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INTERNATIONAL SERIES IN PURE AND APPLIED PHYSICS
G. P. HARNWELL, CONSULTING EDITOR
ADVISORY EDITORIAL COMMITTEE: E. U. Condon, George R. Harrison
Elmer Hutchisson, K. K, Darrow
HEAT CONDUCTION
With Engineering and Geological
Applications
The quality of the materials used in the manufacture
of this book is governed by contin ued postwar shortages.
INTERNATIONAL SERIES IN
PURE AND APPLIED PHYSICS
G. P. HARNWELL, Consulting Editor
BACKER AND GOUDSMIT ATOMIC ENERGY STATES
BITTER INTRODUCTION TO FERROMAGNETISM
BRILLOUIN WAVE PROPAGATION IN PERIODIC STRUCTURES
CADY PIEZOELECTRICITY
CLARK APPLIED XRAYS
CURTIS ELECTRICAL MEASUREMENTS
DAVEY CRYSTAL STRUCTURE AND ITS APPLICATIONS
EDWARDS ANALYTIC AND VECTOR MECHANICS
HARDY AND PERRIN THE PRINCIPLES OF OPTICS
HARNWELL ELECTRICITY AND ELECTROMAGNETISM
HARNWELL AND LIVINGOOD EXPERIMENTAL ATOMIC PHYSICS
HOUSTON PRINCIPLES OF MATHEMATICAL PHYSICS
HUGHES AND DUBRIDGE PHOTOELECTRIC PHENOMENA
HUND HIGHFREQUENCY MEASUREMENTS
PHENOMENA IN HIGHFREQUENCY SYSTEMS
INGERSOLL, ZOBEL, AND INGERSOLL HEAT CONDUCTION
KEMBLE PRINCIPLES OF QUANTUM MECHANICS
KENNARD KINETIC THEORY OF GASES
ROLLER THE PHYSICS OF ELECTRON TUBES
MORSE VIBRATION AND SOUND
PAULING AND GOUDSMIT THE STRUCTURE OF LINE SPECTRA
RICHTMYER AND KENNARD INTRODUCTION TO MODERN PHYSICS
RUARK AND UREY ATOMS, MOLECULES AND QUANTA
SEITZ THE MODERN THEORY OF SOLIDS
SLATER INTRODUCTION TO CHEMICAL PHYSICS
MICROWAVE TRANSMISSION
SLATER AND FRANK ELECTROMAGNETISM
INTRODUCTION TO THEORETICAL PHYSICS
MECHANICS
SMYTHB STATIC AND DYNAMIC ELECTRICITY
STRATTON ELECTROMAGNETIC THEORY
WHITE INTRODUCTION TO ATOMIC SPECTRA
WILLIAMS MAGNETIC PHENOMENA
Dr. Lee A. DuBridge was consulting editor of the series from 1939
to 1946.
HEAT CONDUCTION
With Engineering and Geological
Applications
By Leonard R. Ingersoll
Professor of Physics
University of Wisconsin
Otto J. Zobel
Member of the Technical Staff
Bt>ll Telephone Laboratories, Inc., New York
and Alfred C. Ingersoll
Instructor in Civil Engineering
University of Wisconsin
FIRST EDITION
NEW YORK TORONTO LONDON
MCGRAWHILL BOOK COMPANY, INC.
1948
HEAT CONDUCTION
Copyright, 1948, by the McGrawHill Book Company, Inc. Printed in the
United States of America. All rights reserved. This book, or parts thereof,
may not be reproduced in any form without permission of the publishers.
THE MAPLE PRESS COMPANY, YORK, PA.
PREFACE
The present volume is the successor to and, in effect, a
revision of the Ingersoll and Zobel text of some years ago. To
quote from the earlier preface: ". . . the theory of heat con
duction is of importance, not only intrinsically 'but also because
its broad bearing and the generality of its methods of analysis
make it one of the best introductions to more advanced mathe
matical physics.
"The aim of the authors has been twofold. They have
attempted, in the first place, to develop the subject with special
reference to the needs of the student who has neither time nor
mathematical preparation to pursue the study at great length.
To this end, fewer types of problems are handled than in the
larger treatises, and less stress has been placed on purely mathe
matical derivations such as uniqueness, existence, and con
vergence theorems.
"The second aim has been to point out . . . the many
applications of which the results are susceptible .... It is
hoped that in this respect the subject matter may be of interest
to the engineer, for the authors have attempted to select appli
cations with special reference to their technical importance, and
in furtherance of this idea have sought and received suggestions
from engineers in many lines of work. While many of these
applications have doubtless only a small practical bearing and
serve chiefly to illustrate the theory, . . . the results in some
cases . . . may be found worthy of note. The same may be
said of the geological problems.
"While a number of solutions are here presented for the first
time ... no originality can be claimed for the underlying
mathematical theory which dates back, of course, to the time of
Fourier."
Since the above was written there has been a steady increase
vi PREFACE
in interest in the theory of heat conduction, largely along prac
tical lines. The geologist and geographer are interested in a
new tool which will help them in explaining many thermal
phenomena and in establishing certain time scales. The engi
neer, whose use of the theory was formerly limited almost
entirely to the steady state, has developed many useful tables
and curves for the solution of more general cases and is interested
in finding still other methods of attack. The physicist and
mathematician have done their part in treating problems which
have hitherto resisted solution.
The present volume carries out and extends the aims of
the earlier one. Most of the old material has been retained,
although revised, and almost an equal amount of new has been
added. The geologist, geographer, and engineer will find many
new applications discussed, while the mathematician, physicist,
and chemist will welcome the addition of a little Bessel
function and conjugate function theory, as well as the several
extended tables in the appendixes. Some of these are new
and have had to be specially evaluated. The number of refer
ences has also been greatly enlarged and threequarters of them
are of more recent date than the older volume. A special
feature is the extended treatment, particularly as regards
applications, of the theory of permanent sources. This is
carried out for all three dimensions, but most of the applications
center about the twodimensional case, the most interesting of
these being the theory of groundpipe heat sources for the heat
pump. Other features of the revision are a modernized nomen
clature, many new problems and illustrations, and the segre
gation of descriptions of methods of measuring heatconduction
constants in a special chapter.
A feature of particular importance to those whose interests
are largely on the practical side is the discussion in Chapter 11
of auxiliary graphical and other approximation methods by
which many practical heat conduction problems may be solved
with only the simplest mathematics. It is believed that many
will appreciate this and in particular the discussion of pro
cedures by which it is possible to handle simply, and with
sufficient accuracy for practical purposes, many problems whose
PREFACE vil
solution would be almost impossible by classical methods. As
regards the book as a whole, the only mathematical prerequisite
necessary for reading it is a reasonable knowledge of calculus.
Despite occasional appearances to the contrary, the mathe
matical theory is not difficult and falls into a pattern which is
readily mastered. The authors have tried, in general, to reduce
mathematical difficulties to a minimum, and in some cases have
deliberately chosen the simpler of two alternate methods of
solving a problem, even at a small sacrifice of accuracy.
The authors acknowledge again their indebtedness to the
several standard treatises referred to in the preface to the earlier
edition, and in particular to Carslaw's " Mathematical Theory
of the Conduction of Heat in Solids " ; also Carslaw and Jaeger's
" Conduction of Heat in Solids." It is hard to single out for
special credit any of the hundredodd other books and papers
to which they are indebted and which are listed at the end of
this volume, but perhaps particular reference should be made
to Me Adams' "Heat Transmission" and to papers by Emmons,
Newman, and Olson and Schultz.
The authors are glad to acknowledge assistance from many
friends. These include: 0. A. Hougen, D. W. Nelson, F. E.
Volk, and M. 0. Withey of the College of Engineering, Uni
versity of Wisconsin; J. D. MacLean of the Forest Products
Laboratory; J. H. Van Vleck of Harvard University, W. J. Mead
of Massachusetts Institute of Technology, and A. C. Lane of
Cambridge; C. E. Van Orstrand, formerly of the U.S. Geo
logical Survey; H. W. Nelson of Oak Ridge, Tennessee; C. C.
Furnas of the CurtissWright Corp., B. Kelley of the Bell
Aircraft Corp., and G. H. Zenner and L. D. Potts of the Linde
Air Products Laboratory, in Buffalo; A. C. Crandall of the
Indianapolis Light and Power Co. ; M. S. Oldacre of the Utilities
Research Commission in Chicago ; and a large number of others
who have given help and suggestions. The authors are par
ticularly indebted to F. T. Adler of the Department of Physics
of the University of Wisconsin and to H. W. March of the
Department of Mathematics for much assistance; also to K. J.
Arnold of the same department and to Mrs. M. H. Glissendorf
and Miss R. C. Bernstein of the university computing service
viii PREFACE
for the recalculation and correction of many of the tables; to
Miss Frances L. Christison and Mrs. Doris A. Bennett, librar
ians; to H. J. Plass and other graduate students for helping in
the elimination of errors; and to Mrs. L. R. Ingersoll and Mrs.
A. C. Ingersoll for assistance in many ways,
THE AUTHORS
January, 1948
CONTENTS
PREFACE
Chapter 1
INTRODUCTION
Symbols. Historical. Definitions. Fields of Application. Units;
Dimensions. Conversion Factors. Thermal Constants.
Chapter 2
THE FOURIER CONDUCTION EQUATION . 11
Differential Equations. Derivation of the Fourier Equation. Bound
ary Conditions. Uniqueness Theorem.
Chapter 3
STEADY STATE ONE DIMENSION 18
Steady State Defined. Onedimensional Flow of Heat. Thermal Re
sistance. Edges and Corners. Steady Flow in a Long Thin Rod.
APPLICATIONS: Furnace Walls; Refrigerator and Furnace Insulation;
Airplanecabin Insulation; Contact Resistance. Problems.
Chapter 4
STEADY STATE MORE THAN ONE DIMENSION 30
Flow of Heat in a Plane. Conjugate Function Treatment. Radial
Flow in Sphere and Cylinder. Simple Derivation of Sphere and Cylinder
Heatflow Equations. APPLICATIONS: Covered Steam Pipes; Solid and
Hollow Cones; Subterranean Temperature Sinks and Power Develop
ment; Geysers; Gasturbine Cooling. Problems.
Chapter 5
PERIODIC FLOW OF HEAT IN ONE DIMENSION 45
Generality of Application. Solution of Problem. Amplitude, Range,
Lag, Velocity, Wavelength. Temperature Curve in the Medium. Flow
x CONTENTS
of Heat per Cycle through the Surface. APPLICATIONS: Diurnal Wave;
Annual Wave; Cold Waves; Temperature Waves in Concrete; Periodic
Flow and Climate; "Ice Mines"; Periodic Flow in Cylinder Walls;
Thermal Stresses. Problems.
Chapter 6
FOURIER SERIES 58
General conditions. Development in Sine Series and Cosine Series.
Complete Fourier Series. Change of Limits. Fourier's Integral. Har
monic Analyzers. Problems.
Chapter 7
LINEAR FLOW OF HEAT, 1 78
Case /. Infinite Solid. Solution with Initial Temperature Distribution
Given. Discontinuities. APPLICATIONS: Concrete Wall; Thermit Weld
ing. Problems.
Case II. Semiinfinite Stilid. Solution for Boundary at Zero Tempera
ture. Surface and Initial Temperature of Body Constant. Law of
Times. Rate of Flow of Heat. Temperature of Surface of Contact.
APPLICATIONS: Concrete; Soil; Thawing of Frozen Soil; Removal of
Shrink Fittings; Hardening of Steel; Cooling of Lava under Water;
Cooling of the Earth, with and without Radioactive Considerations and
Estimates of Its Age. Problems.
Chapter 8
LINEAR FLOW OF HEAT, II 109
Case III. Heat Sources. Solution for Instantaneous and Permanent
Plane Sources. Use of Doublets; Solution for Semiinfinite Solid with
Temperature of Face a Function of Time. APPLICATIONS: Heat Sources
for Heat Pumps; Electric Welding; Casting; Temperatures in Decom
posing Granite; Ground Temperature Fluctuations and Cold Waves;
Postglacial Time Calculations. Problems.
Case IV. The Slab. Both Faces at Zero. Simplification for Surface
and Initial Temperature of Body Constant. Adiabatic Case. APPLI
CATIONS: Fireproof Wall Theory; Heat Penetration in Walls of Various
Materials; Experimental Considerations; Molten Metal Container;
Optical Mirrors; Vulcanizing; Fireproof Containers; Annealing Castings.
Problems.
CONTENTS ri
Case V. Radiating Rod. Initial Temperature Distribution Given. One
End at Zero. Initial Temperature of Rod Zero. Problems.
Chapter 9
FLOW OF HEAT IN MORE THAN ONE DIMENSION 139
Case I. Radial Flow. APPLICATIONS: Cooling of Laccolith. Problems.
Case II. Heat Sources and Sinks. Point Source. Line Source. Point
Source in a Plane Sheet. Source and Sink Equations. APPLICATIONS:
Subterranean Sources and Sinks; Geysers; Groundpipe Heat Sources and
Spherical and Plane Sources for the Heat Pump; Electric Welding;
Electrical Contacts; Cooling of Concrete Dams. Problems.
Case III. Sphere with Surface at Constant Temperature. Calculation of
Center and Average Temperature. APPLICATIONS: Mercury Ther
mometer; Spherical Safes of Steel and Concrete; Hardening of Steel
Shot; Household Applications. Problems.
Case IV. Cooling of a Sphere by Radiation. Transcendental Equation.
General Sine Series Development. Final Solution. Special Cases.
APPLICATIONS: Terrestrial Temperatures; Mercury Thermometer.
Problems.
Case V. Infinite Circular Cylinder. Bessel Functions. Surface at Zero.
Simplification for Constant Initial Temperature. APPLICATIONS: Heating
of Timbers; Concrete Columns. Problems.
Case VI. General Case of Heat Flow in an Infinite Medium. Special
Formulas for Various Solids. APPLICATIONS: Canning Process; Brick
Temperatures; Drying of Porous Solids. Problems.
Chapter 10
FORMATION OF ICE 190
Neumann's Solution. Stefan's Solution. Thickness of Ice Propor
tional to Time. Solution for Thin Ice. Formation of Ice in Warm
Climates. APPLICATIONS: Frozen Soil. Problems.
Chapter 11
AUXILIARY METHODS OF TREATING HEATCONDUCTION
PROBLEMS 200
/. Method of Isothermal Surfaces and Flow Lines. Solutions for Square
Edge, Nonsymmetrical Cylindrical Flow, Wall with Internal Ribs,
and Cylindricaltank Edge Loss.
xh CONTENTS
II. Electrical Methods. Eccentric Spherical and Cylindrical Flow.
///. Solutions from Tables and Curves.
IV. The Schmidt Method. Application to Cooling of Semiinfinite Solid
and Plate.
V. The Relaxation Method. Edge Losses in a Furnace.
VI. The Step Method. Ice Formation about Pipes; Ice Cofferdam;
Warming of Soil; Cooling of Armor Plate; Heating of Sphere; Other
Applications.
Chapter 12
METHODS OF MEASURING THERMALCONDUCTIVITY CON
STANTS 234
General Discussion and References. Linear Flow, Poor Conductors.
Linear Flow, Metals. Radial Flow. Diffusivity Measurements.
Liquids and Gases.
APPENDIX A. Table A.I. Values of the Thermal Conductivity Con
stants 241
Table A.2. Values of the Heat Transfer Coefficient h . . 246
APPENDIX B. Indefinite Integrals 247
APPENDIX C. Definite Integrals . . 248
APPENDIX D. Values of the Probability Integrals (x) m ^= f x e~Pdfi . 249
APPENDIX E. Values of e* . . 252
APPENDIX F. Values of I(x) m f p^* dp . 253
APPENDIX G. Values of S(x) = * ( e~***   e**'* + \ <r 28ir2 *  . . ) 255
APPENDIX H. Values of B(x)  2(  e~ 4a; + e  ) .... 257
and B.(x) m ^ (e f \ e~** f g  +  )
APPENDIX I. Table I.I. Bessel functions J Q (x) and J i(x) 258
Table 1.2. Roots of J n (x) =0 259
APPENDIX J. Values of C(x) m 2 [~^\ + g^\ + z^lz] + ' ' ' 1 26
APPENDIX K. Miscellaneous Formulas 261
APPENDIX L. The Use of Conjugate Functions for Isotherms and Flow
Lines 262
APPENDIX M. References 264
INDEX 271
CHAPTER 1
INTRODUCTION
1.1. Symbols. The following table lists the principal sym
bols and abbreviations used in this book. They have been
chosen in agreement, so far as practicable, with the recommenda
tions of the American Standards Association and with general
scientific practice.
TABLE 1.1. NOMENCLATURE
A Area, cm 2 or ft 2 .
a Thermal diffusivity, cgs or fph (Sees. 1.3, 1.5, Appendix A).
B(x) 2(e~*  e** + e~>*  ) (Sec. 9.17, Appendix H).
B a (x) ^ (e~* + \ e~** +  e~** + ) (Sec. 9.18, Appendix H).
3, 7 Variables of integration; also constants.
X Variable of integration; also a constant; also wave length.
Btu British thermal unit, 1 Ib water 1F (Sec. 1.5).
c Specific heat (constant pressure), cal/(gm)(C), or Btu/(lb)(F); also
a constant.
cal Calorie, 1 gm water 1C (Sec. 1.5).
cgs Centimetergramsecond system; used here only with centigrade tem
perature scale and calorie as unit of heat.
(Sec  9  38 ' Appendix J) 
exp x e*.
fph Footpoundhour system, used here only with Fahrenheit temperature
scale and Btu as heat unit.
h Coefficient of heat transfer between a surface and its surroundings,
cal/ (sec) (cm 2 ) (C) or Btu/(hr)(ft 2 )(F); sometimes called "emis
sivity" or "exterior conductivity" (Sec. 2.5, Appendix A).
1
11 2V5"
I(x) f x ** dft (Sec. 9.8, Appendix F).
Jn(x) Bessel function (Sec. 9.36).
k Thermal conductivity, cgs or fph (Sees. 1.3, 1.5, Appendix A).
In x log* x.
I
2 HEAT CONDUCTION [CHAP. 1
TABLE 1.1. NOMENCLATURE (Continued)
$(x) Probability integral, p I* e~P d& (Appendix D).
Q Quantity of heat, cal or Btu (sometimes taken per unit length or unit
area; see Q').
q Rate of heat flow, cal/sec or Btu/hr (sometimes also used for rate of
heat production).
Q' Rate of heat production or withdrawal in permanent sources or sinks,
cal/sec or Btu/hr for threedimensional case; cal/sec per cm length
or Btu/hr per ft length for twodimensional case; cal/(sec)(cm 2 ) or
Btu/(hr)(ft 2 ) for onedimensional case (Sees. 8.2, 9.9).
p Density, gm/cm 3 , or lb/ft 3 .
R .Thermal resistance TT (Sec. 3.3).
KA.
S Strength of instantaneous source, (Sees. 8.2, 9.9).
Q'
S' Strength of permanent source, (Sees. 8.2, 9.9).
S(x)  (e***  I e 9 *** + 4 e 25 * 2 *  ) (Sec. 8.16, Appendix G).
W V u O /
t Time, seconds or hours.
T* Temperature, C or F.
if Rate of flow of heat per unit area, ^; cal/(sec)(cm 2 ) or Btu/ (hr) (ft 2 )
(Sec. 1.3)
1.2. Historical. The mathematical theory of heat conduc
tion in solids, the subject of principal concern in this book, is
due principally to Jean Baptiste Joseph Fourier (17681830)
and was set forth by him in his "Th6orie analytique de la
chaleur." 42 f While Lambert, Biot, and others had developed
some more or less correct ideas on the subject, it was Fourier
who first brought order out of the confusion in which the experi
mental physicists had left the subject. While Fourier treated
a large number of cases, his work was extended and applied
to more complicated problems by his contemporaries Laplace
and Poisson, and later by a number of others, including Lam6,
Sir W. Thomson 146  147 (Lord Kelvin), and Riemann. 160 To the
* The use of $ for temperature, as in the former edition of this book, has been
discontinued here, partly because many modem writers attach the significance of
time to it and partly because of the increasing adoption of T. It is suggested that,
to avoid confusion, this be always pronounced "captee."
t Superscript figures throughout the text denote references in Appendix M.
SBC. 1.3] INTRODUCTION 3
last mentioned writer all students of the subject should feel
indebted for the very readable form in which he has put much
of Fourier's work. The most authoritative recent work on the
subject is that of Carslaw and Jaeger. 27a
1.3. Definitions. When different parts of a solid body are at
different temperatures, heat flows from the hotter to the colder
portions by a process of electronic and atomic energy transfer
known as " conduction," The rate at which heat will be trans
ferred has been found by experiment to depend on a number of
conditions that we shall now consider.
To help visualize these ideas imagine in a body two parallel
planes or laminae of area A and distance x apart, over each of
which the temperature is constant, being T\ in one case and T\
in the other. Heat will then flow from the hotter of these iso
thermal surfaces to the colder, and the quantity Q that will be
conducted in time t will be given by
m _ m
Q  k ~ At (a)
V
'f*^' 1 "
where k is a constant for any given material known as the
thermal conductivity of the substance. It is then numerically
equal to the quantity of heat that flows in unit time through
unit area of a plate of unit thickness having unit temperature
difference between its faces.
The limiting value of (T% T\)/x or dT/dx is known as
the temperature gradient at any point. If due attention is paid
to sign, we see that if dT/dx is taken in the direction of heat
flow it is intrinsically negative. Hence, if we wish to have a
positive value for the rate at which heat is transferred across an
isothermal surface in a positive direction, we write
dT
or w fc gj (d)
where w (== q/A) is called the "flux" of heat across the surface
4 HEAT CONDUCTION [CHAP. 1
at that point. If instead of an isothermal surface we consider
another, making an angle <t> with it, we can see that both the
flux across the surface and the temperature gradient across the
normal to such surface will be diminished, the factor being cos <,
so that we may write in general for the flux across any surface
, dT
where the derivative is taken along the outward drawn normal,
i.e., in the direction of decreasing temperature. This shows
that the direction of (maximum) heat flow is normal to the
isotherms.
While the rate at which heat is transferred in a body, e.g.,
along a thermally insulated rod, is dependent only on the con
ductivity and other factors noted, the rise in temperature that
this heat will produce will vary with the specific heat c and the
density p of the body. We must then introduce another con
stant a whose significance will be considered later, determined
by the relation
The constant a has been termed by Kelvin the thermal diffusivity
of the substance, and by Maxwell its thermometric conductivity.
Equations (a) and (e) express what is sometimes referred to
as the fundamental hypothesis of heat conduction. Its justi
fication or proof rests on the agreement of calculations made on
this hypothesis, with the results of experiment, not only for the
very simple but for the more complicated cases as well.
1.4. Fields of Application. From (1.3a) we may infer in
what field the results of our study will find application. We
may conclude first that our derivations will hold good for any
body in which heat transfer takes place according to this law,
if k is the same for all parts and all directions in the body.
This includes all homogeneous isotropic solids and also liquids
and gases in cases where convection and radiation are negligible.
The equation also shows that, since only differences of tempera
ture are involved, the actual temperature of the system is
SEC. 1.4] INTRODUCTION 5
immaterial. We shall have cause to remember this statement
frequently; for, while many cases are derived on the supposition
that the temperature at the boundary is zero, the results are
made applicable to cases in which this is any other constant
temperature by a simple shift of the temperature scale.
But the results of the study of heat conduction are not
limited in their application to heat alone, for parts of the theory
find application in certain gravitational problems, in static and
current electricity, and in elasticity, while the methods devel
oped are of very general application in mathematical physics.
As an example of such relationship to other fields it may be
pointed out that, if T in (1.3a) is interpreted as electric potential
and k as electric conductivity, we have the law of the flow of elec
tricity and all our derivations may be interpreted accordingly.
Another field of application is in drying of porous solids,
e.g., wood. It is found that for certain stages of drying the
moisture flow is fairly well represented* by the heatconduction
equation. In this case Q represents the amount of water (or
other liquid) transferred by diffusion, T is the moisture content
in unit volume of the (dry) solid, k is the rate of moisture flow
per unit area for unit concentration gradient. The quantity
cp, which normally represents the amount of heat required to
raise the temperature of unit volume of the substance by one
degree, is here the amount of water required to raise the moisture
content of unit volume by unit amount. This is obviously unity,
so k and a are the same in this case; k is here called the " diffusion
constant." The passage of liquid through a porous solid, as in
drying, is a more complicated process than heat flow, and the
application of conduction theory has definite limitations, as
pointed out by Hougen, McCauley, and Marshall. 68 It may
be added that in all probability the diffusion of gas in a metal is
subject to the same general theory as water diffusion in porous
materials.
Lastly, we may mention the work of Biot 15 on settlement and
consolidation of soils. This indicates that the conduction
* Bateman, Hohf and Stamm, 8 Ceaglske and Hougen, 29 Gilliland and Sher
wood, 45 Lewis, 86 McCready and McCabe," Newman, 101 Sherwood, 127  128 and
Tuttle. 180
6 HEAT CONDUCTION [CHAP. 1
equation may play an important part in the theory of these
phenomena.
1.5. Units; Dimensions. Two consistent systems of con
ductivity units are in common use, having as units of length,
mass, time, and temperature, respectively, the centimeter,
gram, second, and centigrade degree, on the one hand and the foot,
pound, hour, and Fahrenheit degree on the other. The former
unit will be referred to as cgs and the latter as fph as regards
system. This gives as the unit of heat in the first case the
(small) calorie, or heat required to raise the temperature of 1 gm
of water 1C, frequently specified at 15C; and in the second
the Btu, or heat required to raise 1 Ib of water 1F, sometimes
specified at 39.1F* and sometimes at GOF. The cgs thermal
conductivity unit is the calorie per second, per square centimeter
of area, for a temperature gradient of 1C per centimeter, which
shortens to cal/(sec)(cm)(C), while the fph conductivity unit
is the Btu/(hr)(ft)(F). Similarly, the units of diffusivity come
out cm 2 /sec and ft 2 /hr. The unit in frequent use in some
branches of engineering having areas in square feet but tempera
ture gradients expressed in degrees per inch will not be used here
because of difficulties attendant on the use of two different units
of length.
In converting thermal constants from one system to another
and in solving many problems Table 1.2 will be found useful.
Conversion factors other than those listed above may be
readily derived from a consideration of the dimensions of the
units. From (1.3a)
lr  Q (n\
K ~ T l  2' 2 At (a)
Since putting the matter as simply as possible the unit of
heat is that necessary to raise unit mass of water one degree,
its dimensions are mass and temperature; thus, the dimensions
of Q/(Ti T 2 ) are simply M. Hence, K the unit of conduc
tivity is the unit of mass M divided by the units of length L
* The matter of whether heat units are specified for the temperature of maxi
mum density of water or for a slightly higher temperature may result in dis
crepancies of the order of half a percent, but this is of little practical importance
since this is below the usual limit of error in thermal conductivity work.
SEC. 1.5] INTRODUCTION 7
TABLE 1.2. CONVERSION FACTORS AND OTHER CONSTANTS
1 m 39.370 in.  3.2808 ft 1.0936 yd
1 in.  2.540 cm
1 f t 30.48 cm
1 m 2  10.764 ft 2 1.196 yd 2
1 hi. 2  6.452 cm 2
1 ft 2 = 929.0 cm 2
1 m 3 = 61,023 in. 3  35.314 ft 3  1.308 yd 3
1 in, 3  16.387 cm 3
1 ft 3 = 28,317 cm 8
1 kg 2.2046 Ib
1 Ib = 453.6 gm
1 gm/cm 3 62.4 ib/ft 3
1 Btu  252 cal  1055 joules * 777.5 ftlb
1 watt 0.2389 cal/sec
1 kw 56.88 Btu/min 3413 Btu/hr
1 cai = 4.185 joules
1 cal/cm 2 = 3.687 Btu/ft*
1 cal/sec = 14.29 Btu/hr
1 watt/ft 2 = 3.413 Btu/(ft 2 )(hr)
1 cai/(cm 2 )(sec) = 318,500 Btu/ (ft 2 ) (day)
1 Btu/hr * 0.293 watts = 0.000393 hp
1 yr = 3.156 X 10 7 sec = 8,766 hr
k in fph = 241.9 k in cgs
k in cgs 0.00413 k in fph
a in fph = 3.875 a in cgs
a in cgs * 0.2581 a hi fph
Temp C *= %(tempF  32)
e = 2.7183 = 1/0.36788
*  3.1416 1/0.31831
T 2 = 9.8696  1/0.10132
VZ  1.7725  1/0.56419
g (45 lat) = 980.6 cm/sec 2  32.17 ft/sec 2
and time 6. If we have another system in which the units are
M', I/, and 0', the number k' that represents the conductivity
in this system is related to the number k that represents the
conductivity in the first system, through the equation
*Z0~*'z7F (&)
,, M L'6 f ( .
Or K AC TTJ/ "y / (C)
M. Lt v
8 HEAT CONDUCTION [CHAP. 1
Similarly, it is easily shown that for diffusivity
L * Q '
a =<* m (d)
1.6. Values of the Constants. In Appendix A is given a
table of the conductivity coefficients, or " constants/' as they
are called even if they show considerable variation with tem
perature and other factors for a considerable number of sub
stances, in both cgs and fph units. Thermal conductivities of
different solids at ordinary temperatures range in value some
20,000 fold. Of ordinary materials silver (k = 0.999 cgs or 242
Fph) is the best conductor,* with copper only slightly inferior
and iron hardly more than onetenth as good. Turning to the
poor conductors or insulators, we have materials ranging from
certain rocks with conductivities around 0.005 cgs vs. 1.2 fph,
down to silica aerogel, whose conductivity of 0.00005 cgs vs.
0.012 fph is actually a little less than that measured for still air.
A considerable number of building insulators have values in the
neighborhood of 0.0001 cgs vs. 0.024 fph. Loosely packed
cotton and wool are also in this category. Because of density
and specificheat considerations the diffusivities follow the order
of conductivities only in a general way, in some cases being
strikingly out of line. The range is smaller, running from 1.7 cgs
vs. 6.6 fph for silver, down to about 0.0008 cgs vs. 0.003 fph
For soft rubber.
Of the factors affecting conductivity one of the most impor
tant for porous, easily compressible materials such as cotton,
wool, and many building insulators is the degree of compression
or bulk density. The ideal of such insulators is to break down
the air spaces to a point where convection is negligible, in other
words to approach the conductivity of air itself as closely as
possible and with a minimum of heat transmitted by radiation.
Many building insulators come within a factor of two or three
of this, for suitable bulk densities, and silica aerogel is actually
below air as a conductor as already indicated. The question
of density is one of the reasons why wool is, in practice, a better
* The remarkable substance liquid helium II has an apparent conductivity
many thousands of times greater than silver; see Powell. "*
SEC. 1.6] INTRODUCTION 9
insulating material than cotton for clothing, bedding, etc. The
difference between the two when new is small, but in use cotton
tends to compact while wool keeps its porosity even in the
presence of moisture.
Most metals show a small and nearly linear decrease of con
ductivity with increase of temperature, of the order of a few
per cent per 100C, but a few (e.g., aluminum and brass) show
the reverse effect as do also many alloys. The conductivity of
nonmetallic substances increases in general with temperature
(there are, however, many exceptions such as most rocks). 16
The diffusivity for such substances, however, usually shows a
smaller change, as the specific heat in most cases also increases
with temperature while the density change is small. When
possible, the change of thermal constants with temperature
should be taken into account in calculations, and this may be
done approximately by using the conductivity and diffusivity
for the average temperature involved. When k is linear with
temperature, as is often the case, its arithmetic mean value for
the two extreme temperatures can usually be used. If k is not
linear, we can use a mean value k m defined by
 TO = ' k dT (a)
In the more complicated cases of heat flow involving other than
the steady state, it may be difficult to take into account tem
perature changes of thermal constants in a satisfactory manner.*
The modern theory of heat conduction in solids f involves
the transmission of thermal agitation energy from hot to cold
regions by means of the motion of free electrons and also through
vibrations of the crystal lattice structure at whose lattice points
the atoms (or ions) are located. The first part, or electronic
contribution, is the most important for metals, and the second
part for nonmetallic solids.
Because of the predominantly electronic nature of metallic
conduction it might be expected that there would be a relation
between the thermal and electrical conductivities of metals,
and this fact is expressed in the law of Wiedemann and Franz
* See Sec. 11.20 for the solution of a special problem involving such changes.
t See, e.g., Austin, 2 HumeRothery, 69 and Seitz. 1 * 6
10 HEAT CONDUCTION [CHAP. 1
that states that one is proportional to the other. While this
holds in a general way where different metals are under consid
eration, it does not express the facts when a single metal at
several different temperatures is concerned; for the electrical
conductivity decreases with rise of temperature, while the
thermal conductivity is more nearly constant. Lorenz 86 took
account of this fact and expressed it in the law that the ratio of
thermal divided by electrical conductivity increases for any
given metal proportionately to the absolute temperature. It
holds only for pure metals with any degree of approximation
and only for very moderate temperature ranges. Griffiths, 60
however, finds that this law holds also for certain aluminum
and bronze alloys.
CHAPTER 2
THE FOURIER CONDUCTION EQUATION
2.1. Differential Equations. In any mathematical study of
heat conduction use must continually be made of differential
equations, both ordinary and partial. These occur, however,
only in a few special forms whose solutions can be explained as
they appear, so only a brief general discussion of the subject is
necessary here.
Differential equations are those involving differentials or dif
ferential coefficients and are classified as ordinary or partial,
according as the differential coefficients have reference to one,
or to more than one, independent variable. A solution of
such an equation is a function of the independent variables
that satisfies the equation for all values of these variables. For
example,
y = sin x + c (a)
is a solution of the simple differential equation
dy = cos x dx (6)
The general solution, as its name implies, is the most general
function of this sort that satisfies the differential equation and
will always contain arbitrary, i.e., undetermined, constants or
functions. A particular solution may be obtained by substi
tuting particular values of the constants or functions in the
general solution. But while this is theoretically the method of
obtaining the particular solution, we shall find in practice that
in many cases where it would be almost impossible to obtain
the general solution of the differential equation, we are still able
to arrive at the desired result by combining particular solutions
that can be obtained directly by various simple expedients.
2.2. A differential equation is linear when it is of the first
degree with respect to the dependent variable and its deriva
11
12
HEAT CONDUCTION
[CHAP. 2
tives. It is also homogeneous if, in addition, there is no term
that does not involve this variable or one of its derivatives.
Practically all the differential equations we shall have occasion
to use are both linear and homogeneous, as are indeed a large
share of those occurring in all work in mathematical physics.
As examples we may mention the following partial differential
equations that are both linear and homogeneous :
Laplace's equation, of constant use in the theory of potential,
~w W "a?
also the equation of the vibrating cord,
& 2  dx*
and the Fourier conduction equation,
dT _
dt ~ a
*T\
J
2.3. The Fourier Equation. We shall now derive this last
equation. Choose three mutually rectangular axes of reference
OX, 07, and OZ (Fig. 2.1) in any
isotropic body and consider a small
rectangular parallelepiped of edges Ax,
Ay, and Az parallel, respectively, to
these three axes. Let T denote the
temperature at the center of this ele
ment of volume; then, since the tem
perature will in general be variable
throughout the body, we may express
its value on any face of the parallele
piped this being so small that the
temperature is effectively uniform over
 , . .
an y OM B f ace as being greater or less
than this mean temperature T by a
small amount. The magnitude of this small amount for the case
of the Aj/Az faces we may readily show to be
1 dT A / N
2 aS A * (a)
FIG. 2.1. Elementary
parallelepiped in medium
through which heat is flowing.
SEC. 2.3] THE FOURIER CONDUCTION EQUATION 13
since the temperature gradient dT/dx measures the change of
temperature per unit length along OX, and the distance of AyA2
from the center is evidently J^Az. Then the temperature of
the left and righthand faces may be written
Using (1.3c), q = kAdT/dx, we see that the flow of heat
per second in the positive x direction through the lefthand face
At/As is
(c)
and through the righthand face in the same direction
(d)
the negative sign being used, since a positive flow of heat evi
dently requires a negative temperature gradient. The differ
ence between these two quantities is evidently the gain in heat
of the element due to the x component of flow alone ; then, since
similar expressions hold for the other two pairs of faces, the
sum of the differences of these three pairs of expressions, or
k fr^ AzAyAz + k ^ AzAi/As + k ^ AzAyAs (e)
represents the difference between the total inflow and total out
flow of heat, or the amount by which the heat of the element is
being increased per second. If the specific heat of the material
of the body is c and its density p, this sum must equal
A A A dT ,
cpAxAyAz ^ (/)
Hence, we may write
or, since a m k/cp,
ar
dt
14 HEAT CONDUCTION [CHAP, 2
which is usually written
_ XV
at ~ a
This is known as Fourier's equation. It expresses the con
ditions that govern the flow of heat in a body, and the solu
tion of any particular problem in heat conduction must first
of all satisfy this equation, either as it stands or in a modified
form.
In the general case, where the thermal conductivity varies
from point to point, the corresponding equation isf
dT 1 d dT\ d /, dT\ d d
Its solution would be more difficult than that of the previous
one.
2.4. If a linear and homogeneous equation such as the Fourier
equation is written so that all the terms are on the left side, the
righthand member being consequently reduced to zero, a very
useful proposition can be deduced at once as follows : Any value
of the dependent variable that satisfies the equation must reduce
the lefthand member to zero. Thus, if such particular solution
is multiplied by a constant, it will still reduce this member to
zero, as this is merely equivalent to multiplying each term by
the constant. In the same way it can be seen that the sum of
any number of particular solutions will still be a solution. We
may then state as a general proposition that, in the case of the
linear, homogeneous differential equation (ordinary or partial),
any combination formed by adding particular solutions, with or
without multiplication by arbitrary constants, is still a solution.
We shall have frequent occasion to make application of this law.
2.6. Boundary Conditions. The solution of practically all
heatconduction problems involves the determination of the tem
perature I 7 as a function of the time and space coordinates.
Such value of T is assumed to be a finite and continuous function
of x,y,z and t and must satisfy not only the general differential
equation, which in one modification or another is common to all
* V is frequently called "nabla."
t See Bateman, 9 * lto Carslaw and
SBC. 2.5] THE FOURIER CONDUCTION EQUATION 15
heatconduction problems, but also certain equations of condi
tion that are characteristic of each particular problem. Such
are
Initial Conditions. These express the temperature through
out the body at the instant that is chosen as the origin of the
time coordinate, as a function of the space coordinates, i.e.,
T = f(x,y,z) when t = (a)
Boundary or Surface Conditions. These are of several sorts
according as they express
1. The temperature on the boundary surface as a function
of time, position, or both, i.e.,
T = t(x,y,z,t) (b)
2. That at the surface of separation of two media there is
continuity of flow of heat, expressed by the relation

l dn ~ 2 dn . c
3. That the boundary surface is impervious to heat, expressed
4. That radiation and convection losses take place at the
surface, in which case we have, for surroundings at zero,
In (e) h is the coefficient of heat transfer between the surface
and surroundings (sometimes referred to as the emissivity or
*See (1.3e).
t This assumes Newton's law of cooling, which states that the rate of loss of
heat is proportional to the temperature above the surroundings, for small tem
perature differences. That this is not inconsistent with Stefan's law of radiation
is shown by the following simple reasoning: Stefan's law states that radiation
q r C(K* K Q), where K and K o are the absolute temperatures of the radiating
body and of the surrounding walls, respectively. For small values of K KQ we
have K*  K 4 Q  A(#<) PI 4KI&K, or q r  4CK* Q &K, which agrees with (e) if we
remember that Alf is here equivalent to T.
16 HEAT CONDUCTION [CHAP. 2
as the exterior or surface conductivity*), i.e., the rate of loss of
heat by radiation and convection per unit area of surface per
degree above the temperature of the surroundings, h is a con
stant only for relatively small temperature differences.
There are also other possible boundary conditions, which we
shall have frequent occasion to use and shall treat more at
length when they occur. Following a common practice, we
shall hereafter refer to both initial and surface conditions as
simply " boundary conditions. "
2.6. Uniqueness Theorem. Our task in general, then, in
solving any given heatconduction problem is to attempt, by
building up a combination of particular solutions of the general
conduction equation, to secure one that will satisfy the given
boundary conditions. It is easy to see that such a result is one
solution of our problem and it may be shown that it is also the
only solution. The reader is referred to the larger treatises
(e.g., Carslaw 27 ) for a rigorous proof of this uniqueness theorem,
but the following simple physical discussion is satisfactory for
our purposes :
Consider a solid body with the Fourier equation (2.3i) hold
ing everywhere inside, with the initial condition
for t = (a)
and the boundary condition
T = \(/(x,y,z,t) at the surface (6)
Assume that there are two solutions T\ and T 2 of these equations,
and let = TI  IV Then 6 satisfies
and, since Ti and T 2 are obviously equal under the conditions
(a) and again of (6),
= for t = in the solid (d)
and = at the .surface (e)
We shall now visualize these last three equations as tempera
ture equations applying to some body. The two boundary
* See Carslaw and Jaeger. 270  > l3
SEC. 2.6J THE FOURIER CONDUCTION EQUATION 17
conditions mean that the temperature is initially everywhere
zero inside the body and that it is at all times zero at the surface.
Now it is physically impossible for an isolated body whose initial
temperature is everywhere zero and whose surface is kept at
zero ever to be other than zero at any point radiation and
selfgeneration of heat, of course, excluded. In other words,
6 = throughout the volume and for any time, which means
that the two assumed solutions T\ and Ti are the same.
CHAPTER 3
STEADY STATE ONE DIMENSION
3.1. A body in which heat is flowing is said to have reached a
steady state when the temperatures of its different parts do not
change with time. Such a state occurs in practice only after
the heat has been flowing for a long while. Each part of the
body then gives up on one side as much heat as it receives on the
other, and the temperature is therefore independent of the time t,
although it varies from point to point in the body, being a func
tion of the coordinates x, y, and z. For the steady state, then,
Fourier's equation (2.3/i) becomes
We shall investigate a few applications of this equation for the
case of flow in the x direction only.
3.2. Onedimensional Flow of Heat. This includes the com
mon cases of flow of heat through a thin plate or along a rod, the
two faces of the plate, or ends of the rod, being at constant
temperatures T\ and T^ and in the latter case the surface of the
rod being protected so that heat can enter or leave only at the
ends. It also includes the case of the steady flow of heat in
any body such that the isothermal surfaces, or surfaces of equal
temperature, are parallel planes.
For these cases the general equation of conduction reduces to
the ordinary derivative being written instead of the partial,
since in the case of only a single independent variable a partial
derivative would have no particular significance. This inte
grates into
T  Bx + C (6)
18
SBC. 3.31 STEADY STATE ONE DIMENSION 19
The constants B and C are determined from the boundary
conditions for this case, which are that the temperature is TI
at the face of the plate (or end of the bar) whose distance from
the yz plane may be called Z, and T* for the face at distance m;
or, as these conditions may be simply expressed,
T = Ti at x = 1} T = T 2 at x m (c)
Therefore, Ti = Bl + C and T z = Bm + C. Evaluating B and
C, we get as the temperature at any point in a plate distant x
from the yz plane
l  IT, (Ti ~ Tt)x
This, with the aid of (1.3d), gives
^WM^TVT,
m I u ^ '
where u is the thickness of the plate or length of the rod. This,
of course, also follows directly from (1.36).
3.3. Thermal Resistance. The close relationship between
thermal and electrical equations suggests at once that the con
cept of thermal resistance may be useful Thus, (1.36) may be
written (overlooking the minus sign)
, AAT Ar AT 7
= * "T"  570  IT (a >
X
where R ss rr (6)
is called the thermal resistance.* It is particularly useful in
the case of steady heat flow through several layers of different
thickness and conductivity in series (Fig. 3.1a). Here (again
overlooking sign)
* Some engineers use the concept of thermal resistivity, the reciprocal of con
ductivity. It is numerically equal to the resistance of a unit cube. In this case,
however, the heat rate is usually measured in watts instead of cal/sec.
20 HEAT CONDUCTION
from which we get by addition
7 T 4 TI = q(R a \ Rb ~\ R
or a =
[CHAP. 3
 T l
* R (x a /k a A a ) + (x b /k b A b ) + (x c /k c A c )
This takes the general form
q =
T T
* n J n
f n dx_
J m kA
(a)
With the aid of (/) and (d) the temperatures T% and T$ as in
Fig. 3. la may be readily computed. For a plane wall the areas
FIG. 3. la. Temperature distri FIG. 3.16. Wall with "through metal";
bution in a composite wall; thermal thermal resistances in parallel,
resistances in series. (The heat flow
is obviously to the left here.)
A aj AI, etc., are equal, but in many cases this will not be true,
e.g., when these considerations are applied to spherical or cylin
drical flow (see Sec. 4.7).
The resistance concept is also useful when conductors,
instead of being in series as above, are in parallel, as in an insu
lated wall with " through metal/ ' e.g., bolts extending from one
side to the other (Fig. 3.16). In this case
SEC. 3.5] STEADY STATEONE DIMENSION 21
nn ffi rji rn
1 I ~~ 1 \ 1 2 "~ <* 1 ,, x
q l ._ ^ ; g 2 ._ _ ^
jT 2  !Fi
or 9 = 31 + 92 = p (t)
where p ^ p~ + p~ 0')
It XV /tfc
Thus, an insulated wall of thickness x and conductivity of insula
tion 0.03 fph, with 0.2 per cent of its area consisting of iron bolts
of conductivity 35 fph, may be readily shown from (i) to have
no more insulation value than a wall without such bolts and of
thickness only 0.3x; i.e., the heat loss is more than tripled by the
presence of the bolts, Paschkis and Heisler find that the heat
loss may be even more than that calculated in this way.
3.4. Edges and Corners.* If, in calculating the heat loss or
gain from a furnace or refrigerator, we use A as the inside area,
it is evident that the results will be much too low because of the
loss through the edges and corners. The situation is no better
if we use the outside area or even the arithmetic mean area, for
in this case the calculated values are too high. If the lengths y
of the inside edges are each greater than about onefifthf the
thickness x of the walls, the work of Langmuir, Adams, and
Meikle 81 gives this equation for the average area A m to be used:
A m = A + 0.54xSy + 1.2x 2 (a)
where A is the actual inside area. For a cube whose inside
dimensions are each twice the thickness, the edge and corner
terms in (a) account for 37 per cent of the whole loss. If the
inside dimensions are each five times the wall thickness, this
drops to 18 per cent.
3.5. Steady Flow of Heat in a Long Thin Rod. This case
differs from the one in Sec. 3.2 in that losses of heat by radiation
and convection are supposed to take place from the sides of the
bar and must be taken into account in our calculation. To do
this we must add to the Fourier equation (2.3/t), written for one
dimension, a term that will represent this loss of heat. Now by
* See also Sec. 11.2 and Carslaw and Jaeger. 270 ' p  m
t For cases where the inside dimensions are less than onefifth the wall thick
ness, see Me Adams. &0 ' l4
22 HEAT CONDUCTION [CHAP. 3
Newton's law of cooling the rate of this loss will be proportional
to the excess of temperature (if not too large) of the surface
element over that of the surrounding medium, which we shall
assume to be at zero, and hence may be represented by 6 2 T
where 6 2 is a constant. Fourier's equation for this case then
becomes
and, when the steady state has been reached, this reduces to
(b)
v '
a
This is readily solved by the usual process of substituting e mx
for T y which gives
6 2
raV* =  e mx (c)
^ '
a
from which we get ra = 7= (d)
Va
and hence T = BePM" + Ce~ bx/v (e)
as the sum of two particular solutions.
3.6. The significance of the constant b is most easily shown
by considering the problem entirely independently of Fourier's
equation. For when the steady state has been reached in such
a bar, the flow of heat per unit of time across any area of cross
section A of the bar will be, at the point x y
kA (a)
and, at the point x + A#,
and consequently the excess of heat left in the bar between
these two points A# apart is
z (c)
This must escape by loss from the surface, and such loss per
SEC. 3.7] STEADY STATE ONE DIMENSION 23
unit of time will be given by hTp&x, where h* is the socalled
surface emissivity of the bar (see Sec. 2.5), and where p&x is
the product of the perimeter p of the bar and the length Az of
the element, i.e., the element of surface. Hence, we have
kA g = hTp (d)
d*T hp m . ,
By comparison with (3.56) we then see that
" >
Writing for convenience, hp/kA == /x 2 , our general solution (3.5e)
takes the form
T = Be* + CV** (jr)
3.7. We may use this solution to investigate the state of tem
perature in a long bar, whose far end has the same temperature
as the surrounding medium, while the near end is at TI, say, the
temperature of the furnace. If the area, perimeter, conduc
tivity, and emissivity were all known or readily calculable to
give /z, no further condition would be required to obtain a com
plete solution. In lieu of any or all of these, however, a single
further condition will suffice, i.e. y that the point at which an
intermediate temperature T* is reached be also known. The
boundary conditions are then
(1) T = at x = oo
(2) T = Ti at x = (a)
(3) T = T* at x *= I
From condition (1) we get
= Be + Ce" 1 "* (6)
so that Be 00 = or B = (c)
Condition (2) then gives
T l a Ce~* or C = Ti (d)
* For values of h, see Appendix A.
24 HEAT CONDUCTION [CHAP. 3
and (3) means that
T 2 = Tie*' or M Z = In (e)
(rp
Y
2
x/l
For different bars subject to the same conditions (1) and (2)
and having the same temperature T 2 at points Ii 9 / 2 , I* ... we
have
T
In Tfr = MI^I = M2^2 = Ms^s = a constant (g)
1 2
which, from the definition of ju, means that
_i _H _ ^ fM
12 "~ 72 ~~~ 72 ~ 72 W
*1 *2 ^3 fc n
providing the several bars have each the same perimeter, cross
section, and coefficient of emission.
3.8. This is the fundamental equation underlying the
socalled IngenHausz experiment for comparing the conduc
tivities of different metals. The metals, in the form of rods
of the same size and character of surface, are coated thinly with
beeswax (melting point T 2 ) and are placed with one end in a
bath of hot oil at temperature TV After standing for some time
the wax is found to be melted for a certain definite distance (I)
on each bar, and the conductivities are therefore in the ratio of
the squares of these distances.
Another application* of (3.60) is found in the solution for
the case of the bar, heated as above, with the temperatures
known at three equally spaced points.
APPLICATIONS
3.9. There could be pointed out an almost unlimited number
of practical applications of these deductions for the steady flow
of heat in one dimension, particularly of (3.2e), but since these
are treated at length in general physics and engineering works,
and especially in texts on furnaces, boilers, refrigeration, and
the like, we shall be content with a few common examples.
* See Preston. I.P.WI
SEC. 3.11]
STEADY STATEONE DIMENSION
3.10. Furnace .Walls. What is the loss of heat through a
furnace wall 45.7 cm (18 in.) thick if the two faces are at 800C
and 60C (1472F and 140F), assuming an average conductivity
of 0.0024 cgs for the wall?
Here we have
w =
0.0024 X 740
45.7
= 0.0389 cal/(cm 2 )(sec) or 151 watts/ft 2
3.11. Refrigerator or Furnace Insulation. Equation (3.4a)
can be effectively used in studying the relation between heat
gain or loss in a refrigerator or furnace, and insulation thickness.
The curves of Fig. 3.2 have been calculated for the case of an
2.00
0.25
25
30
10 15 20
Insulation thickness, in
FIG. 3.2. Curves showing the relation between insulation thickness and the
corresponding heat transfer and insulation cost for a rectangular refrigerator or
furnace of inside dimensions 2 by 2 by 4 ft.
26 HEAT CONDUCTION [CHAP. 3
insulated refrigerator or frozenfood locker of inside dimensions
2 by 2 by 4 ft. They would hold equally well for a furnace of
these dimensions. The heat transfer and insulation cost (i.e.,
volume of insulation) are each taken as unity for 6 in. insulation
thickness. The curves show that to reduce the heat transfer
to onehalf its value for 6 in. of insulation would require a thick
ness of 16 in., necessitating over four times the original amount
of insulating material. In other words, if one were to increase
materially the customary insulation thickness (4 to 6 in.) of
small frozenfood lockers, the law of diminishing returns would
soon come into account.
We shall make use of (3.4a) and (3.3/) in calculating the
heat inflow for a frozenfood locker of inside dimensions 1.5 by
1.5 by 4 ft, with 4 in. (0.333 ft) of glasswool insulation
(k = 0.022 fph), outside of which is the box of % in. (0.062 ft)
thickness pine (k = 0.087 fph). The inside and outside surface
temperatures will be assumed at 10F and 70F, respectively.
From (3.4a) the effective area of the insulation is
Ai = 28.5 + 0.54 X 0.333 X 28 + 1.2 X 0.33 2 = 33.66 ft 2
Then R >  0.022X33.7  ' 448
Similarly, for the box (inside dimensions 2.17 by 2.17 by
4.67 ft)
A 2 = 50.03 + 0.54 X 0.062 X 36.04 + 1.2 X 0.062 2 = 51.25 ft 2
and B 2 = X 51.2 "  014
Then R = RI + 72 2 = 0.462
80
and q = 173 Btu / hr " 50  7 watts
Note the relatively small effect of the pine box in the matter of
insulation.
3.12. Airplane cabin Insulation. Because of the wide varia
tion of temperature encountered by highflying allseason planes
the matter of cabin insulation may be of vital importance.
The construction involves, in general, the use of two or more
layers of material, with perhaps some "through metal."
SEC. 3.13] STEADY STATE ONE DIMENSION 27
Consider a cabin of cylindrical form this can be treated as
essentially a case of linear flow because of the relatively small
wall thickness with internal radius of 4 ft. Assume the wall
to be 2.5 in. thick and to consist of layers as follows, starting
from the inside: 0.5 in. of thickness of material of A? = 0.11 fph;
1.8 in. of k = 0.02; and 0.2 in. of k = 0.06; with 0.1 per cent
of the wall area taken up by throughmetal bolts, etc., of
k = 20. The two outer layers may be of composition sheathing
material, while the center one is of glass wool or other high
grade insulator. For each foot of cabin length the average areas
are A l = 25.3 ft 2 ; A 2 = 25.8 ft 2 ; A 3  26.4 ft 2 . Then, from
Sec. 3.3 the individual resistances are
Rl = 0.11 X 25.3 X 0.999 = ' 151
0.15
" J 0.02 X 25.8 X 0.999
Rz = 0.06 X 26.4 X 0.999 = * 0107
and R w = Ri + Rz + Rz = 0.317. The resistance of the through
metal is 0.208/(20 X 25.8 X 0.001) = 0.403. Then,
1 ' + '
R ~ 0.317 ^ 0.403
or R = 0.177.
