Bird R.B., Stewart W.E., Lightfoot E.N. Transport Phenomena
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Problems
367
examined. The wax droplets leave the atomizer only slightly above their melting point. Esti-
mate the time tf required for a drop of radius R to freeze completely, if the drop is initially at
its melting point To and the surrounding air is at T,. Heat is lost from the drop to the sur-
rounding air according to Newton's law of cooling, with a constant heat-transfer coefficient h.
Assume that there is no volume change in the solidification process. Solve the problem by
using a quasi-steady-state method.
(a)
First solve the steady-state heat conduction problem in the solid phase in the region be-
tween
r
=
Rf
(the liquid-solid interface) and
r
=
R (the solid-air interface). Let
k
be the ther-
mal conductivity of the solid phase. Then find the radial heat flow
Q
across the spherical
surface at
r
=
R.
(b)
Then write an unsteady-state energy balance, by equating the heat liberation at r
=
Rf(t)
resulting from the freezing of the liquid to the heat flow
Q
across the spherical surface at r
=
R.
Integrate the resulting separable, first-order differential equ9tion between the limits
0
and
R, to obtain the time that it takes for the drop to solidify. Let AHf be the latent heat of freezing
(per unit mass).
h
.
4nR2(T0
-
T,)
Answers:
(a)
Q
=
[I
-
(hR/k)]
+
(h~~/k~$);
(b)
llB.ll.
Temperature rise in
a
spherical catalyst pellet
(Fig. 11B.11).
A
catalyst pellet has a radius
R
and a thermal conductivity
k
(which may be assumed constant). Because of the chemical reac-
tion occurring within the porous pellet, heat is generated at a rate of
S,
cal/cm3. s. Heat is lost
at the outer surface of the pellet to a gas stream at constant temperature T, by convective heat
transfer with heat transfer coefficient
h.
Find the steady-state temperature profile, assuming
that
S,
is constant throughout.
(a)
Set up the differential equation by making a shell energy balance.
(b)
Set up the differential equation by simplifying the appropriate form of the energy equation.
(c)
Integrate the differential equation to get the temperature profile. Sketch the function T(r).
(dl
What is the limiting form of T(r) when h
+
a?
(e)
What is the maximum temperature in the system?
(f)
Where in the derivation would one modify the procedure to account for variable
k
and
variable
S,?
llB.12.
Stability of an exothermic reaction ~ystern.~
Consider
a
porous slab of thickness
2B,
width
W,
and length
L,
with
B
<<
W
and
B
<<
L.
Within the slab an exothermic reaction occurs,
with a temperature-dependent rate of heat production
S,(T)
=
Sco exp A(T
-
To).
(a)
Use the energy equation to obtain a differential equation for the temperature in the slab.
Assume constant physical properties, and postulate a steady-state solution T(x).
(b)
Write the differential equation and boundary conditions in terms of these dimensionless
quantities:
8
=
x/B,
O
=
A(T
-
To), and
A
=
Sc,,AB2/k; here
A
is a constant.
(c)
Integrate the differential equation (hint: first multiply by 2dO/d() to obtain
in which On is an auxiliary constant representing the value of O at
(
=
0.
-7-
R
-1-
Fig.
llB.ll.
Sphere with internal heat generation.
368
Chapter 11 The Equations of Change for Nonisothermal Systems
(d)
Integrate the result of (c) and make use of the boundary conditions to obtain the relation
between the slab thickness and midplane temperature
(el
Calculate
A
at Oo
=
0.5, 1.0, 1.2, 1.4, and
2.0;
graph these results to find the maximum
value of
A
for steady-state conditions. If this value of
h
is exceeded, the system will explode.
Laminar annular flow with constant wall heat flux.
Repeat the development of s10.8 for
flow in an annulus of inner and outer radii
KR
and R, respectively, starting with the equations
of change. Heat is added to the fluid through the inner cylinder wall at
a
rate
qo
(heat per unit
per unit time), and the outer cylinder wall is thermally insulated.
Unsteady-state heating of a sphere.
A
sphere of radius R and thermal diffusivity
a
is initially
at a uniform temperature To. For
t
>
0
the sphere is immersed in a well-stirred water bath main-
tained at a temperature TI
>
To. The temperature within the sphere
is
then a function of the
ra-
dial coordinate
r
and the time
t.
The solution to the heat conduction equation is given by?
