiling , we know what its influence is. Both qmax and qmin increase directly as g 1/4 in finite bodies, and there is an additional gravitational influence through the parameter L . However, when gravity is small enough to reduce R below about 0.15, the hydrody- Transition boiling and system influences §9.6 Figure 9.16 The influence of subcooling on the boiling curve. namic transitions deteriorate and eventually vanish altogether. Although Rohsenow’s equation suggests that q is proportional to g 1/2 in the nucleate boiling regime, other evidence suggests that the influence of gravity on the nucleate boiling curve is very slight, apart from an indirect effect on the onset of boiling. Forced convection The influence of superposed flow on the pool boiling curve for a given heater (e.g., Fig. 9.2) is generally to improve heat transfer everywhere. But flow is particularly effective in raising qmax . Let us look at the influence of flow on the different regimes of boiling. 493 494 Heat transfer in boiling and other phase-change configurations §9.6 Influences of forced convection on nucleate boiling. Figure 9.17 shows nucleate boiling during the forced convection of water over a flat plate. Bergles and Rohsenow [9.36] offer an empirical strategy for predicting the heat flux during nucleate flow boiling when the net vapor generation is still relatively small. (The photograph in Fig. 9.17 shows how a substantial buildup of vapor can radically alter flow boiling behavior.) They suggest that q = qFC 1 + qB qFC qi 1− qB 2 (9.37) where • qFC is the single-phase forced convection heat transfer for the heater, as one might calculate using the methods of Chapters 6 and 7. • qB is the pool boiling heat flux for that liquid and that heater from eqn. (9.4). • qi is the heat flux from the pool boiling curve evaluated at the value of (Tw −Tsat ) where boiling begins during flow boiling (see Fig. 9.17). An estimate of (Tw − Tsat )onset can be made by intersecting the forced convection equation q = hFC (Tw − Tb ) with the following equation [9.37]: (Tw − Tsat )onset = 8σ Tsat q ρg hfg kf 1/2 (9.38) Equation (9.37) will provide a first approximation in most boiling configurations, but it is restricted to subcooled flows or other situations in which vapor generation is not too great. Peak heat flux in external flows. The peak heat flux on a submerged body is strongly augmented by an external flow around it. Although knowledge of this area is still evolving, we do know from dimensional analysis that qmax = fn WeD , ρf ρg ρg hfg u∞ (9.39) Transition boiling and system influences §9.6 Figure 9.17 Forced convection boiling on an external surface. where the Weber number, We, is WeL ≡ ρg u2 L inertia force L ∞ = σ surface force L and where L is any characteristic length. Kheyrandish and Lienhard [9.38] suggest fairly complex expressions of this form for qmax on horizontal cylinders in cross flows. For a cylindrical liquid jet impinging on a heated disk of diameter D , Sharan and 495 496 Heat transfer in boiling and other phase-change configurations §9.7 Lienhard [9.39] obtained qmax = 0.21 + 0.0017ρf ρg ρg hfg ujet djet D 1/3 1000ρg /ρf WeD A (9.40) where, if we call ρf /ρg ≡ r , A = 0.486 + 0.06052 ln r − 0.0378 (ln r )2 + 0.00362 (ln r )3 (9.41) This correlation represents all the existing data within ±20% over the full range of the data. The influence of fluid flow on film boiling. Bromley et al. [9.40] showed that the film boiling heat flux during forced flow normal to a cylinder should take the form q = constant kg ρg hfg ∆T u∞ 1/2 D (9.42) for u2 /(gD) ≥ 4 with hfg from eqn. (9.29). Their data fixed the constant ∞ at 2.70. Witte [9.41] obtained the same relationship for flow over a sphere and recommended a value of 2.98 for the constant. Additional work in the literature deals with forced film boiling on plane surfaces and combined forced and subcooled film boiling in a variety of geometries [9.42]. Although these studies are beyond our present scope, it is worth noting that one may attain very high cooling rates using film boiling with both forced convection and subcooling. 9.7 Forced convection boiling in tubes Flowing fluids undergo boiling or condensation in many of the cases in which we transfer heat to fluids moving through tubes. For example, such phase change occurs in all vapor-compression power cycles and refrigerators. When we use the terms boiler, condenser, steam generator, or evaporator we usually refer to equipment that involves heat transfer within tubes. The prediction of heat transfer coefficients in these systems is often essential to determining U and sizing the equipment. So let us consider the problem of predicting boiling heat transfer to liquids flowing through tubes. Figure 9.18 The development of a two-phase flow in a vertical tube with a uniform wall heat flux (not to scale). 497 498 Heat transfer in boiling and other phase-change configurations §9.7 Relationship between heat transfer and temperature difference Forced convection boiling in a tube or duct is a process that becomes very hard to delineate because it takes so many forms. In addition to the usual system variables that must be considered in pool boiling, the formation of many regimes of boiling requires that we understand several boiling mechanisms and the transitions between them, as well. Collier and Thome’s excellent book, Convective Boiling and Condensation [9.43], provides a comprehensive discussion of the issues involved in forced convection boiling. Figure 9.18 is their representation of the fairly simple case of flow of liquid in a uniform wall heat flux tube in which body forces can be neglected. This situation is representative of a fairly low heat flux at the wall. The vapor fraction, or quality, of the flow increases steadily until the wall “dries out.” Then the wall temperature rises rapidly. With a very high wall heat flux, the pipe could burn out before dryout occurs. Figure 9.19, also provided by Collier, shows how the regimes shown in Fig. 9.18 are distributed in heat flux and in position along the tube. Notice that, at high enough heat fluxes, burnout can be made to occur at any station in the pipe. In the subcooled nucleate boiling regime (B in Fig. 9.18) and the low quality saturated regime (C ), the heat transfer can be predicted using eqn. (9.37) in Section 9.6. But in the subsequent regimes of slug flow and annular flow (D , E , and F ) the heat transfer mechanism changes substantially. Nucleation is increasingly suppressed, and vaporization takes place mainly at the free surface of the liquid film on the tube wall. Most efforts to model flow boiling differentiate between nucleateboiling-controlled heat transfer and convective boiling heat transfer. In those regimes where fully developed nucleate boiling occurs (the later parts of C ), the heat transfer coefficient is essentially unaffected by the mass flow rate and the flow quality. Locally, conditions are similar to pool boiling. In convective boiling, on the other hand, vaporization occurs away from the wall, with a liquid-phase convection process dominating at the wall. For example, in the annular regions E and F , heat is convected from the wall by the liquid film, and vaporization occurs at the interface of the film with the vapor in the core of the tube. Convective boiling can also dominate at low heat fluxes or high mass flow rates, where wall nucleate is again suppressed. Vaporization then occurs mainly on entrained bubbles in the core of the tube. In convective boiling, the heat transfer coefficient is essentially independent of the heat flux, but it is §9.7 Forced convection boiling in tubes Figure 9.19 The influence of heat flux on two-phase flow behavior. strongly affected by the mass flow rate and quality. Building a model to capture these complicated and competing trends has presented a challenge to researchers for several decades. One early effort by Chen [9.44] used a weighted sum of a nucleate boiling heat transfer coefficient and a convective boiling coefficient, where the weighting depended on local flow conditions. This model represents water data to an accuracy of about ±30% [9.45], but it does not work well with most other fluids. Chen’s mechanistic approach was substantially improved in a more complex version due to Steiner and Taborek [9.46]. Many other investigators have instead pursued correlations built from dimensional analysis and physical reasoning. To proceed with a dimensional analysis, we first note that the liquid and vapor phases may have different velocities. Thus, we avoid intro- 499 500 Heat transfer in boiling and other phase-change configurations §9.7 ducing a flow speed and instead rely on the the superficial mass flux, G, through the pipe: G≡ ˙ m Apipe (kg/m2 s) (9.43) This mass flow per unit area is constant along the duct if the flow is steady. From this, we can define a “liquid only” Reynolds number Relo ≡ GD µf (9.44) which would be the Reynolds number if all the flowing mass were in the liquid state. Then we may use Relo to compute a liquid-only heat transfer cofficient, hlo from Gnielinski’s equation, eqn. (7.43), using liquid properties at Tsat . Physical arguments then suggest that the dimensional functional equation for the flow boiling heat transfer coefficient, hfb , should take the following form for saturated flow in vertical tubes: hfb = fn hlo , G, x, hfg , qw , ρf , ρg , D (9.45) It should be noted that other liquid properties, such as viscosity and conductivity, are represented indirectly through hlo . This functional equation has eight dimensional variables (and one dimensionless variable, x ) in five dimensions (m, kg, s, J, K). We thus obtain three more dimensionless groups to go with x , specifically qw ρg hfb = fn x , , hlo Ghfg ρf (9.46) In fact, the situation is even a bit simpler than this, since arguments related to the pressure gradient show that the quality and the density ratio can be combined into a single group, called the convection number : Co ≡ 1−x x 0.8 ρg ρf 0.5 (9.47) The other dimensionless group in eqn. (9.46) is called the boiling number : Bo ≡ qw Ghfg (9.48) Forced convection boiling in tubes §9.7 501 Table 9.4 Fluid-dependent parameter F in the Kandlikar correlation for copper tubing. Additional values are given in [9.47]. Fluid Water Propane R-12 R-22 R-32 F 1.0 2.15 1.50 2.20 1.20 Fluid R-124 R-125 R-134a R-152a R-410a F 1.90 1.10 1.63 1.10 1.72 so that hfb = fn (Bo, Co) hlo (9.49) When the convection number is large (Co 1), as for low quality, nucleate boiling dominates. In this range, hfb /hlo rises with increasing Bo and is approximately independent of Co. When the convection number is smaller, as at higher quality, the effect of the boiling number declines and hfb /hlo increases with decreasing Co. Correlations having the general form of eqn. (9.49) were developed by Schrock and Grossman [9.48], Shah [9.49], and Gungor and Winterton [9.50]. Kandlikar [9.45, 9.47, 9.51] refined this approach further, obtaining good accuracy and better capturing the parametric trends. His method is to calculate hfb /hlo from each of the following two correlations and to choose the larger value: hfb hlo nbd hfb hlo cbd = (1 − x)0.8 0.6683 Co−0.2 fo + 1058 Bo0.7 F (9.50a) = (1 − x)0.8 1.136 Co−0.9 fo + 667.2 Bo0.7 F (9.50b) where “nbd” means “nucleate boiling dominant” and “cbd” means “convective boiling dominant”. In these equations, the orientation factor, fo , is set to unity for vertical tubes4 and F is a fluid-dependent parameter whose value is given 4 The value for horizontal tubes is given in eqn. (9.52). 502 Heat transfer in boiling and other phase-change configurations §9.7 in Table 9.4. The parameter F arises here for the same reason that fluiddependent parameters appear in nucleate boiling correlations: surface tension, contact angles, and other fluid-dependent variables influence nucleation and bubble growth. The values in Table 9.4 are for commercial grades of copper tubing. For stainless steel tubing, Kandlikar recommends F = 1 for all fluids. Equations (9.50) are applicable for the saturated boiling regimes (C through F ) with quality in the range 0 < x ≤ 0.8. For subcooled conditions, see Problem 9.21. Example 9.9 0.6 kg/s of saturated H2 O at Tb = 207◦ C flows in a 5 cm diameter vertical tube heated at a rate of 184,000 W/m2 . Find the wall temperature at a point where the quality x is 20%. Solution. Data for water are taken from Tables A.3–A.5. We first compute hlo . G= and Relo = ˙ m 0.6 = 305.6 kg/m2 s = Apipe 0.001964 GD (305.6)(0.05) = = 1.178 × 105 µf 1.297 × 10−4 From eqns. (7.42) and (7.43): f= NuD = 1 1.82 log10 (1.178 × 105 ) − 1.64 2 = 0.01736 (0.01736/8) 1.178 × 105 − 1000 (0.892) = 236.3 1 + 12.7 0.01736/8 (0.892)2/3 − 1 Hence, kf 0.6590 236.3 = 3, 115 W/m2 K 0.05 D Next, we find the parameters for eqns. (9.50). From Table 9.4, F = 1 for water, and for a vertical tube, fo = 1. Also, hlo = Co = 1−x x Bo = 0.8 NuD = ρg ρf 0.5 = 1 − 0.20 0.2 0.8 9.014 856.5 0.5 = 0.3110 qw 184, 000 = 3.147 × 10−4 = Ghfg (305.6)(1, 913, 000) Forced convection boiling in tubes §9.7 Substituting into eqns. (9.50): hfb nbd = (3, 115)(1 − 0.2)0.8 0.6683 (0.3110)−0.2 (1) + 1058 (3.147 × 10−4 )0.7 (1) = 11, 950 W/m2 K hfb cbd = (3, 115)(1 − 0.2)0.8 1.136 (0.3110)−0.9 (1) + 667.2 (3.147 × 10−4 )0.7 (1) = 14, 620 W/m2 K Since the second value is larger, we use it: hfb = 14, 620 W/m2 K. Then, Tw = Tb + qw 184, 000 = 220◦ C = 207 + hfb 14, 620 The Kandlikar correlation leads to mean deviations of 16% for water and 19% for the various refrigerants. The Gungor and Winterton correlation [9.50], which is popular for its simplicity, does not contain fluidspecific coefficients, but it is somewhat less accurate than either the Kandlikar equations or the more complex Steiner and Taborek method [9.45, 9.46]. These three approaches, however, are among the best available. Two-phase flow and heat transfer in horizontal tubes The preceding discussion of flow boiling in tubes is largely restricted to vertical tubes. Several of the flow regimes in Fig. 9.18 will be altered as shown in Fig. 9.20 if the tube is oriented horizontally. The reason is that, especially at low quality, liquid will tend to flow along the bottom of the pipe and vapor along the top. The patterns shown in Fig. 9.20, by the way, will also be observed during the reverse process—condensation—or during adiabatic two-phase flow. Which flow pattern actually occurs depends on several parameters in a fairly complex way. While many methods have been suggested to predict what flow pattern will result for a given set of conditions in the pipe, one of the best is that developed by Dukler, Taitel, and their coworkers. Their two-phase flow-regime maps are summarized in [9.52] and [9.53]. For the prediction of heat transfer, the most important additional parameter is the Froude number, Frlo , which characterizes the strength of the flow’s inertia (or momentum) relative to the gravitational forces 503 504 Heat transfer in boiling and other phase-change configurations §9.7 Figure 9.20 The discernible flow regimes during boiling, condensation, or adiabatic flow from left to right in horizontal tubes. that drive the separation of the liquid and vapor phases: Frlo ≡ G2 2 ρf gD (9.51) When Frlo < 0.04, the top of the tube becomes relatively dry and hfb /hlo begins to decline as the Froude number decreases further. Kandlikar found that he could modify his correlation to account for gravitational effects in horizontal tubes by changing the value of fo in eqns. (9.50): fo = ⎧ ⎨1 ⎩(25 Frlo ) for Frlo ≥ 0.04 0.3 for Frlo < 0.04 (9.52) Peak heat flux We have seen that there are two limiting heat fluxes in flow boiling in a tube: dryout and burnout. The latter is the more dangerous of the two since it occurs at higher heat fluxes and gives rise to more catastrophic temperature rises. Collier and Thome provide an extensive discussion of the subject [9.43], as does Hewitt [9.54]. §9.8 Forced convective condensation heat transfer One effective set of empirical formulas was developed by Katto [9.55]. He used dimensional analysis to show that qmax = fn Ghfg ρg σ ρf L , , ρf G2 L D where L is the length of the tube and D its diameter. Since G2 L σ ρf is a Weber number, we can see that this equation is of the same form as eqn. (9.39). Katto identifies several regimes of flow boiling with both saturated and subcooled liquid entering the pipe. For each of these regions, he and Ohne [9.56] later fit a successful correlation of this form to existing data. Pressure gradients in flow boiling Pressure gradients in flow boiling interact with the flow pattern and the void fraction, and they can change the local saturation temperature of the fluid. Gravity, flow acceleration, and friction all contribute to pressure change, and friction can be particularly hard to predict. In particular, the frictional pressure gradient can increase greatly as the flow quality rises from the pure liquid state to the pure vapor state; the change can amount to more than two orders of magnitude at low pressures. Data correlations are usually used to estimate the frictional pressure loss, but they are, at best, accurate to within about ±30%. Whalley [9.57] provides a nice introduction such methods. Certain complex models, designed for use in computer codes, can be used to make more accurate predictions [9.58]. 9.8 Forced convective condensation heat transfer When vapor is blown or forced past a cool wall, it exerts a shear stress on the condensate film. If the direction of forced flow is downward, it will drag the condensate film along, thinning it out and enhancing heat transfer. It is not hard to show (see Problem 9.22) that 4 4µk(Tsat − Tw )x = δ4 + ghfg ρf (ρf − ρg ) 3 τδ δ3 (ρf − ρg )g (9.53) where τδ is the shear stress exerted by the vapor flow on the condensate film. Equation (9.53) is the starting point for any analysis of forced convection condensation on an external surface. Notice that if τδ is negative—if 505 506 Heat transfer in boiling and other phase-change configurations §9.9 the shear opposes the direction of gravity—then it will have the effect of thickening δ and reducing heat transfer. Indeed, if for any value of δ, τδ = − 3g(ρf − ρg ) 4 δ, (9.54) the shear stress will have the effect of halting the flow of condensate completely for a moment until δ grows to a larger value. Heat transfer solutions based on eqn. (9.53) are complex because they require that one solve the boundary layer problem in the vapor in order to evaluate τδ ; and this solution must be matched with the velocity at the outside surface of the condensate film. Collier and Thome [9.43, §10.5] discuss such solutions in some detail. One explicit result has been obtained in this way for condensation on the outside of a horizontal cylinder by Shekriladze and Gomelauri [9.59]: ⎧ ⎡ ⎨ ρ u∞ D ghfg µf D f ⎣1 + 1 + 1.69 NuD = 0.64 ⎩ µf u2 kf (Tsat − Tw ) ∞ 1/2 ⎤⎫1/2 ⎬ ⎦ (9.55) ⎭ where u∞ is the free stream velocity and NuD is based on the liquid conductivity. Equation (9.55) is valid up to ReD ≡ ρf u∞ D µf = 106 . Notice, too, that under appropriate flow conditions (large values of u∞ , for example), gravity becomes unimportant and NuD → 0.64 2ReD (9.56) The prediction of heat transfer during forced convective condensation in tubes becomes a different problem for each of the many possible flow regimes. The reader is referred to [9.43, §10.5] or [9.60] for details. 9.9 Dropwise condensation An automobile windshield normally is covered with droplets during a light rainfall. They are hard to see through, and one must keep the windshield wiper moving constantly to achieve any kind of visibility. A glass windshield is normally quite clean and is free of any natural oxides, so the water forms a contact angle on it and any film will be unstable. The water tends to pull into droplets, which intersect the surface at the contact angle. Visibility can be improved by mixing a surfactant chemical into the window-washing water to reduce surface tension. It can also be §9.9 Dropwise condensation improved by preparing the surface with a “wetting agent” to reduce the contact angle.5 Such behavior can also occur on a metallic condensing surface, but there is an important difference: Such surfaces are generally wetting. Wetting can be temporarily suppressed, and dropwise condensation can be encouraged, by treating an otherwise clean surface (or the vapor) with oil, kerosene, or a fatty acid. But these contaminants wash away fairly quickly. More permanent solutions have proven very elusive, with the result the liquid condensed in heat exchangers almost always forms a film. It is regrettable that this is the case, because what is called dropwise condensation is an extremely effective heat removal mechanism. Figure 9.21 shows how it works. Droplets grow from active nucleation sites on the surface, and in this sense there is a great similarity between nucleate boiling and dropwise condensation. The similarity persists as the droplets grow, touch, and merge with one another until one is large enough to be pulled away from its position by gravity. It then slides off, wiping away the smaller droplets in its path and leaving a dry swathe in its wake. New droplets immediately begin to grow at the nucleation sites in the path. The repeated re-creation of the early droplet growth cycle creates a very efficient heat removal mechanism. It is typically ten times more effective than film condensation under the same temperature difference. Indeed, condensing heat transfer coefficients as high as 200,000 W/m2 K can be obtained with water at 1 atm. Were it possible to sustain dropwise condensation, we would certainly design equipment in such a way as to make use of it. Unfortunately, laboratory experiments on dropwise condensation are almost always done on surfaces that have been prepared with oleic, stearic, or other fatty acids, or, more recently, with dioctadecyl disulphide. These nonwetting agents, or promoters as they are called, are discussed in [9.60, 9.61]. While promoters are normally impractical for industrial use, since they either wash away or oxidize, experienced plant engineers have sometimes added rancid butter through the cup valves of commercial condensers to get at least temporary dropwise condensation. Finally, we note that the obvious tactic of coating the surface with a 5 A way in which one can accomplish these ends is by wiping the wet window with a cigarette. It is hard to tell which of the two effects the many nasty chemicals in the cigarette achieve. 507 a. The process of liquid removal during dropwise condensation. b. Typical photograph of dropwise condensation provided by Professor Borivoje B. Miki´. Notice the dry paths c on the left and in the wake of the middle droplet. Figure 9.21 Dropwise condensation. 508 The heat pipe §9.10 thin, nonwetting, polymer film (such as PTFE, or Teflon) adds just enough conduction resistance to reduce the overall heat transfer coefficient to a value similar to film condensation, fully defeating its purpose! (Sufficiently thin polymer layers have not been found to be durable.) Noble metals, such as gold, platinum, and palladium, can also be used as nonwetting coating, and they have sufficiently high thermal conductivity to avoid the problem encountered with polymeric coatings. For gold, however, the minimum effective coating thickness is about 0.2 µm, or about 1/8 Troy ounce per square meter [9.62]. Such coatings are far too expensive for the vast majority of technical applications. 9.10 The heat pipe A heat pipe is a device that combines the high efficiencies of boiling and condensation. It is aptly named because it literally pipes heat from a hot region to a cold one. The operation of a heat pipe is shown in Fig. 9.22. The pipe is a tube that can be bent or turned in any way that is convenient. The inside of the tube is lined with a layer of wicking material. The wick is wetted with an appropriate liquid. One end of the tube is exposed to a heat source that evaporates the liquid from the wick. The vapor then flows from the hot end of the tube to the cold end, where it is condensed. Capillary action moves the condensed liquid axially along the wick, back to the evaporator where it is again vaporized. Placing a heat pipe between a hot region and a cold one is thus similar to connecting the regions with a material of extremely high thermal conductivity—potentially orders of magnitude higher than any solid material. Such devices are used not only for achieving high heat transfer rates between a source and a sink but for a variety of less obvious purposes. They are used, for example, to level out temperatures in systems, since they function almost isothermally and offer very little thermal resistance. Design considerations in matching a heat pipe to a given application center on the following issues. • Selection of the right liquid. The intended operating temperature of the heat pipe can be met only with a fluid whose saturation temperatures cover the design temperature range. Depending on the temperature range needed, the liquid can be a cryogen, an organic 509 510 Heat transfer in boiling and other phase-change configurations §9.10 Figure 9.22 A typical heat pipe configuration. liquid, water, a liquid metal, or, in principle, almost any fluid. However, the following characteristics will serve to limit the vapor mass flow per watt, provide good capillary action in the wick, and control the temperature rise between the wall and the wick: i) High latent heat ii) High surface tension iii) Low liquid viscosities iv) High thermal conductivity Two liquids that meet these four criteria admirably are water and mercury, although toxicity and wetting problems discourage the use of the latter. Ammonia is useful at temperatures that are a bit too low for water. At high temperatures, sodium and lithium have good characteristics, while nitrogen is good for cryogenic temperatures. Fluids can be compared using the merit number, M = hfg σ /νf (see Problem 9.36). • Selection of the tube material. The tube material must be compatible with the working fluid. Gas generation and corrosion are particular considerations. Copper tubes are widely used with water, methanol, and acetone, but they cannot be used with ammonia. Stainless steel §9.10 The heat pipe tubes can be used with ammonia and many liquid metals, but are not suitable for long term service with water. In some aerospace applications, aluminum is used for its low weight; however, it is compatible with working fluids other than ammonia. • Selection and installation of the wick. Like the tube material, the wick material must be compatible with the working fluid. In addition, the working fluid must be able to wet the wick. Wicks can be fabricated from a metallic mesh, from a layer of sintered beads, or simply by scoring grooves along the inside surface of the tube. Many ingenious schemes have been created for bonding the wick to the inside of the pipe and keeping it at optimum porosity. • Operating limits of the heat pipe. The heat transfer through a heat pipe is restricted by i) Viscous drag in the wick at low temperature ii) The sonic, or choking, speed of the vapor iii) Drag of the vapor on the counterflowing liquid in the wick iv) Ability of capillary forces in the wick to pump the liquid through the pressure rise between evaporator and condenser v) The boiling burnout heat flux in the evaporator section. These items much each be dealt with in detail during the design of a new heat pipe [9.63]. • Control of the pipe performance. Often a given heat pipe will be called upon to function over a range of conditions—under varying evaporator heat loads, for example. One way to vary its performance is through the introduction of a non-condensible gas in the pipe. This gas will collect at the condenser, limiting the area of the condenser that vapor can reach. By varying the amount of gas, the thermal resistance of the heat pipe can be controlled. In the absence of active control of the gas, an increase in the heat load at the evaporator will raise the pressure in the pipe, compressing the noncondensible gas and lowering the thermal resistance of the pipe. The result is that the temperature at the evaporator remains essentially constant even as the heat load rises as falls. Heat pipes have proven useful in cooling high power-density electronic devices. The evaporator is located on a small electronic component 511 512 Chapter 9: Heat transfer in boiling and other phase-change configurations Figure 9.23 A heat sink for cooling a microprocessor. Courtesy of Dr. A. B. Patel, Aavid Thermalloy LLC. to be cooled, perhaps a microprocessor, and the condenser is finned and cooled by a forced air flow (in a desktop or mainframe computer) or is unfinned and cooled by conduction into the exterior casing or structural frame (in a laptop computer). These applications rely on having a heat pipe with much larger condenser area than evaporator area. Thus, the heat fluxes on the condenser are kept relatively low. This facilitates such uncomplicated means for the ultimate heat disposal as using a small fan to blow air over the condenser. One heat-pipe-based electronics heat sink is shown in Fig. 9.23. The copper block at center is attached to a microprocessor, and the evaporator sections of two heat pipes are embedded in the block. The condenser sections of the pipes have copper fins pressed along their length. A pair of spring clips holds the unit in place. These particular heat pipes have copper tubes with water as the working fluid. The reader interested in designing or selecting a heat pipe will find a broad discussion of such devices in the book by Dunn and Reay [9.63]. Problems 513 Problems 9.1 A large square tank with insulated sides has a copper base 1.27 cm thick. The base is heated to 650◦ C and saturated water is suddenly poured in the tank. Plot the temperature of the base as a function of time on the basis of Fig. 9.2 if the bottom of the base is insulated. In your graph, indicate the regimes of boiling and note the temperature at which cooling is most rapid. 9.2 Predict qmax for the two heaters in Fig. 9.3b. At what percentage of qmax is each one operating? 9.3 A very clean glass container of water at 70◦ C is depressurized until it is subcooled 30◦ C. Then it suddenly and explosively “flashes” (or boils). What is the pressure at which this happens? Approximately what diameter of gas bubble, or other disturbance in the liquid, caused it to flash? 9.4 Plot the unstable bubble radius as a function of liquid superheat for water at 1 atm. Comment on the significance of your curve. 9.5 In chemistry class you have probably witnessed the phenomenon of “bumping” in a test tube (the explosive boiling that blows the contents of the tube all over the ceiling). Yet you have never seen this happen in a kitchen pot. Explain why not. 9.6 Use van der Waal’s equation of state to approximate the highest reduced temperature to which water can be superheated at low pressure. How many degrees of superheat does this suggest that water can sustain at the low pressure of 1 atm? (It turns out that this calculation is accurate within about 10%.) What would Rb be at this superheat? 9.7 Use Yamagata’s equation, (9.3), to determine how nucleation site density increases with ∆T for Berenson’s curves in Fig. 9.14. (That is, find c in the relation n = constant ∆T c .) 9.8 Suppose that Csf for a given surface is high by 50%. What will be the percentage error in q calculated for a given value of ∆T ? [Low by 70%.] 514 Chapter 9: Heat transfer in boiling and other phase-change configurations 9.9 Water at 100 atm boils on a nickel heater whose temperature is 6◦ C above Tsat . Find h and q. 9.10 Water boils on a large flat plate at 1 atm. Calculate qmax if the 1 plate is operated on the surface of the moon (at 6 of gearth−normal ). What would qmax be in a space vehicle experiencing 10−4 of gearth−normal ? 9.11 Water boils on a 0.002 m diameter horizontal copper wire. Plot, to scale, as much of the boiling curve on log q vs. log ∆T coordinates as you can. The system is at 1 atm. 9.12 Redo Problem 9.11 for a 0.03 m diameter sphere in water at 10 atm. 9.13 Verify eqn. (9.17). 9.14 Make a sketch of the q vs. (Tw − Tsat ) relation for a pool boiling process, and invent a graphical method for locating the points where h is maximum and minimum. 9.15 A 2 mm diameter jet of methanol is directed normal to the center of a 1.5 cm diameter disk heater at 1 m/s. How many watts can safely be supplied by the heater? 9.16 Saturated water at 1 atm boils on a ½ cm diameter platinum rod. Estimate the temperature of the rod at burnout. 9.17 Plot (Tw − Tsat ) and the quality x as a function of position x for the conditions in Example 9.9. Set x = 0 where x = 0 and end the plot where the quality reaches 80%. 9.18 Plot (Tw − Tsat ) and the quality x as a function of position in an 8 cm I.D. pipe if 0.3 kg/s of water at 100◦ C passes through it and qw = 200, 000 W/m2 . 9.19 Use dimensional analysis to verify the form of eqn. (9.8). 9.20 Compare the peak heat flux calculated from the data given in Problem 5.6 with the appropriate prediction. [The prediction is within 11%.] Problems 9.21 515 The Kandlikar correlation, eqn. (9.50a), can be adapted subcooled flow boiling, with x = 0 (region B in Fig. 9.19). Noting that qw = hfb (Tw − Tsat ), show that qw = 1058 hlo F (Ghfg )−0.7 (Tw − Tsat ) 1/0.3 in subcooled flow boiling [9.47]. 9.22 Verify eqn. (9.53) by repeating the analysis following eqn. (8.47) but using the b.c. (∂u/∂y)y =δ = τδ µ in place of (∂u/∂y)y =δ = 0. Verify the statement involving eqn. (9.54). 9.23 A cool-water-carrying pipe 7 cm in outside diameter has an outside temperature of 40◦ C. Saturated steam at 80◦ C flows across it. Plot hcondensation over the range of Reynolds numbers 0 ReD 106 . Do you get the value at ReD = 0 that you would anticipate from Chapter 8? 9.24 (a) Suppose that you have pits of roughly 0.002 mm diameter in a metallic heater surface. At about what temperature might you expect water to boil on that surface if the pressure is 20 atm. (b) Measurements have shown that water at atmospheric pressure can be superheated about 200◦ C above its normal boiling point. Roughly how large an embryonic bubble would be needed to trigger nucleation in water in such a state. 9.25 Obtain the dimensionless functional form of the pool boiling qmax equation and the qmax equation for flow boiling on external surfaces, using dimensional analysis. 