Page 163 - Physical Principles of Sedimentary Basin Analysis

6.11 Forced convective heat transfer 145

2 ˆ
dT ˆ d T
v 0 ˆz − = 0 (6.156)
dˆz dˆz 2
where the velocity term is v 0 ˆz. The fluid flux is not constant but it decreases from v 0 at
ˆ
the top (ˆz = 1) towards zero at the base (ˆz = 0). Setting f = dT /dˆz in the temperature
equation (6.156) gives after integration that
dT ˆ 1 2
= c 1 exp v 0 ˆz (6.157)
dˆz 2
and a second time of integration gives

ˆ z 1
ˆ 2
T (ˆz) = c 1 exp v 0 μ dμ + c 2 . (6.158)
0 2
ˆ
The boundary conditions are dT /dˆz =ˆq at ˆz = 0 and T = 0at ˆz = 1. The first boundary
ˆ
condition gives c 1 =ˆq and the second boundary condition gives
1 1
c 2 =−ˆq exp v 0 μ 2 dμ (6.159)
0 2
and the solution for temperature becomes
1 1
ˆ
T (ˆz) =−ˆq exp v 0 μ 2 dμ. (6.160)
ˆ z 2
It is possible to rewrite solution (6.160) for the case where v 0 < 0 using the error-function
defined as
2 u 2
erf(u) = √ exp(−μ ) dμ. (6.161)
π 0
The velocity v 0 is written as v 0 =−v 1 (where v 1 is positive) and the solution becomes

π v 1 v 1
T (ˆz) =ˆq erf ˆ z − erf . (6.162)
ˆ
2v 1 2 2
The scaled heat flux is ˆq =−1 in Figure 6.20 and the optimal values for v 1 are then
v 1 = 1.07 and v 1 = 4.5. The parameter v 1 is the maximum of the linearly decreasing fluid
flow term, and its optimal values are similar to the Pe-numbers obtained from the maximum
◦
◦
vertical Darcy flow. Other numbers used in Figure 6.20 are T 1 = 0 C, T 2 = 33 C and
z = 1000ˆz − 750 in units of meters. The temperature solution with units is therefore

z + 750
T (z) = T 1 + (T 2 − T 1 ) T ˆ (6.163)
1000
which is the analytical temperature plotted in Figure 6.20. The optimal value for v 1 can
be used to estimate the Darcy flux, and we get that v 1 = 1.1 gives the Darcy flux v D =
4
4
1 · 10 mMa −1 and that v 1 = 4.5gives v D = 4 · 10 mMa −1 , assuming that l 0 = 1000 m
2 −1
and κ = 3 · 10 −7 m s .