T
-
To
m
-- -
1
+
2
2
(-1)" exp (-c~n~~~t/~~) (llB.14-1)
TI
-
TO
n=l
It is desired to verlfy that this equation satisfies the differential equation, the boundary condi-
tions, and the initial condition.
(a)
Write down the differential equation describing the problem.
(b)
Show that Eq. llB.14-1 for T(r,
t)
satisfies the differential equation in (a).
(c)
Show that the boundary condition at r
=
R is satisfied.
(dl
Show that T is finite at r
=
0.
(el
To show that Eq. 118.14-1 satisfies the initial condition, set t
=
0 and T
=
To and obtain the
following:
To show that this is true, multiply both sides by (r/R)sin(rnm/R), where
rn
is any integer
from
1
to
m,
and integrate from r
=
0 to
r
=
R.
In the integration all terms with
rn
#
n
vanish
on the right side. The term with
m
=
n,
when integrated, equals the integral on the left side.
Dimensionless variables for free con~ection.~
The dimensionless variables in Eqs. 11.4-39 to
43 can be obtained by simple arguments. The form of
O
is dictated by the boundary condi-
tions and that of
5
is suggested by the geometry. The remaining dimensionless variables may
be found as follows:
(a)
Set
77
=
y/yo,
4z
=
vz/vzo, and
4,
=
vY/v@, the subscript-zero quantities being constants.
Then the differential equations in Eqs. 11.4-33 to 35 become
with the boundary conditions given in Eqs. 11.4-47 to
49.
H.
S.
Carslaw and
J.
C.
Jaeger,
Conduction
of
Heat
in Solids,
2nd edition, Oxford University Press
(1959),
p.
233,
Eq.
(4).
The procedure used here is similar
to
that suggested
by
J.
D.
Hellums
and
S.
W.
Churchill,
AIChE
Journal,
10,110-114
(1964).
Problems
369
(b)
Choose appropriate values of
v, v,,,
and yo to convert the equations
in
(a) into Eqs. 11.4-44
to 46, and show that the definitions in Eqs. 11.441 to
43
follow directly.
(c)
Why is the choice of variables developed in (b) preferable to that obtained by setting the
dimensionless groups in Eqs. llB.15-1 and
2
equal to unity?
llC.l.
The speed of propagation
of
sound waves. Sound waves are harmonic compression waves
of very small amplitude traveling through a compressible fluid. The velocity of propagation
of such waves may be estimated by assuming that the momentum flux tensor
7
and the heat
flux vector
q
are zero and that the velocity v of the fluid is small.6 The neglect of
T
and
q
is
equivalent to assuming that the entropy is constant following the motion of a given fluid ele-
ment (see Problem 11D.1).
(a)
Use equilibrium thermodynamics to show that
in which
y
=
CJC,.
(b)
When sound is being propagated through a fluid, there are slight perturbations in the
pressure, density,
and
velocity from the rest state: p
=
po
+
p', p
=
po
+
p', and
v
=
vo
+
v',
the subscript-zero quantities being constants associated with the rest state (with vo being
zero), and the primed quantities being very small. Show that when these quantities are substi-
tuted into the equation of continuity and the equation of motion (with the
I
and
g
terms omit-
ted) and products of the small primed quantities are omitted, we get
Equation of continuity
dp
-
=
-p0(V
.
v)
dt
(11C.l-2)
Equation of motion
dv
p,,
-
=
-Vp
dt
(11C.l-3)
(c) Next use the result in (a) to rewrite the equation of motion as
in which
v:
=
y(dp/dp),.
(d)
Show how Eqs. llC.1-2 and 4 can be combined to give
(el
Show that a solution of Eq. llC.1-5 is
p
=
po[l
+
A sin
(F
o
-
v,t))]
This solution represents a harmonic wave of wavelength
h
and amplitude
pd
traveling in the
z
direction at
a
speed v,. More general solutions may be constructed by a superposition of
waves of different wavelengths and directions.
11C.2.
Free convection in
a
slot.
A
fluid of constant viscosity, with density given by Eq. 11.3-1, is
confined in a rectangular slot. The slot has vertical walls at
x
=
+
B,
y
=
+
W,
and a top and
bottom at
z
=
_tH, with H
>>
W
>>
B.
The walls are nonisothermal, with temperature dis-
tribution
T,
=
T
+
Ay, so that the fluid circulates by free convection. The velocity profiles
are to be predicted, for steady laminar flow conditions and small deviations from the mean
density
p.
See
L.