9.26 A chemist produces a nondegradable additive that will increase σ by a factor of ten for water at 1 atm. By what factor will the additive improve qmax during pool boiling on (a) infinite flat plates and (b) small horizontal cylinders? By what factor will it improve burnout in the flow of jet on a disk? 9.27 Steam at 1 atm is blown at 26 m/s over a 1 cm O.D. cylinder at 90◦ C. What is h? Can you suggest any physical process within the cylinder that could sustain this temperature in this flow? 9.28 The water shown in Fig. 9.17 is at 1 atm, and the Nichrome heater can be approximated as nickel. What is Tw − Tsat ? 516 Chapter 9: Heat transfer in boiling and other phase-change configurations 9.29 For film boiling on horizontal cylinders, eqn. (9.6) is modified to −1/2 √ g (ρf − ρg ) 2 + λd = 2π 3 . σ (diam.)2 If ρf is 748 kg/m3 for saturated acetone, compare this λd , and the flat plate value, with Fig. 9.3d. 9.30 Water at 47◦ C flows through a 13 cm diameter thin-walled tube at 8 m/s. Saturated water vapor, at 1 atm, flows across the tube at 50 m/s. Evaluate Ttube , U , and q. 9.31 A 1 cm diameter thin-walled tube carries liquid metal through saturated water at 1 atm. The throughflow of metal is increased until burnout occurs. At that point the metal temperature is 250◦ C and h inside the tube is 9600 W/m2 K. What is the wall temperature at burnout? 9.32 At about what velocity of liquid metal flow does burnout occur in Problem 9.31 if the metal is mercury? 9.33 Explain, in physical terms, why eqns. (9.23) and (9.24), instead of differing by a factor of two, are almost equal. How do these equations change when H is large? 9.34 A liquid enters the heated section of a pipe at a location z = 0 ˆ with a specific enthalpy hin . If the wall heat flux is qw and the pipe diameter is D , show that the enthalpy a distance z = L downstream is πD L ˆˆ qw dz. h = hin + ˙ m0 ˆˆ Since the quality may be defined as x ≡ (h − hf ,sat ) hfg , show that for constant qw x= 9.35 ˆ ˆ hin − hf ,sat hfg + 4qw L GD Consider again the x-ray monochrometer described in Problem 7.44. Suppose now that the mass flow rate of liquid nitrogen is 0.023 kg/s, that the nitrogen is saturated at 110 K when it enters the heated section, and that the passage horizontal. Estimate the quality and the wall temperature at end of the References heated section if F = 4.70 for nitrogen in eqns. (9.50). As before, assume the silicon to conduct well enough that the heat load is distributed uniformly over the surface of the passage. 9.36 Use data from Appendix A and Sect. 9.1 to calculate the merit number, M , for the following potential heat-pipe working fluids over the range 200 K to 600 K in 100 K increments: water, mercury, methanol, ammonia, and HCFC-22. If data are unavailable for a fluid in some range, indicate so. What fluids are best suited for particular temperature ranges? References [9.1] S. Nukiyama. The maximum and minimum values of the heat q transmitted from metal to boiling water under atmospheric pressure. J. Jap. Soc. Mech. Eng., 37:367–374, 1934. (transl.: Int. J. Heat Mass Transfer, vol. 9, 1966, pp. 1419–1433). [9.2] T. B. Drew and C. Mueller. Boiling. Trans. AIChE, 33:449, 1937. [9.3] International Association for the Properties of Water and Steam. Release on surface tension of ordinary water substance. Technical report, September 1994. 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Mass Transfer, 22:629–632, 1979. [9.63] P. D. Dunn and D. A. Reay. Heat Pipes. Pergamon Press Ltd., Oxford, UK, 4th edition, 1994. Part IV Thermal Radiation Heat Transfer 523 10. Radiative heat transfer The sun that shines from Heaven shines but warm, And, lo, I lie between that sun and thee: The heat I have from thence doth little harm, Thine eye darts forth the fire that burneth me: And were I not immortal, life were done Between this heavenly and earthly sun. Venus and Adonis, Wm. Shakespeare, 1593 10.1 The problem of radiative exchange Chapter 1 described the elementary mechanisms of heat radiation. Before we proceed, you should reflect upon what you remember about the following key ideas from Chapter 1: • • • • • • • • Electromagnetic wave spectrum Heat radiation & infrared radiation Black body Absorptance, α Reflectance, ρ Transmittance, τ α+ρ+τ =1 e(T ) and eλ (T ) for black bodies • • • • • • • The Stefan-Boltzmann law Wien’s law & Planck’s law Radiant heat exchange Configuration factor, F1–2 Emittance, ε Transfer factor, F1–2 Radiation shielding The additional concept of a radiation heat transfer coefficient was developed in Section 2.3. We presume that all these concepts are understood. The heat exchange problem Figure 10.1 shows two arbitrary surfaces radiating energy to one another. The net heat exchange, Qnet , from the hotter surface (1) to the cooler 525 526 Radiative heat transfer §10.1 Figure 10.1 Thermal radiation between two arbitrary surfaces. surface (2) depends on the following influences: • T1 and T2 . • The areas of (1) and (2), A1 and A2 . • The shape, orientation, and spacing of (1) and (2). • The radiative properties of the surfaces. • Additional surfaces in the environment, whose radiation may be reflected by one surface to the other. • The medium between (1) and (2) if it absorbs, emits, or “reflects” radiation. (When the medium is air, we can usually neglect these effects.) If surfaces (1) and (2) are black, if they are surrounded by air, and if no heat flows between them by conduction or convection, then only the The problem of radiative exchange §10.1 527 first three considerations are involved in determining Qnet . We saw some elementary examples of how this could be done in Chapter 1, leading to 4 4 Qnet = A1 F1–2 σ T1 − T2 (10.1) The last three considerations complicate the problem considerably. In Chapter 1, we saw that these nonideal factors are sometimes included in a transfer factor F1–2 , such that 4 4 Qnet = A1 F1–2 σ T1 − T2 (10.2) Before we undertake the problem of evaluating heat exchange among real bodies, we need several definitions. Some definitions Emittance. A real body at temperature T does not emit with the black body emissive power eb = σ T 4 but rather with some fraction, ε, of eb . The same is true of the monochromatic emissive power, eλ (T ), which is always lower for a real body than the black body value given by Planck’s law, eqn. (1.30). Thus, we define either the monochromatic emittance, ελ : ελ ≡ eλ (λ, T ) eλb (λ, T ) (10.3) or the total emittance, ε: ∞ ε≡ e(T ) = eb (T ) 0 ∞ eλ (λ, T ) dλ σT4 = 0 ελ eλb (λ, T ) dλ σT4 (10.4) For real bodies, both ε and ελ are greater than zero and less than one; for black bodies, ε = ελ = 1. The emittance is determined entirely by the properties of the surface of the particular body and its temperature. It is independent of the environment of the body. Table 10.1 lists typical values of the total emittance for a variety of substances. Notice that most metals have quite low emittances, unless they are oxidized. Most nonmetals have emittances that are quite high— approaching the black body limit of unity. One particular kind of surface behavior is that for which ελ is independent of λ. We call such a surface a gray body. The monochromatic emissive power, eλ (T ), for a gray body is a constant fraction, ε, of ebλ (T ), as indicated in the inset of Fig. 10.2. In other words, for a gray body, ελ = ε. Table 10.1 Total emittances for a variety of surfaces [10.1] Metals Surface Nonmetals ◦ Temp. ( C) Aluminum Polished, 98% pure 200−600 Commercial sheet 90 Heavily oxidized 90−540 Brass Highly polished 260 Dull plate 40−260 Oxidized 40−260 Copper Highly polished electrolytic 90 Slightly polished to dull 40 Black oxidized 40 Gold: pure, polished 90−600 Iron and steel Mild steel, polished 150−480 Steel, polished 40−260 Sheet steel, rolled 40 Sheet steel, strong 40 rough oxide Cast iron, oxidized 40−260 Iron, rusted 40 Wrought iron, smooth 40 Wrought iron, dull oxidized 20−360 Stainless, polished 40 Stainless, after repeated 230−900 heating Lead Polished 40−260 Oxidized 40−200 Mercury: pure, clean 40−90 Platinum Pure, polished plate 200−590 Oxidized at 590◦ C 260−590 Drawn wire and strips 40−1370 Silver 200 Tin 40−90 Tungsten Filament 540−1090 Filament 2760 528 ε 0.04–0.06 0.09 0.20–0.33 0.03 0.22 0.46–0.56 0.02 0.12–0.15 0.76 0.02–0.035 0.14–0.32 0.07–0.10 0.66 0.80 0.57–0.66 0.61–0.85 0.35 0.94 0.07–0.17 0.50–0.70 0.05–0.08 0.63 0.10–0.12 0.05–0.10 0.07–0.11 0.04–0.19 0.01–0.04 0.05 0.11–0.16 0.39 Surface Asbestos Brick Red, rough Silica Fireclay Ordinary refractory Magnesite refractory White refractory Carbon Filament Lampsoot Concrete, rough Glass Smooth Quartz glass (2 mm) Pyrex Gypsum Ice Limestone Marble Mica Paints Black gloss White paint Lacquer Various oil paints Red lead Paper White Other colors Roofing Plaster, rough lime Quartz Rubber Snow Water, thickness ≥0.1 mm Wood Oak, planed Temp. (◦ C) ε 40 0.93–0.97 40 980 980 1090 980 1090 0.93 0.80–0.85 0.75 0.59 0.38 0.29 1040−1430 40 40 0.53 0.95 0.94 40 260−540 260−540 40 0 0.94 0.96–0.66 0.94–0.74 0.80–0.90 0.97–0.98 400−260 40 40 0.95–0.83 0.93–0.95 0.75 40 40 40 40 90 40 40 40 40−260 100−1000 40 10−20 40 40 20 0.90 0.89–0.97 0.80–0.95 0.92–0.96 0.93 0.95–0.98 0.92–0.94 0.91 0.92 0.89–0.58 0.86–0.94 0.82 0.96 0.80–0.90 0.90 §10.1 The problem of radiative exchange 529 Figure 10.2 Comparison of the sun’s energy as typically seen through the earth’s atmosphere with that of a black body having the same mean temperature, size, and distance from the earth. (Notice that eλ , just outside the earth’s atmosphere, is far less than on the surface of the sun because the radiation has spread out over a much greater area.) No real body is gray, but many exhibit approximately gray behavior. We see in Fig. 10.2, for example, that the sun appears to us on earth as an approximately gray body with an emittance of approximately 0.6. Some materials—for example, copper, aluminum oxide, and certain paints—are actually pretty close to being gray surfaces at normal temperatures. Yet the emittance of most common materials and coatings varies with wavelength in the thermal range. The total emittance accounts for this behavior at a particular temperature. By using it, we can write the emissive power as if the body were gray, without integrating over wavelength: e(T ) = ε σ T 4 (10.5) We shall use this type of “gray body approximation” often in this chapter. 530 Radiative heat transfer Specular or mirror-like reflection of incoming ray. Reflection which is between diffuse and specular (a real surface). §10.1 Diffuse radiation in which directions of departure are uninfluenced by incoming ray angle, θ . Figure 10.3 Specular and diffuse reflection of radiation. (Arrows indicate magnitude of the heat flux in the directions indicated.) In situations where surfaces at very different temperatures are involved, the wavelength dependence of ελ must be dealt with explicitly. This occurs, for example, when sunlight heats objects here on earth. Solar radiation (from a high temperature source) is on visible wavelengths, whereas radiation from low temperature objects on earth is mainly in the infrared range. We look at this issue further in the next section. Diffuse and specular emittance and reflection. The energy emitted by a non-black surface, together with that portion of an incoming ray of energy that is reflected by the surface, may leave the body diffusely or specularly, as shown in Fig. 10.3. That energy may also be emitted or reflected in a way that lies between these limits. A mirror reflects visible radiation in an almost perfectly specular fashion. (The “reflection” of a billiard ball as it rebounds from the side of a pool table is also specular.) When reflection or emission is diffuse, there is no preferred direction for outgoing rays. Black body emission is always diffuse. The character of the emittance or reflectance of a surface will normally change with the wavelength of the radiation. If we take account of both directional and spectral characteristics, then properties like emittance and reflectance depend on wavelength, temperature, and angles of incidence and/or departure. In this chapter, we shall assume diffuse §10.1 The problem of radiative exchange 531 behavior for most surfaces. This approximation works well for many problems in engineering, in part because most tabulated spectral and total emittances have been averaged over all angles (in which case they are properly called hemispherical properties). Experiment 10.1 Obtain a flashlight with as narrow a spot focus as you can find. Direct it at an angle onto a mirror, onto the surface of a bowl filled with sugar, and onto a variety of other surfaces, all in a darkened room. In each case, move the palm of your hand around the surface of an imaginary hemisphere centered on the point where the spot touches the surface. Notice how your palm is illuminated, and categorize the kind of reflectance of each surface—at least in the range of visible wavelengths. Intensity of radiation. To account for the effects of geometry on radiant exchange, we must think about how angles of orientation affect the radiation between surfaces. Consider radiation from a circular surface element, dA, as shown at the top of Fig. 10.4. If the element is black, the radiation that it emits is indistinguishable from that which would be emitted from a black cavity at the same temperature, and that radiation is diffuse — the same in all directions. If it were non-black but diffuse, the heat flux leaving the surface would again be independent of direction. Thus, the rate at which energy is emitted in any direction from this diffuse element is proportional to the projected area of dA normal to the direction of view, as shown in the upper right side of Fig. 10.4. If an aperture of area dAa is placed at a radius r and angle θ from dA and is normal to the radius, it will see dA as having an area cos θ dA. The energy dAa receives will depend on the solid angle,1 dω, it subtends. Radiation that leaves dA within the solid angle dω stays within dω as it travels to dAa . Hence, we define a quantity called the intensity of radiation, i (W/m2 ·steradian) using an energy conservation statement: radiant energy from dA dQoutgoing = (i dω)(cos θ dA) = that is intercepted by dA a 1 (10.6) The unit of solid angle is the steradian. One steradian is the solid angle subtended by a spherical segment whose area equals the square of its radius. A full sphere therefore subtends 4π r 2 /r 2 = 4π steradians. The aperture dAa subtends dω = dAa r 2 . 532 Radiative heat transfer §10.1 Figure 10.4 Radiation intensity through a unit sphere. Notice that while the heat flux from dA decreases with θ (as indicated on the right side of Fig. 10.4), the intensity of radiation from a diffuse surface is uniform in all directions. Finally, we compute i in terms of the heat flux from dA by dividing eqn. (10.6) by dA and integrating over the entire hemisphere. For convenience we set r = 1, and we note (see Fig. 10.4) that dω = sin θ dθdφ. 2π qoutgoing = π /2 φ=0 θ =0 i cos θ (sin θ dθdφ) = π i (10.7a) Kirchhoff’s law §10.2 533 In the particular case of a black body, ib = σT4 eb = = fn (T only) π π (10.7b) For a given wavelength, we likewise define the monochromatic intensity iλ = 10.2 eλ = fn (T , λ) π (10.7c) Kirchhoff’s law The problem of predicting α The total emittance, ε, of a surface is determined only by the physical properties and temperature of that surface, as can be seen from eqn. (10.4). The total absorptance, α, on the other hand, depends on the source from which the surface absorbs radiation, as well as the surface’s own characteristics. This happens because the surface may absorb some wavelengths better than others. Thus, the total absorptance will depend on the way that incoming radiation is distributed in wavelength. And that distribution, in turn, depends on the temperature and physical properties of the surface or surfaces from which radiation is absorbed. The total absorptance α thus depends on the physical properties and temperatures of all bodies involved in the heat exchange process. Kirchhoff’s law2 is an expression that allows α to be determined under certain restrictions. Kirchhoff’s law Kirchhoff’s law is a relationship between the monochromatic, directional emittance and the monochromatic, directional absorptance for a surface that is in thermodynamic equilibrium with its surroundings ελ (T , θ, φ) = αλ (T , θ, φ) exact form of Kirchhoff’s law (10.8a) Kirchhoff’s law states that a body in thermodynamic equilibrium emits as much energy as it absorbs in each direction and at each wavelength. If 2 Gustav Robert Kirchhoff (1824–1887) developed important new ideas in electrical circuit theory, thermal physics, spectroscopy, and astronomy. He formulated this particular “Kirchhoff’s Law” when he was only 25. He and Robert Bunsen (inventor of the Bunsen burner) subsequently went on to do significant work on radiation from gases. 534 Radiative heat transfer §10.2 this were not so, for example, a body might absorb more energy than it emits in one direction, θ1 , and might also emit more than it absorbs in another direction, θ2 . The body would thus pump heat out of its surroundings from the first direction, θ1 , and into its surroundings in the second direction, θ2 . Since whatever matter lies in the first direction would be refrigerated without any work input, the Second Law of Thermodynamics would be violated. Similar arguments can be built for the wavelength dependence. In essence, then, Kirchhoff’s law is a consequence of the laws of thermodynamics. For a diffuse body, the emittance and absorptance do not depend on the angles, and Kirchhoff’s law becomes diffuse form of Kirchhoff’s law ελ (T ) = αλ (T ) (10.8b) If, in addition, the body is gray, Kirchhoff’s law is further simplified ε (T ) = α (T ) diffuse, gray form of Kirchhoff’s law (10.8c) Equation (10.8c) is the most widely used form of Kirchhoff’s law. Yet, it is a somewhat dangerous result, since many surfaces are not even approximately gray. If radiation is emitted on wavelengths much different from those that are absorbed, then a non-gray surface’s variation of ελ and αλ with wavelength will matter, as we discuss next. Total absorptance during radiant exchange Let us restrict our attention to diffuse surfaces, so that eqn. (10.8b) is the appropriate form of Kirchhoff’s law. Consider two plates as shown in Fig. 10.5. Let the plate at T1 be non-black and that at T2 be black. Then net heat transfer from plate 1 to plate 2 is the difference between what plate 1 emits and what it absorbs. Since all the radiation reaching plate 1 comes from a black source at T2 , we may write qnet = ∞ 0 ελ1 (T1 ) eλb (T1 ) dλ − emitted by plate 1 ∞ 0 αλ1 (T1 ) eλb (T2 ) dλ (10.9) radiation from plate 2 absorbed by plate 1 From eqn. (10.4), we may write the first integral in terms of total emit4 tance, as ε1 σ T1 . We define the total absorptance, α1 (T1 , T2 ), as the sec- Kirchhoff’s law §10.2 535 Figure 10.5 Heat transfer between two infinite parallel plates. 4 ond integral divided by σ T2 . Hence, qnet = 4 ε1 (T1 )σ T1 emitted by plate 1 4 − α1 (T1 , T2 )σ T2 (10.10) absorbed by plate 1 We see that the total absorptance depends on T2 , as well as T1 . Why does total absorptance depend on both temperatures? The dependence on T1 is simply because αλ1 is a property of plate 1 that may be temperature dependent. The dependence on T2 is because the spectrum of radiation from plate 2 depends on the temperature of plate 2 according to Planck’s law, as was shown in Fig. 1.15. As a typical example, consider solar radiation incident on a warm roof, painted black. From Table 10.1, we see that ε is on the order of 0.94. It turns out that α is just about the same. If we repaint the roof white, ε will not change noticeably. However, much of the energy arriving from the sun is carried in visible wavelengths, owing to the sun’s very high temperature (about 5800 K).3 Our eyes tell us that white paint reflects sunlight very strongly in these wavelengths, and indeed this is the case — 80 to 90% of the sunlight is reflected. The absorptance of 3 Ninety percent of the sun’s energy is on wavelengths between 0.33 and 2.2 µm (see Figure 10.2). For a black object at 300 K, 90% of the radiant energy is between 6.3 and 42 µm, in the infrared. Radiative heat transfer 536 §10.3 white paint to energy from the sun is only 0.1 to 0.2 — much less than ε for the energy it emits, which is mainly at infrared wavelengths. For both paints, eqn. (10.8b) applies. However, in this situation, eqn. (10.8c) is only accurate for the black paint. The gray body approximation Let us consider our facing plates again. If plate 1 is painted with white paint, and plate 2 is at a temperature near plate 1 (say T1 = 400 K and T2 = 300 K, to be specific), then the incoming radiation from plate 2 has a wavelength distribution not too dissimilar to plate 1. We might be very comfortable approximating ε1 α1 . The net heat flux between the plates can be expressed very simply 4 4 qnet = ε1 σ T1 − α1 (T1 , T2 )σ T2 4 4 ε1 σ T1 − ε1 σ T2 4 4 = ε1 σ T1 − T2 (10.11) In effect, we are approximating plate 1 as a gray body. In general, the simplest first estimate for total absorptance is the diffuse, gray body approximation, eqn. (10.8c). It will be accurate either if the monochromatic emittance does not vary strongly with wavelength or if the bodies exchanging radiation are at similar absolute temperatures. More advanced texts describe techniques for calculating total absorptance (by integration) in other situations [10.2, 10.3]. One situation in which eqn. (10.8c) should always be mistrusted is when solar radiation is absorbed by a low temperature object — a space vehicle or something on earth’s surface, say. In this case, the best first approximation is to set total absorptance to a value for visible wavelengths of radiation (near 0.5 µm). Total emittance may be taken at the object’s actual temperature, typically for infrared wavelengths. We return to solar absorptance in Section 10.6. 10.3 Radiant heat exchange between two finite black bodies Let us now return to the purely geometric problem of evaluating the view factor, F1–2 . Although the evaluation of F1–2 is also used in the calculation §10.3 Radiant heat exchange between two finite black bodies Figure 10.6 Some configurations for which the value of the view factor is immediately apparent. of heat exchange among diffuse, nonblack bodies, it is the only correction of the Stefan-Boltzmann law that we need for black bodies. Some evident results. Figure 10.6 shows three elementary situations in which the value of F1–2 is evident using just the definition: F1–2 ≡ fraction of field of view of (1) occupied by (2). When the surfaces are each isothermal and diffuse, this corresponds to F1–2 = fraction of energy leaving (1) that reaches (2) A second apparent result in regard to the view factor is that all the energy leaving a body (1) reaches something else. Thus, conservation of energy requires 1 = F1–1 + F1–2 + F1–3 + · · · + F1–n (10.12) where (2), (3),…,(n) are all of the bodies in the neighborhood of (1). Figure 10.7 shows a representative situation in which a body (1) is surrounded by three other bodies. It sees all three bodies, but it also views 537 538 Radiative heat transfer §10.3 Figure 10.7 A body (1) that views three other bodies and itself as well. itself, in part. This accounts for the inclusion of the view factor, F1–1 in eqn. (10.12). By the same token, it should also be apparent from Fig. 10.7 that the kind of sum expressed by eqn. (10.12) would also be true for any subset of the bodies seen by surface 1. Thus, F1–(2+3) = F1–2 + F1–3 Of course, such a sum makes sense only when all the view factors are based on the same viewing surface (surface 1 in this case). One might be tempted to write this sort of sum in the opposite direction, but it would clearly be untrue, F(2+3)–1 ≠ F2–1 + F3–1 , since each view factor is for a different viewing surface—(2 + 3), 2, and 3, in this case. View factor reciprocity. So far, we have referred to the net radiation from black surface (1) to black surface (2) as Qnet . Let us refine our notation a bit, and call this Qnet1–2 : 4 4 Qnet1–2 = A1 F1–2 σ T1 − T2 (10.13) Likewise, the net radiation from (2) to (1) is 4 4 Qnet2–1 = A2 F2–1 σ T2 − T1 (10.14) Radiant heat exchange between two finite black bodies §10.3 Of course, Qnet1–2 = −Qnet2–1 . It follows that 4 4 4 4 A1 F1–2 σ T1 − T2 = −A2 F2–1 σ T2 − T1 or A1 F1–2 = A2 F2–1 (10.15) This result, called view factor reciprocity, is very useful in calculations. Example 10.1 A jet of liquid metal at 2000◦ C pours from a crucible. It is 3 mm in diameter. A long cylindrical radiation shield, 5 cm diameter, surrounds the jet through an angle of 330◦ , but there is a 30◦ slit in it. The jet and the shield radiate as black bodies. They sit in a room at 30◦ C, and the shield has a temperature of 700◦ C. Calculate the net heat transfer: from the jet to the room through the slit; from the jet to the shield; and from the inside of the shield to the room. Solution. By inspection, we see that Fjet–room = 30/360 = 0.08333 and Fjet–shield = 330/360 = 0.9167. Thus, 4 4 Qnetjet–room = Ajet Fjet–room σ Tjet − Troom = π (0.003) m2 m length (0.08333)(5.67 × 10−8 ) 22734 − 3034 = 1, 188 W/m Likewise, 4 4 Qnetjet–shield = Ajet Fjet–shield σ Tjet − Tshield = π (0.003) m2 m length (0.9167)(5.67 × 10−8 ) 22734 − 9734 = 12, 637 W/m The heat absorbed by the shield leaves it by radiation and convection to the room. (A balance of these effects can be used to calculate the shield temperature given here.) To find the radiation from the inside of the shield to the room, we need Fshield–room . Since any radiation passing out of the slit goes to the 539 540 Radiative heat transfer §10.3 room, we can find this view factor equating view factors to the room with view factors to the slit. The slit’s area is Aslit = π (0.05)30/360 = 0.01309 m2 /m length. Hence, using our reciprocity and summation rules, eqns. (10.12) and (10.15), Fslit–jet = Ajet π (0.003) (0.0833) = 0.0600 Fjet–room = Aslit 0.01309 Fslit–shield = 1 − Fslit–jet − Fslit–slit = 1 − 0.0600 − 0 = 0.940 0 Fshield–room Aslit = Fslit–shield Ashield 0.01309 (0.940) = 0.08545 = π (0.05)(330)/(360) Hence, for heat transfer from the inside of the shield only, 4 4 Qnetshield–room = Ashield Fshield–room σ Tshield − Troom π (0.05)330 (0.08545)(5.67 × 10−8 ) 9734 − 3034 360 = 619 W/m = Both the jet and the inside of the shield have relatively small view factors to the room, so that comparatively little heat is lost through the slit. Calculation of the black-body view factor, F1–2 . Consider two elements, dA1 and dA2 , of larger black bodies (1) and (2), as shown in Fig. 10.8. Body (1) and body (2) are each isothermal. Since element dA2 subtends a solid angle dω1 , we use eqn. (10.6) to write dQ1 to 2 = (i1 dω1 )(cos β1 dA1 ) But from eqn. (10.7b), i1 = 4 σ T1 π Note that because black bodies radiate diffusely, i1 does not vary with angle; and because these bodies are isothermal, it does not vary with position. The element of solid angle is given by dω1 = cos β2 dA2 s2 Radiant heat exchange between two finite black bodies §10.3 Figure 10.8 Radiant exchange between two black elements that are part of the bodies (1) and (2). where s is the distance from (1) to (2) and cos β2 enters because dA2 is not necessarily normal to s . Thus, dQ1 to 2 = 4 σ T1 π cos β1 cos β2 dA1 dA2 s2 4 σ T2 π cos β2 cos β1 dA2 dA1 s2 By the same token, dQ2 to 1 = Then 4 4 Qnet1–2 = σ T1 − T2 A1 A2 cos β1 cos β2 dA1 dA2 π s2 (10.16) The view factors F1–2 and F2–1 are immediately obtainable from eqn. 4 4 (10.16). If we compare this result with Qnet1–2 = A1 F1–2 σ (T1 − T2 ), we get F1–2 = 1 A1 A1 A2 cos β1 cos β2 dA1 dA2 π s2 (10.17a) 541 542 Radiative heat transfer §10.3 From the inherent symmetry of the problem, we can also write F2–1 = 1 A2 A2 A1 cos β2 cos β1 dA2 dA1 π s2 (10.17b) You can easily see that eqns. (10.17a) and (10.17b) are consistent with the reciprocity relation, eqn. (10.15). The direct evaluation of F1–2 from eqn. (10.17a) becomes fairly involved, even for the simplest configurations. Siegel and Howell [10.4] provide a comprehensive discussion of such calculations and a large catalog of their results. Howell [10.5] gives an even more extensive tabulation of view factor equations, which is now available on the World Wide Web. At present, no other reference is as complete. We list some typical expressions for view factors in Tables 10.2 and 10.3. Table 10.2 gives calculated values of F1–2 for two-dimensional bodies—various configurations of cylinders and strips that approach infinite length. Table 10.3 gives F1–2 for some three-dimensional configurations. Many view factors have been evaluated numerically and presented in graphical form for easy reference. Figure 10.9, for example, includes graphs for configurations 1, 2, and 3 from Table 10.3. The reader should study these results and be sure that the trends they show make sense. Is it clear, for example, that F1–2 → constant, which is < 1 in each case, as the abscissa becomes large? Can you locate the configuration on the right-hand side of Fig. 10.6 in Fig. 10.9? And so forth. Figure 10.10 shows view factors for another kind of configuration— one in which one area is very small in comparison with the other one. Many solutions like this exist because they are a bit less difficult to calculate, and they can often be very useful in practice. Example 10.2 A heater (h) as shown in Fig. 10.11 radiates to the partially conical shield (s) that surrounds it. If the heater and shield are black, calculate the net heat transfer from the heater to the shield. Solution. First imagine a plane (i) laid across the open top of the shield: Fh−s + Fh−i = 1 But Fh−i can be obtained from Fig. 10.9 or case 3 of Table 10.3, Table 10.2 View factors for a variety of two-dimensional configurations (infinite in extent normal to the paper) Configuration Equation 1. F1–2 = F2–1 = h w 1+ 2 − h w 2. F1–2 = F2–1 = 1 − sin(α/2) 3. F1–2 ⎡ h 1⎣ 1+ − = 2 w h 1+ w 2 ⎤ ⎦ 4. F1–2 = (A1 + A2 − A3 ) 2A1 5. F1–2 = 6. r b a tan−1 − tan−1 b−a c c Let X = 1 + s/D. F1–2 = F2–1 = 7. F1–2 = 1, F2–2 Then: 1 π X 2 − 1 + sin−1 1 −X X r1 , and r2 r1 = 1 − F2–1 = 1 − r2 F2–1 = 543 Table 10.3 View factors for some three-dimensional configurations Configuration 1. Equation Let X = a/c and Y = b/c . Then: F1–2 ⎧ (1 + X 2 )(1 + Y 2 ) 2⎨ ln = ⎩ π XY 1 + X2 + Y 2 1/2 − X tan−1 X − Y tan−1 Y + X 1 + Y 2 tan−1 2. ⎫ ⎬ X Y √ + Y 1 + X 2 tan−1 √ 1 + Y2 1 + X2 ⎭ Let H = h/ and W = w/ . Then: F1–2 ⎧ ⎪ 1⎨ 1 W tan−1 − H 2 + W 2 tan−1 H 2 + W 2 = ⎪ πW ⎩ W ⎧ 1 1 ⎨ (1 + W 2 )(1 + H 2 ) + H tan−1 + ln H 4⎩ 1 + W 2 + H2 × W (1 + W + H ) (1 + W 2 )(W 2 + H 2 ) 2 2 2 W2 H (1 + H + W ) (1 + H 2 )(H 2 + W 2 ) 2 2 2 −1/2 ⎫⎫ H 2 ⎬⎪ ⎬ ⎭⎪ ⎭ 3. 