Page 190 - Physical Principles of Sedimentary Basin Analysis

172 Heat flow

a latent heat L = 350 kJ kg −1 , a heat capacity c = 1.3kJ kg −1 K −1 and a melting tem-
◦
◦
perature T m = 1261 C. The background temperature is T 0 = 0 C, and the η m -parameter
2 −1
becomes 0.8. The thermal diffusivity of both the rock and the melt is κ = 1 · 10 −6 m s .
The characteristic time for the process becomes t 0 = 79.2 years, and the sill has solidified
after t s = 0.39 t 0 = 31 years. The cooling of the sill after solidification can be modeled
with the temperature solution (6.228). But we have to make an adjustment of the initial
sill temperature with respect to the energy liberated by solidification. The energy released
when a volume V solidifies is E m = LV , and the energy stored in the same volume due
to the temperature difference T m −T 0 is E = c (T m −T 0 )V . The total energy E = E m +E

can be represented by an increase in the temperature from T m to T = T m + L/c.Using

m
ˆ
the temperature T with the T -function (6.228) the temperature solution for cooling of the

m
ˆ
sill becomes T (z, t) = T T (z/a, t/t 0 ). This is not the exact temperature solution for the
m
cooling of a sill after it has solidified, but it is a good approximation. Figure 6.34bshows a
comparison of this solution and a numerical solution for solidification and cooling. It turns
out that the approximation is accurate, except for a short time interval right after solidifi-
cation, where there is a slight mismatch (see ˆ t = 1). The numerical solution is based on
◦
an “effective heat capacity” method (see Note 6.9), where the sill solidifies over a 25 C
◦
◦
temperature interval from 1236 C to 1261 C.
Note 6.9 Effective heat capacity. Solidification of a volume V liberates the energy E =
LV , where L is the latent heat of fusion. (The term specific latent heat can also be used
for L because it is energy per mass.) If the solidification takes place over a temperature
interval from T 1 to T 2 the energy liberated, when the temperature decreases by T ,is
LV
E = T. (6.263)
T 2 − T 1
The temperatures T 1 and T 2 are the solidus and liquidus, respectively. The rate at which
energy is released can be added as a source term in the energy equation

2
∂T ∂ T L ∂T
c − λ =− (6.264)
∂t ∂z 2 T 2 − T 1 ∂t
where the minus sign is needed before the source term in order to make it positive. (∂T/∂t
is negative when the temperature is decreasing.) The heat source term can be brought over
to the left-hand side, where the heat capacity is replaced by effective heat capacity
⎧
⎨ c, T ≤ T 1
L
c eff (T ) = c + , T 1 < T < T 2 (6.265)
T 2 −T 1
c, T 2 ≤ T.
⎩
Water goes from liquid to solid at 0 C, but most rocks solidify over a temperature interval
◦
that may be several hundred degrees. The effective heat capacity method is not suited for
water, unless the melting temperature can be approximated by a temperature interval.