For a 60F temperature difference between the outside and
inside surfaces the heat flow q = 60/0.177 = 339 Btu/hr per ft
length of cabin. This means that for a 30ft cabin the heating
(or cooling) input to compensate for the cylindrical wall loss
would have to be 3 kw, not allowing for windows or other open
ings. Contact resistance (Sec. 3.13) might diminish this some
what but only slightly in view of the high insulating value of
the central layer.
3.13. Contact Resistance. In any practical consideration of
heat transfer it is disastrous to overlook the contact resistance
that is offered to tlie heat flow by any discontinuity of material.
Thus, brick masonry, as in a wall, shows a somewhat smaller
conductivity than the brick itself, while powdered brick dust
may have many times the insulation value of the solid material.
28 HEAT CONDUCTION [CHAP. 3
The thermal insulation afforded by multiple layers of paper is
another illustration.
While this thermal contact resistance is not unlike its elec
trical analogue and in some cases might require a similar explana
tion, based, at least partly, on electronic considerations, it is
probable that the cause in most cases lies in the intrinsic resist
ance of a gassolid interface. Here we have a phenomenon,
known in kinetic theory as thermal slip, which is really a
temperature discontinuity at the gassolid boundary and which
greatly increases the resistance. This resistance varies with
the gas, and Birch and Clark 16 have corrected for it in their
rock conductivity determinations by making measurements
with nitrogen and again with helium (which has some six times
greater conductivity) as the interpenetrating gas at the rock
metal boundary.
The insulating value of porous materials has been referred to
(Sec. 1.6) and explained on the basis of the low conductivity of
air when in such small cells that convection is excluded. One
can reason, from considerations based on thermal slip, that it
should not be impossible to produce porous or cellular insulators
that have lower conductivity than air itself.*
3.14. Problems
1. Compute the heat loss per day through 100 m 2 of brick wall (k = 0.0020
cgs) 30 cm thick, if the inner face is at 20C and the outer at 0C. How much
coal must be burned to compensate this loss if the heat of combustion is
7,000 cal/gm and the efficiency of the furnace 60 per cent?
Ans. 11.5 X 10 7 cal; 27.4 kg
2. Calculate the rate of heat loss through a pane of glass (k = 0.0021 cgs)
4 mm thick and 1 m 2 if the two surfaces differ in temperature by 1.5C.
(NOTE: Because of the small value of h the heat transfer coefficient between
glass and air, which may be of the order of only 10~ 4 cgs for still air, the differ
ence between the two surface temperatures of the glass is much less than that
of the two air temperatures.) Ans. 78.7 cal/sec
3. A 5in. wall is composed of 1H in. thickness of pine wood (k = 0.06 fph)
on the outside and } in. of asbestos board (k = 0.09 fph) on the inside with
3 in. of mineral wool (k = 0.024 fph) in between. Neglecting contact resist
ance, calculate the rate of heat loss through the wall if the outside surface is at
* Silica aerogel is an example, although it is not certain that the cause is that
indicated above.
SEC. 3.14] STEADY STATE ONE DIMENSION 29
10F and the inside at 70F. Also, calculate the temperature drop through
each of the three layers.
Ans. 4.63 Btu/(hr)(ft 2 ). Temperature drops: 9.6F through the wood,
48.2F through the mineral wool, 2.2F through the asbestos board
4. A small electric furnace is 15 by 15 by 20 cm inside dimensions and has
firebrick (k = 0.0021 cgs) walls 12 cm thick. If the surface temperature of
the walls is 1000C inside and 200C outside, what is the rate of heat loss in
watts? Arts. 1,828 watts
6. What is the rate of heat flow in Btu/hr into a refrigerator of inside
dimensions 1.5 by 2 by 3 ft with walls insulated with ground cork (k = 0.025
fph) 4 in. thick? Neglect the sheathing of the walls that hold the cork and
assume a temperature difference of 30F. Ans. 71.6 Btu/hr
6. A steam boiler with shell of K in. thickness evaporates water at the rate
of 3.45 lb/hr per ft 2 of area. Assuming a heat of evaporation of 970 Btu/lb
and a conductivity for the steel boiler plate of 23 fph, calculate the tempera
ture drop through the shell. Ans. 6.06F
CHAPTER 4
STEADY STATE MORE THAN ONE DIMENSION
In this chapter we shall discuss several cases of heat flow
in more than one dimension, including the important examples
of spherical and cylindrical flow.
4.1. Flow of Heat in a Plane. We shall first solve Fourier's
problem of the permanent state of temperatures in a thin rec
tangular plate of infinite length, whose surfaces are insulated.
Call the width of the plate TT and suppose that the two long
edges are kept constantly at the temperature zero, while the
one short edge, or base, is kept at temperature unity. Heat
will then flow out from the base to the two sides and toward
the infinitely distant end, and our problem will be to find the
temperature at any point.
Take the plate as the xy plane with the base on the x axis
and one side as the y axis. Then (2.3/0 becomes
" (a)
To solve this problem, then, we must find a value for the tem
perature at any point that will not only be a solution of (a)
but will also satisfy the boundary conditions for this case, which
are
(1) T = at x =
(2) T = at x = TT
(3) T = 1 at y = (6)
(4) T = at y = oo
We shall attempt to find a simple particular solution of (a)
that will satisfy all the conditions of (6), but, failing this, it
may still be possible to combine several particular solutions, as
explained in Sec. 2.4, to secure one that will do this.
4.2. Of the several ways of arriving at such a particular solu
tion we may outline two. The first is with the aid of a device
30
SBC. 4.2] STEADY STATEMORE THAN ONE DIMENSION 31
that always succeeds when the equation is linear and homo
geneous with constant coefficients. This is to assume that
T = e **** (a)
where a and b are constants. Substituting this in (4. la), we
find at once that
a 2 + & 2 = (b)
which is then the condition to be satisfied in order that T = e av+bx
may be a solution. But this means that
T = e ayaxi (c)
for any value of a, is a solution, which is equivalent to saying
that T = ff* (d)
and T = e av e~ axi (e)
are solutions, and by Sec. 2.4 their sum or difference divided by
any constant must be a solution also. Then, since*
e i* + er* = 2 cos <f> (/)
and e** ~* = 2i sin <f> (g)
we get, upon adding (d) and (e) and dividing by 2,
T = e a " cos ax (h)
and, upon subtracting and dividing by 2i,
T = e ay sin ax (i)
Now obviously (ti) does not satisfy condition (1) of (4.16).
Thus, we turn to (i), which can be seen at once to satisfy condi
tions (1) and (2), also (4) if a is negative. As it stands, (i) fails
to meet condition (3), but it may still be possible to combine a
number of particular solutions of the type of (i) that will do
X 3 X 6
while sinsz jfg
X 1 X*
and cos*l jj+jj
Putting x t>, where i is written for \A 1, we see from these that
e** m cos < f t sin 0, and e"" 1 '* cos ^ i sin 0, from which (/) and (0) follow at
once,
32 HEAT CONDUCTION [CHAP. 4
this. For if n is any positive integer,
T = Be""* sin nx (j)
fulfills the first, second, and last of the above conditions, as will
also the sum
T = Erf* sin x + B 2 e~ 2y sin 2x + B^e" zv sin 3z + (k)
where Bi, B 2 , . . . are constant coefficients. For y = this
becomes
T = Bi sin x + B z sin 2x + 5 3 sin 3z + (I)
and if it is possible to develop unity in such a series, we may
still be able to satisfy condition (3) of (4.16). Now, as we shall
discuss at length in Chap. 6, Fourier showed that such a devel
opment in a trigonometric series is possible, the expression in
this case being
1 =  ( sin x + o sin 3x + = sin 5x + 1
(m)
for all values of x between and TT. Therefore, our required
solution is
4/. 1 _ . 1_. \ / x
T =  I e~^ sm x + o e sm 3x + = e * y sin ox + ' ' ' 1 (n)
7T \ o O /
which satisfies (4. la) as well as all the boundary conditions of
(4.16).
4.3. In the second method of solving (4. la) we shall separate
the variables at once by assuming that T == XY where X is a
function of x only, and Y of y only. Substituting, we obtain
1 &L  1 d * x M
or Y dy*  " X ~Atf (b)
Since the two sides of this equation are functions of entirely
independent variables, they can be equal only if each is equal to
a constant that we may call X 2 . The solution of the partial
differential equation (4. la) is thus reduced to that of the two
SBC. 4.3] STEADY STATE MORE THAN ONE DIMENSION 33
ordinary differential equations
/72V
and + XZ  (d)
These may be solved by substitutions similar to (4.2a) but
somewhat simpler, viz.,
Y = & and JSf = e ax respectively (e)
The first gives b = X; therefore,
Y  Be x " + C<r x " (/)
The second gives a = iX, so that
X  BV** + C'e~ ix * (g)
which, from the note to Sec. 4.2, reduces, if we call
(B f  C")i D
and B' + C' = #, to
X = D sin Xx + E cos Xz (/&)
Choosing B = # = to satisfy (1), (2), and (4) of (4.1fe), the
solution resulting from the product of (/) and (h) reduces at
once to (4.2j), and the remainder of the process is the same as
before.
It may be noted that this same sort of solution will hold
even if the temperature T of the base of the plate is other than
unity, indeed even if it ceases to be constant and is instead a
function of x, provided it can be expressed also in this latter
case by a Fourier series. In case we wish to have the values
of x run from to / instead of from to TT, we must introduce
as a variable the quantity irx/l, and the expressions will other
wise be the same as before. We shall discuss this at length in
Chap. 6.
It is also of interest to note that our solution is entirely inde
pendent of the physical constants of the medium, so that the
temperature at any point is independent of what material is
used, so long as the steady state exists.
34 HEAT CONDUCTION [CHAP. 4
4.4. The reader who wishes to make a further study of the
solution (4.2n) will find that the sum of the infinite series can be
expressed in closed form to give finally
, . x\*
tan l [ . i ) (a)
y/ ^ J
sn
. i
\sinh
That this compact function satisfies the fundamental differential
equation (4. la) can be verified by straight forward differentiation.
Obviously, it also satisfies the boundary conditions (4.16).
This form clearly shows that x and y vary along any iso
therm according to the equation
fr
= tan ~ T = a constant (6)
i UO/Il rt
smh y 2
By letting T take on a series of constant values from T = to
T = 1 in this equation, we can obtain a family of isotherms
which covers the infinite plate. They all terminate at the
corners (x = 0, y = 0) and (x = TT, y = 0).
A corresponding family of lines of heat flow must everywhere
be orthogonal to these isotherms as we know from Sec. 1.3.
Such a family can be obtained from a function U which is
conjugate to T in the analytic function U + iT of z = x + iy,
as treated in the theory of functions of a complex variable.
Conjugate functions have the general property of giving orthog
onal families of twodimensional curves for constant values of
the functions. The derivation of the conjugate function U from
the known function T in (a) is given in Appendix L. It has the
similar form
TT 2 / cos x\ , .
U =  tanh * I r ) (c)
TT \cosh yj v '
Lines of heat flow in the plate then correspond to constant
values of U and satisfy the equation
cos x , TT TT f N
r = tanh K U = a constant (a)
It is obvious that the line of heat flow f or U = is a straight
line parallel to the y axis at x = Tr/2, i.e. 9 along the center line
* See Byerly" Art. 58.
SEC. 4.5] STEADY STATEMORE THAN ONE DIMENSION 35
of the plate, parallel to the two sides. This checks with the
physical symmetry of the external temperatures.
If x is allowed to extend indefinitely in both directions, the
above solution corresponds to the physical case in which T on
the boundary y = is kept alternately equal to +1 and 1
over ranges of x = TT.
Problem 1 of Sec. 4.12 calls for a graph of the case con
sidered in these last four sections, while in Sees. 11.2 to 11.5
there are a number of other isotherm and flowline diagrams.
4.5. Flow of Heat in a Sphere. To investigate the radial
flow of heat in a sphere, we must first replace the rectilinear coor
dinates, x y y, and z in (2.3/0 by the single variable r. This is done
by means of the following transformation :
L = $T dr ^ dT x
dx ~ dr dx ~~ dr r ^
dr x
because, since r 2 = x 2 + J/ 2 + z 2 ,
also
d*Tx 2 . dTl dTx*
with similar expressions for the derivatives with respect to y
and z. We thus obtain
^
V T = ~M "" dr 2 r dr
Since, however,
we have V 2 T =  ^ (e)
The Fourier equation for steady radial heat flow thus becomes
r dr 2 ** ^
and its integral may at once be written
T  B +  (flO
36 HEAT CONDUCTION (CHAP. 4
For boundary conditions we may take
(1) T = T l at r = n
(2) T = T 2 at r  r 2 w
where ri and r 2 are, respectively, the internal and external radii
of the hollow sphere. These conditions give, on substitution in
(0r), after the elimination of B and <7,
_ r 2 T 2  rig*!
* "
r,  n r(r 2  n)
This expresses the temperature for any point of the sphere
and also shows that the isothermal surfaces are concentric
spheres. The rate of flow of heat per unit area in the direction
r is given by
w ._ fc ^.fc(rir.)rtr.
dr r 2 (r 2 n) Vi/ '
and the total quantity that flows out in unit time is
q , 4^ = *" k(T r \ ~ ^ (k)
If g* units of heat are released per unit of time at a point
(i.e., in a region of small spherical volume) in an infinite medium,
at zero initial temperature, the steady state of the temperature
in the medium can be calculated at once from (gr), (/?,), and (k).
Boundary condition (2) of (K) becomes T = at r = oo ; thus
(0) becomes
We can get T\ from (fc) by writing
= g; T 2 = 0; r 2 = oo (m)
Thus, q = 47rfc2Vi (n)
r '
Then ' r
Compare this with (9.5m).
* In Chaps. 8 and 9 the symbol Q' is, in general, used for the rate of heat
generation.
SEC. 4.7] STEADY STATE MORE THAN ONE DIMENSION
37
4.6. Radial Flow of Heat in a Cylinder. Let the axis of the
cylinder be the z axis. Then, the problem is similar to that for
the sphere, save that now we are concerned with only two dimen
sions and may put r 2 = x z + y*. By a process similar to that
by which (4.5c) and (4.5e) were obtained we then get
v *  dr*
The integral of this is
d*T 1 dT ^ 1 d(rdT/dr)
r dr ~~ r
dr
(a)
T = B In r + C
which gives, by the use of boundary conditions quite similar to
those of (4.5/i),
r 2 = B In r 2 + C (c)
Ti = B In ri + C;
and from these we obtain
M (7\  T 2 ) In r
In r 2  T, In
In ri In r 2 In r 2 In
The rate of flow per unit area is given by
k(T l  r,)
_ .\ 
(d)
w " r(ln r 2  In n) v ;
and the quantity of heat that flows out through unit length of
the cylinder per second by
2irrw =
In r 2 In r\
4.7. The results of the two
preceding sections may be very
simply obtained from the linear
flow equation, for the flow in
any element of small angle is
essentially in one direction.
However, the crosssectional
area is continually increasing,
being obviously proportional to
the distance from the center in
the cylindrical case and to the square of this distance in the spher
ical. From (1.3c) we get at once as the rate of flow q through
any spherical shell of area 4?rr 2 and tliickness dr,
FIG. 4.1.
Section of a sphere or
cylinder.
38 HEAT CONDUCTION [CHAP. 4
q = Mirr 2 ^ (a)
Writing this as dT = 
we have, on integration,
=
r 2 n x '
which is identical with (4.5&)
Similarly, for unit length of a cylinder,
dT
q = k2irr^ (e)
* T =
which gives, on integration,
^ In (r 2 /ri) v/v
which is essentially the same as (4.6/). By integrating (6) and
(/) between Ti (or T 2 ) and T, and correspondingly between ri
(or r 2 ) and r, we can obtain at once (4.5t) and (4.6d) on substi
tuting values of q from (d) and (h).
Carrying a step further our treatment of spherical and
cylindrical flow with the aid of the fundamental linearflow equa
tion, we may write from (d),
4ark(Ti 
where A m is the mean value of the area to be used in the spherical
case. This gives
A m = 47irir 2 = VAiA 2 (j)
which means that the average area to be taken if we use the
simple linearflow equation for the hollow sphere is the geometric
mean of the inner and outer surfaces.
SKO. 4.8] STEADY STATE MORE THAN ONE DIMENSION 39
For a cylinder of length I/,
In (r t /n) ~ w r 2  r,
, _ 2
or A m 
) " In (A 2 /A X )
__ A 2 At
~ 2.303 logic (A 2 /AO w
If A i and A 2 are not far different, we can frequently use the
arithmetic mean value for A m instead of the logarithmic mean
as given by (Z) and still keep within prescribed limits of error.
Thus, if A*/ A i = 2, the arithmetic mean is only 4 per cent
larger than the logarithmic; while if A*/Ai does not exceed
1.4, the difference is less than 1 per cent.
Thermalresistance equations, in particular (3.3/), may be
applied to a series of concentric spherical or cylindrical shells
if the areas A a , A^ etc., of (3.3/) are evaluated from (j) or (Z).
APPLICATIONS
4.8. Covered Steam Pipes. Some of the best applications of
the theory of Sees. 4.5 and 4.6 are the various radialflow meth
ods of measuring thermal conductivity described in Sec. 12.5.
We shall confine ourselves here, however, to applications of a
slightly different sort. As an example of the use of (4.6f) let
us investigate the heat loss per unit length of a 2in. steam pipe
(outside diameter 2.375 in. or 6.04 cm), protected by a covering
1 in. (2.54 cm) thick of conductivity 0.0378 fph (0.000156 cgs).
Assume the inner surface of the covering to be at the pipe tem
perature of 365F (185C) and the outer at 117F (47.2C).
Then from (4.6/)
X 0.0378 X 248
= 96.6 Btu/hr per ft of pipe length
= 0.222 cal/sec per cm length
It is of interest to note that double this thickness of covering
would still allow a loss of 59.8 Btu/hr per ft length for the same
temperature range, or only 38 per cent decrease in loss for 158
per cent added covering material. That the proportional sav
40
HEAT CONDUCTION
[CHAP. 4
ing* is greater for a larger pipe is shown by the curves of Fig.
4.2.
The temperature of c.urrontc.arrymg wires as affected by
the insulation is also a question that might be studied with the
2610
217.5
114.0
30.5
GQ
87.0 .E
43.5
2.3 4
Thickness of covering, inches
Fio. 4.2. Curves showing the relation of heat loss to thickness of covering,
for two sizes of steam pipe, with temperature drop through the covering of 248F
or 138C. Conductivity of covering, 0.0378 fph or 0.000156 cgs.
aid of the foregoing equations. It can easily be shown that a
wire insulated with a covering of not too low thermal conduc
tivity may run cooler, for a given current, than the same wire
* For a discussion of the most economical thickness for* pipe coverings see
Walker, Lewis, and McAdams. 167  pm
SEC. 4.9] STEADY STATE MORE THAN ONE DIMENSION
41
(or cold)
liquid
if bare; the insulation in this case produces, effectively, so much
more cooling surface. A similar case for steam pipes would
occur under special circumstances of small pipe and very poor
insulation.
4.9. Flow of Heat in Solid and Hollow Cones. A truncated
solid cone of not too large angle is in effect part of a hollow
sphere, the fraction being the ratio of its solid angle to 4?r. The
rate of flow down such a cone may be determined at once from
(4.7^) The hollow cone, if of uniform thickness, is made from
the sector of a circle. The heat flow may be found with the aid
of (4.70, using for A 2 and Ai the sectional areas (metal only)
for the large and small ends. A hollow cone is frequently used
to connect the outlet pipe of a
vessel (Fig. 4.3) containing very
hot or very cold liquids with a
base or surface at room tempera
ture. Assume that such a cone
of metal of low conductivity
(e.g., "inconeP'jfc = 0.036 cgs)
0.5 mm in thickness connects a
pipe of 3 cm diameter with the
exterior metal sheath of the
insulated vessel, the base of the
cone being 10 cm in diameter
and its length, measured along the cone, 12 cm. If the pipe
is at 200C and the base of the cone at 0C, what is the rate of
heat loss through the cone?
Such a cone is equivalent to a sector of a circle with
7*2 ri = 12 cm
If p represents its fraction of a circle, 27rr 2 p = TT X 10 and
2irrip = TT X 3. From these relations we find at once TI = 5.14
cm; r 2 = 17.14 cm; p = 0.292. From (4.6/) we then have as
the flow of heat down the cone
CoM
for warm)
surface
/
.Cone
/; }: If 'ns u/a f/'on ^f:v
FIG. 4.3. Hollow cone used in con
nection with insulated vessel
q = 2irrpw X 0.05
2ir X 0.292 X 0.036 X 200 X 0.05
2.303 logic (17.14/5.14)
= 0.55 cal/sec (a)
If the pipe is directly connected with the outer sheath as the
42 HEAT CONDUCTION [CHAP. 4
center of a 10cm diameter circle of this same metal 0.5 mm
thick, and if it is assumed that the circumference of this circle
is at 0C as was the case for the cone, the loss will now be
2?r X 0.036 X 200 X 0.05 1 QQ , , /JA
2.303 lo glo W = L88 Cal/S6C (&)
It is evident that the cone lessens the heat waste, the ratio of
the losses under these conditions being the fraction p.
4.10. Subterranean Temperature Sinks and Powerdevelop
ment; Geysers. The question is sometimes raised as to the
possibility of power development from large areas of heated
rock, e.g., old lava beds, etc. Its answer forms an interesting
application of (4.5k) and (4.5p). Assume that an old buried
lava bed (k = 1.2 fph) at temperature 500F has a deep hole
ending in a spherical cavity of 4 ft radius. Water is fed into
this and the resultant steam used for power purposes. When
a steady state has been reached, what steady power develop
ment might be expected? Assume that the temperature of the
interior of the cavity must not fall below 300F.
We shall treat this problem as a point sink (negative source)
and consider temperature conditions at r = 4 f t where the tem
perature is 200F below that of the lava. We may then use
(4.5p) with the understanding that we are not concerned with
the temperature distribution inside r = 4 f t providing that the
temperature for this radius is kept steadily 200 below the initial
value.
Then 200 =  A ^ n g v , or q = 12,050 Btu/hr (a)
TcTT X 1.^5 XT:
This means that only 4.73 hp could be developed. Conditions
while the steady state is being approached, and the time involved
in reaching the steady state, will be studied later (Sees. 9.4
and 9.10).
It is evident that these same principles would apply to a
study of geysers if conditions are such that the heat is supplied
at or near the bottom of the tube. In general, however, the
inflow of heat is probably along a considerable length of tube,
and accordingly it is a case of cylindrical rather than spherical
flow. We shall treat this case in Sec. 9.10.
SEC. 4.11] STEADY STATE MORE THAN ONE DIMENSION
43
4.11. Gasturbine Cooling. A major problem in gasturbine
design is that of keeping the tempera
tures of the parts from running too high.
The cooling of the rotor is principally
due to gas convection, but it is impor
tant to know how large a part conduc
tion cooling may play. It is possible
to make a simple approximate calcula
tion of the conduction cooling, assuming
that the heat flows radially in from the
bladed periphery of the rotor disk and
is carried away at the center by conduc
tion along the axle or perhaps by
liquid cooling in the axle. FlG 4 4 gection of gas .
Such a rotor is shown in section in turbine rotor: (a) hollow
Fig. 4.4. Let u c be the thickness of the axle, (&) biading.
disk at the center and u c pr the thickness at radius r, where
p ss (U G Uo)/R, UQ being the thickness where the biading
begins and R the corresponding radius. From (3.30),
(a)
(b)
(c)
(d)
dr
J ri 2irr(u c  pr)
But since (Appendix B)
dx 1 , x
f
J
x(a + bx) a n a + bx
u f ri dr ! i r *( u * 
we have / 7 r = In 7
J ri r(u c  pr) u c ri(u c 
Then q = r 2 /^ _! m \
2.303 logic ;
e  pr 2 )
Note that for a disk of uniform unit thickness, (d) reduces to
(4.6/) or (4.7/0, as it should.
We shall calculate the rate of radial heat flow from biading
to center for a turbine rotor of dimensions R = 25 cm (9. 84 in.);
UQ 2 cm (0.79 in.) ; u c = 7 cm (2.76 in.). Assume the material
of conductivity 0.09 cgs (22 fph) for the average temperatures
involved, and take the temperatures as 600C (1112F) at
r 2  R = 25 cm, and 320C (608F) at n  5 cm (1.97 in.).
44 HEAT CONDUCTION [CHAP. 4
Then from (d) we calculate the rate of heat flow from periphery
to center as q = 409 cal/sec = 5846 Btu/hr. For a disk of
2 cm uniform thickness we can calculate from (4.6/) or (4.7/0
that, for the same temperatures as used above,
q = 197 cal/sec = 2810 Btu/hr
The smallness of these figures shows clearly the inadequacy of
conduction cooling alone.
It is evident at once that, having calculated q for tempera
tures Ti and T^ (d) can be used to find the temperature for
any other radius of the disk, assuming conduction cooling alone
as operative.
4.12. Problems
1. Plot the temperatures for a dozen points in a plane such as is treated in
Sees. 4.1 to 4.4, and draw the isotherms and lines of flow.
2. A wire whose resistance per cm length is 0.1 ohm is embedded along
the axis of a cylindrical cement tube of radii 0.05 cm and 1.0 cm. A current
of 5 amp is found to keep a steady difference of 125C between the inner and
outer surfaces. What is the conductivity of the cement and how much heat
must be supplied per cm length? Ans. 0.0023 cgs; 0.597 cal/sec
3. A hollow lead (k 0.083 cgs) sphere whose inner and outer diameters
are 1 cm and 10 cm is heated electrically with the aid of a 10ohm coil placed
in the cavity. What current will keep the two surfaces at a steady difference
of temperature of 5C? Also, at what rate must heat be supplied?
Ans. 1.10 amp; 2.90 cal/sec
4. Calculate the rate of heat loss from a 10in. (actual diameter 10.75 in.)
steam pipe protected with a 2in. covering of conductivity 0.04 fph if the
inner and outer surfaces of the covering are at 410F and 90F.
Ans. 254 Btu/hr per ft length
6. A 60watt lamp is buried in soil (k = 0.002 cgs) at 0C and burned
until a steady state of temperature is reached. What is the temperature
30 cm away? Ans. 19C
6. Calculate the rate of heat flow for the following cases, the metal being
nickel (k = 0.142 cgs) with surfaces insulated: (a) a circular disk 1 mm thick
and 10 cm in diameter with a central hole 1 cm in diameter and with 100C tem
perature difference between hole and edge; (6) a cone of the same thickness of
sheet nickel, 20 cm long, 1 cm mean diameter at the small end, and 10 cm
diameter at the large, and with 100C temperature difference between the
ends; (c) a solid cone* of the same dimensions and same temperature difference.
Measure cone lengths on the element.
Ans. 3.87 cal/sec; 0.87 cal/sec; 5.65 cal/sec
* It can be readily shown that a cone of half angle $ has a solid angle of
 cos 0).
CHAPTER 5
PERIODIC FLOW OF HEAT IN ONE DIMENSION
5.1. We shall now take up the problem of the flow of heat in
one dimension that takes place in a medium when the boundary
plane, normal to the direction of flow, undergoes simply periodic
variations in temperature. This problem occupies in a way an
intermediate place between those of the steady state already
considered and the more general cases that can be treated only
after a familiarity has been gained with Fourier's series; for in
the former cases the temperature at any point has been constant,
while in the latter it is a more or less complicated function of
the time, rarely reaching the same value twice at a given point;
but in the present case the temperature at each point in the
medium varies in a simply periodic manner with the time, and
while the temperature condition is by no means "steady," as we
have defined this term, it duplicates itself in each complete
period.
The problem derives its interest and importance from its very
practical applications. The surface of the earth undergoes daily
and annual changes of temperature that are nearly simply
periodic, and it is frequently desirable to know at just what time
a maximum or minimum of temperature will be reached at any
point below the surface, as well as the actual value of this tem
perature. Such knowledge would be of value, e.g., in deter
mining the necessary depth for water pipes, to avoid danger
of freezing, or in giving warning of just when to anticipate such
danger after the appearance of a "cold wave," i.e., one of those
roughly periodic variations of temperature that frequently
characterize a winter.
5.2. Solution. Our fundamental equation for this case is the
Fourier conduction equation
(a)
45
46 HEAT CONDUCTION [CHAP. 5
written in one dimension
dT ...
IT" " (5)
and tne solution must fit the boundary condition
T  To sin cot at x = (c)
As the equation (b) is linear and homogeneous with constant
coefficients, we can arrive at a particular solutten by the same
device used in Sec. 4.2, viz., by the assumption that
T = Be"*** (d)
where b and c are constants. Substitution in (6) shows that
this is a solution, provided only that
6 = ac 2 (e)
Thus, we have as a solution
If 6 is replaced by 17, this becomes
(/^ \
ijt x J VTt)
x
But Vt  
and v^i = + (t)
so that (0) becomes
T = B exp [ i7< x ^ (1 + t)] (j)
or 2 1 = B exp (a; ^) exp [ t ( 7 f x ^)] (A;)
From the several solutions contained in (k) other particular
solutions may be built up by addition, such as
n:r\f f / Pv
, ^)(exp [f (yt  x ^
B exp
f t)  1 4 2 ~1  2 /. V7  r
V2
SEC. 5.3] PERIODIC FLOW OF HEAT IN ONE DIMENSION 47
and from Sec. 4.2 this may be written
sin (yt  x
T = Ce*VT /2a s i n ( y t  x JJ I (m)
\ \ LOL/
Other solutions may be formed in the same way, care being
taken to note, however, that, from the manner of its formation
[see (/)], the sign before i in each term of (j) must be the same.
This will be found equivalent to saying that the same sign must
be used before x Vy/2a in each .term of equations like (I).
With this in mind we may write at once as other particular
solutions
. / /v\
T C'e*v Y/2 sin I yt + x J~ ) (n)
\ \ 2<x/
/ rz\
T = Pe"" 3 ^ 7 / 2 " cos 17* x \/o~" ) (o)
, / nr\
and T = D'^ /2 cos ( yt + x Jl ) (p)
\ \ AOL/
Of these four solutions, (n) and (p) demand that the tem
perature increase indefinitely as x increases, which is evidently
absurd, while (o) is excluded by (c). Equation (ra) will satisfy
this condition if C is put equal to T Q and 7 to co. Making these
changes, we then have as the solution
sin (ut  x
T = Toe"*^/^ sin ut  x <J (q)
which expresses the temperature at any time t at any distance
x from the surface.
5.3. Amplitude, Range. The equation (5.2<?) is that of a wave
motion whose rapidly decreasing amplitude is given by the factor
Toe*^" 7 *". The range of temperature, or maximum variation,
for any point below the surface is given by
T R  22V aVZ;7 ^ = 2Tve~*^^* (a)
putting for co its value 2ir/JP, where P is the period. To is the
amplitude, or half range, at the surface. This shows at once
that the slower the variation of temperature the greater the
range in the interior of Jfhe body.
48 HEAT CONDUCTION [CHAP. 5
6.4. Lag, Velocity, Wave Length. The time at which a maxi
mum or minimum of temperature will occur at any point is
evidently that for which
co*  x ^ = (2n + 1) \ (a)
x Vco/2a + (2n + 1W2
or = (6 )
odd values of n giving minima, and even, maxima. Fixing our
attention on the minimum that occurs at the surface when, say,
wt = 3?r/2, we see that if x and t are both supposed to increase
so that
~~ 37T
, = y (c)
we may think of this particular minimum being propagated into
the medium and reaching any point x at the time given by this
equation. This is later than its occurrence at the surface by an
amount
which may be called the "lag" of the temperature wave. The
same reasoning holds for the maximum, or zero, or any other
phase.
The apparent velocity of such a wave in the medium is given
from (d) by
"E'Vr W
but this is merely the rate at which a given maximum or mini
mum may be said to travel and has nothing to do with the actual
speed with which the heat energy is transmitted from particle to
particle.
From (e) we may deduce as the expression for the wave
length of such a wave
X VP = 2 VT^P (/)
Equations (d) to (/) may be used to measure the diffusivity
SBC. 5.6] PERIODIC FLOW OF HEAT IN ONE DIMENSION 49
of any medium from determinations of the lag, velocity, or
wave length.
5.5. Temperature Curve in the Medium. The form of this
curve at any given time may be conveniently investigated by
differentiating (5.2g) with respect to x to find the maxima and
minima of the curve, which, of course, will be distinguished from
the maxima and minima above treated. Then, writing
we have tan (< M&) = 1 ( a )
T/4 + at 57T/4 + <at 9ir/4 + ut ..
*=  ,  _,.._._,... (b )
This shows that the minima and maxima are equally spaced,
and if we note that the corresponding minima and maxima of
the pure sine curve
y = sin (co px) (c)
7T/2 + C0 37T/2 + COf , , N
occur at x =  >  > (a)
r* M
they are seen to be nearer the surface than these latter by an
amount 7r/4ju. This means that, when t = nP [or (n + >^)P],
the first minimum (or maximum) is found at just half the dis
tance of the corresponding minimum (or maximum) for the sine
curve. This is illustrated in the solid line curve in Fig. 5.1,
which gives the temperatures for different depths for the diurnal
wave in soil of diffusivity = 0.0049 cgs. The broken line is the
curve of amplitudes for an amplitude, or half range, of 5 at
the surface.
5.6. Flow of Heat per Cycle through the Surface. This is
readily computed by forming the temperature gradient from
(5.2g) and then integrating it over a half period in which the
gradient is of one sign, i.e., going from zero to zero. Thus,
cos
(0,1  x ^)] (a)
50 HEAT CONDUCTION [CHAP. 5
and
C\ f3P/S //)T\ /3w/4a) //JT^
1  * / (ir) *  * / (^) <tt
A y _p /8 \ da; /,<) J r/4 V&c A=o
= fcr ^^ cal/cm 2 , or Btu/ft 2 (&)
The limits of integration in (6) are determined by the fact that
dT/dx is not in phase with T but, for x = 0, has a minimum
at t = P/8 = 7r/4co and is zero at t = P/8 = 7r/4co and
J = 3P/8 = 37r/4a>. The amount of heat given by (6) flows
through the surface into the material during one half the cycle
in which dT/dx is negative and out again during the other half.
APPLICATIONS
6.7. With the aid of the foregoing equations we may investi
gate the penetration of periodic temperature waves into the
earth. The questions of interest and importance in this connec
tion are (1) the range or variation of temperature at various
depths for the diurnal and annual changes; and (2) the velocity
of penetration of such waves, and hence the time at which the
maximum or minimum may be expected to occur at various
depths.
6.8. Diurnal Wave. First consider the diurnal or daily
wave. If the surface of the soil varies daily, at a certain season,
from +16 to 4C (60.8 to 24.8F), what is the range at 30 cm
(11.8 in.) and 1 m (39.4 in.)? The mean of the above tempera
tures is +6C; and as condition (5.2c) supposes a mean tem
perature of zero, our temperature scale must be reduced by the
subtraction of 6, which can be added again later if necessary.
In this case, then, TO is 10C and P = 86,400 sec. Using 'the
constants for ordinary moist soil (a = 0.0049 cgs), (5.3a) shows
that the range is reduced from 20 at the surface to only 0.07 of
this, or 1.4C (2.5F), at 30 cm below, and to less than 0.004C
at 1 m below. Since a range of 12 would just be sufficient in
this case assuming an average temperature of 6C in the soil
to reach a freezing temperature, we conclude that a layer of
soil 6 cm thick will be enough to prevent freezing under these
conditions. Dry soil will afford even smaller penetration than
SBC. 5.9) PERIODIC FLOW OF HEAT IN ONE DIMENSION
51
this, and in the damp soil we have neglected the latent heat of
freezing of the soil, which, while nearly negligible for small
water content, would still reduce the penetration of the freezing
temperature somewhat. We may also deduce from (5.4d) that
the maximum or minimum temperature at 30 cm would lag
10
20
30
40 50
Depth, cm
60
80
FIG. 5.1. Curves showing the penetration of the diurnal temperature wave in
soil of diffusivity 0.0049 cgs. Solid line is curve of temperatures at time
t = (n + y^)p (i.e., in the early evening). Broken Jine is curve of amplitudes for
an amplitude, or half range, of 5 at the surface.
some 35,000 sec, or 9.7 hr, behind that at the surface. In a
series of soil temperature measurements by MacDougal 92 the
lag of the maximum at 30 cm depth was found to be from 8 to 12 hr,
and the range generally less than onetenth of the range in air,
both figures being in substantial agreement with the above
deductions.
6.9. Annual Wave. For the annual wave the variation for
temperate latitudes may be taken as 22 to 8C (71.6 to
17.6F). The range at 1 m will then be reduced to 19C, while
at 10 m below the surface it will be only 0.33C. The freezing
52
HEAT CONDUCTION
[CHAP. 5
temperature will penetrate to a depth of less than 170 cm
(5.6 ft).* From (5Ae) the velocity of penetration of such a
wave is 0.000045 cm/sec, or 3.9 cm per day. For a soil of this
diffusivity, then, the minimum temperature at a depth of about
7 m (23 ft) would occur in July and the maximum in January.
Table 5.1 is compiled from measurements of underground
temperatures in Japan, cited by Tamura. 144 The computed
temperature range and lag were calculated for a diffusivity of
0.0027 cgs by (5.3a) and (5 Ad).
a
TABLE 5.1
Depth,
cm
Observed
annual range,
C
Calculated
annual range,
C
Observed lag,
days
Calculated lag,
days
28.2
28.2
30
22.7
23.4
2.5
10.6
60
18.7
19.5
9.0
21.6
120
14.0
13.5
35.0
42.3
300
5.2
4.6
93.5
106.0
500
1.3
1.3
177.5
176.5
700
0.4
0.4
267.0
247.0
Fitton and Brooks 40 have published a series of soil tem
peratures in the United States f that give much material for
calculations on lag, range, diffusivity, etc. Thus, a series of
measurements at Bozeman, Mont., at depths from 1 to 10 ft
give an annual temperature range at the greater depth of
only 0.416 that at the shallower and a lag for the greater depth
of 55 days behind the other. Using (5.3a), we have
0.416 =
(a)
and, putting # 2 x\ 9 ft = 274 cm and
P * 1 year 3.156 X 10 7 sec
we get a = 0.0097 cgs, a high value for soil. Computing from
* In reality, considerably less than this, because of the latent heat of freezing,
t See also Smith. 134
7.5 = exp (22.9 ^5
SEC. 5.10] PERIODIC FLOW OF HEAT IN ONE DIMENSION 53
the lag with the aid of (5.4d), we use
274 IP
t 2  ti  55 days  4.75 X 10 6 sec = ^r\ (b)
L MTTOC
from which we get a = 0.0083 cgs. Similarly, in feandy loam
at New Haven, Conn., a series of readings at depths from 3 to
12 in. give an average daily range at the former depth 7.5 times
that at the latter. From (5.3a) we then have
(c)
where P = 86,400 sec. This gives a = 0.0047 cgs.
Birge, Juday, and March 17 have made a study of the tem
peratures in the mud at the bottom of a lake (Mendota) by
means of a special resistance thermometer that could be driven
into the mud to a depth of 5 m. From a large series of measure
ments the amplitude and lag of the annual temperature wave
could be determined. This allowed the computation of the
diffusivity of the mud and (with auxiliary data) of the annual
heat flow [see (5.66)] into and out of the lake through the
bottom.
6.10. Cold Waves. While the preceding formulas were
developed on the assumption of a simply periodic temperature
wave that continues indefinitely, they are still applicable with
a fair degree of approximation to temporary variations of a
roughly periodic nature, such as cold waves. A good example
of this is furnished by observations on underground tempera
tures by Rambaut. 116 The curve of temperatures for March,
1899, shows a marked drop, or cold wave, of about 10 days'
duration whole period 20 days the lowering (jP ) amounting
to about 8.6C. The magnitude of the temperature fall and lag
of the minimum, as observed by platinum thermometers at
various depths, is given in Table 5.2, and also the computed
values. These latter were obtained by using the value of
a = 0.0074 cgs computed by Rambaut from the annualwave
curve. The computed temperature fall is of course half the
range as determined from (5.3a).
More accurate calculations will be possible with the aid of
the theory of Sec. 8.6.
54
HEAT CONDUCTION
TABLE 5.2
[CHAP. 5
Depth,
cm
Observed
temperature
fall, C
Computed
temperature
fall, C
Observed lag,
days
Computed lag,
days
0.0
8.6
8.6
16.5
5.9
6.7
1.4
0.8
45.7
3.4
4.2
2.5
2.3
107.9
1.3
1.6
4.9
5.4
174.0
0.33
0.57
8.0
8.7
6.11. Temperature Waves in Concrete. The above discus
sions may be applied at once to a mass of concrete as in a dam.
Taking the diffusivity, e.g., as 0.0058 cgs we may conclude that
a cold wave of 3 days' duration (period 6 days), of minimum
temperature 20C (4F), might cause the freezing tempera
ture to penetrate a concrete mass at 4C (39.2F), a depth of
some 56 cm (22 in.), while the annual variation of temperature
at a depth of 2 m (6.6 ft) would be only 0.43 of what it is at the
surface.
6.12. Periodic Flow and Climate ; "Ice Mines." The annual
periodic heat flow into the earth's surface in spring and summer
and out in fall and winter tends to cause the seasons to lag
behind the sun in phase and also may moderate slightly the
annual temperature extremes. When we come to calculate
this from (5.66), however, we find that for soil (k = 0.0022,
a = 0.0038 cgs) it amounts to only about 1920 cal/cm 2 for the
season, and for rock (k = 0.006, a = 0.010 cgs), 3260 cal/cm 2 ,
assuming an average annual surface temperature amplitude of
12C. This would have its greatest effect in deep canyons where
the large area of rock walls results in a marked reduction in the
annual temperature range.
There are a number of wellknown "ice mines " in the world.
These are small regions, perhaps excavations, where the order
of nature is reversed and ice forms in summer, while in winter
the region is warmer than the surrounding locality. There
seems to be no generally accepted explanation of this phe
nomenon, but it is undoubtedly connected with periodic heat
inflow and outflow. It is doubtful if this explanation can be
SBC. 5.13] PERIODIC FLOW OF HEAT IN ONE DIMENSION 55
sufficient in itself unless there is some way of increasing greatly
the area of surface involved. This can happen if the local geo
logic structure, as in an immediately adjoining hill, is of a very
porous character. In this case the whole hill might act like a
gigantic calorimeter or regenerator, cooled by the winter winds
to a considerable depth. This "cold" coming out in the form
of cold air in summer could produce the freezing effects. It is
suggested that this may be the explanation of the ice mine at
the foot of a hill at Coudersport, Pa.*
6.13. Periodic Flow in Cylinder Walls. As another instance
of periodic flow may be mentioned the heat penetration in the
walls of a steamengine cylinder. Callendar and Nicolson 24 f
found that for 100 rpm the temperature range of the inner sur
face of the cylinder wall (cylinder head) during a cycle was 2.8C
(5.1F). Using a = 0.121 and k = 0.108 cgs, we find from
(5.3a) that this variation is reduced at a depth of 0.25 cm (0.1
in.) to
and at three times this depth to only 0.021C (0.04F). The
heat flow into and out of the walls that takes place each cycle
is given from (5.66) as
Q ^2.8 X 0.108 rW AOAO w 2
A = ^ 2 X 0.348 ViOO^ = ' 269 cal/cm
= 0.99 Btu/ft 2 (6)
This results in a loss of efficiency since it subtracts from the
available energy during the power part of the stroke. To
remedy this the "uniflow" engine is specially designed so that
the steam enters at the ends and exhausts from the middle of
the (long) cylinder. This involves smaller cyclical temperature
changes of the cylinder walls and hence lessens the wasteful
inflow and outflow of heat.
* See Lautensach 82 'for an apparently similar case of coldair storage but with a
smaller temperature range.
t For a discussion of several of the other factors involved here see Janeway. 6 *
Also, see Meier. 96
56 HEAT CONDUCTION [CHAP. 5
5.14. Thermal Stresses.* If a body or a portion of it is
heated or cooled and at the same time constrained so that it
cannot expand or contract, it will be subject to stress. Such
stresses may be computed on the basis of the forces necessary
to compress or extend the body from the dimensions it would
take if allowed to expand or contract freely, back to its original
ones.
If a bar of length L has its temperature raised from T\ to
7^2, it will, if allowed to expand freely, increase in length by an
amount
AL = eL(T z  Tx) (a)
where is the coefficient of expansion. The stress, or force per
unit area, necessary to compress the bar back to its original
dimensions is
P = = Ee(T>  T l ) (b)
where E is the modulus of elasticity. This is then the stress
required to keep it from expanding, in other words, the thermal
stress in a constrained bar.
As an example, consider the stresses in tramway rails that
have been welded together at a temperature of 40F, if the rails
are warmed to 95F. If we take E = 3 X 10 7 lb/in. 2 and
= 6.4 X 10~~ 6 /F for steel, we can compute at once from (6)
that the stress would be 10,560 lb/in. 2 compression. The
customary burying of the body of the rail so that only its top
surface is exposed affords some protection from the severity
of daily temperature changes although very little for the annual,
as the preceding theory readily shows.
In unconstrained bodies thermal stresses are produced by
nonuniform temperature distribution. Examples of such occur
in the warming up of steam turbine rotors and in the periodic
heating and cooling of engine cylinder walls, or in the daily
variation of surface temperatures in rocks, concrete structures,
and the like. Such stresses may be taken as largely determined
by the temperature gradient at the point. Differentiating
*See Timoshenko, 148 '*" 3 Timoshenko and MacCullough, 149 '" 20 Kent, 76 and
Roark. 118
SEC. 5.15] PERIODIC FLOW OF HEAT IN ONE DIMENSION 57
(5.2g), we have the expression, similar to (5.6a)
dT
*\ " X *%/ ITkC/ I OiiJL  WV X
OX
(o)t  x ^^p) J (c)
+ cos
which shows that temperature stresses due to periodic variation
are greatest for the surface layers of the material. It can be
shown (Timoshenko 148  p  212 ) that for not too slow cyclical variations
the stress is approximately given by the quantity eET P /(l  v),
where T P is the amplitude of the temperature variation at the
point and v is Poisson's ratio. For the cylinder wall of a diesel
engine subject to surfacetemperature fluctuations of 20F
we find, using the above value of e and E for steel and putting
v = 0.3, a stress of 5,500 lb/in. 2 It is evident from (5.3o) that
this would fall off rapidly below the surface, but that the rate
of decrease would be less for a slowrunning engine.
5.15. Problems
1. If the daily range of temperature at the surface of a soil of diffusivity
0.0049 cgs is 20C, what is the range at 10 cm and 1 m below the surface?
Ans. 8.4C; 0.0036C
2. Solve the preceding problem for an annual range of 30C and for depths
of 10 cm, 1 m, and 10 m. Ans. 28.7C; 19.1C; 0.33C
3. Compute the periodic heat flow into and out of the surface for the two
preceding problems. (Use k = 0.0037 cgs.) Ans. 124 cal/cm 2 ; 3550 cal/cm 2
4. A long copper (a = 1.14 cgs) rod is carefully insulated throughout its
length and one end is alternately heated and cooled through the range to
100C every halfhour. Plot the temperatures along the bar for such time
as will make the temperature of the heated end 50C. Determine the wave
length and velocity for this case; also, for the case in which the period is one
quarter hour.
Ans.\ = 161 cm, V = 0.089 cm/sec, forP = }hr;X 114cm, V 0.126
cm/sec, for P = 1 4 hr
5. A cold wave of 2 weeks' duration (P = 4 weeks) brings a temperature
fall (amplitude) at the surface of 20C. What will be the fail at a depth of
1 m in soil of diffusivity 0.0031 cgs and also of 0.0058 cgs? Also compute the
time lag of the minimum in these cases.
Ans. 2.6C, 4.5C; 9.1, 6.7 days
CHAPTER 6
FOURIER SERIES
6.1. Before we can proceed further with our study of heat
conduction problems, we shall be obliged to take up the develop
ment of functions in trigonometric series. The necessity for this
was apparent in Chap. 4 and could indeed be foreseen in the
last chapter; for it was evident that, if the boundary condition
had been expressed by other than a simple sine or cosine function,
as it was, it could not have been satisfied by any of the solutions
obtained, unless it should be of such a nature that it could be
developed as a series of sine or cosine terms, in which case it
might be possible to build up particular solutions to fit it.
Such a development was shown by Fourier to be possible for
all functions that fulfill certain simple conditions. For example,
the curve y = f(x) may be represented between the limits x =
and x = TT, by adding a series of sine curves, thus:
f(x) = y = ai sin x + a 2 sin 2x + a 3 sin 3x + (a)
or by a similar cosine series. The f(x) can be represented in this
way if it meets the following conditions within the range
considered :
1. The/(x) is singlevalued: i.e., for every value of # there is
one and only one value of y (save at discontinuities).
2. The f(x) is finite. For example, f(x) = tan x cannot be
expanded in a Fourier series.
3. There are only a finite number of maxima and minima.
For example, f(x) = sin 1/x cannot be so expanded.
4. The f(x) is continuous, or at least has only a finite number
of finite discontinuities.
The function that represents the initial state of temperature
will satisfy these conditions, for there can be but a single value
of the temperature at each point of a body, and this value must
be finite. Furthermore, while there may exist initial discon
58
SEC. 6.2] FOURIER SERIES 59
tinuities, as at a surface of separation between two bodies, such
discontinuities will always be finite. This indicates the applica
bility as well as the importance of Fourier's series in the theory
of heat conduction.
6.2. Development in Sine Series. To accomplish this devel
opment it is necessary to find the values of the coefficients
oil 2 , 8 , . . . f t* 16 ser ies (6. la). It is possible to find the
value of a finite number, n, of these by solving n equations of
the type
y p = ai sin x p + a 2 sin 2x p + + a n sin nx p (a)
where x p is one of n particular values of x chosen between and
TT. This process also has the merit of making plausible the pos
sibility of expanding a function in such a series; for with n terms
the curve made up by summing the trigonometrical series coin
cides with the curve y = f(x) at the n points and can be made
identical with it if we take n large enough. But while this
method is possible, it is not the simplest way, for the results
may be obtained by a much shorter procedure, as follows:
We shall proceed on the assumption that the expansion (6.1a)
is possible, and consider this assumption justified if we can find
values for the coefficients. Multiply both sides of (6. la) by
sin mx dx, where m is the number of the coefficient we wish to
determine ; then integrate from to TT :*
/ f(x) sin mxdx = a\ I sin mxsmxdx +
Jo Jo
+ a m I sin 2 mxdx +
Jo
+ a p I sin mx sin px dx + . (6)
Jo
f* 1 f r
Now / sin mx sin pxdx = o / cos (p m)xdx
Jo * Jo
 \ fl cos (P + m ^ xdx  \ [j^t sin (P 
 \ [pTS 8in (p + W)
* It can be shown that this procedure is essentially the same as that employed
above if n is large. See Byerly. 28 * p  *
(50 HEAT CONDUCTION (CHAP. 6
Hence, the only term remaining on the righthand side of (6) is
Therefore,
a m
a m
sin 2 mxdx = a m ~
I si
Jo
2 /**
=  / /(#) si
u JO
(d)
sin mxdx
and the complete series may be written
2 ff /* 1
/(x) sin x dx sin re
J
+ / f(x) sin 2# da: sin 2x +
+
/ f(x) sin nx dx sin nx + 
6.3. As examples of the application of this series let us
develop a few simple functions in this way.
(1) f(x) = c, any constant (Figs. 6. la to d).
(a)
(b)
(c)
FIG. 6.1. The approximation curves for the sine series for y /(), where
f(x) a constant, (0 < x < *). (a) One term, (6) two terms, (c) three terms,
(d) four terms.
tec. 6.3] FOURIER SERIES 61
2 f r . 2c /* . , , x
4^ SB  / c sin wxdx / sin mxdx (a)
1* Jo K Jo
= [l ~ ^^ (6)
=* if m is even (c)
4c
= if m is odd (d)
mJyUli[[[
Ttm
::i;:::::::::::::::
a
ll m ffnfn
(a) W
FIG. 6.2. The approximation curves for the sine series for y ~ /(x), where
f(x) *= x, (Q < x < 7T/2); f(x) ** *  x, (ir/2 < z < TT). (a) One term, (6) two
terms.
Hence, the even terms will be lacking, and we get
 4c /sin x . sin ?>x . sin 5x
\
/
For x == 7T/2, this enables us to write the expansion for Tr/4
thus:
(2) Let us reproduce the curve (Figs. 6.2a and b)
. . . _ 7T
/(x) = x from x = to x = H
/(x) = TT x from x = o to x = TT
2 r v2 , N . j , 2 r* N . ,
"* ~ / /0*0 sln ^^o^ H / /W Sln ^w^d^
IT Jo * J*/2
2 r^ 2 2 /"*
=  / x sin mx dx H / (TT x) sin mx dx
IT Jo ft J*/2
0)
(t)
62 HEAT CONDUCTION [CHAP. 6
2 ["/sin mx x cos m#\ T/2 / 1 V
I  5  ) + TT I  cos m# 1
7rL\ m 2 m /o \ m A /2
(s
sn mx x cos
TT
7T
If m = 1, or 4p + 1, sin m 7j = 1
TT
m = 2, or 4p + 2, sin m ^ =
m = 3, or 4p + 3, sin m ~ = 1
m = 4, or 4p + 4, or 4p, sin m ^ =
where p is any integer
Again, the even terms are absent, and
, x 4 /sin x sin 3x , sin 5x
For x = 7T/2 this gives
"g = ! + 32 + 52 + 72 + ' ' ' (n)
(3) Finite discontinuity (Figs. 6.3a to/).
7T
/(a:) = a: from a: = to x =*= (o)
= from x = o to x = TT (p)
Breaking up a m into two parts and substituting the values for
/(*) we get
2 T T
sin mx cte H / sin mx dx
2 C* /2
=  / x si
IT JO
2 fir/2
 I x sin mx dx (q)
" JO
_ 2 /sin ma; a; cos mx\ w/2
~~ ~
SBC. 6.3]
FOURIER SERIES
63
2 / m7r\ . r 4,4
" I ~~ o 2 ) if w = 4p + 4
7r\ 2m 2 / ^
2 /sin a; , TT sin 2a: sin 3o: 2?r sin
sin 5x Sir sin Qx
oH
25
36
(0
I
(d)
FIG. 6.3. The approximation curves for the sine series for y = /(x), where
/(*)  x, (0 < x < ir/2); /(*)  0, (r/2 < * < T). (a) One term, (6) two
64 HEAT CONDUCTION [CHAP. 6
It may be noted that at the point of discontinuity,* x = ir/2,
the value of the series is
11 J_
+ 9 + 25 + 49 +
2/V\ 7T . .
iFVTJi (w)
which is the mean of the values approached by the function as
x approaches Tr/2 from opposite sides.