Landau and
E.
M.
Lifshitz,
Fluid
Mechanics,
2nd
edition,
Pergamon,
Oxford
(1987),
Chapter
VIII;
R.
J.
Silbey
and
R.
A.
Alberty,
Physical Chernisfy,
3rd edition,
Wiley,
New
York
(2001),§17.4.
370
Chapter
11
The Equations of Change for Nonisothemal Systems
(a) Simplify the equations of continuity, motion, and energy according to the postulates:
v
=
6,v,(x,
y),
d2v,/d$
<<
d2vZ/dx2, and
T
=
T(y). These postulates are reasonable for slow
flow, except near the edges
y
=
-C
Wand
z
=
?H.
(b) List the boundary conditions to be used with the problem as simplified in (a).
(c)
Solve for the temperature, pressure, and velocity profiles.
(d)
When making diffusion measurements in closed chambers, free convection can be a seri-
ous source of error, and temperature gradients must be avoided. By way of illustration, com-
pute the maximum tolerable temperature gradient,
A,
for an experiment with water at 20°C in
a chamber with
B
=
0.1 mm, W
=
2.0 mm, and
H
=
2
cm, if the maximum permissible convec-
tive movement is 0.1% of
H
in a one-hour experiment.
-
PSPA
Answers: (c) v,(x, y)
=
-
(B2
-
x2)y; (dl 2.7
X
K/cm
2~
Tangential annular flow of a highly viscous liquid. Show that Eq. 11.4-13 for flow in an
an-
nular region reduces to Eq. 10.4-9 for plane slit flow in the limit as
K
approaches unity. Com-
parisons of this kind are often useful for checking results.
The right side of Eq. 11.4-13 is indeterminate at
K
=
1, but its limit as
K
+
1
can be ob-
tained by expanding in powers of
E
=
1
-
K.
To do this, set
K
=
1
-
s
and
5
=
1
-
s[l
-
(x/b)l; then the range
K
5
8
%
1 in Problem 11.4-2 corresponds to the range
0
%
x
5
b
in 310.4.
After making the substitutions, expand the right side of Eq.
11.4-13
in powers of
s
(neglecting
terms beyond
8')
and show that Eq. 10.4-9 is obtained.
Heat conduction with variable thermal conductivity.
(a) For steady-state heat conduction in solids, Eq. 11.2-5 becomes
(O
-
q)
=
0, and insertion
of
Fourier's law gives
(V
.
kVT)
=
0. Show that the function
F
=
JkdT
-+
const, satisfies the
Laplace equation V2F
=
0, provided that
k
depends only on
T.
(b) Use the result in (a) to solve Problem 108.12 (part a), using an arbitrary function
k(7').
Effective thermal conductivity of
a
solid with spherical inclusions (Fig. 11C.5). Derive
Eq.
9.6-1 for the effective thermal conductivity of a two-phase system by starting with
Eqs.
llB.8-1 and 2. We construct two systems both contained within a spherical region of radius
R':
(a) the "true" system, a medium with thermal conductivity
ko,
in which there are embed-
ded
n
tiny spheres of thermal conductivity
k,
and radius
R;
and (b) an "equivalent" system,
which is a continuum, with an effective thermal conductivity
k,,,.
Both of these systems are
placed in a temperature gradient A, and both are surrounded by a medium with thermal
conductivity ko.
Medium
0
with
thermal conductivity
ko
Medium
0
n
spheres of
material
1
of radius
R
and thermal
conductivity
k,
Sphere of radius
R'
Medium
0
with
thermal conductivity
ko
Sphere of radius
R'
of a hypothetical
"smoothed out"
material equivalent
to the granular
material in (a)
Thermal conductivity is
keff
Fig.
llC.5.
Thought experiment used by Maxwell to get the thermal conductiv-
ity of a composite solid:
(a)
the "true" discrete system, and
(b)
the "equivalent"
continuum system.
Problems
371
(a)
For the "true" system we know that at a large distance
L
from the system (i.e.,
L
>>
R'),
the temperature field will be given
by
a slight modification of Eq. llB.8-2, provided that the
tiny occluded spheres are very "dilute" in the true system:
Explain carefully how this result is obtained.
(b) Next, for the "equivalent system," we can write from Eq. llB.8-2
(c)
Next derive the relation n~~
=
4Rf3, in which
4
is the volume fraction of the occlusions in
the "true system."