2 Let R1 = r1 /h, R2 = r2 /h, and X = 1 + 1 + R2 F1–2 = 2 R1 . Then: 1 X − X 2 − 4(R2 /R1 )2 2 4. Concentric spheres: F1–2 = 1, 544 F2–1 = (r1 /r2 )2 , F2–2 = 1 − (r1 /r2 )2 545 Figure 10.9 The view factors for configurations shown in Table 10.3 Figure 10.10 The view factor for three very small surfaces A2 ). “looking at” three large surfaces (A1 546 Radiant heat exchange between two finite black bodies §10.3 Figure 10.11 Heat transfer from a disc heater to its radiation shield. for R1 = r1 /h = 5/20 = 0.25 and R2 = r2 /h = 10/20 = 0.5. The result is Fh−i = 0.192. Then Fh−s = 1 − 0.192 = 0.808 Thus, 4 4 Qneth−s = Ah Fh−s σ Th − Ts π = (0.1)2 (0.808)(5.67 × 10−8 ) (1200 + 273)4 − 3734 4 = 1687 W Example 10.3 Suppose that the shield in Example 10.2 were heating the region where the heater is presently located. What would Fs −h be? Solution. From eqn. (10.15) we have As Fs −h = Ah Fh−s But the frustrum-shaped shield has an area of As = π (r1 + r2 ) h2 + (r2 − r1 )2 = π (0.05 + 0.1) 0.22 + 0.052 = 0.09715 m2 547 Radiative heat transfer 548 §10.3 and Ah = π (0.1)2 = 0.007854 m2 4 so Fs −h = 0.007854 (0.808) = 0.0653 0.09715 Example 10.4 Find F1–2 for the configuration of two offset squares of area A, as shown in Fig. 10.12. Solution. Consider two fictitious areas 3 and 4 as indicated by the dotted lines. The view factor between the combined areas, (1+3) and (2+4), can be obtained from Fig. 10.9. In addition, we can write that view factor in terms of the unknown F1–2 and other known view factors: (2A)F(1+3)–(4+2) = AF1–4 + AF1–2 + AF3–4 + AF3–2 2F(1+3)–(4+2) = 2F1–4 + 2F1–2 F1–2 = F(1+3)–(4+2) − F1–4 And F(1+3)–(4+2) can be read from Fig. 10.9 (at φ = 90, w/ and h/ = 1/2) as 0.245 and F1–4 as 0.20. Thus, F1–2 = (0.245 − 0.20) = 0.045 Figure 10.12 Radiation between two offset perpendicular squares. = 1/2, Heat transfer among gray bodies §10.4 10.4 549 Heat transfer among gray bodies Electrical analogy for gray body heat exchange An electric circuit analogy for heat exchange among diffuse gray bodies was developed by Oppenheim [10.6] in 1956. It begins with the definition of two new quantities: flux of energy that irradiates the H (W/m2 ) ≡ irradiance = surface and total flux of radiative energy B (W/m2 ) ≡ radiosity = away from the surface The radiosity can be expressed as the sum of the irradiated energy that is reflected by the surface and the radiation emitted by it. Thus, B = ρH + εeb (10.18) We can immediately write the net heat flux leaving any particular surface as the difference between B and H for that surface. Then, with the help of eqn. (10.18), we get qnet = B − H = B − B − εeb ρ (10.19) This can be rearranged as qnet = ε 1−ρ eb − B ρ ρ (10.20) If the surface is opaque (τ = 0), 1 − ρ = α, and if it is gray, α = ε. Then, eqn. (10.20) gives qnet A = Qnet = eb − B eb − B = ρ/εA (1 − ε) εA (10.21) Equation (10.21) is a form of Ohm’s law, which tells us that (eb − B) can be viewed as a driving potential for transferring heat away from a surface through an effective surface resistance, (1 − ε)/εA. Now consider heat transfer from one infinite gray plate to another parallel to it. Radiant energy flows past an imaginary surface, parallel to the first infinite plate and quite close to it, as shown as a dotted line 550 Radiative heat transfer §10.4 Figure 10.13 The electrical circuit analogy for radiation between two gray infinite plates. in Fig. 10.13. If the gray plate is diffuse, its radiation has the same geometrical distribution as that from a black body, and it will travel to other objects in the same way that black body radiation would. Therefore, we can treat the radiation leaving the imaginary surface — the radiosity, that is — as though it were black body radiation travelling to an imaginary surface above the other plate. Thus, by analogy to eqn. (10.13), Qnet1–2 = A1 F1–2 (B1 − B2 ) = B1 − B2 (10.22) 1 A1 F1–2 where the final fraction shows that this is also a form of Ohm’s law: the radiosity difference (B1 − B2 ), can be said to drive heat through the geometrical resistance, 1/A1 F1–2 , that describes the field of view between the two surfaces. When two gray surfaces exchange radiation only with each other, the net radiation flows through a surface resistance for each surface and a geometric resistance for the configuration. The electrical circuit shown in Fig. 10.13 expresses the analogy and gives us means for calculating Qnet1–2 from Ohm’s law. Recalling that eb = σ T 4 , we obtain Qnet1–2 = eb1 − eb2 resistances = 4 4 σ T1 − T2 1−ε εA + 1 1 A1 F1–2 + (10.23) 1−ε εA 2 For the particular case of infinite parallel plates, F1–2 = 1 and A1 = A2 Heat transfer among gray bodies §10.4 551 (Fig. 10.6), and, with qnet1–2 = Qnet1–2 /A1 , we find qnet1–2 = 1 1 ε1 + 1 ε2 4 4 σ T1 − T2 (10.24) −1 Comparing eqn. (10.24) with eqn. (10.2), we may identify F1–2 = 1 1 1 + −1 ε1 ε2 (10.25) for infinite parallel plates. Notice, too, that if the plates are both black (ε1 = ε2 = 1), then both surface resistances are zero and F1–2 = 1 = F1–2 which, of course, is what we would have expected. Example 10.5 One gray body enclosed by another Evaluate the heat transfer and the transfer factor for one gray body enclosed by another, as shown in Fig. 10.14. Solution. The electrical circuit analogy is exactly the same as that shown in Fig. 10.13, and F1–2 is still unity. Therefore, with eqn. (10.23), Qnet1–2 = A1 qnet1–2 = 4 4 σ T1 − T2 1 − ε1 ε1 A1 + 1 A1 + 1 − ε2 (10.26) ε2 A2 Figure 10.14 Heat transfer between an enclosed body and the body surrounding it. 552 Radiative heat transfer §10.4 The transfer factor may again be identified by comparison to eqn. (10.2): Qnet1–2 = A1 1 A1 1 + ε1 A2 1 −1 ε2 4 4 σ T1 − T2 (10.27) =F1–2 This calculation assumes that body (1) does not view itself. Example 10.6 Transfer factor reciprocity Derive F2–1 for the enclosed bodies shown in Fig. 10.14. Solution. Qnet1–2 = −Qnet2–1 4 4 4 4 A1 F1–2 σ T1 − T2 = −A2 F2–1 σ T2 − T1 from which we obtain the reciprocity relationship for transfer factors: A1 F1–2 = A2 F2–1 (10.28) Hence, with the result of Example 10.5, we have F2–1 = Example 10.7 A1 1 F1–2 = 1 1 A2 A2 + −1 ε1 A1 ε2 (10.29) Small gray object in a large environment Derive F1–2 for a small gray object (1) in a large isothermal environment (2), the result that was given as eqn. (1.35). Solution. We may use eqn. (10.27) with A1 /A2 F1–2 = 1 A1 1 + ε1 A2 1 −1 ε2 ε1 1: (10.30) 1 Note that the same result is obtained for any value of A1 /A2 if the enclosure is black (ε2 = 1). A large enclosure does not reflect much radiation back to the small object, and therefore becomes like a perfect absorber of the small object’s radiation — a black body. Heat transfer among gray bodies §10.4 553 Additional two-body exchange problems Radiation shields. A radiation shield is a surface, usually of high reflectance, that is placed between a high-temperature source and its cooler environment. Earlier examples in this chapter and in Chapter 1 show how such a surface can reduce heat exchange. Let us now examine the role of reflectance (or emittance: ε = 1 − ρ ) in the performance of a radiation shield. Consider a gray body (1) surrounded by another gray body (2), as discussed in Example 10.5. Suppose now that a thin sheet of reflective material is placed between bodies (1) and (2) as a radiation shield. The sheet will reflect radiation arriving from body (1) back toward body (1); likewise, owing to its low emittance, it will radiate little energy to body (2). The radiation from body (1) to the inside of the shield and from the outside of the shield to body (2) are each two-body exchange problems, coupled by the shield temperature. We may put the various radiation resistances in series to find (see Problem 10.46) Qnet1–2 = 4 4 σ T1 − T2 1 − ε1 ε1 A1 + 1 A1 + 1 − ε2 ε2 A2 +2 1 − εs εs As + 1 (10.31) As added by shield assuming F1–s = Fs–2 = 1. Note that the radiation shield reduces Qnet1–2 more if its emittance is smaller, i.e., if it is highly reflective. Specular surfaces. The electrical circuit analogy that we have developed is for diffuse surfaces. If the surface reflection or emission has directional characteristics, different methods of analysis must be used [10.2]. One important special case deserves to be mentioned. If the two gray surfaces in Fig. 10.14 are diffuse emitters but are perfectly specular reflectors — that is, if they each have only mirror-like reflections — then the transfer factor becomes F1–2 = 1 1 1 + −1 ε1 ε2 for specularly reflecting bodies (10.32) This result is interestingly identical to eqn. (10.25) for parallel plates. Since parallel plates are a special case of the situation in Fig. 10.14, it follows that eqn. (10.25) is true for either specular or diffuse reflection. 554 Radiative heat transfer §10.4 Example 10.8 A physics experiment uses liquid nitrogen as a coolant. Saturated liquid nitrogen at 80 K flows through 6.35 mm O.D. stainless steel line (εl = 0.2) inside a vacuum chamber. The chamber walls are at Tc = 230 K and are at some distance from the line. Determine the heat gain of the line per unit length. If a second stainless steel tube, 12.7 mm in diameter, is placed around the line to act as radiation shield, to what rate is the heat gain reduced? Find the temperature of the shield. Solution. The nitrogen coolant will hold the surface of the line at essentially 80 K, since the thermal resistances of the tube wall and the internal convection or boiling process are small. Without the shield, we can model the line as a small object in a large enclosure, as in Example 10.7: 4 Qgain = (π Dl )εl σ Tc − Tl4 = π (0.00635)(0.2)(5.67 × 10−8 )(2304 − 804 ) = 0.624 W/m With the shield, eqn. (10.31) applies. Assuming that the chamber area Al ), is large compared to the shielded line (Ac Qgain = 4 σ Tc − Tl4 1 − εl εl Al + 1 Al + 1 − εc ε2 Ac +2 1 − εs εs As + 1 As neglect = π (0.00635)(5.67 × 10−8 )(2304 − 804 ) 1 − 0.2 0.2 +1 + 0.00635 2 0.0127 1 − 0.2 0.2 +1 = 0.328 W/m The radiation shield would cut the heat gain by 47%. The temperature of the shield, Ts , may be found using the heat loss and considering the heat flow from the chamber to the shield, with the shield now acting as a small object in a large enclosure: 4 4 Qgain = (π Ds )εs σ Tc − Ts 4 0.328 W/m = π (0.0127)(0.2)(5.67 × 10−8 ) 2304 − Ts Solving, we find Ts = 213 K. Heat transfer among gray bodies §10.4 The electrical circuit analogy when more than two gray bodies are involved in heat exchange Let us first consider a three-body transaction, as pictured in at the bottom and left-hand sides of Fig. 10.15. The triangular circuit for three bodies is not so easy to analyze as the in-line circuits obtained in twobody problems. The basic approach is to apply energy conservation at each radiosity node in the circuit, setting the net heat transfer from any one of the surfaces (which we designate as i) Qneti = ebi − Bi 1 − εi εi Ai (10.33a) equal to the sum of the net radiation to each of the other surfaces (call them j ) ⎛ ⎞ Bi − Bj ⎝ ⎠ (10.33b) Qneti = 1 Ai Fi−j j For the three body situation shown in Fig. 10.15, this leads to three equations Qnet1 , at node B1 : eb1 − B1 B1 − B2 B1 − B3 = + 1 − ε1 1 1 ε1 A1 A1 F1–2 A1 F1–3 Qnet2 , at node B2 : eb2 − B2 B2 − B1 B2 − B3 = + 1 − ε2 1 1 ε2 A2 A1 F1–2 A2 F2–3 Qnet3 , at node B3 : eb3 − B3 B3 − B1 B3 − B2 = + 1 − ε3 1 1 ε3 A3 A1 F1–3 A2 F2–3 (10.34a) (10.34b) (10.34c) If the temperatures T1 , T2 , and T3 are known (so that eb1 , eb2 , eb3 are known), these equations can be solved simultaneously for the three unknowns, B1 , B2 , and B3 . After they are solved, one can compute the net heat transfer to or from any body (i) from either of eqns. (10.33). Thus far, we have considered only cases in which the surface temperature is known for each body involved in the heat exchange process. Let us consider two other possibilities. 555 Radiative heat transfer 556 §10.4 Figure 10.15 The electrical circuit analogy for radiation among three gray surfaces. An insulated wall. If a wall is adiabatic, Qnet = 0 at that wall. For example, if wall (3) in Fig. 10.15 is insulated, then eqn. (10.33b) shows that eb3 = B3 . We can eliminate one leg of the circuit, as shown on the right-hand side of Fig. 10.15; likewise, the left-hand side of eqn. (10.34c) equals zero. This means that all radiation absorbed by an adiabatic wall is immediately reemitted. Such walls are sometimes called “refractory surfaces” in discussing thermal radiation. The circuit for an insulated wall can be treated as a series-parallel circuit, since all the heat from body (1) flows to body (2), even if it does so by travelling first to body (3). Then Qnet1 = eb1 − eb2 1 − ε1 + ε1 A1 1 1 1 /(A1 F1–3 ) + 1 /(A2 F2–3 ) + 1 + 1 − ε2 ε2 A2 1 /(A1 F1–2 ) (10.35) §10.4 Heat transfer among gray bodies A specified wall heat flux. The heat flux leaving a surface may be known, if, say, it is an electrically powered radiant heater. In this case, the lefthand side of one of eqns. (10.34) can be replaced with the surface’s known Qnet , via eqn. (10.33b). For the adiabatic wall case just considered, if surface (1) had a specified heat flux, then eqn. (10.35) could be solved for eb1 and the unknown temperature T1 . Example 10.9 Two very long strips 1 m wide and 2.40 m apart face each other, as shown in Fig. 10.16. (a) Find Qnet1–2 (W/m) if the surroundings are black and at 250 K. (b) Find Qnet1–2 (W/m) if they are connected by an insulated diffuse reflector between the edges on both sides. Also evaluate the temperature of the reflector in part (b). Solution. From Table 10.2, case 1, we find F1–2 = 0.2 = F2–1 . In addition, F2–3 = 1 − F2–1 = 0.8, irrespective of whether surface (3) represents the surroundings or the insulated shield. In case (a), the two nodal equations (10.34a) and (10.34b) become B1 − B3 B1 − B2 1451 − B1 = + 2.333 1/0.2 1/0.8 B2 − B3 B2 − B1 459.3 − B2 = + 1 1/0.2 1/0.8 Equation (10.34c) cannot be used directly for black surroundings, since ε3 = 1 and the surface resistance in the left-hand side denominator would be zero. But the numerator is also zero in this case, since eb3 = B3 for black surroundings. And since we now know 4 B3 = σ T3 = 221.5 W/m2 K, we can use it directly in the two equations above. Figure 10.16 Illustration for Example 10.9. 557 Radiative heat transfer 558 §10.4 Thus, B1 − 0.14 B2 −0.56(221.5) = 435.6 −B1 +10.00 B2 −4.00(221.5) = 2296.5 or B1 − 0.14 B2 = 559.6 −B1 + 10.00 B2 = 3182.5 so B1 = 612.1 W/m2 B2 = 379.5 W/m2 Thus, the net flow from (1) to (2) is quite small: Qnet1–2 = B1 − B2 = 46.53 W/m 1 /(A1 F1–2 ) Since each strip also loses heat to the surroundings, Qnet1 ≠ Qnet2 ≠ Qnet1–2 . For case (b), with the adiabatic shield in place, eqn. (10.34c) can be combined with the other two nodal equations: 0= B3 − B1 1/0.8 + B3 − B2 1/0.8 The three equations can be solved manually, by the use of determinants, or with a computerized matrix algebra package. The result is B1 = 987.7 W/m2 B2 = 657.4 W/m2 B3 = 822.6 W/m2 In this case, because surface (3) is adiabatic, all net heat transfer from surface (1) is to surface (2): Qnet1 = Qnet1–2 . Then, from eqn. (10.33a), we get Qnet1–2 = 987.7 − 657.4 987.7 − 822.6 + 1/(1)(0.2) 1/(1)(0.8) = 198 W/m Of course, because node (3) is insulated, it is much easier to use eqn. (10.35) to get Qnet1–2 : Qnet1–2 = 5.67 × 10−8 4004 − 3004 0.7 0.3 + 1 1 1/0.8 + 1/0.8 + + 0.2 0.5 0.5 = 198 W/m §10.4 Heat transfer among gray bodies 559 The result, of course, is the same. We note that the presence of the reflector increases the net heat flow from (1) to (2). The temperature of the reflector (3) is obtained from eqn. (10.33b) with Qnet3 = 0: 4 0 = eb3 − B3 = 5.67 × 10−8 T3 − 822.6 so T3 = 347 K Algebraic solution of multisurface enclosure problems An enclosure can consist of any number of surfaces that exchange radiation with one another. The evaluation of radiant heat transfer among these surfaces proceeds in essentially the same way as for three surfaces. For multisurface problems, however, the electrical circuit approach is less convenient than a formulation based on matrices. The matrix equations are usually solved on a computer. An enclosure formed by n surfaces is shown in Fig. 10.17. As before, we will assume that: • Each surface is diffuse, gray, and opaque, so that ε = α and ρ = 1−ε. • The temperature and net heat flux are uniform over each surface (more precisely, the radiosity must be uniform and the other properties are averages for each surface). Either temperature or flux must be specified on every surface. • The view factor, Fi−j , between any two surfaces i and j is known. • Conduction and convection within the enclosure can be neglected, and any fluid in the enclosure is transparent and nonradiating. We are interested in determining the heat fluxes at the surfaces where temperatures are specified, and vice versa. The rate of heat loss from the ith surface of the enclosure can conveniently be written in terms of the radiosity, Bi , and the irradiation, Hi , from eqns. (10.19) and (10.21) qneti = Bi − Hi = εi σ Ti4 − Bi 1 − εi (10.36) Radiative heat transfer 560 §10.4 Figure 10.17 An enclosure composed of n diffuse, gray surfaces. where Bi = ρi Hi + εi ebi = (1 − εi ) Hi + εi σ Ti4 (10.37) However, Ai Hi , the irradiating heat transfer incident on surface i, is the sum of energies reaching i from all other surfaces, including itself n n Ai Hi = Aj Bj Fj −i = j =1 Bj Ai Fi−j j =1 where we have used the reciprocity rule, Aj Fj −i = Ai Fi−j . Thus n Hi = Bj Fi−j (10.38) j =1 It follows from eqns. (10.37) and (10.38) that n Bj Fi−j + εi σ Ti4 Bi = (1 − εi ) (10.39) j =1 This equation applies to every surface, i = 1, . . . , n. When all the surface temperatures are specified, the result is a set of n linear equations for the n unknown radiosities. For numerical purposes, it is sometimes convenient to introduce the Kronecker delta, ⎧ ⎨1 for i = j (10.40) δij = ⎩0 for i ≠ j Heat transfer among gray bodies §10.4 561 and to rearrange eqn. (10.39) as n δij − (1 − εi )Fi−j Bj = εi σ Ti4 j =1 for i = 1, . . . , n (10.41) ≡Cij The radiosities are then found by inverting the matrix Cij . The rate of heat loss from the ith surface, Qneti = Ai qneti , can be obtained from eqn. (10.36). For those surfaces where heat fluxes are prescribed, we can eliminate the εi σ Ti4 term in eqn. (10.39) or (10.41) using eqn. (10.36). We again obtain a matrix equation that can be solved for the Bi ’s. Finally, eqn. (10.36) is solved for the unknown temperature of surface in question. In many cases, the radiosities themselves are of no particular interest. The heat flows are what is really desired. With a bit more algebra (see Problem 10.45), one can formulate a matrix equation for the n unknown values of Qneti : n j =1 δij εi − (1 − εj ) εj Aj n 4 Ai Fi−j σ Ti4 − σ Tj Ai Fi−j Qnetj = (10.42) j =1 Example 10.10 Two sides of a long triangular duct, as shown in Fig. 10.18, are made of stainless steel (ε = 0.5) and are maintained at 500◦ C. The third side is of copper (ε = 0.15) and has a uniform temperature of 100◦ C. Calculate the rate of heat transfer to the copper base per meter of length of the duct. Solution. Assume the duct walls to be gray and diffuse and that convection is negligible. The view factors can be calculated from configuration 4 of Table 10.2: F1–2 = A1 + A2 − A3 0.5 + 0.3 − 0.4 = 0.4 = 2A1 1.0 Similarly, F2–1 = 0.67, F1–3 = 0.6, F3–1 = 0.75, F2–3 = 0.33, and F3–2 = 0.25. The surfaces cannot “see” themselves, so F1–1 = F2–2 = F3–3 = 0. Equation (10.39) leads to three algebraic equations for the three 562 Radiative heat transfer §10.4 Figure 10.18 Illustration for Example 10.10. unknowns, B1 , B2 , and B3 . B1 = 1 − ε1 0.85 B2 = 1 − ε2 0.5 B3 = 1 − ε3 0.5 4 F1–1 B1 + F1–2 B2 + F1–3 B3 + ε1 σ T1 0 0.4 0.6 0.15 4 F2–1 B1 + F2–2 B2 + F2–3 B3 + ε2 σ T2 0.67 0 0.33 0.5 4 F3–1 B1 + F3–2 B2 + F3–3 B3 + ε3 σ T3 0.75 0.25 0 0.5 It would be easy to solve this system numerically using matrix methods. Alternatively, we can substitute the third equation into the first two to eliminate B3 , and then use the second equation to eliminate B2 from the first. The result is 4 4 4 B1 = 0.232 σ T1 + 0.319 σ T2 + 0.447 σ T3 Equation (10.36) gives the rate of heat loss by surface (1) as ε1 4 σ T1 − B1 1 − ε1 ε1 4 4 4 4 σ T1 − 0.232 T1 − 0.319 T2 − 0.447 T3 = A1 1 − ε1 Qnet1 = A1 Gaseous radiation §10.5 = (0.5) 563 0.15 (5.67 × 10−8 ) 0.85 × (373)4 − 0.232(373)4 − 0.319(773)4 − 0.447(773)4 W/m = −1294 W/m The negative sign indicates that the copper base is gaining heat. Enclosures with nonisothermal, nongray, or nondiffuse surfaces The representation of enclosure heat exchange by eqn. (10.41) or (10.42) is actually quite powerful. For example, if the primary surfaces in an enclosure are not isothermal, they may be subdivided into a larger number of smaller surfaces, each of which is approximately isothermal. Then either equation may be used to calculate the heat exchange among the set of smaller surfaces. For those cases in which the gray surface approximation, eqn. (10.8c), cannot be applied (owing to very different temperatures or strong wavelength dependence in ελ ), eqns. (10.41) and (10.42) may be applied on a monochromatic basis, since the monochromatic form of Kirchhoff’s law, eqn. (10.8b), remains valid. The results must, of course, be integrated over wavelength to get the heat exchange. The calculation is usually simplified by breaking the wavelength spectrum into a few discrete bands within which radiative properties are approximately constant [10.2, Chpt. 7]. When the surfaces are not diffuse — when emission or reflection vary with angle — a variety of other methods can be applied. Among them, the Monte Carlo technique is probably the most widely used. The Monte Carlo technique tracks emissions and reflections through various angles among the surfaces and estimates the probability of absorption or rereflection [10.4, 10.7]. This method allows complex situations to be numerically computed with relative ease, provided that one is careful to obtain statistical convergence. 10.5 Gaseous radiation We have treated every radiation problem thus far as though radiant heat flow in the space separating the surfaces of interest were completely unobstructed by any fluid in between. However, all gases interact with photons to some extent, by absorbing or deflecting them, and they can 564 Radiative heat transfer §10.5 even emit additional photons. The result is that fluids can play a role in the thermal radiation to the the surfaces that surround them. We have ignored this effect so far because it is generally very small, especially in air and if the distance between the surfaces is on the order of meters or less. When other gases are involved, especially at high temperatures, as in furnaces, or when long distances are involved, as in the atmosphere, gas radiation can become an important part of the heat exchange process. How gases interact with photons The photons of radiant energy passing through a gaseous region can be impeded in two ways. Some can be “scattered,” or deflected, in various directions, and some can be absorbed into the molecules. Scattering is a fairly minor influence in most gases unless they contain foreign particles, such as dust or fog. In cloudless air, for example, we are aware of the scattering of sunlight only when it passes through many miles of the atmosphere. Then the shorter wavelengths of sunlight are scattered (short wavelengths, as it happens, are far more susceptible to scattering by gas molecules than longer wavelengths, through a process known as Rayleigh scattering ). That scattered light gives the sky its blue hues. At sunset, sunlight passes through the atmosphere at a shallow angle for hundreds of miles. Radiation in the blue wavelengths has all been scattered out before it can be seen. Thus, we see only the unscattered red hues, just before dark. When particles suspended in a gas have diameters near the wavelength of light, a more complex type of scattering can occur, known as Mie scattering. Such scattering occurs from the water droplets in clouds (often making them a brilliant white color). It also occurs in gases that contain soot or in pulverized coal combustion. Mie scattering has a strong angular variation that changes with wavelength and particle size [10.8]. The absorption or emission of radiation by molecules, rather than particles, will be our principal focus. The interaction of molecules with radiation — photons, that is — is governed by quantum mechanics. It’s helpful at this point to recall a few facts from molecular physics. Each photon has an energy hco /λ, where h is Planck’s constant, co is the speed of light, and λ is the wavelength of light. Thus, photons of shorter wavelengths have higher energies: ultraviolet photons are more energetic than visible photons, which are in turn more energetic than infrared photons. It is not surprising that hotter objects emit more visible photons. §10.5 Gaseous radiation Figure 10.19 Vibrational modes of carbon dioxide and water. Molecules can store energy by rotation, by vibration (Fig. 10.19), or in their electrons. Whereas the possible energy of a photon varies smoothly with wavelength, the energies of molecules are constrained by quantum mechanics to change only in discrete steps between the molecule’s allowable “energy levels.” The available energy levels depend on the molecule’s chemical structure. When a molecule emits a photon, its energy drops in a discrete step from a higher energy level to a lower one. The energy given up is carried away by the photon. As a result, the wavelength of that photon is determined by the specific change in molecular energy level that caused it to be emitted. Just the opposite happens when a photon is absorbed: the photon’s wavelength must match a specific energy level change available to that particular molecule. As a result, each molecular species can absorb only photons at, or very close to, particular wavelengths! Often, these wavelengths are tightly grouped into so-called absorption bands, outside of which the gas is essentially transparent to photons. The fact that a molecule’s structure determines how it absorbs and emits light has been used extensively by chemists as a tool for deducing 565 566 Radiative heat transfer §10.5 molecular structure. A knowledge of the energy levels in a molecule, in conjunction with quantum theory, allows specific atoms and bonds to be identified. This is called spectroscopy (see [10.9, Chpt. 18 & 19] for an introduction; see [10.10] to go overboard). At the wavelengths that correspond to thermal radiation at typical temperatures, it happens that transitions in the vibrational and rotation modes of molecules have the greatest influence on radiative absorptance. Such transitions can be driven by photons only when the molecule has some asymmetry.4 Thus, for all practical purposes, monatomic and symmetrical diatomic molecules are transparent to thermal radiation. The major components of air—N2 and O2 —are therefore nonabsorbing; so, too, are H2 and such monatomic gases as argon. Asymmetrical molecules like CO2 , H2 O, CH4 , O3 , NH3 , N2 O, and SO2 , on the other hand, each absorb thermal radiation of certain wavelengths. The first two of these, CO2 and H2 O, are always present in air. To understand how the interaction works, consider the possible vibrations of CO2 and H2 O shown in Fig. 10.19. For CO2 , the topmost mode of vibration is symmetrical and has no interaction with thermal radiation at normal pressures. The other three modes produce asymmetries in the molecule when they occur; each is important to thermal radiation. The primary absorption wavelength for the two middle modes of CO2 is 15 µm, which lies in the thermal infrared. The wavelength for the bottommost mode is 4.3 µm. For H2 O, middle mode of vibration interacts strongly with thermal radiation at 6.3 µm. The other two both affect 2.7 µm radiation, although the bottom one does so more strongly. In addition, H2 O has a rotational mode that absorbs thermal radiation having wavelengths of 14 µm or more. Both of these molecules show additional absorption lines at shorter wavelengths, which result from the superposition of two or more vibrations and their harmonics (e.g., at 2.7 µm for CO2 and at 1.9 and 1.4 µm for H2 O). Additional absorption bands can appear at high temperature or high pressure. Absorptance, transmittance, and emittance Figure 10.20 shows radiant energy passing through an absorbing gas with a monochromatic intensity iλ . As it passes through an element of thick4 The asymmetry required is in the distribution of electric charge — the dipole moment. A vibration of the molecule must create a fluctuating dipole moment in order to interact with photons. A rotation interacts with photons only if the molecule has a permanent dipole moment. Gaseous radiation §10.5 567 Figure 10.20 The attenuation of radiation through an absorbing (and/or scattering) gas. ness dx , the intensity will be reduced by an amount diλ : diλ = −ρκλ iλ dx (10.43) where ρ is the gas density and κλ is called the monochromatic absorption coefficient. If the gas scatters radiation, we replace κλ with γλ , the monochromatic scattering coefficient. If it both absorbs and scatters radiation, we replace κλ with βλ ≡ κλ + γλ , the monochromatic extinction coefficient.5 The dimensions of κλ , βλ , and γλ are all m2/kg. If ρκλ is constant through the gas, eqn. (10.43) can be integrated from an initial intensity iλ0 at x = 0 to obtain iλ (x) = iλ0 e−ρκλ x (10.44) This result is called Beer’s law (pronounced “Bayr’s” law). For a gas layer of a given depth x = L, the ratio of final to initial intensity defines that layer’s monochromatic transmittance, τλ : τλ ≡ iλ (L) = e−ρκλ L iλ0 (10.45) Further, since gases do not reflect radiant energy, τλ + αλ = 1. Thus, the monochromatic absorptance, αλ , is αλ = 1 − e−ρκλ L (10.46) Both τλ and αλ depend on the density and thickness of the gas layer. The product ρκλ L is sometimes called the optical depth of the gas. For very small values of ρκλ L, the gas is transparent to the wavelength λ. 5 All three coefficients, κλ , γλ , and βλ , are expressed on a mass basis. They could, alternatively, have been expressed on a volumetric basis. 568 Radiative heat transfer §10.5 Figure 10.21 The monochromatic absorptance of a 1.09 m thick layer of steam at 127◦ C. The dependence of αλ on λ is normally very strong. As we have seen, a given molecule will absorb radiation in certain wavelength bands, while allowing radiation with somewhat higher or lower wavelengths to pass almost unhindered. Figure 10.21 shows the absorptance of water vapor as a function of wavelength for a fixed depth. We can see the absorption bands at wavelengths of 6.3, 2.7, 1.9, and 1.4 µm that were mentioned before. A comparison of Fig. 10.21 with Fig. 10.2 readily shows why radiation from the sun, as viewed from the earth’s surface, shows a number of spikey indentations at certain wavelengths. Several of those indentations occur in bands where atmospheric water vapor absorbs incoming solar radiation, in accordance with Fig. 10.21. The other indentations in Fig. 10.2 occur where ozone and CO2 absorb radiation. The sun itself does not have these regions of low emittance; it is just that much of the radiation in these bands is absorbed by gases in the atmosphere before it can reach the ground. Just as αλ and ελ are equal to one another for a diffuse solid surface, they are equal for a gas. We may demonstrate this by considering an isothermal gas that is in thermal equilibrium with a black enclosure that contains it. The radiant intensity within the enclosure is that of a black body, iλb , at the temperature of the gas and enclosure. Equation (10.43) shows that a small section of gas absorbs radiation, reducing the intensity by an amount ρκλ iλb dx . To maintain equilibrium, the gas must therefore emit an equal amount of radiation: diλ = ρκλ iλb dx (10.47) Now, if radiation from some other source is transmitted through a nonscattering isothermal gas, we can combine the absorption from Gaseous radiation §10.5 569 eqn. (10.43) with the emission from eqn. (10.47) to form an energy balance called the equation of transfer diλ = −ρκλ iλ + ρκλ iλb dx (10.48) Integration of this equation yields a result similar to eqn. (10.44): iλ (L) = iλ0 e−ρκλ L +iλb 1 − e−ρκλ L =τλ (10.49) ≡ελ The first righthand term represents the transmission of the incoming intensity, as in eqn. (10.44), and the second is the radiation emitted by the gas itself. The coefficient of the second righthand term defines the monochromatic emittance, ελ , of the gas layer. Finally, comparison to eqn. (10.46) shows that ελ = αλ = 1 − e−ρκλ L (10.50) Again, we see that for very small ρκλ L the gas will neither absorb nor emit radiation of wavelength λ. Heat transfer from gases to walls We now see that predicting the total emissivity, εg , of a gas layer will be complex. We have to take account of the gases’ absorption bands as well as the layer’s thickness and density. Such predictions can be done [10.11], but they are laborious. For making simpler (but less accurate) estimates, correlations of εg have been developed. Such correlations are based on the following model: An isothermal gas of temperature Tg and thickness L, is bounded by walls at the single temperature Tw . The gas consists of a small fraction of an absorbing species (say CO2 ) mixed into a nonabsorbing species (say N2 ). If the absorbing gas has a partial pressure pa and the mixture has a total pressure p , the correlation takes this form: εg = fn pa L, p, Tg (10.51) The parameter pa L is a measure of the layer’s optical depth; p and Tg account for changes in the absorption bands with pressure and temperature. Radiative heat transfer 570 §10.5 Hottel and Sarofim [10.12] provide such correlations for CO2 and H2 O, built from research by Hottel and others before 1960. The correlations take the form εg pa L, p, Tg = f1 pa L, Tg × f2 p , pa , pa L (10.52) where the experimental functions f1 and f2 are plotted in Figs. 10.22 and 10.23 for CO2 and H2 O, respectively. The first function, f1 , is a correlation for a total pressure of p = 1 atm with a very small partial pressure of the absorbing species. The second function, f2 , is a correction factor to account for other values of pa or p . Additional corrections must be applied if both CO2 and H2 O are present in the same mixture. To find the net heat transfer between the gas and the walls, we must also find the total absorptance, αg , of the gas. Despite the equality of the monochromatic emittance and absorptance, ελ and αλ , the total values, εg and αg , will not generally be equal. This is because the absorbed radiation may come from, say, a wall having a much different temperature than the gas with a correspondingly different wavelength distribution. Hottel and Sarofim show that αg may be estimated from the correlation for εg as follows:6 αg = Tg Tw 1/2 · ε pa L Tw , p, Tw Tg (10.53) Finally, we need to determine an appropriate value of L for a given enclosure. The correlations just given for εg and αg assume L to be a one-dimensional path through the gas. Even for a pair of flat plates a distance L apart, this won’t be appropriate since radiation can travel much farther if it follows a path that is not perpendicular to the plates. For enclosures that have black walls at a uniform temperature, we can use an effective path length, L0 , called the geometrical mean beam length, to represent both the size and the configuration of a gaseous region. The geometrical mean beam length is defined as L0 ≡ 4 (volume of gas) boundary area that is irradiated (10.54) Thus, for two infinite parallel plates a distance apart, L0 = 4A /2A = 2 . Some other values of L0 for gas volumes exchanging heat with all points on their boundaries are as follow: 6 Hottel originally recommended replacing the exponent 1/2 by 0.65 for CO2 and 0.45 for H2 O. Theory, and more recent work, both suggest using the value 1/2 [10.13]. Figure 10.22 Functions used to predict εg = f1 f2 for water vapor in air. 571 Figure 10.23 Functions used to predict εg = f1 f2 for CO2 in air. All pressures in atmospheres. 572 Gaseous radiation §10.5 • For a sphere of diameter D , L0 = 2D/3 • For an infinite cylinder of diameter D , L0 = D • For a cube of side L, L0 = 2L/3 • For a cylinder with height = D , L0 = 2D/3 For cases where the gas is strongly absorbing, better accuracy can be obtained by replacing the constant 4 in eqn. (10.54) by 3.5, lowering the mean beam length about 12%. We are now in position to treat a problem in which hot gases (say the products of combustion) radiate to a black container. Consider an example: Example 10.11 A long cylindrical combustor 40 cm in diameter contains a gas at 1200◦ C consisting of 0.8 atm N2 and 0.2 atm CO2 . What is the net heat radiated to the walls if they are at 300◦ C? Solution. Let us first obtain εg . We have L0 = D = 0.40 m, a total pressure of 1.0 atm, pCO2 = 0.2 atm, and T = 1200◦ C = 2651◦ R. 1, so Then Fig. 10.23a gives f1 as 0.098 and Fig. 10.23b gives f2 εg = 0.098. Next, we use eqn. (10.53) to obtain αg , with Tw = 1031◦ R, pH2 O LTw /Tg = 0.031: αg = 1200 + 273 300 + 273 0.5 (0.074) = 0.12 Now we can calculate Qnetg-w . For these problems with one wall surrounding one gas, the use of the mean beam length in finding εg and αg accounts for all geometrical effects, and no view factor is required. The net heat transfer is calculated using the surface area of the wall: 4 4 Qnetg-w = Aw εg σ Tg − αg σ Tw = π (0.4)(5.67 × 10−8 ) (0.098)(1473)4 − (0.12)(573)4 = 32 kW/m Total emissivity charts and the mean beam length provide a simple, but crude, tool for dealing with gas radiation. Since the introduction 573 Radiative heat transfer 574 §10.6 of these ideas in the mid-twentieth century, major advances have been made in our knowledge of the radiative properties of gases and in the tools available for solving gas radiation problems. In particular, band models of gas radiation, and better measurements, have led to better procedures for dealing with the total radiative properties of gases (see, in particular, References [10.11] and [10.13]). Tools for dealing with radiation in complex enclosures have also improved. The most versitile of these is the previously-mentioned Monte Carlo method [10.4, 10.7], which can deal with nongray, nondiffuse, and nonisothermal walls with nongray, scattering, and nonisothermal gases. An extensive literature also deals with approximate analytical techniques, many of which are based on the idea of a “gray gas” — one for which ελ and αλ are independent of wavelength. However, as we have pointed out, the gray gas model is not even a qualitative approximation to the properties of real gases.7 Finally, it is worth noting that gaseous radiation is frequently less important than one might imagine. Consider, for example, two flames: a bright orange candle flame and a “cold-blue” hydrogen flame. Both have a great deal of water vapor in them, as a result of oxidizing H2 . But the candle will warm your hands if you place them near it and the hydrogen flame will not. Yet the temperature in the hydrogen flame is higher. It turns out that what is radiating both heat and light from the candle is soot — small solid particles of almost thermally black carbon. The CO2 and H2 O in the candle flame actually contribute relatively little to radiation. 