6.4. Development in Cosine Series. In a manner quite simi
lar to the foregoing we are also able to develop such functions as
fulfill the conditions we have mentioned, in cosine series between
the limits x = and x = TT. Thus,
f(x) = b' + 61 cos x + 6 2 cos 2x + 6 3 cos 3x + (a)
The constant term that appears here, though not in the sine
series, may be thought of as the coefficient of a term 6J cos (0 x),
which shows at once why the corresponding term for the sine
series is lacking.
To find the value of any coefficient b m , we proceed as before,
multiplying both sides of (a) by cos mx dx and integrating from
to TT; then, since terms of the type
b p cos px cos mxdx (6)
vanish just as did similar terms in (6.2c), we have remaining on
the righthand side only
b m I cos 2 mxdx = ^ [(mx + cos mx sin mx)]l (c)
= ~n b m if m T
2 /"*
/. b m =  / /(re) cos mxdx (d)
TT Jo
To get &' we must multiply (a) by dx only and integrate
from to TT ; then,
\f(x)dx= lV Q dx+ f*b l cosxdx+
Jo Jo Jo
(e)
* It is seen that the representation of the curve (see Figs. 6.3/ and 6. Id) is not
as perfect near the discontinuities as elsewhere. This is known as the "Gibbs'
phenomenon." See Carslaw," Churchill.
SEC. 6.5] FOURIER SERIES
since all terms but the first vanish. Therefore,
65
This is just half the value that (d) would give if m = were
substituted; therefore, to save an extra formula, (a) is generally
written
/0*0 =
+ bi cos x + 6 2 cos 2x + 6 3 cos 3x +
(ff)
where the value of any coefficient, including the first, is given
by (d). The complete cosine series may then be written
+
f(x) dx + I / J(x) cos x dx I cos x
LJo J
/ /G*0 cos 2x dx cos 2x +
[I
#) cos mx dx cos mx +
o J
(h)
(a)
FIG. 6.4. The approximation curves for the cosine series for y /(a;), where
f(x) = x, (0 <x < 7T/2); f(x) = ^r  x, (7T/2 < x < x). (a) Constant term and
next term, (6) constant term and next two terms.
6.5. As an example take the same function as we developed
in a sine series under (2) in Sec. 6.3 (see Figs. 6.2a and 6 and
6.4a and 6) :
7T
/(#) = x from x == to x = ~
6
7T
/(#) = TT o; from x = * x "" ^
2 f f w/2 /* T 1
Then, 6 m =  / x cos rnxdx + / (ir x) cos wzcte (a)
w LJo J*/z J
66 HEAT CONDUCTION [CHAP. 6
2 /cos mx + mx sin mx\* /2 2 TT
2 7T / . V
I sin T?W/ I
* m \ J r/2
TT \ m 2 /o
2 /cos mx + mx sin mo/ % , . /t x
;v ^ A/2 whenw ^ w
2 /COS W7T/2 , 7T . mTT 1 7T . W7T
_ f _ 1_ sin 7: ; sm ^r
TT V m 2 2m 2 m 2 m 2
cos WTT , cos m7r/2 , TT . m7r\ . .
^+ m 2 + 2^ sm TJ (c)
( \
2 cos o cos m?r 1 J (d)
If m = 1 or 4p + 1, bracket =
/. 6m =
m = 2 or 4p + 2, bracket = 4
= _ 2 1_
'* bm ~ IT (m/2) 2
m == 3 or 4p + 3, bracket =
.'. b m =
m = 4 or 4p + 4, bracket =
A b m 
To get 60, substitute m = in (a) and integrate Then,
/2 2 / ^ /** \
(~ " I TT / dx / x dx )
^ \ J*/2 J*/2 /
37T 2 7T
So, finally, we have
xv TT 2 /cos 2a: cos 6# , cos Wx , \ e s
/W  4  J ^p^ + ~3^~~ + ~^~ + ' ' J (0
to represent the same curve as is given by the sine series (6.3m).
6.6. The Complete Fourier Series. It is possible to combine
the sine series and the cosine series so as to expand any function
satisfying our original conditions (Sec. 6.1) between TT and TT.
This gives the true or complete Fourier series
/(#) = M&o + &i cos x + 6 2 cos 2x + + ai sin x
+ a 2 sin 2x + (a)
The coefficients 01, a 2 . . . 6 , Z>i, 6 2 , . . . may be determined
SBC. 6.6J FOURIER SERIES 67
in much the same way as before. Multiply both sides of (a)
by sin mxdx and integrate from TT to TT. Then,
f* 1 [*
I f(x) sin mx dx = ^ &o / sin mx dx
+ bi I sinmxcosxdx + + b p I sinmxcospxdx +
J T J W
+ ai / sin mx sin xdx + + a m I sin 2 mxdx +
J T 7 T
+ a p I sin rax sin pxdx + (6)
Now / sin mxdx = and / sin mx cos mxdx = (c)
y T j if
Also (see Appendix B)
cos (m p)o? cos (m + p^x'Y
2(m + p) J,,
sm * cos
2(m  p) 2(m + p)
= (d)
and
/*
^
sin (m p)x sin (m + p)x
sm mx sm pxx = 2(m _ p)  2 (m + p)
= (6)
Hence, the only term remaining on the righthand side of (6) is
a m I sin 2 mx dx = a m ir (/)
J v
1 /"*
Therefore, o^ =  / /(x) sin mx dx (gO
^"JT
In the same fashion we can, after multiplication of (a) by
cos mx dx and integrating, determine
/(&) cos mxdx (ft)
T
which also holds for m = 0.
Since x will generally refer in our conduction problems to
some particular point or plane in a body, it is better to use some
variable of integration such as X in writing (g) and (ft), which
then become
68
and
HEAT CONDUCTION
1
b m = 
sin m\d\
cos m\d\
[CHAP. 6
(J)
6.7. It is instructive to get expressions (6.6i,j) by another
method. We have seen that any function of the kind consid
ered can be represented by either a sine or cosine development
for all values of x between and TT. We may now question
what such series would give at and beyond these limits. Obvi
ously, the sine series can hold at the limits x = and x = TT
only when the f(x) is itself zero at these points, although it will
(a)
FIG. 6.5. Curves showing the results of extending the limits beyond and *.
The cosine development for (a) gives a curve like (6), while the sine series for (a)
gives (c).
hold for points infinitesimally near these limits for any value
of f(x). For example, it breaks down at the limits in the case
of f(x) = c already given.
Both series are periodic and afford curves that must repeat
themselves whenever x is changed by 27r; and, as both series
give the same curve between and TT, the difference, if any,
between the curves given by the two series must come between TT
and 27r, or, what amounts to the same thing, between and TT.
This difference is at once evident if we consider that the values
of the sine terms will change sign with change to negative angle,
while the cosine terms will not. Thus, the sine and cosine devel
opments, when extended beyond the limits and TT, give curves
of the type shown in Fig. 6.5. We may conclude from this,
SEC. 6.7) FOURIER SERIES 69
then, that if f(x) is an even function, i.e., if f(x) =/( #), it
may be represented by a cosine series from TT to TT. Similarly,
an odd function [f(x) = /( x)] will be given by a sine series
for these same limits. Not all functions are either odd or even,
e.g., e x , but it is possible to express any function as the sum of
an odd and an even function ; thus,
.
'
the first term being even, since it does not change sign with x,
while the second does and is therefore odd. To expand any
function satisfying our primitive conditions, then, between
x = TT and TT, we may write (6.6a) where the coefficients are
determined from (6.2e) and (6.4d) as
2 f*f(x) f(x) .
a m =  / ^ ^   sin mxdx (o)
and b m =  cos mxdx (c)
Since the values of definite integrals are functions only of the
limits and not of the variable of integration, we may replace x in
these expressions by any other variable X; thus,
2 /"'
a m =  /
TT Jo
. x ,, M .
sm mX dX (d)
/(X) +/(X) . ,. . ,
and b m =  / '^ ^y^   cos mXdX (e)
2 /"'
 /
K JO
We can simplify expressions (d) and (e) somewhat, for the
former is equivalent to
in mXdx ^ /(X) sin mXdx (/)
7o J
and if we replace X by X' in the second integral, it is trans
formed into
sinmX'dX' (g)
o
This is equal to
/(V) sinmX'dX' (A)
70 HEAT CONDUCTION [CHAP. 6
which, since it is immaterial what symbol is used for the integra
tion variable, may as well be written
ro
+ \ /(X)sin m\d\ (i)
1 f v
Hence, we have a m =  I /(X) sin wXdX (j)
7T J T
In a similar way we obtain
b m = (' /(X) cosmXdX (*)
" J v
6.8. Change of the Limits. While our expansion as hereto
fore considered holds only for the region x = TT to x = TT, we
can, by a simple change of variable, make it hold from x = I
to L For let
***; then/C*) 
/. f(x) = FO)  ^6 + &i cos 2 + 6 2 cos 2z +
+ cti sin z + o>2 sin 2z
for values of z from TT to TT, and
7rX i L
COS  + ?>2 COS
. wx . . 2wx ,
sin j + a 2 sin j h
for values of a: from I to Z, where
k 1 /"* ET/ \ ^ 1 /*' r/ v m7ra: j / \
&m =  / /^() cos mzdz = j f(x) cos T dx (c)
since 2 = ir/Z, and dz = ir dx/l. This may also be written
b m = T / /(X) cos y dX (d)
Similarly, ^ = ~ T /(X) sin ^ dX (e)
" J I '
In the same way the sine series (6. la) may be written
SBC. 6.9] FOURIER SERIES 71
ir
,, N . irx , . ,
f(x) = ai sin t a 2 sin + * ' (/)
2 [ l , /XN . rmrX  , ,
where a m = 7 / /(X) sm 7 a A yjf;
t yo *
while (6.40) becomes
f(x) =  6 + 61 cos j (62 cos j h (h)
Z L L
where 6 m = / / /(X) cos ^T~ dX (t)
6 jo *
While series (6) applies generally, (/) and (h) hold only from
x = to /, unless f(x) is an even function, in which case the
cosine series will be good from / to Z, while if odd, the sine
series will hold over this range.
6.9. Fourier's Integral. In the foregoing we have developed
f(x) into a Fourier's series that represented the function from
/ to I where I may have any value whatever. We shall now
proceed to express the sum of such a series in the form of an
integral and, by allowing the limits to extend indefinitely,
obtain an expression that holds for all values of x. Write the
series (6.86) with the aid of (6.8d) and (6.8e).
> , f l *f\ \ nX TTX ,^
X + / /(X) cos y cos y aX
, f l .... 27rX 2wx
+ / /(X) cos y cos y aX + *
+ / /(X) sin j siri y dX
J i ii
+ y^/(X) sin ?y sin ?y dX + ] (a)
When terms are collected, this becomes
00
f (x) = T / /(X) dX ( o + / cos T cos
J J i V 4 (>
ml
/ sin r sin r )
Zy I I /
m* 1
72 HEAT CONDUCTION [CHAP. 6
But since cos r cos s + sin r sin s cos (r s), this may be
written
40
cos T" (x ~ x}
or, if we remember that cos (<p) = cos ( <p),
30
cos "j (X  x)
+ V cos^CX  x)l (d)
since cos (Ox//) (A a?) = 1. As / increases indefinitely, we
may write 7 s rajr// and ^7 = ?r/Z, and the expression in
braces in (e) then becomes
/ cos y(\  x) dj (/)
7
i r r
Therefore, f(x) ^ / /(X) d!X / cos T(\  a?) cfry (gr)
an expression holding for all values of x and for the same class
of functions as previously defined. It is known as "Fourier's
integral."
6.10. Equation (6.9gr) can be given a slightly different form
by means of the following deduction, which will prove of use:
For any function,
f 1
J i
\ (a)
o i
In the last term substitute X' for X; then,
d\ =  <p(  X') dX' (6)
(c)
SBC. 6.11J FOURIER SERIES 73
since its value is independent of the integration variable [see
(6.7t)]. If <f>(\) is even, i.e., if <p(X) = <p( X), (c) means that
r <p(\) dx  r ^(x) dx = r
J i Ji Jo
so that / <p(X) d\ = 2 / <p(X) dX (e)
J i Jo
while if <p(X) is odd,
T ?(X) dX = f ^(X) dX  T ^(X) dX = (/)
J i Jo Jo
Since the cosine is an even function, we may write at once,
instead of (6.90),
I I ^ /(X) dX ( cos T (X  a?) d T (g)
" J  oo J
6.11. Again, if f(x) is either odd or even, we may put (6.100)
in somewhat simpler form. Since the limits of integration in
(6.10gr) do not contain either X or 7, the integration may be per
formed in either of two possible orders; i.e.,
/(X)dX
yo
/ " dy I " /(X) cos 7(X  x) d\ (a)
Jo yo
Now f 80 /(X) cos 7(X  x)dX = /*/(X) cos 7(X  x)d\
J  *> Jo
/(X)cos7(X x)dX (6)
and, following the general methods of the previous section, we
may write the last term
/(X) cos T(\  x) dX
ro
=  / /(X') cos 7(X'  x)d\' (c)
= /""/( X') COB y(X' + *)(iX' (d)
yo
"/( X) cos y(\ + x)d\ (e)
/
yo
74 HEAT CONDUCTION [CHAP. 6
 f " /(X) cos 7(X + *) d\ if /(X) is odd (/)
yo
/(X) cos 7(X + x)d\ if /(X) is even (0)
Therefore, if /(x) is odd, (6.100) becomes, for all values of #,
 *) ~ cos T(X + x)] dX (fc)
2 /" /" "
=  / dX / /(X) sin 7X sin 7x^7 (i)
IT yo yo
while, if it is even, we have, instead,
f(x) = (" dy I" /(X)[cos 7(X  x) + cos 7(X + a:)] dX ( j)
TT 70 yo
2 /"" T 00
=  / dX / /(X) cos 7X cos 70: d7 (k)
ft Jo Jo
Equations (i) and (k) hold for all positive values of x in the
case of any function.
6.12. Harmonic Analyzers. The analytical development of
a function in a Fourier's series, with the determination of a large
number of coefficients, is wellnigh impossible in many cases, and
in any event involves considerable computation. To eliminate
this there have been invented several machines that are designed
to compound automatically a limited number of sine or cosine
terms into the resulting curve, or to perform the more difficult
inverse process of analyzing a given function into its component
Fourier's series. One of the earliest of these has become well
known because of its great simplicity, as well as from the fame
of its designer, Lord Kelvin.* A long cord or tape is passed
over a series of fixed and movable pulleys, to each of which
a simple harmonic motion of appropriate period and amplitude
is given. The end of the cord will then have a displacement at
each instant equal to double the sum of the displacements of
the movable pulleys. This principle has been extensively
developed! in machines of 40 or more elements, and Michelson
and Stratton 97 have devised a machine of 80 elements using a
* See Thomson and Tail. 147 ' 1 * 44
t See Kranz" and Miller. '
SEC. 6.12]
FOURIER SERIES
75
spring arrangement instead of the cord. Various electrical
methods have also been developed.
In such a machine of 40 elements the frequencies of the ele
ments are 1,2,3 ... 40 times that of the fundamental. The
process of combining sine or cosine terms is that of giving each
element an amplitude of the proper magnitude and sign. The
sum of all the terms appears in the displacement of a pen draw
FIG. 6.6. Section of one element of the Michelson and Stratton harmonic
analyzer. The adjustable displacement d of the rod R from the center of the
oscillating arm B determines the amplitude of the motion. The sum of all the
effects is transmitted to the pen P.
ing on a sheet of paper that advances as the instrument is
operated.
One such element, for the Michelson and Stratton analyzer,
is shown in Fig. 6.6. The wheel D is of such size as to give the
eccentric A the proper frequency, and the desired amplitude is
secured by adjusting the rod R on the lever B. The corre
sponding harmonic stretching of the spring s causes, along with
that of all the other elements, a pull on the cylinder C. This
gives a vertical motion to the pen P,that writes on paper carried
on a plate moving horizontally as the machine is operated so as
76 HEAT CONDUCTION . [CHAP. 6
to trace a curve that represents the sum of the contributions of
all the elements.
6.13. The method of reversing this process and finding for
any given function the coefficients of the corresponding Fourier's
series may be seen from the following considerations:
Suppose we wish to develop a function in terms of the sine
series. Then,
f(x) = ai sin x + a 2 sin 2x + a 3 sin 3x + ' * * (#)
2 f*
where a p =  / f(x) sin px dx (6)
7T ./o
=  t/Oi) sin pxi + /(x 2 ) sin px 2
7T
+ ' ' ' + /foe) sin pxw] (c)
if we replace the integral by a series and consider that we have
a 40element machine. Now, let x 2 = 2xi, x 3 = 3#i, . . .
x 40 = 40xi. Then, (c) becomes
2dx
a =  Oi) sin pxi + /(2xi) sin
+ /(40*i) sin 40?^!] (d)
To analyze a curve divide it into 40 equal parts whose abscissas
have the values Tr/40, 27T/40, . . . TT and adjust the amplitudes
of the 40 elements of the machine proportionally to the 40 ordi
nates of these parts. As the analyzer is operated, the slowest
turning or fundamental element will, at any instant, have
turned through an angle pxi, and the paper will have advanced
a distance proportional to p, say, p cm. We see then from (d)
that for p = 1 the coefficient a\ is given by the ordinate of the
curve drawn by the analyzer at a distance of 1 cm (i.e., p = 1)
from the origin. Similarly, a 2 is the ordinate at 2 cm, and the
other ordinates are obtained in the same fashion. When the
curve has been completed, it is evident that the slowest element
has rotated TT radians and the fastest 407T.
Such instruments are of great usefulness in analyzing sound
waves, alternatingcurrent waves, and various other curves.*
* For a simple graphical method of analysis see Slichter. 138
SEC. 6.14] FOURIER SERIES 77
6.14. Problems
1. Develop the sine series that gives y = for x between and T/2; and
y = c for x between Tr/2 and IT. Plot and add the first four or five terms.
2c /sin x 2 sin 2x sin 3x sin 5x 2 sin 6x \
Aru.y =  (1  2~ + ~T~ + ~5 6~~ + /
2. Do this for the corresponding cosine series.
2c /TT cos x , cos 3x cos bx . \
Ans  y  7 vi  nr + ~3 s~ + ' * /
~ /sin a: sin 2rr , sin 3x \ f , .
3. Show that x = 2  "" H   for z between
and TT.
4. Develop f(x) in a sine series if f(x) = c/3 for a; = to //3; f(x) = 0,
for x = Z/3 to 2Z/3; f(x) = c/3, for x = 2Z/3 to L
/ ._ %* x , ! _ *** . 1 :![? _i_ 1 ;.
7T
5. Verify
4 ^ x . , . . .
Ans.f(x) =  \sm + 3 sin ~ + 4 sm ~f + 5 sm
, * **  ! et +  1 ^ _i_ et " !
^  21 ( TT ^ /2 . 2 COS y + f2 + 47r2 COS y
, cos
7 h ' * ' ) from x to x I
6. If f(x) = from a; = TT to 0; and /(x) = x from a? = to TT, show that
?! ? /cos a; cos 3a; cos 5a;
sin a; sin 2a; , sin 3a;
1. Develop c + sin x in a cosine series between and ?r; and in a complete
Fourier's series between ic and TT.
2/2 2 2 \
Ans. y = c +  (^1  ^ cos 2z  ^ cos 4z  gg cos 6a; + J;
?/ = c + sin a:
8. Outline the curve between ir and TT, formed by the addition of series
(6.3m) and (6.5i).
9. With the aid of (6.7a) graph the two functions, even and odd, whose
sum is the curve /(a?) = x for x positive and /(a;) = c for x negative.
CHAPTER 7
LINEAR FLOW OF HEAT, I
7.1. In Chaps. 3 to 5 we have already discussed a number
of the simpler problems of heat flow. These have included the
case of the steady state for several different conditions, and the
simplest case in which the temperature varies with time, viz.,
the periodic flow. With the single exception of the steady state
for a plane, in which we were forced to assume one of the
results derived later in the study of Fourier's series, these prob
lems could all be solved without the use of this analysis; but
we now come to a class of problems, at once more interesting
and more difficult, in which continual use is made of Fourier's
series and integrals.
In the present chapter and the following one we shall take
up a number of cases of the flow of heat in one dimension.
These will include the problem of the infinite solid, in which the
heat is supposed to have a given initial distribution i.e., the
initial temperature is known for every point and starts to flow
at time t = 0; the socalled "semiinfinite solid" that has one
plane bounding face, usually under a given condition of tem
perature; the slab with its two plane bounding faces; also, the
case of the long rod with radiating surface; and the problem of
heat sources. In these several cases the solutions hold equally
well for the onedimensional flow of heat in an infinite solid,
or for the flow along a rod whose surface, save in the fourth case
above mentioned, is supposed to be impervious to heat. In all
the problems discussed in this chapter, save that of the radiating
rod, the solutions must first of all satisfy the Fourier conduction
equation, which becomes for one dimension
dT
As we saw in Sec. 3.5, this must be modified for the case of the
radiating rod by the addition of a third term.
78
SBC. 7.2] LINEAR FLOW OF HEAT, I 79
CASE I. INFINITE SOLID. INITIAL TEMPERATURE
DISTRIBUTION GIVEN
7.2. Take the x direction as that of the flow of heat. Then,
all planes parallel to the yz plane will be isothermal surfaces,
and the initial temperature of these surfaces is given as a
function of their x coordinates. The problem is to determine
their temperatures at any subsequent time.
The solution must satisfy (7. la) and the condition
T = f(x) when t = (a)
We shall solve (7. la) by a process that is, at the outset, the
same as that employed in Sec. 5.2, viz., the substitution in
(7.1a) of
T = e bt+cx (b)
b and c being parameters. This gives
6 = ac 2 (c)
Putting now c = iy (d)
instead of fc = iy as before, we get
T = Le^e^ (e)
and T = Me^^e^^ (/)
But since e iyx = cos yx i sin yx (g)
we get, on combination of (e) and (/) by addition or subtraction
choosing suitable values for L and M the particular solutions
T = e"*' cos yx (h)
and T = <T aY " sin yx (i)
These are particular solutions of (7. la) for any value of 7, the
latter being a function of neither x nor t. Now we can multiply
these by B and C, any functions of y, and obtain the sum of an
infinite series of terms represented by
(B cos yx + C sin yx)e"^ H dy 3)
also as a solution of (7. la) by the proposition of Sec. 2.4.
The functions B and C must be so determined that f or t =
(j) becomes equal to /(#). Now Fourier's integral (6.100) gives
80 HEAT CONDUCTION [CHAP. 7
/(*)  I I " dy I" /(X) cos 7 (X  *) d\ (k)
TT Jo J _
and from (j) this must equal
/ (B cos 72 + C sin 70;) dy (I)
7o
J / oo
Hence, B =  / /(X) cos 7\d\ (m)
7T J _ oo
1 /*
and C Y =  / /(X sin 7XdX (n)
TT J_ x
and if these values are substituted in (j), we finally have
T =  f " 6""^ f d7 f /(X) cos 7(X  x) dX (o)
^T JO 7  eo
This is then the required solution, for it satisfies (7. la) and
reduces for ( = to (&), i.e., to f(x). It gives the value of T
for any chosen values of x or t.
7.3. This equation can be simplified and put in a more useful
form by changing the order of integration and evaluating one
of the integrals. For
T =  [ " /(X) d\ [ " ei H cos 7(X  x) dy (a)
IT J  Jo
But since (see Appendix C)
Q '*' cos nydy =
we have
f e~^ s< cos 7(X  x) dy = ~
putting rj = l/(2Va?). Hence
r = 7
V TT y 
/Q
Bj^ putting ]8 = (X  x)ry or X =  + a; (e)
we secure the still shorter form
SBC. 7.4] LINEAR FLOW OF HEAT, 1 81
We may regard this as our final solution, since it is much
easier to handle than the other forms. If f(x) = C, a constant,
then f( + x) = C, and the integral reduces to the " proba
bility integral " (see Appendix D). If f(x) = a; 2 , say, then the
equation (/) becomes
x being a constant as regards this integration, these three inte
grals can be readily evaluated (see Appendixes B, C, and D).
Also, for many other forms of f(x) the integration is not difficult.
7.4. If f(x) is of more than one form, or possesses discon
tinuities, it may be necessary to split the integral (7.3/) into
two or more parts. For example, suppose that f(x) = T Q
between the limits x = I and x = m, and that f(x) = outside
these limits, a condition that would correspond to the sudden
introduction of a slab at temperature T Q between two infinite
blocks of the same material and at zero temperature. We write
the integral (7.3/)
r 7; />"* +^
(a)
7T c
In determining the limits 6 and c it must be remembered that x
(as well as t) is a constant for each particular evaluation of the
integral, and that the initial temperature condition is really
expressed as a function of the variable of integration X, i.e.,
TO = /(X). The limits of 6 and c will then be the values of
corresponding to X = I and X = ra; and from (7.3e) these are
seen to be (I x)y and (m rr)r;, respectively. Equation (a)
then reduces to
(mx)r,
f(m
d*
f+dft (6)
w
This solution may be readily applied to the case in which
f(x) = TQ for x > 0, and f(x) = for x < 0, for in this event
the limits are seen at once to be xij and o.
82 HEAT CONDUCTION [CHAP. 7
APPLICATIONS
7.5. Concrete Wall. While perhaps not having the variety
of applications that we shall find for Case II, next to be con
sidered, the foregoing equations admit of the solution of many
interesting problems. For example, suppose a concrete wall
60 cm (23.6 in.) thick is to be formed by pouring concrete in a
trench cut in soil at a temperature of 4C (24.8F), the con
crete being poured at 8C (46.4F). It is desired to know how
long it will be before the freezing temperature will penetrate the
wall to a depth of 5 cm (2 in.). In other words, will the wall
as a whole have time to "set" before it is frozen?
To apply the foregoing equations we must first assume >that
the soil has the same diffusivity (we shall use a = 0.0058 cgs)
as the concrete, as would be approximately true in many cases,
and that latentheat considerations can be neglected. The solu
tion then follows at once from the equation of the last section.
Taking the origin at the center of the wall, we have I = 30 cm,
m = 30 cm, and x  25 cm. Choosing, say, the positive
value for #, and shifting our temperature scale so that the initial
soil temperature is brought to zero, while the freezing tempera
ture becomes 4C and the initial wall temperature 12C, (7.4&)
becomes
12 f**
~ (a)
71V 55.j
To find t we must determine the limit p (=677) so that
9 /> / 9 fp 9 flip
* / <**& ( =  r \ f+dft + 4= / **
VTT J HP \ V7T./0 VTTJO
From the probabilityintegral table (Appendix D) we readily
find p to be about 0.055, or t] = 0.011, which gives
= 356 > 000 sec = 4J days ^
If we are interested in knowing the temperature at the center
of the wall at the end of this 4.1day interval, we put t = 356,000
sec (i.e., i) = 0.011) in the equation
SBC. 7.8] LINEAR FLOW OF HEAT, I 83
''d/3  4.31C (d)

V7T 7 30,
Subtracting the 4C that was added to shift the temperature
scale so as to make the initial temperature of the soil zero, we
have T c = 0.31C. This indicates that the whole wall is near
the freezing point.
7.6. It may be remarked that in solving this problem we have
also accomplished the solution of another that, at first sight,
appears by no means identical with it. Suppose the same tem
perature conditions to exist, but the wall to be only half as thick,
and one face in contact, not with earth, but with some material
practically impervious to heat, or at least a very much poorer
conductor than cement; e.g., cork or concrete forms of dry wood.
To see the similarity of the two problems, notice that in the
first one conditions of symmetry* show that there would be
no transference of heat across a middle plane in the wall; hence,
this plane could be made of material impervious to heat without
altering the conditions. We could then remove half of the wall
without affecting the half on the other side of this impermeable
plane, in which case we should have our present problem.
7.7. In the above solutions we have omitted consideration of
three important factors which would generally be present in any
practical case, and which would serve to retard to a considerable
extent the freezing of the wall. These are the latent heat of
freezing of the water of the concrete, the heat of reaction that
accompanies the setting of concrete, and the insulating effect of
wooden forms that are frequently used for such a wall. The
theoretical treatment of these factors would be beyond the aims
of the present work.
7.8. Thermit Welding. As a further application let us take
another and more difficult problem. Suppose two sections of a
steel (a = 0.121 cgs) shaft 30 cm (11.8 in.) in diameter are to
be welded end to end by the thermit process. The crevice
between the ends is 8 cm (3.1 in.) wide, and the pouring 1^n
perature of the molten steel is assumed to be about 3000C,
* It is to be noted that this point of view demands a temperature condition sym
metrical about the middle plane of the wall. That this is satisfied in the present
case, i.e., f(\) To, a constant, is evident.
84 HEAT CONDUCTION [CHAP. 7
while the shaft is heated to 500C (i.e., some preheating). It
is found that a temperature much above 700C (the "recales
cence point") modifies to some extent the character of the steel
of the shaft, and it is desired to know, then, to what depth this
temperature will penetrate, or, in other words, how far back
from the ends this overheating will extend.
We shall attempt only an approximate solution of this prob
lem, neglecting any changes that the thermal constants undergo
at higher temperatures, also radiation losses and other compli
cating factors, and shall interpret it as that of the introduction of
a "slab" of steel at 3000C between two infinite masses of steel
at 500C. Taking the origin in the middle and putting I = 4
and m = 4, (7.46) becomes, after shifting the temperature scale
500C,
200 =
V7T
t "
J(4
Our problem is then to find the largest value of x that will
satisfy the above relation, i.e. y that will afford a value of the
above integral equal to 20 <K250> or 016.
We can most conveniently arrive at a solution by the method
of trial and error. Thus, if x = 5, i.e., 1 cm from the original
end of the shaft, the limits of the above integral may be called
9rj and 17, and a little inspection of the table in Appendix D
shows that to give the integral the value 0.16, 17 must be either
0.018 or 0.994. For x = 10 the limits are 14ij and 617,
which necessitates 17 being either 0.019 or 0.165; and a few more
trials show that if x = 24.3, with corresponding limits of
28.3)7 and 20.3^, there is only a single value to be found for
77, and this is approximately equal to 0.029.
This, then, is the key to the solution, for the second and
larger of the two 77 values in the above pairs will evidently give
the shorter time, or, in other words, the time at which the point
first reaches this temperature. For the smaller values of x the
temperature goes higher than this value of 700C and later falls
to this point at a time afforded by the first value of 17. When
the two values are just equal, it means that the temperature just
reaches this value, and the time in this case will be crivfin hv
SEC. 7.10] LINEAR FLOW OF HEAT, I 85
t   4 X 0.121 X 0.029*  2 ' 46 S6C (6)
The overheating then extends in to 20.3 cm (8.0 in.) from the
end and reaches this point in 41 min.*
7.9. It is well to note in these, as in any other applications,
how the results would be affected by changes in the conditions
that enter. In the first case, for instance, it is readily seen
that the time will come out the same for any two temperatures
of the soil and concrete that have the same ratio; e.g., 2 and
+4, or 15 and +30. Moreover, a consideration of the
limits shows that the time is inversely proportional to the
diffusivity a. In the last illustration this same inverse propor
tionality of time and diffusivity also holds, and we can in addi
tion draw the rather striking conclusion that the depth to which
a given temperature will penetrate under such conditions is
independent of the thermal constants of the medium, f The
time it takes to reach this depth however, depends, as just men
tioned, on the diffusivity.
7.10. Problems
1. Show that if the initial temperature is everywhere To, a constant, the
temperature must always be TV
In this case T = ~ ~ eP dfl  T, (a)
(See Appendix D for values of the probability integral.)
2. Show that, if T is initially equal to x, it must always be equal to x\
and, if it is initially equal to x* t it will be x 2 + 2at at any time later.
3. In the application of Sec. 7.5 determine when the freezing temperature
will reach the center of the wall. Ans 4.8 days
4. A slab of molten lava at 1000C and 40 m, thick is intruded in the midst
of rock at 0C. What will be the temperatures at the center and sides of the
slab after cooling for 1 day and for 100 years? Use a = 0.0118 for both lava
and surrounding rock.
Ans. Center, 1000C and 183C; sides, 500C and 178C
*
* It is obvious that a more exact solution of this problem might be obtained by
a process of differentiation. This is left as an exercise for the ambitious reader.
t This is only true, of course, when the heated material introduced is of the
same character as the body itself.
86 HEAT CONDUCTION [CHAP. 7
6. Frozen soil at 6C is to be thawed by spreading over the surface a
15cm layer of hot ashes and cinders at 800C and then covering the surface
of this layer with insulating material to prevent heat loss. Taking the
diffusivity of soil and ashes as 0.0049 cgs and assuming that the latent heat of
fusion of the water content may be taken account of by supposing that the
soil has to be raised to, say, 5C instead of merely to zero, to produce melting,
how far will the thawing proceed in half a day?
SUGGESTION: Try x = 50 cm, 60 cm, etc. Note that the problem is equiva
lent to that for a slab of twice the thickness with ground on each side.
Arts. 45 cm, or x 60 cm
6. A metal bar (a = 0.173 cgs) I cm long, in which the temperatures
have reached a steady state with one end at 0C and the other at 100C, is
placed in endtoend contact between two very long similar bars at 0C.
Assuming that the surfaces of the bars are insulated to prevent loss of heat,
and taking the origin at the zero end of the middle bar, work out the formula
for the temperature at any point and apply it to a bar 100 cm long after 15 min
of cooling. Find the temperatures at the center, at the hot end, and at the
cold end. Ans. 49.75C, 42.95C, 7.05C
7. A great pile of soil (a = 0.0031 cgs) at 30 C is deposited on similar
soil at +2C. Latentheat considerations neglected, how long will it take the
zero temperature to penetrate to a depth of 1 m? Ans. 7.9 days
8. In the application of Sec. 7.8 compute the distance to which the tem
perature 1300C will penetrate. Ans. 2 cm
CASE II. SEMIINFINITE SOLID WITH ONE PLANE BOUNDING
FACE AT CONSTANT TEMPERATURE. INITIAL TEMPERATURE
DISTRIBUTION GIVEN
7.11. This is the case of the body extending to infinity in the
positive x direction only, and bounded by the yz plane, which is
kept at a constant temperature. The temperature for every
point (plane) of the body is given for the time t = 0.
7.12. Boundary at Zero Temperature. We have here to seek
a solution of
dT
~dt *
subject to the conditions T = at x = (a)
and T = f(x) when t = (6)
It is possible to treat this as a special form of Case I (Sees. 7.2
to 7.4) by imagining that for every positive (or negative) tem
perature at distance x there is an equal negative (or positive)
SEC. 7.12] LINEAR FLOW OF HEAT, 1 87
temperature at distance &. In other words, if there should
be a distribution of heat on the side of the negative x identical
with, but opposite in sign to, that on the positive side, the
flow of heat would be such as to keep the temperature of the yz
plane continually zero. A little thought on the symmetry of
such a temperature distribution will suffice to show that this
conclusion is sound; for there is no more reason for the boun
dary surface to take positive temperatures under these condi
tions than negative, and hence its temperature will be zero.
To express this condition mathematically, let us suppose that
for points on the positive side of the origin X = Xi, and on the
negative side X = X 2 . Then, Xi and X 2 are each essentially
positive, and the temperature /(X) can be expressed as /(Xi) for
the positive region and /(X 2 ) for the negative. Equation
(7.3d) can then be written for this case
T =
VTT LJo
(c)
the lower limit of the second integral being +00 instead of
oo , as it would be if X were the variable. But since the
value of a definite integral is independent of the variable of
integration (cf. Sees. 6.7 and 6.10), we can substitute X (or any
other symbol) for Xi and X 2 in the above equation, which can
then be reduced to
= d\ (d)
V7T JO
Making substitutions similar to (7.3e), viz.,
 (X  x)ri 0' s (X + z)i? ()
this becomes
T  U'AI + *
or, what amounts to the same thing,
88 HEAT CONDUCTION [CHAP. 7
It is well to assure ourselves that (g) is the required solution.
From the manner of its formation, t.e., originally from (7.2h)
and (7.2t), it must be a solution of (7. la), while for x = the
two integrals are evidently equal and opposite in sign; thus,
condition (a) is fulfilled. As to condition (&) we see that for
t = the second integral vanishes, and the whole expression
reduces to
"vTT J
7.13. Surface at Zero; Initial Temperature of Body TV An
interesting special case is that in which the initial temperature
is To throughout the body except at the yz surface, which is
kept at zero. /(X)  = / f + x\ or / (&  x j I then reduces to
To, so that (7.12?) becomes
v dp]
/
(a)
OT C
= i? /
V7T JO
7T xi
,
** d^ (c)
O
since e~*' is an even function (Sec. 6.10). Equation (c) will be
commonly written
T =
7.14. Surface at TV, Initial Temperature of Body Zero. By
an extension of (7.13d) we can handle this case at once. For
if (7.13d) is written for a negative initial temperature T 8 , we
have
(a)
and, if T 8 is then added to each side, we get
T  T, + T,  r.[l  *(*i)r W
* Those familiar with electric circuit theory will recognize that, for T, 1,
T IBB, sort of "indicial temperature," corresponding to the "indicial voltage" at a
point in a circuit due to unit voltage applied at the terminals.
SBC. 7.15]
LINEAR FLOW OF HEAT, I
89
This process is, of course, merely equivalent to shifting the
temperature scale, as we have had frequent occasion to do in
previous problems.
We can replace (6) and (7.13d) by a single equation of more
general usefulness than either, which applies to a body initially
T S T
Time Time
(a) (b)
FIG. 7.1. Cooling and heating curves.
at T and with surface at T 8 . Write (7.13d) for a surface tem
perature T 8 different from zero, i.e., shift the temperature scale.
Then
T  T 8 = (To  T t )*(n) (c)
T  T 8
Or jjn n?T = *(!?) (d)
This holds for either heating or cooling of the body. The quan
T T 8 / T 8 T\
tity m __ m ( = rp* _ m 1 is readily visualized from Fig. 7.1
as the fraction, at any time , of the maximum temperature
change that still remains to be completed. It is sometimes
useful to think of this as a new temperature scale that is inde
pendent of the magnitude of the degree in the scale used for T Q
and T 8 .
7.16. Law of Times. An interesting fact can be deduced
from (7.13c) and (7.14d), for it is easily seen that any particular
temperature T is attained at distances x\ and x 2 from the bound
90 HEAT CONDUCTION [CHAP. 7
ary surface in times ti and < 2 conditioned by the relation
(a)
 
This gives the law that the times required for any two points to
reach the same temperature are proportional to the squares of their
distances from the boundary plane, a statement that is true
whether the body is initially at a uniform temperature and the
surface at zero, or initially at zero and the surface heated, pro
vided only that the surface keeps its temperature constant in
each case.
It can also be at once deduced that the time required for any
point to reach a given temperature is inversely proportional to
the diffusivity a. Both these relations are of wide application,
and the one or the other of them holds good for a large number
of cases of heat conduction. We have already noted a case in
which the second law holds in Sec. 7.9.
7.16. Rate of Flow of Heat. We can now determine the rate
at which heat flows into or out of a body, initially at T Q and with
surface at T 8y through any unit of area of plane surface parallel
to the boundary. To do this differentiate (7.14c), using Appen
dix K. Then,
AT AT
V JL U JL \S\+sllJ *J\JL Q JL S ) 'I X*H* / \
dx d(xrj) dx V/TT
The rate of flow of heat into the body through any unit area
parallel to the yz boundary plane is then
or for the boundary plane x =
_ .r )ty _ k(T. Jo)
VTT Virat
To get the total heat inflow at the surface between times t\
SBC. 7.17) LINEAR FLOW OF HEAT, I 91
and <a we integrate (c) and get
7.17. Temperature of Surface of Contact. Suppose two
infinite bodies B and C of conductivities and diffusivities fci, i,
and & 2 , 2, respectively, each with a single plane surface and
with these surfaces placed in contact. Assume that B and C
are initially at temperatures T\ and T*, respectively, and imag
ine for the moment that the boundary surface is kept, either
by the continuous addition or subtraction of heat, at the con
stant temperature T 8 , where TI > T 8 > T^ We shall deter
mine what conditions must be fulfilled that this surface of
contact may receive as much heat from one body as it loses to
the other and hence will require no gain or loss of heat from
the outside to keep constantly at T; in other words, we shall
determine this temperature of the surface of contact.
Each unit of area of surface of contact receives heat from B
at the rate [see (7.16c)]
while it loses to C at the rate
Then, if these two are equal, the boundary plane will neither
gain nor lose heat permanently and hence will remain constant
in temperature. Thus,
or
If ki fc 2 and a v  a 2 , T 9 = (Ti + T 2 )/2, as we should
expect. The same holds if
92 HEAT CONDUCTION [CHAP. 7
APPLICATIONS
7.18. Concrete. In a fire test the surface of a large mass of
concrete (a = 0.030 fph) was heated to 900F; how long should
it take the temperature 212F to penetrate 1 ft if the initial
temperature of the mass was 70F?
From (7.14d) we have
212  900 , , N . /2.89\ , ,
70  900 " *<*">  * (vj) (a)
from which we get, using Appendix D, t = 8.9 hr.
7.19. Soil. How far will the freezing temperature penetrate
in 24 hr in soil (a = 0.0049 cgs) at 5C if the surface is lowered
to 10C?
Using (7.14d), H "j" ^ = *(ij) = * ( ~ ) (a)
from which we get x = 28.2 cm. For twice this depth it would
take 4 days, three times, 9 days, etc.
If the initial temperature of soil is 2C (35.6F) and the
surface is cooled to 24C ( 11F), how long will it be before
the temperature will fall to zero at the depth of 1 m?
2^ 6 = $(3^) ; t = 326,000 sec = 3.8 days (6)
Since no account has been taken of the latent heat of freezing
for the moisture of the soil in the last two problems, the distance
in the first problem is undoubtedly too large, and the time in the
second too small, for the actual case. Even in the case of con
crete, unless it is old and thoroughly dry, there is a considerable
lag in the heating effect as the boiling point is passed, showing
latentheat effects.
An exact treatment of these latentheat considerations must
be reserved for Chap. 10, but in the following problem an
approximate solution for a particular case is suggested.
7.20. The Thawing of Frozen Soil. Soil at 6C (21F), of
diffusivity 0.0049 cgs and moisture content 3 per cent, is to be
thawed by heating the surface with a coke fire to 800C (1472F).
The question is: How far will the thawing proceed in a given
time?
SBC. 7.21] LINEAR FLOW OF HEAT, I 93
To take account of the latent heat of fusion of the 3 per cent
moisture we note that, since the specific heat of such soil is taken
as 0.45 (undoubtedly, however, this is a rather high figure for
such small moisture content), the heat required to thaw this
moisture per gram of soil would be the same as that which
would raise this soil 0.03 X 80 4 0.45, or about 5C in tempera
ture. This is nearly equivalent to saying that the soil must be
raised to 5C (41F) to produce thawing, i.e., a total rise of
11C. Then,
11 = 806[1  $(si)] (a)
and we find that xi) *& x/(2 Vat) must be about 1.74, or
t = 0^95 = 16 ' 8 * (6)
Then for a thawing of 45 cm (1.5 ft), t = 34,000 sec, or 9.5 hr;
and for 90 cm (3 ft) 38 hr, etc.
While local conditions (varying diffusivities and moisture
contents) would alter these figures considerably, the law that
the time for thawing would vary as the square of the depth
holds good in any case in which the soil is initially at sensibly
the same temperature throughout. If it is not as cold below,
the thawing will proceed faster than this law would indicate.
7.21. Shrink Fittings. As a problem of a somewhat different
type from the preceding let us consider the thermal principles
involved in the removal by heating of a ring or collar that
has been shrunk on to a cylinder or wheel. If the thickness
is small compared with the diameter, it may be treated as a case
of onedimensional transmission, and as a very good example we
may cite the case of the locomotive tire. Suppose such a tire
7.62 cm (3 in.) thick is to be removed by heating its outer sur
face; let us question at what time the differential expansion of
tire and rim would be a maximum and hence the tire be most
readily removed. We shall assume that this differential expan
sion is determined by the magnitude of the temperature gradient
across the boundary of tire and rim. From (7.16a), putting
r.o,
<^=  *L
ax Viral
94
HEAT CONDUCTION
[CHAP. 7
To find when this is a maximum, differentiate with respect to t
and equate to zero. Then,
(^\
T,
(c)
So in this case (a = 0.121 cgs), t = 240 sec, or 4 min.
The above discussion of the problem is based on the condi
tions of Sec. 7.14, viz., for the surface heated suddenly to the
0.8
gO.6
.
2.
0.2
" 5 10 15 20 25 30
Time, minutes
FIG. 7.2. A type of theoretical temperaturetime curve obtained on the
assumptions of Sec. 7.21. (The more nearly the actual heating curve of the sur
face approaches this type, the better the case can be handled theoretically.)
temperature T 8 , as by immersion in a bath of molten metal. As
a matter of fact, the surface heating in the practical case would
generally be a more gradual process, brought about in many
cases by a gas flame. A rigorous solution of this complicated
problem is very difficult, but the following is offered as being a
SEC. 7.21] LINEAR FLOW OF HEAT, I 95
good approximate solution. Imagine in the case of the locomo
tive tire just considered that 5 cm thickness is added to the tire
and that the outer surface is, as before, suddenly raised to tem
perature T 8 . The temperature of the original surface will then
be given by (7.146) and will be found to rise gradually (see
Fig. 7.2), increasing more rapidly at first and more slowly
later, just as would be the case if this surface were flame heated.
By varying the thickness of metal that we are to assume added
(the 5 cm added in this case yields a very plausible curve) and
plotting the temperaturetime curve as in Fig. 7.2 for each case,
a result may be obtained very nearly like the actual heating
conditions.*
The problem is then reduced to the preceding, save that the
tire is imagined to be 5 cm thicker. The time comes out 11 min.
For a slower rate of heating the time would be correspondingly
longer.
A point of interest in this connection is a comparison of the
actual maximum temperature gradients for the rapid and slow
heating, for these are the measure of the ease 1 or the possibility
of removal of such a shrunk fitting. Putting t = 240 sec in
(c), we get (dT/dx)^  0.064 T 8 C/cm, while for t = 660 sec
[which is the case for the maximum gradient under the slower
heating (see Fig. 7.3)], the gradient is only O.OSSTVC/cm.
This shows that when difficulty is expected in the removal of
any shrunkon collar, the surface heating should be done as
quickly as possible, perhaps with the use of molten metal or
even thermit. The above calculations would also serve to show
the time for which it is desirable to continue this heating. From
* The reasoning involved here is as follows: If the outer surface A of this
imaginary 12.62cm (i.e., 7.62 + 5) tire is suddenly heated to T,, the initial tem
perature of tire and wheel being zero and the whole treated as a case of onedimen
sional flow (which is justifiable since we are concerned with only a relatively small
depth below the surface), the temperature of the original surface B will be some
t(t) as indicated in Fig. 7.2. This may be thought of as a boundary condition for
this original boundary J5. According to the uniqueness theorem (Sec. 2.6), then,
the temperatures inside i.e., at the "plane" across which we are getting the
temperature gradient, where the tire joins the rim are determined by this ^(p
irrespective of how it is brought about. It is therefore immaterial whether this
\KO is produced by gas heating at the original surface B of the 7.62cm tire or by a
sudden rise of temperature of the surface A of the 12.62cm tire.
96
HEAT CONDUCTION
[CHAP. 7
the shape of the curve in Fig. 7.3 it is evident that it is much
better to continue the heating too long than to cut it too short.
The considerations of this section would also apply to the
socalled "thermal test" of car wheels, which consists in heating
the rim of the wheel with molten metal for a given time. The
temperature gradient might reasonably be taken as a measure of
the stresses introduced in this way, and it could be determined
at once from (a).
0.04
u
CP
8
0.03
I 0.02
E0.01
10
20
40
50
60
30.
Ti'me, minutes
FIG. 7.3. Curve showing the variation of temperature gradient with time, at a
distance of 12.6 cm below a surface of steel suddenly heated to T 6 \ or 7.6 cm below a
surface heated as in Fig. 7.2. (The best time to attempt to remove the fitting
would be when the gradient sign is neglected here is a maximum.)
7.22. Hardening of Steel. A large ingot of steel (a = 0.121
cgs) at To has its surface suddenly chilled to T 8 . Let us discuss
the rate of cooling as a function of the time and of the depth in
the metal.
We shall do this by differentiating (7.14c) (see also Sec.
7.16), and we find
d r T 2 / x \ f jP  T }x6"~ x *^ a *
which is the formula from which the curves of Fig. 7.4 have
SBC. 7.22]
LINEAR FLOW OF HEAT, 1
97
been computed for depths of 0.3 and 1 cm. To apply to a
specific problem let us question what the rates of cooling are at
these depths if the initial temperature is 800C (1472F) and
the chilling temperature 20C (68F), the times being chosen
as those at which the metal is just cooling below the recalescence
point (about 700C or 1292F).
123.45
Time, seconds
FIG. 7.4. Curves showing rate of cooling at depths of 3 mm and 1 cm below the
surface of a steel ingot that is suddenly chilled. T\ is here T Q TV
To find the times, we put from (7.14d)
(6)
which gives t = 0.16 sec for x = 0.3 cm (0.12 in.), and t = 1.8
sec for x = 1 cm (0.39 in.). From (a) or from the curves we
then find the rates of cooling to be 920 and 82C/sec, respec
tively (1656 and 148F).
While it might be impossible in practice to attain as sudden
a chilling of the surface as the above theory supposes, the curves
98 HEAT CONDUCTION [CHAP. 7
of Fig. 7.4 will still serve to give a qualitative explanation of a
wellknown fact, iriz. 9 that the deeper it is desired to have
the metal hardened, the hotter it must be before quenching; but
that a comparatively small proportional increase in the initial
temperature may produce a considerable increase in the depth
of the hardening. To explain this it must be noted that one of
the factors in hardening is the rate of cooling past the recales
cence point. Now from the curves it may be seen that this
rate increases to a maximum and then falls off again; hence,
for maximum hardness at any given depth the initial tempera
ture should, if possible, be high enough so that the recalescence
point will not be passed until the rate of cooling has reached its
maximum value.
The rapid chilling of large ingots introduces temperature
stresses that frequently result in cracks. Taking the tempera
ture gradient as a measure of this tendency to crack, the subject
might be studied theoretically with the equations of the last
article.
7.23. Cooling of Lava. We now turn to some applications of
a geological nature, the first of which is the cooling of lava under
water. Suppose a thickness of, say, 20 m of lava at T Q (about
1000C) is flowed over rock at zero and immediately covered
with water perhaps it is ejected under water; what will be
its rate of cooling?
Since the water will soon cool the surface at least well below
the boiling point, the problem is that of the cooling of a semi
infinite medium with boundary at zero and initial temperature
conditions of T Q as far as x = Z, and zero from there on to
infinity. Formula (7.13a) is for the case where the initial con
dition is TO to infinity, and we may use it by splitting each
integral into two, according to the principles explained in Sec.
7.4, the second integral vanishing in each case, since the initial
temperature for it would be zero. We have as the formula,
then,
"<**> /<*+*)*
(a)
/ /"<
(
\y
xi,
Putting Kelvin's value of a = 0.0118 cgs for both lava and
underlying rock, the accompanying curves (Fig. 7.5) are com
SEC. 7.24]
LINEAR FLOW OF HEAT, I
99
puted for I 20 m. From the relationship between x and t
in the above limits we readily conclude that these same curves
apply to a layer n times as thick if the times are taken n 2 times
as large, and the distances n times as large.*
5 10 15 20 25 30
Depth, meters
FIG. 7.5. Temperature curves for a layer of lava. 20 m thick, after cooling under
water for various times.
7.24. The Cooling of the Earth. The problem of the cooling
of the earth and the estimate of its age based on such cooling
has been discussed by Kelvinf and others J as a special case of
* See Boydell," Berry, 13 and Levering 88 for more extensive treatments of this
problem.
t "Mathematical and Physical Papers," III, p. 295; Smithsonian Report, 1897,
p. 337.
% For a good r<$sum6 of the subject see Becker, 12 also Slichter, 188 Van Or
strand, 164 and Carslaw and Jaeger. 27 *
100 HEAT CONDUCTION [CHAP. 7
the solid with one plane bounding face; for it has been shown
that the error introduced in neglecting the curvature is quite
negligible. For this purpose the age of the earth is counted
from the assumed epoch of Leibnitz's consistentior status, when
the globe, or rather the crust, had attained a " state of greater
consistency " and the formation of the oceans became possible.
Kelvin's assumption for this state was an earth whose tempera
ture was in round numbers 3900C (7000F.) throughout. He
took the average value of the diffusivity as 0.01178 cgs,* and of
the present surface gradient of temperature as 1C in 27.76 m.f
The problem is then to find how long it would take for the
sarth at the assumed initial temperature, and with the surface
it a constant temperature approximately zero, to cool until the
jeothermal gradient at the surface has its present measured
vralue, viz., 1C in 27.76 m.
Differentiate (7.13c) (see Sec. 7.16). Then,
ar = 2To e~* 2/4q *
dx V*2Vrt (a)
/ AT\
and at x = i = i =
r
Putting in the constants given above, Kelvin got a value of
100 million years for the age of the earth, but because of the
uncertainty of the assumptions and data he placed the limits
at 20400 million years, later modifying them to 2040 million
years.
7.26. If the initial temperature of the earth, i.e., its tem
perature condition at the consistentior status, instead of being
Adams, 1 in his discussion of temperatures at moderate depths within the
earth, concludes that a 0.010 cgs is the best average for the surface rocks and
0.007 cgs for the deepseated material.
t Van Orstrand, 1 " 1 " who has made most extensive studies of crustal tempera
ture gradients, places the average for the United States between 1F in 60 ft and
1F in 110 ft (1C in 32.9 and 60.4 m). He states that, for a considerable portion
of the sedimentary areas of the globe, an average gradient of 1P in 50 ft (1C in
27.4 m) is found either at the surface or at depths of one or two miles.
SBC. 7.26] LINEAR FLOW OF HEAT, I 101
uniform throughout, increased with the depth, obeying the law*
T = /(*) mx+T. (a)
where T 8 is the initial surface temperature and m the initial
1200
10
20 30
70 60 90 100
40 50 60
Depth, kilometers
FIG. 7.6. Temperature curves for the earth, after cooling for the specified
number of years, assuming the initial conditions of Sec. 7.25. The smaller of the
two times is for a diffusivity of 0.0118 cgs (Kelvin), and the larger for 0.0064 cgs.
