Page 139 - Physical Principles of Sedimentary Basin Analysis

6.4 Stationary 1D temperature solutions with heat generation 121

Exercise 6.10 Let the heat production in the crust be linearly dependent on depth;
S(z) = az + b. (6.78)

(a) Show that the geotherm in the crust (z < z m ) becomes

q m 1 1 2 1 3 1 2
T (z) = T 0 + z + az z + bz m z − az − bz . (6.79)
m
λ c λ c 2 6 2
(b) Show that the geotherm in the mantle (z > z m )is
q m
T (z) = T m + (z − z m ) (6.80)
λ m
where
q m 1 1 3 1 2
T m = T 0 + z c + az + bz m . (6.81)
m
λ m λ c 3 2
(c) Show that T m in the case of the three distributions (6.54)to(6.56) of heat production
becomes

q m 1 S 0 2
T m,1 (z) = T 0 + z m + z m (6.82)
λ c 3 λ c
q m 1 S 0 2
T m,2 (z) = T 0 + z m + z m (6.83)
λ c 2 λ c
q m 2 S 0 2
T m,3 (z) = T 0 + z m + z m (6.84)
λ c 3 λ c
1
2
which differ by S 0 z /λ c .
6 m
Exercise 6.11 Measurements of heat flow in continental areas like mountains sometimes
suggest a linear relationship between the surface heat flow and the heat generation in the
surface rocks. Figure 6.6 shows an example. An exponentially decreasing heat production
with depth

S(z) = S 0 e −z/z 0 (6.85)
can explain these observations. This depth-dependent crustal heat production is still used,
although recent compilations of data show that it is not valid in general.
(a) Show that the heat generation (6.85) gives the surface heat flow

q s ≈ z 0 S 0 + q m (6.86)
for |z m |
|z 0 |, where q m is the mantle heat flow and z m the thickness of the crust.
(b) Show that the geotherm is

2
q m z S 0
0
T (z) = T 0 + z + 1 − e −z/z 0 (6.87)
λ λ