10.6 Solar energy The sun The sun continually irradiates the earth at a rate of about 1.74×1014 kW. If we imagine this energy to be distributed over a circular disk with the earth’s diameter, the solar irradiation is about 1367 W/m2 , as measured by satellites above the atmosphere. Much of this energy reaches the ground, where it sustains the processes of life. 7 Edwards [10.11] describes the gray gas as a “myth.” He notes, however, that spectral variations may be overlooked for a gas containing spray droplets or particles [in a range of sizes] or for some gases that have wide, weak absorption bands within the spectral range of interest [10.3]. Some accommodation of molecular properties can be achieved using the weighted sum of gray gases concept [10.12], which treats a real gas as superposition of gray gases having different properties. §10.6 Solar energy The temperature of the sun varies from tens of millions of kelvin in its core to between 4000 and 6000 K at its surface, where most of the sun’s thermal radiation originates. The wavelength distribution of the sun’s energy is not quite that of a black body, but it may be approximated as such. A straightforward calculation (see Problem 10.49) shows that a black body of the sun’s size and distance from the earth would produce the same irradiation as the sun if its temperature were 5777 K. The solar radiation reaching the earth’s surface is always less than that above the atmosphere owing to atmospheric absorption and the earth’s curvature and rotation. Solar radiation usually arrives at an angle of less than 90◦ to the surface because the sun is rarely directly overhead. We have seen that a radiant heat ux arriving at an angle less than 90◦ is reduced by the cosine of that angle (Fig. 10.4). The sun’s angle varies with latitude, time of day, and day of year. Trigonometry and data for the earth’s rotation can be used to find the appropriate angle. Figure 10.2 shows the reduction of solar radiation by atmospheric absorption for one particular set of atmospheric conditions. In fact, when the sun passes through the atmosphere at a low angle (near the horizon), the path of radiation through the atmosphere is longer, providing relatively more opportunity for atmospheric absorption and scattering. Additional moisture in the air can increase the absorption by H2 O, and, of course, clouds can dramatically reduce the solar radiation reaching the ground. The consequence of these various effects is that the solar radiation received on the ground is almost never more than 1200 W/m2 and is often only a few hundred W/m2 . Extensive data are available for estimating the ground level solar irradiation at a given location, time, and date [10.14, 10.15]. The distribution of the Sun’s energy and atmospheric irradiation Figure 10.24 shows what becomes of the solar energy that impinges on the earth if we average it over the year and the globe, taking account of all kinds of weather. Only 45% of the sun’s energy actually reaches the earth’s surface. The mean energy received is about 235 W/m2 if averaged over the surface and the year. The lower left-hand portion of the figure shows how this energy is, in turn, all returned to the atmosphere and to space. The solar radiation reaching the earth’s surface includes direct radiation that has passed through the atmosphere and diffuse radiation that has been scattered, but not absorbed, by the atmosphere. Atmospheric 575 Radiative heat transfer 576 45% reaches the earth’s surface §10.6 45% is transmitted to the earth directly and by diffuse radiation 33% is reflected back to space 22% is absorbed in the atmosphere Sensible heat transfer to atmosphere Radiation that reaches the outer atmosphere from the sun Net radiation from surface Evaporation The flow of energy from the earth's surface back to and through - the earth's atmosphere Figure 10.24 The approximate distribution of the flow of the sun’s energy to and from the earth’s surface [10.16]. gases also irradiate the surface. This irradiation is quite important to maintaining the temperature of objects on the surface. In Section 10.5, saw that the energy radiated by a gas depends upon the depth of the gas, its temperature, and the molecules present in it. The emissivity of the atmosphere has been characterized in detail [10.16, 10.17, 10.18]. For practical calculations, however, it is often convenient to treat the sky as a black radiator having some appropriate temperature. Solar energy §10.6 577 This effective sky temperature usually lies between 5 and 30 K below the ground level air temperature. The sky temperature decreases as the amount of water vapor in the air goes down. For cloudless skies, the sky temperature may be estimated using the dew-point temperature, Tdp , and the hour past midnight, t : Tsky = Tair 0.711 + 0.0056 Tdp 2 + 7.3 × 10−5 Tdp + 0.013 cos(2π t/24) 1/4 (10.55) where Tsky and Tair are in kelvin and Tdp is in ◦ C. This equation applies for dew points from −20◦ C to 30◦ C [10.19]. It is fortunate that sky temperatures are relatively warm. In the absence of an atmosphere, not only would more of the sun’s radiation reach the ground during the day, but at night heat would radiate directly into the bitter cold of outer space. Such conditions prevail on the Moon, where average daytime surface temperatures are about 110◦ C while average nighttime temperatures plunge to about −150◦ C. Selective emitters, absorbers, and transmitters We have noted that most of the sun’s energy lies at wavelengths near the visible region of the electromagnetic spectrum and that most of the radiation from objects at temperatures typical of the earth’s surface is on much longer, infrared wavelengths (see pg. 535). One result is that materials may be chosen or designed to be selectively good emitters or reflectors of both solar and infrared radiation. Table 10.4 shows the infrared emittance and solar absorptance for several materials. Among these, we identify several particularly selective solar absorbers and solar reflectors. The selective absorbers have a high absorptance for solar radiation and a low emittance for infrared radiation. Consequently, they do not strongly reradiate the solar energy that they absorb. The selective solar reflectors, on the other hand, reflect solar energy strongly and also radiate heat efficiently in the infrared. Solar reflectors stay much cooler than solar absorbers in bright sunlight. Example 10.12 In Section 10.2, we discussed white paint on a roof as a selective solar absorber. Consider now a barn roof under a sunlit sky. The solar radiation on the plane of the roof is 600 W/m2 , the air temperature is 35◦ C, and a light breeze produces a convective heat transfer 578 Radiative heat transfer §10.6 Table 10.4 Solar absorptance and infrared emittance for several surfaces near 300 K [10.4, 10.15]. Surface αsolar εIR Aluminum, pure Carbon black in acrylic binder Copper, polished 0.09 0.94 0.3 0.1 0.83 0.04 Selective Solar absorbers Black Cr on Ni plate CuO on Cu (Ebanol C) Nickel black on steel Sputtered cermet on steel 0.95 0.90 0.81 0.96 0.09 0.16 0.17 0.16 0.14 0.2–0.35 0.7 0.82 0.26 0.12–0.18 0.90 0.93 Selective Solar Reflectors Magnesium oxide Snow White paint Acrylic Zinc Oxide coefficient of h = 8 W/m2 K. The sky temperature is 18◦ C. Find the temperature of the roof if it is painted with either white acrylic paint or a non-selective black paint having ε = 0.9. Solution. Heat loss from the roof to the inside of the barn will lower the roof temperature. Since we don’t have enough information to evaluate that loss, we can make an upper bound on the roof temperature by assuming that no heat is transferred to the interior. Then, an energy balance on the roof must account for radiation absorbed from the sun and the sky and for heat lost by convection and reradiation: 4 4 αsolar qsolar + εIR σ Tsky = h (Troof − Tair ) + εIR σ Troof Rearranging and substituting the given numbers, 4 8 [Troof − (273 + 35)] + εIR (5.67 × 10−8 ) Troof − (273 + 18)4 = αsolar (600) For the non-selective black paint, αsolar = εIR = 0.90. Solving by §10.6 Solar energy iteration, we find Troof = 338 K = 65◦ C For white acrylic paint, from Table 10.4, αsolar = 0.26 and εIR = 0.90. We find Troof = 312 K = 39◦ C The white painted roof is only a few degrees warmer than the air. Ordinary window glass is a very selective transmitter of solar radiation. Glass is nearly transparent to wavelengths below 2.7 µm or so, passing more than 90% of the incident solar energy. At longer wavelengths, in the infrared, glass is virtually opaque to radiation. A consequence of this fact is that solar energy passing through a window cannot pass back out as infrared reradiation. This is precisely why we make greenhouses out of glass. A greenhouse is a structure in which we use glass trap solar energy in a lower temperature space. The atmospheric greenhouse effect and global warming The atmosphere creates a greenhouse effect on the earth’s surface that is very similar to that caused by a pane of glass. Solar energy passes through the atmosphere, arriving mainly on wavelengths between about 0.3 and 3 µm. The earth’s surface, having a mean temperature of 15◦ C or so, radiates mainly on infrared wavelengths longer than 5 µm. Certain atmospheric gases have strong absorption bands at these longer wavelengths. Those gases absorb energy radiated from the surface, and then reemit it toward both the surface and outer space. The result is that the surface remains some 30 K warmer than the atmosphere. In effect, the atmosphere functions as a radiation shield against infrared heat loss to space. The gases mainly responsible for the the atmospheric greenhouse effect are CO2 , H2 O, CH4 , N2 O, O3 , and some chlorofluorcarbons [10.20]. If the concentration of these gases rises or falls, the strength of the greenhouse effect will change and the surface temperature will also rise or fall. With the exception of the chlorofluorocarbons, each of these gases is created, in part, by natural processes: H2 O by evaporation, CO2 by animal respiration, CH4 through plant decay and digestion by livestock, and so on. Human activities, however, have significantly increased the concentrations of all of the gases. Fossil fuel combustion increased the CO2 579 Radiative heat transfer 580 §10.6 0.8 Temperature Anomaly, ˚C 0.6 0.4 0.2 0.0 -0.2 Annual mean 5-year mean -0.4 -0.6 1880 1900 1920 1940 1960 1980 2000 Year Figure 10.25 Global surface temperature change relative to the mean temperature from 1950–1980 (Courtesy of the NASA Goddard Institute for Space Studies [10.21]). concentration by more than 30% during the twentieth century. Methane concentrations have risen through the transportation and leakage of hydrocarbon fuels. Ground level ozone concentrations have risen as a result of photochemical interactions of other pollutants. Chlorofluorocarbons are human-made chemicals. In parallel to the rising concentrations of these gases, the surface temperature of the earth has risen significantly. Over the course of the twentieth century, a rise of 0.6–0.7 K occurred, with 0.4–0.5 K of that rise coming after 1950 (see Fig. 10.25). The data showing this rise are extensive, are derived from multiple sources, and have been the subject of detailed scrutiny: there is relatively little doubt that surface temperatures have increased [10.21, 10.22]. The question of how much of the rise should be attributed to anthropogenic greenhouse gases, however, was a subject of intense debate throughout the 1990’s. Many factors must be considered in examining the causes of global warming. Carbon dioxide, for example, is present in such high concentrations that adding more of it increases absorption less rapidly than might be expected. Other gases that are present in smaller concentrations, such as methane, have far stronger effects per additional kilogram. The con- §10.6 Solar energy centration of water vapor in the atmosphere rises with increasing surface temperature, amplifying any warming trend. Increased cloud cover has both warming and cooling effects. The melting of polar ice caps as temperatures rise reduces the planet’s reflectivity, or albedo, allowing more solar energy to be absorbed. Small temperature rises that have been observed in the oceans store enormous amounts of energy that must accounted. Atmospheric aerosols (two-thirds of which are produced by sulfate and carbon pollution from fossil fuels) also tend to reduce the greenhouse effect. All of these factors must be built into an accurate climate model (see, for example, [10.23]). The current consensus among mainstream researchers is that the global warming seen during the last half of the twentieth century is mainly attributable to human activity, principally through the combustion of fossil fuels [10.22]. Numerical models have been used to project a continuing temperature rise in the twenty-first century, subject to various assumptions about the use of fossil fuels and government policies for reducing greenhouse gas emissions. Regrettably, the outlook is not very positive, with predictions of twenty-first century warming ranging from 1.4–5.8 K. The potential for solar power One alternative to the continuing use of fossil fuels is solar energy. With so much solar energy falling upon all parts of the world, and with the apparent safety, reliability, and cleanliness of most schemes for utilizing solar energy, one might ask why we do not generally use solar power already. The reason is that solar power involves many serious heat transfer and thermodynamics design problems and may pose environmental threats of its own. We shall discuss the problems qualitatively and refer the reader to [10.15], [10.24], or [10.25] for detailed discussions of the design of solar energy systems. Solar energy reaches the earth with very low intensity. We began this discussion in Chapter 1 by noting that human beings can interface with only a few hundred watts of energy. We could not live on earth if the sun were not relatively gentle. It follows that any large solar power source must concentrate the energy that falls on a huge area. By way of illustration, suppose that we sought to photovoltaically convert 615 W/m2 of solar energy into electric power with a 15% efficiency (which is not pessimistic) during 8 hr of each day. This would correspond to a daily average of 31 W/m2 , and we would need almost 26 square kilometers (10 square miles) of collector area to match the steady output of an 800 MW power plant. 581 582 Radiative heat transfer §10.6 Other forms of solar energy conversion require similarly large areas. Hydroelectric power — the result of evaporation under the sun’s warming influence — requires a large reservoir, and watershed, behind the dam. The burning of organic matter, as wood or grain-based ethanol, requires a large cornfield or forest to be fed by the sun, and so forth. Any energy supply that is served by the sun must draw from a large area of the earth’s surface. Thus, they introduce their own kinds of environmental complications. A second problem stems from the intermittent nature of solar devices. To provide steady power—day and night, rain or shine—requires thermal storage systems, which add both complication and cost. These problems are minimal when one uses solar energy merely to heat air or water to moderate temperatures (50 to 90◦ C). In this case the efficiency will improve from just a few percent to as high as 70%. Such heating can be used for industrial processes (crop drying, for example), or it can be used on a small scale for domestic heating of air or water. Figure 10.26 shows a typical configuration of a domestic solar collector of the flat-plate type. Solar radiation passes through one or more glass plates and impinges on a plate that absorbs the solar wavelengths. The absorber plate would be a selective solar absorber, perhaps blackened copper or nickel. The glass plates might be treated with anti-reflective coatings, raising their solar transmissivity to 98% or more. Once the energy is absorbed, it is reemitted as long-wavelength infrared radiation. Glass is almost opaque in this range, and energy is retained in the collector by a greenhouse effect. Multiple layers of glass serve to reduce both reradiative and convective losses from the absorber plate. Water flowing through tubes, which may be brazed to the absorber plate, carries the energy away for use. The flow rate is adjusted to give an appropriate temperature rise. If the working fluid is to be brought to a fairly high temperature, the direct radiation from the sun must be focused from a large area down to a very small region, using reflecting mirrors. Collectors equipped with a small parabolic reflector, focused on a water or air pipe, can raise the fluid to between 100 and 200◦ C. In any scheme intended to produce electrical power with a conventional thermal cycle, energy must be focused in an area ratio on the order of 1000 : 1 to achieve a practical cycle efficiency. A question of over-riding concern as we enter the 21st century is “How much of the renewable energy that reaches Earth, can we hope to utilize?” Of the 1.74×1014 kW arriving from the sun, 33% is simply §10.6 Solar energy Figure 10.26 A typical flat-plate solar collector. reflected back into outer space. If we were able to collect and use the remainder, 1.16×1014 kW, before it too was reradiated to space, each of the 6 billion or so people on the planet would have 19 MW at his or her disposal. Of course, the vast majority of that power must be used to sustain natural processes in the world around us. In the USA, total energy consumption in 2002 averaged roughly 3.2 × 109 kW, and, dividing this value into a population of 280 million people gives a per capita consumption of roughly 11 kW. Worldwide, energy was consumed at a rate just over 1010 kW. That means that world energy consumption was just under 0.01% of the renewable energy passing into and out of Earth’s ecosystem. Since many countries that once used very little energy are moving toward a life-style which requires much greater energy consumption, this percentage is rising at an estimated rate of 2%/y. We must also bear in mind two aspects of this 0.01% figure. First, it is low enough that we might aim, ultimately, to take all of our energy from renewable sources, and thus avoid consuming irreplaceable terrestrial resources. Second, although 0.01% is a small fraction, the absolute 583 Chapter 10: Radiative heat transfer 584 amount of power it represents is enormous. It is, therefore, unclear just how much renewable energy we can claim before we create new ecological problems. There is little doubt that our short-term needs can be met by fossil fuel reserves. However, continued use of those fuels will clearly amplify the now-well-documented global warming trend. Our long-term hope for an adequate energy supply may be at least partially met with solar power and other renewable sources. Nuclear fission remains a promising option if one or more means for nuclear waste disposal is deemed acceptable. Nuclear fusion—the process by which we might manage to create minisuns upon the earth—may also be a hope for the future. Under almost any scenario, however, we will surely be forced to limit the continuing growth of energy consumption. Problems 10.1 What will ελ of the sun appear to be to an observer on the earth’s surface at λ = 0.2 µm and 0.65 µm? How do these emittances compare with the real emittances of the sun? [At 0.65 µm, ελ 0.77.] 10.2 Plot eλb against λ for T = 300 K and 10, 000 K with the help of eqn. (1.30). About what fraction of energy from each black body is visible? 10.3 A 0.6 mm diameter wire is drawn out through a mandril at 950◦ C. Its emittance is 0.85. It then passes through a long cylindrical shield of commercial aluminum sheet, 7 cm in diameter. The shield is horizontal in still air at 25◦ C. What is the temperature of the shield? Is it reasonable to neglect natural convection inside and radiation outside? [Tshield = 153◦ C.] 10.4 A 1 ft2 shallow pan with adiabatic sides is filled to the brim with water at 32◦ F. It radiates to a night sky whose temperature is −18◦ F, while a 50◦ F breeze blows over it at 1.5 ft/s. Will the water freeze or warm up? 10.5 A thermometer is held vertically in a room with air at 10◦ C and walls at 27◦ C. What temperature will the thermometer read if everything can be considered black? State your assumptions. 10.6 Rework Problem 10.5, taking the room to be wall-papered and considering the thermometer to be nonblack. Problems 585 10.7 Two thin aluminum plates, the first polished and the second painted black, are placed horizontally outdoors, where they are cooled by air at 10◦ C. The heat transfer coefficient is 5 W/m2 K on both the top and the bottom. The top is irradiated with 750 W/m2 and it radiates to the sky at 250 K. The earth below the plates is black at 10◦ C. Find the equilibrium temperature of each plate. 10.8 A sample holder of 99% pure aluminum, 1 cm in diameter and 16 cm in length, protrudes from a small housing on an orbital space vehicle. The holder “sees” almost nothing but outer space at an effective temperature of 30 K. The base of the holders is 0◦ C and you must find the temperature of the sample at its tip. It will help if you note that aluminum is used, so that the temperature of the tip stays quite close to that of the root. [Tend = −0.7◦ C.] 10.9 There is a radiant heater in the bottom of the box shown in Fig. 10.27. What percentage of the heat goes out the top? What fraction impinges on each of the four sides? (Remember that the percentages must add up to 100.) Figure 10.27 Configuration for Prob. 10.9. 10.10 With reference to Fig. 10.12, find F1–(2+4) and F(2+4)–1 . 10.11 Find F2–4 for the surfaces shown in Fig. 10.28. [0.315.] 10.12 What is F1–2 for the squares shown in Fig. 10.29? 10.13 A particular internal combustion engine has an exhaust manifold at 600◦ C running parallel to a water cooling line at 20◦ C. If both the manifold and the cooling line are 4 cm in diameter, their centers are 7 cm apart, and both are approximately black, how much heat will be transferred to the cooling line by radiation? [383 W/m.] Chapter 10: Radiative heat transfer 586 Figure 10.28 Configuration for Prob. 10.11. Figure 10.29 Configuration for Prob. 10.12. 10.14 Prove that F1–2 for any pair of two-dimensional plane surfaces, as shown in Fig. 10.30, is equal to [(a + b) − (c + d)]/2L1 . This is called the string rule because we can imagine that the numerator equals the difference between the lengths of a set of crossed strings (a and b) and a set of uncrossed strings (c and d). Figure 10.30 Configuration for Prob. 10.14. 10.15 Find F1–5 for the surfaces shown in Fig. 10.31. 10.16 Find F1–(2+3+4) for the surfaces shown in Fig. 10.32. 10.17 A cubic box 1 m on the side is black except for one side, which has an emittance of 0.2 and is kept at 300◦ C. An adjacent side is kept at 500◦ C. The other sides are insulated. Find Qnet inside the box. [2494 W.] Problems 587 Figure 10.31 Configuration for Prob. 10.15. Figure 10.32 Configuration for Prob. 10.16. 10.18 Rework Problem 10.17, but this time set the emittance of the insulated walls equal to 0.6. Compare the insulated wall temperature with the value you would get if the walls were black. 10.19 An insulated black cylinder, 10 cm in length and with an inside diameter of 5 cm, has a black cap on one end and a cap with an emittance of 0.1 on the other. The black end is kept at 100◦ C and the reflecting end is kept at 0◦ C. Find Qnet inside the cylinder and Tcylinder . 10.20 Rework Example 10.2 if the shield has an inside emittance of 0.34 and the room is at 20◦ C. How much cooling must be provided to keep the shield at 100◦ C? Chapter 10: Radiative heat transfer 588 10.21 A 0.8 m long cylindrical burning chamber is 0.2 m in diameter. The hot gases within it are at a temperature of 1500◦ C and a pressure of 1 atm, and the absorbing components consist of 12% by volume of CO2 and 18% H2 O. Neglect end effects and determine how much cooling must be provided the walls to hold them at 750◦ C if they are black. 10.22 A 30 ft by 40 ft house has a conventional 30◦ sloping roof with a peak running in the 40 ft direction. Calculate the temperature of the roof in 20◦ C still air when the sun is overhead (a) if the roofing is of wooden shingles and (b) if it is commercial aluminum sheet. The incident solar energy is 670 W/m2 , Kirchhoff’s law applies for both roofs, and the effective sky temperature is 22◦ C. 10.23 Calculate the radiant heat transfer from a 0.2 m diameter stainless steel hemisphere (εss = 0.4) to a copper floor (εCu = 0.15) that forms its base. The hemisphere is kept at 300◦ C and the base at 100◦ C. Use the algebraic method. [21.24 W.] 10.24 A hemispherical indentation in a smooth wrought-iron plate has an 0.008 m radius. How much heat radiates from the 40◦ C dent to the −20◦ C surroundings? 10.25 A conical hole in a block of metal for which ε = 0.5 is 5 cm in diameter at the surface and 5 cm deep. By what factor will the radiation from the area of the hole be changed by the presence of the hole? (This problem can be done to a close approximation using the methods in this chapter if the cone does not become very deep and slender. If it does, then the fact that the apex is receiving far less radiation makes it incorrect to use the network analogy.) 10.26 A single-pane window in a large room is 4 ft wide and 6 ft high. The room is kept at 70◦ F, but the pane is at 67◦ F owing to heat loss to the colder outdoor air. Find (a) the heat transfer by radiation to the window; (b) the heat transfer by natural convection to the window; and (c) the fraction of heat transferred to the window by radiation. 10.27 Suppose that the windowpane temperature is unknown in Problem 10.26. The outdoor air is at 40◦ F and h is 62 W/m2 K on the Problems 589 outside of the window. It is nighttime and the effective temperature of the sky is 15◦ F. Assume Fwindow−sky = 0.5. Take the rest of the surroundings to be at 40◦ F. Find Twindow and draw the analogous electrical circuit, giving numerical values for all thermal resistances. Discuss the circuit. (It will simplify your calculation to note that the window is opaque to infrared radiation but that it offers very little resistance to conduction. Thus, the window temperature is almost uniform.) 10.28 A very effective low-temperature insulation is made by evacuating the space between parallel metal sheets. Convection is eliminated, conduction occurs only at spacers, and radiation is responsible for what little heat transfer occurs. Calculate q between 150 K and 100 K for three cases: (a) two sheets of highly polished aluminum, (b) three sheets of highly polished aluminum, and (c) three sheets of rolled sheet steel. 10.29 Three parallel black walls, 1 m wide, form an equilateral triangle. One wall is held at 400 K, one is at 300 K, and the third is insulated. Find Q W/m and the temperature of the third wall. 10.30 Two 1 cm diameter rods run parallel, with centers 4 cm apart. One is at 1500 K and black. The other is unheated, and ε = 0.66. They are both encircled by a cylindrical black radiation shield at 400 K. Evaluate Q W/m and the temperature of the unheated rod. 10.31 A small-diameter heater is centered in a large cylindrical radiation shield. Discuss the relative importance of the emittance of the shield during specular and diffuse radiation. 10.32 Two 1 m wide commercial aluminum sheets are joined at a 120◦ angle along one edge. The back (or 240◦ angle) side is insulated. The plates are both held at 120◦ C. The 20◦ C surroundings are distant. What is the net radiant heat transfer from the left-hand plate: to the right-hand side, and to the surroundings? 10.33 Two parallel discs of 0.5 m diameter are separated by an infinite parallel plate, midway between them, with a 0.2 m diameter hole in it. The discs are centered on the hole. What is the view factor between the two discs if they are 0.6 m apart? Chapter 10: Radiative heat transfer 590 10.34 An evacuated spherical cavity, 0.3 m in diameter in a zerogravity environment, is kept at 300◦ C. Saturated steam at 1 atm is then placed in the cavity. (a) What is the initial flux of radiant heat transfer to the steam? (b) Determine how long it will take for qconduction to become less than qradiation . (Correct for the rising steam temperature if it is necessary to do so.) 10.35 Verify cases (1), (2), and (3) in Table 10.2 using the string method described in Problem 10.14. 10.36 Two long parallel heaters consist of 120◦ segments of 10 cm diameter parallel cylinders whose centers are 20 cm apart. The segments are those nearest each other, symmetrically placed on the plane connecting their centers. Find F1–2 using the string method described in Problem 10.14.) 10.37 Two long parallel strips of rolled sheet steel lie along sides of an imaginary 1 m equilateral triangular cylinder. One piece is 1 1 m wide and kept at 20◦ C. The other is 2 m wide, centered in an adjacent leg, and kept at 400◦ C. The surroundings are distant and they are insulated. Find Qnet . (You will need a shape factor; it can be found using the method described in Problem 10.14.) 10.38 Find the shape factor from the hot to the cold strip in Problem 10.37 using Table 10.2, not the string method. If your instructor asks you to do so, complete Problem 10.37 when you have F1–2 . 10.39 Prove that, as the figure becomes very long, the view factor for the second case in Table 10.3 reduces to that given for the third case in Table 10.2. 10.40 Show that F1–2 for the first case in Table 10.3 reduces to the expected result when plates 1 and 2 are extended to infinity. 10.41 In Problem 2.26 you were asked to neglect radiation in showing that q was equal to 8227 W/m2 as the result of conduction alone. Discuss the validity of the assumption quantitatively. 10.42 A 100◦ C sphere with ε = 0.86 is centered within a second sphere at 300◦ C with ε = 0.47. The outer diameter is 0.3 m and the inner diameter is 0.1 m. What is the radiant heat flux? Problems 591 10.43 Verify F1–2 for case 4 in Table 10.2. (Hint: This can be done without integration.) 10.44 Consider the approximation made in eqn. (10.30) for a small gray object in a large isothermal enclosure. How small must A1 /A2 be in order to introduce less than 10% error in F1–2 if the small object has an emittance of ε1 = 0.5 and the enclosure is: a) commerical aluminum sheet; b) rolled sheet steel; c) rough red brick; d) oxidized cast iron; or e) polished electrolytic copper. Assume both the object and its environment have temperatures of 40 to 90◦ C. 10.45 Derive eqn. (10.42), starting with eqns. (10.36–10.38). 10.46 (a) Derive eqn. (10.31), which is for a single radiation shield between two bodies. Include a sketch of the radiation network. (b) Repeat the calculation in the case when two radiation shields lie between body (1) and body (2), with the second shield just outside the first. 10.47 Use eqn. (10.32) to find the net heat transfer from between two specularly reflecting bodies that are separated by a specularly reflecting radiation shield. Compare the result to eqn. (10.31). Does specular reflection reduce the heat transfer? 10.48 Some values of the monochromatic absorption coefficient for liquid water, as ρκλ (cm−1 ), are listed below [10.4]. For each wavelength, find the thickness of a layer of water for which the transmittance is 10%. On this basis, discuss the colors one might see underwater and water’s infrared emittance. λ (µm) 0.3 0.4 0.5 0.6 0.8 1.0 2.0 2.6–10.0 ρκλ (cm−1 ) 0.0067 0.00058 0.00025 0.0023 0.0196 0.363 69.1 > 100. Color violet green orange Chapter 10: Radiative heat transfer 592 10.49 The sun has a diameter of 1.391 × 106 km. The earth has a diameter of 12,740 km and lies at a mean distance of 1.496 × 108 km from the center of the sun. (a) If the earth is treated as a flat disk normal to the radius from sun to earth, determine the view factor Fsun–earth . (b) Use this view factor and the measured solar irradiation of 1367 W/m2 to show that the effective black body temperature of the sun is 5777 K. References [10.1] E. M. Sparrow and R. D. Cess. Radiation Heat Transfer. Hemisphere Publishing Corp./McGraw-Hill Book Company, Washington, D.C., 1978. [10.2] M. F. Modest. Radiative Heat Transfer. McGraw-Hill, New York, 1993. [10.3] D. K. Edwards. Radiation Heat Transfer Notes. Hemisphere Publishing Corp., Washington, D.C., 1981. [10.4] R. Siegel and J. R. Howell. Thermal Radiation Heat Transfer. Taylor and Francis-Hemisphere, Washington, D.C., 4th edition, 2001. [10.5] J. R. Howell. A Catalog of Radiation Heat Transfer Configuration Factors. University of Texas, Austin, 2nd edition, 2001. Available online at http://www.me.utexas.edu/∼howell/. [10.6] A. K. Oppenheim. Radiation analysis by the network method. Trans. ASME, 78:725–735, 1956. [10.7] W.