Page 165 - Physical Principles of Sedimentary Basin Analysis

6.12 Transient convective heat flow 147
√
erf( v 1 /2 ˆz)
ˆ
T (ˆz) = 1 − √ (6.165)
erf( v 1 /2)
for downwards flow.

Exercise 6.18 Show that the temperature equation (6.150) is linear. (Hint: show that if T 1
and T 2 are solutions of the temperature equation then so also is any linear combination
c 1 T 1 + c 2 T 2 .)

6.12 Transient convective heat flow
Solving the time-dependent temperature equation (6.150) gives how fast the temperature
will reach a stationary state. We need an initial condition before we can do that. It is simply
ˆ
T (ˆz, ˆ t=0) = 1 −ˆz in this case, which is the stationary temperature solution in the absence
of heat convection. Notice that the initial condition has to respect the boundary conditions,
T (ˆz=0, ˆ t) = 1 and T (ˆz=1, ˆ t) = 0. The temperature solution is now written as a sum of
ˆ
ˆ
two parts
T (ˆz, ˆ t) = T stat (ˆz) + T trans (ˆz, ˆ t) (6.166)
ˆ
ˆ
ˆ
where the first part, T stat (ˆz), is the stationary temperature solution (6.154), and the second
ˆ
part T trans (ˆz, ˆ t) is the transient. The temperature solution approaches the stationary part
ˆ
(time-independent part) for ˆ t →∞, when the transient decays to zero. The transient part
at ˆ t = 0 has to be
ˆ
T trans (ˆz, ˆ t = 0) = 1 −ˆz − T stat (ˆz), (6.167)
ˆ
in order for the temperature solution to fulfill the initial condition. The transient solution
becomes the Fourier series (as shown in Note 6.5)

∞
1

ˆ
T trans (ˆz, ˆ t) = exp Peˆz a n exp −k n t sin(nπ ˆz), (6.168)
ˆ
2
n=1
where each term decays to zero with the exponential factor exp(−k n t), and where
ˆ
1
2 2 2
k n = n π + Pe . (6.169)
4
The initial condition (6.167) gives the Fourier coefficients
2nπ Pe
n Pe/2
a n = (−1) e − 1 . (6.170)
k n 2