Page 186 - Physical Principles of Sedimentary Basin Analysis

168 Heat flow

where the solidification boundary z m (t) is given by
z m (t) √
√ = η m or z m (t) = 2η m κt. (6.248)
2 κt
We already know that the error function solution solves the temperature equation, and it
is straightforward to verify that the boundary conditions are fulfilled. However, it remains
to be shown that it is possible to find a (constant) number η m , which controls the growth
of the solidified layer. We will see that energy conservation at the base of the solid layer
gives the number η m . The latent heat of fusion (L) is the energy released per unit mass
being solidified, and the energy released per unit volume during solidification becomes
L. The energy released per unit area when the depth interval dz m is frozen is Ldz m .
This energy has to be transported away by conduction in the time interval dt of freezing.
Energy conservation requires that the energy released by solidification is exactly the energy
transported away by heat conduction

∂z m ∂T
L = λ (6.249)
∂t ∂z
η m
where λ is the heat conductivity of the solid part. The left-hand side of boundary
condition (6.249) is obtained from equation (6.248) for the liquid/solid boundary

∂z m κ
L = Lη m (6.250)
∂t t
and the right-hand side is obtained from the temperature solution (6.247)

∂T e −η 2 m 1
λ = λ(T m − T s ) √ . (6.251)
∂z erf(η m ) πκt
When the parts (6.250) and (6.251) are inserted into the boundary condition (6.249) we get

e −η m 2 √ π L
= , (6.252)
η m erf(η m ) c(T m − T 0 )

which is the equation for η m . Notice that the thermal diffusivity is κ = λ/( c), where c is
the specific heat capacity. It is not possible to solve equation (6.252) exactly, but it is simple
to estimate η m by using the plot of the function f (x) = e −x 2 /(x erf(x)) in Figure 6.32.
It is interesting to know what the number η m might be. Typical parameters for magma
are L = 400 kJ kg −1 , T m − T s = 1000 C, and c = 1.0kJ kg −1 K −1 , which makes the
◦
function f (η m ) = 0.7. Figure 6.32 gives that η m ≈ 0.85. Figure 6.33 shows an example of
the temperature solution (6.247) through the solidified magma using these parameter values
◦
and that T s = 0 C. In the case of freezing water typical parameters are T s =−10 C,
◦
L = 320 kJ kg −1 and c = 4kJ kg −1 K −1 , which makes f (η m ) = 14. From Figure 6.32 we
then get that η m = 0.25.