Page 138 - Physical Principles of Sedimentary Basin Analysis

120 Heat flow

The solution for the coefficients (from the boundary conditions) is
2
(λ m /2λ c ) S 0 z − S 0 z m (z m − z a ) + λ m (T a − T 0 )
m
a 1 = (6.69)
λ m z m − λ c (z m − z a )
λ c 1
a 2 = a 1 − S 0 z m (6.70)
λ m λ m
b 1 = T 0 (6.71)
b 2 = T a − a 2 z a . (6.72)

Coefficient a 1 is evaluated first, and it is then used to obtain coefficient a 2 , before b 1 and
b 2 . It is left as an exercise to check that T m = a 2 z m + b 2 is the same as T m (6.65) and that
q m = λ m a 2 is the same as q m (6.66).

Exercise 6.6 Replace the heat production S 0 by the average heat production S av =
(q s − q m )/z m and show that:
(a) the temperature at the base of the crust is
z m
T m = T 0 + (q s + q m ); (6.73)
2λ c
(b) the depth of the base of the lithosphere is
1
λ m λ m q s
z a = (T a − T 0 ) + z m − z m 1 + . (6.74)
q m 2 λ c q m
Exercise 6.7 Check that the solutions (6.59) and (6.60) give the same temperature and the
same heat flow at the point z = z m where they are glued together.

Exercise 6.8 Assume that the surface temperature T 0 and the surface heat flow q s are
known. Show that the solution of the temperature equation (6.49)is

z du z 1 v
T (z) = T 0 + q s − S(u) du dv. (6.75)
0 λ(u) 0 λ(v) 0
Exercise 6.9 (a) Use equation (6.59) to show that the thickness of the crust has to be
q m
z m = (6.76)
S
for the mantle heat flow and a constant crustal heat generation to contribute equally to the
surface heat flow.
(b) Show that the contribution from the linear term in the temperature solution (6.59) con-
tributes the same as the heat source term at the base of the crust (at z = z m ), when the crust
has the thickness
2q m
z m = . (6.77)
S
(c) What is z m in (a) and (b) when q m = 0.02 W m −2 and S = 10 −6 Wm −3 ?