Page 393 - Bird R.B. Transport phenomena

§12.1 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 §4.1.

EXAMPLE 12.1-1 A solid material occupying the space from у = 0 to у = °° is initially at temperature T . At
o
time t = 0, the surface at у = 0 is suddenly raised to temperature 7^ 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

дв (12.1-3)
Sy 2
Here a dimensionless temperature difference 0 = (Г — Т )/{Т Л - T ) has been introduced.
o
О
The initial and boundary conditions are then
I.C.: at t < 0, 0 = 0 for all у (12.1-4)
B.C.1: at у = 0, 0 = 1 for alH > 0 (12.1-5)
B.C. 2: at у = oo, for alH > 0 (12.1-6)
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:
ry/V4at
0 = 1 - exp(-7] )dr) (12.1-7)
2
or
T-T
o
= 1 - erf (12.1-8)
The solution shown in Fig. 4.1-2 describes the temperature profiles when the ordinate is la-
beled (T - T )/(T^ - Г ) and the abscissa y/Viai.
0
о
Since the error function reaches a value of 0.99 when the argument is about 2, the thermal
penetration thickness 8 is
T
8 = 4Vrt (12.1-9)
T
That is, for distances у > 8 , the temperature has changed by less than 1% of the difference
T
T] - Г . If it is necessary to calculate the temperature in a slab of finite thickness, the solution
о
in Eq. 12.1-8 will be a good approximation when 8 is small with respect to the slab thickness.
T
However, when 8 is of the order of magnitude of the slab thickness or greater, then the series
T
solution of Example 12.1-2 has to be used.
The wall heat flux can be calculated from Eq. 12.1-8 as follows:
дТ
=0 ~ * : (Т, " Го) (12.1-10)
у=о
Hence, the wall heat flux varies as t , whereas the penetration thickness varies as f 1 .