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Full text of "Heat Conduction"

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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 X-RAYS 

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 HIGH-FREQUENCY MEASUREMENTS 

PHENOMENA IN HIGH-FREQUENCY 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 

MCGRAW-HILL BOOK COMPANY, INC. 

1948 



HEAT CONDUCTION 

Copyright, 1948, by the McGraw-Hill 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 three-quarters 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 two-dimensional case, the most interesting of 
these being the theory of ground-pipe 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 heat-conduction 
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 hundred-odd 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 Curtiss-Wright 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. One-dimensional Flow of Heat. Thermal Re- 
sistance. Edges and Corners. Steady Flow in a Long Thin Rod. 
APPLICATIONS: Furnace Walls; Refrigerator and Furnace Insulation; 
Airplane-cabin 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 
Heat-flow Equations. APPLICATIONS: Covered Steam Pipes; Solid and 
Hollow Cones; Subterranean Temperature Sinks and Power Develop- 
ment; Geysers; Gas-turbine 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; Ground-pipe 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 HEAT-CONDUCTION 

PROBLEMS 200 

/. Method of Isothermal Surfaces and Flow Lines. Solutions for Square 
Edge, Nonsymmetrical Cylindrical Flow, Wall with Internal Ribs, 
and Cylindrical-tank 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 THERMAL-CONDUCTIVITY 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 Centimeter-gram-second system; used here only with centigrade tem- 

perature scale and calorie as unit of heat. 

(Sec - 9 - 38 ' Appendix J) - 



exp x e*. 

fph Foot-pound-hour 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 three-dimensional case; cal/sec per cm length 
or Btu/hr per ft length for two-dimensional case; cal/(sec)(cm 2 ) or 
Btu/(hr)(ft 2 ) for one-dimensional 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 (1768-1830) 
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 heat-conduction 
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 ft-lb 
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 one-tenth 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 specific-heat 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 Hume-Rothery, 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 right-hand 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 left-hand face 

At/As is 



(c) 
and through the right-hand 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 
right-hand 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 left-hand 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 
heat-conduction 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 

heat-conduction 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 heat-conduction 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 
self-generation 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. One-dimensional 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 

^W-M^TV-T, 

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 r-r (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 one-fifthf 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 one-fifth 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 so-called 
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 
so-called Ingen-Hausz 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 frozen-food 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 one-half 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 frozen-food 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 frozen-food locker of inside dimensions 1.5 by 
1.5 by 4 ft, with 4 in. (0.333 ft) of glass-wool 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 high-flying all-season 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 through-metal 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 30-ft 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 gas-solid interface. Here we have a phenomenon, 
known in kinetic theory as thermal slip, which is really a 
temperature discontinuity at the gas-solid 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 5-in. 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 
fire-brick (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 j-fg| 

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 two-dimensional 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 flow-line 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(ri-r.)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 cross-sectional 
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 linear-flow 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 linear-flow 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. 

Thermal-resistance 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 radial-flow 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 2-in. 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.urront-c.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 10-cm 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 Power-develop- 
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. Gas-turbine Cooling. A major problem in gas-turbine 
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 10-ohm 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 10-in. (actual diameter 10.75 in.) 
steam pipe protected with a 2-in. 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 60-watt 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 one-tenth 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 annual-wave 
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 well-known "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 steam-engine 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 cold-air 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 surface-temperature 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 slow-running 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 half-hour. 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 single-valued: 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 right-hand 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) 


|mJyU|l|i|-||[[|[| 


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 . . 

iFVTJ-i (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 right-hand 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 right-hand side of (6) is 



a m I sin 2 mx dx = a m ir (/) 

J v 

1 /"* 
Therefore, o^ = - / /(x) sin mx dx (gO 

^"J-T 

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 (> 

m-l 



/ 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 y-o 

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 well-nigh 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 40-element 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, alternating-current 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 so-called "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 one-dimensional 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\ [ " e-i 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 

(m-x)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 latent-heat 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 probability-integral 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 4.1-day 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 0-16. 

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 = ~ ~ e-P 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 
15-cm 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 end-to-end 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. Latent-heat 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 
latent-heat effects. 