It is to be noted that the temperature state at depths greater than 100 km would
be very little affected by cooling for even 50 million years.
gradient, we can solve the problem with the aid of (7.120);
for substitution of (a) in this gives, after some simplification,
T = mx + TsQfr'n) (b)
Differentiating,
or
= 7<
dx
ira/dT
(c)
(d)
When m and x are zero, this reduces, as it should, to Kelvin's
solution (7.24c). As it stands, (d) affords a value for the age of
the earth, t y in terms of the geothermal gradient dT/dx at any
depth x, under the conditions that the initial temperature of
the earth increased uniformly toward the center from some
* Barus. 7
102 HEAT CONDUCTION [CHAP. 7
value T 8 at the surface, and that since that time the surface has
been kept at the constant temperature zero.
7.26. Effect of Radioactivity on the Cooling of the Earth.
Since the discovery of the continuous generation of heat by dis
integrating radioactive compounds, much speculation has been
indulged in as to the possible effect of such heat on the earth's
temperature.* Surface rocks show traces of radioactive mate
rials, and while the quantities thus found are very minute, the
aggregate amount is sufficient, if scattered with this density
throughout the earth, to supply, many times over, the present
yearly loss of heat. In fact, so much heat could be developed
in this way that it has been practically necessary to make the
assumption that the radioactive materials are limited in occur
rence to a surface shell only a few kilometers in thickness.
While a satisfactory mathematical treatment of this prob
lem is impossible with the meager data now available, it can
be seen at once that radioactivity would tend to retard the
cooling of the earth and hence increase our estimate of its age.
A rough idea of the extent to which this is true may be had by
assuming that one fourth of the present annual loss of heat is
due to this cause, and that the radioactive substances are con
tained in a very thin outer shell. The geothermal gradient at
the bottom of this shell will then be only three fourths of its
observed value on the surface, because only three fourths of the
heat that passes out from the earth crosses the lower surface.
Then, since from (7.25d) the age of the earth is inversely pro
portional to the square of the present gradient at x = Z, the
depth of the radioactive shell (if ra = 0, and I is small), this
would nearly double the calculated age of the earth.
7.27. The Effect of Radioactivity on Earth Temperatures;
Mathematical Treatment of a Special Case. While, as remarked
above, we know too little of the actual conditions as regards the
extent of distribution of radioactive substances in the earth to
attempt any rigorous or complete treatment of their effect on
the age and temperature of the earth, we can still solve the
problem for specially assumed conditions. The assumptions we
shall make are at least as consistent as any others with the
* Becker 11 and references in footnotes to Sec. 7.27.
SBC. 7.27] LINEAR FLOW OF HEAT, I 103
facts as we now know them. The first is that only a fraction,
1/n, of the total annual heat lost by the earth is due to radioac
tive causes. The rate of liberation of heat by the disintegration
of such substances is supposed to be independent of the time,
and the density of distribution of these heatproducing sub
stances is assumed to fall off exponentially with increasing depth
below the surface. It was mentioned above that some such
assumption as this is practically necessary, for if these sub
stances were scattered throughout the earth with their surface
density of distribution, vastly more heat would be generated
per year than is actually being conducted through the surface.
The second assumption concerns the initial temperature state
of the earth; i.e., its temperature distribution at the time of the
consistentior status. Instead of supposing, as in Kelvin's original
calculation, that the earth was at a constant temperature at this
time, we shall make the more reasonable assumption of Sec.
7.25, which is based upon data obtained by Barus,* showing the
relation of melting point to pressure to be nearly linear for a
considerable depth.
In solving the problem we must first modify our funda
mental conduction equation so as to take account of this con
tinuous internal generation of heat. We found in Chap. 2 that
the rate at which heat is added by conduction to any element
of volume dxdydz is kV 2 Tdxdydz. If in addition heat sources,
such as these radioactive products, produce an amount of heat
per second represented by iA(#,3/,3,0 dx dy dz, then the tempera
ture of this element will be raised at a rate dT/dt such that
n/77
kV*T dx dy dz + $(x,y,z,t) dx dydz =  cp dx dy dz (a)
Therefore, = VT + (6)
This is our fundamental equation. For linear flow it takes
the form
dT
dt
* See King.
104 HEAT CONDUCTION [CHAP. 7
In the present case the assumption is made that
*(*,0  Bf** (d)
where B is the quantity of heat generated per unit volume per
second at the surface. Separate determinations of this quantity
vary greatly, but the average result will be taken at 0.47 X 10~~ 12
cal/(cm 3 )(sec) for crustal rocks. The total amount of heat gen
erated in this way per second, and escaping through each square
centimeter of the earth's surface, is
* Wr = f~Be bx dx = ~ (a)
Jo o
But if w 8 is the total amount of heat lost by the surface per
square centimeter per second,
When n is assumed, this enables us to determine 6, since both
B and w 8 are known; i.e.,
i, nB ^
b = (a)
W 9 VJ/y
Our fundamental equation (c) then becomes
where C is written for B/cp. The solution of this equation must
satisfy the boundary conditions
T = at x = (t)
T = mx + T 8 when < = (j)
We shall first change (h) 9 by substitution, into a form that
is homogeneous and linear. Assume that
where u is some function of x and <. Then,
dT _ du 6 Z T d*u C
~dt = ITt' ~dx* = dx* ~ a'
SEC. 7.27) LINEAR FLOW OF HEAT, 1 106
and (h) becomes
du d*u *
The boundary conditions then become
u = at x = (n)
n
u = raz + T 8 + e~* x when / = (6)
Since the problem would be much easier to handle if the
first boundary condition were u = at x = 0, we shall make
the further substitution
. v  u  Wa (p)
which gives us, in place of (m),
dv d 2 v . .
and for boundary conditions
v = at x =
v =f(x) ^nu + JT. + f** when<=0 (r)
This now becomes the problem of Sec. 7.12, where was
obtained the solution
Substituting for/f + x\ and/f x ) from (r), this may
be written
mx _ p .
\ Xlj
f
, J^
"' ^  ^ r ^"'
106 HEAT CONDUCTION [CHAP. 7
Of the above four terms the first two can readily be shown
to equal
mx and r.  *(aij) (u)
respectively, while the third vanishes. In evaluating the fourth
we note that
/bft ., / 1 v r f b , v
~ 2 jo __ >,\2ny I 0~\2n ) rl R (n\
Q CvjJ ~~~ 6 i o ^ Q/B It/ y
/
Making use of this fact and of the substitution
7 == + p (w)
we have, finally, since
and
T =
k[ l ~
5;
When B = 0, i.e., when there is no radioactive material
present, this solution reduces, as it should, to (7.256).
A computation of the age of the earth has been made on the
basis of (z) for the following assumed conditions :
B = 0.47 X 10~ 12
w a = 1.285 X 10~~ 6 ; n = 4, i.e., onefourth of the present heat
loss is due to radioactivity; A; = 0.0045; c = 0.25; p = 2.8;
m = 0.00005; and T 8 = 995C. Then, the time required to cool
from the initial conditions* of surface at 995C and temperature
* Strictly speaking, the conditions are really for a temperature of 1000C at a
depth of 5 km below the surface, the surface itself being, in accordance with the
idea of the consistentior status, at or near zero in temperature. The above assump
SEC. 7.28] LINEAR FLOW OF HEAT, I 107
gradient of 5C per kilometer to a present surface gradient of
1C in 35 m comes out to be 45.85 X 10 6 years. Without radio
activity the same initial conditions give 22.0 X 10 6 years, so
we see that in this case the continuous generation of heat under
these conditions increases the computed age of the earth by over
100 per cent.
It may be added that since the estimates of the earth's age
based purely on refrigeration are of the same order of magnitude
as those arrived at from geological considerations, such as stratig
raphy, sodium denudation, etc., some geologists are inclined to
believe that radioactivity is not as important in this connection
as might be supposed; i.e., that it contributes not more than
about onefourth of the present total annual heat loss. If some
such small fraction of the total heat loss is attributed to radio
active causes, estimates of the earth's age based on cooling will
be in fair agreement with certain older geological estimates
although far short of the 2 X 10 9 years which represents the
present trend of thought.*
7.28. Problems
1. Show that, under the conditions of Sec. 7.12, if T is initially equal to x,
it will always be equal to x\ and if it is initially x z , its value at any time later
will be given by
2* . e*>*
\T
2. If the surface of dry soil (a 0.0031 cgs), initially at 2C throughout,
is lowered to 30C, how long will it be before the zero temperature will
penetrate to a depth of 10 cm? 1 m? (Cf. Problem 7, Sec. 7.10.)
Ans. 77 min; 5.4 days
3. An enormous mass of steel (fc 0.108, a = 0.121 cgs) at 100C, with
one plane face, is dropped into water at 10C. Assuming no convection
currents in the water (these would be minimized by choosing the face hori
zontal and on the under side), what will be the temperature of the surface of
tion of a surface initially at 995 Q C, which is then suddenly cooled to and thereafter
kept at zero, is made to render the problem mathematically simpler. That this
would not substantially affect the result may be concluded from the curves of Fig.
7.6.
* For more recent discussions of this subject the reader is referred to Slichter, 1 ' 8
Van Orstrand, 164 and Holmes, 66 all with good bibliographies. See also Lowan, 89
Bullard," Jeffreys, 70 and Joly. 7 *
108 EEA.T CONDUCTION [CHAP. 7
contact? How long will it be before a point 2 m inside the surface will fall in
temperature to 95C? Assume for water, k 0.00143, a  0.00143 cgs.
Ans. 90,2C; 4.4 days
4. In the preceding problem calculate at what rate heat is passing out
through each square meter of the boundary surface after 10 min.
Ans. 699 cal/sec
5. A 3,000lb motor car traveling 30 mph is stopped in 5 sec by four
brakes with brake bands of area 40 in. 2 each, pressing against steel (k = 26,
a = 0.48 fph) drums, each of the above area. Assuming that the brake lining
and drum surfaces are at the same temperature and that the heat is dissipated
by flowing through the surface of the drums (assumed very thick), what
maximum temperature rise might be expected?
SUGGESTION: Assume that this energy is converted into heat at a uniform
rate and that this heat flows into the drum from the surface at a rate given by
(7.16c). Compute the surface temperature T s for the largest value of t, i.e.,
5 sec. Ans. 132F*
6. Show by a method of reasoning similar to that of Sec. 7.12, that if the
plane surface of the solid is made impervious to heat, instead of being kept
at constant temperature, then
T = ~=. [ /(X) (e<*>'*' + <*+*>V) rf\
7. Water pipes are buried 1 m below the surface in concrete masonry
(a = 0.0058 cgs), the whole being at 8C. If the surface temperature is
lowered to 20C, how long will it be before the pipes are in danger of freez
ing? Ans. 9 days
8. If the initial temperature of the earth was 3900C. throughout and
it has been cooling 100 million years since then, with the surface at zero,
plot its present state of temperatures as a function of the distance below the
surface. (Use Kelvin's constants; i.e., a = 0.0118 cgs and k = 0.0042.)
9. Under the conditions of the previous problem compute the present
loss of heat per square centimeter of surface per year. How thick a layer of
ice would this melt? Ans. 47.8 cal; 0.65 cm
10. In some modern heating installations the heat is supplied by pipes in
the floor, e.g., in a concrete slab on the ground. Assuming that such floor is
in intimate contact (no insulation) with soil (A; = 0.5, a = 0.015 fph) initially
at a uniform temperature 20F lower than that of the pipe, calculate (Sec.
7.16) the rate of heat loss to the ground per square foot of floor area 100 hr
and also 10,000 hr after the start of heating. Also, calculate the total loss at
the end of these times. A large enough floor area to ensure linear flow is
assumed. Ans. 4.61 and 0.461 Btu/hr; 921 and 9210 Btu
* This is obviously too high since our calculation assumes this temperature
throughout the 5 sec. A somewhat better treatment is indicated in Problem 7
of Sec. 8.14.
CHAPTER 8
LINEAR FI,OW OF HEAT, H
In this chapter we shall continue the discussion of one
dimensional heat flow, taking up first the important matter of
heat sources and following this with a treatment of the slab or
plate and the radiating rod.
CASE III. HEAT SOURCES
8.1. We shall now make use of the conception of a heat
source, an idea that has been used very successfully by Lord
Kelvin 146 * and other writers in handling problems in heat flow.
If a certain amount of heat is suddenly developed in each unit
of area of a plane surface in a body, this surface becomes an
instantaneous source of heat, while if the heat is developed
continuously instead of suddenly, it is known as a continuous
source or permanent source.^
8.2. Let Q units of heat be suddenly generated on each unit
area of a plane in an infinite body, or on each unit area in some
cross section of a long rod whose surface is impervious to heat.
If the material is of specific heat c and density p, the unit of
heat will raise the unit volume of the material 1/cp degrees.
The quantity
is called the strength of this instantaneous source. If Q' units are
produced in each unit of time, then S' = Q'/cp is the strength of
the permanent source.
8.3. Plane Source. Regard the plane x = X over which the
instantaneous source of heat is spread as of thickness AX; then its
* "Mathematical and Physical Papers," II, p. 41 ff.
t The problem of Sec. 7.27 involved a special case of permanent sources with a
volume distribution.
110 HEAT CONDUCTION [CHAP. 8
temperature when the heat is suddenly generated will be raised by
5Zx  A degrees <>
and we have a case to be handled by (7.3d). The temperature
at point x will be given by
o /X+AX
T = A V / <T< x * )V dX  (6)
AX V7T J\
since /(X) = outside these limits of integration. Now let the
mean value of e~~ (X ~~ a:)21 ' 1 between the above limits be e( x '*) 2l *
where X < X' < (X + AX). Then,
T =  <r< x '*>'< 2 (c)
V7T
which, as AX > 0, approaches the limit
T = ^ 6 <**>v (d)
V7T
where the heat source is at a plane X distant from the origin.
Shifting this to the origin, (d) becomes
T  ^ <r*' f (e)
7T
If we have a permanent source of constant strength S'
located in a plane distant X from the origin, which begins to
liberate heat in a body initially at zero at time t = 0, we have
at any time t later the summation of each effect S = S' dr that
acted at a time t T previously, T being the time variable with
limits and t. Then, from (d)
If the permanent source is at the origin, the expression is
S'
2 Vwa
r
Jo
(9)
SBC. 8.51 LINEAR FLOW OF HEAT, II 111
Putting /3 ss x/2 \/a(t T), this becomes, for positive values
of x f
S'x f e~* 3 Q'x
T = ~
For the evaluation of this integral see Appendix B. See also
(9.12d) and (9.12e). For negative values of x the upper limit
is oo , giving the same value of T as for positive x.
8.4. Equation (8.3e) gives us temperatures at any point for
any time if we have a linear flow of heat from an instantaneous
source of strength S at the origin, the temperature of all other
parts being initially zero. It is well to test the correctness
of this solution by seeing if we can derive from it what is an
inevitable conclusion from the conditions given, viz., that the
total amount of heat in the material at any time is just equal
to the original amount Q (per unit area of section). From
(8.3e) the quantity of heat in any element dx is
Tcpdx = ^e x ^dx (a)
V7T
whence the total amount present in the body at any time is
represented by
/"" Tcpdx = % I * e x '*dx (6)
J   v TT J  
Since the additive effect of any number of such sources could
be obtained by a summation of such terms as (8.3d), the formula
(7.3d) may be regarded as applying to the case in which we
start with an instantaneous source of strength /(X) dX in each
element of length dX of the solid or bar in the x direction.
8.5. Since it appears on expanding (8.3e) in a series that
T = (x i& 0) when t = and also when t = oo , it must have
a maximum value at some time t\. To get this, differentiate
(8.3e) and equate to zero,
from which ^ ~ 2~
112 HEAT CONDUCTION [CHAP. 8
Putting this value of t in (8.3e), we get for the value of this
maximum
Ti  7= (c)
x V2ire
8.6. Use of Doublets. Semiinfinite Solid, Initially at Zero,
with Plane Face at Temperature F(t). We shall now solve,
with the aid of the concept of heat sources, an important prob
lem in linear flow. This is the case of the semiinfinite solid
initially at zero, whose boundary plane surface, instead of being
at a constant temperature as in Sec. 7.14, is now a function of
time.
We must find a solution of the conduction equation
dT d*T
subject to the conditions
T = when t = (a)
and T = F(t) at x = (6)
We shall solve this problem by the use of a concept known
as a "doublet." If a source and sink (negative source) of equal
strength S are made to approach each other, while keeping
constant the product of S and the distance 26 between them,
this combination, in the limit, is called a doublet of strength
Sd ss 2bS. With the aid of (8.3d) we may write at once the
expression for the temperature at any point x due to an instan
taneous doublet placed at the origin, i.e., with the two sources
at distance b on each side. This is
2 Vwat
.
(e 4at  e 4 < ) (c)
bx
= 7=e ** (e 2at  e 2 (d)
46 Vwat
Expanding e bx/2at and 6~ 6a?/2a< in a series (Appendix K) and divid
ing by 6, we find at once that the term in parentheses, divided
by 6, becomes x/at as 6 approaches zero. Then,
SBC. 8.6] LINEAR FLOW OF HEAT, II 113
For a permanent doublet of constant strength S' d located
at the origin, with its axis in the x direction, we have the sum
mation of the effects of each doublet element S' d dr that acted
at a time t r previously, r being the time variable (limits
and t) and t the time since the doublet was started. In this
case we have
r i
/
yo
4^) ( t _
For a permanent doublet of variable strength \l/(t) this becomes
x f l ~ x *
T = / \l/(r)e 4ia ^~ r ^ (t r)""^dr (g)
4 VTra 3 yo
which becomes, on writing
V ~2
18Z7
 r)
/ ~2 \
I ^ 2 d/8 (i)
This expression holds for positive values of x\ for negative values
the upper limit should be <*> .
Now if we suppose a permanent doublet of strength
^ = 2oF(Q placed at the origin, we have
We have in (j) an expression that, from the manner of its
formation, must be a solution of (7. la) a fact that can also
be readily proved by direct differentiation. It also satisfies
boundary conditions (a) and (6) and hence is the solution of our
problem. It is to be noted that we are here interested only in
positive values of x. If F(f) = T 8 , a constant, (j) reduces at
once to (7.146) as it should.
If the initial temperature of the semiinfinite solid is f(x)
instead of zero, the solution may be obtained! by adding to
(j) the equation (7.120), the solution for the case of initial tem
perature f(x) with boundary at zero.
* See Carslaw 27 ' pp  17  48 for a treatment of this problem by Duhamel's theorem.
t Carslaw and Jaeger. 17a ' < 6
114 HEAT CONDUCTION [CHAP. 8
APPLICATIONS
8.7. Electric Welding. Two round iron
(* = 0.15, c = 0.105, p  7.85 cgs)
bars 8 cm (3.1 in.) in diameter are being electrically welded end
to end. If a current of 30,000 amp at 4 volts is required for 4
sec and if this energy is supposed to be all developed at the
plane of contact, how far from the end will the temperature of
1200C (2192F) penetrate, if the initial temperature of the
bars is taken to be 0C?
The total heat developed will be
30,000 X 4 X 4 joules
480,000 . 480,000 .. _ , .
"
i.e., from (8.2a), S = 2760 cgs (6)
Hence, we have, from (8.5c),
120 o _ _=  2760
X ' 4.13
or x = 0.56 cm; i.e., the temperature of 1200C will penetrate
to a depth not greater than 0.56 cm (0.22 in.) somewhat less,
in fact, since the generation of heat is not instantaneous as the
solution assumes.
8.8. Casting. A large flat plate of ferrous metal (use k = 22,
c  0.15, p  480, heat of fusion = 90 fph) 1 in. (0.083 ft)
thick is being cast in a sand (k = 0.25, c = 0.24, p = 105,
a = 0.010 fph) mold. Assuming that the pouring temperature
is 2800F while the mold is at 80F, what will be the maximum
temperature rise in the mold 6 in. from the plate, and when will
this occur?
Because of the relatively high conductivity of the plate we
can neglect its thickness and consider it a plane source. Then,
Q 0.083 X 480 X 0.15 X 2720 + 0.083 X 480 X 90
19,830 Btu/ft 2 (a)
This gives a source in the sand of strength
, *** i *
S  0.24X105 ~ 787 fph
SBC. 8.10] LINEAR FLOW OF HEAT, II 115
Then from (8.5c)
787
Ti = n E xx A 1Q = 381F temperature rise (c)
U.O /\ 4r.lt>
giving a temperature in the sand of 461F. From (8.56) this
will occur at
'  rlfoi  12  5 to <
For half this distance away from the plate the temperature rise
would be twice as much and the corresponding time a quarter
as large as before.
The solution of the problem of Sec. 8.7 gives an idea of how
far from the welded joint one might expect to find the grain
of the material altered by overheating. From the second we
could draw some conclusion as to how near such a casting, wood,
say, might be safely located in the mold.
8.9. Temperatures in Decomposing Granite. We shall now
take up a problem involving permanent sources with a volume
distribution. While of some interest from the geological stand
point, it is difficult, and the solution of only one or two par
ticular cases will be attempted.*
It has been noted in some instances that areas of granite
undergoing decomposition are several degrees warmer than the
surrounding rock. It is known that granite gives out heat
during decomposition, the total amount being of the order of
100 cal/gm, but it is an extremely slow process, and our problem
is to see if any reasonable assumption of the rate at which
such heat is given off would serve to explain this increased
temperature.
8.10. To be able to treat the case as a specific problem we
shall assume first that the decomposing granite is in the form
of a wall of thickness Z, whose faces are kept at zero. Then if
q v cal/(sec)(cm 3 ) of the decomposing material are generated,
we have for our fundamental equation
ar
* Attention is called to the "step method" (Sees. 11.16 to 11.22) for the
approximate solution of problems like this, or even more complicated ones, by very
simple mathematics.
1 1 6 HE A T CONDUCTION [CHAP. 8
with boundary conditions
T = at x = and x = I (b)
and T when t = (c)
Let ^ T + (a?) (d)
where ^f(x) is a function of x (only), yet to be determined.
Replacing T by u  V(x) in (a),
f 
But if we determine ^(x) so that
*"(*)  f
or *(*) = + 6* + d
xl_ dtJ
then, dT "
To satisfy (6) and also make t* = at x = and x = ,
^(o:) must vanish at a: == and x = l\ therefore,
d = and 6 =  ^ (z)
^Qf
Then ^(x) = ^ (x 2  te) (j)
and w  T + ^ (a; 2  Zx) (fc)
or T = u + ~ (te  x 2 ) 0)
The solution of the problem is then merely a question of
determining u under the following conditions:
Fundamental equation, ^ = a % t (m)
Boundary conditions,
u *= at x and x = I
D
w /(x) =: (o: 2  /x) when / == fn)
iQL
This is nothing but the problem of the slab with faces at zero,
SEC. 8.10]
LINEAR FLOW OF HEAT, II
117
which will be treated in Case IV, next to be considered. While
in this particular example the form of f(x) makes the determi
nation of u a rather lengthy process, it offers no special difficul
ties and gives us as a final solution of tKe problem
L(
sin
mirx
r)
(o)
tn2pf 1
The curve of Fig. 8.1 has been computed with the use of the
equation above, the rate q v of heat generation being chosen so
120
100
80
Q.
E
Q>
20
~0 1 2 3 t 4 5 6 7
Time , yea rs
FIG. 8.1. Curve showing the relation between the filial temperature in the
center of a granite layer or wall 915 cm (30 ft) thick and the total time necessary
to effect its decomposition, computed for the conditions of Sec. 8.10.
that the entire process of decomposition with the resultant gen
eration of 100 cal/gm takes place in n years. The thickness of
granite is taken as 915 cm (30 ft), and the time chosen as that
for the completion of the process. The diffusivity is taken as
0.0155 cgs.
118 HEAT CONDUCTION [CHAP. 8
8.11. A second hypothetical case, much simpler than the
above, is as follows: Suppose that this wall or slab of decom
posing granite I cm thick is in contact on each side with ordi
nary granite. Suppose also that this slab is initially heated to
some temperature To about 50C above that of the surrounding
rock and allowed to cool for a year. This gives a temperature
at the center, as may be readily computed from (7.46), of 0.355 TO,
or about 17.7C above that of the surrounding rock at some
distance away. Now by differentiation of (7.46) with respect
to x and multiplication by 0.0081, the conductivity used here
for granite, we get the rate of heat flow out through each face
of this slab as
(1  er"') = 0.000057 cal/(cm 2 )(sec) (a)
V7T
for I = 915 cm.
So far we have taken no account cf the heat cf decomposition,
for the above discussion is merely to find a reasonable assump
tion for the temperature distribution in this slab and the sur
rounding rock as we find it at present. We may now question
at what rate decomposition would have to take place in order to
furnish heat at just the rate required to maintain this tempera
ture state steady for some time, and at once compute this rate
as such that the 100 cal would be liberated, i.e., the process
finished, in about sixtyeight years.
The preceding discussion should enable the geologist to form
some idea of the temperatures that might be caused by or
explained by decomposition. Since the rate of such decom
position is generally supposed to be very much slower than
that taken above, it is evident that a large thickness of such
decomposing granite would be required to cause even a few
degrees of excess temperature.*
8.12. Effect of Groundtemperature Fluctuations; Cold
Waves. Equation (8.6j) enables a more accurate calculation of
the effect of surface temperature fluctuations than is possible
on the assumption that they are simple sine variations as was
done in Sec. 5.10. As an example, suppose that a period of
* See Van Orstrand 162 in a discussion of a somewhat similar problem.
SBC. 8.13] LINEAR FLOW OF HEAT, II 119
uniform ground temperature, say 0C, is broken by a 3day
cold snap that causes a soil surface temperature of 12C for
this period, followed by a quick rise to the original 0C. What
is the temperature at a depth of 80 cm 5 days after the beginning
of the cold snap? Assume a = 0.006 cgs and neglect any
latentheat considerations.
Using (8.6j), put t = 432,000 sec and x = 80 cm. Note that
r[= t (# 2 /4a/3 2 )] is the time variable and that
save in the interval between r = and T = 259,200 sec when
it has the value 12C. For r = we have
x '* 
which gives ft = 0.786. Similarly, for r = 259,200 sec,
ft = 1.24
Our solution then is
2 /*l24
fp 1O v I y~/3 8 x7/Q O O/IO/^
f l* A 7= / e p ap = Z.2Q U
V7T .70.786
For cold or warm waves that are more complicated functions
of time the solution is most readily arrived at by using a block
curve for this function and evaluating the integral for the
various limits involved.
Note that for any value of the time less than 3 days in the
preceding problem the formula gives the same results as (7.14c),
as it should.
8.13. Postglacial Time Calculations. A question of con
siderable interest to geologists is the matter of time that has
elapsed since the last glacial sheet withdrew from any region.
Calculations of such have been carried out by Hotchkiss and
Ingersoll 57 with the aid of a series of carefully made temperature
measurements in the deep Calumet and Hecla copper mines at
Calumet, Mich.
Just as cold or hot waves produce an effect, though very
limited in depth, on subsurface temperatures, so the retreat
120 HEAT CONDUCTION [CHAP. 8
of the ice sheet many thousands of years ago was followed by a
warming of the surface that has produced a slight change in the
geothermal curve of temperature plotted against depth. This
change extends to thousands of feet below the surface. The
problem then is to calculate from the magnitude of this change
for various depths the time when the ice left and also the
general surface temperature changes that have taken place since
this time, i.e., the thermal history of the region.
It is assumed that the last ice sheet lasted so long that the
geothermal curve at its conclusion was a straight line and that
the surface temperature was the freezing point of water. We
shall show later how its slope is deduced. The present geo
thermal curve was determined by temperature measurements
made with special thermometers and under special conditions
at various depths reaching to nearly 6,000 ft below the surface.
It was necessary, in order to secure virginrock temperatures
unaffected by mining operations, to make measurements in
special drill holes run many feet deep into the sides of newly
made tunnels or "drifts" in which the rock surface had been
exposed for only a few days. The curve as finally obtained is
shown in the solid line of Fig. 8.2. The dashed line is the
assumed geothermal curve at the end of the ice age.
Equation (8.6,;) as it stands will not fit the boundary condi
tions of this problem, which are
T = F(f) at x = (a)
and T = Cx when t = (6)
x being the depth below the surface. However, the addition
of a term Cx to (8.6,7) gives the equation
~* 2 d/? (c)
which is readily seen to satisfy the conduction equation (7. la)
quite as well as (8.6j) and also the conditions (a) and (6). The
problem will be solved, then, when the form of F(t) is deter
mined, which, when inserted in (c), gives the best approxima
tion to the present form of the geothermal curve.
It is obvious that it is much simpler to evaluate the integral
SEC. 8.13]
LINEAR FLOW OF HEAT, II
121
in (c) if the F f t T~o2) is taken as a constant between certain
limits. This merely means the use of a block curve instead of
a smooth curve. For example, if it is assumed that the glacial
age ended 24,000 years ago and that the average surface tem
8
<u
CL
I
*
O
o
IUW 1
90F
80F
70F
60'F
50F
<
40F
j
30tf
d
.A
'/
^
&
*
^\/^'
W
/
/,,
//
^
/
/
f /
/
/
4)C
ft 1000ft 2000ft 3000ft 4000ft 5000ft GOOOfi
Depth 1829m
FIG. 8.2. Calumet and Hecla geo thermal curves.
perature was 8C for 18,000 years, followed by 6.83C (its
present value at this location) for the remaining 6,000 years
to the present time, (c) would read
(d)
2v / 24,000na
2\/6,OOOna
where n is the number of seconds in a year.
After a had been determined for two samples of the rock by
122
HEAT CONDUCTION
[CHAP. 8
the method of Sec. 12.6, nearly fifty assumed thermal histories
were tested by calculating values of T for each 500 ft (152 m)
in depth, using equations of the type of (d). The constant C
or slope of the assumed initial geothermal curve was determined
by substituting the observed value of T at 5,500 ft (1,676 m)
depth. This automatically makes the computed and observed
10
Time, years
30.000 20.000 10.000
Depth, ft
2000 4000 6000
10
i
emp
o
10
40.5
A
o o
0.5
ft
o
40.5
o
V
o
0.5 1
o.

40.5 1
C..,,,
o o

0.5
1 r
I
_
40.5
r>
n n
8
w w "000
i

0.5
1
i
1 1
30,000 20.000 10,000 2000 4000 6000
Time, years Depth, ft
FIG. 8.3. Four assumed thermal histories and resultant deviations from the
observed geothermal curve. Rock diffusivity taken as 0.0075 cgs.
value of T agree for the 5,500ft point, and they must also agree
at the surface, for one would naturally use 6.83C, the present
observed surface value, for the last part, at any rate, of the
thermal history. There will be slight but entirely inconse
quential variations in C, dependent on the thermal history used.
Four sample thermal histories are shown graphically in Fig.
8.3, as well as the resultant deviations from the observed geo
thermal curve. These are the differences between the values
of T calculated by an equation of the type of (d) for each thermal
history, and the observed values. In historv A the
SBC. 8.15] LINEAR FLOW OF HEAT, II 123
ice sheet was supposed to melt away from this region some
14,000 years ago with the present average surface temperature
of 6.83C dating from that time. In B the date was 26,000
years ago, and in C 20,000 years. In D the assumption is that
the ice ended 20,000 years ago and for 10,000 years the surface
averaged 10C in temperature, i.e., the climate was somewhat
warmer than at present. This was followed by 8,000 years at
5C, and then for 2,000 years to the present time the tempera
ture was 6.83C. This value of F(t) gave about the smallest
deviations of any tested and accordingly represents the best
conclusions one can draw from this work.
8.14. Problems
1. Derive (7.3d) and (7.12d) on the basis of heat sources (see Sec. 8.4).
2. In electrically welding two large iron (k = 0.15, c 0.105, p = 7.85
cgs) bars 2640 cal is suddenly developed in each square centimeter of contact
plane. If the initial temperature is 30 C, when will the maximum occur at
15 cm from this plane and what will be its value? Ans. 618 sec; 81.7C
3. A plate of lead (k = 0.083, c = 0.030, p = 11.3, latent heat of fusion
6 cgs) is cast in a sand (k = 0.0010, c = 0.25, p = 1.7 cgs) mold. If the mold
is initially at 25C while the lead is poured at 400C, what will be the maximum
temperature 3 cm away and when will this occur? The plate is 1 cm thick.
Ans. 62C; 1,913 sec
4. Show from (8.6t) that, if we have a permanent doublet of strength 2a T
at the origin, we get at once the solution of the case treated in Sec. 7.14 [Equa
tion (7.146)].
5. Soil (a = 0.015 fph) initially at 34F has its surface chilled to 16F for
two days, after which the surface returns to its original temperature. What
is the temperature 2 ft underground 3 days after the cold wave began?
Ans. 31.2F
6. A steel (a = 0.121 cgs) rod at 0C, whose sides are thermally insulated,
has its end suddenly heated by an electric arc to 1400C for 1 min and then
chilled again to 0C. What is the temperature 5 cm from the end 3 min
after the heating was started? Ans. 133C
7. Solve Problem 5 of Sec. 7.28 by the method of heat sources, using (8.3/)
or (8.3/0 an d assuming that the heat is generated at a uniform rate over the
5 sec. (Note that, since these equations assume heat flow in both directions,
we must use double the present rate of heat generation.) Ans. 84F
CASE IV. SOLID WITH Two PARALLEL BOUNDING PLANES
THE SLAB OK PLATE
8.15. In this case we have to deal with a body bounded by
two parallel planes distant I apart, with the initial temperature
124 HEAT CONDUCTION [CHAP. 8
condition of the body given. The problem is to find the subse
quent temperature for any point. The solution will of course
fit equally well the case of a short rod with protected surface.
8.16. Both Faces at Zero. The boundary conditions here are
T = at x = (a)
T = at x = I (V)
T = f(x) when t = (c)
Now we have already seen (Sec. 7.2) that
T = e~^ H sin yx (d)
and T = e~" yH cos yx (e)
are particular solutions of the fundamental equation
dT
m , .
o (7. la)
dt 2
Form (d) satisfies (a) for any value of 7, and also (6) if 7 = rmr/l
where m is a whole number. It does not, as it stands, fulfill (c),
but it may be possible to combine a number of terms like (d)
and secure an expression that will be a solution of (7. la) and
that satisfies (c). For
sin + B 2 e (1 sin
y
9W
3 e~^~ sin + (/)
is still a solution of (7. la), satisfying (a) and (6), which reduces,
when t = 0, to
m r , D . r> .
T = 5i sin y + jB 2 sin y h #3 sin i h ' ' ' (g)
and from Sec. 6.8 this equals f(x) if the function fulfills the
conditions of Sec. 6.1 between and l } and if
D 2 f r/x\ j\
= ] / /( x ) sm ~~
The solution of our problem then is
T
i
ml
2VF ::: ^r^ m
llf[ e smy
SEC. 8.171 LINEAR FLOW OF HEAT, II 125
If /(X) = To, a constant, and if the surfaces are at T t , we
may write from (t),
T  T. 
2 V f =sg I mirx] . ..
(T  T.) j 2, [ * m ~ (1  cos mr) sin yJ (;)
rn= 1
which holds for either heating or cooling. Only odd terms in m
are present; so we have, for the middle of the slab,
T T 4 / " T2a * 1 9* 2 << 1  25r crf
""
The series
 IT^ + 6""' 1 '  ([)
is evaluated in Appendix G (z obviously equals at // 2 ). For a
slab initially at zero, heated by surfaces at T 8 , (k) becomes
T c  T.[l  S(z)] (m)
while, for cooling from an initial temperature T G with surfaces
at zero, the equation is simply
T c = T Q S(z) (n)
8.17. Adiabatic Cases Slab with Nonconducting Faces. If
the faces instead of being kept at constant temperature are
impervious to heat, we shall have the same differential equation
but quite different boundary conditions; viz.,
n/p
^ = at x  (a)
*\m
fa = at x = I (b)
T = /(x) when t = (c)
Conditions (a) and (b) are fulfilled by solution (8.16e) if
rmr
^ = T
just as before, and (c) may be satisfied by combining a number
126 HEAT CONDUCTION [CHAP. 8
of terms of this type. This gives
8.18. If only one face is nonconducting, the other being kept
at zero, the solution is contained in equation (8.16z). This may
be shown by the same considerations that were used in Sec. 7.6,
i.e., by imagining a nonconducting plane cutting through the
center of a slab of double thickness, parallel to its faces, where
the temperature conditions are supposed perfectly symmetrical
on each side of such a plane. There would then be no tendency
to a flow of heat across such a surface, and hence placing a
nonconducting division plane there and removing half of the slab
will not affect the solution in any way. Therefore, in handling
a problem of this nature, i.e., one face impervious to heat,
we solve it as a case of a slab of twice the thickness, and the
temperatures of the nonconducting face would be found as those
at the middle of the slab of double thickness.
APPLICATIONS
8.19. The Theory of the Fireproof Wall. With the aid of the
foregoing deductions we can now develop a theory that finds
immediate application to a large number of practical problems,
viz., that of heat penetration into a slab or wall, one side of
which is subjected to sudden heating, as by fire; or, as we shall
call it for brevity, the " theory of the fireproof wall." It is to be
understood that this theory applies only to the purely thermal
aspects of the question of fireprotecting walls and floors and
not at all to the very important considerations of strength,
ability to withstand heating and quenching, and other questions
that must be largely determined by experiment.
We shall treat the problem for four cases of somewhat differ
ing conditions. It is assumed in all cases that the wall is rela
tively homogeneous in structure, a condition that would be
fulfilled by practically all masonry or concrete walls, floors, or
SEC. 8.21] LINEAR FLOW OF HEAT, II 127
chimneys. For hollow tiling or other cellular structure the
theory would not apply directly but would still afford at least
an indication of the laws for these cases. It is also assumed
that the wall is initially at about the same temperature through
out its thickness, as would be true in almost every practical
example. All temperatures are measured from the initial
"zero" of the wall.
8.20. Case A. The conditions assumed for this case are that
the front face of the wall is suddenly raised to the temperature
T 8 and maintained there, while the rear face is protected so that
it suffers no loss of heat. It is desired to know the rise in tem
perature of the rear face for various intervals of time. The
latter condition is fulfilled sufficiently well by a wall that is
backed by wood, i.e., door casing, or better by a concrete or
masonry floor on which is piled poorly conducting (e.g., com
bustible) material.
As explained in Sec. 8.18, such a case as this, involving an
impervious surface, can be treated as that of a slab of twice the
thickness, the rear (impervious) face of the wall corresponding
to the middle of the slab (x = %l). Accordingly (8.16m) gives
the expression for the rear face temperature, for a wall initially
at zero, i.e.,
T = T,[l  S(z)] (a)
where z = at/I 2 . Note that I in this case is twice the wall
thickness. Values of S(z) are given in Appendix G.
8.21. Case B. This differs from the preceding in that the
temperature of the front face is supposed to rise gradually
instead of suddenly. If the rise is rapid at first, as it would be
in most cases e.g., if the wall were exposed to a flame an
approximate solution may be arrived at by the device suggested
in discussing the removal of shrunkon fittings (Sec. 7.21), i.e.,
the assumption of an added thickness whose outer surface is sud
denly raised to, and kept at, a constant temperature T' t . By
properly choosing T' t as well as the thickness to be added, a
temperaturetime curve can be found for the plane representing
the original surface, nearly like many actual heating curves;
the computation is then carried out accordingly. The .results
128 HEAT CONDUCTION [CHAP. 8
obtained, however, are generally only slightly different from
those for Case A if the mean value of T s is used.
8.22. Case C. We have here an important difference to take
account of in the conditions. While the front surface is sup
posed to be suddenly brought to the temperature T 8 as in Case
A, the rear surface in the present case is supposed to lose heat
by radiation and convection instead of being protected, and
hence will not rise to as high a temperature as in Case A.
The rigorous handling of this problem is extremely difficult
and would be well beyond the limits of the present work, but,
as in many previous cases, it is still possible to reach a solution
accurate enough for all practical purposes, and at not too great
an expense of labor. This may be done as follows : In the treat
ment of the semiinfinite solid with boundary at zero (Sec. 7.12)
we found that the equations could be deduced from those for
the infinite solid by a suitable assumption for the temperatures
on the negative side of the origin, i.e., for /( X), the latter
being so determined that the boundary should remain constantly
at zero. Now if the boundary instead of being at zero radiates
with an emissivity ft, this condition can be introduced* by put
ting into the relation [identical with (7.3d)]
w
the condition that
/(X) = /(X)  2 r* x *f(y)<Fdv (b)
This gives the temperatures for a semiinfinite medium with
radiating surface and initial temperature conditions determined
for /(X). Now let us make the assumption that /(X) has the
value zero for a distance b from the radiating face, and 2T.
from there to infinity. This gives the somewhat complicated
equation
2T
T  =*
V7T
+ 2T.e (b+I t + & { 1  * [(b + x + 2 I at") ,] } (c)
* See WeberRiemann. 160  Art  8fl
SEC. 8.22]
LINEAR FLOW OF HEAT, II
129
and if we investigate with the aid of this equation the tempera
ture in the plane distant b from the radiating face, we find that,
for small values of h and not too small values of 6, this is almost
constant for a considerable time and has the value T 8 .
We have, then, the solution of our problem in the above
equation. This plane that is kept at T 8 corresponds to the
VJ.lt
J
x
/
^
y
/
fi 10
^
/
i
/
^
/
/
t
/
n in
*
>
> 0,10
/
y
y
/
c
1
V
/
2? AfiQ
v>
/
^_ u.uo
A
/
r
/
o
V
?
>
t
^
>
\j
r
e A flfi
/
/
o>
/
/
a.
1
*
1
y
/
F:
^
/
flfiA
/
/
.
s
/
/
/
.
n no
/
t
y
.
/
^
/
L/
^<
b
*
n
*
^
0.5
1.0
15
3.0
3.5
4.0
45
2.0 2.5
Time, hours
FIG. 8.4. Temperatures of the rear face of a concrete wall 20.3 cm (8 in.) thick,
whose front face is heated to T e ; computed for the conditions of Cases A and C.
Ordinates are fractions of T,.
front face of the wall whose thickness is 6, and the temperatures
of the rear or radiating face will be given by putting x = in
this equation. The value of the constant h may be taken for
small ranges of temperature at about 0.0003 cal/(sec)(cm 2 )(C)
above the temperature of the surroundings, for an average sur
face such as a wall (see Appendix A). Strong convection such
as a wind, or higher temperature differences, will increase this
figure considerably; in some cases, however, it may be even less
than the above value.
To gain some idea of the difference of the results for this case
130
HEAT CONDUCTION
[CHAP. 8
and for Case A, a few computations have been carried out with
(c) and plotted in Fig. (8.4). These are for a wall of concrete
(a = 0.0058 cgs) 20.3 cm (8 in.) thick, whose front face is
heated to T 8 . For 2 hr, under these conditions, the tempera
tures of the rear face for Case C are lower than they would be
for Case A in the ratio of 35 to 53.
0.5r
10 12 14 16
Time, hours
FIG. 8.5. Computed curves showing the rise in temperature of the rear faces
of walls of concrete (a = 0.0058 cgs), whose front faces are suddenly heated to,
and afterwards maintained at, T a . See Sees. 8.24 and 8.25. Ordinates are
fractions of TV
8.23. Case D. This differs from the last only in the supposi
tion that the temperature rises gradually instead of suddenly.
No attempt* will be made at treating this case mathematically,
but from the conclusions reached for Case B we are reasonably
safe in handling it as Case C, using a mean value for the tem
perature T 8 .
8.24. Discussion of the General Principles. Having treated
in detail the several cases, we may now draw some general con
clusions in regard to thermal insulation under fire conditions.
From the preceding discussion we see that Case A is the one
from which we can most safely make these deductions; for B and
* For a fairly approximate treatment the method used for Case B might be
followed; i.e., the assumption of a small added thickness.
SEC. 8.24]
LINEAR FLOW OF HEAT, II
131
D are more or less minor modifications, while C would invariably
lead to lower results. Hence, for a margin of safety we shall
make our deductions largely from (the ideal) Case A.
The first conclusion to be drawn from (8.20a) is that the tem
perature of the rear face is a function of a rather than of k. In
other words, the insulating value of material for such a wall is
dependent not alone on its conductivity, but rather on its con
ductivity divided by the product of its specific heat and density,
20
2 4 6 / 8
Time, hours
FIG. 8.6. Computed temperaturetime curves for the rear faces of walls of cinder
10 12
the rear faces
concrete (a = 0.0031 cgs). Ordinates are fractions of T 8 .
i.e., its diffusivity. Material for such purpose should there
fore have as low a conductivity and as high a density and specific
heat as possible, for if the density happens to be low, it may
prove no better insulator than something of higher conductivity
but of correspondingly higher density.
The second conclusion from (8.20a) is that any change that
alters t and I 2 in the same proportion does not affect the tem
perature T of the rear surface of the wall. In other words, for a
given temperature rise of the rear face the time will vary as the
square of the thickness. Since one measure of the effectiveness
of such a fireproof wall or floor would be the time to which it
would delay the penetration of a dangerously high temperature
to the rear face, this makes the efficiency of such a wall or floor
proportional to the square of its thickness (cf. the "law of
times" in Sec. 7.15).
132
HEAT CONDUCTION
[CHAP. 8
These conclusions are represented graphically in the curves
of Figs. 8.5 to 8.7. The temperature T of the rear face of a
wall whose front face is at T 8 is expressed for various times and
thicknesses of wall in fractions of T 8 .
14
16
2 4 6 8 10
Time , hours
FIG. 8.7. Computed temperaturetime curves for the rear faces of walls of building
2 4 6.8 10 12
npi
brick (a = 0.0050 cgs). Ordinates are fractions of jP.
8.26. Experimental. The following simple experimental
check on the preceding conclusions was tried by the authors:
A plate of hard unglazed porcelain 0.905 cm thick was heated
on one surface by the sudden application of hot mercury and
the temperature rise of the other surface, which was protected
from loss of heat by loose cotton wrappings, was measured with
a small thermoelement. The process was repeated for a similar
plate of thickness 1.780 cm, the temperatures being plotted in
Fig. 8.8. Since the diffusivity of the pprcelain was not known,
it was computed from the determination for the thinner plate
that T = y 2 T 3 at time 52 sec. This gives a = 0.0060 cgs,
and the two theoretical curves were computed from this value.
Two plates of each thickness were tested, and it is to be noted
that the agreement with the theoretical curve is at least as
close as that between the two sets of observations. The whole
is in reasonable agreement with the "law of times."
On a larger scale there are available the fire tests on various
walls made by R. L. Humphrey. 60 These were 2hr tests,
mostly on 8in. walls, the temperature T 8 of the front faces
SEC. 8.26]
LINEAR FLOW OF HEAT, II
133
being in the neighborhood of 700C. His results have been
plotted, where possible, in the curves of Figs. 8.5 to 8.7 being
denoted by the symbol H. The agreement, overlooking radia
tion losses, for the case of concrete is good.
1.2
1,0
0.8
 0.6
o.4
0.2
Theoretical.
I
8
10
234 567
Time, minutes
FIG. 8.8. Theoretical and observed temperaturetime curves for the rear
faces of miniature walls of porcelain (a = 0.0060 cgs), initially at zero, the tem
perature of whose front faces was suddenly raised to T 8 and maintained there
daring the experiment.
8.26. Molten metal Container; Firebrick. We may make
brief mention of a number of other problems to which the fore
going principles apply more or less directly. For example, take
the case of a container lined with magnesia firebrick 30.5 cm
thick, in which molten metal at an average temperature of
1300C is kept for two or three hours. How hot may the out
side of the brick be expected to get if the radiation from the
surface is small? Using a = 0.0074 cgs and I = 61 cm, we
find, with the aid of (8.16m) that the temperature of the out
side would be expected to rise only 8C in 2 hr while in 4 hr
it should not exceed 95C.
In a number of practical cases it is desirable to know to
what extent and how rapidly the temperature in the inside of
a brick follows that of the outside. This is of particular interest
in connection with the burning of brick and also in the case of
134 HEAT CONDUCTION [CHAP. 8
the "regenerator," where heat from flue gases is stored up in a
checkerwork wall of firebrick, to be utilized shortly in heating
other gases. Using a = 0.0074 cgs, we find that the center of
such a brick 6*35 cm (2.5 in.) thick the larger dimensions
being of little influence if the two flat sides are exposed (but
see Sec. 9.44) will rise in 5 min to 0.26 of the temperature of
the faces, in 10 min to 0.57, and in 20 min to 0.85. For building
brick of perhaps twothirds this diffusivity the figures would be
0.12 for 5 min, 0.38 for 10 min, and 0.70 for 20 min.
8.27. Optical Mirrors. In the process of finishing huge
telescopic mirrors it is necessary that they be allowed to remain
in a constanttemperature room before testing, until the glass is
at sensibly the same temperature throughout. For such a glass
(a = 0.0057 cgs) mirror 25 cm thick we can calculate from
(8.16m) that if the surface temperature is changed by T s the
change at the center is 90 per cent of this after 7.8 hr. For 14.2
hr the figure would be 98.7 per cent.
8.28. Vulcanizing. The process of vulcanizing tires lends
itself to some theoretical treatment along the preceding lines,
in spite of the fact that the "slab" involved here, i.e., the carcass
of the tire, is sharply curved, with radius of only a few inches
in some cases. We may question how long it would take for
the central layer of a tire initially at 30C to reach 120C if
the steam temperature in the forms on each side is 140C.
Assume a tire thickness of 16 mm and a diffusivity of 0.001 cgs.
Then, from (8.16A;) we have
. 120  140 = /Q.001A
30 140 * \ 2.56 /
Using Appendix G, we find t = 506 sec.
8.29. Fireproof Containers; Annealing Castings. While a
large number of other applications of the foregoing theory might
be mentioned, such as numerous cases of fireplace insulation,
resistorfurnace insulation, fireproofsafe construction, and the
like, we shall content ourselves with only one or two more
examples.
The first is the matter of a fireproof container made with a
thickness of 3 in. of special cement (use a = 0.012 fph). If
the front surface is raised to 500F, how long would it be before
SBC. 8.30] LINEAR FLOW OF HEAT, II 136
the inside surface, considered as adiabatic, would reach 300F,
assuming an initial temperature of 70F? Using (8.16&), we
have at once (300  500)/(70  500) =S(cd/l*). From Appen
dix G we have 0.012f/0.5 2 = 0.102, or t = 2.1 br.
A second problem is that of annealing castings; i.e., the
question of how long the heating must continue to bring the
interior to the desired temperature. We may readily compute
that for a metal casting (a = 0.173 cgs) in the form of a plate
30.5 cm or 1 ft in thickness it would take 23 min for the center
to rise to within 90 per cent of the temperature of the faces,
provided these were quickly raised to their final temperature.
For a plate of half this thickness it would take only onequarter
the time. If the faces were gradually heated, the process
would take longer, but the difference between the outside and
inside temperatures would be lessened.
8.30. Problems
1. A plate of steel (a = 0.121 cgs) of thickness 2.54 cm and temperature
0C is to be tempered by immersion in a bath of stirred inolten metal at T 9 .
How long should it be left to assure that the steel is throughout within 98 per
cent of this higher temperature? Ans. 23 sec
2. A fireplace is insulated from wood by 15 cm of firebrick (a = 0.0074 cgs).
If the face is kept for some time at 425C, how long will it be before the wood
at the rear will char, supposing this to occur at 275C? Initial temperature is
25C. How long for a thickness of 25 cm? Ans. 4.2 hr; 11.6 hr
3. A 2cm thick rubber (a 0.001 cgs) tire is to be vulcanized at 150C,
initial temperature being 20C. How long will it be before the center will
attain 145C? Ans. 1,420 sec
4. Compare the results for the three following problems based on Cases I
and II of Chap. 7 and Case IV of this chapter. A plate of copper (k = 0.918,
c = 0.0914, p = 8.88, a = 1.133 cgs) 10 cm thick and at T Q is placed between
two large slabs of similar material at zero; how long will it be before the center
will fall in temperature to H^o? If instead of a plate we have a large mass
originally at To, while the surface is afterward kept at zero, how long will it be
before the temperature 5 cm in from the surface will fall to H^o? If the slab
is of the same thickness as in the first case, but the faces are kept at zero, solve
this problem for the center. Ans. 24.3 sec; 24.3 sec; 8.3 sec
5. A sheet of ice (k = 0.0052, c = 0.502, p = 0.92, a 0.0112 cgs) 5 cm
thick, in which the temperature varies uniformly from zero on one face to
20C on the other, has its faces protected by an impervious covering. What
will be the temperature of each face after 10 min?
Ans. 10.56C and 9.44C
136 HEAT CONDUCTION [CHAP. 8
CASE V. LONG ROD WITH RADIATING SURFACE
8.31. This differs from Cases I and II of Chap. 7 in that
there is a continual loss of heat by radiation from the surface of
the rod. We have already handled the steady state for this case
in Sees. 3.5 to 3.8, where we found that the Fourier equation had
to be modified by the addition of a term taking account of the
radiation and became
AT /J2T
^7 =<*jrt b * T < a >
dt dx 2
We shall assume as before that the rod is so thin that the
temperature is sensibly uniform over the cross section, and that
the surroundings are at zero.
8.32. Initial Temperature Distribution Given. We must
seek a solution of (8.31a), subject to the conditions
T = f(x) when t = (a)
T = when 2 =00 (&)
Now the substitution T = ue~ bn (c)
reduces (8.31a) at once to ^ = a ~^ 2 ^
where u fulfills the condition
u = f(x) when t = (e)
and indirectly (6), since u is finite. But this is identical with
Case I; thus, the solution for u is given by (7.3/). Using this,
we may write at once
T = L^ t * f( x
V7T J  *
2/3
In other words, this differs from the nonradiating case only
by the factor e~ w .
8.33. One End of Rod at Zero; Initial Temperature Distribu
tion Given. The boundary conditions are
T = at x = (a)
T = f( x ) when t  (6)
SEC. 8.34] LINEAR FLOW OF HEAT, II 137
If we make the substitution (8.32c), then u must satisfy (8.32d)
and also the conditions
u == at x = (c)
u = f(x) when t = (d)
Since this is the case already treated in Sec. 7.12, we may write,
using (7.120),
T  %
8.34. End of Rod at Constant Temperature T 8 ; Initial Tem
perature of Rod Zero. We cannot solve this problem directly,
like the two preceding, as an extension of cases already worked
out; for the boundary condition T = T 8 at x = would mean
u = T s e bH at x = 0, which would not fit any case we have
treated. But we can handle this case with (8.33e) by the aid
of an ingenious device* whereby we first solve the problem for
the boundary conditions
T = at x = (a)
T = TVwVa when t = (6)
Applying (8.336) to this case, we get, on simplifying,
rr / *>x r oo fop / oo \
T = ~^(e^ I e( b ^+wdft  e^ I e^+wdp) (c)
V T \ Jxn J Xr, /
Now T = T.e**rf* (d)
is a particular solution of (8.31a), as is also (c) above. Thus,
the sum of (c) and (d),
T = T 8
bx bx
br
ex/a f * eV<* f
 / e~( b vt+0>*dp p /
V7T 7an V7T J a
is still a solution of (8.31a), which, moreover, fits our present
boundary conditions, viz.,
T  T 8 at x = (/)
T 7 = when t = (flr)
*Cf.
