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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 ©2006 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, c2006
Includes bibliographic references and index.
1. Heat—Transmission 2. Mass Transfer
I. Lienhard, John H., V, 1961– II. Title
TJ260.L445 2006 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 1.24 dated January 22, 2006 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 .
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. 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 .
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141
143
150
159
163
183
190
193
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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 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
7.1 393 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
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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
= mcp
Q=
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
ρc ∂T
∂2T
≡
=
∂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 Figure 1.9 19 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≡ m3 kg·K
J
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= K
3600 s (0.3048 m)2
0.0009478 Btu
·
·
·
h
J
1.8◦ F
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= 6000
q
=
= 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∞ )
=−
ρcV
dt (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
capacity for convection from surface
hAt (J/◦ C)
t
=
=
heat capacity of the body
T
ρcV (J/◦ C) (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
α ∂t
∂x
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
Tb0 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 Table 1.2 §1.3 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
o /kB T λ) − 1] λ5 [exp(hc (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 Figure 2.1 §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).
˙
1 ∂T
∂2T q
∂2T
∂2T
+
+=
+
2
2
2
∂z
k
α ∂t
∂y
∂x
=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. ˙
q
d2 T
=−
k
dx 2
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 + C 2
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 + C 2
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
=
˙
q L2 k
2 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−
2k
k 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
x
T2 − T 1
x ; or
=
L
T2 − T 1
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 T1 − T 2
d
dT
x
= −k
T1 −
L
dx
dx
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. Thermal resistance and the electrical analogy §2.3 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 Figure 2.9
current. §2.3 The one-dimensional ﬂow of 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, we can write
Rtotal = 1
L
1
L
+
+
=
kA hc A kA
A L
1
L
+
+
18 hc
18 Since hc is about 3,000 W/m2 K,
2L
1
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 + ˙
∂2T
q
1 ∂2T
+
+
=
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
⎪
⎪ C 2 = Ti +
ln ri
⎩
ln(ro /ri ) ∆T
(ln r − ln ri ) or
ln(ro /ri )
ln(r /ri )
T − Ti
=
To − T i
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 r − ri
T − Ti
=
To − T i
ro − r i
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 l∆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 = K
W ln(ro /ri )
2π lk (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∞ − T i
1/Bi + ln(ro /ri ) Then
C 2 = Ti − T∞ − Ti
ln ri
1/Bi + ln(ro /ri ) Step 6.
T= T∞ − T i
ln(r /ri ) + Ti
1/Bi + ln(ro /ri ) This can be rearranged in fully dimensionless form:
T − Ti
ln(r /ri )
=
T∞ − T i
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 ∞
T i − 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
2Tm or (∆T /Tm )2 /4
where the last step assumes that (∆T )2 /2
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,
Requiv = 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, as 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 example, 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= 4 1
ri ln(ro /ri )
+
k
hro
ri (2.34) 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. 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 . 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
L
1
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 Figure 2.19 §2.4 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. total thermal resistance of the circuit is
1 Rttotal = Rtconv + 1
Rtpine + + Rtconv 1
Rtsawdust Thus
Q= ∆T
=
Rttotal
1
hA T∞1 − T∞r
1 + kp Ap
L + k s 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 Rttotal , 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 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
† §2.4 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 Figure 2.21 + ri ln(rp /ri )
kpipe + The fouling of a pipe. ri
ro h o + Rf 83 84 Heat conduction, thermal resistance, and the overall heat transfer coeﬃcient Table 2.3 §2.4 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 diusion 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 Figure 3.1 §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 3.3 102 The three basic types of heat exchangers. §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
− 1−
Tc
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
Tc + Thin
− 1+
Tc +
Ch
Ch in
dTc
Cc
Cc
Tc + Thin
Tc −
− 1−
Ch out
Ch (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
Tc + Thin ⎥
Tcout +
⎢− 1 +
Cc
UA
Ch
Ch in
⎢
⎥
⎥ =−
1+
parallel: ln ⎢
Cc
Cc
⎣
⎦
Cc
Ch
− 1+
Tc + Thin
Tcin +
Ch
Ch in
⎤
⎡
Cc
Cc
Tcout + Thin ⎥
Tcout −
− 1−
⎢
Cc
UA
Ch
Ch
⎥
⎢
⎥ =−
1−
counter: ln ⎢
Cc
Cc
⎦
⎣
Ch
Cc
− 1−
Tc + Thin
Tcin −
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 ∆T a
(−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 − r i 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
˙
mH2 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:
− Cmin
Cc
Cmin
Cmin
+
+1
NTU = ln − 1 +
ε
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
, NTU only
= fn
Cmax
1 + Cmin /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: Cc = 20, 000 W/K
hot ﬂow enters at 150◦ C: Ch = 10, 000 W/K
A = 30 m2 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
= 72.78◦ C
+
= 40 +
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
=
α ∂t
k (2.11) Cartesian coordinates: Steady conduction in the y-direction.