An exact treatment of these latent-heat 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 one-dimensional 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 temperature-time 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 temperature-time 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 shrunk-on 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 12.62-cm (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 one-dimen- 
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 7.62-cm tire or by a 
sudden rise of temperature of the surface A of the 12.62-cm 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 
so-called "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 
well-known 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 20-400 million years, later modifying them to 20-40 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 deep-seated 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 heat-producing 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., one-fourth 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 one-fourth 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 3,000-lb 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 

18-Z-7 



- 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) 



tn2p-f 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 sixty-eight 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 Ground-temperature 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 3-day 
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 
latent-heat 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 /*l-24 
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 virgin-rock 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 5,500-ft 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- 

-9-W 

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 

m-l 



2VF ::: ^r^ m 
llf[ e sm-y- 



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 -y-J (;) 

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 fire-protecting 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 shrunk-on 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 
temperature-time 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 Weber-Riemann. 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 
































































/ 






















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fi 10 


























































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/ 








































































t 
















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n in 






















































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> 0,10 




















































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c 














































1 


V 














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2? AfiQ 
















































v> 












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^_ u.uo 














































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r 


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s 




















































































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t 




















































































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b 


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n 


















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^ 







































































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 temperature-time 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 temperature-time 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 2-hr tests, 
mostly on 8-in. 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 temperature-time 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 two-thirds 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 constant-temperature 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, 
resistor-furnace insulation, fireproof-safe 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 one-quarter 
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 2-cm 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 = -TV-wVa 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 7a-n 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- . / p-i* 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 wrought-iron (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 
cylindrical-flow problems; also, the general case of three-dimen- 
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 boundary-surface 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 mineral-bearing 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. Steady-state 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 heat-source 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 linear-flow case (Sec. 8.3), 2 for 
the two-dimensional case (Sec. 9.8), and 3 for the three-dimen- 
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. 9.11-9.12. Q' is expressed in 
Btu/hr or cal/sec for the three-dimensional 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 8-in.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 
2-in. 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 2-in. 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 high-thermal-conductivity 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 8-in. pipe), we get 
Q' = 23.9 X 47r X 25 = 7510 Btu/hr 

for the cavity. This corresponds to 150 ft of 8-in. pipe or 600 
ft of 2-in. 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 point-heat- 
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= / e-e*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 one-sided 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 > *> - 4-X1HT8 = 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 
1-in. 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 50-gm 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 spot-welding 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 air-conditioned 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 
rock-wall temperature might be expected after 10 years of such cooling? 

Ans. 85F 

8. In a heat-pump installation using a 1-in. 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 

m-dX (h) 



m-l 

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 

m-l 
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 spherical-bulb 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 fire-protecting 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 heat-conduction 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 30-cm 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 
-fc-gj: = 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 



0-1 

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 mercury-in-glass 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 wrought-iron 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) 2-4(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) 



m-1 

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 , 



tn-1 



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 

m-1 

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 heat-conduction 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 




^ X 






CM 


CM 




^ 
3 


^ 




cC 


"els* 




* 


"V 




^ 

^ 






e 


CO 




X 


X 




r 






X 


X 


1 "^5 


'T^ 


^"^ 




^ 




^^^^ 






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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SH 


1 

CO 


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S & 




1|| 


a 
H 


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o 

d 




g 


d 


o 

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II 


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o 


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I 


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o f 1 ^ *"w 


S? 


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8 


GO 


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H 


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fe - 


d 


3 
< 

c 

5 
4 
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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 

I-H 


Semiinfinite 
inder 


d 

% 

5 

!s 


Semiinfinite 


Quarter-infii] 
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 


J|J 


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 ^ 


^ r-d 


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^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 time-temperature 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., heat-conduction) 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 moisture-content 
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 heat-conduction 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 

Ca-C.( .. 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 
6-in. 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 * u-i ( 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. 

* Weber-Riemann. 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 -fi-f 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 left-hand 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 right-hand 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 HEAT-CONDUCTION PROBLEMS 

11.1. In this chapter we shall consider various methods of 
solving particular heat-conduction 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 steady-state 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 steady-state heat conduction in two dimensions, involv- 
ing the construction of an isotherm and flow-line 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 heat-flow 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 two-thirds 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 one-sixth the wall thick- 
ness, this means that such a rib, whose length is two-thirds 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. Three-dimensional Cases; Cylindrical-tank Edge Loss. 

