Page 152 - Physical Principles of Sedimentary Basin Analysis

134 Heat flow

and the maximum temperature at the center of the sphere (r = 0) becomes
1
2
S
T max = r + T 0 . (6.124)
0
6 λ
It would be interesting to see what this simple model predicts for the temperature at the
center of the Earth. Before we can do that we need to know the heat production per unit
volume, S. One way to obtain S is to replace it by an expression involving the surface heat
flow, which is

Sr 0
dT
q 0 =−λ = . (6.125)
dr 3
r=r 0
The heat generation is therefore S = 3q 0 /r 0 , when expressed using the surface heat flow,
and the maximum temperature at the center (r = 0) becomes
q 0 r 0
T max = + T 0 . (6.126)
3λ
Using the values r 0 = 6378 km, q 0 = 0.05 W m −2 , λ = 2.5W m −1 K −1 and T 0 = 0 C
◦
we get that S = 2.4 · 10 −8 Wm −3 and T max = 63780 C. This temperature is probably at
◦
least one order of magnitude too large. Heat conduction alone cannot therefore explain the
thermal state of the Earth.

6.9 Transient cooling of a sphere ∗
The purpose of this section is to show that the cooling of the Earth by conduction through
its history is not a good model.
We have so far only looked at stationary (time-independent) solutions of the temperature
equation. The simplest time-dependent version of the full temperature equation (6.15)is

∂T
b c b −∇ (λ∇T ) = 0 (6.127)
∂t
where the convective term and heat source term are both zero. (The material derivative is
replaced by an ordinary derivative because the rock matrix does not move relative to the
coordinate system.)
We will now solve temperature equation (6.127) for a cooling sphere. Although it is too
simplified as a model for the cooling of the Earth, it is nevertheless an interesting exercise in
heat conduction. Realistic models for the cooling of the Earth have to account for heat lost
by mantle convection. It follows from equation (6.121) that temperature equation (6.127)
becomes

∂T 1 ∂ 2 ∂T
b c b − 2 r λ = 0, (6.128)
∂t r ∂r ∂r
using spherical symmetry, T = T (r). The boundary conditions for a cooling sphere with
radius r = r 0 will be a zero temperature at the surface of the sphere, T (r=r 0 , t) = 0, and
a zero temperature gradient at the center of the sphere, ∂T/∂r(r =0, t) = 0. (Symmetry