Page 129 - Physical Principles of Sedimentary Basin Analysis

6.2 Stationary 1D temperature solutions 111

(b) Let the first boundary conditions be the same as in (a), but use a constant heat flow q m
at the depth z m as the second boundary condition. Find A and B and show that the solution
can be written
q m
T (z) = (z − z 0 ) + T 0 (6.30)
λ
where the temperature at the depth z m becomes T m = T (z m ).
Exercise 6.2 Assume that the heat conductivity decreases linearly with porosity as
λ(φ) = λ s · (1 − c φ) (6.31)

where λ s is the heat conductivity of the pure solid (at φ = 0). The coefficient c controls
how fast the heat conductivity decreases with the porosity. Assume furthermore that the
porosity decreases with depth as
z

φ = φ 0 exp − (6.32)
z 0
where z 0 is a characteristic depth for porosity loss. There is little porosity reduction for
depths z
z 0 and there is a substantial reduction for z
z 0 . When the porosity–depth
relationship (6.32) is inserted into the linear expression (6.31) for the bulk heat conductivity
we get an expression for heat conductivity as a function of depth, λ(z) = λ(φ(z)).Find
the 1D stationary temperature solution with this z-dependent heat conductivity, when the
surface temperature is zero.
Solution: The temperature is given by expression (6.22),
z

q 0 dz
T (z) = . (6.33)
λ s 0 1 − c φ 0 exp(−z/z 0 )
A change of variable from z to u = φ 0 exp(−z/z 0 ) leads to

φ 0
q 0 z 0 du q 0 z 0 φ 0
T = = lnu − ln(1 − cu) (6.34)
λ s φ (1 − cu) u λ s φ
and replacing u by the porosity gives

q 0 q 0 1 − c φ 0 exp(−z/z 0 )
T (z) = z + z 0 ln . (6.35)
λ s λ s 1 − c φ 0
The first part of the solution is the temperature in the case of constant heat conductivity of
solid rock (for φ = 0). The second term is therefore the difference between the wanted tem-
perature solution and the linear solution for a constant heat conductivity. Equation (6.35)
shows that the difference is increasing with depth, and that the maximum difference at
infinite depth is
q 0 z 0
T max =− ln(1 − c φ 0 ). (6.36)
λ s
The solution (6.35) is plotted in Figure 6.2 for φ 0 = 0.5, z 0 = 1000 m, c = 1.0, q 0 =
0.05 W m −2 and λ s = 2.5W m −1 K −1 . One sees that the temperature difference is close
◦
to T max = 14.0 Cfor z > 1500 m.