The preceding cases are essentially two-dimensional 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 three-dimensional 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 wedge-shaped 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 
cylindrical-tank 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 two-thirds 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 steady-state heat-flow 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 heat-conduction measurements 
very simple arrangements will suffice. For a two-dimensional 
case a flat, level glass-plate 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 1,000-cycle 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 Wheatstone-bridge 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- 
stone-bridge 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 building-up) 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 Gurney-Lurie, 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 three-dimensional heat flow may 
be obtained directly from the one-dimensional case. Thus, for 
the case of the brick-shaped 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 short-cylinder 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 unsteady-state heat-conduction 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 one-dimensional 
unsteady-state heat flow in a semiinfinite solid whose initial temperature is given 
by the dashed line, with surface at temperature T 8 . 

sider one-dimensional 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 

23J--1 i-e.,*--^ (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 simple-slab 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 three-dimensional 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 two-dimensional 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 mid-points 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, 

y-50 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 mid-temperature 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 heat-conduction 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 open-pit mining and dam-construction 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 4-in. 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 specific-heat 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 frozen-soil cylinder (assumed conductivity 0.0045 cgs) to 
the central pipe. 

Call r 2 the radius of the frozen-soil 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(71-71) 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 


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rH 


8 




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g 00 Tt< 


CO OS OO 
OS CO C^ 


OS C^ *-O 

00 rH C^l 


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C^ CO CO 


t- t- t- 


t^ 00 00 


00 t* 

rH rH 







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ss^ 


t^ IO 00 

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C^ Q IO 
00 iO CO 


00 O* 




^ 


o o o 


rH rH rH 


CO CO CO 


CO CO CO 


rH rH 




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II II II 


II II II 


II II II 


II II II 


II II 




N rH 


^^^ 


^^^ 


B-< s S^ 


S-s 6s ^ 


6s 6s 




3 
1! 


r-T CNP rH 
II II II 

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II II II 


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II II II 


ocT r^ t>^ 

rH rH rH 
II II II 


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II II 






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to 


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if 1 





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& 



AUXILIARY METHODS 

sis * 



CO O 



II II 



89 
n n 



oo 
o 



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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 one-half 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 specific-heat 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 one-dimensional 
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 



s 



5? 

3 
5 

g 

W 
O 

o 



I 
3. 

H * 

J p 
3 

< 5? 



o 
o 
O 



I 

CC 

r? 
O 



o 



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40 8 






1 if f 

" c8 ^ 



ift 






3 



10 00 
CO 



0000 



CO lO 






00^ CO 

rHCOCN 



81OCOOO 
CMOOG5 
t^ O^ O) Oi 



SOC 
^O 



x 

CM 



00 CO 00 

CO C7i O^ 



O 
C^ 



t-OOQcO 
CO *O O5 O^ 
oO O^ O^ O5 



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G5 I s " O^ 



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CO 



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CNC 



SOJ rH CO 
Q -^ 00 
^ 00 Gfi O5 



8 CO CO ' i i 
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OOOCOrHt- 

CN CO *O O5 Oi 

*-O 00 Ol O5 O5 



^ 

fl fl of 
O-- > 



9- 
o 



8 =5 

O 



0- 



SBC. 11.19] 



AUXILIARY METHODS 



225 



SCft r 
J> (f 






O OOrH rH rH 



^ 



^ 
V 

^ 



oo^o 

COrH 



S 
C 



CO CO CO CO 



Gi 

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"tf c 
COC 



CO iO O 00 
CO *O CO 00 
t^OOCO 



<Nr 



<X> Q ^ <N 

rH O CO >O 

oo o5 o co 



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rHOlOO 
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C^ rHrH 



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____ 



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3 ____ 



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 three-dimensional 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 quasi-linear 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 2-cm 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 
3-cm 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 



8 



w o 

W ""* 
u II 

g ft^ 

1 

CO 

^ 

I 



s 

o 

a. 