(2.11) reduces as follows:
˙
∂2T q
∂2T ∂2T
+
+
+=
2
2
2
k
∂x
∂y
∂z
=0 =0 Equation 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 + C 2
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 + C 2
(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 Figure 4.2 145 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
=
α ∂t
k =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
for λ≠0
λ=0 (4.9) The general solution §4.2 147 and
2t T (t) = Ce−αλ
T (t) = C λ≠0
λ=0 for
for (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
+
∂y 2
∂x 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∞ − T i
dependent variable ⎢
= fn⎣ r ri , r o 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 , , (T∞ − Ti ) k 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∞ − T i 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 − T 1
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 = ˙
qL2
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 + C 2
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 − T 2 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 +
˙
qL
2
2 (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 4.6 164 Some of the many varieties of ﬁnned tubes. 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 Figure 4.8 §4.5 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
d2 T
dT /dx |x +δx − dT /dx |x
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
C 1 + 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− Ts − Troot
hr
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 Figure 4.11 179 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 Figure 4.12 §4.5 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
x
d(T − T∞ )
2hb
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 4.13 182 The eﬃciency of several ﬁns with variable cross section. 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/m2 K) covered on the inside, outside, and edges
with a continuous aluminum skin 3.2 mm thick (kAl = 165
W/m2 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
αt
hk(V /A)t
h(V /A)
t
·
=
=
= Bi Fo
2k
k
(V /A)2
T
ρc(V /A) (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 dx
d2 x
+ k x = F (t)
+c
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 = b T 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
slab 1 :
slab 2 : dT1
= hc A(T1 − T2 )
dt
dT2
= hA(T2 − T∞ ) − hc A(T1 − T2 )
−(ρcV )2
dt
−(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
T 2 = 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 dT1
dT1
d2 T 1
dT1
hc
T1
+ T1 − T ∞ +
= T 1 T2
− T2
2
dt
dt
dt
dt
h or
d 2 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
dθ
d2 θ
+ 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 − C1 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:
⎡ θ
b/2 +
T1 − T ∞
−c
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
2 π 2 Fo/4 Θ = Gn e−n
2 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
∞ 2 π 2 Fo/4 Gn e−n Θ= 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 mπ
2
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 ∞ nπ
1 −(nπ /2)2 Fo
ξ
e
sin
2
n
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
⎧
⎪
⎨
3π
1
π2
Fo − exp −
exp −
⎪
2
2
3
⎩
= 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 5π
1
exp −
2
5 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
J 0 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
= 1 − Φ.
=1−
Θ≡
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
=
0.00000343
α 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 Bi One-term coeﬃcients for convective cooling [5.1].
