Page 172 - Physical Principles of Sedimentary Basin Analysis

154 Heat flow

ˆ
ˆ
The Laplace equation is solved for T b by separation of variables, where T b is written as
T b (ˆx, ˆz) = U(ˆx) V (ˆz). When the product UV is inserted into the Laplace equation we get
ˆ
U V 2
=− = k (6.197)
U V
2

where k is a constant. We have that U /U is a function only of ˆx and that V /V is a
function only of ˆz, and because these two expressions are equal then they must be equal to
2
a constant. The equation for U is U − k U = 0, which has as solution any linear com-

x
x
x
bination of U(ˆ) = exp(−k ˆ) and U(ˆ) = exp(k ˆ). Since we cannot allow the solution
x
to grow exponentially with increasing ˆ the solution has to be U 0 exp(−k ˆx) where U 0 is a
x
2

constant. The equation for V is V + k V = 0 which has solutions any linear combina-
tion of sin(kˆz) and cos(kˆz). The boundary conditions T b = 0 along the upper and lower
ˆ
horizontal boundaries are fulfilled for every sin(kˆz) where k = nπ. There is therefore one
k n = nπ for each (positive) integer n, and there is also a U n and a V n for each (positive) n.
ˆ
Each product U n (ˆx)V n (ˆz) is a solution of the Laplace equation for T b , and the solution is
the Fourier series
∞

x
ˆ
T b (ˆx, ˆz) = a n exp(−k n ˆ) sin(k n ˆz). (6.198)
n=1
The boundary condition T b (ˆx=0, ˆz) =ˆz, which becomes
ˆ
∞

a n sin(k n ˆz) =ˆz (6.199)
n=1
leads to the Fourier coefficients a n . Using that V n are orthogonal with respect to the inner
1 1
product (V n , V m ) = V n V m dˆz = δ mn ,wehave
0 2
1 1 1 (−1) n
a n = ˆ z sin(nπ ˆz) dˆz =− , (6.200)
2 2 0 nπ
which gives the Fourier coefficients.
Exercise 6.19 Use the data given in Exercise 6.15.
(a) Estimate the time needed for the fracture to heat up its surroundings.
(b) Assume that a fracture in impermeable rock drains the same amount of fluid as in
Exercise 6.15. Show that the width of the area with vertical fluid flow in Exercise 6.15 has
3
to be W = w /12k, when the permeability of the rock in Exercise 6.15 is k, and the width
2
of the fracture is w. Hint: give that the permeability of the fracture is w /12.
(c) What is the width W when w = 0.1 mm?
6.14 Instantaneous heating or cooling of semi-infinite half-space

The change in temperature of surface rocks caused by lava flows (flood basalt) or glacier
flows can be calculated by taking the temperature change at the surface to be instantaneous
and the crust to be infinitely deep. It also turns out that the temperature solution for the