^ 


*4 


5^o 

S 










iO O O iO 

O CO* CO* O 

Tj< rH 








iO 


b -l 


CO iO iO 


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CO Tt< IO CO 




E-, 


o 

o 


00 


10 ^ 

-H 


J,^0 


CO 00 rH O 


O 00 O rH 




s 

t 1 


-x 

u 


uO 


H r-t 


O5 ^ vo 


CO CO rH CO 


t* CO O5 CO 


K^ 


e-< I 
40 | 
O 

1 

^- 


^p 

1 

~s 


00 


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rH 


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CO O CO rH 


















M 


U~ 


it 

CO 


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gs 


00 C^ CM 
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rH 


00 CM CM Oi 
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rH 


00 W CM O 

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rH 


s 


C? 
1 

Q> 


1 


05 
10 

rfi 


g* 


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1 

x^ 


-N 
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5 


c^ S 

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AUXILIARY METHODS 

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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 brick-shaped 
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 trial-and-error 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 THERMAL-CONDUCTIVITY 

CONSTANTS 

12.1. From the similarity between the flow of heat and of 
electricity it might be supposed that heat-conductivity 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 thermal-conduction 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 heat-conductivity 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 THERMAL-CONDUCTIVITY 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.) Radial-flow methods 
eliminate most of these losses but have difficulties of their own. 
Periodic or other variable-state (as regards temperature-time 
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 water-cooled 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 small-scale 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 THERMAL-CONDUCTIVITY 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 hot-water bath and 
the other into a 6-cm 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 
radial-flow 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 spherical-flow 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 heat-pump 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 unsteady-state 
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 sheet-copper container 
with insulated edges. This is handled just as the slab above. 

Diffusivity can also be measured by the periodic-flow 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 THERMAL-CONDUCTIVITY 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 hot-wire 
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 heat-conduction measurements 
in general, even in the case of solids. See Johnston and Ruehr. 71 



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241 



242 



HEAT CONDUCTION 



(App. A 



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APP. A] VALUES OF THERMAL CONDUCTIVITY CONSTANTS 243 



do 



p o 
odd 



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244 



HEAT CONDUCTION 



[Apr. A 



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APP. A] VALUES OF THERMAL CONDUCTIVITY CONSTANTS 245 



- 



a 



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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) 



0.2-8 
1.3-1.7 
1.8-2.5 
10-300 

50-3000 
300-9000 



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 
0.04-1.2 



* 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 e-f dp - -4= / e-* dp 

-vAr JV -\/ir-f~~ 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-' e-P 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 well-known 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 two-dimensional problems in other fields than heat conduction, in 
particular that of electrical potential. 

Let us now derive the conjugate function U for the heat-conduction 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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132. SLIGHTER, C. S., Elec. World, 54, 146 (1909). [6.13] 

133. SLIGHTER, L. B., Bull. Geol. Soc. Am., 62, 561-600 (1941). [7.24, 7.27] 

134. SMITH, A., Hilgardia, 4, 77-112, 241-272 (1929). [5.9] 

135. SMITH, L. P., /. Appl Phys., 8, 441 (1937). [11.22] 

136. "Smithsonian Physical Tables," Smithsonian Institution, Washington, 

1934. [Apps. A, E] 

137. SOUTHWELL, R. V., "Relaxation Methods in Engineering Science," 

Clarendon Press, Oxford, 1940. [11.14] 

138. STEFAN, J., Wied. Ann., 42, 269 (1891). [10.2] 

139. STOEVER, H. J., "Applied Heat Transmission," McGraw-Hill Book 

Company, Inc., New York, 1941. [9.34] 

140. STOEVER, H. J., "Heat Transfer/' Chem. & Met. Eng., New York, 

1944. [9.34] 

141. "Tables of Probability Functions" (N.Y. Math. Proj.), Bureau of 

Standards, Washington, 1941. [App. D] 

142. "Tables of Sine, Cosine and Exponential Integrals" (N.Y. Math. Proj.), 

Bureau of Standards, Washington, 1940. [App. F] 

143. TAMURA, S. T., Monthly Weather Rev., 33, 55 (1905). [10.2, 10.13] 

144. TAMTJRA, S. T., Monthly Weather Rev., 33, 296 (1905). [5.9] 



270 HEAT CONDUCTION [Apr. M 

145. THOM, A., Proc. Roy. Soc. (London), A141, 651-669 (1933). [11.16] 

146. THOMSON, W. (Lord Kelvin), "Mathematical and Physical Papers," 

Cambridge University Press, London, 1882-1911. 