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
1 ∂T
∂2T
=
2
∂x
α ∂t
as follows, where we call Ti − T∞ ≡ ∆T :
∂T
∂ Θ ∂ζ
x
∂Θ
= (Ti − T∞ )
= ∆T
= ∆T − √
∂t
∂t
∂ζ ∂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 Table 5.3 §5.6 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
T w = Ti + qw
k T w = 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 − T 2
=
T1 − T 2 (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 5.21 238 The evolution of a ﬂux plot. §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 Figure 5.24
ent size. §5.7 The shape factor for two similar bodies of diﬀer- 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
+ C 2 = Ti
ri and C1
+ C 1 = To
ro Therefore,
C1 = Ti − To
ri ro
ro − r i and C2 = Ti − Ti − T o
ro
ro − r i Putting C1 and C2 in the general solution, and calling Ti − To ≡ ∆T ,
we get
T = Ti + ∆T ro
r i ro
−
r (ro − ri ) ro − ri Then
4π (ri ro )
dT
=
k∆T
dr
ro − r i
4π (ri ro )
m
S=
ro − r i 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
4π (0.06/2) m
(35 − 21)K 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 − r i 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 + R i − L 2
2Ro Ri 4R 4π b 1 − a2 b2
tanh−1 1 − a2 b 2 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 5.29 258 Conﬁgurations for Problem 5.22 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 m/s2 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 uctuate 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 Figure 6.1 §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:
−kf ∂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 and
y 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
+
+
∂z
∂x
∂y 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 ∂ρuv
∂ ρu2
+
∂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 ∂ρuv
∂ ρu2
+
∂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
1 dp
∂2u
∂uv
∂u2
=−
+ν
+
∂y
ρ dx
∂y 2
∂x (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
1 dp
∂2u
∂u
+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∞
dx
ρ dx
so from eqn. (6.12):
∂2u
∂(uv)
du∞
∂u2
= u∞
+ν
+
∂y
dx
∂y 2
∂x (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
∂x ∂y 2
∂y 3
∂y ∂y∂x (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
1.566 × 10−5
ν 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
∂2u
∂u
=ν
+v
∂y
∂y 2
∂x
=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
δ
u∞
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
1
Solving the middle two equations (above) for b and d, we obtain d = − 2
3 and b = + 2 , so
1
3y
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):
− 39
d
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
ρu2
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 1.328
0.664
√
=√
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
= 0.0063 m
= 600
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= ∂T
q
k
=−
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
u
u
∂
∂2
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∞ − T w 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= ∂(T − Tw )
k
∂y
T∞ − T w =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 u∞
dx
νx
√ =2 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 = mass ﬂux of molecules
change in ﬂuid
·
from y − /2 to y + /2
velocity 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,
q y = C2 ρ c v 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 = C 2 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.
y
y
T − T∞
=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
b = −3
2 a=1 c=0 d= 1
2 so the temperature proﬁle is
1
3y
T − T∞
+
=1−
2
Tw − T ∞
2 δt 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 1
3
ηφ − η3 φ3
2
2 1
3
3α
⎥
1 − η + η3 dη ⎦ =
2
2
2u∞ (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
δt
1
=
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≡ q
3k
3k δ
=
=
∆T
2δt
2 δ δt (6.57) and substituting eqns. (6.54) and (6.31b) for δ/δt and δ, we obtain
Nux ≡ 3 Rex
hx
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
k∆T
axial heat conductance of the b.l. 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 (6.72) 1/3 Pr 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∞ y
u
−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∞
δ= 1−
δt
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∞ y
u
−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 = actual heat ﬂux to the ﬂuid
h∆T
=
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
0.332
Nux
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 Figure 6.18 317 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∗ y
u
= fn
ν
u∗ (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 τw y
dy
=
ν
ρν
(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
u∗ y
1
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 εm
ν
+
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∞
T w − 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 Pr0.097 (Retrans )1/2
trans
L
(6.118)
Retrans , this result reduces to eqn. (6.116).
If ReL
Whitaker [6.16] suggested setting Pr0.097 ≈ 1 and Retrans ≈ 200, 000
in eqn. (6.118):
NuL = 0.037 Pr 0.43 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+ ⎫1/2
⎪
⎬ 3/5 2/5 ⎪ ⎭ (6.120a) where
0.0468
1+
Pr 2/3 φ ≡ Rex Pr 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 ⎧
⎪
⎨ (φ/12, 500) ⎪
⎩ 1 + (φum /φ)7/2 3/5 1+ ⎫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, 821 so
h= 1821(0.02623)
1821 k
=
= 23.88 W/m2 K
L
2.0 It follows that the average plate temperature is
T w = 290 K + 103 W/m2
= 332 K.
23.88 W/m2 K The ﬁlm temperature is (332 + 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◦ C.
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
=
∂y 2
ν ∂t
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. Heat transfer to and from laminar ﬂows in pipes §7.2 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 − T b =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
˙
mcp
∂x
dx (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 “Poiseuilles 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
2qw α
∂T
dTb
qw P
qw (2π R)
=
=
=
=
2)
˙p
ρcp uav (π R
uav Rk
mc
∂x
dx
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 − T b = 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 ≡ 48
qw D
= 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) + 11 qw D
4qw α
11 qw D
L+
= 20 +
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:
0.046
f
= 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
where n =
(7.44)
NuD = NuD
⎩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
f =f
×K
where K =
(7.46)
⎪
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∗ = τw /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
4(0.25)
11.6
= 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
=
b
uav
2 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 Heat transfer during cross ﬂow over cylinders §7.6 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 0.62(0.71) 1/3 ⎫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 Heat transfer during cross ﬂow over cylinders §7.6 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
∂y
∂x 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
ν2
dx
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∞ |
x
C1
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
Nux dx =
3
x 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
0.7112
L 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 )
5/4 (Tb − T ) 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 − T i
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 k∆T (S)D
4π (D/2)
QD
=
=
=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
(8.46)
NuL = fn ⎝
, Pr, Ja ⎠
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 ∂v
∂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 − ρ g g
d2 u
=−
2
dy
ρf
ν
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 simplication 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
dx
δ (8.51) Substituting eqn. (8.50) in eqn. (8.51), we obtain a ﬁrst-order diﬀerential equation for δ:
k hfg ρf (ρf − ρg )
Tsat − Tw
dδ
=
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 Figure 8.11 §8.5 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 ⎡
⎢ ρ f ρf − ρ g 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,
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[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.
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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 Figure 9.1 §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 Figure 9.3 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. 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 Figure 9.4 9.2 §9.2 Enlarged sketch of a typical metal surface. 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 − ρ g 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 − ρ g
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 − ρ g
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 − ρ g
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 Figure 9.10 §9.3 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 − ρ g 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 − ρ g 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
= 2π 3
Π2 =
≡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 1/2 1/4 1.734/(R ) 0.84 0.94/(R ) 0.90 0.89 + 2.27e 1.14(λd1 /Aheater ) 1.14 qmax /qmaxz 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 Predictions of the peak pool boiling heat ﬂux 10. Any large ﬁnite body 9. 1 side insulated 8. plain Small horizontal ribbon
oriented vertically Inﬁnite ﬂat heater 1. Situation Table 9.3 (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 Transition boiling and system inﬂuences §9.6 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 Forced convection boiling in tubes §9.7 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
D
0.05
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,
T w = 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 for Frlo ≥ 0.04 ⎩(25 Frlo ) 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 G 2 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 τ δ 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 9.21 508 Dropwise condensation. 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 Figure 9.22 §9.10 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
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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 Figure 10.1 §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 Figure 10.4 §10.1 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 Figure 10.7
as well. §10.3 A body (1) that views three other bodies and itself 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 =
0.01309
Aslit 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. ⎫
⎬
Y
X
√
+ Y 1 + X 2 tan−1 √
1 + X2 ⎭
1 + Y2 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 + W 2 )(1 + H 2 )
1
+ ln
+ H tan−1
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 − T 2 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 − T 2 (10.24) −1 Comparing eqn. (10.24) with eqn. (10.2), we may identify
F1–2 = 1
1
1
+
−1
ε2
ε1 (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 A 1 + 1
A1 + 1 − ε2 (10.26) ε2 A 2 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 A 1
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 A 1 + 1
A1 + 1 − ε2
ε2 A 2 +2 1 − εs
εs A s + 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 A l + 1
Al + 1 − εc
ε2 A c +2 1 − εs
εs A s + 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 A i (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
1
1 − ε1
A1 F1–3
ε1 A 1
A1 F1–2 Qnet2 , at node B2 : eb2 − B2
B2 − B1 B2 − B3
=
+
1
1
1 − ε2
A1 F1–2
A2 F2–3
ε2 A 2 Qnet3 , at node B3 : eb3 − B3
B3 − B1 B3 − B2
=
+
1 − ε3
1
1
ε3 A 3
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 A 2 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
B 1 − B3
B1 − B 2
1451 − B1
=
+
2.333
1/0.2
1/0.8
B 2 − B3
B2 − B 1
459.3 − B2
=
+
1/0.8
1
1/0.2
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 + B 3 − 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 Figure 10.17 §10.4 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 H i = 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 A j 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 Figure 10.18 §10.4 Illustration for Example 10.10. unknowns, B1 , B2 , and B3 .
