Page 171 - Physical Principles of Sedimentary Basin Analysis

6.13 Heat flow in fractures 153

gives the maximum impact on the surroundings of the fracture. It is sufficient to solve the
Laplace equation along the positive x-axis (x ≥ 0) and in the interval z 1 < z < z 2 because
of symmetry (see Figure 6.22). The boundary condition along the fracture is T (x=0, z) =
T 2 , and the boundary conditions along the horizontal sides are T (x, z = z 1 ) = T 1 and
T (x, z = z 2 ) = T 2 for x ≥ 0.
A dimensionless version of the Laplace equation is made by scaling x and z with the
characteristic length of the system, l 0 = z 2 − z 1 , and temperature is scaled with the
temperature difference T 2 − T 1 . The dimensionless variables are then
x z 2 − z T − T 1
ˆ
ˆ x = , ˆ z = and T = . (6.193)
l 0 l 0 T 2 − T 1
ˆ
x
x
ˆ
The boundary conditions in dimensionless form are T (ˆ=0, ˆz) = 1, T (ˆ, ˆz = 0) = 1
ˆ
x
and T (ˆ, ˆz = 1) = 0. Note 6.6 shows that the solution of the dimensionless Laplace
equation with these boundary conditions is
∞

ˆ
x
x
T (ˆ, ˆz) = 1 −ˆz + a n exp(−nπ ˆ) sin(nπ ˆz), (6.194)
n=1
where the Fourier coefficients a n are
2 (−1) n
a n =− . (6.195)
nπ
The first part of the solution (6.194), 1−ˆz, represents conductive heat flow, and the Fourier
series is therefore the thermal impact of the “hot” vertical fracture. The solution (6.194)is
plotted in Figure 6.23, where we see how isotherms bend up along the fracture. We also
see that the isotherms are almost horizontal a distance x = l 0 (or ˆx = 1) away from the
fracture. The observation that the temperature solution is T (ˆ, ˆz) ≈ 1 −ˆz for ˆ
1
x
x
ˆ
x
follows from the exponential decay of each Fourier component with ˆ.
The characteristic time (6.148) associated with the length l 0
2
l c b b
0
t 0 = (6.196)
λ
estimates the time span needed for the fracture to heat up its surroundings. The system will
be close to the stationary state when t
t 0 , which is when the stationary solution (6.194)
applies. As an example we see that t 0 becomes 3000 years for l 0 = 100 M and the thermal
2 −1
diffusivity λ/ b c b = 10 −6 M s .
Note 6.6 The solution (6.194) for the Laplace equation with the boundary conditions
x
ˆ
x
ˆ
x
T (ˆ =0, ˆz) = 1, T (ˆ, ˆz = 0) = 1 and T (ˆ, ˆz = 1) = 0 is written as the sum of two
ˆ
ˆ
x
ˆ
parts T (ˆ, ˆz) = 1 −ˆz + T b (where both parts are solutions of the Laplace equation).
The first part, 1 −ˆz, is the temperature solution for heat conduction between the lower
horizontal boundary at T = 1 and the upper horizontal boundary at T = 0. The second
ˆ
ˆ
part, T b (ˆx, ˆz), is the temperature perturbation by the fracture, and we see that the boundary
ˆ
ˆ
ˆ
ˆ
conditions for T b (ˆx, ˆz) are T b (ˆx =0, ˆz) =ˆz, T b (ˆx, ˆz = 0) = 0 and T b (ˆx, ˆz = 1) = 0.
ˆ