Page 168 - Physical Principles of Sedimentary Basin Analysis

150 Heat flow

part is therefore kept, but the factor i is dropped.) There is also a U n associated with each
n because there is one k n for each n. Any linear combination of the products U n (ˆ t)V n (ˆz)
solves the transient equation with the given boundary conditions, because the tempera-
ture equation is linear. The transient temperature can therefore be written as the Fourier
series
∞

T trans (ˆz, ˆ t) = exp(Pe ˆz/2) a n exp(−k n t) sin(πnˆz) (6.182)
ˆ
ˆ
n=1
where the Fourier coefficients a n have to be obtained from the initial condition. The Fourier
series (6.168)for ˆ t = 0 must therefore be equal to the initial condition (6.167), which
implies that

∞
exp(Pe ˆz) − exp(Pe)
exp(Pe ˆz/2) a n sin(πnˆz) = 1 −ˆz − (6.183)
1 − exp(Pe)
n=1
or when rewritten

∞
sinh(Pe(1 −ˆz)/2)

a n sin(πnˆz) = (1 −ˆz) exp(−Pe ˆz/2) + . (6.184)
sinh(Pe/2)
n=1
The sine functions are orthogonal with respect to the inner product defined by integration
of ˆz from 0 to 1:
0, n = m
1
sin(πnˆz) sin(πmˆz) dˆz = 1 (6.185)
0 2 n = m.
The Fourier coefficients a n are therefore obtained by multiplication of equation (6.184)
with sin(πnˆz) followed by an integration of ˆz from 0 to 1. The following definite integrals
simplify the remaining task of finding the coefficients a n :

z = z 2
T 2

fracture
T 1
z = z 1

x = 0
Figure 6.22. A fracture connecting two aquifers.