Page 180 - Physical Principles of Sedimentary Basin Analysis

162 Heat flow

The time span 1000 × t 0 is then ∼0.1 Ma for a sill of thickness 2a = 100 m.
The temperature solution (6.228) is approximated by a simpler expression for distances
far away from the sill or at the center of the sill. The solution becomes
1 η 1 2 η 2 2
1 η 2 2
ˆ −x −x −x
T (ˆz, ˆ t) = √ e dx − e dx = √ e dx (6.229)
π 0 0 π η 1
√ √
where η 1 = (ˆz − 1)/(2 ˆ t) and η 2 = (ˆz + 1)/(2 ˆ t). The last integral is approximated as

2 2
1 η 1 + η 2 1 ˆ z
ˆ
T (ˆz, ˆ t) ≈ √ exp − (η 1 − η 2 ) = √ exp − (6.230)
π 2 π ˆ t 4ˆ t
for ˆz
1.
√ √
At ˆz = 0wehavethat η 1 = 1/(2 t) and η 2 =−1/(2 t), and for ˆ t
1 we get that
η 1
1 and η 2
1. The temperature solution (6.229) can therefore be approximated as
1 η 2 1
ˆ
T (ˆz, ˆ t) ≈ √ dx = √ (6.231)
π π ˆ t
η 1
for ˆ t
1, because the integrand becomes e −x 2 ≈ 1for x ≈ 0.
The sill will be felt as an increase in the temperature followed by a decrease in the
temperature at a position ˆz outside the sill. Solving the equation ∂T /∂ ˆ t = 0for t gives
ˆ
the time when the temperature impact is at its maximum at a position ˆz. A straightforward
differentiation of temperature equation (6.229) yields

∂T ˆ 1 −η 2 −η 2 −3/2
=− e 1 (ˆz + 1) − e 2 (ˆz − 1) ˆ t = 0 (6.232)
∂ ˆ t 4π
which implies that

e −η 2 1 (ˆz + 1) = e −η 2 2 (ˆz − 1). (6.233)
Further simplification leads to

ˆ z ˆ z + 1
exp = (6.234)
ˆ t ˆ z − 1
or
ˆ z
ˆ t max =
. (6.235)
ln ˆ z+1
ˆ z−1
Equation (6.235)for ˆ t max is approximated by the simple expression
ˆ z 2
ˆ t max ≈ (6.236)
2
for ˆz
1 (see Exercise 6.26). Inserting ˆ t max into the temperature solution (6.228)gives
the temperature maximum T max . Using the approximation (6.236) ˆ t max gives
ˆ

2 1
ˆ
T max ≈ . (6.237)
πe ˆz