Page 183 - Physical Principles of Sedimentary Basin Analysis

6.15 Cooling sills and dikes 165

and thermal convection in its surrounding water saturated sediments when it is pressed
into the rock. This could be an explanation for the mismatch between the modeled and
the observed vitrinite reflectance. A model for sill cooling by conduction is therefore best
suited for dry rocks.
Figure 6.30a shows that the distance away from the sill with increased vitrinite
reflectance is roughly the thickness of the sill. This observation holds over several orders
of magnitude as seen from Figure 6.30b.
Note 6.8 A simple way to derive the temperature solution (6.228) is to begin with the error
√
function solution erf(ˆz/(2 ˆ t)). We have already seen in Section 6.14 that it is a solution of
the temperature equation (6.226). A translation of this error function solution at a distance
±1 is also a solution. We will now define the following two functions:

ˆ z + 1 ˆ z − 1
f (ˆz) = erf √ and g(ˆz) =−erf √ (6.239)
2 ˆ t 2 ˆ t
+
for t = 0 . The argument of the error functions are either −∞ or ∞, and the func-
tions f (ˆz) and g(ˆz) become a step function from −1to1and from1to −1, respectively.
The two step functions are both 1 in the ˆz-interval from −1 to 1. The linear combination
(1/2)( f (ˆz) + g(ˆz)) will therefore fulfill the initial condition. See Figure 6.28 where f (ˆz)
and g(ˆz) are plotted. The temperature solution (6.228) is therefore the wanted solution.
The method of adding together linear combinations of error-function-solutions, where the
+
sum at t = 0 fulfills both the initial condition and the boundary conditions, is a simple
and powerful way to solve the temperature equation.
Exercise 6.23
(a) How much time is needed for the temperature at the center of a 2 m, 20 m and 200 m
2 −1
thick sill to be reduced by half? Use κ = 10 −6 m s .
(b) How much time is needed for the temperature at the center of a 2 m, 20 m and 200 m
thick sill to be reduced to 5%?

Exercise 6.24 How far away from the sill will the maximum temperature increase by 10%,
5% and 2.5% of the initial sill temperature?

Exercise 6.25
(a) At what time will the temperature maximum take place at a distance 1000 m away from
2 −1
a 2 m, 20 m and 200 m thick sill? Use κ = 10 −6 m s .
(b) What is the temperature at the maximum in (a)?

Exercise 6.26 Derive approximation (6.236).
Solution: We have

1
ˆ z + 1 1 + ˆ z 1 1 2
= ≈ 1 + 1 + ≈ 1 + (6.240)
ˆ z − 1 1 − 1 ˆ z ˆ z ˆ z
ˆ z