(d) Equate the right sides of Eqs. 11C.5-1 and 2 to get Maxwell's equation7 in Eq. 9.6-1.
11C.6.
Interfacial boundary conditions. Consider a nonisothermal interfacial surface S(t) be-
tween pure phases
I
and
I1
in a nonisothermal system. The phases may consist of two im-
miscible fluids (so that no material crosses S(t)), or two different pure phases of a single
substance (between which mass may be interchanged by condensation, evaporation, freez-
ing, or melting). Let n1 be the local unit normal to S(t) directed into phase I.
A
superscript
I
or
I1
will be used for values along S in each phase, and a superscript s for values in the in-
terface itself. The usual interfacial boundary conditions on tangential velocity
v,
and tem-
perature
T
on
S
are
v~
- -
v
II
(no slip)
(11C.6-1)
T'
=
TI1 (continuity of temperature) (llC.6-2)
In addition, the following simplified conservation equations are suggesteds for surfactant-free
interfaces:
Interfacial mass balance
(nl. {p'(vl
-
J)
-
p'l(v'l
-
J)})
=
0
(1 lC.6-3)
Interfacial momentum balance
Interfacial internal energy balance
(n'
.
pl{v'
-
$})[(I?
-
9)
+
i(v"
-
vU2)]
+
(d
.
{cf
-
qll})
=
o(VS
.
vS)
(1
1C.6-5)
The momentum balance of Eq. 3C.5-1 has been extended here to include the surface gradient
VSu
of the interfacial tension; the resulting tangential force gives rise to a variety of interfacial
flow phenomena, known as Marangoni
effect^.^,'^
Equation llC.6-5 is obtained in the manner
of 511.2, from total and mechanical energy balances on
S,
neglecting interfacial excess energy
Ij",
heat flux
qS,
and viscous dissipation (.rS:VV); fuller results are given elsewhere.'
J.
C. Maxwell,
A
Treatise
on
Electricity
and
Magnetism,
Vol.
1,
Oxford University Press
(1891,
reprinted
1998), 5314.
J.
C.
Slattery,
Advanced
Transport
Phenomena,
Cambridge University Press
(1999),
pp.
58,435;
more
complete conditions are given in Ref.
10.
C.
G.
M.
Marangoni,
Ann.
Phys. (Poggendorf), 3,337-354 (1871);
C. V.
Sternling and
L.
E.
Scriven,
AIChE
Jou~~l, 5,514-523 (1959).
lo
D. A.
Edwards,
H.
Brenner, and
D.
T.
Wasan,
interfacial
Transport Processes
and
Rheology,
Butterworth-Heinemann, Stoneham, Mass.
(1991).
372
Chapter 11 The Equations of Change for Nonisothermal Systems
(a) Verify the dimensional consistency of each interfacial balance equation.
(b) Under what conditions are
v1
and
v"
equal?
(c) Show how the balance equations simplify when phases
I
and
I1
are two pure immiscible
liquids.
(d)
Show how the balance equations simplify when one phase is a solid.
Effect of surface-tension gradients on a falling film.
(a) Repeat the determination of the shear-stress and velocity distributions of Example 2.1-1
in
the presence of a small temperature gradient
dT/dz
in the direction of flow. Assume that this
temperature gradient produces a constant surface-tension gradient
du/dz
=
A
but has
no
other effect on system physical properties. Note that this surface-tension gradient will pro-
duce a shear stress at the free surface of the film (see Problem llC.6) and, hence, will require
a
nonzero velocity gradient there. Once again, postulate a stable, nonrippling, laminar film.
(b) Calculate the film thickness as a function of the net downward flow rate and discuss the
physical significance of the result.
Answer:
(a)
T,,
=
pgx
cos
p
+
A; v,
=
Equation of change for entropy.
This
problem
is
an introduction to the thermodynamics
of
irreversible processes.
A
treatment of multicomponent mixtures is given in 5524.1 and
2.
(a) Write an entropy balance for the fixed volume element
Ax Ay Az.
Let s be the
entropy
flux
vector,
measured with respect to the fluid velocity vector
v.
Further, let the
rafe of entropy
pro-
duction
per unit volume be designated by
gs.
Show that when the volume element
Ax Ay
Az
is
allowed to become vanishingly small, one finally obtains an
equation of change for entropy
in ei-
ther of the following two forms:"
in which
2
is the entropy per unit mass.