-J. Yang, H. Taniguchi, and K. Kudo. Radiative heat transfer by the Monte Carlo method. In T.F. Irvine, Jr., J. P. Hartnett, Y. I. Cho, and G. A. Greene, editors, Advances in Heat Transfer, volume 27. Academic Press, Inc., San Diego, 1995. [10.8] H. C. van de Hulst. Light Scattering by Small Particles. Dover Publications Inc., New York, 1981. [10.9] P. W. Atkins. Physical Chemistry. W. H. Freeman and Co., New York, 3rd edition, 1986. [10.10] G. Herzberg. Molecular Spectra and Molecular Structure. Kreiger Publishing, Malabar, Florida, 1989. In three volumes. References [10.11] D. K. Edwards. Molecular gas band radiation. In T. F. Irvine, Jr. and J. P. Hartnett, editors, Advances in Heat Transfer, volume 12, pages 119–193. Academic Press, Inc., New York, 1976. [10.12] H. C. Hottel and A. F. Sarofim. Radiative Transfer. McGraw-Hill Book Company, New York, 1967. [10.13] D. K. Edwards and R. Matavosian. Scaling rules for total absorptivity and emissivity of gases. J. Heat Transfer, 106(4):684–689, 1984. [10.14] M. Iqbal. An Introduction to Solar Radiation. Academic Press, Inc., New York, 1983. [10.15] J. A. Duffie and W. A. Beckman. Solar Engineering of Thermal Processes. John Wiley & Sons, Inc., New York, 2nd edition, 1991. [10.16] H. G. Houghton. Physical Meteorology. MIT Press, Cambridge, MA, 1985. [10.17] P. Berdahl and R. Fromberg. The thermal radiance of clear skies. Solar Energy, 29:299–314, 1982. [10.18] A. Skartveit, J. A. Olseth, G. Czeplak, and M. Rommel. On the estimation of atmospheric radiation from surface meteorological data. Solar Energy, 56:349–359, 1996. [10.19] P. Berdahl and M. Martin. The emissivity of clear skies. Solar Energy, 32:663–664, 1984. [10.20] J. A. Fay and D. S. Gollub. Energy and Environment. Oxford University Press, New York, 2002. [10.21] J. Hansen, R. Ruedy, M. Sato, M. Imhoff, W. Lawrence, D. Easterling, T. Peterson, and T. Karl. A closer look at United States and global surface temperature change. J. Geophysical Research, 106: 23947, 2001. [10.22] R. T. Watson, editor. Climate Change 2001: Synthesis Report. Third assessment report of the Intergovernmental Panel on Climate Change. Cambridge University Press, New York, 2002. Also available at http://www.ipcc.ch. 593 594 Chapter 10: Radiative heat transfer [10.23] P. A. Stott, S. F. B. Tett, G. S. Jones, M. R. Allen, J. F. B. Mitchell, and G. J. Jenkins. External control of 20th century temperature by natural and anthropogenic forcings. Science, 290:2133–2137, 2000. [10.24] F. Kreith and J. F. Kreider. Principles of Solar Engineering. Hemisphere Publishing Corp./McGraw-Hill Book Company, Washington, D.C., 1978. [10.25] U.S. Department of Commerce. Solar Heating and Cooling of Residential Buildings, volume 1 and 2. Washington, D.C., October 1977. Part V Mass Transfer 595 11. An introduction to mass transfer The edge of a colossal jungle, so dark-green as to be almost black, fringed with white surf, ran straight, like a ruled line, far, far away along a blue sea whose glitter was blurred by a creeping mist. The sun was fierce, the land seemed to glisten and drip with steam. Heart of Darkness, Joseph Conrad, 1902 11.1 Introduction We have, so far, dealt with heat transfer by convection, radiation, and diffusion (which we have been calling conduction). We have dealt only with situations in which heat passes through, or is carried by, a single medium. Many heat transfer processes, however, occur in mixtures of more than one substance. A wall exposed to a hot air stream may be cooled evaporatively by bleeding water through its surface. Water vapor may condense out of damp air onto cool surfaces. Heat will flow through an air-water mixture in these situations, but water vapor will diffuse or convect through air as well. This sort of transport of one substance relative to another is called mass transfer ; it did not occur in the single-component processes of the preceding chapters. In this chapter, we study mass transfer phenomena with an eye toward predicting heat and mass transfer rates in situations like those just mentioned. During mass transfer processes, an individual chemical species travels from regions where it has a high concentration to regions where it has a low concentration. When liquid water is exposed to a dry air stream, its vapor pressure may produce a comparatively high concentration of water vapor in the air near the water surface. The concentration difference between the water vapor near the surface and that in the air stream will drive the diffusion of vapor into the air stream. We call this evaporation. 597 598 An introduction to mass transfer §11.1 Figure 11.1 Schematic diagram of a natural-draft cooling tower at the Rancho Seco nuclear power plant. (From [11.1], courtesy of W. C. Reynolds.) In this and other respects, mass transfer is analogous to heat transfer. Just as thermal energy diffuses from regions of high temperature to regions of low temperature (following the temperature gradient), the mass of one species diffuses from regions high concentration to regions of low concentration (following its concentration gradient.) Just as the diffusional (or conductive) heat flux is directly proportional to a temperature gradient, so the diffusional mass flux of a species is often directly proportional to its concentration gradient; this is called Fick’s law of diffusion. Just as conservation of energy and Fourier’s law lead to equations for the convection and diffusion of heat, conservation of mass and Fick’s law lead to equations for the convection and diffusion of species in a mixture. The great similarity of the equations of heat convection and diffusion to those of mass convection and diffusion extends to the use of convective mass transfer coefficients, which, like heat transfer coefficients, relate convective fluxes to concentration differences. In fact, with simple modifications, the heat transfer coefficients of previous chapters may often be applied to mass transfer calculations. §11.1 Introduction Figure 11.2 A mechanical-draft cooling tower. The fans are located within the cylindrical housings at the top. Air is drawn in through the louvres on the side. Mass transfer, by its very nature, is intimately concerned with mixtures of chemical species. We begin this chapter by learning how to quantify the concentration of chemical species and by defining rates of movement of species. We make frequent reference to an arbitrary “species i,” the ith component of a mixture of N different species. These definitions may remind you of your first course in chemistry. We also spend some time, in Section 11.4, discussing how to calculate the transport properties of mixtures, such as diffusion coefficients and viscosities. Consider a typical technology that is dominated by mass transfer processes. Figure 11.1 shows a huge cooling tower used to cool the water leaving power plant condensers or other large heat exchangers. It is essentially an empty shell, at the bottom of which are arrays of cement boards or plastic louvres over which is sprayed the hot water to be cooled. The hot water runs down this packing, and a small portion of it evaporates into cool air that enters the tower from below. The remaining water, having been cooled by the evaporation, falls to the bottom, where it is collected and recirculated. The temperature of the air rises as it absorbs the warm vapor and, in 599 An introduction to mass transfer 600 §11.2 the natural-draft form of cooling tower shown, the upper portion of the tower acts as an enormous chimney through which the warm, moist air buoys, pulling in cool air at the base. In a mechanical-draft cooling tower, fans are used to pull air through the packing. Mechanical-draft towers are much shorter and can sometimes be seen on the roofs of buildings (Fig. 11.2). The working mass transfer process in a cooling tower is the evaporation of water into air. The rate of evaporation depends on the temperature and humidity of the incoming air, the feed-water temperature, and the air-flow characteristics of the tower and the packing. When the air flow is buoyancy-driven, the flow rates are directly coupled. Thus, mass transfer lies at the core of the complex design of a cooling tower. 11.2 Mixture compositions and species fluxes The composition of mixtures A mixture of various chemical species displays its own density, molecular weight, and other overall thermodynamic properties. These properties depend on the types and relative amounts of the component substances, which may vary from point to point in the mixture. To determine the local properties of a mixture, we must identify the local proportion of each species composing the mixture. One way to describe the amount of a particular species in a mixture is by the mass of that species per unit volume, known as the partial density. The mass of species i in a small volume of mixture, in kg, divided by that volume, in m3 , is the partial density, ρi , for that species, in kg of i per m3 . The composition of the mixture may be describe by stating the partial density of each of its components. The mass density of the mixture itself, ρ , is the total mass of all species per unit volume; therefore, ρ= ρi (11.1) i The relative amount of species i in the mixture may be described by the mass of i per unit mass of the mixture, which is simply ρi /ρ . This ratio is called the mass fraction, mi : mi ≡ ρi mass of species i = ρ mass of mixture (11.2) Mixture compositions and species fluxes §11.2 This definition leads to the following two results: mi = i ρi /ρ = 1 and 0 mi 1 (11.3) i The molar concentration of species i in kmol/m3 , ci , expresses concentration in terms of moles rather than mass. If Mi is the molecular weight of species i in kg/kmol, then ci ≡ ρi moles of i = Mi volume (11.4) The molar concentration of the mixture, c , is the total number of moles for all species per unit volume; thus, c= ci . (11.5) i The mole fraction of species i, xi , is the number of moles of i per mole of mixture: xi ≡ ci moles of i = c mole of mixture (11.6) Just as for the mass fraction, it follows for mole fraction that xi = i ci /c = 1 and 0 xi 1 (11.7) i The molecular weight of the mixture is the number of kg of mixture per kmol of mixture: M ≡ ρ/c . By using eqns. (11.1), (11.4), and (11.6) and (11.5), (11.4), and (11.2), respectively, M may be written in terms of either mole or mass fraction M= xi Mi or i 1 = M i mi Mi (11.8) Mole fraction may be converted to mass fraction using the following relations (derived in Problem 11.1): mi = xi Mi = M xi Mi k xk Mk and xi = Mmi = Mi mi /Mi k mk /Mk (11.9) In some circumstances, such as kinetic theory calculations, one works directly with the number of molecules of i per unit volume. This number density, Ni , is given by Ni = NA ci (11.10) where NA is Avogadro’s number, 6.02214 × 1026 molecules/kmol. 601 602 An introduction to mass transfer §11.2 Ideal gases The relations we have developed so far involve densities and concentrations that vary in as yet unknown ways with temperature or pressure. To get a more useful, though more restrictive, set of results, we now combine the preceding relations with the ideal gas law. For any individual component, i, we may write the partial pressure, pi , exterted by i as: pi = ρi Ri T (11.11) In eqn. (11.11), Ri is the ideal gas constant for species i: Ri ≡ R◦ Mi (11.12) where R ◦ is the universal gas constant, 8314.472 J/kmol· K. Equation (11.11) may alternatively be written in terms of ci : pi = ρi Ri T = (Mi ci ) R◦ T Mi = ci R ◦ T (11.13) Equations (11.5) and (11.13) can be used to relate c to p and T p pi =◦ R◦ T RT ci = c= i i (11.14) Multiplying the last two parts of eqn. (11.14) by R ◦ T yields Dalton’s law of partial pressures,1 p= pi (11.15) i Finally, we combine eqns. (11.6), (11.13), and (11.15) to obtain a very useful relationship between xi and pi : xi = pi ci pi = = ◦T c cR p (11.16) in which the last two equalities are restricted to ideal gases. 1 Dalton’s law (1801) is an empirical principle (not a deduced result) in classical thermodynamics. It can be deduced from molecular principles, however. We built the appropriate molecular principles into our development when we assumed eqn. (11.11) to be true. The reason that eqn. (11.11) is true is that ideal gas molecules occupy a mixture without influencing one another. Mixture compositions and species fluxes §11.2 Example 11.1 The most important mixture that we deal with is air. It has the following composition: Species Mass Fraction N2 O2 Ar trace gases 0.7556 0.2315 0.01289 < 0.01 Determine xO2 , pO2 , cO2 , and ρO2 for air at 1 atm. Solution. To make these calcuations, we need the molecular weights, which are given in Table 11.2 on page 616. We can start by checking the value of Mair , using the second of eqns. (11.8): mN2 mO2 mAr + + MN2 MO2 MAr Mair = −1 0.2315 0.01289 0.7556 + + 28.02 kg/kmol 32.00 kg/kmol 39.95 kg/kmol = −1 = 28.97 kg/kmol We may calculate the mole fraction using the second of eqns. (11.9) xO2 = mO2 M (0.2315)(28.97 kg/kmol) = 0.2095 = MO2 32.00 kg/kmol The partial pressure of oxygen in air at 1 atm is [eqn. (11.16)] pO2 = xO2 p = (0.2095)(101, 325 Pa) = 2.123 × 104 Pa We may now obtain cO2 from eqn. (11.13): pO2 R◦ T = (2.123 × 104 Pa) (300 K)(8314.5 J/kmol·K) c O2 = = 0.008510 kmol/m3 Finally, eqn. (11.4) gives the partial density ρO2 = cO2 MO2 = (0.008510 kmol/m3 )(32.00 kg/kmol) = 0.2723 kg/m3 603 604 An introduction to mass transfer §11.2 Velocities and fluxes Each species in a mixture undergoing a mass transfer process will have an species-average velocity, vi , which can be different for each species in the mixture, as suggested by Fig. 11.3. We may obtain the mass-average velocity,2 v , for the entire mixture from the species average velocities using the formula ρv = ρi vi . (11.17) i This equation is essentially a local calculation of the mixture’s net momentum per unit volume. We refer to ρ v as the mixture’s mass flux, n, ˙ and we call its scalar magnitude m ; each has units of kg/m2 ·s. Likewise, the mass flux of species i is ni = ρi vi (11.18) and, from eqn. (11.17), we see that the mixture’s mass flux equals the sum of all species’ mass fluxes ni = ρ v n= (11.19) i Since each species diffusing through a mixture has some velocity relative to the mixture’s mass-average velocity, the diffusional mass flux, ji , of a species relative to the mixture’s mean flow may be identified: ji = ρi vi − v . (11.20) The total mass flux of the ith species, ni , includes both this diffusional mass flux and bulk convection by the mean flow, as is easily shown: ni = ρi vi = ρi v + ρi vi − v = ρi v + ji = mi n convection + (11.21) ji diffusion 2 The mass average velocity, v , given by eqn. (11.17) is identical to the fluid velocity, u, used in previous chapters. This is apparent if one applies eqn. (11.17) to a “mixture” composed of only one species. We use the symbol v here because v is the more common notation in the mass transfer literature. Mixture compositions and species fluxes §11.2 605 Figure 11.3 Molecules of different species in a mixture moving with different average velocities. The velocity vi is the average over all molecules of species i. Although the convective transport contribution is fully determined as soon as we know the velocity field and partial densities, the causes of diffusion need further discussion, which we defer to Section 11.3. Combining eqns. (11.19) and (11.21), we find that ρi v + ni = n= i ji = ρ v + i i ji = n + i ji i Hence ji = 0 (11.22) i Diffusional mass fluxes must sum to zero because they are each defined relative to the mean mass flux. Velocities may also be stated in molar terms. The mole flux of the ith species, Ni , is ci vi , in kmol/m2 · s. The mixture’s mole flux, N , is obtained by summing over all species ci vi = c v ∗ Ni = N= i (11.23) i where we define the mole-average velocity, v ∗ , as shown. The last flux ∗ we define is the diffusional mole flux, Ji : ∗ Ji = ci vi − v ∗ (11.24) An introduction to mass transfer 606 §11.2 It may be shown, using these definitions, that ∗ Ni = xi N + Ji (11.25) Substitution of eqn. (11.25) into eqn. (11.23) gives Ni = N N= i ∗ Ji = N + xi + i i ∗ Ji i so that ∗ Ji = 0. (11.26) i ∗ Thus, both the Ji ’s and the ji ’s sum to zero. Example 11.2 At low temperatures, carbon oxidizes (burns) in air through the surface reaction: C + O2 → CO2 . Figure 11.4 shows the carbon-air interface in a coordinate system that moves into the stationary carbon at the same speed that the carbon burns away—as though the observer were seated on the moving interface. Oxygen flows toward the carbon surface and carbon dioxide flows away, with a net flow of carbon through the interface. If the system is at steady state and, if a separate analysis shows that carbon is consumed at the rate of 0.00241 kg/m2 ·s, find the mass and mole fluxes through an imaginary surface, s , that stays close to the gas side of the interface. For this case, concentrations at the s -surface turn out to be mO2 ,s = 0.20, mCO2 ,s = 0.052, and ρs = 0.29 kg/m3 . Solution. The mass balance for the reaction is 12.0 kg C + 32.0 kg O2 → 44.0 kg CO2 Since carbon flows through a second imaginary surface, u, moving through the stationary carbon just below the interface, the mass fluxes are related by nC,u = − 12 12 nO2 ,s = nCO2 ,s 32 44 The minus sign arises because the O2 flow is opposite the C and CO2 flows, as shown in Figure 11.4. In steady state, if we apply mass Mixture compositions and species fluxes §11.2 607 Figure 11.4 Low-temperature carbon oxidation. conservation to the control volume between the u and s surfaces, we find that the total mass flux entering the u-surface equals that leaving the s -surface nC,u = nCO2 ,s + nO2 ,s = 0.00241 kg/m2 ·s Hence, nO2 ,s = − nCO2 ,s = 32 (0.00241 kg/m2 ·s) = −0.00643 kg/m2 ·s 12 44 (0.00241 kg/m2 ·s) = 0.00884 kg/m2 ·s 12 To get the diffusional mass flux, we need species and mass average speeds from eqns. (11.18) and (11.19): vO2 ,s = vCO2 ,s = vs = nO2 ,s ρO2 ,s = −0.00643 kg/m2 ·s 0.2 (0.29 kg/m3 ) = −0.111 m/s nCO2 ,s ρCO2 ,s = 0.00884 kg/m2 ·s 0.052 (0.29 kg/m3 ) = 1 ρs ni = i (0.00884 − 0.00643) kg/m2 ·s = 0.29 kg/m3 Thus, from eqn. (11.20), ji,s = ρi,s vi,s − vs ⎧ ⎨−0.00691 kg/m2 ·s for O2 = ⎩ 0.00871 kg/m2 ·s for CO2 0.586 m/s 0.00831 m/s An introduction to mass transfer 608 §11.3 The diffusional mass fluxes, ji,s , are very nearly equal to the species mass fluxes, ni,s . That is because the mass-average speed, vs , is much less than the species speeds, vi,s , in this case. Thus, the convective contribution to ni,s is much smaller than the diffusive contribution, and mass transfer occurs primarily by diffusion. Note that jO2 ,s and jCO2 ,s do not sum to zero because the other, nonreacting species in air must diffuse against the small convective velocity, vs (see Section 11.7). One mole of carbon surface reacts with one mole of O2 to form one mole of CO2 . Thus, the mole fluxes of each species have the same magnitude at the interface: NCO2 ,s = −NO2 ,s = NC,u = nC,u = 0.000201 kmol/m2 ·s MC ∗ The mole average velocity at the s -surface, vs , is identically zero by eqn. (11.23), since NCO2 ,s + NO2 ,s = 0. The diffusional mole fluxes are ⎧ ⎨−0.000201 kmol/m2 ·s for O2 ∗ ∗ Ji,s = ci,s vi,s − vs = Ni,s = ⎩ 0.000201 kmol/m2 ·s for CO2 =0 These two diffusional mole fluxes sum to zero themselves because there is no convective mole flux for other species to diffuse against ∗ (i.e., for the other species Ji,s = 0). The reader may calculate the velocity of the interface from nc,u . That calculation would show the interface to be receding so slowly that the velocities we calculate are almost equal to those that would be seen by a stationary observer. 11.3 Diffusion fluxes and Fick’s law When the composition of a mixture is nonuniform, the concentration gradient in any species, i, of the mixture provides a driving potential for the diffusion of that species. It flows from regions of high concentration to regions of low concentration—similar to the diffusion of heat from regions of high temperature to regions of low temperature. We have already noted in Section 2.1 that mass diffusion obeys Fick’s law ji = −ρ Dim ∇mi (11.27) Diffusion fluxes and Fick’s law §11.3 609 which is analogous to Fourier’s law. The constant of proportionality, ρ Dim , between the local diffusive mass flux of species i and the local concentration gradient of i involves a physical property called the diffusion coefficient, Dim , for species i diffusing in the mixture m. Like the thermal diffusivity, α, or the kinematic viscosity (a momentum diffusivity), ν , the mass diffusivity Dim has the units of m2/s. These three diffusivities can form three dimensionless groups, among which is the Prandtl number: The Prandtl number, Pr ≡ ν/α The Schmidt number,3 Sc ≡ ν/Dim The Lewis number,4 (11.28) Le ≡ α/Dim = Sc/Pr Each of these groups compares the relative strength of two different diffusive processes. We make considerable use of the Schmidt number later in this chapter. When diffusion occurs in mixtures of only two species—so-called binary mixtures —Dim reduces to the binary diffusion coefficient, D12 . In fact, the best-known kinetic models are for binary diffusion.5 In binary diffusion, species 1 has the same diffusivity through species 2 as does species 2 through species 1 (see Problem 11.5); in other words, D12 = D21 3 (11.29) Ernst Schmidt (1892–1975) served successively as the professor of thermodynamics at the Technical Universities of Danzig, Braunschweig, and Munich (Chapter 6, footnote 3). His many contributions to heat and mass transfer include the introduction of aluminum foil as radiation shielding, the first measurements of velocity and temperature fields in a natural convection boundary layer, and a once widely-used graphical procedure for solving unsteady heat conduction problems. He was among the first to develop the analogy between heat and mass transfer. 4 Warren K. Lewis (1882–1975) was a professor of chemical engineering at M.I.T. from 1910 to 1975 and headed the department throughout the 1920s. He defined the original paradigm of chemical engineering, that of “unit operations”, and, through his textbook with Walker and McAdams, Principles of Chemical Engineering, he laid the foundations of the discipline. He was a prolific inventor in the area of industrial chemistry, holding more than 80 patents. He also did important early work on simultaneous heat and mass transfer in connection with evaporation problems. 5 Actually, Fick’s Law is strictly valid only for binary mixtures. It can, however, often be applied to multicomponent mixtures with an appropriate choice of Dim (see Section 11.4). 610 An introduction to mass transfer §11.3 A kinetic model of diffusion Diffusion coefficients depend upon composition, temperature, and pressure. Equations that predict D12 and Dim are given in Section 11.4. For now, let us see how Fick’s law arises from the same sort of elementary molecular kinetics that gave Fourier’s and Newton’s laws in Section 6.4. We consider a two-component dilute gas (one with a low density) in which the molecules A of one species are very similar to the molecules A of a second species (as though some of the molecules of a pure gas had merely been labeled without changing their properties.) The resulting process is called self-diffusion. If we have a one-dimensional concentration distribution, as shown in Fig. 11.5, molecules of A diffuse down their concentration gradient in the x -direction. This process is entirely analogous to the transport of energy and momentum shown in Fig. 6.13. We take the temperature and pressure of the mixture (and thus its number density) to be uniform and the mass-average velocity to be zero. Individual molecules move at a speed C , which varies randomly from molecule to molecule and is called the thermal or peculiar speed. The average speed of the molecules is C . The average rate at which molecules cross the plane x = x0 in either direction is proportional to N C , where N is the number density (molecules/m3 ). Prior to crossing the x0 -plane, the molecules travel a distance close to one mean free path, —call it a , where a is a number on the order of one. The molecular flux travelling rightward across x0 , from its plane of origin at x0 − a , then has a fraction of molecules of A equal to the value of NA /N at x0 − a . The leftward flux, from x0 + a , has a fraction equal to the value of NA /N at x0 + a . Since the mass of a molecule of A is MA /NA (where NA is Avogadro’s number), the net mass flux in the x -direction is then jA x0 = η NC MA NA NA N x0 −a − NA N x0 +a (11.30) where η is a constant of proportionality. Since NA /N changes little in a distance of two mean free paths (in most real situations), we can expand the right side of eqn. (11.30) in a two-term Taylor series expansion about Diffusion fluxes and Fick’s law §11.3 611 Figure 11.5 One-dimensional diffusion. x0 and obtain Fick’s law: jA x0 MA NA = η NC −2a dmA = −2ηa(C )ρ dx d(NA /N ) dx x0 (11.31) x0 (for details, see Problem 11.6). Thus, we identify DAA = (2ηa)C (11.32) and Fick’s law takes the form jA = −ρ DAA dmA dx (11.33) The constant, ηa, in eqn. (11.32) can be fixed only with the help of a more detailed kinetic theory calculation [11.2], the result of which is given in Section 11.4. The choice of ji and mi for the description of diffusion is really some∗ what arbitrary. The molar diffusion flux, Ji , and the mole fraction, xi , are often used instead, in which case Fick’s law reads ∗ Ji = −c Dim ∇xi (11.34) Obtaining eqn. (11.34) from eqn. (11.27) for a binary mixture is left as an exercise (Problem 11.4). 612 An introduction to mass transfer §11.3 Typical values of the diffusion coefficient Fick’s law works well in low density gases and in dilute liquid and solid solutions, but for concetrated liquid and solid solutions the diffusion coefficient is found to vary with the concentration of the diffusing species. In part, the concentration dependence of those diffusion coefficients reflects the inadequacy of the concentration gradient in representing the driving force for diffusion in nondilute solutions. Gradients in the chemical potential actually drive diffusion. In concentrated liquid or solid solutions, chemical potential gradients are not always equivalent to concentration gradients [11.3, 11.4, 11.5]. Table 11.1 lists some experimental values of the diffusion coefficient in binary gas mixtures and dilute liquid solutions. For gases, the diffusion coefficient is typically on the order of 10−5 m2 /s near room temperature. For liquids, the diffusion coefficient is much smaller, on the order of 10−9 m2 /s near room temperature. For both liquids and gases, the diffusion coefficient rises with increasing temperature. Typical diffusion coefficients in solids (not listed) may range from about 10−20 to about 10−9 m2 /s, depending upon what substances are involved and the temperature [11.6]. The diffusion of water vapor through air is of particular technical importance, and it is therefore useful to have an empirical correlation specifically for that mixture: DH2 O,air = 1.87 × 10−10 T 2.072 p for 282 K ≤ T ≤ 450 K (11.35) where DH2 O,air is in m2 /s, T is in kelvin, and p is in atm [11.7]. The scatter of the available data around this equation is about 10%. Coupled diffusion phenomena Mass diffusion can be driven by factors other than concentration gradients, although the latter are of primary importance. For example, temperature gradients can induce mass diffusion in a process known as thermal diffusion or the Soret effect. The diffusional mass flux resulting from both temperature and concentration gradients in a binary gas mixture is then [11.2] ji = −ρ D12 ∇m1 + M1 M2 kT ∇ ln(T ) M2 (11.36) Diffusion fluxes and Fick’s law §11.3 Table 11.1 Typical diffusion coefficients for binary gas mixtures at 1 atm and dilute liquid solutions [11.4]. Gas mixture air-carbon dioxide air-ethanol air-helium air-napthalene air-water T (K) 276 313 276 303 313 D12 (m2/s) 1.42×10−5 1.45 6.24 0.86 2.88 argon-helium 295 628 1068 8.3 32.1 81.0 (dilute solute, 1)-(liquid solvent, 2) T (K) D12 (m2/s) ethanol-benzene benzene-ethanol water-ethanol carbon dioxide-water ethanol-water 288 298 298 298 288 2.25×10−9 1.81 1.24 2.00 1.00 methane-water 275 333 0.85 3.55 pyridene-water 288 0.58 where kT is called the thermal diffusion ratio and is generally quite small. Thermal diffusion is occasionally used in chemical separation processes. Pressure gradients and body forces acting unequally on the different species can also cause diffusion. Again, these effects are normally small. A related phenomenon is the generation of a heat flux by a concentration gradient (as distinct from heat convected by diffusing mass), called the diffusion-thermo or Dufour effect. In this chapter, we deal only with mass transfer produced by concentration gradients. 613 An introduction to mass transfer 614 11.4 §11.4 Transport properties of mixtures6 Direct measurements of mixture transport properties are not always available for the temperature, pressure, or composition of interest. Thus, we must often rely upon theoretical predictions or experimental correlations for estimating mixture properties. In this section, we discuss methods for computing Dim , k, and µ in gas mixtures using equations from kinetic theory—particularly the Chapman-Enskog theory [11.2, 11.8, 11.9]. We also consider some methods for computing D12 in dilute liquid solutions. The diffusion coefficient for binary gas mixtures As a starting point, we return to our simple model for the self-diffusion coefficient of a dilute gas, eqn. (11.32). We can approximate the average molecular speed, C , by Maxwell’s equilibrium formula (see, e.g., [11.9]): C= 8kB NA T πM 1/2 (11.37) where kB = R ◦ /NA is Boltzmann’s constant. If we assume the molecules to be rigid and spherical, then the mean free path turns out to be = kB T 1 √ =√2 2 π 2N d π 2d p (11.38) where d is the effective molecular diameter. Substituting these values of C and in eqn. (11.32) and applying a kinetic theory calculation that shows 2ηa = 1/2, we find DAA = (2ηa)C = (kB /π )3/2 d2 NA M 1/2 T 3/2 p (11.39) The diffusion coefficient varies as p −1 and T 3/2 , based on the simple model for self-diffusion. To get a more accurate result, we must take account of the fact that molecules are not really hard spheres. We also have to allow for differences in the molecular sizes of different species and for nonuniformities 6 This section may be omitted without loss of continuity. The property predictions of this section are used only in Examples 11.11, 11.14, and 11.16, and in some of the end-of-chapter problems. Transport properties of mixtures §11.4 Figure 11.6 The Lennard-Jones potential. in the bulk properties of the gas. The Chapman-Enskog kinetic theory takes all these factors into account [11.8], resulting in the following formula for DAB : DAB = (1.8583 × 10−7 )T 3/2 AB p ΩD (T ) 1 1 + MA MB where the units of p , T , and DAB are atm, K, and m2/s, respectively. The AB function ΩD (T ) describes the collisions between molecules of A and B . It depends, in general, on the specific type of molecules involved and the temperature. The type of molecule matters because of the intermolecular forces of attraction and repulsion that arise when molecules collide. A good approximation to those forces is given by the Lennard-Jones intermolecular potential (see Fig. 11.6.) This potential is based on two parameters, a molecular diameter, σ , and a potential well depth, ε. The potential well depth is the energy required to separate two molecules from one another. Both constants can be inferred from physical property data. Some values are given in Table 11.2 together with the associated molecular weights (from [11.10], with values for calculating the diffusion coefficients of water from [11.11]). 