[1.2, 7.24, 8.1, 12.6] 

147. THOMSON, W., and P. G. TAIT, "Treatise on Natural Philosophy," 

Cambridge University Press, London, 1890. [1.2, 6.12] 

148. TIMOSHENKO, S., "Theory of Elasticity," McGraw-Hill Book Company, 

Inc., New York, 1934. [5.14] 

149. TIMOSHENKO, S., and G. H. MACCULLOTJGH, "Elements of Strength of 

Materials," D. Van Nostrand, Inc., New York, 1940. [5.14] 

150. TTJTTLE, F., J. Franklin Inst., 200, 609-614 (1925). [1.4] 

151. VAN DUSEN, M. S., /. Optical Soc. Am., 6, 739-743 (1922). [12.3] 

152. VAN ORSTRAND, C. E., /. Wash. Acad. /Set., 22, 529-539 (1932). 

[8.11, 11.22] 

153. VAN ORSTRAND, C. E., Trans. Am. Geophys. Union, 18, 21-33 (1937). 

[7.24] 

154. VAN ORSTRAND, C. E., Geophysics, 6, 57-59 (1940). [7.24, 7.27] 

155. VAN ORSTRAND, C. E., "Internal Constitution of the Earth" (ed. by 

Gutenberg), pp. 125-151, McGraw-Hill Book Company, Inc., New 
York, 1939. [7.24] 

156. VILBRANDT, F. C., et al, "Heat Transfer Bibliography," Virginia Poly. 

Inst. Eng. Exp. Sta., Series 53, No. 5, 1943. [9.34] 

157. WALKER, W. H., W. K. LEWIS, and W. H. McADAMs, "Principles of 

Chemical Engineering," McGraw-Hill Book Company, Inc., New 
York, 1927. [4.8] 

158. WALKER, W. H., W. K. LEWIS, W. H. MCADAMS, and E. R. GILLILAND, 

"Principles of Chemical Engineering," McGraw-Hill Book Company, 
Inc., New York, 1937. 

159. WATSON, G. N., "Theory of Bessel Functions," Cambridge University 

Press, London, 1944. [9.36] 

160. WEBER, H., "Differential Gleichungen" (Riemann), Braunschweig, 

1910. [1.2, 8.22, 10.2] 

161. WILLIAMSON, E. D., and L. H. ADAMS, Phys. Rev., 14, 99-114 (1919). 

[11.10] 

162. WINKELMANN, A., "Handbuch d. Physik," III, Leipzig, Verlag Johann 

Ambrosius Barth, 1906. [12.1] 

163. WORTHING, A. G., and D. HALLIDAY, "Heat," John Wiley & Sons, Inc., 

New York, 1948. [12.1] 



INDEX 



Adams, 21, lOOn., 201, 206 
Adiabatic cases, 108, 125 
Adler, 142rc. 

Air conditioning of mine, 162 
Air space, insulation effectiveness of, 8 
Airplane-cabin insulation, 26-27 
Amplitude in periodic flow, 47 
Annealing castings, 135 
Annual wave in soil, 51-52, 57 
Applications, airplane-cabin insulation, 
26-27 