(b) If one assumes th$ the thermodynamic qua;tities can be defined locally in a nonequilib-
riym sityation,;hen
U
can be related to
S
and
V
according to the thermodynamic relation
d
LI
=
TdS
-
pdV.
Combine this relation with Eq. 11.2-2 to get
(c)
The local entropy flux is equal to the local energy flux divided by the local
that is,
s
=
q/T.
Once this relation between
s
and q is recognized, we can compare Eqs. llD.l-2
and
3
to get the following expression for the rate of entropy production per unit volume:
"
G.
A.
J.
Jaurnann,
Sitzungsbeu. der Math.-Natuvwiss. Klasse der Kaiserlichen Ahd. der Wissenschaften
(Wien),
102,
Abt. IIa, 385-530 (1911).
l2
Carl
Henry
Eckart
(1902-1973), vice-chancellor of the University of California at San Diego
(1965-1969), made fundamental contributions to quantum mechanics, geophysical hydrodynamics,
and the thermodynamics of irreversible processes; his key contributions to transport phenomena
are
in
C.
H.
Eckart,
Phys. Rev.,
58,267-268,269-275 (1940).
l3
C.
F.
Curtiss and
J.
0.
Hirschfelder,
I.
Chem. Phys.,
18,171-173 (1950).
l4
J.
G.
Kirkwood and
B.
L.
Crawford, Jr.,
I.
Phys. Chem.
56,1048-1051 (1952).
l5
S.
R.
de Groot and P. Mazur,
Non-Equilibrium Thermodynamics,
North-Holland, Amsterdam (1962).
Problems
373
The first term on the right side is the rate of entropy production associated with heat trans-
port, and the second is the rate of entropy production resulting from momentum transport.
Equation llD.l-4 is the starting point for the thermodynamic study of the irreversible
processes in a pure fluid.
(d)
What conclusions can be drawn when Newton's law of viscosity and Fourier's law of
heat conduction are inserted into
Eq.
11D.1-4?
11D.2.
Viscous heating in laminar tube
flow.
(a)
Continue the analysis begun in Problem IlB.2-namely, that of finding the temperature
profiles in a Newtonian fluid flowing in a circular tube at a speed sufficiently high that vis-
cous heating effects are important. Assume that the velocity profile at the inlet
(z
=
0)
is fully
developed, and that the inlet temperature is uniform over the cross section. Assume all physi-
cal properties to be constant.
(b) Repeat the analysis for
a
power law non-Newtonian viscosity.16
llD.3.
Derivation of the energy equation using integral theorems. In
s11.1
the energy equation is
derived by accounting for the energy changes occurring in a small rectangular volume ele-
ment
Ax
Ay
Az.
(a)
Repeat the derivation using an arbitrary volume element
V
with a fixed boundary
S
by
following the procedure outlined in Problem 3D.1. Begin by writing the law of conservation
of energy as
Then use the Gauss divergence theorem to convert the surface integral into a volume integral,
and obtain Eq. 11.1-6.
(b)
Do the analogous derivation for a moving "blob" of fluid.
l6
R.
B.
Bird,
Soc.
Plastics
Engrs.
Journal,
11,3540
(1955).
Chapter
12
Temperature Distributions with
More Than One Independent
Variable
512.1
Unsteady heat conduction in solids
512.2'
Steady heat conduction in laminar, incompressible flow
512.3'
Steady potential flow of heat in solids
512.4O
Boundary layer theory for nonisothermal flow
In Chapter 10 we saw how simple heat flow problems can be solved by means of shell
energy balances. In Chapter 11 we developed the energy equation for flow systems,
which describes the heat transport processes in more complex situations. To illustrate
the usefulness of the energy equation, we gave in 511.4 a series of examples, most
of
which required no knowledge of solving partial differential equations.
In this chapter we turn to several classes of heat transport problems that involve
more than one dependent variable, either two spatial variables, or one space variable
and the time variable. The types of problems and the mathematical methods parallel
those given
in
Chapter
4.
512.1
UNSTEADY HEAT CONDUCTION IN SOLIDS
For solids, the energy equation of
Eq.
11.2-5, when combined with Fourier's law of
heat
conduction, becomes
If
the thermal conductivity can be assumed to be independent of the temperature
and
position, then Eq. 12.1-1 becomes
in which
a
=
k/&
is the thermal diffusivity of the solid. Many solutions to this equa-
tion have been worked out. The treatise of Carslaw and Jaeger' contains a thorough dis-
H.
S.
Carslaw and
J.