615 An introduction to mass transfer 616 §11.4 Table 11.2 Lennard-Jones constants and molecular weights of selected species. Species σ (Å) ε/kB (K) Al Air Ar Br2 C CCl2 F2 CCl4 CH3 OH CH4 CN CO CO2 C2 H6 C2 H5 OH CH3 COCH3 C6 H6 Cl2 F2 2.655 3.711 3.542 4.296 3.385 5.25 5.947 3.626 3.758 3.856 3.690 3.941 4.443 4.530 4.600 5.349 4.217 3.357 2750 78.6 93.3 507.9 30.6 253 322.7 481.8 148.6 75.0 91.7 195.2 215.7 362.6 560.2 412.3 316.0 112.6 a b M kg kmol 26.98 28.96 39.95 159.8 12.01 120.9 153.8 32.04 16.04 26.02 28.01 44.01 30.07 46.07 58.08 78.11 70.91 38.00 Species σ (Å) ε/kB (K) H2 H2 O H2 O H2 O2 H2 S He Hg I2 Kr Mg NH3 N2 N2 O Ne O2 SO2 Xe 2.827 2.655a 2.641b 4.196 3.623 2.551 2.969 5.160 3.655 2.926 2.900 3.798 3.828 2.820 3.467 4.112 4.047 59.7 363a 809.1b 289.3 301.1 10.22 750 474.2 178.9 1614 558.3 71.4 232.4 32.8 106.7 335.4 231.0 M kg kmol 2.016 18.02 34.01 34.08 4.003 200.6 253.8 83.80 24.31 17.03 28.01 44.01 20.18 32.00 64.06 131.3 Based on mass diffusion data. Based on viscosity and thermal conductivity data. AB An accurate approximation to ΩD (T ) can be obtained using the Lennard-Jones potential function. The result is AB 2 ΩD (T ) = σAB ΩD kB T εAB where, the collision diameter, σAB , may be viewed as an effective molecular diameter for collisions of A and B . If σA and σB are the cross-sectional diameters of A and B , in Å,7 then σAB = (σA + σB ) 2 (11.40) The collision integral, ΩD is a result of kinetic theory calculations calculations based on the Lennard-Jones potential. Table 11.3 gives values of 7 One Ångström (1 Å) is equal to 0.1 nm. Transport properties of mixtures §11.4 617 ΩD from [11.12]. The effective potential well depth for collisions of A and B is √ (11.41) εAB = εA εB Hence, we may calculate the binary diffusion coefficient from DAB = (1.8583 × 10−7 )T 3/2 2 pσAB ΩD 1 1 + MA MB (11.42) where, again, the units of p , T , and DAB are atm, K, and m2/s, respectively, and σAB is in Å. Equation (11.42) indicates that the diffusivity varies as p −1 and is independent of mixture concentrations, just as the simple model indicated that it should. The temperature dependence of ΩD , however, increases the overall temperature dependence of DAB from T 3/2 , as suggested by eqn. (11.39), to approximately T 7/4 . Air, by the way, can be treated as a single substance in Table 11.2 owing to the similarity of its two main constituents, N2 and O2 . Example 11.3 Compute DAB for the diffusion of hydrogen in air at 276 K and 1 atm. Solution. Let air be species A and H2 be species B . Then we read from Table 11.2 εA εB = 78.6 K, = 59.7 K σA = 3.711 Å, σB = 2.827 Å, kB kB and calculate these values σAB = (3.711 + 2.827)/2 = 3.269 Å εAB kB = (78.6)(59.7) = 68.5 K Hence, kB T /εAB = 4.029, and ΩD = 0.8822 from Table 11.3. Then DAB = (1.8583 × 10−7 )(276)3/2 (1)(3.269)2 (0.8822) 1 1 + m2 /s 2.016 28.97 = 6.58 × 10−5 m2 /s An experimental value from Table 11.1 is 6.34 × 10−5 m2 /s, so the prediction is high by 5%. Table 11.3 Collision integrals for diffusivity, viscosity, and thermal conductivity based on the Lennard-Jones potential. kB T /ε 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40 1.45 1.50 1.55 1.60 1.65 1.70 1.75 1.80 1.85 1.90 1.95 2.00 2.10 2.20 2.30 2.40 2.50 2.60 618 ΩD 2.662 2.476 2.318 2.184 2.066 1.966 1.877 1.798 1.729 1.667 1.612 1.562 1.517 1.476 1.439 1.406 1.375 1.346 1.320 1.296 1.273 1.253 1.233 1.215 1.198 1.182 1.167 1.153 1.140 1.128 1.116 1.105 1.094 1.084 1.075 1.057 1.041 1.026 1.012 0.9996 0.9878 Ωµ = Ωk 2.785 2.628 2.492 2.368 2.257 2.156 2.065 1.982 1.908 1.841 1.780 1.725 1.675 1.629 1.587 1.549 1.514 1.482 1.452 1.424 1.399 1.375 1.353 1.333 1.314 1.296 1.279 1.264 1.248 1.234 1.221 1.209 1.197 1.186 1.175 1.156 1.138 1.122 1.107 1.093 1.081 kB T /ε 2.70 2.80 2.90 3.00 3.10 3.20 3.30 3.40 3.50 3.60 3.70 3.80 3.90 4.00 4.10 4.20 4.30 4.40 4.50 4.60 4.70 4.80 4.90 5.00 6.00 7.0 8.0 9.0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0 100.0 200.0 300.0 400.0 ΩD Ωµ = Ωk 0.9770 0.9672 0.9576 0.9490 0.9406 0.9328 0.9256 0.9186 0.9120 0.9058 0.8998 0.8942 0.8888 0.8836 0.8788 0.8740 0.8694 0.8652 0.8610 0.8568 0.8530 0.8492 0.8456 0.8422 0.8124 0.7896 0.7712 0.7556 0.7424 0.6640 0.6232 0.5960 0.5756 0.5596 0.5464 0.5352 0.5256 0.5170 0.4644 0.4360 0.4172 1.069 1.058 1.048 1.039 1.030 1.022 1.014 1.007 0.9999 0.9932 0.9870 0.9811 0.9755 0.9700 0.9649 0.9600 0.9553 0.9507 0.9464 0.9422 0.9382 0.9343 0.9305 0.9269 0.8963 0.8727 0.8538 0.8379 0.8242 0.7432 0.7005 0.6718 0.6504 0.6335 0.6194 0.6076 0.5973 0.5882 0.5320 0.5016 0.4811 §11.4 Transport properties of mixtures 619 Limitations of the diffusion coefficient prediction. Equation (11.42) is not valid for all gas mixtures. We have already noted that concentration gradients cannot be too steep; thus, it cannot be applied in, say, the interior of a shock wave when the Mach number is significantly greater than unity. Furthermore, the gas must be dilute, and its molecules should be, in theory, nonpolar and approximately spherically symmetric. Reid et al. [11.4] compared values of D12 calculated using eqn. (11.42) with data for binary mixtures of monatomic, polyatomic, nonpolar, and polar gases of the sort appearing in Table 11.2. They reported an average absolute error of 7.3 percent. Better results can be obtained by using values of σAB and εAB that have been fit specifically to the pair of gases involved, rather than using eqns. (11.40) and (11.41), or by constructing AB a mixture-specific equation for ΩD (T ) [11.13, Chap. 11]. The density of the gas also affects the accuracy of kinetic theory predictions, which require the gas to be dilute in the sense that its molecules interact with one another only during brief two-molecule collisions. Childs and Hanley [11.14] have suggested that the transport properties of gases are within 1% of the dilute values if the gas densities do not exceed the following limiting value ρmax = 22.93M (σ 3 Ωµ ) (11.43) Here, σ (the collision diameter of the gas) and ρ are expressed in Å and kg/m3 , and Ωµ —a second collision integral for viscosity—is included in Table 11.3. Equation (11.43) normally gives ρmax values that correspond to pressures substantially above 1 atm. At higher gas densities, transport properties can be estimated by a variety of techniques, such as corresponding states theories, absolute reaction-rate theories, or modified Enskog theories [11.13, Chap. 6] (also see [11.4, 11.8]). Conversely, if the gas density is so very low that the mean free path is on the order of the dimensions of the system, we have what is called free molecule flow, and the present kinetic models are again invalid (see, e.g., [11.15]). Diffusion coefficients for multicomponent gases We have already noted that an effective binary diffusivity, Dim , can be used to represent the diffusion of species i into a mixture m. The preceding equations for the diffusion coefficient, however, are strictly applicable only when one pure substance diffuses through another. Different equations are needed when there are three or more species present. 620 An introduction to mass transfer §11.4 If a low concentration of species i diffuses into a homogeneous mix∗ 0 for j ≠ i, and one may show (Probture of n species, then Jj lem 11.14) that D−1 = im n j =1 j ≠i xj Dij (11.44) where Dij is the binary diffusion coefficient for species i and j alone. This rule is sometimes called Blanc’s law [11.4]. If a mixture is dominantly composed of one species, A, and includes only small traces of several other species, then the diffusion coefficient of each trace gas is approximately the same as it would be if the other trace gases were not present. In other words, for any particular trace species i, Dim DiA (11.45) Finally, if the binary diffusion coefficient has the same value for each pair of species in a mixture, then one may show (Problem 11.14) that Dim = Dij , as one might expect. Diffusion coefficients for binary liquid mixtures Each molecule in a liquid is always in contact with several neighboring molecules, and a kinetic theory like that used in gases, which relies on detailed descriptions of two-molecule collisions, is no longer feasible. Instead, a less precise theory can be developed and used to correlate experimental measurements. For a dilute solution of substance A in liquid B , the so-called hydrodynamic model has met some success. Suppose that, when a force per molecule of FA is applied to molecules of A, they reach an average steady speed of vA relative to the liquid B . The ratio vA /FA is called the mobility of A. If there is no applied force, then the molecules of A diffuse as a result of random molecular motions (which we call Brownian motion ). Kinetic and thermodynamic arguments, such as those given by Einstein [11.16] and Sutherland [11.17], lead to an expression for the diffusion coefficient of A in B as a result of Brownian motion: DAB = kB T (vA /FA ) Equation (11.46) is usually called the Nernst-Einstein equation. (11.46) §11.4 Transport properties of mixtures 621 To evaluate the mobility of a molecular (or particulate) solute, we may make the rather bold approximation that Stokes’ law [11.18] applies, even though it is really a drag law for spheres at low Reynolds number (ReD < 1) FA = 6π µB vA RA 1 + 2µB /βRA 1 + 3µB /βRA (11.47) Here, RA is the radius of sphere A and β is a coefficient of “sliding” friction, for a friction force proportional to the velocity. Substituting eqn. (11.47) in eqn. (11.46), we get kB DAB µB = T 6π RA 1 + 3µB /βRA 1 + 2µB /βRA (11.48) This model is valid if the concentration of solute A is so low that the molecules of A do not interact with one another. For viscous liquids one usually assumes that no slip occurs between the liquid and a solid surface that it touches; but, for particles whose size is on the order of the molecular spacing of the solvent molecules, some slip may very well occur. This is the reason for the unfamiliar factor in parentheses on the right side of eqn. (11.47). For large solute particles, there should be no slip, so β → ∞ and the factor in parentheses tends to one, as expected. Equation (11.48) then reduces to8 kB DAB µB = T 6π RA (11.49a) For smaller molecules—close in size to those of the solvent—we expect that β → 0, leading to [11.19] DAB µB kB = T 4π RA (11.49b) The most important feature of eqns. (11.48), (11.49a), and (11.49b) is that, so long as the solute is dilute, the primary determinant of the group Dµ T is the size of the diffusing species, with a secondary dependence on intermolecular forces (i.e., on β). More complex theories, such 8 Equation (11.49a) was first presented by Einstein in May 1905. The more general form, eqn. (11.48), was presented independently by Sutherland in June 1905. Equations (11.48) and (11.49a) are commonly called the Stokes-Einstein equation, although Stokes had no hand in applying eqn. (11.47) to diffusion. It might therefore be argued that eqn. (11.48) should be called the Sutherland-Einstein equation. 622 An introduction to mass transfer §11.4 Table 11.4 Molal specific volumes and latent heats of vaporization for selected substances at their normal boiling points. Substance Methanol Ethanol n-Propanol Isopropanol n-Butanol tert -Butanol n-Pentane Cyclopentane Isopentane Neopentane n-Hexane Cyclohexane n-Heptane n-Octane n-Nonane n-Decane Acetone Benzene Carbon tetrachloride Ethyl bromide Nitromethane Water Vm (m3 /kmol) 0.042 0.064 0.081 0.072 0.103 0.103 0.118 0.100 0.118 0.118 0.141 0.117 0.163 0.185 0.207 0.229 0.074 0.096 0.102 0.075 0.056 0.0187 hfg (MJ/kmol) 35.53 39.33 41.97 40.71 43.76 40.63 25.61 27.32 24.73 22.72 28.85 33.03 31.69 34.14 36.53 39.33 28.90 30.76 29.93 27.41 25.44 40.62 as the absolute reaction-rate theory of Eyring [11.20], lead to the same dependence. Moreover, experimental studies of dilute solutions verify that the group Dµ/T is essentially temperature-independent for a given solute-solvent pair, with the only exception occuring in very high viscosity solutions. Thus, most correlations of experimental data have used some form of eqn. (11.48) as a starting point. Many such correlations have been developed. One fairly successful correlation is due to King et al. [11.21]. They expressed the molecular size in terms of molal volumes at the normal boiling point, Vm,A and Vm,B , and accounted for intermolecular association forces using the latent heats of §11.4 Transport properties of mixtures 623 Figure 11.7 Comparison of liquid diffusion coefficients predicted by eqn. (11.50) with experimental values for assorted substances from [11.4]. vaporization at the normal boiling point, hfg,A and hfg,B . They obtained DAB µB = (4.4 × 10−15 ) T Vm,B Vm,A 1/6 hfg,B hfg,A 1/2 (11.50) which has an rms error of 19.5% and for which the units of DAB µB /T are kg·m/K·s2 . Values of hfg and Vm are given for various substances in Table 11.4. Equation (11.50) is valid for nonelectrolytes at high dilution, and it appears to be satisfactory for both polar and nonpolar substances. The difficulties with polar solvents of high viscosity led the authors to limit eqn. (11.50) to values of Dµ/T < 1.5 × 10−14 kg·m/K·s2 . The predictions of eqn. (11.50) are compared with experimental data in Fig. 11.7. Reid et al. [11.4] review several other liquid-phase correlations and provide an assessment of their accuracies. An introduction to mass transfer 624 §11.4 The thermal conductivity and viscosity of dilute gases In any convective mass transfer problem, we must know the viscosity of the fluid and, if heat is also being transferred, we must also know its thermal conductivity. Accordingly, we now consider the calculation of µ and k for mixtures of gases. Two of the most important results of the kinetic theory of gases are the predictions of µ and k for a pure, monatomic gas of species A: µA = 2.6693 × 10−6 MA T 2 σA Ωµ (11.51) and kA = 0.083228 2 σA Ωk T MA (11.52) where Ωµ and Ωk are collision integrals for the viscosity and thermal conductivity. In fact, Ωµ and Ωk are equal to one another, but they are different from ΩD . In these equations µ is in kg/m·s, k is in W/m·K, T is in kelvin, and σA again has units of Å. The equation for µA applies equally well to polyatomic gases, but kA must be corrected to account for internal modes of energy storage— chiefly molecular rotation and vibration. Eucken (see, e.g., [11.9]) gave a simple analysis showing that this correction was k= 9γ − 5 4γ µ cp (11.53) for an ideal gas, where γ ≡ cp /cv . You may recall from your thermodynamics courses that γ is 5/3 for monatomic gases, 7/5 for diatomic gases at modest temperatures, and approaches unity for very complex molecules. Equation (11.53) should be used with tabulated data for cp ; on average, it will underpredict k by perhaps 10 to 20% [11.4]. An approximate formula for µ for multicomponent gas mixtures was developed by Wilke [11.22], based on the kinetic theory of gases. He introduced certain simplifying assumptions and obtained, for the mixture viscosity, n µm = i=1 xi µi n j =1 xj φij (11.54) Transport properties of mixtures §11.4 625 where φij 1 + (µi /µj )1/2 (Mj /Mi )1/4 = √ 1/2 2 2 1 + (Mi /Mj ) 2 The analogous equation for the thermal conductivity of mixtures was developed by Mason and Saxena [11.23]: n km = i=1 xi ki (11.55) n j =1 xj φij (We have followed [11.4] in omitting a minor empirical correction factor proposed by Mason and Saxena.) Equation (11.54) is accurate to about 2 % and eqn. (11.55) to about 4% for mixtures of nonpolar gases. For higher accuracy or for mixtures with polar components, refer to [11.4] and [11.13]. Example 11.4 Compute the transport properties of normal air at 300 K. Solution. The mass composition of air was given in Example 11.1. Using the methods of Example 11.1, we obtain the mole fractions as xN2 = 0.7808, xO2 = 0.2095, and xAr = 0.0093. We first compute µ and k for the three species to illustrate the use of eqns. (11.51) to (11.53), although we could simply use tabled data in eqns. (11.54) and (11.55). From Tables 11.2 and 11.3, we obtain Species σ (Å) ε/kB (K) N2 O2 Ar 3.798 3.467 3.542 71.4 106.7 93.3 M Ωµ 28.02 32.00 39.95 0.9599 1.057 1.021 Substitution of these values into eqn. (11.51) yields 626 An introduction to mass transfer §11.4 Species µcalc (kg/m·s) µdata (kg/m·s) N2 O2 Ar 1.767 × 10−5 2.059 × 10−5 2.281 × 10−5 1.80 × 10−5 2.07 × 10−5 2.29 × 10−5 where we show data from Appendix A (Table A.6) for comparison. We then read cp from Appendix A and use eqn. (11.52) and (11.53) to get the thermal conductivities of the components: Species cp (J/kg·K) kcalc (W/m·K) N2 O2 Ar 1041. 919.9 521.5 kdata (W/m·K) 0.02500 0.02569 0.01782 0.0260 0.02615 0.01787 The predictions thus agree with the data to within about 2% for µ and within about 4% for k. To compute µm and km , we use eqns. (11.54) and (11.55) and the tabulated values of µ and k. Identifying N2 , O2 , and Ar as species 1, 2, and 3, we get φ12 = 0.9894, φ21 = 1.010 φ13 = 1.043, φ31 = 0.9445 φ23 = 1.058, φ32 = 0.9391 and φii = 1. The sums appearing in the denominators are ⎧ ⎪0.9978 for i = 1 ⎪ ⎨ xj φij = 1.008 for i = 2 ⎪ ⎪ ⎩ 0.9435 for i = 3 When they are substituted in eqns. (11.54) and (11.55), these values give µm,calc = 1.861 × 10−5 kg/m·s, µm,data = 1.857 × 10−5 kg/m·s km,calc = 0.02596 W/m·K, km,data = 0.02623 W/m·K so the mixture values are also predicted within 0.3 and 1.0%, respectively. The equation of species conservation §11.5 Finally, we need cpm to compute the Prandtl number of the mixture. This is merely the mass weighted average of cp , or i mi cpi , and it is equal to 1006 J/kg·K. Then Pr = (µcp /k)m = (1.861 × 10−5 )(1006)/0.02596 = 0.721. This is 1% above the tabled value of 0.713. The reader may wish to compare these values with those obtained directly using the values for air in Table 11.2 or to explore the effects of neglecting argon in the preceding calculations. 11.5 The equation of species conservation Conservation of species Just as we formed an equation of energy conservation in Chapter 6, we now form an equation of species conservation that applies to each substance in a mixture. In addition to accounting for the convection and diffusion of each species, we must allow the possibility that a species may be created or destroyed by chemical reactions occuring in the bulk medium (so-called homogeneous reactions ). Reactions on surfaces surrounding the medium (heterogeneous reactions ) must be accounted for in the boundary conditions. We consider, in the usual way, an arbitrary control volume, R , with a boundary, S , as shown in Fig. 11.8. The control volume is fixed in space, with fluid moving through it. Species i may accumulate in R , it may travel in and out of R by bulk convection or by diffusion, and it may be created within R by homogeneous reactions. The rate of creation of species i is ˙ denoted as ri (kg/m3 ·s); and, because chemical reactions conserve mass, ˙ ˙ the net mass creation is r = ri = 0. The rate of change of the mass of species i in R is then described by the following balance: d dt R ρi dR = − S ni · dS + R ˙ ri dR rate of increase of i in R =− S ρi v · dS rate of convection of i out of R − S ji · dS + diffusion of i out of R R ˙ ri dR rate of creation of i in R (11.56) 627 An introduction to mass transfer 628 §11.5 Figure 11.8 Control volume in a fluid-flow and mass-diffusion field. This species conservation statement is identical to our energy conservation statement, eqn. (6.36) on page 293, except that mass of species i has taken the place of energy and heat. We may convert the surface integrals to volume integrals using Gauss’s theorem [eqn. (2.8)] and rearrange the result to find: R ∂ ρi ˙ + ∇ · (ρi v) + ∇ · ji − ri dR = 0 ∂t (11.57) Since the control volume is selected arbitrarily, the integrand must be identically zero. Thus, we obtain the general form of the species conservation equation: ∂ρi ˙ + ∇ · (ρi v) = −∇ · ji + ri ∂t (11.58) We may obtain a mass conservation equation for the entire mixture by summing eqn. (11.58) over all species and applying eqns. (11.1), (11.17), and (11.22) and the requirement that there be no net creation of mass: i ∂ ρi + ∇ · (ρi v) = ∂t ˙ (−∇ · ji + ri ) i so that ∂ρ + ∇ · (ρ v) = 0 ∂t (11.59) The equation of species conservation §11.5 This equation applies to any mixture, including those with varying density (see Problem 6.36). Incompressible mixtures. For an incompressible mixture, ∇ · v = 0 (see Sect. 6.2 or Problem 11.22), and the second term in eqn. (11.58) may therefore be rewritten as ∇ · (ρi v) = v · ∇ρi + ρi ∇ · v = v · ∇ρi (11.60) =0 We may compare the resulting, incompressible species equation to the incompressible energy equation, eqn. (6.37) ∂ρi ˙ + v · ∇ρi = −∇ · ji + ri ∂t Dρi = Dt ρcp DT =ρcp Dt ∂T + v · ∇T ∂t ˙ = −∇ · q + q (11.61) (6.37) ˙ In these equations: the reaction term, ri , is analogous to the heat gener˙ ation term, q; the diffusional mass flux, ji , is analogous to the heat flux, q; and dρi is analogous to ρcp dT . We can use Fick’s law to eliminate ji in eqn. (11.61). The resulting equation may be written in different forms, depending upon what is assumed about the variation of the physical properties. If the product ρ Dim is independent of (x, y, z)—if it is spatially uniform—then eqn. (11.61) becomes D ˙ mi = Dim ∇2 mi + ri /ρ Dt (11.62) where the material derivative, D/Dt , is defined in eqn. (6.38). If, instead, ρ and Dim are both spatially uniform, then Dρi ˙ = Dim ∇2 ρi + ri Dt (11.63) The equation of species conservation and its particular forms may ∗ also be stated in molar variables, using ci or xi , Ni , and Ji (see Problem 11.24.) Molar analysis sometimes has advantages over mass-based analysis, as we discover in Section 11.7. 629 630 An introduction to mass transfer §11.5 Figure 11.9 Absorption of ammonia into water. Interfacial boundary conditions We are already familiar with the general issue of boundary conditions from our study of the heat equation. To find a temperature distribution, we specified temperatures or heat fluxes at the boundaries of the domain of interest. Likewise, to find a concentration distribution, we must specify the concentration or flux of species i at the boundaries of the medium of interest. Temperature and concentration behave differently at interfaces. At an interface, temperature is the same in both media as a result of the Zeroth Law of Thermodynamics. Concentration, on the other hand, need not be continuous across an interface, even in a state of thermodynamic equilibrium. Water in a drinking glass, for example, shows discontinous changes in the concentration of water at both the glass-water interface on the sides and the air-water interface above. In another example, gaseous ammonia is absorbed into water in some types of refrigeration cycles. A gas mixture containing some particular mass fraction of ammonia will produce a different mass fraction of ammonia just inside an adjacent body of water, as shown in Fig. 11.9. To characterize the conditions at an interface, we introduce imaginary surfaces, s and u, very close to either side of the interface. In the §11.5 The equation of species conservation ammonia absorption process, then, we have a mass fraction mNH3 ,s on the gas side of the interface and a different mass fraction mNH3 ,u on the liquid side. In many mass transfer problems, we must find the concentration distribution of a species in one medium given only its concentration at the interface in the adjacent medium. We might wish to find the distribution of ammonia in the body of water knowing only the concentration of ammonia on the gas side of the interface. We would need to find mNH3 ,u from mNH3 ,s and the interfacial temperature and pressure, since mNH3 ,u is the appropriate boundary condition for the species conservation equation in the water. Thus, for the general mass transfer boundary condition, we must specify not only the concentration of species i in the medium adjacent to the medium of interest but also the solubility of species i from one medium to the other. Although a detailed study of solubility and phase equilibria is far beyond our scope (see, for example, [11.5, 11.24]), we illustrate these concepts with the following simple solubility relations. Gas-liquid interfaces. For a gas mixture in contact with a liquid mixture, two simplified rules dictate the vapor composition. When the liquid is rich in species i, the partial pressure of species i in the gas phase, pi , can be characterized approximately with Raoult’s law, which says that pi = psat,i xi for xi ≈ 1 (11.64) where psat,i is the saturation pressure of pure i at the interface temperature and xi is the mole fraction of i in the liquid. When the species i is dilute in the liquid, Henry’s law applies. It says that pi = H xi for xi 1 (11.65) where H is a temperature-dependent empirical constant that is tabulated in the literature. Figure 11.10 shows how the vapor pressure varies over a liquid mixture of species i and another species, and it indicates the regions of validity of Raoult’s and Henry’s laws. For example, when xi is near one, Raoult’s law applies to species i; when xi is near zero, Raoult’s law applies to the other species. If the vapor pressure were to obey Raoult’s law over the entire range of liquid composition, we would have what is called an ideal solution. When xi is much below unity, the ideal solution approximation is usually very poor. 631 632 An introduction to mass transfer §11.5 Figure 11.10 Typical partial and total vapor-pressure plot for the vapor in contact with a liquid solution, illustrating the regions of validity of Raoult’s and Henry’s laws. Example 11.5 A cup of tea sits in air at 1 atm total pressure. It starts at 100◦ C and cools toward room temperature. What is the mass fraction of water vapor above the surface of the tea as a function of the surface temperature? Solution. We’ll approximate tea as having the properties of pure water. Raoult’s law applies almost exactly in this situation, since it happens that the concentration of air in water is virtually nil. Thus, by eqn. (11.64), pH2 O,s = psat,H2 O (T ). We can read the saturation pressure of water for several temperatures from a steam table or from Table A.5 on pg. 713. From the vapor pressure, pH2 O,s , we can compute the mole fraction with eqn. (11.16), xH2 O,s = pH2 O,s patm = psat,H2 O (T ) (101, 325 Pa) (11.66) The mass fraction can be calculated from eqn. (11.9), noting that xair = 1 − xH2 O and substituting MH2 O = 18.02 kg/kmol and Mair = The equation of species conservation §11.5 Mass fraction of water vapor 1 0.8 0.6 0.4 0.2 0 0 20 40 60 80 100 Temperature (C) Figure 11.11 Mass fraction of water vapor in air above liquid water surface as a function of surface temperature (1 atm total pressure). 28.96 kg/kmol mH2 O,s = (xH2 O,s )(18.02) [(xH2 O,s )(18.02) + (1 − xH2 O,s )(28.96)] (11.67) The result is plotted in Fig. 11.11. Note that the mass fraction is less than 10% if the surface temperature is below about 54◦ C. Gas-solid interfaces. When a solid is exposed to a gas, some amount of it will vaporize. This process is quite visible, for example, when dry ice (solid CO2 ) is placed in air. For other materials and temperatures, the vaporization rate may be indetectably tiny. We call a direct solid-to-vapor phase transition sublimation. The solubility of most gases in most solids is so small that solids are often treated as pure substances when finding their concentration in an adjacent vapor phase. Most data for the solubility of solids into the gas phase is written in the form of the vapor pressure of the solid as a function of surface temperature. Many such relationships are available in the literature (see, e.g., [11.25]). 633 634 An introduction to mass transfer §11.5 Although only small amounts of gas are absorbed into most inorganic solids, the consequences can be quite significant. Material properties may be altered by absorbed gases, and, through absorption and diffusion, gases may leak through metal pressure-vessel walls. The process of absorption may include dissociation of the gas on the solid surface prior to its absorption into the bulk material. For example, when molecular hydrogen gas, H2 , is absorbed into iron, it first dissociates into two hydrogen atoms, 2H. At low temperatures, the dissociation reaction may be so slow that equilibrium conditions cannot be established between the bulk metal and the gas. Solubility relationships for gases entering solids are thus somewhat complex, and they will not be covered here (see [11.26]). One important technical application of gas absorption into solids is the case-hardening of low-carbon steel by a process called carburization. The steel is exposed to a hot carbon-rich gas, such as CO or CO2 , which causes carbon to be absorbed on the surface of the metal. The elevated concentration of carbon within the surface causes carbon to diffuse inward. A typical goal is to raise the carbon mass fraction to 0.8% over a depth of about 2 mm (see Problem 11.27). Example 11.6 Ice at −10◦ C is exposed to 1 atm air. What is the mass fraction of water vapor above the surface of the ice? Solution. To begin, we need the vapor pressure, pv , of water above ice. A typical local curve-fit is ln pv (kPa) = 21.99 − 6141 (T K) for 243 K ≤ T ≤ 273 K At T = −10◦ C = 263.15 K this yields pv = 0.260 kPa. The remainder of the calculation follows exactly the approach of Example 11.5. xH2 O,s = 0.260/101.325 = 0.00257 mH2 O,s = (0.00257)(18.02) [(0.00257)(18.02) + (1 − 0.00257)(28.96)] = 0.00160 Mass transfer at low rates §11.6 11.6 635 Mass transfer at low rates We have seen that mass transfer processes generate flow in mixtures. When the mass transfer rates are sufficiently low, the velocities caused by mass transfer are negligible. Thus, a stationary medium will remain at rest and a flowing fluid will have the same velocity field as if there were no mass transfer. More generally, when the diffusing species is dilute, its total mass flux is principally carried by diffusion. In this section, we examine diffusive and convective mass transfer of dilute species at low rates. These problems have a direct correspondence to the heat transfer problems that we considered Chapters 1 through 8. We refer to this correspondence as the analogy between heat and mass transfer. We will focus our attention on nonreacting systems, for which ˙ ri = 0 in the species conservation equation. Steady mass diffusion in stationary media Equations (11.58) and (11.21) show that steady mass transfer without reactions is described by the equation ∇ · (ρi v) + ∇ · ji = ∇ · ni = 0 (11.68) or, in one dimension, d d dni = ρi v + ji = mi n + ji = 0 dx dx dx (11.69) that is, the mass flux of species i, ni , is independent of x . When the convective mass flux of i, ρi v = mi n, is small, the transport of i is mainly by the diffusional flux, ji . The following pair of examples show how this situation might arise. Example 11.7 A thin slab, made of species 1, separates two volumes of gas. On one side, the pressure of species 2 is high, and on the other it is low. Species 2 two is soluble in the slab material and thus has different concentrations at each inside face of the slab, as shown in Fig. 11.12. What is the mass transfer rate of species 2 through the slab if the concentration of species 2 is low? 636 An introduction to mass transfer §11.6 Figure 11.12 One-dimensional, steady diffusion in a slab. Solution. The mass transfer rate through the slab satisfies eqn. (11.69) dn2 =0 dx If species 2 is dilute, with m2 small 1, the convective transport will be n2 = m2 n + j2 j2 With Fick’s law, we have dn2 dx If ρ D21 dj2 d dm2 = −ρ D21 dx dx dx =0 constant, the mass fraction satisfies d2 m2 =0 dx 2 Integrating and applying the boundary conditions, m2 (x = 0) = m2,0 and m2 (x = L) = m2,L , we obtain the concentration distribution: m2 (x) = m2,0 + m2,L − m2,0 x L Mass transfer at low rates §11.6 637 The mass flux is then n2 j2 = −ρ D21 dm2 ρ D21 =− m2,L − m2,0 dx L (11.70) This, in essence, is the same calculation we made in Example 2.2 in Chapter 2. Example 11.8 Suppose that the concentration of species 2 in the slab were not small in the preceding example. How would the total mass flux of species 1 differ from the diffusional flux? Solution. As before, the total mass flux each species would be constant in the steady state, and if the slab material is not transferred into the gas its mass flux is zero n1 = 0 = ρ1 v + j1 Therefore, the mass-average velocity in the slab is v=− j1 j2 = ρ1 ρ1 since j1 + j2 = 0. The mass flux for species 2 is n2 = ρ2 v + j2 = j2 = j2 ρ2 +1 ρ1 m2 1 + 1 = j2 m1 1 − m2 since m1 + m2 = 1. 1, the diffusional flux will approximate n2 . On the When m2 other hand, if, say, m2 = 0.5, then n2 = 2j2 ! In that case, the convective transport ρ2 v is equal to the diffusive transport j2 . In the second example, we see that the stationary material of the slab had a diffusion velocity, j1 . In order for the slab to remain at rest, the opposing velocity v must be present. For this reason, an induced velocity of this sort is sometimes called a counterdiffusion velocity. From these two examples, we see that steady mass diffusion is directly analogous to heat conduction only if the convective transport is 638 An introduction to mass transfer §11.6 negligible. That can generally be ensured if the transferred species is dilute. When the transferred species has a high concentration, nonnegligible convective transport can occur, even in a solid medium. Unsteady mass diffusion in stationary media Similar conclusions apply to unsteady mass diffusion. Consider a medium at rest through which a dilute species i diffuses. From eqn. (11.58) with ri = 0, ∂ρi = −∇ · ρi v + ji ∂t = −∇ · mi n + ji (11.71) If mi 1, only diffusion contributes significantly to the mass flux of i, and we may neglect mi n ∂ρi ≈ −∇ · ji = ∇ · (ρ Dim ∇mi ) ∂t With small mi , the density ρ and the diffusion coefficient Dim will not vary much, and we can factor ρ through the equation ∂mi = Dim ∇2 mi (11.72) ∂t This is called the mass diffusion equation. It has the same form as the equation of heat conduction. Solutions for the unsteady diffusion of a dilute species in a stationary medium are thus entirely analogous to those for heat conduction when the boundary conditions are the same. Example 11.9 A semi-infinite stationary medium (medium 1) has an initially uniform concentration, mi,0 of species i. From time t = 0 onward, we place the end plane at x = 0 in contact with a second medium (medium 2) with a concentration mi,s . What is the resulting distribution of species in medium 1 if species 1 remains dilute? Solution. Once mi,s and the solubility data are known, the mass fraction just inside the solid surface, mi,u , can be determined (see Fig. 11.13). This concentration provides the boundary condition at x = 0 for t > 0. Our mathematical problem then becomes ∂ 2 mi ∂mi = Dim1 ∂t ∂x 2 (11.73) Mass transfer at low rates §11.6 Figure 11.13 Mass diffusion into a semi-infinite stationary medium. with mi = mi,0 for t = 0 (all x) mi = mi,u for x = 0 (t > 0) mi → mi,0 for x → ∞ (t > 0) This math problem is identical to that for transient heat conduction into a semi-infinite region (Section 5.6), and its solution is completely analogous to eqn. (5.50): ⎛ ⎞ mi − mi,u x ⎠ = erf ⎝ mi,0 − mi,u 2 Dim1 t The reader can solve all sorts of unsteady mass diffusion problems by direct analogy to the methods of Chapters 4 and 5 when the concentration of the diffusing species is low. At higher concentrations of the diffusing species, however, counterdiffusion velocities can be induced, as in Example 11.8. Counterdiffusion may be significant in concentrated metallic alloys, as, for example, during annealing of a butt-welded junction between two dissimilar metals. In those situations, eqn. (11.72) is sometimes modified to use a concentration-dependent, spatially varying interdiffusion coefficient (see [11.6]). 