annealing castings, 135 

annual wave in soil, 51-52, 57 

armor-plate cooling, step treatment 
of, 224-228 

billiard balls, temperature in, 166 

brick wall, temperature in, 132 

canning process, 186-187 

casting, 114, 123 

climate and periodic flow, 54 

cofferdam, ice, 217-220 

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, 158-162 

concrete wall, freezing of, 82-83 
temperatures in, 132-133 

cones, heat flow in, 41-42, 44 

contact resistance, 27-28 

contacts, electric, 158 

covered steam pipes, loss of heat 
from, 39-41 

cylinder wall, periodic flow in, 55 

cylindrical-tank edge loss, 204r-205 

decomposing granite, temperatures 
in, 115-118 

diurnal wave in soil, 50-51, 57 



Applications, drying of porous solids, 

187-188 
earth, cooling of, 99-107 

estimate of age of, 100-107 
eccentric spherical and cylindrical 

flow, 207-208 
edge and corner losses, furnace or 

refrigerator, 21, 201 
edge losses, relaxation treatment, 213 
edges and corners, effect of, 21 
electric welding, 114, 123, 157-158 
electrical methods, 205-208 
fireproof container, 134-135 
fireproof wall, theory of, 126-133 
freezing problems, 190-199 
frozen-soil cofferdam, 217-220 
furnace walls, flow of heat through, 25 
gas-turbine cooling, 43-44 
geysers, 42, 149-151 
ground-temperature fluctuations, 118 
hardening of steel, 96-98 
heat pump, heat sources for, 151-157, 

162 

household applications, 165-167 
ice formation, 190-199 

about pipes, step treatment of, 

217-220 
"ice mines,' 1 54 

insulation, airplane-cabin, 26-27 
refrigerator, thickness vs. effective- 
ness of, 25-26 

laccolith, cooling of, 141-142 
lava, cooling of, under water, 98 
locomotive tires, removal of, 93-96 
mercury thermometers, heating and 

cooling of, 164 
molten-metal container, 133 
optical mirrors, 134 
plate, cooling of, Schmidt treatment 

of, 211-213 
postglacial time calculations, 119-123 



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, 25-26 
regenerator, storage of heat in, 134 
safes, steel and concrete, 164-165 
shrink fittings, removal of, 93-96 
soil, annual wave in, 51-^2 
diurnal wave in, 50-51 
penetration of freezing tempera- 
tures in, 92 
temperatures in, 52 
thawing of frozen, 92-93 
sources of heat for heat pump, 151- 

157, 162 
sphere, heating, step treatment of, 

228-232 

spot welding, 157-158 
steel, tempering of, 96-98 
steel shot, tempering of, 165 
stresses, thermal, 56-57 
subterranean heat sources, 149-151 
subterranean power development, 42 
thawing of frozen soil, 86, 92-93 
thermit welding, 83-85 
" through metal," effect of, in wall, 

20-21 

timbers, heating of, 179 
uranium " piles," 162 
various solids, heating of, 182-186 
vulcanizing, 134 

wall, composite, heat flow through, 20 
fireproof, theory of, 126-133 
with rib, flow of heat through, 203 
warming of soil, step treatment of, 

220-223 
welding, electric, 114, 157-158 

thermit, 83-85 
Approximation curves, for Fourier 

series, 60-61, 63, 66 
Armor-plate cooling, step treatment of, 

224-228 
Auxiliary methods, 200-233 



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, 186-187 

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, Gurney-Lurie, 208 

for heat-conduction 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, 217-220 
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, 158-162 
Concrete wall, freezing of, 82-83 

temperatures in, 132-133 
Conductivity, factors affecting, 8 

theory of, 9 

thermal, definition of, 3 

values of, 241-245 
Cones, heat flow in, 41-42, 44 
Conjugate functions, 34, 189, 262-263 
Consistentior status, 100, 103, 106 
Contact resistance, 27-28 
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, 224-228 
Cosine series, 64 
Coudersport ice mine, 54-55 
Covered steam pipes, loss from, 39-41 
Croft, 175 

Cyclical flow of heat, 49-50 
Cylinder, heat flow in, 175-179 

steady state of radial flow in, 37-39 
Cylinder walls, periodic flow in, 55 
Cylindrical flow, nonsymmetrical, 202 
Cylindrical-tank edge loss, 204-205 



Dams, concrete, cooling of, 158-162 
Decomposing granite, temperatures in, 

115-118 

Definite integrals, 248 
Definitions, 3-6 
Density, values of, 241-245 
Diesselhorst, 237 

Differential equations, boundary con- 
ditions of ,14-15 

examples of, 12 

linear and homogeneous, definition 
of, 11-12 

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, 241-245 
Dimensions, 6 

Diurnal wave in soil, 50-51, 57 
Doublets, use of, 112-113 
Drying of porous solids, 5, 187-188 
DuhamePs theorem, 113n. 
Dusinberre, 216n. 