C. Jaeger,
Conduction of Heat in Solids,
2nd edition, Oxford University Press
(1959).
g12.1 Unsteady Heat Conduction in Solids
375
cussion of solution methods as well as a very comprehensive tabulation of solutions for
a wide variety of boundary and initial conditions. Many frequently encountered heat
conduction problems may be solved just by looking up the solution in this impressive
reference work.
In this section we illustrate four important methods for solving unsteady heat con-
duction problems: the method of combination of variables, the method of separation of
variables, the method of sinusoidal response, and the method of Laplace transform. The
first three of these were also used in
54.1.
A
solid material occupying the space from
y
=
0 to y
=
is initially at temperature To. At
time t
=
0, the surface at
y
=
0
is suddenly raised to temperature TI and maintained at that
Heating
a
Semi-Infinite
temperature for t
>
0.
Find the time-dependent temperature profiles T(y,
t).
Slab
SOLUTION
For this problem, Eq. 12.1-2 becomes
Here a dimensionless temperature difference O
=
(T
-
To)/(T,
-
To) has been introduced.
The initial and boundary conditions are then
LC.:
B.C.
1:
B.C.
2:
attsO, 0=0 forally
at
y
=
0, O
=
1
for all t
>
0
aty=
m,
0=0 forallt>O
This problem is mathematically analogous to that formulated in Eqs. 4.1-1 to 4. Hence the so-
lution in Eq. 4.1-15 can be taken over directly
by
appropriate changes
in
notation:
T
-
To
--
Y
-
1
-
erf-
Tl
-
To
a
The solution shown in Fig. 4.1-2 describes the temperature profiles when the ordinate is la-
beled
(T
-
To)/(Tl
-
To) and the abscissa y/*.
Since the error function reaches a value of 0.99 when the argument is about
2,
the thermal
penetration thickness
6T
is
That is, for distances
y
>
6,
the temperature has changed by less than
1%
of the difference
T,
-
To. If it is necessary to calculate the temperature in a slab of finite thickness, the solution
in Eq. 12.1-8 will be a good approximation when
6,
is small with respect to the slab thickness.
However, when
8,
is of the order of magnitude of the slab thickness or greater, then the series
solution of Example 12.1-2 has to be used.
The wall heat flux can be calculated from Eq. 12.1-8 as follows:
Hence, the wall heat flux varies as t-"2, whereas the penetration thickness varies as t1I2
b
376
Chapter 12 Temperature Distributions with More Than One Independent Variable
EXAMPLE
12.1-2
Heating
of
a Finite Slab
A
solid slab occupying the space between
y
=
-b
and
y
=
+
b
is initially at temperature
To.
At
time
t
=
0
the surfaces at
y
=
?b
are suddenly raised to
T,
and maintained there. Find
T(y,
t).
SOLUTION
For this problem we define the following dimensionless variables:
TI
-
T
Dimensionless temperature
0
=-
T,
-
To
Dimensionless coordinate
Y
'7
=i
(12.1-12)
Dimensionless time
With these dimensionless variables, the differential equation and boundary conditions are
I.C.:
at
7
=
0,
0
=
1
B.C. 1 and
2:
at q
=
+I,
O
=
0
for^>
0
Note that no parameters appear when the problem
is
restated thus.
We can solve this problem by the method of separation of variables. We start by postulat-
ing that a solution of the following product form can be obtained:
Substitution of this trial function into Eq. 12.1-14 and subsequent division by the product
f
(q)g(d gives
The left side is a function of
7
alone, and the right side is a function of
'7
alone. This can be
true only if both sides equal a constant, which we call -c2. If the constant is called +c2, +c,
or
-c,
the same final result is obtained, but the solution is a bit messier. Equation 12.1-18
can
then be separated into two ordinary differential equations
These equations are of the form of Eq. C.l-1 and
3
and may be integrated to give
g
=
A
exp (-c27)
f
=
Bsincq
+
Ccoscv
in which
A,
El,
and C are constants of integration.
Because of the symmetry about the xz-plane, we must have @(q,
T)
=
@(-'7,
r),
and thus
f
(7)
=
f
(-7).
Since the sine function does not have this kind of behavior, we have to require
that
B
be zero. Use of either of the two boundary conditions gives
Ccos
c
=
0 (12.1-23)
Clearly
C
cannot be zero, because that choice leads to a physically inadmissible solution.
However, the equality can be satisfied by many different choices of
c,
which we call c,: