Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 332 / 2 / a

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- Solution a

Solution

Let

κ = \frac{k}{ρ c _{p}}, κ_{s} = \frac{k _{s}}{ρ c _{p}}, h (t) = 2 λ κ t .

The Neumann solution of the Stefan problem in the ice and substrate is

T_{i} (z, t) = T_{0} + (T_{m} - T_{0}) \frac{erf ( z / ( 2 κ t ))}{erf λ} (0 < z < h),

and

T_{s} (z, t) = T_{s} + (T_{0} - T_{s}) erfc (\frac{- z}{2 κ _{s} t}) (z < 0) .

Continuity of heat flux at the contact gives, with

r = k_{s} / k

,

\frac{T _{m} - T _{0}}{erf λ} = r (T_{0} - T_{s}),

and therefore

T_{0} = \frac{T _{m} + r erf ( λ ) T _{s}}{1 + r erf ( λ )} .

At the ice–water interface, the water is isothermal at

T_{m}

. The Stefan condition

ρ L \dot{h} = k T_{i, z} (h^{-})

gives

S λ = \frac{T _{m} - T _{0}}{T _{m} - T _{s}} \frac{e ^{- λ^{2}}}{π erf λ}, S = \frac{L}{c _{p} ( T _{m} - T _{s} )} .

Eliminating

T_{0}

yields the implicit equation

π S λ e^{λ^{2}} (1 + r erf λ) = r,

which determines

λ

and hence

h (t) = 2 λ κ t

.

If

k_{s} ≫ k

, then

r ≫ 1

,

T_{0} \sim T_{s}

, and

S λ \sim \frac{e ^{- λ^{2}}}{π erf λ} .

The highly conducting substrate acts as a reservoir fixed near its initial cold temperature, giving the usual one-phase Stefan problem.

If

k_{s} ≪ k

, then

r ≪ 1

,

T_{0} \sim T_{m}

, and

λ \sim r / (π S)

. Consequently

h (t) \sim \frac{2 c _{p} ( T _{m} - T _{s} )}{L} \frac{κ _{s} t}{π} .

Here heat removal through the poorly conducting substrate is rate limiting, and only a small temperature drop is needed across the much more conducting ice.

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