138 HEAT CONDUCTION [CHAP. 8
We may simplify this somewhat by writing
7 . 6 Vt + ft (h)
and hence dy = d@ (i)
in (e). This gives
bx bx
T = T, (e^ + * [ " e>'d>Y  ~ f " e ?'d T ) (j)
\ VlT JbVt+xn V7T J bVt^xr, /
8.36. A careful examination of this expression is worth while
to be sure that it is the desired solution. For t = (i.e.,
77 = oo ) and x ^ the lower limit of the first integral becomes
oo , hence the integral vanishes; in the second integral it becomes
oo , giving a value of VTT to the integral. Hence, f or t =
we have T = 0, as it should be for all cross sections of the rod
except the heated end. Since both integrals have the same
limiting value as x > 0, this gives the right temperature for the
end, viz., T = T 8 . Both integrals vanish for t = oo, and thus,
for the steady state, we have the result deduced in Sees. 3.5
and 3.7,
T = T 9 e~ M ^ (a)
From the value for 6 2 given in (3.6/), viz., ahp/kA, we see that
6 2 is very small if the emissivity is very small. Setting fe 2 =
in (8.34J), we get
I" e~*dy
VTTjxr, VlT J 
T T I 1 4 . / pi* d^v / t>~"** (\^ I (h}
j. ^ 8 i i "t~ / I & u i /~ I ** u i I \v)
\ V7T Jxr, V7T J XT, /
which is readily seen to be identical with the results of Sec. 7.14
for the linear flow of heat in an infinite body.
8.36. Problems
1. A wroughtiron (k =* 0.144, a = 0.173 cgs) rod 1 cm in diameter and
1 m long is shielded with an impervious covering and subjected to tempera
tures 0C and 100C at its ends, until a steady state is reached. The covering
is then removed and the rod placed in close contact at its ends with two long
similar rods at zero, the temperature of the air being zero also. If h is 0.0003
cgs, what will be the temperature at the middle of the meter rod after 15 min
(cf. Problem 6, Sec. 7.10)? Ans. 13.5C
2. Show that Case IV can also be applied to this problem of the radiating
rod.
CHAPTER 9
FLOW OF HEAT IN MORE THAN ONE DIMENSION
In this chapter we shall consider a few of the many heat
conduction problems involving more than one dimension. In
particular we shall take up the case of the radial flow of
heat, including heat sources, "cooling of the sphere," and
cylindricalflow problems; also, the general case of threedimen
sional conduction.
CASE I. RADIAL FLOW. INITIAL TEMPERATURE GIVEN AS A
FUNCTION OF THE DISTANCE FROM A FIXED POINT
9.1. This is the case analogous to the first discussed under
linear flow in Chap. 7, but with the essential difference that the
isothermal surfaces instead of being plane are here spherical.
In the discussion of the steady state for radial flow (Sec. 4.5), we
had occasion to express Fourier's equation in terms of the
variable r, finding that
V 2 ? 7 = ^ ' (a)
r 5r 2
the partial notation being used here to show differentiation with
respect to r alone, T now depending on t as well; thus, the
fundamental equation becomes
dT ct d*(rT)
Qt = r ~^~ (6)
= (c)
or a Z (c)
The solution of our problem must satisfy this equation, and the
boundary condition
T = /(r) when t  (d)
Let u = rT ()
and our differential equation (c) reduces to
139
140 HEAT CONDUCTION [CHAP. 9
du
where u = rf(r) when t = (g)
and M = at r = (A)
u being always positive if T is taken as positive. But the solu
tion of (/) under these conditions will be identical to that for
the case of linear flow with one face at zero, treated in Sec. 7.12.
Using, as in this case, X as the variable of integration, and
remembering that when t =
u = X/(X) (t)
we have the temperature at any distance r from the point, given,
from (7.12d), by the equation
(j)
u = rT = 4= [ [ " X/(X)<r< x '>'"'dX  [ * X/(X)e< x + r >*' i dxl (j
VTrUo Jo J
With the substitutions
ft m (X  r)i7 or X = ^ + r
and ft' s (X + r)77 or X = ^  r (*)
77
this becomes
9.2. If the initial temperature is a constant, T Q , within a
sphere of radius R in the infinite solid, and zero everywhere out
side, the subsequent temperatures are given from (9.1j) by
 t R X<
Jo
(a)
or, from (9.1Z), by
This gives T 7 directly for all points save r = 0, where it becomes
SEC. 9.3] FLOW OF HEAT IN MORE THAN ONE DIMENSION 141
indeterminate and must then be evaluated by differentiation.
This gives for the center
APPLICATIONS
9.3. The Cooling of a Laccolith. By means of equation
(9.26) we can solve a problem of interest to geologists, viz., that
200
400
1400
1600
600 800 1000
Dfstonce from center, meters
FIG. 9.1. Computed temperature curves for a laccolith 1,000 m in radius, which
has been cooling from an initial temperature To for various periods of time. A
point 5 m from the boundary surface would reach its maximum temperature in
about 100 years, while at 100 m the maximum would not be reached for over 1,000
years.
of the cooling of a laccolith. This is a huge mass of igneous rock,
more or less spherical or lenticular in shape, which has been
intruded in a molten condition into the midst of a sedimentary
rock, e.g.y limestone. The importance of the formation, from a
geological standpoint, lies in the fact that ores are frequently
found in the region immediately adjoining the original surface
of the laccolith, and the conditions and time of cooling of the
142 HEAT CONDUCTION [CHAP. 9
igneous mass would naturally have a bearing on any explanation
of the deposit of such ores.
The temperature curves given in Fig. 9.1 were computed for
the following conditions: radius R of laccolith, 1,000 m; diffu
sivity = 0.0118 cgs. (Kelvin's estimate. This is also not far
from the mean of the values for granite and limestone; the
medium must here be assumed to be uniform.) The initial
temperature of the igneous rock is taken as T Q , probably between
1000 and 2000C, while the surrounding rock is assumed at zero.
The conclusions to be drawn from the curves are (1) that the
cooling is a very slow process, occupying tens of thousands of
years; (2) that the boundarysurface temperature quickly falls
to half* the initial value and then cools only slowly, and also
that for a hundred or more years there is a large temperature
gradient over only a few meters and a very slow progress of the
heat wave; (3) that the maximum temperature in the limestone,
or the crest (so to speak) of the heat wave, travels outward only
a few centimeters a year. The mass behind it will then suffer
a contraction as soon as it begins to cool, and the cracking and
introduction of mineralbearing material! is doubtless a con
sequence of this.
9.4. Problems
1. Molten copper at 1085C is suddenly poured into a spherical cavity in a
large mass of copper at 0C. If the radius of the cavity is 20 cm, find the
temperature at a point 10 cm from the center after 5 min. Also, solve for
center. Neglect latent heat of fusion and assume k = 0.92, a = 1.133 cgs.
Ans. 103C; center, 109C
2. Show that
T  = U  *l(r  B)iiH t forr^B (a)
is a solution of the problem of the temperature in an infinite medium, initially
at zero, which has a spherical cavity of radius R with surface kept at T, from
time t = 0.
SUGGESTION: Show that u = rT is a solution of (9. 1/) and satisfies the
boundary conditions: u RT, at r = #; w = at r = oo ; u = when t = 0.
* The temperature of the boundary surface for the first hundred years or so
could best be estimated from (7.l7d). The error introduced by assuming the
diffusivities to be the same becomes less and less as the cooling proceeds.
f See Leith and Harder 84 and Jones. 78
t We are indebted to Professor Felix Adler for pointing out certain features of
this solution. See Carslaw and Jaeger. 87a  p  20 *
SEC. 9.5] FLOW OF HEAT IN MORE THAN ONE DIMENSION 143
3. Show, by evaluating dT/dr from (a), that the rate of heat inflow into
the medium at r = R in Problem 2 is
(6)
4. In the application x>f Sec. 4.10 find approximately how long it will take
for the steady state to be established. In doing this, calculate the rate of
heat inflow after 1 week, 1 month, 3 months, 1 year, and 10 years, assuming a
constant surface temperature of 200F below the initial lava (k = 1.2,
a = 0.03 fph) temperature.
Am. 24,200, 17,870, 15,450, 13,750, 12,600 Btu/hr. Steadystate rate is
12,050 Btu/hr
CASE II. HEAT SOURCES AND SINKS
9.5. Point Source. If Q units of heat are suddenly developed
at a point in the interior of a solid that is everywhere else at zero,
a radial flow will at once take place and the temperature at any
point for any subsequent time can be found in terms of the time
and the distance from this center. This case is analogous to that
discussed in Sec. 8.3, where we had a linear flow from an instan
taneous heat source located in a plane of infinitesimal thick
ness. Just as in this case, too, we can deduce the solution by a
special application of a more general one. For if in (9.2a) we let
the radius R of the spherical region, which is initially at constant
temperature T Q , become vanishingly small, while its initial tem
perature is correspondingly increased so as to make the amount
of heat finite, we shall have a solution of the present problem.
To get this, put
Q  ToCptfrR* (a)
as the amount of heat in a very small sphere of radius R, and
substitute the value of T deduced from this in (9.2a). Then,
 [
7o
(6)
Now we may write
e *ri* e
 X
(d)
144 HEAT CONDUCTION [CHAP. 9
since
<f 1 + a; + ~ + ' ' ' (6)
We can see by inspection the similar expression for e~~ (X + r ^\
Since X is a very small quantity in this integration, being confined
to the limits and R, (d) simplifies to
the effect of the other terms vanishing in the limit as R > 0, as
may be readily seen on inspection of (0) following:
Then, (&) becomes
By the same reasoning used in deriving (8.30) we can write
with the aid of (h) and ({) the expression for the temperature at a
distance r from a permanent source releasing Q' units of heat
per second (or hour if in fph), starting t sec (hr) ago, as
which reduces, on putting /3 = r/2 Va(t r), to
f M
T = \ \ e*dp =   <T**dft (K)
* ar J rrt K
_
or, writing S' = Q'/cp,
T. s '
If we put t = oo in the last equations, we have
_ a 1 _ Q' _ Q'
(m)
SBC. 9.6] FLOW OF HEAT IN MORE THAN ONE DIMENSION 145
as the temperature for the steady state in an infinite solid where
Q f units of heat are released per unit of time, at a point [ef.
(4.5p), noting that here q and Q' have the same value].
If the permanent source, instead of being of constant strength
Q'/cp, is of variable strength /(O, (j) becomes
T =
Equation (9.5i) shows that 5P has a value different from zero
in all parts of space even when t is exceedingly small, or, in other
words, that heat is propagated apparently with an infinite
velocity. As a matter of fact, the heat disturbance is undoubt
edly transmitted with great rapidity through the medium,
although it is continually losing so much energy to this medium,
which it has to heat up as it passes through, that the actual
amount of heat traveling any appreciable distance from the
source in a very short time is very small.
9.6. With (9.5t) derived, it may be instructive to reverse the
process and show that it is our desired solution. To do this we
must show that it satisfies (9.1c) and the boundary conditions
T = when = <*> (a)
T = when t = save at r = (6)
and also the condition that the total amount of heat at any time
shall equal Q.
Differentiation gives
W2 / 3 j
dt ~\ 2t + a t
<
3 /
dr* ~\
showing that (9.1c) is satisfied. That conditions (a) and (b)
are fulfilled may be shown if we rewrite that part of (9.5t) con
146 HEAT CONDUCTION (CHAP. 9
taining I,
1 1
The denominator is seen to be infinite for t = or > ; hence,
(9.5i) vanishes for each of these values. As to the last condition,
the total amount of heat is given by
f " P cT4irr*dr  f "
7o jo
Q

TT/
If we put 7 s rq (ft)
the second member becomes
* (i)
which (Appendix C) is equal to Q.
9.7. The time t\ at which T reaches its maximum value is
given by differentiating (9.5z) and equating to zero. This gives
The corresponding temperature is
Ti _ / IV Q S
9,8. Line Source. Point Source in a Plane Sheet. A line
source may be thought of as a continuous series of point sources
along an infinite straight line. The magnitude of each such
point source would be Q dz, where Q is the heat released per unit
length of line. Similarly, the strength is S dz. To get the effect
of such an instantaneous line source in an infinite medium,
initially at zero, at a point distant r from the line, we sum the
effects of terms like (9.5i) and get
 s (/=)' *" rv / " < r "" dz  ***** ( a )
\ V7T/ J  *
* It will appear in Sec. 9.41 that (a) and also (8.3e) and (9.6t) are special cases
of (9.41c). It may also be pointed out that (8.3e) is readily obtainable from (a) as
the summation of the effects of a continuous distribution of line sources in a plane.
SEC. 9.9] FLOW OF HEAT IN MORE THAN ONE DIMENSION 147
The flow of heat from a point source in a thin plane sheet or
lamina, if there is no radiation or other loss from the sides, may
be considered as a special case of line source, perpendicular to
the plane, since the heat flow is all normal to such line source,
i.e., radially in the plane. Equation (a) applies if we divide the
actual amount of heat released at the point by the thickness of
the sheet, so as to get Q (or S) for unit thickness, i.e., per unit
length of line source.
If the line source, or the point source in a plane, is a perma
nent one starting at zero time, and if the plane or medium is
everywhere initially at zero temperature, the temperature at
any later time t at any point may be written at once as
T =
or, putting ft m ^ / '^ _ (c)
we have T    /(n,) = 7(r,) (d)
where Q' is the number of heat units released per unit of time per
unit length of the line source. For values of this integral see
Appendix F.
It is of interest to calculate the rate of heat outflow for any
radius r\. To do this we must first differentiate (d), using
Appendix K, and get
d(rri) dr /
Then, the rate of heat outflow per unit length of cylinder at any
radius r\ would be
9.9. Synopsis of Source and Sink Equations. From Sees.
8.3, 9.5, and 9.8 we may write the general heatsource equation
(a)
* See Jahnke and Emde M ** 47 ~" "Mend. f or graphs of this function.
148 HEAT CONDUCTION (CHAP. 9
where T is the temperature in a medium initially at zero at dis
tance r from an instantaneous source of strength S at time t
after its release, n = 1 f or the linearflow case (Sec. 8.3), 2 for
the twodimensional case (Sec. 9.8), and 3 for the threedimen
sional case (Sec. 9.5). The three equations (a) are sometimes
referred to as the fundamental solutions of the heat conduction
equation.
For a permanent source the temperature at time t after its
start is given by
For the evaluation of this integral see Appendixes B, D, and
F.* Many illustrations of its use will be found in the following
applications, particularly in Sees. 9.119.12. Q' is expressed in
Btu/hr or cal/sec for the threedimensional case; in Btu/hr per
ft length or cal/sec per cm length for the line source or sink; and
in Btu/(hr)(ft 2 ) or cal/(sec)(cm 2 ) for the plane source or sink.
An inspection of the three integrals involved in (6) will show
that the only case in which there is a steady state is for n = 3.
For the other two cases, as t approaches infinity, T increases
indefinitely. For points very close to the plane source the tem
perature is roughly proportional to the square root of the time,
as shown in (9.12e), while for the line source the rise is slower.
Further study of (6) will show that the plane source is the only
case of the three that gives a finite temperature for r = 0.
If there are a number of sources in an infinite medium, the
temperature at any point is the sum of the effects due to each
source separately, making use of a principle we have already
applied many times.
An inspection of the way in which (9.5n) and (9.5o) are
obtained from (9.5j) and (9.5Z) will show at once how to modify
(6) to fit the case where a permanent source, instead of having
a constant strength S', is of variable strength /(i).
For an instantaneous source the time ti at which the maxi
mum temperature is reached at a point r distant, is, as deter
* See also (9.12d) for the integration for the plane source.
SEC. 9.10] FLOW OF HEAT IN MORE THAN ONE DIMENSION 149
mined by methods similar to those of Sec. 9.7,
while the corresponding value of this maximum temperature is
Tl = s
where n in all cases has the values given above.
APPLICATIONS
9.10. Subterranean Sources and Sinks; Geysers. The
foregoing source and sink equations have many interesting
applications, of which we shall consider a few in this and the
following sections.
1. Suppose heat is applied electrically or otherwise at the
bottom of a drill hole or well perhaps in an attempt to increase
the flow of oil at the rate of 360,000 Btu/hr. Take the thermal
constants of the rock as k = 1.2, c = 0.22, p = 168, a = 0.032
fph. What temperature rise might be expected at a distance of
15 ft from the source after 1,000 hr of heating? Using (9.5&)
or (9.96), we have
360,000
2ir* X 1.2 X 157 15/2V 32 M
= 1,592[1  $(1.33)] = 96F (a)
2. It was indicated in Sec. 4.10 how calculations could be
made on geysers, assuming that all the heat was supplied at the
bottom of the tube. It is probable, however, that cylindrical
flow more nearly fits the average case, and we shall make use in
this connection of (9.8d) or (9.96). Assume that in an old lava
bed (use k = 4.8 X 10~ 3 , c = 0.22, p = 2.7, a = 8.1 X 10~ 3
cgs) at 400C we have a geyser tube equivalent to a circular hole
of 30 cm radius and of such depth that the average water tem
perature at eruption is 140C. Equation (9.8d) gives the rela
tion between the temperature T, in a medium at zero, at a
distance r from a permanent line source or sink of strength S f
150 HEAT CONDUCTION [CHAP. 9
(per unit length) and the time t since it started. In handling
the problem we shall shift the temperature scale by 400C and
overlook the minus signs this involves.
We need not inquire for the moment what happens inside
r = 30 cm but will merely ask what constant strength of source
S' will result in a temperature T of 260C (i.e., 400 140) at
r = 30 cm, after a specified time that we shall take in this case
as 100 years, or 3.156 X 10 9 sec. Then, r/2 Vat = 2.96 X 10~ 3 ;
thus, we have
S' f " ~*
260 ~ o^ v n nnai / "~/T
ZTT X U.UUol 72.96x10' P
From Appendix F the integral evaluates as 5.54; thus, S' = 2.39.
This gives Q'(= S'cp) = 1.42 cal/sec per cm length of tube.
If the water enters the geyser tube at 20C, the heat required
per cm length of tube to start an eruption would be approxi
mately TT X 30 2 X 120 = 3.39 X 10 5 cal, giving a period of
2.4 X 10 6 sec or 67 hr. For 10,000 years this would work out
to 94 hr.*
We must now examine a little more closely just what we
have done in this solution. Equation (9. Be) gives the tempera
ture gradient at a distance r\ from the line source at time t,
and (9.8/) the rate of heat outflow or inflow through the cylin
drical surface of radius r\. It is evident then that the problem
of the line source emitting or absorbing Q' heat units per
unit time per unit length of source is, for values cf r equal to
or greater than r\, equivalent to that of a cylindrical source of
radius r\ emitting Q'e~ ri '" 8 heat units per unit time per unit length
of cylinder. In other words, we may regard (9.8e) and (9.8/)
as a boundary condition! for the medium (r 5 TI) that is the
* It is to be noted that these two calculations of period really apply to two
different geysers. The equations apply only to a permanent source or sink of
constant strength, and so what has been calculated here is not the increase in
period of a single geyser but the period of another of such constant strength of
sink (somewhat smaller than the other) that after 10,000 years the temperature at
r 30 cm is 260C below the initial value. The increase in period of a single
geyser would certainly be of this order of magnitude, but the exact calculations
would be difficult.
t Somewhat this same reasoning has already been used in the footnote of Sec.
7.21.
SBC. 9.11] FLOW OF HEAT IN MORE THAN ONE DIMENSION 151
same for either the line or cylindrical source. (The other
boundary conditions are T = everywhere in the medium at
t = 0, and T = at infinity.)
We see then that we have really solved the problem for an
ideal geyser whose rate of heat inflow from the surrounding
medium is determined by (9.8/). However, if we calculate
(9.8/) for 1 year we have, since here ty 2 = 1/(1.02 X 10 6 ),
q = Q' 6 9oon = Q'(l  9 X 10~ 4 ) (c)
This means that for r = 30 cm and for values of t greater than
1 year the rate of heat inflow would differ from Q' by less than
0.1 per cent.
3. As a third example of the use of source and sink equations
we shall inquire in connection with the application of Sec. 4.10
approximately how long before the condition indicated there,
i.e., the steady state, might be reached. Accordingly, we shall
calculate with the aid of (9.5fc) or (9.96) what temperatures
would be found 4 ft away from a permanent source (or sink)
generating (or absorbing) 12,050 Btu/hr after 1,000 hr. Using
k = 1.2, a = 0.03 fph, we have
T  rxTlfcprxw*/** ' 2 !1  * (0 ' 365)1
= 121F (d)
This means that the temperature at 4 ft distance is 121F cooler
than the original rock temperature of 500F. In 100,000 hr
the value is 192F or within 8F of the final temperature. We
may then conclude that anything approaching the steady state
in this case would take ten years or more. It is to be noted
that, until the steady state is reached, the same type of (justi
fiable) approximation is involved here as was investigated in the
preceding paragraph.*
9,11. Heat Sources for the Heat Pump. The heat pump is
one of the newest and most interesting developments in air
conditioning; it serves the dual purpose of heating a building
* See Sec. 9.4, Problem 4, for a treatment of this problem under slightly differ
ent assumptions.
152 HEAT CONDUCTION [CHAP.
in winter and cooling it in summer. Working in the reversed
thermodynamic cycle, like the ordinary electric refrigerator, it
absorbs heat from a cold body or region, adds to it by virtue
of the energy that must be supplied to operate the machinery,
and supplies this augmented energy to the building that is being
heated (winter operation). This energy may be three or four
times the electrical energy required and its operation is accord
ingly cheaper, in this ratio, than plain electric heating.
In the operation of the heat pump for heating in winter it is
necessary to have some outside medium from which heat can
be absorbed. In some installations the outside air is used, in
others well water or running water; but in an increasing number
of cases arrangement is made to abstract the heat from the
ground* itself. This means the installation of a considerable
length of pipe, small or large, in good thermal contact with the
ground below frost line or with the underlying rock, in which
fluid can be circulated. It is highly desirable to be able to cal
culate the temperatures that might be expected in such circu
lating fluid as dependent on the rate of heat withdrawal, the
time since the start of the operation, and the thermal constants
of the soil or rock, which is initially at a known temperature
assumed uniform but actually varying slightly, of course, with
depth.
This is essentially the problem of the line sink, and we shall
solve two special cases. The first is to calculate the tempera
tures that might be expected in an 8in.f diameter pipe if 50
Btu/hr per linear ft of pipe is abstracted from it. We shall use
as constants for the soil or rock k = 1.5 (high!), c = 0.45,
p = 103, a = 0.0324 fph. Temperatures are to be calculated
after 1 week, 1 month, and 6 months of operation at this average
rate of heat withdrawal.
Using (9.8d), we have for 1 week or 168 hr
205
f
}
0.333
2V0.0324X168
* See E. N. Kemler. 74
f The pipe dimensions given in this and the following sections are outside
diameters. For simplicity, round numbers, rather than standard pipe sizes, are
used in the illustrations.
SBC. 9.11] FLOW OF HEAT IN MORE THAN ONE DIMENSION 153
This gives, with the aid of Appendix F, T = 12.5F below the
initial soil temperature of perhaps 50F. The values for 1
month and 6 months are 16.4 and 21.4F, respectively. For a
2in. pipe four times as long (i.e., same surface) with the same
total heat withdrawal we have, for 1 week,
12.5
2w X 1.5
r ^^ = 1.337(0.0179) (6)
J 0.0833 P
2V0.0324X168
This gives a value of 5.02F below initial ground temperatures,
with values of 5.96 and 7.15F for 1 month and 6 months. Since
it is desirable to have a heat source that is no colder than neces
sary, it is evident that, for a given exposed surface, the long
small pipe is better than the shorter large one.
In applying the line source equation (9.8d) to this problem
we are making certain assumptions:
1. The pipe must be long enough so that the heat flow is all
normal to its length, i.e., radial. This would probably be
approximately true in most cases.
2. Since we really have a cylindrical source of radius n
instead of a true line source, we must, according to the consider
ations brought out in the latter part of Sec. 9.10, No. 2, assume
that the heat is absorbed, not at the rate Q', but at Q r e~ r ^\
For the 2in. pipe above treated this means that the absorption
rate should start at zero, rise to 0.8Q' in a quarter of an hour,
0.95Q' in one hour, and 0.99Q' in five hours. The difference
between the effect of this and a uniform rate Q' from the start
is inconsequential after the first half day's run with a small pipe,
but this period would be considerably longer for a large one.
Subject to the above conditions, (9.8d) would give, for
r 5 n, temperatures due to a single pipe in an infinite medium
initially everywhere at zero. If the medium is, say, 30 above
zero initially, this amount should be added to all temperatures
calculated with this equation; i.e., shift the scale as indicated in
the above examples. If the initial temperature varies with the
distance from the pipe, the effect of the pipe should be added to
the changes which would take place with time due to the initial
gradients, i.e., we use the sum of two separate solutions. If
154 HEAT CONDUCTION [CHAP. 9
there is more than one pipe the temperature at any point, e.g.,
the surface of % a pipe, would be the sum of the effects at that
point of each pipe.
If the pipe or pipes are near a ground surface kept at zero,
the problem may be solved by assuming, in addition, a (negative)
image of the pipe(s) above the ground surface. [This is essen
tially the principle used in deriving (7.12c).] If instead the
surface is impervious to heat, the solution would involve the
assumption of a positive image (see Sec. 7.28, Prob. 6). If, as
is usually the case, the surface undergoes seasonal temperature
variations, the temperature at any point would be the sum of the
effects due to the pipes with ground surface held at zero, plus
the effect of the seasonal variation at the point.
If Q' is not constant but varies from month to month, the
integral (9.8d) may be split into parts. If the effect is desired
at the end of 3 months of operation, we use the sum of three
integrals, with Q' in each case taken as the average for the cor
responding month. The limits in each case would be deter
mined by the times since the particular interval under consideration
began and ended. A study of (8.13d) will aid in clarifying this
point. Cases where the temperature varies markedly along the
pipe would present some special difficulties. It is possible that
the rigorous calculations of Kingston 77 on the cooling of con
crete dams (Sec. 9.14) could be applied to this problem.
Some of these same considerations may be applied to the
heat dissipation from underground power cables. However,
the relatively shallow depth, as well as other conditions, may
bring about an approximately steady state after a comparatively
short time of operation.
9.12. Spherical and Plane Sources for the Heat Pump.
While the long small buried pipe seems the most feasible
ground source for the heat pump, a number of other forms have
been suggested. One of these is the "buried cistern " or large
roughly spherical cavity deep in the ground. We shall make
some calculations for such a cavity of radius 5 ft, in soil of the
same highthermalconductivity constants (k = 1.5, a = 0.0324
fph) as used above. If we take the same rate of heat absorption
as already used, viz., 23.9 Btu/hr per ft 2 of surface (correspond
SEC. 9.12] FLOW OF HEAT IN MORE THAN ONE DIMENSION 155
ing to 50 Btu/hr per ft length of 8in. pipe), we get
Q' = 23.9 X 47r X 25 = 7510 Btu/hr
for the cavity. This corresponds to 150 ft of 8in. pipe or 600
ft of 2in. pipe. Using (9.96) with n = 3 and t = <*> (i.e.,
rj = 0), we have
m_ Q' 2 / ,o_ Q'
4vrkr
(a)
for the steady state. This is the same as (4.5p), since under
these conditions q and Q' have the same value. This gives, for
r = 5, i.e., the surface of the cavity, T 8 = 79.8F below the
initial temperature.
We shall now investigate conditions before the cavity reaches
a steady temperature state. The exact solution of the problem*
of what temperature on the surface of the cavity, as a function
of time, will give a uniform rate of heat absorption of 7510
Btu/hr is not easy. We can, however, readily solve two prob
lems closely related to this.
The first problem involves a uniform temperature of the
surface of the cavity. Its solution is reached by a simple
application of (9.46). Using this and taking T 8 as 79.8F below
the initial soil temperature, as used above for the steady state,
we have the following values for q, the rate of heat inflow:
16,600 Btu/hr at the end of 1 week; 11,870 Btu/hr at the end
of 1 month; and 9300 Btu/hr at the end of 6 months.
The second solution is somewhat more complicated. Here
we shall use (9.96) with n = 3, and differentiate it with respect
to r to get the temperature gradient and corresponding rate of
heat inflow for any radius r and time t. We must assume a
particular value of Q', which we shall choose the same as that
used above, viz., 7510 Btu/hr. The corresponding value of T
for the radius r in which we are particularly interested, i.e.,
5 ft, is obtained at once from (9.96). The result of this calcu
lation f will be a series of values of T& and g 6 for the cavity sur
face temperature and rate of heat inflow, for various times. If,
* In this connection see Carslaw. 17 * m
t In this connection examine again the reasoning in Example 2 of Sec. 9.10.
156 HEAT CONDUCTION [CHAP. 9
then, the rate of heat inflow is made to vary with time as indi
cated by these values, the surface temperature will take the
corresponding values. It is to be noted that this is a special
series of values of T& and ? 5 that is afforded by our pointheat
source theory. While neither this series nor the one given above
may fit the actual case of course, it must be remembered that
the values can be adjusted to any scale by the proper choice of
Q' the two solutions together should enable one to furnish an
approximate theoretical background for any practical case.
In reaching the second solution we first write (9.96) for n = 3,
which gives
o f 9 r
7= / ee*dp (b)
^TcJn
We then differentiate it [see Appendix K, also (9.8e)] and get
*L e r>* _ i A r e ^
Vx 6 rVTA, 6
dr
For 1 week or 168 hr, ij = 0.215, giving dT/dr = 8.2F/ft
This gives a rate of heat absorption at r = 5 ft of
ffj = 47r X 25 X 1.5 X 8.2 = 3870 Btu/hr
The corresponding temperature is, from (6), T = 10.3F below
the initial one of the surroundings. For 1 month these values
are 6840 Btu/hr and 37.4F, while for 6 months they are 7460
Btu/hr and 61.3F below the initial value.
Another type of heat absorber that has been suggested is the
plane. In its most feasible form this would probably be an
array of pipes looped back and forth in a plane, the spacing
being much less than would allow them to be considered inde
pendently as treated above. Putting n = 1 in (9.96) we have,
using Appendix B,
T, <*
/* p~P
.V*
2k VTT
For r this becomes
T , Q '  , (.)
2krj VV k VTT
SEC. 9.13] FLOW OF HEAT IN MORE THAN ONE DIMENSION 157
With Q r = 23.9 Btu/hr per ft 2 of surface, as used above, this
gives T = 21.0F below the initial temperature at the end of
1 week, 43.8F after 1 month, and 107.3F after 6 months.
If such a plane absorber is located near the surface of the ground
or below a basement floor, as has been suggested at times, the
heat flow might become mostly a onesided matter and, accord
ingly, the above temperatures would have to be almost doubled.
The relatively rapid lowering of temperature with time in
these two latter heat absorbers (not considering the steady state
that is eventually reached for the spherical cavity) is one of the
factors that point to the long small ground pipe perhaps in the
form of one or more vertical "wells" as perhaps the best type
of absorber or heat source that has been suggested.*
9.13. Electric Welding. A welding machine joining the
straight edges of two flat steel (k = 0.11, c = 0.12, p = 7.8,
a = 0.118 cgs) plates 8 mm (0.315 in.) thick uses 2000 cal per
cm length of weld. What maximum temperature will be
reached in the plate 5 cm (1.97 in.) from the weld and when?
Assuming that all the heat is retained in the plate, that half
flows in each direction, and that it is generated effectively
instantaneously, we have
Q (per cm 2 of the weld) = ^^ = 2,500
U.o
or A = 2,670. Then (9.9d), with n = 1, gives
Tl =  = 129 o C
5
25
and, from (9.9c), ti = o y o ifft == "^ sec ^)
As a second example, consider a spot welding operation
where 2,400 watts for 2 sec generates 4,800 joules or 1146 cal at
* Consideration, however, should be given to the fact that, if more heat is taken
from a system of deep vertical pipes in winter than is returned in summer, a pro
gressive lowering of deep earth temperatures may result in the course of years a
situation that might not be remedied by conduction in from the surface in summer.
This effect could be readily calculated for a period of years by using for Q f the aver
age for the year. Because of the slowness with which the integral I(x) increases
this progressive lowering would not be a serious matter for a single pipe. It would,
in any case, be markedly altered by even a small underground water movement.
158 HEAT CONDUCTION [CHAP. 9
a point in a steel plate 1.5 mm (0.059 in.) thick. What maxi
mum temperature is reached 4 cm (1.57 in.) away from the
point and when?
Using the above constants for steel, we find
7,630
(on the basis of unit thickness) and S = 8,170. Then using
n. = 2 in (9.9c) and (9.9d), we have
T >  ^  59 ' 9 c > *>  4X1HT8 = 33 ' 9 sec <>
It is evident that if these calculations are carried out for
points very close to the weld, the temperatures arrived at would
be far above the melting point of the metal. This simply means
that this is not really an instantaneous source of heat, nor is the
heat all delivered strictly at a line or point. Consequently, cal
culations cannot be made for such points with the equations
used above.
From a conduction standpoint the generation of heat in elec
trical contacts may be considered as a special case of spot weld
ing. For an approximate treatment we may assume that such
a contact is frequently, if not generally, shaped like the frustum
of a cone, with the heat generation at the tip. The cone can
be considered as part of a sphere, the fraction being determined
by the ratio of its solid angle to 47r. Temperatures resulting
from the sudden generation of a small amount of heat at the
tip can then be calculated from (9.5i) or (9.9a), or, for maximum
values (9.9d). It is evident, however, that in using these
equations the amount of heat Q must be taken as the heat gen
erated at the contact multiplied by the ratio of 4?r to the solid
angle of the cone. See footnote to Sec. 4.12, Problem 6.
9.14. Cooling of Concrete Dams. Because of the heat
released in the hydration of cement large masses of concrete, as
in dams, will rise many degrees in temperature unless special
cooling is provided. Without such artificial cooling the tern*
perature rise might be 50F or more; the heat would require
years to dissipate and the final inevitable contraction would
Sue. 9.14] FLOW OF HEAT IN MORE THAN ONE DIMENSION 159
cause extensive cracking. Rawhouser 117 has described the
methods used in cooling Boulder, Grand Coulee, and other dams,
and their results. This is accomplished by embedding 1 in.
(o.d.) pipes in the concrete about 5 or 6 ft apart and circulating
cold water through them for a month or two, beginning as soon
as the concrete is poured.
The problems involved in such conduction cooling have been
extensively studied by the U.S. Bureau of Reclamation engi
neers. 22 * The three following calculations are, by comparison,
crude and simple but not without interest since they arrive at
results of the right order of magnitude by relatively simple
means. We shall assume the pipes 6 ft apart and staggered so
that each pipe cools a cylinder of hexagonal section of area
31.2 ft 2 , equivalent to a circle of radius 3,15 ft. Take as thermal
constants of the concrete k = 1.4, c = 0.22, p = 154, a = 0.041
fph, and assume that the heat released by hydration is 6 cal/gm
or 10.8 Btu/lb, which would cause an adiabatic temperature
rise of 49F. This hydration heat amounts to 1663 Btu/ft 3 ;
thus, each foot of pipe must carry away 51,900 Btu.
Our first calculation will be only a rough approximation.
Assume that the heat is released at a uniform rate and carried
away as released (i.e., steady state) and that the mass of con
crete averages 15F in temperature above the cooling pipe.
Furthermore, since for such a small pipe (radius 0.0417 ft) the
temperature gradient is much the largest near the pipe, we
shall arbitrarily assume that the concrete temperature remains
uniformly 15F above the pipe at distances greater than 1 ft
from the pipe. We then get with the aid of (4.6/), as the heat
loss per foot of pipe,
2r X 1.4 X 15 A1 _ _ n
(2.303 log io 1/0.0417) = 4L5 BtU/hr
This would involve a total time of the order, of 1,250 hr or 52
days for the dissipation of all the heat.
Perhaps a better approximation is afforded by the following
treatment: Suppose the hydration heat of 1663 Btu/ft 3 is
released at a uniform rate so that the process is completed in
* See also Glover, 47 Rawhouser, 117 Kingston, 77 and Savage. 121
160 HEAT CONDUCTION [CHAP.
1,250 hr, which means a rate of heat development q v = 1.33
Btu/(ft 8 )(hr). Then for a foot length of pipe the rate of heat
flow through any annulus will be determined by the steady state
equation
/ r>2 2\
q = q v (irR 2  irr 2 } =
This means that for a cylinder of external radius R the heat
that flows through any annulus of average radius r and width A/
and is carried away at the center must be generated outside
the radius r. This heat will flow radially through area 2irr
(for unit length) under a gradient AT/Ar. This gives
/T^ n /V 2 /JD2 \
AT 7 = g I /L _ r \
2k Jn \r /
dr (c)
''
Using ri = 0.0417 ft and r 2 = R = 3.15 ft (see above), this
gives T 2  T l = 18.TF as against 15F for the simpler cal
culation above, for the completion of the process in the same
time.
The fundamental weakness of both the foregoing calculations
is the assumption that the hydration heat is released at a uni
form rate and over a period of a month or more. This, in
general, is not the case ; in fact, most of it may be released in the
first few days. We shall accordingly make another calculation,
based on (9.8d). This will give the temperature at any radius r,
t sec after a permanent line source (or sink) has been started in
a medium of uniform temperature. This assumes that the
medium has been rather quickly raised to this uniform tempera
ture by the release of the heat of hydration and then cools
according to the special conditions we shall lay down. While
these conditions apparently are not closely related to our prob
lem, we can get some information in this way that will be of
interest.
In applying this equation we shall withdraw heat at the
same rate as in the two preceding calculations, viz., 41.5 Btu/hr
SEC. 9.14] FLOW OF HEAT IN MORE THAN ONE DIMENSION 161
per ft of pipe. We shall then calculate the temperature of the
pipe necessary to do this, at the end of 1, 5, 10, 20, 40, and 60
days. Putting S' = 41.5/cp = 1.22 andr = 0.0417 ft in (9.8d),
we have, using Appendix F, T = 16.9F below the initial con
crete temperature at the end of the first day of cooling. The
values for 5, 10, 20, 40, and 60 days are 20.8, 22.4, 24.1, 25.7,
and 26.7F. The temperatures at radii 1, 2, and 3 ft at the end
of 10 days are 7.4, 4.3, and 2.7F. below the initial temperature,
and at the end of 50 days they are 11.2, 7.9, and 6.1F below
this temperature. From these figures we may conclude that a
1in. pipe held for 52 days at a temperature averaging 25F
below that of a large mass of concrete will withdraw some 51,900
Btu for each foot of pipe length. This is equivalent to the
heat of hydration in a cylinder of radius 3.15 ft. During this
time the temperature of the immediate surroundings ranges
from 2.7F below the initial value at 3 ft from the pipe after
10 days, to 11.2F below this value at 1 ft after 50 days averag
ing 15 to 20F above the pipe temperature. These figures are
of the order of magnitude encountered in practice.
When these three methods of calculation are compared, the
first two assume a uniform rate of heat release and the third a
sudden release that raises the mass to its maximum tempera
ture, after which cooling begins. Neither assumption fits the
actual case, which lies somewhere between the two. However,
all give results of the same order of magnitude, indicating that
the largest share of the heat might be withdrawn inside of two
months. As a matter of actual practice artificial cooling is
usually continued for from 1 to 3 months.
There are two obvious defects in the last solution. The
first is the rather trivial one discussed in Sec. 9.10, example 2.
The other and more serious one is the fact that it fails to take
proper account of^the action of neighboring pipes.* In reality
each pipe is in effect the center of a cylindrical column of con
crete of radius 3.15 ft with no heat transfer across the boundary
from one cylinder to another. This and other factors, such as
the inevitable rise in cooling water temperature as it flows
* See, however, the suggestions in Sec. 9.11 for treatment of an array of pipes.
162 HEAT CONDUCTION [CHAP. 9
through the pipes, are taken into account in the elaborate solu
tion of Kingston. 77
Some of these same considerations might be utilized in cool
ing calculations on certain types of uranium (fission) "piles."
9.16. Problems
1. A 50gm lead (c = 0.030; heat of fusion = 5.47 cgs) bullet is cast in an
iron (k = 0.144, c = 0.105, p = 7.85, a = 0.174 cgs) mold. Assuming the
pouring temperature as 350C and the mold at zero, find the temperature
3 cm away from the bullet after 10 sec; also find the maximum temperature.
Neglect dimensions of the bullet. Ans. 2.58C; 2.64C
2. If heat equivalent to the combustion of 10 6 kg of coal with a heat of
combustion of 7000 cal/gm is suddenly generated at a point in the earth, when
will the maximum temperature occur at a point 50 m distant, and what will be
its value? Assume k = 0.0045, a = 0.0064 cgs for the earth concerned.
Ans. 20.6 years ; 5.9C
3. If the coal of the previous problem burns at a rate of 1,000 kg per day,
what will the temperature be at a distance of 10 m from the point in 2 years?
[The use of (9.96) should be considered in connection with this and the follow
ing problems.] Ans. 38C
4. In a geyser of the type described in Sec. 9.10 make the calculation of
the period for t = 1,000 years. t Ans. 80 hr
5. In the second or spotwelding example of Sec. 9.13 assume that 200
cal/sec is generated at a point for a period of 10 sec. Calculate the tempera
ture 3 cm from this point at the end of this period. Ans. 54C
6. In the first illustration of Sec. 9.13 assume that the welcfing machine
generates 800 cal/sec per cm of weld for a period of 12 sec. What will be the
temperature 3 cm from the weld at the end of this period? Ans. 228C
7. A certain deep mine is to be airconditioned by the Abstraction of
60 Btu/hr for each linear foot of a circular shaft or tunnel 7 ft in diameter.
This is driven in rock (fc = 1.2, a 0.032 fph) initially ail at 110F. What
rockwall temperature might be expected after 10 years of such cooling?
Ans. 85F
8. In a heatpump installation using a 1in. diameter ground pipe in soil
(k 1.0, a = 0.02 fph) at a uniform initial temperature of 50F, heat is
withdrawn at an average rate of 10 Btu/hr per linear ft of pipe. What tem
perature might be expected in the pipe after 2 months of operation?
Ans. 42F
CASE III. COOLING OF A SPHEBE WITH SURFACE AT CONSTANT
TEMPERATURE
9.16. Surface at Zero. To solve this problem we must find a
solution of (9.1c) that satisfies the boundary conditions
SEC. 9.17] FLOW OF HEAT IN MORE THAN ONE DIMENSION 163
T  /(r) when i  (a)
T = at r = R (6)
Making the substitution u =E rT (c)
/^ . \ i x dw
(9.1c) reduces to ^ = <
where ti must fulfill the conditions
u = r/(r) when t = (e)
u = at r = # (/)
w = at r = (0)
It will be seen that this makes the problem similar to that of the
slab (Sec. 8.16) with faces at temperature zero and initial tem
perature r/(r). With the aid of (8.16i) we may then write
u 2V m7rr i^g* [ R .,.. . mir\ ^ /LX
mdX (h)
ml
If the initial temperature is a constant, To, we may write (h)
_ 2T V mr ^^ f R ^ ** * ,\
T = Tt; I, sm r e * h Xon g dx w
ml
r> x [ R ^ m7r ^ , x /2 2 / v
But / X sin ~o~ a\ =  cos mir (J)
Jo K wwr
so that (/i) may be written for this case
. 37rr
Following Sec. 7.14, we may write (k) for the case of either
heating or cooling, with surface at T 8 , as
T  T a 2i
9.17. Center Temperature. Equation (9.16Z) is readily
evaluated for the central point if we note that the limit of
(sin mwr/R)/(mirr/R) = 1 as r  0. Then we have, for a sur
164 HEAT CONDUCTION [CHAP. 9
face temperature T 9 ,
ffi fji
L C * *
*"""*
_
= 2 (e  e ' + e ' ) *(*) (a)
where a: = ir*at/R 2 . B(x) is tabulated in Appendix H.
9.18. Average Temperature. The average temperature T a
of the sphere at any time t may be found from (9.16&) by multi
plying each element of volume by its corresponding temperature,
summing such terms for the whole sphere, and dividing by the
volume of the sphere. Thus, since T is a function of r,
6T
sln
/ Q 7rr
] "**%*
1
\
9
or, in general,
where x = T 2 at/R 2 .
APPLICATIONS
9.19. Mercury Thermometer. Equations (9.18c) and (9.18d)
may be applied to a sphericalbulb thermometer immersed in
a stirred liquid. Neglecting the effect of the glass shell, the
temperature of the mercury is given to a close approximation by
the first term of the equation unless t is very small. The rate
of cooling is then
~ "aT " 5 s
9.20. Spherical Safes. Compare the fireprotecting quali
ties of two safes of solid steel (a = 0.121 cgs) and solid concrete
* See Appendix H.
SBC. 9.22] FLOW OF HEAT IN MORE THAN ONE DIMENSION 165
(a = 0.0058 cgs), each spherical in form, of diameter 150 cm
(59 in.) and of very small internal cavity. Assuming that the
surfaces are quickly raised from initial temperatures of 20C
(68F) to 500C (932F), determine the temperatures at the
centers after various times.
Using (9.17a) and Appendix H, we find that the temperature
in the center of the steel safe would be 98C (208F) at the end
of 1 hr and 455C (850F) after 4 hr, while in concrete the tem
peratures would run only 25C (77F) at the end of 10 hr and
not exceed 130C (266F) before 24 hr. Obviously, this com
parison is hardly fair to the steel safe since it would be prac
tically impossible to raise its surface temperature as rapidly as
is assumed here.
9.21. Steel Shot. Such a shot or ball 3 cm (1.18 in.) in
diameter, at 800C (1472F), has its surface suddenly chilled
to 20C (68F) ; what is the temperature 1 cm below the surface
in 1.8 sec? Putting r = 0.5 and R = 1.5, also a = 0.121 in
(9.160, we readily find T to be 501C (934F). It will be noted
that the cooling is much more rapid than in the case treated in
Sec. 7.22.
The rate of cooling may be found by differentiating (9.160
with respect to t. This gives
dT 7r<*(r T s
^ ==  2 Wr
This equation might be used in an investigation of the relation
between rapidity of cooling and hardness for approximately
spherical steel ingots. The preceding equations might also be
applied to a large number of practical problems of somewhat
the same nature as those discussed in previous chapters, by treat
ing all roughly spherical shapes as spheres. The theory might
prove of service in such problems as the annealing of large steel
castings or in a study of the temperature stresses and conse
quent tendency to cracking that accompanies the quenching of
large steel ingots.
9.22. Household Applications. There are numerous every
day examples of the type of heatconduction problem discussed
in these sections. The processes of roasting meats, boiling
166 HEAT CONDUCTION [CHAP. 9
potatoes or eggs, cooling of melons, etc., all involve the heating
or cooling of roughly spherical bodies under conditions of rea
sonably constant surface temperature. As an example, we may
question how long a spherical potato 7 cm in diameter must be
in boiling water before the center attains a temperature of 90C,
assuming an initial temperature of 20C. We may use the
same diffusivity as for water (a = 0.00143 cgs) for this and
other vegetables and fruits. Then, using (9.17a), we have
90  100 = (20  100)B(x), which, from Appendix H, gives
z(= 7r 2 orf/3.5 2 ) = 2.76, or t = 2,400 sec or 40 min. It may be
remarked that unless the potato is in rapidly boiling, i.e., vig
orously stirred, water, the surface will not attain the 100C
rapidly and the cooking process will accordingly take longer.
Tradition requires that ivory billiard balls, after exposure to
violent temperature change, should be allowed to remain in
constant temperature surroundings for a matter of several hours
before being used for play. For such a ball 6.35 cm (2.5 in.)
in diameter we may inquire how long it will be before the center
temperature change is 99 per cent of the surface change. Using
a = 0.002 cgs, we have from (9.17a), 1 = 100 B(x), or, from
Appendix H, x = 5.3. This means that the temperature should
be uniform throughout to within 1 per cent in 2,710 sec, or
considerably less than 1 hr. It would seem then that this tra
dition must be explained on a basis of other than temperature
considerations alone.
9.23. Problems
1. The surface of a sphere of cinder concrete (a = 0.0031 cgs) 30 cm in
diameter is rapidly raised to 1500C and held there. If it is all initially at
zero, what will be the temperature of the center in 1 hr? In 5 hr?
Ans. 49C; 1240C
2. A mercury thermometer, with a spherical bulb 1 cm in diameter, at
40C is immersed in a stirred mixture of ice and water. Neglecting the glass
envelope and assuming that the surface is instantly chilled to zero, determine
how soon the average temperature is within 0.01C of the bath. Use a =
0.044 cgs. Ans. 4.5 sec
3. An egg equivalent to a sphere 4.4 cm in diameter and at 20C is placed
in boiling water. Calculate the center and also average temperatures in
3 min. Solve the same problem for a 30cm diameter melon at 20C in ice
water for 3 hr; for 6 hr. Assume a ** 0.00143 cgs in each case.
SBC. 9.25] FLOW OF HEAT IN MORE THAN ONE DIMJSNtilUN io/
Ana. Egg, 23.6 and 69.7C; melon, center, 17.7 and 10.2C, and average,
6.4 and 3.2C
4. Show that the common rule for roasting meats of allowing so much
time per pound but decreasing somewhat this allowance per pound for the
larger roasts rests on a good theoretical basis.
CASE IV. THE COOLING OF A SPHERE BY RADIATION
9.24. We shall now solve a more difficult problem than any
we have before attempted, viz., that of the temperature state
in a sphere cooled by radiation. The solution will apply to the
case of the sphere either in air or in vacuo, for the only assump
tion made in regard to the loss of heat is that Newton's law of
cooling holds; i.e., that the rate of loss of heat by a surface
is proportional to the difference between its temperature and
that of the surroundings. This does not hold for large tem
perature differences. See Sec. 2.5.
As we shall see, the solution can also be applied to the case
of a sphere of metal or other material of high conductivity,
covered with a thin coating of some poorly conducting sub
stance and placed in a bath at constant temperature. For the
rate of loss of heat by the surface of the metal sphere will be
proportional to the temperature gradient through the surface
coating, i.e., to the difference of temperature between the inner
and outer surfaces of this coating, which, by the conditions of
the problem, is equal to the difference of temperature of the
metal surface and the bath. An example of this latter case is
the mercury thermometer with a spherical bulb, immersed in a
liquid, it being desired to make correction for the glass envelope.
9.25. The differential equation for this case is, as before,
d(rT) a 2 (rr) , ,
\. = a ~ o (a)
dt dr 2 ^ '
with the boundary conditions
T = f(r) when t = (6)
dT
fcgj: = hT atr (c)
The last condition states that the rate at which heat is brought
to unit area of the surface by conduction, viz., k(dT/dr), must
168 HEAT CONDUCTION [CHAP. 9
be the rate at which it is radiated from this area, and this is hT,
where h is the emissivity of the surface. The surroundings
are supposed to be at zero.
As before, put u = rT (d)
Then we have TT = OL ^r (e)
ot or 2
and the conditions u = rf(r) when t = (/)
u = at r = (g)
; = at r = R (h)
where, for short, C is written for h/k.
Now we have already seen in Sec. 7.2 that
u = e~~ amH cos mr (i)
and u = e^ amH sin mr (j)
are particular solutions of (e). Solution (i) is excluded by con
dition (gr), but (j) satisfies this condition for all values of m.
To see if (h) is also fulfilled, we substitute the value of u from
(j) and get
mR cos mR = (1 CR) sin mfl (K)
If Wp is a root of this transcendental equation, then
is a particular solution of (e) satisfying (</) and (K). We must
now endeavor to build up, with the aid of terms of the type (0,
a solution that will also satisfy (/).
Since the sum of a number of particular solutions of a linear,
homogeneous partial differential equation is also a solution, we
note that
u = Bie~ amiH sin mir + B 2 e~~ amzH sin m 2 r
+ B*e am * H sin m 3 r + (m)
where Bi, B^ B*, . . . are arbitrary constants, is a solution of
(e) satisfying (0). It moreover satisfies (h) if mi, 7w 2 , m 3 , . , .
are roots of (k). It evidently reduces f or t = to
BI sin mir + B 2 sin w 2 r + B z sin m s r + (ri)
SBC. 9.28] FLOW OF HEAT IN MORE THAN ONE DIMENSION 169
and if it is possible to develop r/(r), for all values of r between
and R, in terms of such a series, we shall have (/) satisfied as
well.
9.26. The solution of our problem, then, will consist of two
parts: (1) the solution of the transcendental equation (9.25fc),
i.e., the determination of the roots m\, w 2 , w 3 , . . . (we antici
pate a fact shortly to be shown, viz., that there are an infinite
number of such roots); and (2) the expansion of the function
rf(r) in the sine series (9.25n). The second part of the prob
lem is analogous to development in terms of a Fourier's series,
but more complicated because the numbers Wi, w 2 , w 3 , instead
of being the integers 1, 2, 3, as in the regular Fourier's series,
must in the present case be roots of equation (9.25fc).*
9.27. The Solution of the Transcendental Equation. The
roots of (9.25&) are easily obtained by computation, but a study
of their values under various conditions may be most easily
made by graphical methods. If we make the substitutions
7 = mR (a)
and ft ^ 1  CR (6)
(9.25&) becomes
7 cos 7 = ]8 sin 7 (c)
or, more simply, 7 = j8 tan 7 (d)
Then, if we construct the curves
y = tan x (e)
and y =  (/)
their points of intersection will give the values of x for which
= tan x (g)
i.e., the roots of (d) and hence of (9.25A;).
9.28. We may draw some general conclusions as to these
roots. In the first place, there are evidently an infinite number
of positive roots, and the same number of negative, which are
* This is the most general sine development that can be obtained by Fourier's
method. See Byerly. 23  * m
170
HEAT CONDUCTION
[CHAP. 9
equal in absolute value to the positive. The values of the roots
vary between certain limits with the slope of the line y = x//3,
i.e., with the value of C, or h/k. Since
y / / C can have, theoretically at least, any
value between and <*> but must al
ways be positive, the slope
1.
ft ~ 1  CR
(a)
can have any value between 1 and QO
or between and > .
