Page 394 - Bird R.B. Transport phenomena

376 Chapter 12 Temperature Distributions with More Than One Independent Variable

EXAMPLE 12.1-2 A solid slab occupying the space between у = —b and у = +b is initially at temperature T . At
o
time t = 0 the surfaces at у = ±b are suddenly raised to 7^ and maintained there. Find T{y, t).
Heating of a Finite Slab
SOLUTION
For this problem we define the following dimensionless variables:

T -T
Dimensionless temperature 0 = — } — (12.1-11)
у
Dimensionless coordinate ту = - (12.1-12)
b

Dimensionless time r = — (12.1-13)
With these dimensionless variables, the differential equation and boundary conditions are

^ = ?-Ц (12.1-14)
I.C.: at т = 0, 0 = 1 (12.1-15)
B.C. land 2: atTj = ±l, 0 = 0 for т > 0 (12.1-16)

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:
6(TJ, T) = /(Ty)g(r) (12.1-17)
Substitution of this trial function into Eq. 12.1-14 and subsequent division by the product
f(y)gM gives
-i do -t d f 2
(12.1-18)
The left side is a function of r alone, and the right side is a function of ту alone. This can be
2
true only if both sides equal a constant, which we call -c . If the constant is called +c, +c, or
l
—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
dg
2
j- = -c g (12.1-19)
— 2 = -c f (12.1-20)
2
These equations are of the form of Eq. C.l-1 and 3 and may be integrated to give
2
£ = Лехр(-с т) (12.1-21)
/ = В sin сту + С cos сту (12.1-22)
in which A, B, and С are constants of integration.
Because of the symmetry about the xz-plane, we must have 0(ту, т) = ©(-ту, т), and thus
/(ту) = /(-ту). Since the sine function does not have this kind of behavior, we have to require
that В be zero. Use of either of the two boundary conditions gives
Ccosc = 0 (12.1-23)
Clearly С 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 \
n
c n = (п + \)тг n = 0, ±1, ±2, ± 3 . . . , ±oo (12.1-24)