Page 188 - Physical Principles of Sedimentary Basin Analysis

170 Heat flow

6.17 Solidification of sills and dikes
The solidification of a sill is treated in a similar way as the freezing of a lake. The sill is
assumed to be at the melting temperature when it is emplaced, and solidification then starts
at the boundaries against the surrounding rock and proceeds towards the center of the sill.
We have once more to solve the temperature equation for the solid part that is now growing
into the sill. The top surface of the sill is placed at z = 0 with the z-axis pointing upwards.
The solidified part of the sill is then the area z m < z < 0, where z m is the interface between
liquid and solid magma. The solution for the interface z m is
z m √
√ =−η m or z m (t) =−2η m κt (6.253)
2 κt
where η m is the parameter that controls how fast the magma solidifies. Notice the minus
sign on η m which tells us that z m (t) grows downwards from z = 0. (The parameter η m is
a positive number.) Boundary conditions for the temperature equation are the temperature
T 0 far away from the sill (z =∞) and T = T m , the melting temperature at z = z m .The
temperature solution is then once again an erf-function solution
√
erfc(z/2 κt)
T (z, t) = T 0 + (T m − T 0 ) , (6.254)
erfc(−η m )
where the variable η m is obtained from energy conservation at the interface between liquid
and solid magma. Energy conservation requires that the rate of energy released by solidifi-
cation is exactly the rate of energy transported away by conduction, which is written in the
same way as in the previous section:

∂z m ∂T
L = λ . (6.255)
∂t ∂z
η m
The velocity of the solid/liquid interface (6.253) becomes

∂z m κ
=−η m (6.256)
∂t t
and the gradient of the temperature solution (6.254)is
∂T e −η m 2 1
=−(T m − T 0 ) √ . (6.257)
∂z (1 + erf(η m )) πκt
This assumes that the complementary error function is erfc(−x) = 1 − erf(−x), and
that erf(−x) =−erf(x). When expressions for ∂z m /∂t and ∂T/∂z are inserted into the
boundary condition (6.255) we get an equation for the parameter η m (where both z m and t
drop out)
e −η 2 m √ π L
= . (6.258)
η m (1 + erf(η m )) (T m − T 0 )c
This equation is most easily solved graphically using a plot of the function g(x) =
e −x 2 /(x(1 + erf(x))), see Figure 6.32. The time it takes for the sill to solidify is obtained