We can easily show with the aid of
a figure the approximate values of
the roots for the several cases as
follows :
Let C = 0, corresponding to the
case Q f a sp here protected with a
thermally impervious covering. The
roots then correspond to the intersections of the line (1) (Fig. 9.2)
of 45 deg slope. Their values are 0, 71, 72, , where
FIG. 9.2. Curves whose
intersections give the roots for
Sec. 9.28.
3?r
< 7i < ;
rt 5?r
2?r < 72 < ;
.nw < 7n < ( n + 2 J TT (6)
7 n in this case approaches the limit (n + ^)?r as n increases.
Next, let C lie between and 1/R so that < (1  CR") < 1.
The Kne (2) corresponds to this case, and the roots 0, 71, 7 2 ,
7a, ... have the values
7T
37T
< 71 < 2>" TT < 72 < y;
(n  !)TT < 7 n < (n  2) T W
approaching the larger values as C increases. When C
then the roots become
A TT STT STT
0, n> "o>  o ' ' '
& L
SEC. 9.29] FLOW OF HEAT IN MORE THAN ONE DIMENSION 171
Finally, if C lies between l/R and oo , the intersecting straight
line will fall below the axis in some position such as (3), and
the roots 0, 71, 72, ... will have values
7i < T; < 72
" 2)
which become f or C = oo
7i = ir, 72 = 27T, , 7n = nx (/)
Prom these roots 71, 72, 73, the values mi, ra 2 , w 3 , . . . satisfying
(9.25&) are obtained at once with the aid of (9.27a).
9.29. The General Sine Series Development. We shall
arrive at this development by assuming that it is possible to
expand rf(r) in a series
rf(r) = BI sin mir + 2 sin m 2 r +
00
+ B b sin m b r + as B b sin m b r (a)
6*1
just as we assumed before that such a function could be expanded
in an ordinary Fourier's series, and then proceed to find the
values of the coefficients BI, J3 2 , B 3 , . . . , to which this assump
tion leads. The values mi, ra 2 , m 3 , . . . are the roots of equa
tion (9.25Jk) determined above. While zero is a root in each
case, there is no corresponding term in the series since sin 0=0.
The negative roots that occur are included with the positive in
the terms of (a), for since sin ( x) = sin x, we may write
B' b sin m b r + B" sin ( m b r) ~ B b sin m b r (6)
Multiplying each side of (a) by sin m a rdr and integrating
from to R,
00
/ r/(r) sin m a rdr = / B b \ sin m b r sin m a rdr (c)
JQ Jo
rR
Now / sin m b r sin m a rdr
Jo
1 C
2 /
~ cos ^ m6 + m ^ dr
172 HEAT CONDUCTION [CHAP. 9
sin [(m b m a )R] _ sin
2(m b m a ) 2(m b + m a )
(m a sin m b R cos m a R m b cos m b R sin n
= " x 2 2\
But since m a and m b are roots of (9.25&),
m a R = (1 CR) tan w jR; m b R = (1 CR) tan
so that m a tan ra^R = m b tan ra a # (/i)
or ra a sin ra^R cos ra a # = m b sin ra a # cos m b R (i)
[R
Therefore, / sin m b r sin m a rdr = (j)
Jo
when w a and m b are different. If they are equal, we have
[ R 1 [ R
i sin 2 m a rdr == 75 / (1 cos 2m a r) dr (k)
Jo * Jo
__ R __ sin 2ra a JZ
"" "2 i ' '
XT < T> ^ tan ra a /t . .
Now sin 2raJ? = 1 , , n ^ 2 ^ p (m)
i "T" tan 7/i o xi/
 CB) . .
P~2 ()
(Cfi  I) 2 +
Therefore, / sin 2 m a rdr = ^ 2p2 , .^p ^ (o)
JO & fi^a^ I V^ / " / U
Applying this in the series (c), i.e., in
r /
/ i/(r) sin m a rdr = JSi / sin m\r sin m a rdr
Jo Jo
+ B* I sin m 2 r sin m a rdr + (p)
Jo
2 ray? 2 + (CR  I) 2 /"* ,
"\X7fi H fl "\7 A A? ~ "" ~ . . I tff ( /* i CTKI 1TJ V /I'*' ^/>^
we nave x> o p 202 i m}ff~iT> 1\ / 'J\') olJl f'i>ar Wi (O)
it m a t p Cit^u/t ij yo
9.30. Final Solution. Our problem is now solved, for we
have evaluated the coefficients of the series (9.29a) in terms of
the roots of equation (9.25fc), which roots we have shown to
have real values that are easily determined. The solution may
be written
u  B a e am " sin m r (o)
SEC. 9.32] FLOW OF HEAT IN MORE THAN ONE DIMENSION 173
or, evaluating B a from (9.29g) and remembering that u rT 9
2 mlR* + (CB  1)'
1 ~ rR LI mlR* + CR(CR  1) e Sln m r
01
X/(X) sinra XdX (6)
/:
9.31. Initial Temperature T Q . In the case in which the
initial temperature of the sphere is everywhere the same, i.e.,
/(r) = TQ, we find that the above integral reduces to
C
I
Jo
R tp
X sin raX d\ = % (sin w/J mR cos m#) (a)
o m
and, with the use of (9.25A), =
Tfi
Thus, (9.306) becomes for this case
m 2CT Q \ mlR* + (CR  I) 2
 
r lm?K# 2 + CR(CR  1)] c 1 " """ Sm
, mjR 2 + (CR  I) 2 _ am2it _ .
+ 2r zpz _i_ riTxm? TTT e sm m 2/c sm TO 2 r + '
7fl^\Jfl^f\j ~\~ L'/t\ s Uxt lyj
(C)
9.32. Special Cases. If CR is small in comparison with
unity, as it would be in many cases, the problem is greatly sim
plified. For an inspection of Fig. 9.2 shows that in this case
m\R will be very small, while the other values of mR will be
larger than TT, so that only the first term of the series (9.31c)
need be considered. The value of mi is readily determined
from (9.25/c) by developing the sine and cosine in series and
neglecting higher powers of m\R, in which case we obtain
from which it follows that
3C
R
ml = ^ (6)
With the aid of (6), equation (9.31c) may be still further
simplified if it be remembered that miR and m\r are small quan
174 HEAT CONDUCTION ICHAP. 9
titles, and if C*R* is neglected, for it reduces at once to
T = T e* c t/R (c)
= T &**'* (d)
c being the specific heat.
9.33. The assumptions involved in this last formula are that
the sphere is so small or the cooling so slow that the tempera
ture at any time is sensibly uniform throughout the whole
volume. With this assumption it may be derived independently
in a very simple manner; for the quantity of heat that the
sphere radiates in time dt is
4irR*hTdt (d)
This means a change in temperature of the sphere of dT y which
corresponds to a quantity of heat given up equal to
%TrR*cpdT (6)
the negative sign being used, since dT is a negative quantity.
Hence, we have
4irR*hTdt = %7rR*cpdT (c)
the integration of which gives, since the temperature of the
sphere is T at the time t = 0,
T = T<>e* ht/cpR (d)
as above.
9.34. Applications. Equations (9.306) and (9.31c) make
possible the treatment of the problem of the cooling of the earth
by radiation* before the formation of a surface crust, which was
kept, by the evaporation of the water, at a nearly constant
temperature. The solutions of Cases III and IV of the present
chapter would enable one to treat the problem of terrestrial
temperatures with account taken of the spherical shape of the
earth, but as already noted our present data would by no means
warrant such a rigorous solution, which would alter the result
in any case by only a very small fraction. It may be noted
that the solution of the problem cf radiation for the semiinfinite
* However, see Sec. 2.5 in thia connection.
SEC. 9.36] FLOW OF HEAT IN MORE THAN ONE DIMENSION 175
solid is gained from the present case by letting R approach
infinity.
As already suggested, the solution for the present case will
fit another that at first sight seems quite foreign to it, viz. y
the cooling of a mercuryinglass thermometer in a liquid. If
the glass is so thin, as it usually is, that its heat capacity can be
neglected, we have only to set in place of h, in the above equa
tions, k/l, where I is the thickness of the glass and k its conduc
tivity, and we shall have a solution of this problem.
The general case of cooling or heating roughly spherical
bodies by convection or radiation especially in its simpler
phases has many applications. Most of these, however, are
beyond the scope of this book since conduction in many of them
plays a secondary part. Students who are interested in pursuing
the general subject of heat transfer may profitably consult
Brown and Marco, 20 Croft, 34 Grober, 63 Jakob and Hawkins, 68
McAdams, 90 Schack, 122 Stoever, 139  140 Vilbrandt, 156 and similar
books.
9.35. Problems
1. A wroughtiron cannon ball of 10 cm radius and at a uniform tempera
ture of 50C is allowed to cool by radiation in a vacuum to surroundings at
30C. If the value of h for the surface is 0.00015 cal/(sec)(cm 2 )(C), what will
be the temperature at the center and at the surface after 1 hr? Use k = 0.144,
a = 0.173 cgs, for iron. Ans. 46.5, 46.4C
2. A thermometer with spherical mercury bulb of 3.5 mm outside and
2.5 mm inside radius, heated to an initial temperature of 30C, is plunged into
stirred ice water. Find, to a first approximation, how long it will be before
the temperature at its center will fall to within HC of that of the bath.
Neglect the heat capacity but not the conductivity of the glass (use k = 0.0024
cgs). For mercury use c = 0.033, p = 13.6 cgs. Ans. 7.5 sec
3. The initial temperature of an orange 10 cm in diameter is 15C while
the surroundings are at 0C. If the emissivity of the surface is 0.00025 cgs
and the thermal constants of the orange the same as those of water, what
will be the temperature 1 cm below the surface after 8 hr? Ana. 0.38 C
CASE V. FLOW OF HEAT IN AN INFINITE CIRCULAR CYLINDER
9.36. Bessel Functions. In order to solve the problem of the
unsteady state in the cylinder we must gain a slight acquaint
176 HEAT CONDUCTION [CHAP. 9
ance with some of the simpler properties of Bessel functions.*
The function J Q (z) defined by the series
~2 ~4 ~6
T / \ 1 I I f \
JQ(Z) s 1 ^2 + 22 . 42 ~~~ 22 42 . Q2 i W
is called a " Bessel function of order zero. 7 ' If n is zero or a
positive integer, J n (z), of order n, is defined by the series
2(2n +~2) 24(2n + 2)(2n + 4)
2 4 6(2n + 2)(2n + 4)(2n + 6) ^ J w
Putting 0! = 1 (i.e., 1!/1), the above is seen to reduce to (a)
for n = 0. If we write JQ(Z) for the derivative dJ Q (z)/dz ) it
is seen at once that
J f (%\ = Ji(z)^ (c)
It can also be shown that
(d)
9.37. From an inspection of (4.6a) we can write at once for
the Fourier equation in cylindrical coordinates, if T is a function
of r and t only,
dT d*T 1 d
We shall use this in solving the problem of the nonsteady state
in a long cylinder of radius R under conditions of purely radial
flow.
9.38. Surface at Zero. To solve this problem we must
find a solution of (9.37a) that satisfies the boundary conditions
T = /(r) when t = 0, (r ^ R") (a)
T = at r = R (b)
Making the substitution T s ue~~ aftH (c)
where u is a function of r only and ft a number whose value will
* See, e.g., Watson, 169 Carslaw, 27 McLachlan. 93
t Tables of Jo(z) and /i(z) are given in Appendix I.
SEC. 9.38] FLOW OF HEAT IN MORE THAN ONE DIMENSION 177
be investigated later, (9.37a) becomes
9t ~ '
or
which is known as a "Bessel equation of order zero." Now,
as is easily shown by differentiation, u = Jo(fir) is a solution of
(e). Thus,
T = BJo(0r)e" (/)
is a particular solution of (9.37a) suitable for our problem.
To satisfy condition (6) we must have
= (g)
The values of j8i, j8 2 , . . . that satisfy this equation for any
particular value of 7? may be obtained from Appendix I. If
f(r) can be expanded in the series
f(r) = Bi
condition (a) will also be satisfied and the solution of the problem
will be
In evaluating J5i, 5 2 , . . we follow a procedure net unlike
that employed in Sec. 6.2 in determining the Fourier coefficients.
Multiply both sides of (h) by rJ (0 m r) dr and integrate from to
R. Then,
f*rf(r)J Q (l3 m r)dr  B l f* r
Now it can be shownf that
/ %/o(^r)J (/3 p r)dr = (fc)
* This is commonly written
178 HEAT CONDUCTION [CHAP. 9
and also
fR D2
J o r[J Q (!3 m r)]*dr  y [/'oGS,n#)] 2 (Z)
Then, substituting from (9.36c) for Jj, we have
2 A rf(r)J Q (/3 m r) dr
Therefore, the final solution is
T = A V ^r r/(r)Jo( ^ r)dr
When/(r) = T , a constant, we evaluate (n) as follows:
^o / r/oCftnr) rfr = ^ / (/3 m r) J (/3 w r) d^r) (o)
JO Pw JO
and from (9.36eO this equals
'wi
T<>R
Pm
which means that (ri) reduces to
(p)
m1
A more easily usable form is obtained by writing
(r)
where z m is the wth root of J (z) = 0.
Thus, we have finally, for a body at T and surface at T a ,
tn1
which holds for either heating or cooling.
If we are interested only in the temperature T c at the center
SEC. 9.40] FLOW OF HEAT IN MORE THAN ONE DIMENSION 179
wbiere r =0, (s) becomes
00
~^T = 2 2 fjlz
To
m1
where x = cd/R*. Values of this series are tabulated in Appen
dix J.
APPLICATIONS
9.39. Timbers; Concrete Columns. MacLean 95 has made
extensive studies of the heating of various woods, using equa
tions like the preceding in connection with round timbers.
Computations of center temperatures may be very easily made
with the aid of Appendix J. As an example, let us calculate
the temperature at the center (and not near the ends) of a
round oak (a = 0.0063 fph) log 12 in. in diameter, 8 hr after it
has been placed in a steam bath. Initial temperature is 60F
and steam temperature 260F. Using (9.380 and putting
x = at/R 2 = 0.201, we have from Appendix J, (7(0.201) = 0.498,
and therefore T = 161F.
For points not on the axis the calculations are not so simple.
As an example, suppose that a long circular column of concrete
(a = 0.03 fph) 3 ft in diameter and initially at 50F has its
surface suddenly heated to 450F. What will be the tempera
ture at a depth of 6 in. below the surface after 2 hr?
We use (9.38s). The values of z to satisfy (9.380), i.e.,
J (z) = 0, are found from Appendix I, Table 1.2, to be z\ = 2.405;
s 2 = 5.520;z 3 = 8.654;s 4 = 11.79. Using Table I.I of Appendix
I, we find that the corresponding values for J^(z m r/K) are
0.454, 0.398, 0.082, and 0.203; and for Ji(z m \ 0.519, 0.340,
0.271, and 0.232. Putting these values in the various terms
of the series, we finally get T = 123F.
Problems of this type are important in connection with fire
proofing considerations when it is important to know how long
it will take supporting columns to get dangerously hot in a fire.
9.40. Problems
1. In the second application of Sec. 9.39 calculate the temperature after
4 hr at a depth of 6 in. below the surface and also at the center.
Ans. 202F; 57F
180 HEAT CONDUCTION [CHAP. 9
2. A long glass rod (a = 0.006 cgs) of radius 5 cm and at 100C has its
surface suddenly cooled to 20C. What is the temperature at the center
after 8 min? Am. 83.3C
CASE VI. GENERAL CASE OF HEAT FLOW IN AN INFINITE
MEDIUM
9.41. In Case II of this chapter we solved the problem of
the flow of heat from an instantaneous point source. We shall
extend this result to cover the case in which we have an initial
arbitrary distribution of heat, the initial temperature being
given as a function of the coordinates in three dimensions.
Let x,y, and z be the coordinates of any point whose tem
perature we wish to investigate at any time t, while A,JU,J> are
the coordinates of any heated element of volume and become in
general the variables of integration. Then, the initial tempera
ture is
To = /(X,/i,iO (a)
and the quantity of heat initially contained in any volume ele
ment d\dndv is
dQ = f(\,n, v )d\dtJLdv (6)
If this quantity of heat is propagated through the body, it will
produce a rise in temperature which can be obtained at once
from (9.5z), and which is, since
r 2  (X  xY + (M  yY + (v zY (c)
dT = 7' e^**^+<w /(X,M, v) d\dfjidv (d)
The temperature at any point will be the sum of all these
increments of temperature and may be obtained by integrating
(d):
(e)
Making the substitutions
8 s (X  x)rj; 7 s ( M  y)rj; e s (v  z)t\ (/)
Sc. 9.42] FLOW OF HEAT IN MORE THAN ONE DIMENSION 181
this becomes
T 
9.42. It will be instructive to show how this solution may be
obtained independently as a particular integral of the conduc
tion equation
dT /d*T d z T d*
subject to the boundary condition
,J>) when* = (6)
Assume T = XYZ, where X ip a function of # and , and
where F and Z are functions of t/, and z,t, respectively. Then
we have from (a)
d*X
But since X,Y, and Z are essentially independent, being func
tions of the independent variables x,y,z, this can only be true
if the corresponding terms on each side of the equation are
equal, i.e., if
dX d*X
with similar equations for Y and Z.
Now it may be easily shown by differentiation that
is a particular solution of (d), a type of solution already made
use of in Sec. 8.3, so that
T = i, er< x * )V e" ( "^ )lft 4= e~(^> 2t;2 (/)
Vt Vt Vt
182 HEAT CONDUCTION [CHAP. 9
is a solution of (a). Therefore, if C is any constant, and
an arbitrary function of
(g)
is also a solution of (a). By the substitutions (9.41/) this
reduces to
/OO r 00 /* W
/ /
oo y oo J ~~
If we now let = 0, this becomes
and, remembering that
/e~ p *dp = VTT (j)
 00
this becomes T Q = C(2 VOTT) 3 ^ (x,y,z) (k)
From (6) we see that if
and V(x,y,z) = f(x,y,z) =/(X,/i,^) since i =0 (m)
the boundary condition (6) is fulfilled. Putting in (h) these
values of C and ^, we find at once that it reduces to the solution
(9.410) already found.
9.43. Formulas for Various Solids. Since the solution of
the heatconduction equation for three dimensions and with
constant initial temperature can in most cases be considered as
the product of three solutions, each of one dimension, it is
possible* to arrive at once at a solution of a large variety of
simple cases where the initial and surface temperatures are each
constant. Equation (8.16&) gives for the center temperature
of a slab of thickness I, initially at T and with surfaces at T 8 ,
* See Newman 108 and Olson and Schultz. 106
SEC. 9.43] FLOW OF HEAT IN MORE THAN ONE DIMENSION 183
the relation
_
fj^ np
For the center of a rectangular brick of dimensions I, m, and n
we would accordingly have
p
*H
*/>
10
n
FIG. 0.3. Diagrams to accompany Table 9.1.
184
HEAT CONDUCTION
[CHAP. 9
CM

X
3
D
^~~^
^K*""*^
n
E^ ^ 
M '^
** Q?
1 1
^T^
"e ;>
" *>"
r
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CM
CM
^
3
^
cC
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*
"V
^
^
e
CO
X
X
r
X
X
1 "^5
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^"^
^
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1
X?
" \
^3
^s
" ^
^ N
^
CM
"Sl^
/j
^<
e
e
e
^
3
3
s
i ^
i i
m
1
o
CQ O
CO 03
d
s
D
ti
1
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CO
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S &
1
a
H
z;
15*
o
d
g
d
o
L*
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1
D
I
1
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^
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5 8 *
o f 1 ^ *"w
S?
i>
_j
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8
GO
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rs
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<
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o
5
p VH at
S
^
H
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d
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&
d 03
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5
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o
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^
CO PH
fe 
d
3
<
c
5
4
_1
<J
1
1
<
fi
H
4
H
! Physical equivalent
i
Region of a long cylinder of i
R remote from both ends,
finite cylinder with insulated
Region of a long cylinder nea
of the ends
Cylinder whose length Z is o
same order of magnitude i
diameter
ce
*o
t+i
0)
1
a
03
d
d
4) O
Region near the edge or inters^
of two perpendicular faces
large solid
Region near the corner or into
tion of three mutually perper
lar faces of a large solid
Region of a large slab of thick
remote from the edges
r5
u,
*O
i
1
1
.s
S
s
3
H
O
i
d
IH
Semiinfinite
inder
d
%
5
!s
Semiinfinite
Quarterinfii]
solid
1
1
as
a g
i
1
d
HH
^
CO
<*
*>
<o
^
Sue. 9.43] FLOW OF HEAT IN MORE THAN ONE DIMENSION 185
'1
?
fi?
X
V A^'
"^T^
^TH
T*
X
X
"* N
1 OJ
55
JJ
53
*
^
CO
CO
CO
X
X
X
X
X
< T N
T^
C^
^^
IB
^isT
^IcT
iS
i3
CO
CO
CO
CO
CO
oq
OQ 8
*0> S
III
a
1
(H V ~ l O
Q>
g,"^
&T3 g
'ts
^9
(5 o
rj QJ o
S O f>
8
fc
2
0> g
V & **^
S
"S
8,
jl
l!^
.22
8
rf ^
^ rd
.2
^orf
1
^
a
^
ll
lisi
[3 a o
03
I 1
1
1
2
o
>
a
0}
ear one plane
to the faces
ear the inter
adicular sur
5 perpendicu
slab
f rectangular
ind thickness
h ends, or a
asulated ends
f rectangular
e of the ends
whose length
t n are of the
ude
e functions *, iS
Region of a large slab n
surface perpendicular
of the slab
Region of a large slab n
section of two perpej
faces, each of which if
lar to the faces of the
Region of a long rod o:
cross section (width Z i
m), remote from bot
rectangular rod with i]
Region of a long rod o:
cross section, near on
i
Parallelepiped or brick
J, width ra, and heigh
same order of magnit
Sphere of radius R
1
8
1
Q,
a?
*o
3
si
S
2 **
11
gpT
i
.
go
1
5
i
8
m
M
:
1
i
iL
j,
I" 8
1 3
2
e
ll
o c
fcj
i!
S
J3 *
g p
o
*a
JC
ff o
^ ^
5 ^
& J
W
*
*
Jcg
# *
00
o5
T I
^
2
CO
pH
rH
186 HEAT CONDUCTION [CHAP. 9
while for the center of a round cylinder of radius R and length I
the relation would be
Table 9.1 lists the formulas for all the simpler cases.
APPLICATIONS
9.44. Canning Process. Brick Temperatures. The fore
going equations have been made use of in the canning industry
in studying the timetemperature relations in the sterilizing
process. In this connection we may calculate the temperature
at the center of a can of vegetables of length 11.0 cm and radius
4.2 cm, after 30 min in steam at 130C, the initial temperature
being 20C. Using the same diffusivity (0.00143 cgs) as for
water, we have
T ~ * Q = 5(0.0213) X C(0.14G) = 0.65 (a)
or T = 58.5C. It is to be noted in this connection that the
center temperature will cf course continue to rise even after the
can has been removed from the boiler and the surface starts to
cool.
As a second illustration we shall calculate the temperature
at the center of a brick (a = 0.020 fph) of dimensions 2 by 4 by
8 in. What is the temperature after 15 min if the brick is
initially at 3COF and the surface has been chilled to 40F?
We have here
5(0.18) X 5(0.045) X 5(0.011) = 0.174 (6)
300  40
or T = 85F.
In all our previous discussions the expressions infinite plate,
long rod, point remote from end, etc., are of frequent occurrence.
It is natural to question the error involved if the dimensions do
not meet these ideal specifications. The problem of the brick
solved above indicates the answer. It will be noted that the
heat flow in the direction of the largest dimension, which is four
SEC. 9.45] FLOW OF HEAT IN MORE THAN ONE DIMENSION 187
times the smallest, has little effect on the result. If the largest
dimension is half a dozen, or so, times the smallest, the ideal
conditions may in general be considered as fulfilled.
9.45. Drying of Porous Solids. As indicated in Sec. 1.4, the
diffusion of moisture in porous solids follows, within certain
limits, equations similar to those for heat conduction. New
man 101 and others have developed the theory along these lines.
As an example of this application, we shall solve the^ following
problem : A sphere of clay 6 in. in diameter dries from a moisture
content of 18 per cent (i.e., the water is this fraction of the total
weight) down to 12 per cent in 8 hr, under conditions that indi
cate that diffusion (i.e., heatconduction) equations apply in this
case. If the equilibrium moisture content is 4 per cent, how
much more time would be required for drying down to 7 per
cent moisture?
In solving we must first translate the moisturecontent
figures to percentages of dry weight, i.e., pounds of water per
pound of dry clay. This gives
C = total initial moisture content = x % 2 = 0.219
C a = total moisture content at 8 hr = *% 8 = 0.136
Cb = total final moisture content = %3 ~ 0.075
C = equilibrium moisture content = % = 0.042
In applying heatconduction equations to diffusion problems,
liquid concentration corresponds to temperature. We may
accordingly use, in this case, the equations developed in Sees.
9.16 to 9.18. We must note, however, that while <7 and C 8
refer to moisture concentrations that may be assumed to be uni
form throughout the sphere, this is not true for C a and C&,
which are average* concentrations after certain drying periods.
We must accordingly use the equations of Sec. 9.18. We have
then
CaC.( .. x T a  T.\ 0.136  0.042
g% Corresponding to JT^Y.) = 0.219  0.042
= 0.531  B a (x) (a)
* A little thought will show the reason for this. Temperature is readily deter
mined for various points in a body, but this would be difficult for liquid concentra
tions, which are usually measured by weighing and hence are average values.
188 HEAT CONDUCTION [CHAF. 9
This gives, from Appendix H,
x = 0.258 = tjp (6)
from which we get as the diffusion constant in this case,
0.258 X 0.0625
a =
nAAAOn/1 ,., ,
= 0.000204 ft 2 hr
oTT
For the final 7 per cent moisture content we have
=  186 = *<*>' OT * = L
Using the above value of a, we have for the total drying time
t = 36.9 hr 0)
or 28.9 hr beyond the first drying period.
Tests of drying periods on one shape enable calculations of
drying times for other shapes and sizes of solids made of the
same material. Such calculations, however, require curves or
tables (similar to our B a table) for average temperatures or
moisture contents, for such shapes as the slab, cylinder, brick,
etc. For such, as well as for a more complete treatment of the
subject, the reader is referred to Newman's paper. 101
9.46. Problems
1. A square pine (a = 0.0059 fph) post of large dimensions, at 70F, has
.ts surface heated to 250F. What is the temperature 1 in. below the surface
after half an hour? Solve this for a point well away from the edge and also
for one near an edge and 1 in. from each surface. What bearing do these
results have on the form of the isotherms near the edges? (In answering
this question calculate at what equal distance from each face, near the edge,
the temperature is the same as at 1 in. from the surface and well away from the
edge.) Ana. 120F, 156F
2. In the brick (a 0.0074 cgs) of Sec. 8.26 heated for 10 min, what would
the result have been if the other dimensions had been taken into account?
Assume the width to be twice the thickness and the length four times.
Ans. 0.607 7 ,
3. Molten copper (use k = 0.92, c = 0.091, p 8.9 cgs) at 1085C is sud
denly poured into a cubical cavity in a large mass of copper at 0C. If the
edge of the cube is 40 cm, find the temperature at the center after 5 min.
Neglect latent heat of fusion (cf. Problem 1, Sec. 9.4). Ana. 186C
SEC. 9.46] FLOW OF HEAT IN MORE THAN ONE DIMENSION 189
4. A sphere, cylinder (height equal to diameter), and cube of cement
(a = 0.04 fph) are each of the same linear dimensions, viz., 6 in. high. If the
initial temperature is zero and the surface in each case is heated to 100F,
calculate the temperature in the center in each case after K hr. Also, make
the same calculations for all bodies of the same volume, equal to that of the
6in. cube. Ans. 91.4F, 85.5F, 80.7F; 74.3F, 78.5F, 80.7F
6. A clay ball 4 in. in diameter dries from a moisture content of 19 per cent
(i.e., 19 per cent of total weight) down to 11 per cent in 3 hr. Assuming
that diffusion equations apply and that the equilibrium moisture content is
3 per cent, what will be the moisture content after 10 hr of drying?
Ans. 6.3 per cent
6. Consider the steady temperature state in a long rod of radius ft, one
half of whose surface for < < TT is kept at 7\ and the other half, for
TT < < 2?r, at zero. Since T is here a function of the cylindrical coordinates
r and only, the Fourier equation for the steady state is*
_
dr 2 r dr r 2 d0 2 ~ "
Show that the temperature at any point (r, 6) is given by
T. A , / sm9 \
T corl Uh In (R/r))
...
(4)
Show also that the conjugate function to T, of the complex variable
6 + i In (R/r) y which gives the lines of heat flow is
TT T * ui ( cos
SUGGESTIONS. Apply the method of Sec. 4.3 and show that
nB
(r\ n sinjj
R) ~~n~
is a particular solution of the Fourier equation, where n may be any positive
integer. Assume that the desired solution is possible with a series of such
particular solutions having undetermined coefficients as in (4.2fc), including a
possible constant term. Choose these coefficients such that the boundary
conditions at r = R are satisfied, thus giving the first form of the solution
above. Compare this with (4.2w) where y corresponds to In (R/r) and get the
closed forms for T. The conjugate function follows from Appendix L,
* See Churchill. 32  * 13
CHAPTER 10
FORMATION OF ICE
10.1. We shall now take up the study of the formation of ice,
i.e., of the relationship that must exist between the thickness
and rate of freezing or melting of a sheet of ice and the time
when a lake of still water is frozen or a sheet of ice thawed.
In our previous study of the various cases of heat conduction in a
medium we have assumed that the addition or subtraction of
heat from any element of the medium serves only to change its
temperature f and does not in any way alter its conductivity con
stants or other physical properties. In ice formation, however,
we have essentially a more complicated case, for the freezing of
water or thawing of ice results not only in a change from one
medium to another that has entirely different thermal con
stants, but also in the accompanying release or absorption of
the latent heat of fusion.
10.2. We shall treat the problem in two somewhat different
ways, the first following substantially the method of Franz
Neumann* and the second that of J. Stefan. I38 f In each case we
have initially a surface of still water lowered, as by contact with
the air or some other body, to some temperature 7 T , which must
always be below the freezing point. There will then be formed
a layer of ice whose thickness e is a function of the time t.
Take the upper surface of ice as the yz plane, and the positive
x direction as running into the ice. Let T\ apply to tempera
tures in the ice, and T^ to the water; and similarly, let /Ci, c\,
and c*i be the thermal constants for ice, while A* 2 , c 2 , and o? 2 are
those for water. It is assumed that there is no convection in
the water, and the changes of volume that occur on freezing
or melting are neglected.
* WeberRiemann. 180 ' * U7
f Soe also Tamura. 14 *
190
SBC. 10. 3] FORMATION OF ICE 191
10.3. Neumann's Solution. Instead of one fundamental
equation, as in the case of a single homogeneous medium, there
will now be two, applying respectively to the ice and to the
water under the ice. These are
f^HF r)2'T'
r~ = cti fif in the ice (0 < x < e) (a)
AT 1 f^^ r P
and  = <* 2 ~ in the water (e < x) (6)
The temperature of the boundary surface of ice and water
(at x = e) must always be 0C, and there will be continual
formation of new ice. If the thickness increases by de in time dt,
there will be set free for each unit of area an amount of heat
Q = Lpi dt (c)
where L is the latent heat of fusion. This must escape upward
by conduction through the ice, and in addition there will be a
certain amount of heat carried away from the water below, so
that the total amount of heat that flows outward through unit
area of the lower surface of the ice sheet is
Of this amount the quantity
flows up from the water below; hence, we obtain for our first
boundary condition
dT l , dT 2 \ de
  k *  = Lpi ~
The other boundary conditions are to be
T l = T 8 = Ci at x = (0)
T l = T 2 = at x = e (h)
T 2 = C 2 at x = oo (i)
We also have three other boundary conditions derived from the
fact that when t = 0, e is fixed, while TI and T 2 must be given
192 HEAT CONDUCTION [CHAP. 10
as functions of x, the first between the limits and e and the
last between e and <*> . We shall investigate later the particular
form of these functions.
10.4. The general solution of the problem for these condi
tions is not possible as yet, for the condition (10.3/) containing
the unknown function is not linear and homogeneous, and we
cannot then expect to reach a solution by the combination of
particular solutions. Our method of solution then will be to
seek particular integrals of (10.3a) and (10.3&) and, after modify
ing them to fit boundary conditions (10.30), (10.3/0, and (10.3i),
find under what conditions the solution will satisfy (10.3/).
This will then determine the initial values of c, Ti, and TV
Now, as we have seen many times in the previous pages, the
function $(ZT/) is a solution of such differential equations as
(10.3a) and (10.36). Consequently, if J5i, D\, B 2 , #2 are con
stants and if 171 ss 1/2 Vctrf and 17 2 = 1/2 VW,
T l = Bi + D&(xrn) (a)
and T 2 = B 2 + D^(xrj 2 ) (b)
are also solutions. Now, boundary condition (10.3 K) means
that $(i/i) and $(172) must each be constant, which will be
true if = 0, e = , or if e is proportional to VT. The first
two of these assumptions are evidently inconsistent with (10.3A) ;
thus, there remains only the last, which may be put in the form
e = b Vt (c)
where b is a constant we shall determine later, together with
Bi, Di, J5 2 , and D 2 .
From the properties of $(x) we know that $(0) =0 and
$(<*>) = 1, Then fitting boundary conditions (10.30), (10.3/&),
and (10.3t) in (a) and (6) with the use of (c), we find that
Bi = Ci (d)
B, + j  C, (g)
SEC. 10.7] FORMATION OF ICE 193
while (a), (6), and (c) in connection with (10.3/) give
Solving equations (d) to (g) for DI and D 2 , we get
TX
1 ~
$(6/2 V^i) ' 1  $(6/2
and, substituting these values in (h), we have finally
Vai $(6/2 Vai) Va 2 [1  $(6/2 Vo~ 2 )] 2 ' Pl U}
10.6. This transcendental equation can be solved for 6 by
the method employed in Sec. 9.27. Plot the curves
^r * i ^
y = yLpifc (a)
and y  /(6) (6)
where /(&) represents the lefthand side of (10.4J). Then 6 is
given as the abscissa of the intersection of the two curves.
When 6 is found, the problem is solved, for from (10.4c) we can
then express the exact relation between the thickness and time,
and, having solved (10.4d) to (10.40) for Bi, DI, B 2 , and Z) 2 , we
have from (10.4a) and (10.46) the temperatures at any point
in the water or ice.
10.6. We are now able to specify the initial conditions for
which we have solved the problem, and which have up to this
time been indeterminate. It follows from (10.4c) that when
t = 0, = 0, and from (10,46) that T 2 is initially equal to
B 2 + D 2 = C 2 , everywhere except at the point x = 0, where it is
indeterminate. This means that we have taken the instant
t = as that at which the ice just begins to form, the water
being everywhere at the constant temperature C 2 . Inasmuch,
then, as there is no ice at time t = 0, the temperature TI must
be indeterminate, as is shown by (10.4a).
10.7, In the case of freezing as just treated, Ci is necessarily
a negative and C 2 a positive quantity. By reversing the signs
194 HEAT CONDUCTION [CHAP. 10
and making C\ positive and C 2 negative we have equations
applicable to thawing. But thawing in this case means that a
layer of water is formed on the ice and that the heat flows in
from the upper surface of the water, which is then at tem
perature Ci. But this means that the ice and water have just
changed places, so that in the case of thawing, Ci, &i, <*i, and
d apply to the water, while C 2 , 2, 2, and c 2 apply to the ice.
10.8. Stefan's Solution. Stefan simplified the conditions of
the problem by assuming that the temperature of the water was
everywhere constant and equal to zero. The fundamental equa
tion (10. 3a) then becomes
dT, d*T,
for < x < (a)
while the second is missing. Likewise, the boundary conditions
(10.3/) to (10.3i) are simplified to
i = T 8  Ci at x = (c)
Ti = at x = 6 (d)
Since Ti may be expressed as a function of both time and
place, we may write its total differential
1
From (d) we see that this total differential must be zero at
x = e, so that
so that with the aid of (6) we have
a,ci
. ,
8mcek
As a special solution of (a) we shall examine the integral
T  B I" e~* d\ (K)
SBC. 10.8] FORMATION OF ICE 195
and see if the constants B and /3 can be so chosen that this solu
tion is consistent with the conditions (6), (c), (d), and (/). We
need not prove that (h) is a particular integral of (a), for we
have used this type of integral many times as a solution of the
Fourier equation in one dimension. Thus, we can proceed at
once with our attempt at fitting it to these boundary conditions.
Condition (c) demands that
B f Q ft
(i)
which gives one relation between B and /3. Condition (d) means
that the two limits of the integral must be the same for x = ,
so that
cr?1 or e = 2]8 A/cM (j)
2 V ait
This gives the same law of thickness as found by Neumann's
method of (10.4c), viz., that the thickness increases with the
square root of the time. However, we have not yet determined
the constant ]8, and to do this we must use (</). The differential
coefficients STi/dt and dTi/dx are obtained from (h) after the
method described in Sec. 7.16 and are
dt 2t
^  BT^',, (0
If we now put in these expressions x = c = 8/iji and then
apply (fir), we have
Be~"  =  2& B*e'r,l (m)
or, with the use of (i),
/ft
>
and this equation enables us to determine j8. The integral may
be evaluated by expanding e~" x * in the customary power series
and performing the integration. When this result is multiplied
by the series for /V, we get a series whose first two terms are
196 HEAT CONDUCTION [CHAP. 10
To a first approximation, then, (n) gives
Consequently, to the same degree of approximation, (j) means
. A 2 , .
that e 2 =  j (?)
For the second approximation
= 
from which /8 and consequently e are readily determined.
Since Ci is intrinsically negative, the righthand member of
the above equation is a positive quantity.
It should be noted that the same law of freezing holds in each
case, i.e., the proportionality of thickness with the square root
of the time; the proportionality constant only is changed.
Indeed, if* we put C 2 = in Neumann's solution (10.4J), it
reduces at once to Stefan's solution (n), if b = 2/3 \fa\. This
makes the two expressions for the thickness, (10.4c) and (j),
identical and shows that Stefan's solution may be regarded as
only a special case of Neumann's.
10.9. Thickness of Ice Proportional to Time. Stefan also
outlined the solution of one or two special cases that we shalj
find interesting.
Consider the expression
Ti = f (e pt ~ 9X ~ 1) ()
where B, p, and q are constants.
It may be readily seen upon differentiation that if
p = cxitf 2 (6)
(a) is a solution of the fundamental equation (10.8a). Now
Ti *= for pt  qx = (c)
and from (10.8d) Ti = at x = e (d)
from which pt qx = at x = c (e}
or 6 = qa r t (/)
SEC. 10.10) FORMATION OP ICE 197
This shows that the thickness of ice may increase in direct
proportion to the time if T 8 is not a constant, as we have here
tofore taken it. Equation (a) shows that (since TI = T 8 when
x = 0), T s must be a function of the time, and it will be our task
to investigate the form of this function.
Since (10. 80) must hold, we find on substitution of (a) and
(/) that
so that the relation between B and p is
H   T 1 (W
For x = we find from (a) that
) (0
^
"! 2! L 2 3! "
This shows, since JS is negative, that if the thickness of ice is to
increase directly as the time, the surface temperature must
decrease more rapidly than as a linear function of the time.
For any value we wish to give B, the thickness is determinate
from (/).
10.10. Simple Solution for Thin Ice. If we assume that the
ice is thin enough so that the temperature gradient can be con
sidered as uniform from the upper to the lower surface, we can
derive at once a very simple solution; for the quantity of heat
that flows upward per unit area through the ice in time dt will
then be
fci^ift (a)
and this must equal the heat that is released when the ice
increases in thickness by dc. Hence, we have
kiT.dt T , ,..
 = Lpide (6)
198 HEAT CONDUCTION [CHAP. 10
Integrating this and assuming that is zero when t is zero, we
have
f =
(0
which is identical with (10.80). This shows that the approxima
tion involved in (10.8#) amounts to the assumption of a uniform
temperature gradient through the ice.
10.11. With the aid of some of his formulas Stefan calculated
k for polar ice from the measured rates of ice formation at
Assistance Bay, Gulf of Boothia, and other places, and found
* = 0.0042 cgs (a)
This value lies between the values attributed to Neumann
(0.0057) and to Forbes (0.00223), and it is only slightly lower
than that now accepted (0.0053; see Appendix A).
10.12. The fact that the conductivity of ice is considerably
larger than that of water gives rise to an interesting phenomenon
that has been noted by H. T. Barnes. 6 When ice is being frozen
on still water, particularly when the surface is kept very cold as
by liquid air, ice crystals grow out into the water and are found
in the ice with their long axes all pointing normal to the plane of
the surface. It is probable also that their conductivity is greater
along this axis. "See International Critical Tables." 64 ' v " 231
10.13. It may be noted in connection with the study of the
formation of ice that the temperature of the surface, which, as
we have seen, is the controlling factor as regards the rate of
freezing, is determined by a variety of conditions; for, while in
most climates and under most weather conditions this is largely
dependent on the temperature of the surrounding air, in cases
where the air is exceptionally clear so that an appreciable amount
of radiation can take place to the outer space that is nearly at
absolute zero, the surface of the ice may be considerably cooler
than the air. Thus, the natives of Bengal, India, make ice by
exposing water in shallow earthen dishes to the clear night sky,
even when the air temperature is 16 to 20F above the freez
ing point.*
* See Tamura. 148 See also Sec. 5.12 on "ice mines.' 1
SBC. 10.15] FORMATION OF ICE 199
10.14. Applications. While problems involving latent heat
have been handled in the preceding chapters, the solutions have
either neglected this consideration or taken account of it by some
more or less rough approximation method. With the aid of the
deductions of the present chapter many of these problems could
now be treated rigorously, in particular such as relate to the
freezing or thawing of soil. The equations would be directly
applicable to this case if the thermal constants for soil were used
instead of those for ice or water, and if the latent heat of fusion
of ice was modified by a factor depending on the percentage of
moisture in the soil.*
The theory would also apply to many cases of ice forma
tion in still water, for either natural or artificial refrigeration,
while, as already noted, it has been used by Stefan in connection
with polar ice.
10.15. Problems
1. Applying Stefan's formulas, find how long, if T, = 15C, it will take
to freeze 5 cm of ice (a) to the first approximation, and (6) to the second
approximation. Use k = 0.0052, c = 0.50, p = 0.92, a = 0.011, L 80 cgs
for ice. Ans. 3.28 hr; 3.39 hr
2. Using only the first approximation of Stefan's formula, find how long
it would take to thaw 5 cm deep in a cake of ice, supposing that the water
remains on top, and that the top surface of water is at +15C. Use
a = 0.00143 cgs for water. Ans. 12.95 hr
3. Using Stefan's first approximation formula, find how long it would
take for the soil to freeze to a depth of 1 m if the average surface temperature
is 10C and the soil initially at 0C, and if the soil has 10 per cent moisture.
Use c = 0.45, a = 0.0049 cgs for the frozen soil. Ans. 21 days
4. Assume that T s varies with time, so that the rate of freezing of ice is
constant, and that this rate is such that 5 cm will be frozen in the time deter
mined in Problem la. Determine T, for 1 hr, 4 hr, and 10 hr.
Ans. 9.5C; 41C; 123C
5. If Ci = 15C and C 2 = +4C in Neumann's solution, how long
will it take to freeze 5 cm of ice (cf. Problem 1)? Ans. 3.8 hr
* See also Sec. 7.10, Problem 5, and Sees. 7.19, 7.20, and 11.17.
CHAPTER 11
AUXILIARY METHODS
OF TREATING HEATCONDUCTION PROBLEMS
11.1. In this chapter we shall consider various methods of
solving particular heatconduction problems other than by the
classical calculations and experiments already described. Home
of the methods are electrical in character, others graphical or
computational. Some apply to the steadystate flow, others to
the unsteady state. While the principal use of these methods is
to provide a relatively quick answer to problems whose solu
tion by rigorous analytical methods would be difficult, they also
sometimes allow the handling of cases impossible of treatment
by the Fourier analysis. The accuracy is in general limited
mainly by the pains one is willing to take.
METHOD OF ISOTHERMAL SURFACES AND FLOWHWNES
11.2. This is a graphical method* of considerable use in
treating steadystate heat conduction in two dimensions, involv
ing the construction of an isotherm and flowline diagram. As
an illustration we shall apply it to the case of heat flow through
a "square edge/' e.g., one of the 12 edges of a rectangular furnace
or refrigerator. Figure 11.1 represents a section of such edge,
with inner and outer surfaces at temperatures TI and 7%, respec
tively. The five lines roughly parallel to these surfaces, save
where they bend around at the edge, are isotherms that divide
the temperature difference T }  7 7 2 into six equal parts of value
AT each. The heatflow lines are everywhere at right angles
(Sec. 1.3) to the isotherms, and there is a steady rate of flow (/
down any lane between these flow lines. For a wall of height
y normal to the diagram we have for the flow down any lane
across a small portion such as A BCD of average length u and
* Awbery and iSeho field. 5
200
SEC. 11.3]
AUXILIARY METHODS
201
width v, q = kyvkT/u. Then, if u = 0, as is approximately the
case for all the little quadrilaterals (for the diagram is so con
structed, as explained later), the flow down any lane is q kykT.
This is the same for all lanes since AJ 7 is the same between any
two adjoining isotherms. Careful measurement of the diagram
FIG. 11.1. Isotherms and flow lines for steady heat conduction through a wall near
a square edge.
will show that such an edge adds approximately 3.2 lanes to
the number that would be required if the spacing were uniform
and equal to that remote from the edge. This means an added
heat flow due to the edge of
= 3.2ky
(T, 
6
where x is the wall thickness. In other words, to take account
of edge loss we must add to the inside area a term 0.54t/:r, where
y is the (inside) length of the edge. This is in agreement with
the results of Langmuir, Adams, and Meikle 81 (see Sec. 3.4).
11.3. In solving problems by this method one must first
decide on the number of equal parts into which he wishes to
divide the total temperature drop T\ T^ (in this case six is
used although four or five would give fairly satisfactory results)
and then locate by trial the system of isotherms and flow lines
so that they intersect everywhere at right angles to form little quad
202
HEAT CONDUCTION
[CHAP. 11
rilaterals that approximate squares as closely as possible; i.e.,
the sums of the j opposite sides should be equal, or
AB + CD  BC + AD
When this is accomplished, the flow ky&T in each lane is the
same between a given pair of isotherms, and, since the flow
down any lane is the same throughout its length, the value of
AT 7 between any two adjoining pairs of isotherms must be the
(a) (b)
Fio. 11.2. Isotherms and flow lines for a steam pipe with (a) symmetrical and (6)
nonsymmetrical coverings.
same. As explained in Sec. 11.8, a little simple electrical
experimentation is useful in shortening the time required to
locate the isotherms.
11.4. Nonsymmetrical Cylindrical Flow. We shall also
apply this method to the problem of nonsymmetrical or eccentric
cylindrical flow, e.g., as in a steam pipe whose covering is thicker
on one side than the other. Figure 11.2 represents two half
sections of a steam pipe with a covering that in case (a) is sym
metrical, while in (b) it is three times as thick on one side as
SEC. 11.5)
AUXILIARY METHODS
203
the other. Here the number of lanes in the half sections is
21.5 for the concentric case and 24.2 for the eccentric. This
gives a heat loss for the eccentric case of 1.125 times that of the
other, for pipe and covering proportional to the dimensions
shown here, i.e., radius of pipe equal to 0.64 radius of covering
(cf. Sec. 11.9).
11.5. Heat Loss through a Wall with Ribs. As another illus
tration cf this graphical method we shall apply it to the prob
lem* of heat flow through a wall as affected by the presence of
FIG. 1 1.3. Isotherms and flow lines for steady heat conduction through a wall with
internal projecting rib of high conductivity.
internal projecting fins or ribs. It is assumed that the rib has a
high conductivity as compared with the insulating material of
the wall so that it is an isothermal surface taking the tempera
ture TI of the surface of the wall that it joins. Figure 11.3
shows the isotherms and flow lines constructed for the case of
a rib projecting twothirds through the wall thickness. The
graph shows that there are 22 lanes, i.e., 11 on each side, in the
region affected by the rib, while with the normal undisturbed
spacing shown in the extreme left of the diagram there would
be 16.6 lanes in the same length of wall. The difference or 5.4
lanes represents the heat loss due to the rib. Since each of the
undisturbed lanes has a width equal to onesixth the wall thick
ness, this means that such a rib, whose length is twothirds the
wall thickness, causes the same heat loss as a length of wall
5.4/6 or 0.9 the wall thickness, t
* Awbery and Schofield 5 ; see also Carslaw and Jaeger. 270 '* 8 "
t For further references and methods of taking account of change of conduc
tivity with temperature, see McAdams. 90 ' pp  18 ' 17
204 HEAT CONDUCTION [CHAP. 11
11.6. Threedimensional Cases; Cylindricaltank Edge Loss.
The preceding cases are essentially twodimensional in character
in that the third dimension, which is perpendicular to the plane
of the figure, affects the problem only as a constant factor. As
a threedimensional example we may investigate the edge losses
for a heavily insulated cylindrical container with spherically
shaped ends, such as is used in shipping very hot or very cold
liquids, e.g., liquid oxygen. Figure 11.4 represents a section of
such tank covered with thick insulation. In this case the radius
of the spherical end of the tank is equal to the diameter of the
cylinder.
To calculate the heat loss for such a tank we shall imagine
ourselves cutting a thin wedgeshaped slice, perhaps Koo of the
whole tank, by rotating the figure three degrees or so about the
axis of the cylinder; we shall investigate the heat loss for this
wedge. The same condition q = kyvkT/u holds as in the pre
ceding cases, but here y is not constant; thus, u, instead of
being equal to v, must be proportional to yv. The thickness y
of the wedge is obviously proportional to the distance from the
axis, and so for a constant v, as occurs in the cylinder at a point
such as A well away from the ends, the distance u between iso
therms is proportional to this distance from the axis. This
means that the little elements, which are drawn as squares for
the innermost row in the cylindrical insulation, become more
and more elongated rectangles for the outer rows. A little
thought will show that for the spherical ends the distance
between isotherms must vary as the square of the radius of the
sphere.
Figure 11.4 has been constructed to meet these various con
ditions as closely as possible. The proportions for the rectangles
in each row have been preserved, for the cylindrical part or for
the spherical part, as nearly uniform as possible when fitting
around the edge. The flow down each channel that starts at the
cylindricaltank wall is the same, as in the cases previously
considered, but for the spherical end the channels farthest from
the axis evidently count the most because the height y obviously
diminishes toward the axis. Measurement shows that the
SEC. 11.7]
AUXILIARY METHODS
205
edge loss for such an end can be taken account of by adding
33 per cent of the insulation thickness to the cylindrical length
in computing the total heat loss. This means that the spherical
end loss is to be computed as the loss through the fraction of the
A
Axis of cylinder
FIG. 11.4. Construction of isotherms and flow lines to show edge losses at the
spherically shaped ends of a cylindrical tank (Sec. 11.6).
sphere of solid angle determined by the tank end, and the cylin
drical loss computed in the usual way (Sec. 4.7), with the cylin
drical length increased by twothirds the insulation thickness to
take account of the edge losses at the two ends.
ELECTRICAL METHODS
11.7. The fundamental equations for heat flow are identical
with those for the flow of electricity. Ohm's law corresponds
to the conduction law, potential difference to temperature differ
ence, electrical conductivity to heat conductivity, and electrical
capacity to heat capacity. This means that electrical methods
can be used to solve many of the problems of heat conduction
and sometimes with a great saving of time. Perhaps the most
extensive application of electrical methods is in the work of
206 HEAT CONDUCTION [CHAP. 11
Paschkis 107 ' 108 ' 109 and his associates. By means of a network
of resistances and condensers the electrical analogy of a heat
flow problem can be set up and a solution reached.
Much simpler electrical arrangements can be used to solve
certain steadystate heatflow problems, with k constant, such
as the heat flow through the edges (cf. Sees. 3.4 and 11.2) and
corners of a furnace or refrigerator. Langmuir, Adams, and
Meikle 81 made measurements of the resistance of suitably shaped
cells with metal and glass sides filled with copper sulphate solu
tion, to solve these and similar problems.
A less direct method* makes use of a thin sheet of metal or
layer of electrolyte in which the current is led in at one edge or
several edges and out at another. The equipotential lines (cor
responding to the isotherms) can then be determined and the
lines of current flow (heat flow) drawn perpendicular to them.
11.8. One of the present authors has done more or less
experimental work along these lines and finds that if the accuracy
requirements are only moderate i.e., allowable error of a few
per cent as is the case in most heatconduction measurements
very simple arrangements will suffice. For a twodimensional
case a flat, level glassplate cell is used with a layer of tap water
2 or 3 mm deep. Metal electrodes of the desired shape, e.g.,
the outside and inside of a square edge (cf. Fig. 11.1), are con
nected with a 1,000cycle microphone " hummer. 7 ' Two metal
probes or points connected with earphones are used to determine
the equipotential lines. In doing this, one point is fixed and the
other moved until the sound is a minimum. While the con
struction method described in Sec. 11.3 will, if carefully carried
out, locate unambiguously the isotherms and flow lines, time
may be saved by the use of the electrical method to get the form
of these isotherms.
A series of measurements was also made on the resistance of
cells shaped as square edges or corners, and the formulas of
Langmuir (Sec. 3.4) were checked. These cells were made
rather simply of metal and glass and filled with tap water with
a few drops of sulphuric acid. The resistance was measured
* See e.g., Schofield." 4
SBC. 11.9]
AUXILIARY METHODS
207
with a Wheatstonebridge circuit, the hummer being used as a
battery and phones in place of galvanometer.
11.9. Eccentric Spherical and Cylindrical Flow. With the
aid of simple apparatus like this a rather important problem
that presents considerable analytical difficulty was solved.
This is the question already treated graphically in Sec. 11.4
for the cylindrical case of heat flow between eccentric cylin
drical or spherical surfaces. The apparatus consisted of a cell
(for the cylindrical case) with glass bottom, to which was waxed
a brass cylinder of 19.73 cm (7.76 in.) inside diameter. Cylin
ders of outside diameter 0.63, 4.92, 12.70, and 17.83 cm were
used in turn as the inner electrode and the cell were filled to a
depth of 16 cm with tap water. In the case of the sphere the
outer shell was of 25.5 cm inside diameter, and the inner spheres
of 3.81, 11.41, and 15.41 cm outside diameter, respectively.
