Page 184 - Physical Principles of Sedimentary Basin Analysis

166 Heat flow

to first order in 1/ˆz when ˆz
1. The log function can also be approximated to first order,
ln(1 + x) ≈ x,for x ≈ 0. We then get that

ˆ z + 1 2
ln ≈ (6.241)
ˆ z − 1 ˆ z
when ˆz
1.

Exercise 6.27 Derive approximation (6.237).
Solution: The temperature at a given time and position is given by the integral (6.229),
√ √
where the integration limits are η 1 = (ˆz + 1)/(2 ˆ t) and η 2 = (ˆz − 1)/(2 ˆ t). Using that
2
ˆ t max ≈ˆz /2for ˆz
1gives
1 1
1 1
η 1 = √ 1 + and η 2 = √ 1 − . (6.242)
2 z 2 z

The integral (6.229) can then be approximated as
1 −1/2 2
T max = √ e √ (6.243)
ˆ
π 2ˆz
which is the approximation (6.237).

Exercise 6.28 Show that the solution (6.229) for the sill temperature can be approxi-
mated by

1 3ˆz + 1
ˆ
T (ˆz, ˆ t) = √ 1 − (6.244)
π ˆ t 12ˆ t
2
when |ˆz|
ˆ t. Hint: use that e −x 2 ≈ 1 − x for x ≈ 0.
Exercise 6.29 Show that the following temperature solution gives the cooling of a sill
emplaced at a finite depth h with zero temperature at the surface,

T (ˆz, ˆ t) = F(ˆz, ˆ t) − F(2h −ˆz, ˆ t) (6.245)
ˆ
where
1 (ˆz + 1)
(ˆz − 1)
F(ˆz, ˆ t) = erf √ − erf √ . (6.246)
2 2 ˆ t 2 ˆ t

Notice that the center of the sill is at ˆz = 0 and the surface is at ˆz = h. The solution is
shown in Figure 6.31.
Hint: show that the solution has the symmetry T (h + l, ˆ t) =−T (h − l, ˆ t) around z = h.
ˆ
ˆ
The symmetry implies that T = 0at ˆz = h for all ˆ t, which assures the boundary condition
ˆ
at the surface.