E 

Earth, cooling of, 99-107 

effect of radioactivity in, 102 
estimate of age of, 100-107 

Eccentric spherical and cylindrical 
flow, 207-208 

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, 157-158 

Electrical contacts, 158 

Electrical methods of treating conduc- 
tion problems, 205-208 

Emde, 147n. 

Emmons, 213, 216 

Erk, 239 

Error function, values of, 249-251 



274 



HEAT CONDUCTION 



Firebrick regenerator, 133-134 

Fireproof container, 134-135 

Fireproof wall, theory of, 126-133 

Fishenden, 210n. 

Fitton, 52 

Flux of heat, definition of, 3 

Forbes, 198, 238 

Formulas, miscellaneous, 261 

Fourier, 2, 32, 58^. 

Fourier equation, derived, 12-14 

Fourier integral, 71jf., 79 

Fourier series, 33, 58jf., 66, 169 

conditions for development in, 58 
Fourier's problem of heat flow in a 

plane, 30-35 
Fph units, defined, 6 
Freezing problems, 190-199 
Frozen soil, thawing of, 86, 92-93 
Frozen-soil cofferdam, 217-220 
Frocht, 216n. 
Furnace insulation, 25 
Furnace walls, flow of heat through, 

26-26 



G 



Gas-turbine cooling, 43-44 

Gases, measurement of conductivity 

in, 239 

Gemant, 233n. 
Geothermal curve, 121 
Geysers, 42, 149-151 
Gibbs' phenomenon, 64n. 
Gilliland, 5n. 

Glass, loss of heat through, 28 
Glazebrook, 234 
Glover, 159n. 
Granite, decomposing, temperatures in, 

115-118 

Graphical methods, 200/. 
Gray, 237 

Griffiths, 10, 234, 235, 239 
Ground-pipe heat source for heat pump, 

theory of, 151-157 

Ground temperature fluctuations, 118 
Gr6ber, 175 
Gurney, 208 



H 



Halliday, 234 

Hardening of steel, 96-98 

Harder, 142n. 

Harmonic analyzer, 74-75 

Hawkins, 175 

Heat flow, general case of, 180-182 

Heat pump, heat sources for, 151-157, 

162 
Heat sources and sinks, 109-113, 143- 

149 

Heat-transfer 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, 165-167 
Hume-Rothery, 9n. 
Humphrey, 132 
Hyperbolic functions, 261, 263 



Ice formation, 190-199 

about pipes, step treatment of, 217- 
220 

thickness proportional to time, 196 

thin, solution for, 197 
Ice cofferdam, 217-220 
"Ice mines/' 54 
Indefinite integrals, 247 
Indicial temperature, 88n. 
Indicial voltage, 88n. 
Infinite solid, linear heat flow in, 78Jf. 
Ingen-Hausz experiment, 24 
Ingersoll, 119, 234 
Initial conditions, 15 
Instantaneous heat source, 109 
Insulation, airplane-cabin, 26-27 

refrigerator, thickness vs. effective- 
ness of, 25-26 
Integrals, definite, 248 

indefinite, 247 



INDEX 



275 



Isothermal surfaces and flow lines, 200 
Isotherms, cylindrical tank, 204-205 

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, 99-101, 103, 109, 142, 

238n. 

(See also Thomson) 
Kemler, 152w. 
Kent, 56n. 
King, 103n. 

Kingston, 154, 159n., 162 
Kohlrausch, 234 
Kranz, 74n. 



Laccolith, cooling of, 141-142 

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, 93-96 
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, 236-237 

Methods of measuring thermal-con- 
ductivity constants, 234-239 

Michelson, 74 

Miller, 74n. 

Mine, air conditioning of, 162 

Mirrors, optical, temperature uniform- 
ity in, 134 

Molten-metal container, 133 



276 



HEAT CONDUCTION 



N 



Nessi, 210n., 213 

Neumann, 190, 198 

Neumann's solution for ice formation, 

191-194 

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. 
One-dimensional 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, 30-35 
Plane source, 109-112 
Plate, casting of, 114-115 

cooling of, by Schmidt treatment, 
211-213 

heated, problem of, 124# 
Point source, 143-146 
Poisson, 2 

Poor conductors, measurement of con- 
ductivity in, 235-236 
Porous solids, drying of, 187-188 
Postglacial time calculations, 119-123 
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, 249-251 



R 



Radial heat flow, 35, 37, 139jf. 

in conductivity measurements, 237 
in cylinder, 175-179 
Radiating rod, 21-24, 136-138 
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, 25-26 
Regenerator, storage of heat in, 134 
Relaxation method, 213-216 
Resistance, contact, 27-28 

thermal, 19-21 
Resistivity, thermal, 19n. 
Riemann, 2, 128n., 190n. 
Roark, 56n. 
Roberts, 234 
Rocks, measurement of conductivity 

of, 235-236 

Rod, steady flow in, 21-24 
Ruehr, 239n. 