639 An introduction to mass transfer 640 §11.6 Figure 11.14 Concentration boundary layer on a flat plate. Convective mass transfer at low rates Convective mass transfer is analogous to convective heat transfer when two conditions apply: 1. The mass flux normal to the surface, ni,s , must be essentially equal to the diffusional mass flux, ji,s from the surface. In general, this requires that the concentration of the diffusing species, mi , be low.9 2. The diffusional mass flux must be low enough that it does not affect the imposed velocity field. The first condition ensures that mass flow from the wall is diffusional, as is the heat flow in a convective heat transfer problem. The second condition ensures that the flow field will be the same as for the heat transfer problem. As a concrete example, consider a laminar flat-plate boundary layer in which species i is transferred from the wall to the free stream, as shown in Fig. 11.14. Free stream values, at the edge of the b.l., are labeled with the subscript e, and values at the wall (the s -surface) are labeled with the subscript s . The mass fraction of species i varies from mi,s to mi,e across a concentration boundary layer on the wall. If the mass fraction of species i at the wall, mi,s , is small, then ni,s ≈ ji,s , as we saw earlier in this section. Mass transfer from the wall will be essentially diffusional. This is the first condition. In regard to the second condition, when the concentration difference, mi,s − mi,e , is small, then the diffusional mass flux of species i through the wall, ji,s , will be small compared to the bulk mass flow in the stream9 In a few situations, such as catalysis, there is no net mass flow through the wall, and convective transport will be identically zero irrespective of the concentration (see Problems 11.9 and 11.44). Mass transfer at low rates §11.6 641 wise direction, and it will have little influence on the velocity field. Hence, we would expect that v is essentially that for the Blasius boundary layer. These two conditions can be combined into a single requirement for low-rate mass transfer, as will be described in Section 11.8. Specifically, low-rate mass transfer can be assumed if Bm,i ≡ mi,s − mi,e 1 − mi,s 0.2 condition for low-rate mass transfer (11.74) The quantity Bm,i is called the mass transfer driving force. It is written here in the form that applies when only one species is transferred through the s -surface. The evaporation of water into air is typical example of single-species transfer: only water vapor crosses the s -surface. The mass transfer coefficient. In convective heat transfer problems, we have found it useful to express the heat flux from a surface, q, as the product of a heat transfer coefficient, h, and a driving force for heat transfer, ∆T . Thus, in the notation of Fig. 11.14, qs = h (Ts − Te ) (1.17) In convective mass transfer problems, we would therefore like to express the diffusional mass flux from a surface, ji,s , as the product of a mass transfer coefficient and the concentration difference between the s -surface and the free stream. Hence, we define the mass transfer coefficient for species i, gm,i (kg/m2 ·s), as follows: ji,s ≡ gm,i mi,s − mi,e (11.75) We expect gm,i , like h, to be determined mainly by the flow field, fluid, and geometry of the problem. The analogy to convective heat transfer. We saw in Sect. 11.5 that the equation of species conservation and the energy equation were quite similar in an incompressible flow. If there are no reactions and no heat generation, then eqns. (11.61) and (6.37) can be written as ∂ρi + v · ∇ρi = −∇ · ji ∂t ρcp ∂T + v · ∇T ∂t = −∇ · q (11.61) (6.37) 642 An introduction to mass transfer §11.6 These conservation equations describe changes in, respectively, the amount of mass or energy per unit volume that results from convection by a given velocity field and from diffusion under either Fick’s or Fourier’s law. We may identify the analogous quantities in these equations. For the capacity of mass or energy per unit volume, we see that dρi is analogous to ρcp dT (11.76a) or, in terms of the mass fraction, is analogous to ρ dmi ρcp dT (11.76b) The flux laws may be rewritten to show the capacities explicitly ji = −ρ Dim ∇mi = −Dim ρ ∇mi q = −k∇T =− k ρ cp ∇T ρcp Hence, we find the analogy of the diffusivities: Dim is analogous to k =α ρcp (11.76c) It follows that the Schmidt number and the Prandtl number are directly analogous: Sc = ν Dim is analogous to Pr = µcp ν = α k (11.76d) Thus, a high Schmidt number signals a thin concentration boundary layer, just as a high Prandtl number signals a thin thermal boundary layer. Finally, we may write the transfer coefficients in terms of the capacities ji,s = gm,i mi,s − mi,e = qs = h (Ts − Te ) = gm,i ρ ρ mi,s − mi,e h ρcp ρ cp (Ts − Te ) from which we see that gm,i is analogous to h cp (11.76e) Mass transfer at low rates §11.6 643 From these comparisons, we conclude that the solution of a heat convection problem becomes the solution of a low-rate mass convection problem upon replacing the variables in the heat transfer problem with the analogous mass transfer variables given by eqns. (11.76). Convective heat transfer coefficients are usually expressed in terms of the Nusselt number as a function of Reynolds and Prandtl number Nux = (h/cp )x hx = = fn (Rex , Pr) k ρ(k/ρcp ) (11.77) For convective mass transfer problems, we expect the same functional dependence after we make the substitutions indicated above. Specifically, if we replace h/cp by gm,i , k/ρcp by Di,m , and Pr by Sc, we obtain Num,x ≡ gm,i x = fn (Rex , Sc) ρ Dim (11.78) where Num,x , the Nusselt number for mass transfer, is defined as indicated. Num is sometimes called the Sherwood number 10 , Sh. Example 11.10 A napthalene model of a printed circuit board (PCB) is placed in a wind tunnel. The napthalene sublimates slowly as a result of forced convective mass transfer. If the first 5 cm of the napthalene model is a flat plate, calculate the average rate of loss of napthalene from that part of the model. Assume that conditions are isothermal at 303 K and that the air speed is 5 m/s. Also, explain how napthalene sublimation might be used to determine heat transfer coefficients . Solution. Let us first find the mass fraction of napthalene just above the model surface. A relationship for the vapor pressure of napthalene (in mmHg) is log10 pv = 11.450 − 3729.3 (T K). At 303 K, this gives pv = 0.1387 mmHg = 18.49 Pa. The mole fraction of napthalene is thus xnap,s = 18.49/101325 = 1.825 × 10−4 , and with 10 Thomas K. Sherwood (1903–1976) obtained his doctoral degree at M.I.T. under Warren K. Lewis in 1929 and was a professor of Chemical Engineering there from 1930 to 1969. He served as Dean of Engineering from 1946 to 1952. His research dealt with mass transfer and related industrial processes. Sherwood was also the author of very influential textbooks on mass transfer. 644 An introduction to mass transfer §11.6 eqn. (11.9), the mass fraction is, with Mnap = 128.2 kg/kmol, mnap,s = (1.825 × 10−4 )(128.2) (1.825 × 10−4 )(128.2) + (1 − 1.825 × 10−4 )(28.96) = 8.074 × 10−4 The mass fraction of napthalene in the free stream, mnap,s , is zero. With these numbers, we can check to see if the mass transfer rate is low enough to use the analogy of heat and mass transfer, with eqn. (11.74): Bm,nap = 8.074 × 10−4 − 0 1 − 8.074 × 10−4 = 8.081 × 10−4 0.2 The analogy therefore applies. The convective heat transfer coefficient for this situation is that for a flat plate boundary layer. The Reynolds number is ReL = (5)(0.05) u∞ L = = 1.339 × 104 ν 1.867 × 10−5 where we have used the viscosity of pure air, since the concentration of napthalene is very low. The flow is laminar, so the applicable heat transfer relationship is eqn. (6.68) NuL = hL 1/2 = 0.664 ReL Pr1/3 k (6.68) Under the analogy, the Nusselt number for mass transfer is Num,L = gm,i L 1/2 = 0.664 ReL Sc1/3 ρ Dim The diffusion coefficient for napthalene in air, from Table 11.1, is Dnap,air = 0.86 × 10−5 m/s, and thus Sc = 1.867 × 10−5 /0.86 × 10−5 = 2.17. Hence, Num,L = 0.664 (1.339 × 104 )1/2 (2.17)1/3 = 99.5 and, using the density of pure air, ρ Dnap,air Num,L L (1.166)(0.86 × 10−5 ) (99.5) = 0.0200 kg/m2 s = 0.05 gm,nap = Mass transfer at low rates §11.6 645 The average mass flux from this part of the model is then nnap,s = gm,nap mnap,s − mnap,e = (0.0200)(8.074 × 10−4 − 0) = 1.61 × 10−5 kg/m2 s = 58.0 g/m2 h Napthalene sublimation can be used to infer heat transfer coefficients by measuring the loss of napthalene from a model over some length of time. Experiments are run at several Reynolds numbers. The lost mass fixes the sublimation rate and the mass transfer coefficient. The mass transfer coefficient is then substituted in the analogy to heat transfer to determine a heat transfer Nusselt number at each Reynolds number. Since the Schmidt number of napthalene is not generally equal to the Prandtl number under the conditions of interest, some assumption about the dependence of the Nusselt number on the Prandtl number must usually be introduced. Boundary conditions. When we apply the analogy between heat transfer and mass transfer to calculate gm,i , we must consider the boundary condition at the wall. We have dealt with two common types of wall condition in the study of heat transfer: uniform temperature and uniform heat flux. The analogous mass transfer wall conditions are uniform concentration and uniform mass flux. We used the mass transfer analog of the uniform wall temperature solution in the preceding example, since the mass fraction of napthalene was uniform over the entire model. Had the mass flux been uniform at the wall, we would have used the analog of a uniform heat flux solution. Natural convection in mass transfer. In Chapter 8, we saw that the density differences produced by temperature variations can lead to flow and convection in a fluid. Variations in fluid composition can also produce density variations that result in natural convection mass transfer. This type of natural convection flow is still governed by eqn. (8.3), u ∂u ∂2u ∂u +v = (1 − ρ∞ /ρ)g + ν ∂x ∂y ∂y 2 (8.3) but the species equation is now used in place of the energy equation in determining the variation of density. Rather than solving eqn. (8.3) and 646 An introduction to mass transfer §11.6 the species equation for specific mass transfer problems, we again turn to the analogy between heat and mass transfer. In analyzing natural convection heat transfer, we eliminated ρ from eqn. (8.3) using (1 − ρ∞ /ρ) = β(T − T∞ ), and the resulting Grashof and Rayleigh numbers came out in terms of an appropriate β∆T instead of ∆ρ/ρ . These groups could just as well have been written for the heat transfer problem as GrL = g ∆ρL3 ρν 2 and RaL = g ∆ρL3 g ∆ρL3 = ραν µα (11.79) although ∆ρ would still have had to have been evaluated from ∆T . With Gr and Pr expressed in terms of density differences instead of temperature differences, the analogy between heat transfer and low-rate mass transfer may be used directly to adapt natural convection heat transfer predictions to natural convection mass transfer. As before, we replace Nu by Num and Pr by Sc. But this time we also write RaL = GrL Sc = g ∆ρL3 µ D12 (11.80) or calculate GrL as in eqn. (11.79). The densities must now be calculated from the concentrations. Example 11.11 Helium is bled through a porous vertical wall, 40 cm high, into surrounding air at a rate of 87.0 mg/m2 ·s. Both the helium and the air are at 300 K, and the environment is at 1 atm. What is the average concentration of helium at the wall, mHe,s ? Solution. This is a uniform flux natural convection problem. Here gm,He and ∆ρ depend on mHe,s , so the calculation is not as straightforward as it was for thermally driven natural convection. To begin, let us assume that the concentration of helium at the wall will be small enough that the mass transfer rate is low. Since mHe,e = 1, then mHe,s − mHe,e 1 as well. Both conditions for 0, if mHe,s the analogy to heat transfer will be met. The mass flux of helium at the wall, nHe,s , is known, and because low rates prevail, nHe,s ≈ jHe,s = gm,He mHe,s − mHe,e Mass transfer at low rates §11.6 Hence, Num,L = nHe,s L gm,He L = ρ DHe,air ρ DHe,air mHe,s − mHe,e The appropriate Nusselt number is obtained from the mass transfer analog of eqn. (8.44b): Num,L = 6 5 Ra∗ Sc √L 4 + 9 Sc + 10 Sc with Ra∗ = RaL Num,L = L 1/5 g ∆ρ nHe,s L4 µρ D2 ,air He mHe,s − mHe,e The Rayleigh number cannot easily be evaluated without assuming a value of the mass fraction of helium at the wall. As a first guess, we pick mHe,s = 0.010. Then the film composition is mHe,f = (0.010 + 0)/2 = 0.005 From eqn. (11.8) and the ideal gas law, we obtain estimates for the film density (at the film composition) and the wall density ρf = 1.141 kg/m3 and ρs = 1.107 kg/m3 From eqn. (11.42) the diffusion coefficient is DHe,air = 7.119 × 10−5 m2 /s. At this low concentration of helium, we expect the film viscosity to be close to that of pure air. From Appendix A, for air at 300 K µf µair = 1.857 × 10−5 kg/m·s. The corresponding Schmidt number is Sc = (µf /ρf ) DHe,air = 0.2286. Furthermore, ρe = ρair = 1.177 kg/m3 From these values, Ra∗ = L 9.806(1.177 − 1.107)(87.0 × 10−6 )(0.40)4 (1.857 × 10−5 )(1.141)(7.119 × 10−5 )2 (0.010) = 1.424 × 109 647 An introduction to mass transfer 648 §11.7 We may now evaluate the mass transfer Nusselt number 1/5 Num,L 6 (1.424 × 109 )(0.2286) = √ 5 4 + 9 0.2286 + 10(0.2286) 1/5 = 37.73 From this we calculate mHe,s − mHe,e = = nHe,s L ρ DHe,air Num,L (87.0 × 10−6 )(0.40) (1.141)(7.119 × 10−5 )(37.73) = 0.01136 We have already noted that mHe,s − mHe,e = mHe,s , so we have obtained an average wall concentration 14% higher than our initial guess of 0.010. Using mHe,s = 0.01136 as our second guess, we repeat the preceding calculations with revised values of the densities to obtain mHe,s = 0.01142 Since this result is within 0.5% of our second guess, a third iteration is not needed. In the preceding example, concentration variations alone gave rise to buoyancy. If both temperature and density vary in a natural convection problem, the appropriate Gr or Ra may be calculated using density differences based on both the local mi and the local T , provided that the Prandtl and Schmidt numbers are approximately equal (that is, if the Lewis number 1). This is usually true in gases. If the Lewis number is far from unity, the analogy between heat and mass transfer breaks down in those natural convection problems that involve both heat and mass transfer, because the concentration and thermal boundary layers may take on very different thicknesses, complicating the density distributions that drive the velocity field. 11.7 Steady mass transfer with counterdiffusion In 1874, Josef Stefan presented his solution for evaporation from a liquid pool at the bottom of a vertical tube over which a gas flows (Fig. 11.15). This configuration, often called a Stefan tube, is has often been used to §11.7 Steady mass transfer with counterdiffusion Figure 11.15 The Stefan tube. measure diffusion coefficients. Vapor leaving the liquid surface diffuses through the gas in the tube and is carried away by the gas flow across top of the tube. If the gas stream itself has a low concentration of the vapor, then diffusion is driven by the higher concentration of vapor over the liquid pool that arises from the vapor pressure of the liquid. A typical Stefan tube is 5 to 10 mm in diameter and 10 to 20 cm long. If the air flow at the top is not too vigorous, and if density variations in the tube do not give rise to natural convection, then the transport of vapor from the liquid pool to the top of the tube will be a one-dimensional upflow. The other gas in the tube is stationary if it is not being absorbed by the liquid (e.g., if it is insoluble in the liquid or if the liquid is saturated with it). Yet, because there is a concentration gradient of vapor, there must also be an opposing concentration gradient of gas and an associated diffusional mass flux of gas, similar to what we found in Example 11.8. For the gas in the tube to have a net diffusion flux when it is stationary, there must be an induced upward convective velocity — a counterdiffusion velocity — against which the gas diffuses. As in Example 11.8, the counterdiffusion velocity can be found in terms of the diffusional mass fluxes: v = −jgas ρgas = jvapor ρgas 649 650 An introduction to mass transfer §11.7 Figure 11.16 Mass flow across a one-dimensional layer. In this section, we determine the mass transfer rate and concentration profiles in the tube, treating it as the one-dimensional layer shown in Fig. 11.16. The s -surface lies above the liquid and the e-surface lies at the top end of the tube. We allow for the possibility that the counterdiffusion velocity may not be negligible, so that both diffusion and vertical convection may occur. We also allow for the possibility that the gas passes through the liquid surface (N2,s ≠ 0). The results obtained here form an important prototype for our subsequent analyses of convective mass transfer at high rates. The solution of the mass transfer problem begins with an appropriate form of the equation of species conservation. Since the mixture composition varies along the length of the tube, the density may vary as well. If the temperature and pressure are constant, however, the molar concentration of the mixture does not change through the tube [cf. (11.14)]. The system is therefore most accurately analyzed using the molar form of species conservation. For one-dimensional steady mass transfer, the mole fluxes N1 and N2 have only vertical components and depend only on the vertical coordinate, y . Using eqn. (11.69), we get, with ni = Mi Ni , dN1 dN2 = =0 dy dy so that N1 and N2 are constant throughout the layer. They have s -surface values, N1,s and N2,s , everywhere. These constants will be positive for upward mass flow. (For the orientations in Fig. 11.16, N1,s > 0 and N2,s < 0.) These results are a straightforward consequence of steady-state species conservation. Recalling the general expression for Ni , eqn. (11.25), and introducing Steady mass transfer with counterdiffusion §11.7 Fick’s law, eqn. (11.34), we write N1 = x1 N − c D12 dx1 = N1,s dy (11.81) The term xN1 represents vertical convective transport induced by mass transfer. The total mole flux, N , must also be constant at its s -surface value; by eqn. (11.23), this is N = N1,s + N2,s = Ns (11.82) Substituting this result into eqn. (11.81), we obtain a differential equation for x1 : c D12 dx1 = Ns x1 − N1,s dy (11.83) In this equation, x1 is a function of y , the N ’s are constants, and c D12 depends on temperature and pressure. If the temperature and pressure are constant, so too is c D12 . Integration then yields Ns y = ln Ns x1 − N1,s + constant c D12 (11.84) We need to fix the constant and the two mole fluxes, N1,s and Ns . To do this, we apply the boundary conditions at either end of the layer. The first boundary condition is the mole fraction of species 1 at the bottom of the layer x1 = x1,s at y =0 and it requires that constant = − ln(Ns x1,s − N1,s ) (11.85) so Ns y = ln c D12 Ns x1 − N1,s Ns x1,s − N1,s (11.86) The second boundary condition is the mole fraction at the top of the layer x1 = x1,e at y =L 651 An introduction to mass transfer 652 §11.7 which yields Ns L = ln c D12 x1,e − N1,s /Ns x1,s − N1,s /Ns (11.87) or Ns = x1,e − x1,s c D12 ln 1 + L x1,s − N1,s /Ns (11.88) The last boundary condition is the value of N1,s /Ns . Since we have allowed for the possiblity that species 2 passes through the bottom of the layer, N1,s /Ns may not equal unity. The ratio depends on the specific problem at hand, as shown in the two following examples. Example 11.12 Find an equation for the evaporation rate of the liquid in the Stefan tube described at the beginning of this section. Solution. Species 1 is the evaporating vapor, and species 2 be the stationary gas. Only vapor is transferred through the s -surface, since the gas is not significantly absorbed into the already gas-saturated liquid. Thus, N2,s = 0, and Ns = N1,s = Nvapor,s is simply the evaporation rate of the liquid. The s -surface is just above the surface of the liquid. The mole fraction of the evaporating liquid can be determined from solubility data; for example, if the gas is more-or-less insoluble in the liquid, Raoult’s law, eqn. (11.64), may be used. The e-surface is at the mouth of the tube. The gas flow over the top may contain some concentration of the vapor, although it should generally be near zero. The ratio N1,s /Ns is unity, and the rate of evaporation is Ns = Nvapor,s = x1,e − x1,s c D12 ln 1 + L x1,s − 1 (11.89) Example 11.13 What is the evaporation rate in the Stefan tube if the gas is bubbled up to the liquid surface at some fixed rate, Ngas ? Solution. Again, N1,s = Nvapor,s is the evaporation rate. However, the total mole flux is Ns = Ngas + N1,s Steady mass transfer with counterdiffusion §11.7 Thus, Ngas + N1,s = x1,e − x1,s c D12 ln 1 + L x1,s − N1,s /(N1,s + Ngas ) (11.90) This equation fixes N1,s , but it must be solved iteratively. Once we have found the mole fluxes, we may compute the concentration distribution, x1 (y), using eqn. (11.86): x1 (y) = N1,s + x1,s − N1,s Ns exp(Ns y/c D12 ) Ns (11.91) Alternatively, we may eliminate Ns between eqns. (11.86) and (11.87) to obtain the concentration distribution in a form that depends only on the ratio N1,s /Ns : x1 − N1,s /Ns = x1,s − N1,s /Ns x1,e − N1,s /Ns x1,s − N1,s /Ns y /L (11.92) Example 11.14 Find the concentration distribution of water vapor in a helium–water Stefan tube at 325 K and 1 atm. The tube is 20 cm in length. Assume the helium stream at the top of the tube to have a mole fraction of water of 0.01. Solution. Let water be species 1 and helium be species 2. The vapor pressure of the liquid water is approximately the saturation pressure at the water temperature. Using the steam tables, we get pv = 1.341 × 104 Pa and, from eqn. (11.16), x1,s = 1.341 × 104 Pa = 0.1323 101, 325 Pa We use eqn. (11.14) to evaluate the mole concentration in the tube: c= 101, 325 = 0.03750 kmol/m3 8314.5(325) From eqn. (11.42) we obtain D12 (325 K, 1 atm) = 1.067 × 10−4 m2 /s. Then eqn. (11.89) gives the molar evaporation rate: 0.01 − 0.1323 0.03750(1.067 × 10−4 ) ln 1 + 0.20 0.1323 − 1 −6 2 = 2.638 × 10 kmol/m ·s N1,s = 653 An introduction to mass transfer 654 §11.8 This corresponds to a mass evaporation rate: n1,s = 4.754 × 10−5 kg/m2 ·s The concentration distribution of water vapor [eqn. (11.91)] is x1 (y) = 1 − 0.8677 exp(0.6593y) where y is expressed in meters. Stefan tubes have been widely used to measure mass transfer coefficients, by observing the change in liquid level over a long period of time and solving eqn. (11.89) for D12 . These measurements are subject to a variety of experimental errors, however. For example, the latent heat of vaporization may tend to cool the gas mixture near the interface, causing a temperature gradient in the tube. Vortices near the top of the tube, where it meets the gas stream, may cause additional mixing, and density gradients may cause buoyant circulation. Additional sources of error and alternative measurement techniques are described by Marrero and Mason [11.7]. The problem dealt with in this section can alternatively be solved on a mass basis, assuming a constant value of ρ D12 (see Problem 11.33 and Problem 11.34). The mass-based solution of this problem provides an important approximation in our analysis of high-rate convective mass transfer in the next section. 11.8 Mass transfer coefficients at high rates of mass transfer In Section 11.6, we developed an analogy between heat and mass transfer that allowed us to calculate mass transfer coefficients when the rate of mass transfer was low. This analogy required that the velocity field be unaffected by mass transfer and that the transferred species be dilute. When those conditions are not met, the mass transfer coefficient will be different than the value given by the analogy. The difference can be either an increase or a decrease and can range from a few percent to an order of magnitude or more, depending upon the concentrations of the diffusing species. In addition to the diffusive transport represented by the mass transfer coefficient, convective transport can contribute substantially to the total mass flux. §11.8 Mass transfer coefficients at high rates of mass transfer 655 Figure 11.17 The mass concentration boundary layer. In this section, we model mass convection when the transferred species affects the velocity field and is not necessarily dilute. First, we define the mass transfer driving force, which governs the total mass flux from the wall. Then, we relate the mass transfer coefficient at high mass transfer rates to that at low mass transfer rates. The mass transfer driving force Figure 11.17 shows a boundary layer over a wall through which there is ˙ a net mass transfer, ns ≡ m , of the various species in the direction normal to the wall.11 In particular, we will focus on species i. In the free stream, i has a concentration mi,e ; at the wall, it has a concentration mi,s . The mass flux of i leaving the wall is obtained from eqn. (11.21): ˙ ni,s = mi,s m + ji,s (11.93) ˙ We seek to obtain m in terms of the concentrations mi,s and mi,e . As before, we define the mass transfer coefficient for species i, gm,i (kg/m2 ·s), as gm,i = ji,s mi,s − mi,e (11.94) Thus, ˙ ni,s = mi,s m + gm,i mi,s − mi,e (11.95) The mass transfer coefficient is again based on the diffusive transfer from the wall; however, it may now differ considerably from the value for lowrate transport. 11 ˙ In this context, we denote the total mass flux through the wall as m , rather than ns , so as to be consistent with other literature on the subject. An introduction to mass transfer 656 §11.8 Equation (11.95) may be rearranged as mi,e − mi,s ˙ mi,s − ni,s /m ˙ m = gm,i (11.96) ˙ which express the total mass flux of all species through the wall, m , as the product of the mass transfer coefficient and a ratio of concentrations. This ratio is called the mass transfer driving force for species i: Bm,i ≡ mi,e − mi,s ˙ mi,s − ni,s /m (11.97) The ratio of mass fluxes in the denominator is called the mass fraction in the transferred state, denoted as mi,t : ˙ mi,t ≡ ni,s /m (11.98) The mass fraction in the transferred state is simply the fraction of the ˙ total mass flux, m , which is made up of species i. It is not really a mass fraction in the sense of Section 11.2 because it can have any value from ˙ −∞ to +∞, depending on the relative magnitudes of m and ni,s . If, for ˙ −n2,s in a binary mixture, then m is very small and example, n1,s both m1,t and m2,t are very large. Equations (11.96), (11.97), and (11.98) provide a formulation of mass transfer problems in terms of the mass transfer coefficient, gm,i , and the driving force for mass transfer, Bm,i : ˙ m = gm,i Bm,i (11.99) where Bm,i = mi,e − mi,s mi,s − mi,t , ˙ mi,t = ni,s /m (11.100) These relations are based on an arbitrary species, i. The mass transfer rate may equally well be calculated using any species in a mixture; one obtains the same result for each. This is well illustrated in a binary mixture for which one may show that (Problem 11.36) gm,1 = gm,2 and Bm,1 = Bm,2 Mass transfer coefficients at high rates of mass transfer §11.8 In many situations, only one species is transferred through the wall. ˙ If species i is the only one passing through the s -surface, then ni,s = m , so that mt,i = 1. The mass transfer driving force is simply Bm,i = mi,e − mi,s mi,s − 1 one species transferred (11.101) In all the cases described in Section 11.6, only one species is transferred. Example 11.15 A pan of hot water with a surface temperature of 75◦ C is placed in an air stream that has a mass fraction of water equal to 0.05. If the average mass transfer coefficient for water over the pan is gm,H2 O = 0.0170 kg/m2 ·s and the pan has a surface area of 0.04 m2 , what is the evaporation rate? Solution. Only water vapor passes through the liquid surface, since air is not strongly absorbed into water under normal conditions. Thus, we use eqn. (11.101) for the mass transfer driving force. Reference to a steam table shows the saturation pressure of water to be 38.58 kPa at 75◦ C, so xH2 O,s = 38.58/101.325 = 0.381 Putting this value into eqn. (11.67), we obtain mH2 O,s = 0.277 so that Bm,H2 O = 0.05 − 0.277 = 0.314 0.277 − 1.0 Thus, ˙ mH2 O = gm,H2 O Bm,H2 O (0.04 m2 ) = (0.0170 kg/m2 ·s)(0.314)(0.04 m2 ) = 0.000214 kg/s = 769 g/hr 657 658 An introduction to mass transfer §11.8 Figure 11.18 A stagnant film. The effect of mass transfer rates on the mass transfer coefficient We still face the task of finding the mass transfer coefficient, gm,i . The most obvious way to do this would be to apply the same methods we used to find the heat transfer coefficient in Chapters 6 through 8—solution of the momentum and species equations or through correlation of mass transfer data. These approaches are often used, but they are more complicated than the analogous heat transfer problems, owing to the coupling of the flow field and the mass transfer rate. Simple solutions are not so readily available for mass transfer problems. We instead employ a widely used approximate method that allows us to calculate gm,i from the low-rate mass transfer coefficient by applying a correction for the effect of finite mass transfer rates. ˙ To isolate the effect of m on the mass transfer coefficient, we first ∗ define the mass transfer coefficient at zero net mass transfer, gm,i : ∗ gm,i ≡ lim gm,i ˙ m →0 ∗ The value gm,i is simply the mass transfer coefficient for low rates that would be obtained from the analogy between heat and mass transfer, as described in Section 11.6. Although gm,i depends directly on the rate of ∗ mass transfer, gm,i does not: it is determined only by flow configuration and physical properties. In a boundary layer, the fluid near the wall is slowed by the no-slip condition. One way of modeling high-rate mass transfer effects on gm,i is to approximate the boundary layer as a stagnant film —a stationary layer of fluid with no horizontal gradients in it, as shown in Fig. 11.18. The film thickness, δc , is an effective local concentration boundary layer thickness. The presence of a finite mass transfer rate across the film means that vertical convection—counterdiffusion effects—will be present. In fact, §11.8 Mass transfer coefficients at high rates of mass transfer the stagnant film shown in Fig. 11.18 is identical to the configuration dealt with in the previous section (i.e., Fig. 11.16). Thus, the solution obtained in the previous section—eqn. (11.88)—also gives the rate of mass transfer across the stagnant film, taking account of vertical convective transport. In the present mass-based analysis, it is convenient to use the massbased analog of the mole-based eqn. (11.88). This analog can be shown to be (Problem 11.33) ˙ m= mi,e − mi,s ρ Dim ln 1 + ˙ δc mi,s − ni,s /m which we may recast in the following, more suggestive form ˙ m= ln(1 + Bm,i ) Bm,i Bm,i ρ Dim δc (11.102) Comparing this equation with eqn. (11.99), we see that gm,i = ρ Dim δc ln(1 + Bm,i ) Bm,i ˙ and when m approaches zero, ∗ gm,i = lim gm,i = ˙ m →0 lim gm,i = Bm,i →0 ρ Dim δc (11.103) Hence, ∗ gm,i = gm,i ln(1 + Bm,i ) Bm,i (11.104) ∗ The appropriate value gm,i (or δc ) may be found from the solution of corresponding low-rate mass transfer problem, using the analogy of heat ∗ and mass transfer. (The value of gm,i , in turn, defines the effective concentration b.l. thickness, δc .) The group [ln(1 + Bm,i )]/Bm,i is called the blowing factor. It accounts the effect of mass transfer on the velocity field. When Bm,i > 0, we have mass flow away from the wall (or blowing.) In this case, the blowing factor is always a positive number less than unity, so blowing reduces gm,i . When Bm,i < 0, we have mass flow toward the wall (or suction ), and the blowing factor is always a positive number greater than unity. Thus, 659 660 An introduction to mass transfer §11.8 gm,i is increased by suction. These trends may be better understood if we note that wall suction removes the slow fluid at the wall and thins the boundary layer. The thinner b.l. offers less resistance to mass transfer. Likewise, blowing tends to thicken the b.l., increasing the resistance to mass transfer. The stagnant film b.l. model ignores details of the flow in the b.l. and focuses on the balance of mass fluxes across it. It is equally valid for both laminar and turbulent flows. Analogous stagnant film analyses of heat and momentum transport may also be made, as discussed in Problem 11.37. Example 11.16 Calculate the mass transfer coefficient for Example 11.15 if the air speed is 5 m/s, the length of the pan in the flow direction is 20 cm, and the air temperature is 25◦ C. Assume that the air flow does not generate waves on the water surface. Solution. The water surface is essentially a flat plate, as shown in Fig. 11.19. To find the appropriate equation for the Nusselt number, we must first compute ReL . The properties are evaluated at the film temperature, Tf = (75 + 25)/2 = 50◦ C, and the film composition, mf ,H2 O = (0.050 + 0.277)/2 = 0.164 For these conditions, we find the mixture molecular weight from eqn. (11.8) as Mf = 26.34 kg/kmol. Thus, from the ideal gas law, ρf = (101, 325)(26.34) (8314.5)(323.15) = 0.993 kg/m3 From Appendix A, we get µair = 1.949 × 10−5 kg/m·s and µwater vapor = 1.062 × 10−5 kg/m·s. Then eqn. (11.54), with xH2 O,f = 0.240 and xair,f = 0.760, yields µf = 1.75 × 10−5 kg/m·s so νf = (µ/ρ)f = 1.76 × 10−5 m2 /s We compute ReL = 5(0.2)/(1.76 × 10−5 ) = 56, 800, so the flow is laminar. The appropriate Nusselt number is obtained from the mass transfer version of eqn. (6.68): 1/2 Num,L = 0.664 ReL Sc1/3 §11.8 Mass transfer coefficients at high rates of mass transfer Figure 11.19 Evaporation from a tray of water. Equation (11.35) yields DH2 O,air = 2.96 × 10−5 m2 /s, so Sc = 1.76/2.96 = 0.595 and Num,L = 133 Hence, ∗ gm,H2 O = Num,L (ρ DH2 O,air /L) = 0.0195 kg/m2 ·s Finally, ∗ gm,H2 O = gm,H2 O ln(1 + Bm,H2 O ) Bm,H2 O = 0.0195 ln(1.314)/0.314 = 0.0170 kg/m2 ·s In this case, the blowing factor is 0.870. Thus, mild blowing has reduced the mass transfer coefficient by about 13%. Conditions for low-rate mass transfer. When the mass transfer driving force is small enough, the low-rate mass transfer coefficient itself is an adequate approximation to the actual mass transfer coefficient. This is because the blowing factor tends toward unity as Bm,i → 0: lim Bm,i →0 ln(1 + Bm,i ) =1 Bm,i Thus, for small values of Bm,i , gm,i ∗ gm,i . 