In each case the internal cylinder or sphere could be moved from
the concentric position to any other within the limits. Capacity
effects gave little trouble except in the cases of the larger internal
cylinders or spheres. Resistances were measured with a Wheat
stonebridge circuit as mentioned above.
TABLE 11.1. RELATIVE HEAT LOSSES FOR ECCENTRIC CYLINDERS AND
SPHERES*
Cylinders
Spheres
Insulation
thickness on
thin side,
r =
r =
r ~
r =
r =
r
per cent
0.03/2
0.25/2
0.64/2
0.90/2
0.15/2
0.60/2
100
1.00
1.00
1.00
1.00
1.00
1.00
90
1.00
1.00
1.01
1.01
1.00
1.01
80
1.01
1.01
1.02
1.02
1.00
1.02
70
1.02
1.03
1.04
1.04
1.01
1.03
60
1.05
1.05
1.08
1.08
1.02
1.05
50
1.08
1.10
1.13
1.14
1.03
1.08
* Based on resistance measurements.
The results are summarized in Table 11.1, which shows that
if the internal cylinder or sphere is shifted from the concentric
208 HEAT CONDUCTION [CHAP. 11
position (100 per cent) until the insulation thickness on the thin
side is reduced to 50 per cent of its initial value (i.e., is three
times as thick on one side as on the other), the heat loss will be
increased by some 3 to 14 per cent according to the relative
sizes of the internal cylinder or sphere (radius r) and the external
one (radius R). It shows, furthermore, that the effect is less
when the internal cylinder or sphere is small relative to the
external one, and that it is less for the sphere than for the
cylinder. The measured 13 per cent increase in the lowest line
of column 4 (r = 0.64#) is to be compared with the 12.5 per
cent obtained by the graphical solution of the problem in Sec.
11.4.
It may be pointed out that these results may be applied at
once to problems involving electrical capacity, e.g., a coaxial
cable with eccentric core.
SOLUTIONS FROM TABLES AND CURVES
11.10. A number of tables for determining temperatures in
the unsteady (i.e., transient or buildingup) state of heat flow
are available, and one of the most useful, taken from Williamson
and Adams, 161 is reproduced in Table 11.2, which is, in effect,
a brief synopsis of Table 9.1. This allows the determination
of the temperature T at the center of solids of various shapes,
initially at temperature T Q uniform throughout the solid, t sec
(cgs) or hr (fph) after the surface temperature has been changed
to T 8 .
From Table 11.2 we can conclude that if a sphere of gran
ite (a = 0.016 cgs) of radius 15 cm and at a temperature of
T Q = 100C has its surface temperature suddenly lowered to
T 8 = 0C, the center temperature 4,500 sec later
will be T = 8.5C. If T Q = and T s = 100C, the tempera
ture after 4,500 sec will be 91.5C.
11.11. Charts. A large number of charts,* of which the best
known are the GurneyLurie, 64 are available for the ready calcu
* See, e.g., Me Adams, 90 pp  32 '/ Ede. 35
SBC. 11.12]
AUXILIARY METHODS
209
lation of temperatures in slabs, cylinders, spheres, etc. These
apply not only to the case of constant surface temperature but
also for known temperature of surroundings with various surface
coefficients of heat transfer.
TABLE 11.2. VALUES OF (T T,)/(T<> T t ) AT THE CENTER OF SOLIDS OF
VARIOUS SHAPES
at/b**
Slab
Square
bar
Cube
Cylinder
of infinite
Cylinder
of length
Sphere
length
= diam.
1
1
1
1
1
1
0.032
0.9998
0.9997
0.9995
0.9990
0.9988
0.9975
0.080
0.9752
0.9510
0.9274
0.9175
0.8947
0.8276
0.100
0.9493
0.9012
0.8555
0.8484
0.8054
0.7071
0.160
0.8458
0.7154
0.6051
0.6268
0.5301
0.4087
0.240
0.7022
0.4931
0.3462
0.3991
0.2802
0.1871
0.320
0.5779
0.3340
0.1930
0.2515
0.1453
0.0850
0.800
0.1768
0.0313
0.00553
0.0157
0.00277
0.00074
1.600
0.0246
0.00060
0.00015
3.200
0.00047
* 6 is the radius or half thickness.
As was made clear in Table 9.1, there are a number of cases
in which the results for two or threedimensional heat flow may
be obtained directly from the onedimensional case. Thus, for
the case of the brickshaped solid, as shown in Sec. 9.43* the
solution is readily obtained by multiplying together the three
solutions for slabs whose thicknesses are the three dimensions
of the brick. It is to be noted in Table 11.2 that the values for
the square bar are the squares of the slab values, while those for
the cube are the cubes. Also, the shortcylinder values are the
product of those for the long cylinder and the slab.
THE SCHMIDT METHOD
11.12. It is possible to arrive at an approximate solution of
an unsteadystate heatconduction problem by methods, graph
ical or otherwise, involving only the simplest mathematics.
The accuracy depends on the number of steps used in the solu
* See also Newman. 101
210
HEAT CONDUCTION
(CHAP. 11
tion. Many a problem whose exact analytical solution is very
difficult can be solved in this way with an accuracy sufficient
for all practical purposes.
The best known approximation method is the graphical
Schmidt method. 123 * As an illustration of this we shall con
T 2
1 .2 3
Distance from surface
FIG. 11.5. Application of the Schmidt graphical method to onedimensional
unsteadystate heat flow in a semiinfinite solid whose initial temperature is given
by the dashed line, with surface at temperature T 8 .
sider onedimensional nonsteady heat flow in a body whose
plane face is at temperature T 8 (i.e., case of semiinfinite solid).
Imagine a series of planes Ao: apart in the body and let the
initial temperature To be represented by the heavy dashed line
in Fig. 11.5. As a matter of fact, the temperature distribution
might be anything, e.g., T Q = 0, but, for reasons that will appear
in connection with the next illustration, it is somewhat easier
to explain the process with a distribution of the type given here.
The average initial temperature gradient in the first layer
is (T 9 Ti)/&x, and in the second, (T\ TJ)/kx. Then, in
*See also Sherwood and Reed, 129 '" 241 Fishenden and Saunders, 39 * 77
Me Adams, M ' p ' M and Nessi and Nissole. 100 For a precursor of this method see
Binder. 14
SEC. 11.13] AUXILIARY METHODS 211
time Ai the heat flow per unit area from the surface to plane 1
will be kAt(T 9  5Pi)/As heat units, while kkt(Ti  5P 2 )/As
heat units will flow away from plane 1 to plane 2. The differ
ence will remain in the vicinity of plane 1 and will heat a layer
Ax thick that centers on plane 1. Then,
mw^M _ ME. T. _
where T( is the temperature in plane 1 at time A (T" is like
wise the temperature in the same plane at time 2A, f" the
temperature in plane 4 at time 3A, etc.). This gives
,
Tl " (5ri ""
2
Now if Ai is taken of such size that
(Ax) 2 (Ax) 2
23J1 ie.,*^ (c)
we have TJ = ^ * ^ (d)
This means that the temperature in plane 1 at time A is the
arithmetic mean of the temperatures in planes and 2 at time
0. In the same way it can be shown that the temperature in
any plane at any time is the arithmetic mean of the temperatures
in the planes on each side of it that prevailed A previously.
This choice of At as determined by (c) is the principle of the
Schmidt method. Figure 11.5 illustrates how the lines are
drawn to determine the arithmetic means and therefore give the
temperatures in the different planes for various intervals, in
this case for times up to 3A(. Particular care must be taken in
constructing such a diagram to see that the lines connect only
points representing the same time interval, e.g., T" and Ti',
etc. The temperature at time 3A< would be represented approxi
mately by drawing a smooth curve through the points T'".
11.13. Cooling Plate. The Schmidt method lends itself
particularly well to calculations on the slab or plate. As an
illustration the graph is worked out in Fig. 11.6 for a plate
initially at a uniform temperature TQ whose surfaces are sud
212
HEAT CONDUCTION
[CHAP. 11
denly lowered to T B . The plate is considered as divided into
10 layers, but, because of symmetry, only half of it need be
represented. Obviously, the temperatures could be reversed
so that the problem is one of heating instead of cooling.
Two points are to be noted here that did not appear in con
nection with the graph of Fig. 11.5. The first is that here the
23456
Planes
FIG. 11.6. The Schmidt graphical method applied to the cooling of a plate initially
at temperature TQ. The center of the plate is at plane 5.
temperatures change only every other period. A little experi
ence with these graphs will show that this is inherent in the
construction when the initial temperature is uniform throughout
the solid. This is a matter of little moment since a smooth
curve, using a little interpolation, can always be drawn. The
second matter is in connection with the determination of the
center temperatures, plane 5 in this case. Because of symmetry
the temperatures in plane 6 are identical with those in plane 4.
Accordingly, the points in 5 are determined by connecting cor
responding points in 4 and 6; e.g., point 9 in plane 5 is found by
connecting the two points 8 in planes 4 and 6.
SEC 11.14] AUXILIARY METHODS 213
It is of interest to compare the conclusions from Fig. 11.6
with the results of classical theory. Let us use as an example
a large steel plate 1 ft thick at a temperature of 1000F with
surfaces suddenly lowered to 0F; assume average diffusivity
for this temperature range, 0.40 fph. Since each of the 10
layers is 0.1 ft in thickness, the time interval from (11.12c)
is Af = 0.01/0.80 = 0.0125 hr. The time t at the end of
the fifteenth interval is then 0.1875 hr. From (8.16n) we
can at once calculate the temperature of the center of the plate
for this time as 607F, while Fig. 11.6 gives about 575F.
Obviously, division into thinner layers will give more accurate
results.
The Schmidt method is also capable of handling many varia
tions of the simpleslab case,* and Nessi and Nissole 100 have
worked out methods by which it is possible to apply it to cylin
drical and spherical bodies. Its field of greatest usefulness,
however, is the case of linear flow with thermal constants not
dependent on temperature. When applicable it is probably the
simplest approximation method.
THE RELAXATION METHOD
11.14. This method is an ingenious application by
Emmons 36 ' 37 of the relaxation method of Southwell. 31 ' 137 It is
applicable to one, two, or threedimensional problems for either
the steady or unsteady state of heat conduction and, for one
dimension, is practically identical with the Schmidt method.
It is particularly useful in giving quick and reasonably accurate
solutions of problems involving shapes such as edges, etc., not
easily treated by other methods.
We shall illustrate the use of this method by a single simple
example f of steady twodimensional flow. This is the loss from
a square edge already treated in Sees. 3.4, 11.2, and 11.8.
Figure 11.7 represents a section near the square edge of a rec
tangular furnace 24 by 24 in. inside, with a wall 10 in. thick.
The inside surface of the wall is at a temperature of 500F and
the outside at 100F. It is desired to find the temperature at a
* Seo Me Adams, . 42 Sherwood and Reed. 129 * 260
+ tfmrnrm* 37, p. 609
214
HEAT CONDUCTION
[CHAP. 11
series of midpoints A, B, C, an4 D in the wall, and the heat
loss from the furnace.
We shall assume that the heat is effectively conducted not
by the continuous material of the wall but along a series of
"rods " from point to point, forming a square lattice as indicated.
When a steady state of heat transfer is reached, there will be
a balance between the heat flowing to and away from one of the
points A, B y etc., and we shall endeavor to fix the temperatures
s , s~
^.
JS
. 12 in. *
> k S=5/n. H
F *'
/
/
/
/r $'
500
B
T $'
500 1
W.J//7T*
k~<H
C
7 1 5'
Z0/>7.>
^,
^>
7*
*
.c:
$
400o^
/o,
y50 2
/ 30 3
/ 16 5
/ 07
/
300
200, 100,
2
I80 3 20 3
28 4
!S37 05
* 4 '
300
25 2
275 2 1 4
? 5
2s
?&
300
84
292 4 Oe
300
222s
i
/loo
100
10D
100
too
FIG. 11.7. The Emmons relaxation method applied to calculate the steady heat
flow through a square edge of a rectangular furnace.
of these points so that this will be the case. Until this is done,
however, there will be temperature arrangements in which
more heat is conducted to a point than is taken away, in which
case a positive heat sz'nfc of magnitude s'* will be required,
while the reverse means a negative sink. Since each of these
points in the plane is connected with four others, a lowering of
its temperature by 1 means heat coming in from the surround
ing four points with a gradient of 1 in distance 8, requiring a
heat sink of magnitude 4; i.e., the numerical change in the heat
sink at a point is four times the temperature change of the
point.
* The unit here is the amount of heat that would flow along a rodin unit time
with unit temperature difference between its ends.
SEC. 11.15] AUXILIARY METHODS 215
11.16. In explaining Fig. 11.7 we must first give each of the
points A, B, C, and D a midtemperature of 300F. The sub
script indicates that this is the initial step, and the subscripts
1, 2, 3, ... show subsequent successive steps. It is evident
that this initial assumption means a balance between inflow
and outflow for J5, C, and Z), as indicated by s' =0; but wfcle
A is receiving nothing from B or E, it is losing heat to F and G
under a 200F temperature drop, and this means a (negative)
sink of magnitude 400. Accordingly, we " relax" the tem
perature of A by subtracting 100F, which reestablishes the heat
balance so that s' is now 0. But this destroys the balance for
B that now must have a sink of magnitude 100, so that the
second step is to relax B by lowering its temperature 25F.
This, since it applies equally to E, requires a sink of 50 for
A, likewise a sink of 25 for C, but it reduces the sink at B
to 0. The third step is to lower A 20F more, which results
in a positive sink of 30 at A but a negative sink of 20 at B.
The fourth step is a lowering of 8F for C, which raises its sink
to +7 but gives a 8 sink to D and lowers the sink at B to
28. The remaining steps are clearly indicated, and after the
heat sinks are reduced to or negligibly small values, the tem
peratures arrived at are those underlined.
To calculate the heat flow for a section of furnace 1 ft high
we note that each rod, save the one through D, effectively carries
tRe heat from an area 1 ft high and 5 in. wide. The heat trans
ferred in unit time along the rod running from the inside surface
of the wall through B would then be
% y 12 (500 268")
Q = k CAA x 57 " = 232k heat units < a )
144 M2
Thus, the total transfer through one side, including edge, would
be
Q = 2k ^232 + 208 + ^ 202\ = l,244fc units (6)
\ & /
(Note that there is no transfer considered through A since no
rods from the inside pass through A.) The loss through a slab
2 ft long, 1 ft high, and 10 in. thick, with a temperature differ
216 HEAT CONDUCTION [CHAP. 11
ence of 400F would be
Q k 2 * 40 = 960fc units (c)
The edge then increases the loss in the ratio 1,244/960 = 1.296.
The Langmuir formula (Sec. 3.4) gives a ratio in this case of
1.224, which is in satisfactory agreement considering the few
points used. With a finer net, i.e., more points, a greater
accuracy is naturally attained. For further illustrations of this
interesting and useful method the reader is referred to the
Emmons papers. 36 ' 37
THE STEP METHOD
11.16. There are a number of other approximation methods
for the solution of heatconduction or similar problems, all more
or less related to the preceding but in general more complicated.
We shall complete our discussion by describing in some detail
a simple scheme of wide applicability for handling specific
numerical problems, which will be referred to as the "step
method." This consists in imagining the body divided into
layers and the time into discrete intervals. The temperature
throughout any layer is considered uniform and constant
throughout any interval and the heat flow from layer to layer
is computed, and from this the corresponding temperature
change. There is nothing original in the principle of this
method; like the replacement of an integral by a series it is* a
procedure that almost everyone has had to make use of at one
time or another. It involves the same principles as the Schmidt
method but lacks its ingenuity. On the other hand, its field
of application is wider. It will handle problems involving
changes in thermal constants with temperature, release of latent
heat of fusion as in ice formation, etc., which would be difficult
of solution in any other way.
While the step method is exceedingly simple in principle,
there are a number of factors that must be taken into account
in its application if one wants to secure best results. Accord
ingly, we shall illustrate it by using it in solving a variety of
problems.
* Carlson, 88 Dusinberre, 85 Frocht and Leven, 44 Shortley and Weller, 130 and
Thorn.*"
SBC. 11.17] AUXILIARY METHODS 217
APPLICATIONS OF STEP METHOD
11.17. Ice Formation about Pipes; Ice Cofferdam. Our
first and simplest illustration will be a problem in ice formation.*
Certain openpit mining and damconstruction operations 43 ' 48 in
wet soil have been carried on by first driving a circle of pipes into
the soil and then, by introducing cold brine or other coolant,
freezing a cylinder of ice about each pipe until they unite to
form a circular cofferdam. We shall calculate the time required
to freeze cylinders of various sizes. The same principles will
of course apply to almost any case of ice formation about pipes.
Let us assume a long 4in. pipe (outside radius 5.72 cm)
driven into soil of temperature 0C. Assume a 50 per cent
(by volume) water saturation and a latent heat of fusion of
40 cal/cm 3 . The outside of the pipe is kept at 5P C, and,
since specificheat considerations are secondary here to latent
heat, it is assumed f that the temperature distribution and heat
flow are similar to those in the steady state. As the ice is
formed, the latent heat released is conducted radially through
the frozensoil cylinder (assumed conductivity 0.0045 cgs) to
the central pipe.
Call r 2 the radius of the frozensoil cylinder at the beginning
of any time interval A and r 3 the radius at the end. The average
radius r a = (r 2 + r 3 )/2, the volume of the cylindrical layer of
frozen soil, per cm cylinder length, is Tr(r\ rl), and the latent
heat released is 4Qir(rl r%) cal/cm. Applying (4.6/) for the
steady state of radial conduction per cm length of a cylinder,
we have for the heat transfer in A sec,
2.303 lo glo r a /5.72
This gives, if A< is in days,
T M  20(r ~ **> X 2 ' 303 logl r / 5  72
0.0045 X 86,400

= 0.1185(7171) logic f2 (6)
* For the analytical solution of this problem see Pekeris and Slichter. 110
t Pekeris and Slichter. 110 ^ "*
218
HEAT CONDUCTION
[CHAP. 11
s
e
00
S!
~"t
rH
rH
8
CO
*
1
oo
O
CO
^
rH
8
.J^ R
g 00 Tt<
CO OS OO
OS CO C^
OS C^ *O
00 rH C^l
S28
10
3^^
CO rH <N
C^ CO CO
t t t
t^ 00 00
00 t*
rH rH
fei
ss^
t^ IO 00
^f OO t>
C^ Q IO
00 iO CO
00 O*
^
o o o
rH rH rH
CO CO CO
CO CO CO
rH rH
~%
II II II
II II II
II II II
II II II
II II
N rH
^^^
^^^
B< s S^
Ss 6s ^
6s 6s
3
1!
rT CNP rH
II II II
^^ fe;
II II II
OS rH OS
II II II
ocT r^ t>^
rH rH rH
II II II
oT cT
<N CO
II II
1
f^fe R
8
rH
1
oo
CO
00
88
5.73
O
(N
to
00
g &
8
rH
rH
00
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00
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<j n
rH
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rH
n &2
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(N
10
to
3l
rH
10
s
8
o
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fe,
o jC^
B5
rH
CO
s
1
i
^o *\io
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P3
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CM*
00
CO
rH
rH
rH
CO
00^
rT
K.
a
*
od
CO
rH
3
3
if 1
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<N
2
S
*
f a
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B
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s
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00
CQ
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SBC. 1L17]
i
O
q
'58
s
8
 3
z g
f a
&
AUXILIARY METHODS
sis *
CO O
II II
89
n n
oo
o
12
o ^ o
II II
II II
8
10
_8_
"^T"
t^
S8
11 II
lO
00
00
S
10
10
S
S
219
220 HEAT CONDUCTION (CHAP. 11
Two calculations will be made: one for T Q constant and the
other for To = 24 sin (7r^V/195)C* corresponding to a case
where the cooling of the liquid, which is circulated through the
pipes, is provided naturally by winter temperatures. N is the
number of elapsed days since the beginning of the freezing
process, and the angle in brackets is measured in radians.
The step calculations are given in Table 11.3. The first
layers of ice are taken as 5 cm thick, then 10, 20, 30, 40 cm,
respectively. It will be noted in columns / and K that a cer
tain amount of trial and error is involved in arriving at the
value of N and the corresponding A<, for T Q must obviously be
taken as the mean temperature for the period under considera
tion. This means that N must come out as the approximate
average of the initial and final times for the period, or, in other
words, for any layer the acceptable value of N must approxi
mately equal onehalf the At (column K) for that layer plus the
t (column L) of the preceding layer. The last listed is the
accepted value. The values in column M are calculated by
Pekeris and Slichter.
The results are seen to be in satisfactory agreement with the
Pekeris and Slichter calculations. It is to be noted from (a)
and (6) that, for the case of a constant T , the time required
for the freezing of any particular size of cylinder is directly
proportional to the latent heat of fusion, i.e., to the moisture
content of the soil. It is also inversely proportional to the
conductivity of the frozen soil and inversely proportional to TV
By the use of extreme cooling measures the time required for
the production of the largest cylinder here considered might be
reduced to a very few months. This assumes a conductivity
independent of temperature, which in general would not be the
case. A more exact solution, taking account of such variation
in conductivity and also of specificheat considerations, could
be obtained as well by the step method.
11.18. Semiinfinite Solid; Wanning of Soil. The step
method will now be applied to the problem of onedimensional
heat flow from a warm surface into a solid at a cooler uniform
temperature. While one would not in general apply the step
* Pekeris and Slichter. 110 ." 137
SEC. 11.18] AUXILIARY METHODS 221
method to a problem for which the solution is so readily avail
able by analytical means or for that matter by the simple
Schmidt method it nevertheless serves in this case as a good
illustration.
We shall determine the temperatures at various depths and
times in soil (assume k = 0.0037; c = 0.45; p = 1.67; a = 0.0049
cgs) initially at 0C, whose surface is suddenly warmed to 10C.
Plane
FIG. 11.8. Application of the step method to the problem of warming a semi
infinite solid.
Imagine horizontal planes in the soil 5 cm apart (Fig. 11.8) and
let us inquire what will happen in the first 1,000 sec. In this
period it is assumed that heat flows from a surface at 10C
through a layer of soil 5 cm thick to plane 1 at 0C. This
amount of heat per square centimeter of area is
Q = 1,000 X 0.0037 X *% = 7.4 cal
It will go toward warming up a layer (A in Fig. 11.8) 5 cm thick,
centered on plane 1, and its temperature will rise by
^ = 1.97C
ocp
In the second interval, which is taken as 1,500 sec, heat will
flow from plane to plane 1 under an initial temperature dif
ference of 8.03C and from plane 1 to plane 2 under a difference
of 1.97C. The heat delivered to plane 1 in this interval is
8.91 cal, of which 2.18 cal flows on to plane 2, leaving 6.73 cal
to increase the temperature of plane 1 (A) by 1.79C, while
plane 2 (B) rises to 0.58C. This is the temperature in plane 2
at the end of 2,500 sec, while plane 1 is at 1,97 + 1.79 = 3.76C.
The step calculations are given in Table 11.4. Here A< is
the magnitude of the interval and t the total elapsed time at the
end of the interval. AT 7 is the temperature difference between
222
HEAT CONDUCTION
[CHAP. 11
TABLE 11.4. STEP CALCULATIONS FOB LINEAR HEAT FLOW INTO SOIL AT
0C WITH SURFACE AT 10C. As  5 CM, k  0.0037, cp = 0.752 cos
A
B
C
D
E
F
G
H
I
At,
t,
AT,
(mm If A* A 1* / \f\
Q*Q,
ST
/ Q~Qn\
/A
T f ( for
Plane
sec
sec
C
cal/cm*
cal/cm 8
( ^ )
C
mu),
1
1,000
1,000
10
7.40
7.40
1.97
1.97
1.11
1
2
1,500
2,500
8.03
1.97
8.9;
2.18
6.73
2.18
1.79
0.58
3.76
0.58
3.13
0.43
1
2
3
2,000
4,500
6.24
3.18
58
9.23
4.70
0.84
4.47
3.86
0.84
1.19
1.03
0.22
4.95
1.61
0.22
4.51
1.31
24
1
2
3
4
2,500
7,000
5.05
3 34
1 39
22
9 35
6.18
2.57
0.41
3.17
3.61
2.04
0.41
0.84
0.96
0.54
0.11
5.79
2 57
0.76
0.11
5.45
2.28
0.70
0.16
1
2
3
4
5
2,500
9,500
4.21
3.22
1.81
65
0.11
7 80
5.96
3.35
1 20
0.20
1.84
2.61
2.15
1.00
0.20
0.49
0.69
0.57
0.27
0.05
6.28
3 26
1.33
0.38
0.05
6.05
3.02
1.21
0.38
0.10
1
2
3
4
5
6
4,000
13,500
3.72
3.02
1.93
0.95
0.33
05
11 03
8.94
5.72
2.81
0.98
0.15
2.09
3 22
2.91
1 83
0.83
0.15
0.56
0.85
0.77
0.49
0.22
04
6.84
4 11
2.10
87
27
04
6.64
3.86
1.92
0.82
0.29
0.09
1
2
3
4
5
6
7
5,000
18,500
3.16
2.73
2.00
1.23
0.60
0.23
0.04
11.70
10.10
7.40
4.55
2.22
0.85
0.15
1.60
2.70
2.85
2 33
1.37
0.70
15
0.42
0.72
0.76
0.62
0.36
0.19
0.04
7.26
4 83
2.86
1 49
0.63
0.23
0.04
7.11
4.57
2.65
1.37
0.63
0.26
0.09
1
2
3
4
5
6
7
8
7,500
20,000
2 74
2.43
1.97
1.37
86
0.40
0.19
0.04
15.20
13.50
10.94
7.61
4.78
2.22
1.05
22
1.70
2 56
3 33
2.83
2.56
1.17
83
0.22
0.45
0.68
0.89
0.75
0.68
0.31
0.22
06
7.71
5.51*
3.75
2.24*
1.31
0.54*
0.26
0.06*
7.54
5.31
3 48
2.10
1.17
0.60
0.29
0.12
2
4
6
8
10
12,000
38,000
4 49
3.27
1 70
48
0.06
Change to As =
19 93
14.54
7 55
2 13
27
* 10 cm
5 39
6.99
5.42
1.86
27
72
0.93
0.72
25
04
6.23
3.17
1.26
0.31
0.04
6.05
3.00
1.21
0.38
0.09
2
6
8
10
12
20,000
58,000
3.77
3.06
1.91
0.95
27
0.04
27.90
22.63
14.14
7.03
2.00
0.30
5 27
8 49
7.11
5.03
1 70
0.30
70
1 13
0.95
0.67
23
0.04
6 93
4.30
2.21
0.98
0.27
0.04
6 86
4.03
2.08
0.94
0.36
12
SBC. 11.19] AUXILIARY METHODS 223
any two adjoining planes at the beginning of the interval. Q
is the heat transferred from plane to plane and 5T the corre
sponding temperature rise. T is the final temperature at time I ,
and Tf is the temperature calculated from (7.14d).
11.19. The following comments may be made on the step
calculations of Table 11.4:
1. It will be noted that the time intervals are taken progres
sively larger. This is a radical departure from the procedure
of the Schmidt method. When the time interval is uniform
and chosen according to the Schmidt scheme, the results of
the two methods are identical.
2. By occasionally doubling the value of Ax as was done at
t = 38,000 sec it is possible to speed the calculation greatly.
3. The results are in reasonably satisfactory agreement with
those of classical theory. The step method gives results that
are a bit too high for moderate distances from the surface and
too low for greater distances. This is because the temperature
gradient in the middle region is decreasing as time increases, so
that the value used, which is that at the beginning of the
interval, is larger than the average for the interval, which is
the value that should really be used. At greater distances the
reverse is true. A method of remedying this will be explained
in the next problem. Thinner layers and shorter time intervals
will of course give better results.
4. When the points show a tendency toward irregularity,
it may be necessary to " smooth " the curve for any interval and
then proceed from the smoothed curve. This is particularly
necessary when the time intervals are chosen rather large.
5. It will be noted that the first half layer is in effect neg
lected. This is in keeping with the principle of the method
that the temperature of each layer is that of its center, which,
in the first half layer, is the surface. Only in very special
cases, e.g., some cases of spherical heat flow, does this introduce
an error that need be considered.
6. When this is applied to a slab, it will be noted that the
center plane gets heat from both sides; thus, its temperature
rise is doubled. For this central plane we must accordingly
use twice the temperature rise as calculated above assuming,
224
HEAT CONDUCTION
(CHAP. 11
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SBC. 11.19]
AUXILIARY METHODS
225
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226 HEAT CONDUCTION [CHAP. 11
of course, symmetrical heating for the two faces. The next
illustration furnishes an example of this.
7. A little study will show that in cases like this where the
thermal constants are not dependent on temperature (over the
range used), the process may be somewhat shortened by making
more direct use of the diffusivity a.
11.20. Cooling of Armor Plate. We shall now apply the step
method to a problem whose solution by other schemes not
involving electrical or other experimentation would be of
doubtful feasibility. A large plane steel plate 0.8 ft (9.6 in.)
thick and at a uniform temperature of 1000F has its surfaces
cooled to 0F at the rate of 200F/min for the first 3 min and
100F/min for the next 4 min. The thermal constants are
assumed as follows: at 1000F, k = 22, c = 0.16, p = 480 fph;
at 0F, * = 27, c = 0.11, p = 490, with an assumed linear
variation between these temperatures. Temperatures inside
the plate will be calculated for various times. The calculations
would also hold without serious error for the range 1100 to
100F.
We shall divide the plate by planes 0.1 ft apart, and, because
of symmetry, it will be necessary to consider only half the
thickness. To try to avoid the error, which would be rather
serious in this case, mentioned in paragraph 3 of Sec. 11.19 we
shall use the average temperature for any time interval. This
involves no difficulty at the surface, but it is evident that for
any other plane the final temperature calculated for any interval
will be dependent on the average chosen. The best way to
arrive at the estimated final and average temperature for any
time interval is to plot the temperature curve for each interval
as determined and then project it for the next. If the final
values for the interval agree reasonably well with the projected
or estimated values, the results may be considered satisfactory.
The procedure involves trial and error and is in effect a relaxa
tion method.
The step calculations for the first 15 min of cooling are given
in Table 11.5 and some of the curves in Fig. 11.9. Column B
of the table gives the time interval used, and C the total elapsed
time at the end of each interval. AT in column G is the (aver
SEC. 11.20]
AUXILIARY METHODS
227
age) temperature difference between planes, and Q in column /
is the heat loss per square foot from the layer centering on any
plane. J gives the net heat loss and L the temperature change.
The doubled value for the center plane in column M has been
1000
FIG. 11.9. Calculated cooling curves for a steel plate 0.8 ft thick with thermal
coefficients dependent on temperature. See Sec. 11.20.
explained in comment 6, Sec. 11.19. When and not until
when the values in column M agree closely with those in JEJ,
the results for any interval are considered satisfactory. In
arriving at the estimated values for column E various expedients
of the trained calculator may be found useful, such as making
use of differences and, in particular, extrapolation of the curves
of temperature vs. time for each of the planes. Smoothing not
here mav be resorted to, to quicken the calculations.
228
HEAT CONDUCTION
[CHAP. 11
It is evident from a glance at columns H and K that any
calculation by classical methods involving the assumption of
constant thermal coefficients would be considerably in error.
It may also be remarked that cooling of the surfaces by radia
tion or by contact with a fluid, with known surface heat transfer
coefficient, would not present any insuperable difficulty to the
step method.
11.21. Heating of a Sphere. As a last illustration of the
step method we shall calculate the temperatures in a sphere of
FIG. 11.10. Application of the step method to the problem of heating a sphere.
glass, initially at 0C, whose surface is suddenly heated to
100C. This is of interest as a case of threedimensional flow
whose results can be easily compared with classical theory.
Assume (see Fig. 11.10) R = 10 cm, k = 0.0024, c = 0.161,
p = 2.60, a = 0.00573 cgs. Imagine the sphere divided into
layers 2 cm thick by spherical surfaces of radii 8, 6, 4, and 2 cm.
We shall consider the heat flow from the surface to layer A
(r = 8), then to B, and assume that the difference goes to warm
a spherical shell 2 cm thick, centered (as regards thickness) on
A. This shell would have radii 7 and 9 cm.
This case will be treated like the previous ones as essentially
one of quasilinear flow from layer to layer; we must accordingly
find the mean area to use in calculating the heat flow from the
surface to layer A, from this to JS, etc. Consider the equation
SBC. 11.21] AUXILIARY METHODS 229
for linear heat flow
AT 7
A<3 * kA AS A ' (a)
and the equation [see (4.5A;)] for heat flowing radially through a
spherical shell
_
^ ri  r 2 v '
If these two are equated, the average area A m to be used in (a)
is obtained. Considering that n r 2 is equivalent to Ax and
T l  T 2 to AT 7 , we have
X 47rri = A/Ai^4. 2 (c)
Using then the geometric means of the two areas, we have
A' = 47r X 10 X 8 = 47r X 80;5' = 4?r X 48;C" = 4?r X 24;and
D f = 4?r X 8. Likewise, the volumes of the 2cm thick spherical
shells (shaded portion in Fig. 11.10) whose heating we have to
consider are V A = ^7r(9 3  7 3 ) = $fa X 386; V B = %w X 218;
F c = ^TT X 98; Fjr, = ^TT X 27. (Layer D is taken as the
3cm radius core, which is assumed as uniform in temperature.)
4ir may be canceled throughout and the areas taken as 80, 48,
24, and 8 cm 2 , and the volumes as 128.7, 72.7, 32.7, and 9 cm 3 .
The step calculations are listed in Table 11.6, and the result
ant temperature curves are given in Fig. 11.11. As in Table
11.5 the values in column E are the estimated temperatures
for the end of each interval, giving therefore average values for
the temperatures and temperature differences in F and G. Trial
and error, with help from plotting, is used in arriving at values
for column E such that they will be in fair agreement with the
final temperatures as calculated in column M . Any estimated
values for E that do not lead to such agreement must be dis
carded. If less pains are taken than in the present calculations
and a larger departure between E and M allowed, fair results
can still be obtained by smoothing the curves. It is to be noted
that, as in the two previous illustrations, the first half layer is
(effectively) neglected and is supposed to assume the surface
temperature quickly.
The values in column N have been calculated from (9.16Z)
230
HEAT CONDUCTION
[CHAP. 11
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232
HEAT CONDUCTION
[CHAP. 11
and are seen to be in excellent agreement with the results of the
step method. In this formula the values of r used are 8, 6, 4,
and 1.5 for A, B, C, and D, respectively. From a practical
standpoint one would, of course, hardly expect to use the step
method on a problem involving constant thermal coefficients
and conditions as simple as this. It turns out, however, that
for the shorter times the step method may involve less labor
HEATING CURVES FOR GLASS SPHERE,/e/Ocm
Step cct/cu/arfions
* x Calculated
from formula
6.6 4 2
r, Distance from center, cm
FIG. 11.11. Heating curves for a glass sphere of radius 10 cm, initially at 0C,
whose surface is quickly raised to 100C. Calculations by the step method are
seen to give results almost identical with those of formula (9.16/).
than the application of the formula, because of the number of
terms required in the latter. In this case, smaller time inter
vals would have to be used. For the longer times the use of
the formula is much simpler.
11.22. Cylindrical flow may be treated by the same prin
ciples as those used in the last case. The average areas to be
used are logarithmic means defined by
4 A,
A m =
(a)
2.303 Iog 10 (Ai/ A 2 )
(If A i/ A 2 < 1.4, the above value is within 1 per cent of the
arithmetic mean, which may accordingly be used.) The step
SEC. 11.22] AUXILIARY METHODS 233
method has also been used with good success in treating a prob
lem whose analytical solution* presents some difficulty. This
is the case of the heat flow in an infinite solid bounded internally
by a cylindrical surface of controlled temperature a problem
of practical interest in connection with the air conditioning of
deep mines. In applying the step method to brickshaped
solids, rectangular bars, etc., the solution may be approximately
obtained, as indicated in Sec. 11.11, by multiplying together
solutions for the corresponding slab cases.
The step method should be particularly useful to geologists
in making possible the treatment of all sorts of special problems
such as the cooling of intrusions f of various sorts, either with
or without generation of heat (as in the decomposition of
granite). It would allow the treatment of cases where the tem
perature of the intrusion or the rate of heat generation is not
uniform, or even where the intrusion and surrounding rocks are
of different materials. While the step method is simplest to
apply when the boundary temperatures are known, a little
application of the trialanderror principle should give an
approximate solution of almost any problem of this sort, even
if radiation cooling is involved.
* Smith. 135 See also Carslaw and Jaeger. 28 For graphs of the solution of this
problem see Gemant. 44 "
 See Sees. 7.23, 8.9, and 9.3; also Lovering, 87 Boydell, 19 Berry, 13 and Van
Orstrand. 152
CHAPTER 12
METHODS OF MEASURING THERMALCONDUCTIVITY
CONSTANTS
12.1. From the similarity between the flow of heat and of
electricity it might be supposed that heatconductivity meas
urements could be made with an accuracy approaching that
of electrical conductivity. Unfortunately, this is by no means
the case. Temperature difference and heat flow are not as
easily and accurately measurable as their electrical analogues,
potential difference and current. Furthermore, while we have
almost perfect insulators for electricity, we do not have such
for heat. The result is that thermalconduction measurements
are seldom of greater accuracy than one or two per cent probable
error, and indeed the error is likely to be much larger than this
unless great care is taken.
It is not proposed to give here an exhaustive account of
methods of conductivity measurement but rather to limit the
discussion to several standard methods and certain others that
are interesting applications of the preceding theory. Those
who wish to pursue the subject further may consult the articles
dealing with heatconductivity measurement in Glazebrook, 46
Winkelmann, 162 Kohlrausch, 78 or Roberts, 119 the surveys by
Griffiths, 60 Ingersoll, 61 and Jakob, 67 and the modern discussions
by Awbery, 3  4 Powell, 1U ' U2 ' 113 ' U4 , and Worthing and Halliday. 163
12.2. The modern tendency in measuring thermal conduc
tivity is toward greater directness than formerly. All that is
necessary to determine this constant is a knowledge of the rate
of heat flow through a given area of specimen under known tem
perature gradient. The heat is almost always produced elec
trically. The simplest and commonest arrangement involves
flow in only one dimension. The chief difficulties here arise from
heat losses, and these may be minimized by the use of silica
234
SEC. 12.3] MEASURING THERMALCONDUCTIVITY CONSTANTS 235
aerogel for insulation and by the employment of guard rings.
(This means that the heat flow is measured only for a central
portion of the area where it is uniform.) Radialflow methods
eliminate most of these losses but have difficulties of their own.
Periodic or other variablestate (as regards temperaturetime
relations) methods have sometimes been used to give conduc
tivity, but more generally diffusivity.
12.3. Linear Flow; Poor Conductors. The standard method
here is to sandwich a flat electrically heated element between
two similar flat slabs of the material under test, on the farther
side of which are watercooled plates. A guard ring is used to
prevent losses that might otherwise be large. In one form of
this apparatus (Griffiths 50 ' 51 ) usable for specimens up to almost a
foot in thickness, the hot plate is 3 by 3 ft with a similarly
heated guard ring 1 ft wide and separated from the central
plate by a narrow air gap. The two cold plates and specimens
are 5 by 5 ft, and surface temperatures are determined by
thermocouples. The use cf the guard ring assures heat flow
normal to the surface all over the central hot plate, 3 by 3 ft,
whose energy input is measured. Apparatus of this general
type is also used by our National Bureau of Standards.
In a smallscale apparatus of this type developed by Griffiths
and Kaye 52 the specimens are 45 mm in diameter and 0.5 to
4 mm thick arranged on each side of an electrically heated
copper disk, the outer surfaces being in contact with water
cooled copper blocks. Thermoelements give the temperature
gradient. A guard ring is unnecessary. The method is well
adapted to porous materials under definite pressure.
Birch and Clark 16 have measured the conductivity of various
rocks by a variation of the preceding methods in which special
care is taken to avoid certain errors. Instead of using two
similar specimens, one on each side of the heating coil, only a
single specimen is used at a time. To eliminate loss of heat
the heater is surrounded by a "dome" that covers it and is kept
at the same temperature as the heater. The rock specimen is
0.25 in. (6.35 mm) thick and 1.50 in. (38.1 mm) in diameter.
It is surrounded by a guard ring of "isolantite" with outer
diameter of 3 in. The cold plate, heater, and dome are all of
236 HEAT CONDUCTION [CHAP. 12
copper with heating coils in the last two. The temperature
drop through the specimen is about 5C, and the whole apparatus
can be immersed in baths at temperatures up to 400C or more.
The special feature of the method is the use of atmospheres of
nitrogen and helium that give thin films of these gases between
the rock faces and copper plates; through these films the heat
is conducted to or away from the rock faces. By measurement
of the apparent conductivity in each gas it is possible to make
the small correction for temperature discontinuity at the rock
faces.
In a method useful for thin materials such as mica, the
specimen is clamped between the ends of two copper bars, one
of which carries a heating and the other a cooling coil. The
heat flow is determined by measuring with thermocouples
the temperature gradient along the bar, the conductivity of the
copper bars being known. This method has also been developed
so that it can be used at various points on a sheet of continuous
material.
Comparison methods go back to Christiansen. 30 The speci
men under test, which should be thin and a rather poor con
ductor, is placed between two plates of a material, e.g., glass,
whose conductivity is known. Thermocouples placed in thin
copper sheets on each side of the glass plates, and thereby on
each side of the specimen also, allow measurement of tempera
ture gradients. If a steady heat flow is maintained normal to
these surfaces, the conductivities of specimen and glass are
inversely proportional to their temperature gradients. Sieg 131
and Van Dusen 151 have applied this method to small specimens,
and the same principle has been made use of in the heat meter
(Nicholls 104 ). This is a thin plate of cork board or similar mate
rial of known conductivity with an array of thermocouples on
each side, which can be applied to measure the heat loss from a
wall.
12.4. Linear Flow; Bar Method Metals. Of the many
methods used to determine the thermal conductivity of metals
one of the best 51 surrounds the bar with silica aerogel in a guard
cylinder with heating and cooling coils on the ends. These
maintain a temperature gradient in the cylinder the same as
SBC. 12.5] MEASURING THERMALCONDUCTIVITY CONSTANTS 237
that in the bar under test so that the radial and other losses
are reduced to a minimum.
A very simple and usable, but only moderately accurate,
method is that of Gray. 49 The specimen in the form of a bar 4
to 8 cm long and 2 to 4 mm in diameter has one end screwed
into a copper block forming the bottom of a hotwater bath and
the other into a 6cm diameter copper sphere that serves as a
calorimeter. Temperatures are determined by thermometers
in the copper block and ball. Lateral losses are largely elimi
nated by a protective covering. For a description of the more
complicated bar methods such as that of Jaeger and Diessel
horst 65 the reader is referred to the above mentioned surveys.
12.5. Radial Flow. When materials, particularly poor
conductors, can be formed into cylinders or hollow spheres, the
radialflow method may be useful. This has the advantage of
largely or even totally eliminating lateral heat losses, but the
advantage gained may be lost through difficulties in tempera
ture measurement. In the Niven 105 method the specimen is in
the form of two half cylinders 9 cm in diameter and 15 cm or
more long which are fitted together accurately. A known
amount of heat per cm length is supplied by a resistance wire
along the axis, and the temperatures at radial distances of,
say, 1 and 3 cm are determined by thermocouples. From these
data the conductivity is readily computed with the aid of the
cylindrical flow equation. Bering 55 has suggested the use of
hemispherical caps to avoid end losses in the cylindrical method.
A standard method of measuring the conductivity of some
types of insulating material is to wrap the material about an
electrically heated cylinder or pipe. The cylinder has an
extension or guard ring at each end, and only the heat input to
the central section is used in the measurement, thereby eliminat
ing end losses.
In applying the sphericalflow method the material is formed
into two closely fitting hemispherical shells of perhaps 8 cm
internal diameter and 15 cm external, filled with oil or other
liquid and immersed in a bath. In the cavity is a resistance
coil that furnishes a known amount of heat and also a stirrer,
whose energy input must also be taken into account. Thermo
238 HEAT CONDUCTION [CHAP. 12
couples measure the two surface temperatures. In applying
this method to iron, Laws, Bishop, and McJunkin 83 formed the
thermoelements by electroplating the surfaces with copper and
using copper leads.
The British Electrical and Allied Industries Research Asso
ciation 200 has developed a method for determining the thermal
conductivity of soils based on (4.5p). Heat is electrically
supplied at a measured rate to a buried copper sphere 3 to 9 in.
in diameter, and the temperature of its surface measured after
the steady state has been reached. This is useful for determin
ing conductivity with a minimum of disturbance of the soil.
It should also be easily possible to develop methods based on
(9.5/0, using a buried source, for the relatively quick " assay
ing 7 ' of soil in connection with heatpump installations.
12.6. Diffusivity Measurements. Conductivity may be cal
culated from diffusivity measurements if specific heat and
density are also determined. One method of measuring dif
fusivity is to have the material in the form of a plate or slab
with a thermocouple buried in the center midway between the
two faces. The slab is kept at constant temperature until the
temperature is uniform throughout, and then the surfaces are
suddenly chilled (or heated) by immersion in a stirred liquid
bath, the center temperature changes being continuously
recorded. With the aid of the equation for the unsteadystate
linear flow in the slab the diffusivity is readily obtained. The
method has also been applied 63 to measurements on sands or
muds by packing them in a rectangular sheetcopper container
with insulated edges. This is handled just as the slab above.
Diffusivity can also be measured by the periodicflow method,
by use of (5.3a). This involves a knowledge of the period and
range of temperature at a given distance below the surface, the
range at the surface being known. This last condition can be
eliminated if the range is known for two or more distances.
Forbes, 41 * who measured the annual variations of temperature
for different depths of soil and rock near Edinburgh, was one
of the first to determine thermal constants in this way.
* See also Kelvin, 148 "Mathematical and Physical Papers," III, p. 261.
SEC. 12.7] MEASURING THERMALCONDUCTIVITY CONSTANTS 239
12.7. Liquids and Gases. Some of the same methods
applicable to measurements of conductivity in solids, viz., heat
transfer through a specimen from a hot to a cold plate, are also
usable for liquids and gases. Convection* can be minimized by
using small thicknesses and by having the heat flow downward.
Absence of convection is shown if variations in thickness and
temperature gradient have no effect on the final result. In
gases convection may be considered to be eliminated if the
apparent conductivity is independent of pressure.
Erk and Keller 38 in measuring the conductivity of glycerin
water mixtures used disks of fluid 11.7 cm in diameter and only
3 mm thick, but Bates, 10 by using special precautions, including
the equivalent of a guard ring, was able to avoid convection
even when the thickness was as great as 50 mm (diameter of
central area, 12.7 cm). In measurements on gases the hotwire
method has certain advantages over the plate arrangement.
The heat flows radially from a central hot wire to the surround
ing cylinder. Sherratt and Griffiths, 126 in using this method
on air, Freon, and other gases, have avoided some of the diffi
culties associated with it by using a thick platinum wire. This
is arranged so that the energy input can be measured for the
central section only, thus avoiding end effects.
* Radiation effects must be guarded against in heatconduction measurements
in general, even in the case of solids. See Johnston and Ruehr. 71
I
a s ' s
111
S.a6
.2
S fe
&
s
.2 o
s  a
Q o3 ^u
1 IP
H ^3 ST3"
Hid
following
nian Phy
O Q3
II
I 10 s
i^S fH g
1 r >5b
a
388 S ^8 S
3
COOO rH TJ<TjH Tt^O
cfl
Q
O
1
13s 1 is
>
OOOOi tO rHCO O^ 5
O^C^ tO OiOi COO
<NOO O OO O^
O
ooo o oo oo
I
CO Oi Oi !>
p
t^O^O^O^OOicOOOO
THrH T^tO^C^T 1 IrHrHCO
M
rH (N (M ^
^
ec
CO
S
X
iO O t^ O C^ Oi 00
co t** cO "^ cO C5 cC
rt
rH "^ rH ^5 ^^
a
rH rH rH
Q
a
ScOO? CO O5t^ COtO
n
a
<MCOO5 00 0000 O500
rH
1
OOC&O tO rHCO <52?
O^C^ to O5Oi COO
C^OO O OO O 1 '
n
ooo o oo oo
eo
j
rH
w
X
tO "^ CO "^ ^^ ^O t > * 00 O5 CO CO CC
00 '*1^ C^ CO C^ CO C^ cO t^* t** CO CO
^
^^ C^ C^ ^H O5 00 t >
E
rj
S
r
oooooooooooooo
3
H
O* **.
T" 1 CO O C'l T < t^
o
^
QJ
S
Tf
6 55
H~
^^
s
o ^
h
1
H
d
o
t )
o 2
.^'
5 >2
S y,
. . . '*''*
O
' rH
S "^
1
rH
M
5
: & '
 1 o
H
5
yx ) *> ^
33 &S
J;
H  ^ r3 rd
l^lli ^Js i
3!
^
*3 S ' A> "3 *""* 5s 5^ *"O a d
9
3
l^l^fSoooowJcSrS
* 
241
242
HEAT CONDUCTION
(App. A
O <M *O O "
t > lO O5 CO ****
O O O CO O
S 8
S
^ ^ f^ X 00
CO O tO O O5
^ ^* ^O ^ l>
I s * CO CO Tt* ^t^
CO ^f O CO
^O O OO ^f
o i i co
rH r 1 O
C^
OO
SlO O ^ iO CO 1 O rH CO rH iO CO CO O5
O ^ r 1 O O O rH rH rH O O O O
OOOO OOOOOOOOOO OOO O
00 rH CO 00
VO t^ O rH C<i ^*
CO OS <M Oi rH
O O <N
o
X
rH "ttl 1C
CX) CO T<
'^'^^O OOiO
C^l^C^ OirH
rH rH COCO
CO CO
O
COCOOi
rt^ O <N
O ' < CO CO
rH ^ O (N
CO O^ CO C? ^t 1 C\l CO '"^ CNI
COlOO<MOCOiOiHCOrH LOCO
^O^rHOOOrHrHrH OO
OOOO OOOOOOOOOO OOO
O
O
O
rH
X
GO
^
(M CO CO
S3 ^ LO
CO
rH rH rHCOC^COC^I
o ocoooooooooco oo oo ooooo
APP. A] VALUES OF THERMAL CONDUCTIVITY CONSTANTS 243
do
p o
odd
CO
CO
00 iO (Mr 1 11
2
/i O
^ lH
^ CO
^
O O
0* C^
o o
00
r^
d
OOOOOOOOO
dodddoddd
o o o o o o o
2
X
o
X
<N ^ 00
COOO
T ( O
O^OO<N OcO
rH r 1 O TI <
o odd od
d
o> oo
rH O CO t^ CO
OO (M* IH (M
o o o
3 <N S
d d o
O CO rH T (r I rH T IT! O
dddddddod
O iH rH O
d o o o
te
o
T3
^3
: ^ :
tf5
"^
. GO
o
J!
' ^ . !
:::::::&
00
Oj 
03
. . . . . ^J
:
1
2
. G
*i '
 : :
O O O O Q O
PQ CQ
244
HEAT CONDUCTION
[Apr. A
S
o
^^ iO CO CO ^* C^ **t^
O O O O O O O
r^ooco.0^^00
O O O O T 1 O
Jo o
8 8
o o o o o o o
o o o o o q o
a
^ 00 00 00 (M
iO CO CD CD CD
^^ CO O5 rH ^^
O O O rH O5
rH rH rH rH O T
w
82SSS
CO Oi ^^ CO C?
^O rH (M CO CD
CO X
T^ CO
O O O O O
OOOO
O O

rf CD (N CO *O
rH rH rH rH rH
<N O
o o o o o oo o
rH QO CO O <M Ci
OOOO O O
l^ t^ rH > CO *O
O5 O CO 1> ^ C^
O <M
o >o
X
O O3 00 C5 rH 1> O
rH rH T 1 rH
rH (M CO CO ^ ^ (N
rH rH
a
a*
o
rf t^ t^ I"* CO
<N <M <N <N <N
t> >o o oo o
CO CO !> l> lO
rH rH rH rH rH
<M (M Q ^0 TH
00 oo 25 25 co i>
O O O O O O
O O O O O
CO O5 ^ CO O
O rH C^ CO CO
OOOO
^
O O
O
rH
X
GO iO QO O (M
^O CO ^^ *O CO
CO
Q CO CO O <M rH
O GO (N CO TH Tt<
OOOO O O
u 9 o
^
CO CO cO CO Oi Oi
<N C^ (N (M (N <N
'b : ;
i ^ o
 5 ^
^J
S " ^ !
. . . ^ . "
Ills ' : i
. . . R . ^c .
o
z2 ^
_^
o
^ S % ' ': ':
: ~ wj
co 3 g ' . .
c5 ' .S
*!2 ?3 c5 1
rc ... c3
^  CO c> ' . .
' So i P^
1 . : : 3
o : . . g
^ . 0) t . J5
u . cj a ^ g
1 5 3^ 5 1
'3 8  3 &J4
Illsl^i
jiii
'S .S o a? S "33 a>
1
APP. A] VALUES OF THERMAL CONDUCTIVITY CONSTANTS 245

a
$ 8 S 10 ^opoco <o
o o oo co co ^t* c<i c<i "*
o o t^o oooooooo
o o o do ooooo^o
3 8 &
Q. o O
o o o
ooo
T( T t TH T I TH
O O O O
co
T^
O
O
OOOOCOO O
o o o
oi
<NrH
O<N
OO
O
lO
O
O(NO'OOOO^
O*HOOOOOO
S
co
QO oo
coc
T^
tC
O
O O O
C^
<N
OOSrHrl
OO' t '<
ri O <
OOOOOOOO
*0
o
CO
o
8
o
OrHOOT<CO'^<N<NOlOOOOO' IrHrH
O
<S
c
1
a
H
a
S fe &
246 HEAT CONDUCTION [App. A
TABLE A.2. VALUES OF THE COEFFICIENT OF HEAT TRANSFER h*
Air, heating or cooling
Polished surface in still air, small
temp, difference
Blackened surface in still air, small
temp, difference
Surface in contact with oil, heating or
cooling
Surface in contact with water, heat
ing or cooling
Surface in contact with boiling water.
fph units,
Btu/(hr)
(ft 2 )(F)
0.28
1.31.7
1.82.5
10300
503000
3009000
cgs units,
cal/(sec)(cm')(C)
2.7 X 10~ 5 to 1.1 X 10~ 3
1.8 X 10 4 to2 3 X 10 4
2.5 X 10 4 to3.3 X 10<
1.4 X 10~ 8 to4 X 10~ 2
7 X 10 3 to0.4
0.041.2
* From McAdama 90 and other sources.