Safes, steel and concrete, 164-165 

Saunders, 210n. 

Savage, 159n. 

Schack, 175 

Schmidt method, 209-213, 216 

Schofield, 200n., 203n., 206n. 

Schultz, 182n., 185n., 255n., 260n. 

Seitz, 9n. 

Semimfinite solid, linear flow of heat in, 



with plane face at zero, 86-88 
solution of, by step method, 220-223 
with temperature of plane face a 
function of time, 112-113 



INDEX 



277 



Sherratt, 239 

Sherwood, 5n., 210n., 213n. 

Shortley, 216n. 

Shrink fittings, removal of, 93-96 

Sieg, 236 

Sine series, development in, 59-64 

general development in, 171-172 
Sink, heat, 214 
Slab, problem of, 123-126 
Slichter, 76n., 99n., 107n., 217n., 219n., 

220n. 

Slip, thermal, 28 
Smith, 52n., 233n. 
Soil, annual wave in, 51-52 

consolidation of, 5 

diurnal wave in, 50-51 

measurement of conductivity in, 238 

penetration of freezing temperatures 
in, 92 

temperatures in, 50-54 

thawing of frozen, 92-93 
Sources, of heat for heat pump, 151- 
157, 162 

and sinks, 143.fr. 

equations for, 147-149 
Southwell, 213 

Specific heat, values of, 241-245 
Sphere, cooling of, by radiation, 167-175 
with surface at constant tempera- 
ture, 162-166 

heating of, by step treatment, 228- 
232 

steady state of radial flow in, 35-36 
Spherical cavity, problem of, 42, 151, 

155-156 

Spherical flow, eccentric, 207-208 
Spot welding, 157-158 
Stamm, 5n. 
Steady state, definition of, 18 

in more than one dimension, 30^. 

in one dimension, ISff. 
Steam pipes, covered, loss from, 39-41 
Steel, tempering of, 96-98 
Steel shaft, welding of, 83-85 
Steel shot, tempering of, 165 
Stefan, 190, 198 
Stefan's law of radiation, 15n. 
Stefan's solution for ice formation, 
194-197 



Step method, 115n., 216-233 
Stoever, 175 
Stratton, 74 

Strength of heat source, definition of, 109 
Stresses, thermal, 56-57 
Subterranean heat sources, 42, 149-151 
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, 50-54 

Thawing of frozen soil, 92-93 
Thermal conductivity constants, values 

of, 241-245 
Thermal histories, 122 
Thermal resistance, 19-21 
Thermal slip, 28 
Thermal stress, 56-57 
Thermal test of car wheels, 96 
Thermit welding, 83-85 
Thermometric conductivity (see Dif- 

fusivity) 
Thorn, 216n. 
Thomson, 2, 74n. 

(See also Kelvin) 
"Through metal," effect of, in wall, 

20-21 
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, 182-186 
Velocity, in periodic flow, 48 
Vilbrandt, 175 
Vulcanizing, 134-135 

W 

Walker, 40n. 

Wall, composite, heat flow through, 20, 

28 
fireproof, theory of, 126-133 



Wall, with rib, flow of heat through, 203 
temperature distribution in, 20 

Warming of soil, step treatment of, 
220-223 

Watson, 176n. 

Wave length in periodic flow, 48 

Weber-Riemann, 128n., 190n. 

Welding, electric, 114, 157-158 
spot, 157 
thermit, 83-85 

Weller, 216n. 

Wiedemann and Franz, law of, 9 

Williamson, 208 

Winkelmann, 234 

Wires, insulated, cooling of, 40 

Worthing, 234