661 662 An introduction to mass transfer §11.8 The calculation of mass transfer proceeds in one of two ways for ˙ low rates of mass transfer, depending upon how the limit of small m ˙ is reached. The first situation is when the ratio ni,s /m is fixed at a ˙ → 0. This would be the case when only one nonzero value while m ˙ species is transferred, since ni,s /m = 1. Then the mass flux at low rates is ˙ m ∗ gm,i Bm,i (11.105) In this case, convective and diffusive contributions to ni,s are of the same order of magnitude, in general. To reach conditions for which the analogy 1, so of heat and mass transfer applies, it is also necessary that mi,s that convective effects will be negligible, as discussed in Section 11.6. When that condition also applies, and if only one species is transferred, we have ˙ m = ni,s ∗ gm,i Bm,i ∗ = gm,i mi,e − mi,s mi,s − 1 ∗ gm,i (mi,s − mi,e ) ˙ In the second situation, ni,s remains finite while m from eqn. (11.93), ni,s ji,s ∗ gm,i (mi,s − mi,e ) → 0. Then, (11.106) The transport in this case is purely diffusive, irrespective of the size of mi,s . This situation arises is catalysis, where two species flow to a wall and react, creating a third species that flows away from the wall. Since the reaction conserves mass, the net mass flow through the s -surface is zero, even though ni,s is not (see Problem 11.44). An estimate of the blowing factor can be used to determine whether Bm,i is small enough to justify using the simpler low-rate theory. If, for example, Bm,i = 0.20, then [ln(1 + Bm,i )]/Bm,i = 0.91 and an error of only 9 percent is introduced by assuming low rates. This level of accuracy is adequate for many engineering calculations. §11.9 11.9 Simultaneous heat and mass transfer Simultaneous heat and mass transfer Many important engineering mass transfer processes occur simultaneously with heat transfer. Cooling towers, dryers, and combustors are just a few examples of equipment that intimately couple heat and mass transfer. Coupling can arise when temperature-dependent mass transfer processes cause heat to be released or absorbed at a surface. For example, during evaporation, latent heat is absorbed at a liquid surface when vapor is created. This tends to cool the surface, lowering the vapor pressure and reducing the evaporation rate. Similarly, in the carbon oxidation problem discussed in Example 11.2, heat is released when carbon is oxidized, and the rate of oxidation is a function of temperature. The balance between convective cooling and the rate of reaction determines the surface temperature of the burning carbon. Simultaneous heat and mass transfer processes may be classified as low-rate or high-rate. At low rates of mass transfer, mass transfer has only a negligible influence on the velocity field, and heat transfer rates may be calculated as if mass transfer were not occurring. At high rates of mass transfer, the heat transfer coefficient must be corrected for the effect of counterdiffusion. In this section, we consider these two possibilities in turn. Heat transfer at low rates of mass transfer One very common case of low-rate heat and mass transfer is the evaporation of water into air at low or moderate temperatures. An archetypical example of such a process is provided by a sling psychrometer, which is a device used to measure the humidity of air. In a sling psychrometer, a wet cloth is wrapped about the bulb of a thermometer, as shown in Fig. 11.20. This so-called wet-bulb thermometer is mounted, along with a second dry-bulb thermometer, on a swivel handle, and the pair are “slung” in a rotary motion until they reach steady state. The wet-bulb thermometer is cooled, as the latent heat of the vaporized water is given up, until it reaches the temperature at which the rate of cooling by evaporation just balances the rate of convective heating by the warmer air. This temperature, which is called the wet-bulb tem- 663 664 An introduction to mass transfer §11.9 Figure 11.20 The wet bulb of a sling psychrometer. perature, is directly related to the amount of water in the surrounding air.12 The highest ambient air temperatures we normally encounter are fairly low, so the rate of mass transfer should be small. We can test this suggestion by computing an upper bound on Bm,H2 O , under conditions that should maximize the evaporation rate: using the highest likely air temperature and the lowest humidity. Let us set those values, say, at 120◦ F (49◦ C) and zero humidity (mH2 O,e = 0). We know that the vapor pressure on the wet bulb will be less than the saturation pressure at 120◦ F, since evaporation will keep the bulb at a lower temperature: xH2 O,s psat (120◦ F)/patm = (11, 671 Pa)/(101, 325 Pa) = 0.115 12 The wet-bulb temperature for air–water systems is very nearly the adiabatic saturation temperature of the air–water mixture — the temperature reached by a mixture if it is brought to saturation with water by adding water vapor without adding heat. It is a thermodynamic property of an air–water combination. Simultaneous heat and mass transfer §11.9 so, with eqn. (11.67), 0.0750 mH2 O,s Thus, our criterion for low-rate mass transfer, eqn. (11.74), is met: Bm,H2 O = mH2 O,s − mH2 O,e 1 − mH2 O,s 0.0811 Alternatively, in terms of the blowing factor, eqn. (11.104), ln(1 + Bm,H2 O ) Bm,H2 O 0.962 This means that under the worst normal circumstances, the low-rate theory should deviate by only 4 percent from the actual rate of evaporation. We may form an energy balance on the wick by considering the u, s , and e surfaces shown in Fig. 11.20. At the steady temperature, no heat is conducted past the u-surface (into the wet bulb), but liquid water flows through it to the surface of the wick where it evaporates. An energy balance on the region between the u and s surfaces gives ˆ nH2 O,s hH2 O,s − enthalpy of water vapor leaving ˆ = nH2 O,u hH2 O,u qs heat convected to the wet bulb enthalpy of liquid water arriving Since mass is conserved, nH2 O,s = nH2 O,u , and because the enthalpy ˆ ˆ change results from vaporization, hH2 O,s − hH2 O,u = hfg . Hence, nH2 O,s hfg Twet-bulb = h(Te − Twet-bulb ) jH2 O,s , and this equation can be For low-rate mass transfer, nH2 O,s written in terms of the mass transfer coefficient gm,H2 O mH2 O,s − mH2 O,e hfg Twet-bulb = h(Te − Twet-bulb ) (11.107) The heat and mass transfer coefficients depend on the geometry and flow rates of the psychrometer, so it would appear that Twet-bulb should depend on the device used to measure it. The two coefficients are not independent, however, owing to the analogy between heat and mass transfer. For forced convection in cross flow, we saw in Chapter 7 that the heat transfer coefficient had the general form hD = C Rea Prb k 665 666 An introduction to mass transfer §11.9 where C is a constant, and typical values of a and b are a b 1/3. From the analogy, 1/2 and gm D = C Rea Scb ρ D12 Dividing the second expression into the first, we find h D12 = gm cp α Pr Sc b Both α/D12 and Sc/Pr are equal to the Lewis number, Le. Hence, h = Le1−b gm cp Le2/3 (11.108) Equation (11.108) shows that the ratio of h to gm depends primarily on the physical properties of the gas mixture, Le and cp , rather than the geometry or flow rate. The Lewis number for air–water systems is about 0.847; and, because the concentration of water vapor is generally low, cp can often be approximated by cpair . This type of relationship between h and gm was first developed by W. K. Lewis in 1922 for the case in which Le = 1 [11.27]. (Lewis’s primary interest was in air–water systems, so the approximation was not too bad.) The more general form, eqn. (11.108), is another ReynoldsColburn type of analogy, similar to eqn. (6.76). It was given by Chilton and Colburn [11.28] in 1934. Equation (11.107) may now be written as Te − Twet-bulb = hfg Twet-bulb cpair Le2/3 mH2 O,s − mH2 O,e (11.109) This expression can be solved iteratively with a steam table to obtain the wet-bulb temperature as a function of the dry-bulb temperature, Te , and the humidity of the ambient air, mH2 O,e . The psychrometric charts found in engineering handbooks and thermodynamics texts can be generated in this way. We ask the reader to make such calculations in Problem 11.49. The wet-bulb temperature is a helpful concept in many phase-change processes. When a small body without internal heat sources evaporates or sublimes, it cools to a steady “wet-bulb” temperature at which convective heating is balanced by latent heat removal. The body will stay at that temperature until the phase-change process is complete. Thus, the wet-bulb temperature appears in the evaporation of water droplets, the sublimation of dry ice, the combustion of fuel sprays, and so on. If the body is massive, however, steady state may not be reached very quickly. Simultaneous heat and mass transfer §11.9 Stagnant film model of heat transfer at high mass transfer rates The multicomponent energy equation. Each species in a mixture carˆ ries its own enthalpy, hi . In a flow with mass transfer, different species move with different velocities, so that enthalpy transport by individual species must enter the energy equation along with heat conduction through the fluid mixture. For steady, low-speed flow without internal heat generation or chemical reactions, we may rewrite the energy balance, eqn. (6.36), as ⎛ ⎞ − S (−k∇T ) · dS − ⎝ S ˆ ρi hi vi ⎠ · dS = 0 i where the second term accounts for the enthalpy transport by each species in the mixture. The usual procedure of applying Gauss’s theorem and requiring the integrand to vanish identically gives ⎞ ⎛ ˆ ρi hi vi ⎠ = 0 ∇ · ⎝−k∇T + (11.110) i This equation shows that the total energy flux—the sum of heat conduction and enthalpy transport—is conserved in steady flow.13 The stagnant film model. Let us restrict attention to the transport of a single species, i, across a boundary layer. We again use the stagnant film model for the thermal boundary layer and consider the one-dimensional flow of energy through it (see Fig. 11.21). Equation (11.110) simplifies to d dy −k dT ˆ + ρi hi vi dy =0 (11.111) From eqn. (11.69) for steady, one-dimensional mass conservation ni = constant in y = ni,s 13 The multicomponent energy equation becomes substantially more complex when kinetic energy, body forces, and thermal or pressure diffusion are taken into account. The complexities are such that most published derivations of the multicomponent energy equation are incorrect, as shown by Mills in 1998 [11.29]. The main source of error has been the assignment of an independent kinetic energy to the ordinary diffusion velocity. This leads to such inconsistencies as a mechanical work term in the thermal energy equation. 667 668 An introduction to mass transfer §11.9 Figure 11.21 Energy transport in a stagnant film. If we neglect pressure variations and assume a constant specific heat ˆ capacity (as in Sect. 6.3), the enthalpy may be written as hi = cp,i (T −Tref ), and eqn. (11.111) becomes d dy −k dT + ni,s cp,i T dy =0 Integrating twice and applying the boundary conditions T (y = 0) = Ts and T (y = δt ) = Te we obtain the temperature profile of the stagnant film: T − Ts = Te − Ts exp exp ni,s cp,i k ni,s cp,i k y −1 (11.112) δt − 1 The temperature distribution may be used to find the heat transfer coefficient according to its definition [eqn. (6.5)]: −k h≡ dT dy Ts − Te s = ni,s cp,i ni,s cp,i exp δt − 1 k (11.113) We define the heat transfer coefficient in the limit of zero mass transfer, h∗ , as h∗ ≡ lim h = ni,s →0 k δt (11.114) Simultaneous heat and mass transfer §11.9 Substitution of eqn. (11.114) into eqn. (11.113) yields h= ni,s cp,i exp(ni,s cp,i /h∗ ) − 1 (11.115) To use this result, one first calculates the heat transfer coefficient as if there were no mass transfer, using the methods of Chapters 6 through 8. The value obtained is h∗ , which is then placed in eqn. (11.115) to determine h in the presence of mass transfer. Note that h∗ defines the effective film thickness δt through eqn. (11.114). Equation (11.115) shows the primary effects of mass transfer on h. When ni,s is large and positive—the blowing case—h becomes smaller than h∗ . Thus, blowing decreases the heat transfer coefficient, just as it decreases the mass transfer coefficient. Likewise, when ni,s is large and negative—the suction case—h becomes very large relative to h∗ : suction increases the heat transfer coefficient just as it increases the mass transfer coefficient. Condition for the low-rate approximation. When the rate of mass transfer is small, we may approximate h by h∗ , just as we approximated gm ∗ by gm at low mass transfer rates. The approximation h = h∗ may be tested by considering the ratio ni,s cp,i /h∗ in eqn. (11.115). For example, if ni,s cp,i /h∗ = 0.2, then h/h∗ = 0.90, and h = h∗ within an error of only 10 percent. This is within the uncertainty to which h∗ can be predicted in most flows. In gases, if Bm,i is small, ni,s cp,i /h∗ will usually be small as well. ∗ Property reference state. In Section 11.8, we calculated gm,i (and thus gm,i ) at the film temperature and film composition, as though mass transfer were occurring at the mean mixture composition and tempera∗ ture. We may evaluate h∗ and gm,i in the same way when heat and mass transfer occur simultaneously. If composition variations are not large, as in many low-rate problems, it may be adequate to use the freestream composition and film temperature. When large properties variations are present, other schemes may be required [11.30]. 669 670 An introduction to mass transfer §11.9 Figure 11.22 Transpiration cooling. Energy balances in simultaneous heat and mass transfer Transpiration cooling. To calculate simultaneous heat and mass transfer rates, one must generally look at the energy balance below the wall as well as those at the surface and across the boundary layer. Consider, for example, the process of transpiration cooling, shown in Fig. 11.22. Here a wall exposed to high temperature gases is protected by injecting a cooler gas into the flow through a porous section of the surface. A portion of the heat transfer to the wall is taken up in raising the temperature of the transpired gas. Blowing serves to thicken the boundary layer and reduce h, as well. This process is frequently used to cool turbine blades and combustion chamber walls. Let us construct an energy balance for a steady state in which the wall has reached a temperature Ts . The enthalpy and heat fluxes are as shown in Fig. 11.22. We take the coolant reservoir to be far enough back from the surface that temperature gradients at the r -surface are negligible and the conductive heat flux, qr , is zero. An energy balance between the r and u-surfaces gives ˆ ˆ ni,r hi.r = ni,u hi,u − qu (11.116) and between the u- and s -surfaces, ˆ ˆ ni,u hi,u − qu = ni,s hi,s − qs (11.117) §11.9 Simultaneous heat and mass transfer Since there is no change in the enthalpy of the transpired species when it passes out of the wall, ˆ ˆ hi,u = hi,s (11.118) and, because the process is steady, conservation of mass gives ni,r = ni,u = ni,s (11.119) Thus, eqn. (11.117) reduces to qs = qu (11.120) The flux qu is the conductive heat flux into the wall, while qs is the convective heat transfer from the gas stream, qs = h(Te − Ts ) (11.121) Combining eqns. (11.116) through (11.121), we find ˆ ˆ ni,s hi,s − hi,r = h(Te − Ts ) (11.122) This equation shows that, at steady state, the heat convection to the wall is absorbed by the enthalpy rise of the transpired gas. Writing the ˆ enthalpy as hi = cp,i (Ts − Tref ), we obtain ni,s cp,i (Ts − Tr ) = h(Te − Ts ) (11.123) or Ts = hTe + ni,s cp,i Tr h + ni,s cp,i (11.124) It is left as an exercise (Problem 11.47) to show that Ts = Tr + (Te − Tr ) exp(−ni,s cp,i /h∗ ) (11.125) The wall temperature decreases exponentially to Tr as the mass flux of the transpired gas increases. Transpiration cooling may be enhanced by injecting a gas with a high specific heat. 671 672 An introduction to mass transfer §11.9 Sweat Cooling. A common variation on transpiration cooling is sweat cooling, in which a liquid is bled through a porous wall. The liquid is vaporized by convective heat flow to the wall, and the latent heat of vaporization acts as a sink. Figure 11.22 also represents this process. The balances, eqns. (11.116) and (11.117), as well as mass conservation, eqn. (11.119), still apply, but the enthalpies at the interface now differ by the latent heat of vaporization: ˆ ˆ hi,u + hfg = hi,s (11.126) Thus, eqn. (11.120) becomes qs = qu + hfg ni,s and eqn. (11.122) takes the form ni,s hfg + cp,if (Ts − Tr ) = h(Te − Ts ) (11.127) where cp,if is the specific heat of liquid i. Since the latent heat is generally much larger than the sensible heat, a comparison of eqn. (11.127) to eqn. (11.123) exposes the greater efficiency per unit mass flow of sweat cooling relative to transpiration cooling. Thermal radiation. When thermal radiation falls on the surface through which mass is transferred, the additional heat flux must enter the energy balances. For example, suppose that thermal radiation were present during transpiration cooling. Radiant heat flux, qrad,e , originating above the e-surface would be absorbed below the u-surface.14 Thus, eqn. (11.116) becomes ˆ ˆ ni,r hi,r = ni,u hi,u − qu − αqrad,e (11.128) where α is the radiation absorptance. Equation (11.117) is unchanged. Similarly, thermal radiation emitted by the wall is taken to originate below the u-surface, so eqn. (11.128) is now ˆ ˆ ni,r hi,r = ni,u hi,u − qu − αqrad,e + qrad,u (11.129) or, in terms of radiosity and irradiation (see Section 10.4) ˆ ˆ ni,r hi,r = ni,u hi,u − qu − (H − B) (11.130) for an opaque surface. 14 Remember that the s - and u-surfaces are fictitious elements of the enthalpy balances at the phase interface. The apparent space between them need be only a few molecules thick. Thermal radiation therefore passes through the u-surface and is absorbed below it. Problems 673 Chemical Reactions. The heat and mass transfer analyses in this section and Section 11.8 assume that the transferred species undergo no homogeneous reactions. If reactions do occur, the mass balances of Section 11.8 are invalid, because the mass flux of a reacting species will vary across the region of reaction. Likewise, the energy balance of this section will fail because it does not include the heat of reaction. For heterogeneous reactions, the complications are not so severe. Reactions at the boundaries release the heat of reaction released between the s - and u-surfaces, altering the boundary conditions. The proper stoichiometry of the mole fluxes to and from the surface must be taken into account, and the heat transfer coefficient [eqn. (11.115)] must be modified to account for the transfer of more than one species [11.30]. Problems 11.1 Derive: (a) eqns. (11.8); (b) eqns. (11.9). 11.2 A 1000 liter cylinder at 300 K contains a gaseous mixture composed of 0.10 kmol of NH3 , 0.04 kmol of CO2 , and 0.06 kmol of He. (a) Find the mass fraction for each species and the pressure in the cylinder. (b) After the cylinder is heated to 600 K, what are the new mole fractions, mass fractions, and molar concentrations? (c) The cylinder is now compressed isothermally to a volume of 600 liters. What are the molar concentrations, mass fractions, and partial densities? (d) If 0.40 kg of gaseous N2 is injected into the cylinder while the temperature remains at 600 K, find the mole fractions, mass fractions, and molar concentrations. [(a) mCO2 = 0.475; (c) cCO2 = 0.0667 kmol/m3 ; (d) xCO2 = 0.187.] 11.3 Planetary atmospheres show significant variations of temperature and pressure in the vertical direction. Observations suggest that the atmosphere of Jupiter has the following composition at the tropopause level: number density of H2 = 5.7 × 1021 (molecules/m3 ) number density of He = 7.2 × 1020 (molecules/m3 ) number density of CH4 = 6.5 × 1018 (molecules/m3 ) number density of NH3 = 1.3 × 1018 (molecules/m3 ) Chapter 11: An introduction to mass transfer 674 Find the mole fraction and partial density of each species at this level if p = 0.1 atm and T = 113 K. Estimate the number densities at the level where p = 10 atm and T = 400 K, deeper within the Jovian troposphere. (Deeper in the Jupiter’s atmosphere, the pressure may exceed 105 atm.) 11.4 Using the definitions of the fluxes, velocities, and concentrations, derive eqn. (11.34) from eqn. (11.27) for binary diffusion. 11.5 Show that D12 = D21 in a binary mixture. 11.6 Fill in the details involved in obtaining eqn. (11.31) from eqn. (11.30). 11.7 Batteries commonly contain an aqueous solution of sulfuric acid with lead plates as electrodes. Current is generated by the reaction of the electrolyte with the electrode material. At the negative electrode, the reaction is Pb(s) + SO2− 4 PbSO4 (s) + 2e− where the (s) denotes a solid phase component and the charge of an electron is −1.609 × 10−19 coulombs. If the current density at such an electrode is J = 5 milliamperes/cm2 , what is the mole flux of SO2− to the electrode? (1 amp =1 coulomb/s.) 4 What is the mass flux of SO2− ? At what mass rate is PbSO4 4 produced? If the electrolyte is to remain electrically neutral, at what rate does H+ flow toward the electrode? Hydrogen ˙ does not react at the negative electrode. [mPbSO4 = 7.83 × −5 kg/m2 ·s.] 10 11.8 The salt concentration in the ocean increases with increasing depth, z. A model for the concentration distribution in the upper ocean is S = 33.25 + 0.75 tanh(0.026z − 3.7) where S is the salinity in grams of salt per kilogram of ocean water and z is the distance below the surface in meters. (a) Plot the mass fraction of salt as a function of z. (The region of rapid transition of msalt (z) is called the halocline.) (b) Ignoring the effects of waves or currents, compute jsalt (z). Use a value of Problems 675 Dsalt,water = 1.5 × 10−5 cm2 /s. Indicate the position of maximum diffusion on your plot of the salt concentration. (c) The upper region of the ocean is well mixed by wind-driven waves and turbulence, while the lower region and halocline tend to be calmer. Using jsalt (z) from part (b), make a simple estimate of the amount of salt carried upward in one week in a 5 km2 horizontal area of the sea. 11.9 In catalysis, one gaseous species reacts with another on a passive surface (the catalyst) to form a gaseous product. For example, butane reacts with hydrogen on the surface of a nickel catalyst to form methane and propane. This heterogeneous reaction, referred to as hydrogenolysis, is Ni C4 H10 + H2 → C3 H8 + CH4 The molar rate of consumption of C4 H10 per unit area in the ◦ −2 ˙ reaction is RC4 H10 = A(e−∆E/R T )pC4 H10 pH2 .4 , where A = 6.3 × 1010 kmol/m2 ·s, ∆E = 1.9 × 108 J/kmol, and p is in atm. (a) If pC4 H10 ,s = pC3 H8 ,s = 0.2 atm, pCH4 ,s = 0.17 atm, and pH2 ,s = 0.3 atm at a nickel surface with conditions of 440◦ C and 0.87 atm total pressure, what is the rate of consumption of butane? (b) What are the mole fluxes of butane and hydrogen to the surface? What are the mass fluxes of propane and ethane ˙ away from the surface? (c) What is m ? What are v , v ∗ , and vC4 H10 ? (d) What is the diffusional mole flux of butane? What is the diffusional mass flux of propane? What is the flux of Ni? [(b) nCH4 ,s = 0.0441 kg/m2 ·s; (d) jC3 H8 = 0.121 kg/m2 ·s.] 11.10 Consider two chambers held at temperatures T1 and T2 , respectively, and joined by a small insulated tube. The chambers are filled with a binary gas mixture, with the tube open, and allowed to come to steady state. If the Soret effect is taken into account, what is the concentration difference between the two chambers? Assume that an effective mean value of the thermal diffusion ratio is known. 11.11 Compute D12 for oxygen gas diffusing through nitrogen gas at p = 1 atm, using eqns. (11.39) and (11.42), for T = 200 K, 500 K, and 1000 K. Observe that eqn. (11.39) shows large deviations from eqn. (11.42), even for such simple and similar molecules. Chapter 11: An introduction to mass transfer 676 11.12 (a) Compute the binary diffusivity of each of the noble gases when they are individually mixed with nitrogen gas at 1 atm and 300 K. Plot the results as a function of the molecular weight of the noble gas. What do you conclude? (b) Consider the addition of a small amount of helium (xHe = 0.04) to a mixture of nitrogen (xN2 = 0.48) and argon (xAr = 0.48). Compute DHe,m and compare it with DAr,m . Note that the higher concentration of argon does not improve its ability to diffuse through the mixture. 11.13 (a) One particular correlation shows that gas phase diffusion coefficients vary as T 1.81 and p −1 . If an experimental value of D12 is known at T1 and p1 , develop an equation to predict D12 at T2 and p2 . (b) The diffusivity of water vapor (1) in air (2) was measured to be 2.39 × 10−5 m2 /s at 8◦ C and 1 atm. Provide a formula for D12 (T , p). 11.14 Kinetic arguments lead to the Stefan-Maxwell equation for a dilute-gas mixture: ⎛ ⎞ ∗ n ∗ Jj Ji ci cj ⎝ ⎠ − ∇xi = c 2 Dij cj ci j =1 (a) Derive eqn. (11.44) from this, making the appropriate assumptions. (b) Show that if Dij has the same value for each pair of species, then Dim = Dij . 11.15 Compute the diffusivity of methane in air using (a) eqn. (11.42) and (b) Blanc’s law. For part (b), treat air as a mixture of oxygen and nitrogen, ignoring argon. Let xmethane = 0.05, T = 420◦ F, and p = 10 psia. [(a) DCH4 ,air = 7.66 × 10−5 m2 /s; (b) DCH4 ,air = 8.13 × 10−5 m2 /s.] 11.16 Diffusion of solutes in liquids is driven by the chemical potential, µ . Work is required to move a mole of solute A from a region of low chemical potential to a region of high chemical potential; that is, dµA dx dx under isothermal, isobaric conditions. For an ideal (very dilute) solute, µA is given by dW = dµA = µA = µ0 + R ◦ T ln(cA ) Problems 677 where µ0 is a constant. Using an elementary principle of mechanics, derive the Nernst-Einstein equation. Note that the solution must be assumed to be very dilute. 11.17 A dilute aqueous solution at 300 K contains potassium ions, K+ . If the velocity of aqueous K+ ions is 6.61 × 10−4 cm2 /s·V per unit electric field (1 V/cm), estimate the effective radius of K+ ions in an aqueous solution. Criticize this estimate. (The charge of an electron is −1.609 × 10−19 coulomb and a volt = 1J/coulomb.) 11.18 (a) Obtain diffusion coefficients for: (1) dilute CCl4 diffusing through liquid methanol at 340 K; (2) dilute benzene diffusing through water at 290 K; (3) dilute ethyl alcohol diffusing through water at 350 K; and (4) dilute acetone diffusing through methanol at 370 K. (b) Estimate the effective radius of a methanol molecule in a dilute aqueous solution. [(a) Dacetone,methanol = 6.8 × 10−9 m2 /s.] 11.19 If possible, calculate values of the viscosity, µ , for methane, hydrogen sulfide, and nitrous oxide, under the following conditions: 250 K and 1 atm, 500 K and 1 atm, 250 K and 2 atm, 250 K and 12 atm, 500 K and 12 atm. 11.20 (a) Show that k = (5/2)µcv for a monatomic gas. (b) Obtain Eucken’s formula for the Prandtl number of a dilute gas: Pr = 4γ (9γ − 5) (c) Recall that for an ideal gas, γ (D + 2)/D , where D is the number of modes of energy storage of its molecules. Obtain an expression for Pr as a function of D and describe what it means. (d) Use Eucken’s formula to compute Pr for gaseous Ar, N2 , and H2 O. Compare the result to data in Appendix A over the range of temperatures. Explain the results obtained for steam as opposed to Ar and N2 . (Note that for each mode of vibration, there are two modes of energy storage but that vibration is normally inactive until T is very high.) 11.21 A student is studying the combustion of a premixed gaseous fuel with the following molar composition: 10.3% methane, 15.4% ethane, and 74.3% oxygen. She passes 0.006 ft3/s of the Chapter 11: An introduction to mass transfer 678 mixture (at 70◦ F and 18 psia) through a smooth 3/8 inch I.D. tube, 47 inches long. (a) What is the pressure drop? (b) The student’s advisor recommends preheating the fuel mixture, using a Nichrome strip heater wrapped around the last 5 inches of the duct. If the heater produces 0.8 W/inch, what is the wall temperature at the outlet of the duct? Let cp,CH4 = 2280 J/kg·K, γCH4 = 1.3, cp,C2 H6 = 1730 J/kg·K, and γC2 H6 = 1.2, and evaluate the properties at the inlet conditions. 11.22 (a) Work Problem 6.36. (b) A fluid is said to be incompressible if the density of a fluid particle does not change as it moves about in the flow (i.e., if Dρ/Dt = 0). Show that an incompressible flow satisfies ∇ · u = 0. (c) How does the condition of incompressibility differ from that of “constant density”? Describe a flow that is incompressible but that does not have “constant density.” 11.23 Carefully derive eqns. (11.62) and (11.63). Note that ρ is not assumed constant in eqn. (11.62). 11.24 Derive the equation of species conservation on a molar basis, using ci rather than ρi . Also obtain an equation in ci alone, similar to eqn. (11.63) but without the assumption of incompressibility. What assumptions must be made to obtain the latter result? 11.25 Find the following concentrations: (a) the mole fraction of air in solution with water at 5◦ C and 1 atm, exposed to air at the same conditions, H = 4.88 × 104 atm; (b) the mole fraction of ammonia in air above an aqueous solution, with xNH3 = 0.05 at 0.9 atm and 40◦ C and H = 1522 mm Hg; (c) the mole fraction of SO2 in an aqueous solution at 15◦ C and 1 atm, if pSO2 = 28.0 mm Hg and H = 1.42 × 104 mm Hg; and (d) the partial pressure of ethylene over an aqueous solution at 25◦ C and 1 atm, with xC2 H4 = 1.75 × 10−5 and H = 11.4 × 103 atm. 11.26 Use a steam table to estimate (a) the mass fraction of water vapor in air over water at 1 atm and 20◦ C, 50◦ C, 70◦ C, and 90◦ C; (b) the partial pressure of water over a 3 percent-byweight aqueous solution of HCl at 50◦ C; (c) the boiling point at 1 atm of salt water with a mass fraction mNaCl = 0.18. [(c) TB.P . = 101.8◦ C.] Problems 11.27 679 Suppose that a steel fitting with a carbon mass fraction of 0.2% is put into contact with carburizing gases at 940◦ C, and that these gases produce a steady mass fraction, mC,u , of 1.0% carbon just within the surface of the metal. The diffusion coefficient of carbon in this steel is DC,Fe = 1.50 × 10−5 m2 s exp −(1.42 × 108 J/kmol) (R ◦ T ) for T in kelvin. How long does it take to produce a carbon concentration of 0.6% by mass at a depth of 0.5 mm? How much less time would it take if the temperature were 980◦ C? 11.28 (a) Write eqn. (11.62) in its boundary layer form. (b) Write this concentration boundary layer equation and its b.c.’s in terms of a nondimensional mass fraction, ψ, analogous to the dimensionless temperature in eqn. (6.42). (c) For ν = Dim , relate ψ to the Blasius function, f , for flow over a flat plate. (d) Note the similar roles of Pr and Sc in the two boundary layer transport processes. Infer the mass concentration analog of eqn. (6.55) and sketch the concentration and momentum b.l. profiles for Sc = 1 and Sc 1. 11.29 When Sc is large, momentum diffuses more easily than mass, and the concentration b.l. thickness, δc , is much less than the momentum b.l. thickness, δ. On a flat plate, the small part of the velocity profile within the concentration b.l. is approximately u/Ue = 3y/2δ. Compute Num,x based on this velocity profile, assuming a constant wall concentration. (Hint : Use the mass transfer analogs of eqn. (6.47) and (6.50) and note that qw /ρcp becomes ji,s /ρ .). 11.30 Consider a one-dimensional, binary gaseous diffusion process in which species 1 and 2 travel in opposite directions along the z-axis at equal molar rates. (The gas mixture will be at rest, with v = 0 if the species have identical molecular weights). This process is known as equimolar counter-diffusion. (a) What ∗ ∗ are the relations between N1 , N2 , J1 , and J2 ? (b) If steady state prevails and conditions are isothermal and isobaric, what is the concentration of species 1 as a function of z? (c) Write the mole flux in terms of the difference in partial pressure of species 1 between locations z1 and z2 . Chapter 11: An introduction to mass transfer 680 11.31 Consider steady mass diffusion from a small sphere. When convection is negligible, the mass flux in the radial direction is nr ,i = jr ,i = −ρ Dim dmi /dr . If the concentration is mi,∞ far from the sphere and mi,s at its surface, use a mass balance to obtain the surface mass flux in terms of the overall concentration difference (assuming that ρ Dim is constant). Then apply the definition eqns. (11.94) and (11.78) to show that Num,D = 2 for this situation. 11.32 An experimental Stefan tube is 1 cm in diameter and 10 cm from the liquid surface to the top. It is held at 10◦ C and 8.0 × 104 Pa. Pure argon flows over the top and liquid CCl4 is at the bottom. The pool level is maintained while 0.086 ml of liquid CCl4 evaporates during a period of 12 hours. What is the diffusivity of carbon tetrachloride in argon measured under these conditions? The specific gravity of liquid CCl4 is 1.59 and its vapor pressure is log10 pv = 8.004 − 1771/T , where pv is expressed in mm Hg and T in K. 11.33 Repeat the analysis given in Section 11.7 on the basis of mass fluxes, assuming that ρ Dim is constant and neglecting any buoyancy-driven convection. Obtain the analog of eqn. (11.88). 11.34 In Sections 11.5 and 11.7, it was assumed at points that c D12 or ρ D12 was independent of position. (a) If the mixture composition (e.g., x1 ) varies in space, this assumption may be poor. Using eqn. (11.42) and the definitions from Section 11.2, examine the composition dependence of these two groups. For what type of mixture is ρ D12 most sensitive to composition? What does this indicate about molar versus mass-based analysis? (b) How do each of these groups depend on pressure and temperature? Is the analysis of Section 11.7 really limited to isobaric conditions? (c) Do the Prandtl and Schmidt numbers depend on composition, temperature, or pressure? 11.35 A Stefan tube contains liquid bromine at 320 K and 1.2 atm. Carbon dioxide flows over the top and is also bubbled up through the liquid at the rate of 4.4 ml/hr. If the distance from the liquid surface to the top is 16 cm and the diameter is 1 cm, what is the evaporation rate of Br2 ? (psat,Br2 = 0.680 bar at 320 K.) [NBr2 ,s = 1.90 × 10−6 kmol/m2 ·s.] 11.36 Show that gm,1 = gm,2 and Bm,1 = Bm,2 in a binary mixture. Problems 681 11.37 Demonstrate that stagnant film models of the momentum and thermal boundary layers reproduce the proper dependence of Cf ,x and Nux on Rex and Pr. Using eqns. (6.31b) and (6.55) to obtain the dependence of δ and δt on Rex and Pr, show that stagnant film models gives eqns. (6.33) and (6.58) within a constant on the order of unity. [The constants in these results will differ from the exact results because the effective b.l. thicknesses of the stagnant film model are not the same as the exact values—see eqn. (6.57).] 11.38 (a) What is the largest value of the mass transfer driving force when one species is transferred? What is the smallest value? (b) Plot the blowing factor as a function of Bm,i for one species transferred. Indicate on your graph the regions of blowing, suction, and low-rate mass transfer. (c) Verify the two limits ∗ used to show that gm,i = ρ Dim /δc . 11.39 Nitrous oxide is bled through the surface of a porous 3/8 in. O.D. tube at 0.025 liter/s per meter of tube length. Air flows over the tube at 25 ft/s. Both the air and the tube are at 18◦ C, and the ambient pressure is 1 atm. Estimate the mean concentration of N2 O at the tube surface. (Hint : First estimate the concentration using properties of pure air; then correct the properties if necessary.) 11.40 Film absorbtion is a process whereby gases are absorbed into a falling liquid film. Typically, a thin film of liquid runs down the inside of a vertical tube through