Page 128 - Physical Principles of Sedimentary Basin Analysis

110 Heat flow

The temperature increase over a depth interval z becomes T = (q 0 /λ 0 ) z.The
temperature at the base of a basin with several layers each with a constant heat conductivity
λ i and thickness z i is therefore

z i
T (z) = T 0 + q 0 (6.24)
λ i
i

where z = z i is the total thickness of all layers. The temperature (6.24) is the discrete
i
version of the temperature solution (6.22). The harmonic average (2.89) normal to a bed-
ding plane is identified in the temperature solution (6.22). The same averaging applies to
the heat conductivity as to the permeability, and the average heat conductivity in the depth
interval from 0 to z is
1 1 z dv
= (6.25)
λ av (z) z 0 λ(v)
and the temperature solution (6.22) can therefore be written
z
T (z) = T 0 + q 0 . (6.26)
λ av (z)
The average heat conductivity is a useful property because it gives the (constant) heat flux
through a vertical column. In the case that the temperature is T m at the depth z m the heat
flux q 0 becomes
T m
q 0 = λ av (z m ) . (6.27)
z m
The solution (6.26) then gives the temperature at any position in the depth interval 0 to z m .
This is a simpler way to find the temperature in a layered rock mass than the alternative,
which is to glue together piecewise linear solutions.
2
Exercise 6.1 The temperature equation (6.20) becomes d T/dz 2 = 0, when the heat
conductivity is independent of the z-coordinate. This equation has the general solution
T (z) = Az + B, where the two coefficients A and B are given by two boundary condi-
tions.
(a) Let the boundary conditions be the temperature T 0 at the surface z = z 0 and the tem-
perature T m at the depth z .Find A and B and show that the temperature solution can be

m
written
(T m − T 0 )
T (z) = (z − z 0 ) + T 0 (6.28)
(z m − z 0 )
which gives the constant heat flow
(T m − T 0 )
q = λ (6.29)
(z m − z 0 )
in the entire depth interval between z 0 and z m .