APPENDIX B
INDEFINITE INTEGRALS
du
f u dv = uv Iv
I = In x
X m+l
x m dx = rr if wi
e * dx =
1
A ^ax
f a^d
dx
(ax 1)
b In a
r dx 1 , . x
J ^+^ = a tan a
/* (x 2 o)W dx = ^ [x Vx 2 2 2 In (x + Vx r T~a 2 )]
/" (a 2  x*)* dx = ^ (a: V 2  * 2 f 2 sin" 1 ^)
/ sec 2 x dx tan x / x 2 sin x dx = 2x sin x (x 2 2) cos x
I tan x dx = In cos x I x 2 cos x dx = 2x cos x + (x 2 2) sin x
1
~ 2
r , 1 , , N /"
I x cos ax dx = 5 (cos ax + x sin ax) / /
7 a y V a
r . . , 7 sin (a 6)x sin (a + b)x
/ sm ax sin bx dx = ^77  rr"  o/^ . r\ > a
x(a + fa) ~ a
2 Va + bx
/ ,: =
V a 4 ox
2(a  6) 2(o +
sin ax cos bx dx = ~
cos (a 6)x cos (a + b)x
( 6)
b)
. ,
cos ax cos bx dx
sin (a  6)x sin (a f 6)x

r
/ cos a cos = ^7  r\ o7 i k\
J 2(a o) 2(a + o)
j sin 2 ax dx = H~ (ax sin ax cos ax)
/ cos 2 ax dx = 2" ( ax + s ^ n aa; cos ax )
/g a *
e"* sin 6x dx = 2 . , 2 (a sin 6x 6 cos bx)
/e a *
e a * cos 6x dx  a , , 2 (a cos bx + b sin 6x)
 2
dx
xe~
x ' ^ f
+2J e ~*
dx
247
APPENDIX C
DEFINITE INTEGRALS
/V2 . , /W2
/ sm n x dx = I cos n x dx
f * sin 2 x dx _ TT
yo # 2 2
r oo sin ax dx TT . , ^ _ A . . _ TT .
/ = ^> if a > 0; 0, if a = 0; if a <
Jo x L &
f * sin x cos ax dx TT
/ = 0, if a < 1 or > 1; j if a = 1 or +1;
yo X ^t
,if 1 > a > 1
/" * f 1 A/^
/ cos (x 2 ) dx = / sin (x 2 ) dx = ^ \o
/ sin ax sin bx dx = / cos ax cos 6x dx = 0, if a 5^ 6
/ sin 2 ax dx = / cos 2 ax dx = 5
;o yo 2
n!
r ec n
Jo *5^
^
s 2 e*' dx  2
e J *' cos bxdx ^~ e  b */***, if a >
248
APPENDIX D
TABLE D.I. VALUES OF THE PROBABILITY INTEGRAL OR ERROR FUNCTION*
9 Fx 9 fO
*(*)  L ef dp  4= / e* dp
vAr JV \/irf~~ x
X
*(*)
X
*(a
X
*(*)
00
0.00000
0.25
0.27633
0.50
0.52050
01
01128
0.26
0.28690
0.51
0.52924
02
0.02256
0.27
0.29742
0.52
0.53790
03
0.03384
0.28
0.30788
0.53
0.54646
0.04
0.04511
0.29
0.31828
0.54
0.55494
0.05
0.05637
0.30
0.32863
0.55
0.56332
0.06
0.06762
0.31
0.33891
0.56
0.57162
0.07
0.07886
0.32
0.34913
0.57
0.57982
08
0.09008
0.33
0.35928
0.58
0.58792
0.09
0.10128
0.34
0.36936
0.59
0.59594
0.10
0.11246
0.35
0.37938
0.60
0.60386
0.11
0.12362
0.36
0.38933
0.61
0.61168
0.12
. 13476
0.37
0.39921
0.62
0.61941
0.13
0.14587
0.38
0.40901
0.63
0.62705
0.14
0.15695
0.39
0.41874
0.64
0.63459
0.15
0.16800
0.40
0.42839
0.65
0.64203
0.16
0.17901
0.41
0.43797
0.66
0.64938
0.17
. 18999
0.42
0.44747
0.67
0.65663
0.18
0.20094
0.43
0.45689
0.68
0.66378
0.19
0.21184
0.44
0.46623
0.69
0.67084
0.20
0.22270
0.45
0.47548
0.70
0.67780
0.21
0.23352
0.46
0.48466
0.71
0.68467
0.22
0.24430
0.47
0.49375
0.72
0.69143
0.23
0.25502
0.48
0.50275
0.73
0.69810
0.24
0.26570
0.49
0.51167
0.74
0.70468
* From "Tables of Probability Functions," Vol. I, Bureau of Standards, Washington, 1941. Ml
249
250
HEAT CONDUCTION
TABLE D.I. (Continued)
[Apr. D
X
*(*)
X
<*>(*)
X
*(*)
0.75
0.71116
1.10
0.88021
1.45
0.95970
0.76
0.71754
1.11
0.88353
1.46
0.96105
0.77
0.72382
1 12
0.88679
1.47
0.96237
0.78
0.73001
1.13
0.88997
1.48
0.96365
0.79
0.73610
1.14
0.89308
1.49
0.96490
0.80
0.74210
1.15
0.89612
1.50
0.96611
0.81
0.74800
1.16
0.89910
1.51
0.96728
0.82
0.75381
1.17
90200
1.52
0.96841
0.83
0.75952
1.18
0.90484
1.53
0.96952
0.84
0.76514
1.19
0.90761
1.54
0.97059
0.85
0.77067
1.20
0.91031
1.55
0.97162
86
0.77610
1.21
91296
1.56
0.97263
0.87
0.78144
1.22
0.91553
1.57
0.97360
0.88
0.78669
1.23
0.91805
1.58
0.97455
0.89
0.79184
1.24
0.92051
1.59
0.97546
0.90
0.79691
1.25
0.92290
1.60
0.97635
0.91
0.80188
1.26
0.92524
1.61
0.97721
0.92
0.80677
1.27
0.92751
1.62
0.97804
0.93
0.81156
1.28
0.92973
1.63
0.97884
0.94
0.81627
1.29
0.93190
1.64
0.97962
0.95
0.82089
1.30
0.93401
1.65
0.98038
0.96
0.82542
1.31
0.93606
1.66
0.98110
0.97
0.82987
1.32
0.93807
1.67
0.98181
0.98
0.83423
1.33
0.94002
1.68
0.98249
0.99
0.83851
1.34
0.94191
1.69
0.98315
1.00
0.84270
1.35
0.94376
1.70
0.98379
1.01
0.84681
1.36
0.94556
1.71
0.98441
1.02
0.85084
1.37
0.94731
1.72
0.98500
1.03
0.85478
1.38
0.94902
1.73
0.98558
1.04
0.85865
1.39
0.95067
1.74
0.98613
1.05
0.86244
1.40
0.95229
1.75
0.98667
1.06
0.86614
1.41
0.95385
1.76
0.98719
1.07
0.86977
1.42
0.95538
1.77
0.98769
1.08
0.87333
1 43
0.95686
1.78
0.98817
1.09
0.87680
1.44
95830
1.79
0.98864
APP. D] VALUES OF THE PROBABILITY INTEGRAL
TABLE D.I. (Continued}
251
X
*(*)
X
*(s)
X
*(s)
1.80
0.98909
2.10
0.99702 05
2 75
0.99989 94
1.81
0.98952
2.12
0.99728 36
2.80
99992 50
1.82
98994
2.14
0.99752 53
2 85
0.99994 43
1.83
99035
2.16
0.99774 72
2.90
0.99995 89
1.84
0.99074
2.18
0.99795 06
2.95
0.99996 98
1.85
0.99111
2.20
0.99813 72
3.00
0.99997 79095
1.86
0.99147
2.22
0.99830 79
3.10
0.99998 83513
1.87
99182
2.24
99846 42
3.20
0.99999 39742
1.88
0.99216
2.26
0.99860 71
3.30
99999 69423
1.89
0.99248
2.28
0.99873 77
3.40
0.99999 84780
1.90
0.99279
2.30
0.99885 68
3.50.
0.99999 92569
1.91
0.99309
2.32
0.99896 55
3 60
0.99999 96441
1.92
99338
2.34
0.99906 46
3.70
0.99999 98328
1.93
0.99366
2.36
0.99915 48
3.80
99999 99230
1.94
0.99392
2.38
0.99923 69
3.90
0.99999 99652
1.95
99418
2.40
0.99931 15
4.00
0.99999 99846
1.96
0.99443
2.42
0.99937 93
4.20
0.99999 99971
1.97
99466
2.44
0.99944 08
4.40
0.99999 99995
1.98
0.99489
2.46
0.99949 66
4.60
0.99999 99999
1.99
99511
2.48
0.99954 72
00
1.00000
2.00
0.99532 23
2.50
0.99959 30
2.02
0.99571 95
2.55
0.99968 93
2.04
9 99608 58
2.60
0.99976 40
2.06
0.99642 35
2.65
0.99982 15
2.08
99673 44
2.70
0.99986*57
APPENDIX E
TABLE E.I VALUES OF e~ x *
These may be taken at once from an ordinary logarithm table as values of
1/10  4343 *, but the following abbreviated table may prove of occasional
convenience :
X
e~ x
X
e~*
X
e~ x
O.OOf
1 000000
1.00
0.367879
4.00
0.018316
0.05
0.951229
1.10
0.332871
4.20
0.014996
0.10
0.904837
1 20
301194
4.40
0.012277
0.15
0.860708
1.30
0.272532
4.60
0.010052
0.20
0.818731
1.40
0.246597
4.80
0.008230
0.25
. 778801
1.50
0.223130
5.00
0.006738
0.30
0.740818
1.60
0.201897
5.50
0.004087
0.35
0.704688
1.70
0.182684
6.00
002479
0.40
0.670320
1.80
0.165299
6.50
0.001503
0.45
0.637628
1.90
0.149569
7.00
0.000912
50
606531
2 00
0.135335
7.50
0.000553
55
. 576950
2 20
0.110803
8.00
0.000335
0.60
0.548812
2.40
0.090718
8.50
0.000203
0.65
. 522046
2.60
0.074274
9.00
0.000123
0.70
0.496585
2.80
0.060810
9.50
0.000075
0.75
0.472367
3.00
0.049787
10.00
0.000045
0.80
0.449329
3.20
0.040762
0.85
0.427415
3.40
033373
0.90
0.406570
3.60
0.027324
0.95
0.386741
3.80
0.022371
* From "Smithsonian Physical Tables." 1 "
t For very small x, e~* 1 x.
252
APPENDIX F
TABLE F.I. VALUES OF THE INTEGRAL
0' eP dp
X
/(*)
X
/(*)
X
/(*)
0.0001
8.9217
06
2 5266
31
9295
0.0002
8.2286
0.07
2.3731
0.32
9007
0.0003
7.8231
0.08
2.2403
0.33
8731
0004
7.5354
09
2 . 1234
0.34
8464
0.0005
7.3123
0.10
2.0190
0.35
0.8206
0.0006
7.1300
0.11
1.9247
0.36
0.7958
0007
6.9758
0.12
1.8388
0.37
7718
0.0008
6.8423
0.13
1 . 7600
0.38
0.7487
0.0009
6.7245
0.14
1.6873
0.39
0.7263
0.0010
6.6191
0.15
1.6197
0.40
0.7046
0.001
6.6191
0.16
1 . 5567
0.41
0.6836
0.002
5.9260
0.17
1.4977
0.42
0.6634
0.003
5.5205
0.18
1.4423
0.43
6437
0.004
5.2329
0.19
1.3900
0.44
0.6247
0.005
5.0097
0.20
1.3406
0.45
0.6062
0.006
4.8274
0.21
1.2938
0.46
0.5884
007
4.6733
0.22
1.2494
0.47
5711
0.008
4.5397
0.23
1 2072
0.48
0.5543
0.009
4.4220
0.24
1 . 1669
0.49
0.5380
0.010
4.3166
0.25
1 . 1285
0.50
0.5221
0.01
4.3166
0.26
1.0917
0.51
0.5068
0.02
3.6236
0.27
1.0565
0.52
0.4919
03
3.2184
0.28
1.0228
0.53
0.4774
0.04
2.9311
0.29
0.9904
0.54
0.4634
0.05
2.7084
0.30
0.9594
0.55
0.4498
* Computed from "Tables of Sine, Cosine and Exponential Integrals," 148 Vols. I and II, and
^
253
other sources. For x < 0.2, 1(x)  In +    0.2886.
254
HEAT CONDUCTION
TABLE F.I. (Continued)
(APP. F
X
W
X
w
X
/(*)
0.56
0.4365
91
0.1476
1.65
0.009315
0.57
0.4237
0.92
0.1429
1.70
0.007508
0.58
4112
93
0.1383
1.75
0.006027
0.59
0.3990
0.94
0.1339
1.80
0.004818
0.60
0.3872
0.95
0.1295
1.85
0.003837
0.61
0.3758
0.96
0.1253
1,90
0.003042
0.62
3646
0.97
0.1212
1.95
0.002403
0.63
0.3538
0.98
0.1173
2.00
0.001890
0.64
3433
0.99
0.1134
2.05
0.001480
0.65
0.3331
1.00
0.1097
2.10
0.001154
0.66
3231
1.00
0.10969
2.15
8.963 X 10~ 4
0.67
0.3134
1.02
0.10255
2.20
6.930 "
0.68
0.3041
1.04
0.09583
2.25
5.336 "
0.69
0.2949
1.06
0.08950
2.30
4.090 "
0.70
0.2860
1.08
0.08355
2.35
3.122 "'
0.71
0.2774
1.10
0.07796
2.40
2.373 "
0.72
0.2690
1.12
0.07270
2.45
1.795 "
0.73
0.2609
1.14
0.06777
2.50
1.352 "
0.74
0.2529
1.16
0.06313
2.55
1.014 "
0.75
0.2452
1.18
0.05878
2.60
7.573 X 10'
0.76
0.2377
1.20
0.05470
2.65
5.629 "
0.77
0.2305
1.22
0.05088
2.70
4.166 "
0.78
0.2234
1.24
0.04730
2.75
3.069 "
79
0.2165
1.26
0.04394
2.80
2.251 "
0.80
0.2098
1.28
0.04081
2.85
1.643 "
0.81
0.2033
1.30
0.03787
2.90
1.194 "
82
0.1970
1.32
0.03512
2.95
8.641 X 10 6
0.83
0.1909
1.34
0.03256
3.00
6.224 "
0.84
0.1849
1.36
0.03016
3.05
4.462 "
0.85
0.1791
1.38
0.02793
3.10
3.184 "
0.86
0.1735
1.40
0.02585
0.87
0.1680
1.45
0.02123
0.88
0.1627
1.50
0.01738
0.89
0.1575
1.55
0.01417
90
0.1525
1.60
0.01151
APPENDIX G
TABLE G.I. VALUES OP S(x) *s  (**  g e' + ^ e""  )
X
8(x)
X
8(x)
X
S(x)
0.001
1.0000
0.036
0.8752
0.071
0.6310
0.002
1.0000
0.037
0.8679
0.072
0.6249
0.003
1.0000
0.038
0.8605
0.073
0.6188
0.004
1.0000
0.039
0.8532
0.074
0.6128
0.005
1.0000
0.040
0.8458
0.075
6068
0.006
1.0000
0.041
0.8384
0.076
0.6009
0.007
1.0000
0.042
0.8310
0.077
0.5950
0.008
0.9998
0.043
0.8236
0.078
0.5892
0.009
0.9996
0.044
0.8162
0.079
0.5835
0.010
0.9992
0.045
0.8088
0.080
0.5778
0.011
0.9985
0.046
0.8015
0.081
0.5721
0.012
0.9975
0.047
0.7941
0.082
0.5665
0.013
0.9961
0.048
0.7868
0.083
5610
014
0.9944
0.049
0.7796
0.084
0.5555
0.015
0.9922
0.050
0.7723
0.085
0.5500
0.016
0.9896
0.051
0.7651
0.086
0.5447
0.017
0.9866
0.052
0.7579
0.087
0.5393
0.018
0.9832
0.053
0.7508
0.088
0.5340
0.019
0.9794
0.054
0.7437
0.089
0.5288
020
0.9752
0.055
0.7367
0.090
0.5236
0.021
0.9706
0.056
0.7297
0.091
0.5185
022
0.9657
0.057
0.7227
0.092
0.5134
0.023
0.9605
0.058
0.7158
0.093
0.5084
0.024
0.9550
6.059
0.7090
0.094
0.5034
0.025
0.9493
0.060
0.7022
0.095
0.4985
0.026
0.9433
0.061
0.6955
0.096
0.4936
0.027
0.9372
0.062
0.6888
0.097
0.4887
0.028
0.9308
0.063
0.6821
0.098
0.4839
0.029
0.9242
0.064
0.6756
0.099
0.4792
0.030
0.9175
0.065
0.6690
0.100
0.4745
0.031
0.9107
0.066
0.6626
0.102
0.4652
0.032
0.9038
0.067
0.6561
0.104
0.4561
0.033
0.8967
0.068
0.6498
0.106
0.4472
0.034
0.8896
0.069
0.6435
0.108
0.4385
0.035
0.8824
0.070
0.6372
0.110
0.4299
* From OUon and Schultz 1 " and other sources.
255
256
HEAT CONDUCTION
TABLE G.I. (Continued)
(App. G
X
S(x)
X
S(x)
X
8(x)
0.112
0.4215
0.182
0.2113
0.36
0.0365
0.114
0.4133
0.184
0.2071
0.37
0.0330
0.116
0.4052
0.186
0.2031
0.38
0.0299
0.118
0.3973
0.188
0.1991
0.39
0.0271
0.120
0.3895
0.190
0.1952
0.40
0.0246
0.122
0.3819
0.192
0.1914
0.42
0.0202
0.124
0.3745
0.194
0.1877
0.44
0.0166
0.126
0.3671
0.196
0.1840
0.46
0.0136
0.128
0.3600
0.198
0.1804
0.48
0.0112
0.130
0.3529
0.200
0.1769
0.50
0.0092
0.132
0.3460
0.205
0.1684
0.52
0.0075
0.134
0.3393
0.210
0.1602
0.54
0.0062
0.136
0.3326
0.215
0.1525
0.56
0.0051
0.138
0.3261
220
0.1452
58
0.0042
0.140
0.3198
0.225
0.1382
0.60
0.0034
0.142
0.3135
0.230
0.1315
0.62
0.0028
0.144
0.3074
0.235
0.1252
0.64
0.0023
0.146
0.3014
0.240
0.1192
0.66
0.0019
0.148
0.2955
0.245
0.1134
0.68
0.0016
0.150
0.2897
0.250
0.1080
0.70
0.0013
0.152
0.2840
0.255
0.1028
0.72
0.0010
0.154
0.2785
0.260
0.0978
0.74
0.0009
0.156
0.2731
0.265
0.0931
0.76
0.0007
0.158
0.2677
0.270
0.0886
0.78
0006
0.160
0.2625
0.275
0.0844
0.80
0.0005
0.162
0.2574
0.280
0.0803
0.82
0.0004
0.164
0.2523
0.285
0.0764
0.84
0.0003
0.166
0.2474
0.290
0.0728
0.86
0.0003
0.168
0.2426
0.295
0.0693
0.88
0.0002
0.170
0.2378
0.300
0.0659
0.90
0.0002
0.172
0.2332
0.31
0.0597
0.92
0.0001
0.174
0.2286
0.32
0.0541
0.94
0,0001
0.176
0.2241
0.33
0.0490
0.96
0.0001
0.178
0.2198
0.34
0.0444
0.98
0.0001
0.180
0.2155
0.35
0.0402
1.00
0.0001
APPENDIX H
TABLE H.I. VALUES OF B(x)~= 2(e~*  e~'* +
" "
) AND
X
B(x)
B a (x)
X
B(x)
B U (X)
X
B(x)
B.(x)
0.00
1.0000
1.0000
0.70
0.8752
0.3113
2.00
0.2700
0.0823
0.02
1.0000
0.8537
0.72
0.8643
0.3045
2.10
0.2445
0.0745
0.04
1.0000
0.7967
0.74
0.8531
0.2980
2.20
0.2213
0.0674
0.06
1.0000
0.7543
0.76
0.8418
0.2916
2.30
0.2003
0.0610
0.08
1.0000
0.7195
0.78
0.8303
0.2854
2.40
0.1813
0.0552
0.10
1.0000
0.6897
0.80
0.8186
0.2794
2.50
0.1641
0.0499
0.12
1.0000
0.6632
0.82
0.8068
0.2735
2.60
0.1485
0.0452
0.14
1.0000
0.6394
84
0.7950
0.2678
2.70
0.1344
0409
0.16
1.0000
0.6176
0.86
0.7831
0.2622
2.80
0.1216
0.0370
0.18
1.0000
0.5976
0.88
0.7711
0.2567
2.90
0.1100
0.0335
0.20
1.0000
0.5789
0.90
0.7591
0.2513
3.00
0.0996
0.0303
0.22
0.9999
0.5615
0.92
0.7471
0.2461
3.20
0.0815
0.0248
0.24
0.9998
0.5451
0.94
0.7351
0.2410
3.40
0.0667
0.0203
0.26
0.9995
0.5296
0.96
0.7232
0.2360
3.60
0.0546
0.0166
0.28
0.9990
0.5149
0.98
0.7112
0.2312
3.80
0.0447
0.0136
0.30
0.9983
0.5010
1.00
0.6994
0.2264
4.00
0.0366
0.0111
0.32
0.9972
0.4877
1.05
0.6700
0.2150
4.50
0.0222
0.0068
0.34
0.9957
0.4750
1.10
0.6413
0.2042
5.00
0.0135
0.0041
0.36
0.9938
0.4629
1.15
0.6132
0.1940
5.50
0.0082
0.0025
0.38
0.9913
0.4513
1.20
0.5860
0.1844
6.00
0.0050
0.0015
0.40
0.9883
0.4401
1.25
0.5596
0.1752
6.50
0.0030
0.0009
0.42
0.9846
0.4294
1.30
0.5340
0.1665
7.00
0.0018
0.0006
0.44
0.9804
0.4190
1.35
0.5095
0.1583
7.50
0.0011
0.0003
0.46
0.9755
0.4090
1.40
0.4858
0.1505
8.00
0.0007
0.0002
0.48
0.9700
0.3994
1.45
0.4631
0.1431
8.50
0.0004
0.0001
0.50
0.9639
0.3901
1.50
0.4413
0.1360
0.52
0.9573
0.3810
1.55
0.4204
0.1293
0.54
0.9500
0.3723
1.60
0.4005
0.1230
0.56
0.9422
0.3639
1.65
0.3814
0.1170
0.58
0.9339
0.3557
1.70
0.3631
0.1112
0.60
0.9251
0.3477
1.75
0.3457
0.1058
0.62
0.9158
0.3400
1.80
0.3291
0.1006
0.64
0.9062
0.3325
1.85
0.3133
0.0957
0.66
0.8962
0.3252
1.90
0.2981
0.0910
0.68
0.8858
0.3181
1.95
0.2837
0.0866
257
APPENDIX I
TABLE I.I. BESSEL FUNCTIONS
X
Jt(x)
Ji(x)
X
/oM
Ji(x)
X
/o(a)
Ji(x)
0.0
1.00000
0.00000
4.0
0.39715
0.06604
8.C
0.1716
0.23464
0.1
0.99750
0.04994
4.1
0.38867
0.10327
8.1
0.14752
0.24761
2
0.99002
0.09950
4.2
0.37656
0.13865
8.2
0.12222
0.25800
0.3
0.97763
0.14832
4.3
0.36101
0.17190
8.3
0.09601
0.26574
0.4
0.96040
0.19603
4.4
0.34226
0.20278
8.4
0.06916
0.27079
0.5
0.93847
0.24227
4.5
0.32054
0.23106
8.5
0.04194
0.27312
0.6
0.91200
0.28670
4.6
0.29614
0.25655
8.6
0.01462
0.27275
0.7
0.88120
0.32900
4.7
0.26933
0.27908
8.7
0.01252
0.26972
0.8
0.84629
0.36884
4.8
0.24043
0.29850
8.8
0.03923
0.26407
0.9
0.80752
0.40595
4.9
0.20974
0.31469
8.9
0.06525
0.25590
1.0
0.76520
0.44005
5.0
0.17760
0.32758
9.0
0.09033
0.24531
1.1
0.71962
0.47090
5.1
0.14433
0.33710
9.1
0.11424
0.23243
1.2
0.67113
0.49829
5.2
0.11029
0.34322
9.2
0.13675
0.21741
1.3
0.62009
0.52202
5.3
0.07580
0.34596
9.3
0.15766
0.20041
1.4
0.56686
0.54195
5.4
0.04121
0.34534
9.4
0.17677
0.18163
1.5
0.51183
0.55794
5.5
0.00684
0.34144
9.5
0.19393
0.16126
1.6
0.45540
0.56990
5.6
0.02697
0.33433
9.6
0.20898
0.13952
1.7
0.39798
0.57777
5.7
0.05992
0.32415
9.7
0.22180
0.11664
1 8
0.33999
0.58152
5.8
0.09170
0.31103
9.8
0.23228
0.09284
1.9
0.28182
0.58116
5.9
0.12203
0.29514
9.9
0.24034
0.06837
2.0
0.22389
0.57672
6.0
0.15065
0.27668
10.0
0.24594
0.04347
2.1
0.16661
0.56829
6.1
0.17729
0.25586
10.1
0.24903
0.01840
2.2
0.11036
0.55596
6.2
0.20175
0.23292
10.2
0.24962
0.00662
2.3
0.05554
0.53987
6.3
0.22381
0.20809
10.3
0.24772
0.03132
2.4
0.00251
0.52019
6.4
0.24331
0.18164
10.4
0.24337
0.05547
2.5
0.04838
0.49709
6.5
0.26009
0.15384
10.5
0.23665
0.07885
2.6
0.09680
0.47082
6.6
0.27404
0.12498
10.6
0.22764
0.10123
2.7
0.14245
0.44160
6.7
0.28506
0.09534
10.7
0.21644
0.12240
2.8
0.18504
0.40971
6.8
0.29310
0.06522
10.8
0.20320
0.14217
2 9
0.22431
0.37543
6.9
0.29810
0.03490
10.9
0.18806
0.16035
3.0
0.26005
0.33906
7.0
0.30008
0.00468
11.0
0.17119
0.17679
3.1
0.29206
0.30092
7.1
0.29905
0.02515
11.1
0.15277
0.19133
3.2
0.32019
0.26134
7.2
0.29507
0.05433
11.2
0.13299
0.20385
3.3
0.34430
0.22066
7.3
0.28822
0.08257
11.3
0.11207
0.21426
3 4
0.36430
0.17923
7.4
0.27860
0.10963
11.4
0.09021
0.22245
3.5
0.38013
0.13738
7.5
0.26634
0.13525
11.5
0.06765
0.22838
3.6
0.39177
0.09547
7.6
0,25160
0.15921
11.6
0.04462
0.23200
3.7
0.39923
0.05383
7.7
0.23456
0.18131
11.7
0.02133
0.23330
3.8
0.40256
0.01282
7.8
0.21541
0.20136
11.8
0.00197
0.23228
3.9
0.40183
0.02724
7.9
0.19436
0.21918
11.9
0.02505
0.22898
258
Apr, I]
BESSEL FUNCTIONS
TABLE 1.2. ROOTS or J n (x)
259
Root
num
ber
n =
n  1
n = 2
n = 3
n = 4
n  5
1
2.40483
3.83171
5.13562
6.38016
7.58834
8.77148
2
5.52008
7.01559
8.41724
9.76102
11.06471
12.33860
3
8.65373
10.17347
11.61984
13.01520
14.37254
15.70017
4
11.79153
13.32369
14.79595
16.22346
17.61597
18.98013
5
14.93092
16.47063
17.95982
19.40941
20.82693
22.21780
6
18.07106
19.61586
21.11700
22.58273
24.01902
25.43034
7
21.21164
22.76008
24.27011
25.74817
27.19909
28.62662
8
24.35247
25.90367
27.42057
28.90835
30.37101
31.81172
9
27.49348
29.04683
30.56920
32.06485
33.53714
34.98878
10
30.63461
32.18968
33 71652
35.21867
36.69900
38.15987
APPENDIX J
TABLE J.I. VALUES OF C(z)* as 2 T e **
2t, 2 2 , ARE ROOTS OF /o(Zm)
=
WHERE
a;
C(x)
X
C(x)
X
C(x)
0.005
1.0000
0.205
0.4875
0.41
0.1496
0.010
1.0000
0.210
0.4738
0.42
0.1412
0.015
1.0000
0.215
0.4605
0.43
0.1332
0.020
1.0000
0.220
0.4475
0.44
0.1258
0.025
0.9999
0.225
0.4349
0.45
0.1187
0.030
0.9995
0.230
0.4227
0.46
0.1120
0.035
0.9985
0.235
0.4107
0.47
0.1057
0.040
0.9963
0.240
0.3991
0.48
0.0998
0.045
0.9926
0.245
0.3878
0.49
0.0942
0.050
0.9871
0.250
0.3768
0.50
0.0887
0.055
0.9798
0.255
0.3662
0.52
0.0792
0.060
0.9705
0.260
0.3558
0.54
0.0704
0.065
0.9596
0.265
0.3457
0.56
0.0628
0.070
0.9470
0.270
0.3359
0.58
0.0560
0.075
0.9330
0.275
0.3263
0.60
0.0499
0.080
0.9177
0.280
0.3170
0.62
0.0444
0.085
0.9015
0.285
0.3080
0.64
0.0396
0.090
0.8844
0.290
0.2993
0.66
0.0352
0.095
0.8666
0.295
0.2908
0.68
0.0314
0.100
0.8484
0.300
0.2825
0.70
0.0280
0.105
0.8297
0.305
0.2744
0.72
0.0249
0.110
0.8109
0.310
0.2666
0.74
0.0222
0.115
0.7919
0.315
0.2590
0.76
0.0198
0.120
0.7729
0.320
0.2517
0.78
0.0176
0.125
0.7540
0.325
0.2445
0.80
0.0157
0.130
0.7351
0.330
0.2375
0.85
0.0117
0.135
0.7164
0.335
0.2308
0.90
0.0088
0.140
0.6980
0.340
0.2242
0.95
0.0066
0.145
0.6798
0.345
0.2178
1.00
0.0049
0.150
0.6618
0.350
0.2116
a. 05
0.0037
0.155
0.6442
0.355
0.2056
1.10
0.0028
0.160
0.6269
0.360
0.1997
1.15
0.0021
0.165
0.6100
0.365
0.1940
1.20
0.0016
0.170
0.5934
0.370
0.1885
1.25
0.0012
0.175
0.5771
0.375
0.1831
1.30
0.0009
0.180
0.5613
0.380
0.1779
1.35
0.0007
0.185
0.5458
0.385
0.1728
1.40
0.0005
0.190
0.5306
0.390
0.1679
1.50
0.0003
0.195
0.5159
0.395
0.1631
1.60
0.0002
0.200
0.5015
0.400
0.1585
1.70
0.0001
* Mainly from Olson and Sohultz. 108
260
APPENDIX K
MISCELLANEOUS FORMULAS
e = 2.71828
x 2 x z
e* = 1 + x + j + 3j + (x* < oo )
in (i + x) = x  y^ + y&*  yx* +  (** < i)
log, x = logo a; log* a = 2.3026 logio x
x z cc 6 a; 7
flin * = *3I + 5!"7i + ' ' ' (**< o)
cos x = 1  j + j  j + (x < oo)
ei* = cos x + i sin a:
,,,, U /
"8 == P + 32 + 52 + 72+ " ** smh x = 2 ( e * ~~ e ~~ x ^
~ f*f(x) dx = f(b) cosh a; = ~ (e + e~*)
d f b ss \ j tt \ u sinha;
Ta la /(*> ^  /<) tanh *
/(x) dx = (6  a)/(8), where a < 8 < &
/(x + /i) = /(x) + Af(x + 0/0, where < < 1
sin a: sin ?/ = H cos (& j/) M cos (a; + $/)
cos x cos y = H cos (x y) + H cos (a; + T/)
sin a: cos y =* }$ sin (a; y) + y% sin (a? + y)
261
APPENDIX L
THE USE OF CONJUGATE FUNCTIONS FOR ISOTHERMS AND LINES
OF HEAT FLOW IN TWO DIMENSIONS*
In the elementary theory of complex analytic functions it is easily proved
that if f(z) = u(x } y) + iv(x,y) is an analytic function of the complex variable
z = x + iy, and thus has a definite derivative with respect to z, then u and v,
which are the real and imaginary parts of f(z), must be related by the Cauchy
Riemann differential equations
du _ dv_
?~* (a)
du _ dv
By = ~ dx
Because of this interrelation, u and v are called conjugate functions. The pair
have the following interesting properties which can be derived immediately
from (a) :
1. Both u and v satisfy the same differential equation
2. The equations u(x,y) = Ci and v(x,y) = c 2 represent two families of
curves in the xy plane which are orthogonal to each other. For at any point
(x,y) (where the denominators are not zero)
du dv
d _? _ W  ~ i ( c \
~du " dv  (dy\ w
dy dx \dx/c t
dy
which is the wellknown condition for such curves to cross each other orthogo
nally. .That is, the slope of one curve is the negative reciprocal of the slope of
the other.
3. When either u or v is known, its conjugate function can be obtained by
integration. If u is known, we integrate the exact differential expression
dv , , dv , \ du , , du ^ .
W
and if v is known we integrate
du , , du . \ dv dv
See, e.g., Jeans, *.*! Carslaw and Jaeger, a7a ' p  348 and Livens. 86a ' p  104
262
APP. L] THE USE OF CONJUGATE FUNCTIONS 263
The conditions that must be fulfilled to make both of the above exact dif
ferentials are satisfied by (a). Obviously, if the same function is used in one
case for u and in another case for v, the derived conjugate functions in the
respective cases will differ only in sign (neglecting any constant).
The above properties of conjugate functions have been utilized for the
solution of twodimensional problems in other fields than heat conduction, in
particular that of electrical potential.
Let us now derive the conjugate function U for the heatconduction problem
of Sec. 4.4. Put in (e) u = U and v = T, which is known, viz.,
Then we have
,,, 2 f / sin z/cosh y \ f coax sinh y/cosh 2 y\ ~
dU = r L Vl  (cos z/cosh yy) dx + ( 1  (cos */cosh y) ) dy J
which is readily verified. Hence our solution for the conjugate function to T is
T7 2 4. U 1 f COS X \ f '\
U =  tantr 1 I r ) (i)
TT \cosh y/ v '
It may be added that since (i) satisfies (4. la), this function might be taken
to represent temperature, and its conjugate function (/) would then give the
lines of heat flow. But the resulting temperature boundary conditions would
differ accordingly and would represent quite a different physical situation from
the problem treated in Sec. 4.1 Another application of the above results is
found in Problem 6, Sec. 9.46.
APPENDIX M
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112. POWELL, R. W., Reports on Progress in Physics, Proc. Phys. Soc. (Lon
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113. POWELL, R. W., Reports on Progress in Physics, Proc. Phys. Soc.
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114. POWELL, R. W., Reports on Progress in Physics, Proc. Phys. Soc. (Lon
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115. PRESTON, T., "Heat," The Macmillan Company, New York, 1929. [3.8]
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117. RAWHOUSER, C., Proc. Am. Concrete Inst., 41, 305346 (1945). [9.14]
APP. M] REFERENCES 269
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119. ROBERTS, J. K, "Heat and Thermodynamics" (Chap. XI), Blackie,
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120. ROYDS, R., "Heat Transmission by Radiation, Conduction, and Con
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121. SAVAGE, J. L., "Special Cements for Mass Concrete," U.S. Bureau of
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122. SCHACK, A., (trans, by H. Goldschmidt and E. P. Partridge), "Industrial
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124. SCHOFIELD, F. H., Phil. Mag., 12, 329348 (1931). [11.7]
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126. SHERRATT, G. G., and E. GRIFFITHS, Phil Mag., 27, 68 (1939). [12.7]
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INDEX
Adams, 21, lOOn., 201, 206
Adiabatic cases, 108, 125
Adler, 142rc.
Air conditioning of mine, 162
Air space, insulation effectiveness of, 8
Airplanecabin insulation, 2627
Amplitude in periodic flow, 47
Annealing castings, 135
Annual wave in soil, 5152, 57
Applications, airplanecabin insulation,
2627
annealing castings, 135
annual wave in soil, 5152, 57
armorplate cooling, step treatment
of, 224228
billiard balls, temperature in, 166
brick wall, temperature in, 132
canning process, 186187
casting, 114, 123
climate and periodic flow, 54
cofferdam, ice, 217220
cold waves, 53, 57, 118
composite wall, 20
concrete, heat penetration in, 92
temperature waves in, 54
concrete columns, heating of, 179
concrete dams, cooling of, 158162
concrete wall, freezing of, 8283
temperatures in, 132133
cones, heat flow in, 4142, 44
contact resistance, 2728
contacts, electric, 158
covered steam pipes, loss of heat
from, 3941
cylinder wall, periodic flow in, 55
cylindricaltank edge loss, 204r205
decomposing granite, temperatures
in, 115118
diurnal wave in soil, 5051, 57
Applications, drying of porous solids,
187188
earth, cooling of, 99107
estimate of age of, 100107
eccentric spherical and cylindrical
flow, 207208
edge and corner losses, furnace or
refrigerator, 21, 201
edge losses, relaxation treatment, 213
edges and corners, effect of, 21
electric welding, 114, 123, 157158
electrical methods, 205208
fireproof container, 134135
fireproof wall, theory of, 126133
freezing problems, 190199
frozensoil cofferdam, 217220
furnace walls, flow of heat through, 25
gasturbine cooling, 4344
geysers, 42, 149151
groundtemperature fluctuations, 118
hardening of steel, 9698
heat pump, heat sources for, 151157,
162
household applications, 165167
ice formation, 190199
about pipes, step treatment of,
217220
"ice mines,' 1 54
insulation, airplanecabin, 2627
refrigerator, thickness vs. effective
ness of, 2526
laccolith, cooling of, 141142
lava, cooling of, under water, 98
locomotive tires, removal of, 9396
mercury thermometers, heating and
cooling of, 164
moltenmetal container, 133
optical mirrors, 134
plate, cooling of, Schmidt treatment
of, 211213
postglacial time calculations, 119123
271
272
HEAT CONDUCTION
Applications, power cables, under
ground, heat dissipation from,
154
power development, subterranean, 42
radiation heating, loss of, to ground,
108
radioactivity and earth cooling, 102
107
refrigerator insulation, thickness vs.
effectiveness of, 2526
regenerator, storage of heat in, 134
safes, steel and concrete, 164165
shrink fittings, removal of, 9396
soil, annual wave in, 51^2
diurnal wave in, 5051
penetration of freezing tempera
tures in, 92
temperatures in, 52
thawing of frozen, 9293
sources of heat for heat pump, 151
157, 162
sphere, heating, step treatment of,
228232
spot welding, 157158
steel, tempering of, 9698
steel shot, tempering of, 165
stresses, thermal, 5657
subterranean heat sources, 149151
subterranean power development, 42
thawing of frozen soil, 86, 9293
thermit welding, 8385
" through metal," effect of, in wall,
2021
timbers, heating of, 179
uranium " piles," 162
various solids, heating of, 182186
vulcanizing, 134
wall, composite, heat flow through, 20
fireproof, theory of, 126133
with rib, flow of heat through, 203
warming of soil, step treatment of,
220223
welding, electric, 114, 157158
thermit, 8385
Approximation curves, for Fourier
series, 6061, 63, 66
Armorplate cooling, step treatment of,
224228
Auxiliary methods, 200233
Austin, 9w.
Awbery, 200n., 203n., 234
B
Barnes, 198
Barus, lOln., 103
Bateman, 5n., 14n.
Bates, 239
Becker, 99n., 102n.
Berry, 99n., 233n.
Bessel functions, 175
roots of, 259
values of, 258
Bibliography, 264
Billiard balls, temperatures in, 166
Binder, 210n.
Biot, 2, 5
Birch, 28, 235
Birge, 53
Bishop, 238
Boiler, heat flow into, 29
Boundary conditions, 14, 15
Boydell, 99n., 233n.
Brakes, heat dissipation from, 108
Brick temperatures, 186
Brick wall, loss of heat through, 28
temperatures in, 132
British thermal unit (Btu), definition
of, "6
Brooks, 52
Brown, 175
Bullard, 107n.
Buried sphere, conductivity measure
ments with, 238
Byerly, 34n., 59n., 137n., 169n.
Callendar, 55
Calorie, definition of, 6
Calumet and Hecla mine, 119
Canning process, 186187
Carlson, 216n.
Carslaw, 3, 14n., 16w., 64n., 99n.,
113n., 142n., 155n., 176n., 177n.,
233n., 262n.
Casting, 114, 123
Castings, annealing of, 134
INDEX
273
Ceaglske, 5n.
Cgs units, definition of, 6
Charts, GurneyLurie, 208
for heatconduction problems, 208
Christiansen, 236
Churchill, 64rc., 189n.
Clark, 28, 235
Climate and periodic flow, 54
Coefficient of heat transfer, definition
of, 15
values of, 246
Cofferdam, ice, 217220
Cold waves, 53, 57, 118
Comparison methods of measuring
thermal conductivity, 236
Composite wall, heat flow through, 20
Concrete, heat penetration in, 92
temperature waves in, 54
Concrete columns, heating of, 179
Concrete dams, cooling of, 158162
Concrete wall, freezing of, 8283
temperatures in, 132133
Conductivity, factors affecting, 8
theory of, 9
thermal, definition of, 3
values of, 241245
Cones, heat flow in, 4142, 44
Conjugate functions, 34, 189, 262263
Consistentior status, 100, 103, 106
Contact resistance, 2728
Contacts, electric, 158
Container, molten metal, 133
Continuous heat source (see Permanent
heat source)
Conversion factors, 7
Cooling of lava under water, 98
Cooling plate, Schmidt solution of, 211
213
step solution of, 224228
Cosine series, 64
Coudersport ice mine, 5455
Covered steam pipes, loss from, 3941
Croft, 175
Cyclical flow of heat, 4950
Cylinder, heat flow in, 175179
steady state of radial flow in, 3739
Cylinder walls, periodic flow in, 55
Cylindrical flow, nonsymmetrical, 202
Cylindricaltank edge loss, 204205
Dams, concrete, cooling of, 158162
Decomposing granite, temperatures in,
115118
Definite integrals, 248
Definitions, 36
Density, values of, 241245
Diesselhorst, 237
Differential equations, boundary con
ditions of ,1415
examples of, 12
linear and homogeneous, definition
of, 1112
ordinary and partial, definition of, 11
solution of, general and particular, 11
Diffusion constant in drying, 5
Diffusivity, measurement of, 238
thermal, definition of, 4
values of, 241245
Dimensions, 6
Diurnal wave in soil, 5051, 57
Doublets, use of, 112113
Drying of porous solids, 5, 187188
DuhamePs theorem, 113n.
Dusinberre, 216n.
E
Earth, cooling of, 99107
effect of radioactivity in, 102
estimate of age of, 100107
Eccentric spherical and cylindrical
flow, 207208
Ede, 208n.
Edge and corner losses, in furnace or
refrigerator, 21, 201
Edge losses, relaxation treatment of,
214
Edges and corners, allowance for, 21
Eggs, boiling of, 166
Electric furnace, heat loss from, 29
Electric welding, 114, 123, 157158
Electrical contacts, 158
Electrical methods of treating conduc
tion problems, 205208
Emde, 147n.
Emmons, 213, 216
Erk, 239
Error function, values of, 249251
274
HEAT CONDUCTION
Firebrick regenerator, 133134
Fireproof container, 134135
Fireproof wall, theory of, 126133
Fishenden, 210n.
Fitton, 52
Flux of heat, definition of, 3
Forbes, 198, 238
Formulas, miscellaneous, 261
Fourier, 2, 32, 58^.
Fourier equation, derived, 1214
Fourier integral, 71jf., 79
Fourier series, 33, 58jf., 66, 169
conditions for development in, 58
Fourier's problem of heat flow in a
plane, 3035
Fph units, defined, 6
Freezing problems, 190199
Frozen soil, thawing of, 86, 9293
Frozensoil cofferdam, 217220
Frocht, 216n.
Furnace insulation, 25
Furnace walls, flow of heat through,
2626
G
Gasturbine cooling, 4344
Gases, measurement of conductivity
in, 239
Gemant, 233n.
Geothermal curve, 121
Geysers, 42, 149151
Gibbs' phenomenon, 64n.
Gilliland, 5n.
Glass, loss of heat through, 28
Glazebrook, 234
Glover, 159n.
Granite, decomposing, temperatures in,
115118
Graphical methods, 200/.
Gray, 237
Griffiths, 10, 234, 235, 239
Groundpipe heat source for heat pump,
theory of, 151157
Ground temperature fluctuations, 118
Gr6ber, 175
Gurney, 208
H
Halliday, 234
Hardening of steel, 9698
Harder, 142n.
Harmonic analyzer, 7475
Hawkins, 175
Heat flow, general case of, 180182
Heat pump, heat sources for, 151157,
162
Heat sources and sinks, 109113, 143
149
Heattransfer coefficient, values of, 246
Heating of sphere, step solution of, 228
232
Heisler, 21
Helium II, 8
Hering, 237
History of heat conduction theory, 2
Hohf, 5n.
Holmes, 107w.
Hotchkiss, 119
Hougen, 5
Household applications, 165167
HumeRothery, 9n.
Humphrey, 132
Hyperbolic functions, 261, 263
Ice formation, 190199
about pipes, step treatment of, 217
220
thickness proportional to time, 196
thin, solution for, 197
Ice cofferdam, 217220
"Ice mines/' 54
Indefinite integrals, 247
Indicial temperature, 88n.
Indicial voltage, 88n.
Infinite solid, linear heat flow in, 78Jf.
IngenHausz experiment, 24
Ingersoll, 119, 234
Initial conditions, 15
Instantaneous heat source, 109
Insulation, airplanecabin, 2627
refrigerator, thickness vs. effective
ness of, 2526
Integrals, definite, 248
indefinite, 247
INDEX
275
Isothermal surfaces and flow lines, 200
Isotherms, cylindrical tank, 204205
near edge of wall, 201
in rectangular plate, 34
in steam pipe covering, 202
in wall with rib, 203
Jaeger, 3, 14n., 16n., 99n.,
142n., 233n., 237, 262n.
Jahnke, 147n.
Jakob, 175, 234
Janeway, 55n.
Jeans, 262n.
Jeffreys, 107n.
Johnston, 239n.
Joly, 107n.
Jones, 142n.
Juday, 53
Kaye, 235
Keller, 239
Kelvin, 2, 4, 74, 99101, 103, 109, 142,
238n.
(See also Thomson)
Kemler, 152w.
Kent, 56n.
King, 103n.
Kingston, 154, 159n., 162
Kohlrausch, 234
Kranz, 74n.
Laccolith, cooling of, 141142
Lag, in periodic flow, 48
Lambert, 2
Lame", 2
Langmuir, 21, 201, 206, 216
Laplace, 2
Laplace's equation, 12
Lautensach, 55n.
Lava intrusion, cooling of, 85, 98
Law of times, 89
Laws, 238
Leith, 142n.
Leven, 216n.
Lewis, 5n,, 40n.
Limits, change of, in Fourier series, 70
71
Line source, 146
Liquids, measurement of conductivity
of, 239
Livens, 262n.
Locomotive tires, removal of, 9396
Lorenz, 10
Lovering, 99n., 233n.
Lowan, 107n.
Lurie, 208
M
McAdams, 21n., 40n., 175, 203n., 208n.,
210n., 213n., 241, 246n.
McCabe, 5n.
McCauley, 5
McCready, 5n.
MacCullough, 56n.
MacDougal, 51
McJunkin, 238
McLachlan, 176n.
MacLean, 179
March, 53
Marco, 175
Marshall, 5
Mathematical theory of heat conduc
tion, history of, 2
Maxwell, 4
Meats, roasting of, 167
Meier, 55n.
Meikle, 21, 201, 206
Melons, cooling of, 166
Mendota (lake), bottom temperatures
of, 53
Mercury thermometers, heating and
cooling of, 164
Metals, measurement of conductivity
in, 236237
Methods of measuring thermalcon
ductivity constants, 234239
Michelson, 74
Miller, 74n.
Mine, air conditioning of, 162
Mirrors, optical, temperature uniform
ity in, 134
Moltenmetal container, 133
276
HEAT CONDUCTION
N
Nessi, 210n., 213
Neumann, 190, 198
Neumann's solution for ice formation,
191194
Newman, 5n., 182n., 187, 188, 209n.
Newton's law of cooling, 15n., 167
Nicholls, 236
Nicolson, 55
Nissole, 210n., 213
Niven, 237
Nomenclature, 1
O
Olson, 182n., 185n., 255n., 260n.
Onedimensional flow, steady state of,
18jf.
Optical mirrors, 134
Paschkis, 21, 206
Pekeris, 217n., 219n., 220n.
Periodic flow of heat, 45jf.
and climate, 54
in cylinder walls, 55
Permanent heat source, definition of,
109
Pipes, ice formation about, 217
Plane, flow of heat in, 3035
Plane source, 109112
Plate, casting of, 114115
cooling of, by Schmidt treatment,
211213
heated, problem of, 124#
Point source, 143146
Poisson, 2
Poor conductors, measurement of con
ductivity in, 235236
Porous solids, drying of, 187188
Postglacial time calculations, 119123
Potatoes, boiling of, 166
Powell, 8n., 234
Power cables, underground, heat dissi
pation from, 154
Power development, subterranean, 42
Preston, 24n.
Probability integral, values of, 249251
R
Radial heat flow, 35, 37, 139jf.
in conductivity measurements, 237
in cylinder, 175179
Radiating rod, 2124, 136138
Radiation heating, loss of, to ground,
108
Radioactivity, and earth cooling, 102
107
Rambaut, 53
Range of temperature in periodic flow,
47
Rate of heat flow, semimfinite solid, 90
Rawhouser, 159
Reed, 210n., 213n.
References, 264
Refrigerator, heat flow into, 25, 29
Refrigerator insulation, thickness vs.
effectiveness of, 2526
Regenerator, storage of heat in, 134
Relaxation method, 213216
Resistance, contact, 2728
thermal, 1921
Resistivity, thermal, 19n.
Riemann, 2, 128n., 190n.
Roark, 56n.
Roberts, 234
Rocks, measurement of conductivity
of, 235236
Rod, steady flow in, 2124
Ruehr, 239n.
Safes, steel and concrete, 164165
Saunders, 210n.
Savage, 159n.
Schack, 175
Schmidt method, 209213, 216
Schofield, 200n., 203n., 206n.
Schultz, 182n., 185n., 255n., 260n.
Seitz, 9n.
Semimfinite solid, linear flow of heat in,
with plane face at zero, 8688
solution of, by step method, 220223
with temperature of plane face a
function of time, 112113
INDEX
277
Sherratt, 239
Sherwood, 5n., 210n., 213n.
Shortley, 216n.
Shrink fittings, removal of, 9396
Sieg, 236
Sine series, development in, 5964
general development in, 171172
Sink, heat, 214
Slab, problem of, 123126
Slichter, 76n., 99n., 107n., 217n., 219n.,
220n.
Slip, thermal, 28
Smith, 52n., 233n.
Soil, annual wave in, 5152
consolidation of, 5
diurnal wave in, 5051
measurement of conductivity in, 238
penetration of freezing temperatures
in, 92
temperatures in, 5054
thawing of frozen, 9293
Sources, of heat for heat pump, 151
157, 162
and sinks, 143.fr.
equations for, 147149
Southwell, 213
Specific heat, values of, 241245
Sphere, cooling of, by radiation, 167175
with surface at constant tempera
ture, 162166
heating of, by step treatment, 228
232
steady state of radial flow in, 3536
Spherical cavity, problem of, 42, 151,
155156
Spherical flow, eccentric, 207208
Spot welding, 157158
Stamm, 5n.
Steady state, definition of, 18
in more than one dimension, 30^.
in one dimension, ISff.
Steam pipes, covered, loss from, 3941
Steel, tempering of, 9698
Steel shaft, welding of, 8385
Steel shot, tempering of, 165
Stefan, 190, 198
Stefan's law of radiation, 15n.
Stefan's solution for ice formation,
194197
Step method, 115n., 216233
Stoever, 175
Stratton, 74
Strength of heat source, definition of, 109
Stresses, thermal, 5657
Subterranean heat sources, 42, 149151
Subterranean power development, 42
Surface of contact, temperature of, 91
Symbols, 1
Tables and curves, solution from, 208
209
Tait, 74n.
Tamura, 52, 190n., 198n.
Temperature curve in medium, periodic
flow, 49
Temperature gradient, definition of, 3
Temperature waves, in concrete, 54
hi soil, 5054
Thawing of frozen soil, 9293
Thermal conductivity constants, values
of, 241245
Thermal histories, 122
Thermal resistance, 1921
Thermal slip, 28
Thermal stress, 5657
Thermal test of car wheels, 96
Thermit welding, 8385
Thermometric conductivity (see Dif
fusivity)
Thorn, 216n.
Thomson, 2, 74n.
(See also Kelvin)
"Through metal," effect of, in wall,
2021
Timbers, heating or cooling of, 179,
188
Time calculations, postglacial, 119
Timoshenko, 56rc., 57
Transcendental equation, in sphere
problem, 169
Tuttle, 5n.
U
Underground power cables, 154
Uniflow engine, 55
Uniqueness theorem, 16
Uranium "piles," 162
278
HEAT CONDUCTION
Van Dusen, 236
Van Orstrand, 99n., lOOn., 107n., 118. f
233n.
Various solids, heating of, 182186
Velocity, in periodic flow, 48
Vilbrandt, 175
Vulcanizing, 134135
W
Walker, 40n.
Wall, composite, heat flow through, 20,
28
fireproof, theory of, 126133
Wall, with rib, flow of heat through, 203
temperature distribution in, 20
Warming of soil, step treatment of,
220223
Watson, 176n.
Wave length in periodic flow, 48
WeberRiemann, 128n., 190n.
Welding, electric, 114, 157158
spot, 157
thermit, 8385
Weller, 216n.
Wiedemann and Franz, law of, 9
Williamson, 208
Winkelmann, 234
Wires, insulated, cooling of, 40
Worthing, 234