B1 = 1 − ε1
0.85 B2 = 1 − ε 2
0.5 B3 = 1 − ε 3
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 − B 1
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. Gaseous radiation §10.5 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
vapor in air. Functions used to predict εg = f1 f2 for water 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 ﬂux 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 Solar energy §10.6 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
Prob. 10.9. Conﬁguration for 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
Prob. 10.11. Conﬁguration for Figure 10.29
Prob. 10.12. Conﬁguration for 10.14 Figure 10.30
Prob. 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). Conﬁguration for 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
Prob. 10.15. Conﬁguration for Figure 10.32
Prob. 10.16. Conﬁguration for 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 ≡ mass of species i
ρi
=
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 ≡ moles of i
ci
=
c
mole of mixture (11.6) Just as for the mass fraction, it follows for mole fraction that
ci /c = 1 xi =
i 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 M k 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 = N A c i (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:
p i = ρi R i 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
pi
ci
=
=
◦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) cO2 = = 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 v i . (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
n i = ρi v i (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:
j i = ρi v i − 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:
n i = ρi v i = ρi v + ρ i v i − 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 v i = 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 :
∗
J i = ci v i − v ∗ (11.24) An introduction to mass transfer 606 §11.2 It may be shown, using these deﬁnitions, that
∗
N i = xi N + J i (11.25) Substitution of eqn. (11.25) into eqn. (11.23) gives i ∗
Ji = N + xi + Ni = N N= 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 Figure 11.4
oxidation. 607 Low-temperature carbon 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 diusing 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 2d p
π 2N d
√ (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
potential. 615 The Lennard-Jones 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]). 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
C 2 H6
C2 H5 OH
CH3 COCH3
C 6 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
H 2 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 Figure 11.9 §11.5 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 + j i =
mi n + j i = 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 n 2 = m2 n + j 2 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 ρ D21
dm2
=−
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) 1, only diﬀusion contributes signiﬁcantly to the mass ﬂux of i,
If mi
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 Figure 11.14 §11.6 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,
ρcp dT is analogous to ρ dmi (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
= (1 − ρ∞ /ρ)g + ν
+v
∂y
∂y 2
∂x (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 as well. Both conditions for
1, then mHe,s − mHe,e
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 An introduction to mass transfer 658 Figure 11.18 §11.8 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 Figure 11.20 §11.9 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 h i v i ⎠ = 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
+ ρ i 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 Figure 11.21 §11.9 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 − T s 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 − T e s = ni,s cp,i
ni,s cp,i
δt − 1
exp
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 Figure 11.22 §11.9 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
q s = 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 c j
⎝
⎠
−
∇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 which the gas ﬂows. Analyze this process under the following assumptions: The ﬁlm
ﬂow is laminar and of constant thickness, δ0 , with a velocity
proﬁle given by eqn. (8.48); the gas is only slightly soluble in
the liquid, so that it does not penetrate far beyond the liquid surface and so that liquid properties are unaﬀected by it;
and, the gas concentration at the s - and u-surfaces (above and
below the liquid-vapor interface, respectively) does not vary
along the length of the tube. The inlet concentration of gas in
the liquid is m1,0 . Show that the mass transfer is given by
Num,x = u0 x
π D12 1/2 with u0 = (ρf − ρg )gδ2
0
2µf The mass transfer coeﬃcient here is based on the concentration diﬀerence between the u-surface and the bulk liquid at Chapter 11: An introduction to mass transfer 682 m1,0 . (Hint : The small penetration assumption can be used to
reduce the species equation for the ﬁlm to the diﬀusion equation, eqn. 11.72.)
11.41 Benzene vapor ﬂows through a 3 cm I.D. vertical tube. A thin
ﬁlm of initially pure water runs down the inside wall of the tube
at a ﬂow rate of 0.3 liter/s. If the tube is 0.5 m long and 40◦ C,
estimate the rate (in kg/s) at which benzene is absorbed into
water over the entire length of the tube. The mass fraction of
benzene at the u-surface is 0.206. (Hint : Use the result stated
in Problem 11.40. Obtain δ0 from the results in Chapter 8.) 11.42 A mothball consists of a 2.5 cm diameter sphere of naphthalene (C10 H8 ) that is hung by a wire in a closet. The solid naphthalene slowly sublimates to vapor, which drives oﬀ the moths.
The latent heat of sublimation and evaporation rate are low
enough that the wet-bulb temperature is essentially the ambient temperature. Estimate the lifetime of this mothball in
a closet with a mean temperature of 20◦ C. Use the following
data:
σ = 6.18 Å, ε/kB = 561.5 K for C10 H8 , and, for solid naphthalene,
ρC10 H8 = 1145 kg/m3 at 20◦ C