Table C1. Comparison of closed
R
(1 + R / 3)
K1 = 1 +
solution (G = 0) and numerical
2
quadrature (G = 0.0001).
K2 = U(1 + ST )
ST = 0.144, α = 58.89 m2/yr, ∆T1 =
10C, U = 1 mm/yr.
K3 = αR(1 + R).
Freeze
Time
Time
depth
(yr)
(yr)
Percent
(m)
eq C12
eq C17
difference
Note that if U is zero, the phase change interface is
1000
55,867
54,778
1.99
R(1 + R)
2000
206,996
203,653
1.64
X2 =
2αt .
(C18)
3000
437,935
428,448
2.20
1 + R / 2(1 + R / 3)
This is identical to the well known Stefan solution given in Lunardini (1991). We may compare the closed
form solution (for which G = 0) with the numerical quadrature of eq C12 by letting G be very small. Table
C1 shows that the results are quite good even for very long freeze times.
Method 2
We can examine the same problem with a different approximation method by referring to Figure 13. For
region 3, a quasi-steady approach will be used, leading to a linear temperature profile. The basic equations
for heterogenetic growth are valid except that the surface temperature will be replaced by a transient func-
tion Ts′(t) . Equations 15 are valid but the temperature profiles are changed as follows. Quadratic tempera-
ture profiles in regions 1 and 2 and a linear temperature in region 3 that satisfy the boundary conditions are
x - X
x - X
2
+ (a1 X - ∆T3 )
T1 = Tf + a1 X
(C19)
X
X
(x - X)
x-X
2
[
]
T2 = Tf + G(δ + 2 X ) + 2∆T
- (GX + ∆T )
(C20)
δ
δ2
x
T3 = Ts + ∆T1MRX
+ 1
(C21)
Xd
where
α 21(∆T + GX )X
∆T1M
R = σ d k13 (2 - 1 / g) / σ,
a1 X =
g=
+ 1,
[
]
,
δ G(δ + 2 X ) + 2∆T
g
[
]
1
∆T3 = Tf - Ts′(t) = Rσ∆T1M / σ d k13 (2 - 1 / g) .
M=
,
1+ R
Equation 5 can be used to find a relation between X and δ. In nondimensional form this is
2ρ β(g - 1)
βM
[
]
- k21 σ(β + 2) + 2φ = 21
.
(C22)
g
ST
Equation 3, the energy integral equation, can now be written nondimensionally as
σ
τ = ∫ K / Hdσ
(C23)
0
[(
]
)
K = b1 + (b2 - Aσ / 3)β - M /(6g) - Aσ / 2 + σ d M 2k13 (0.5 / g + 1) g′ / g2 - (2 - 1 / g) / σ / 3
[
]
()
+ σ (b2 - Aσ / 3)β′ + Mg′ / 6g2 - A(β / 3 + 0.5) .
(C24)
32
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