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

LetThe Neumann solution of the Stefan problem in the ice and substrate isandContinuity of heat flux at the contact gives, with $r=\sqrt{k_s/k}$,and therefore

At the ice–water interface, the water is isothermal at $T_m$. The Stefan condition $\rho L\dot{h}=k{T}_{i,z}(h^{-})$ givesEliminating $T_0$ yields the implicit equationwhich determines $\lambda$ and hence $h(t)=2\lambda \sqrt{\kappa t}$.

If $k_s\gg k$, then $r\gg 1$, $T_0\sim T_s$, andThe highly conducting substrate acts as a reservoir fixed near its initial cold temperature, giving the usual one-phase Stefan problem.

If $k_s\ll k$, then $r\ll 1$, $T_0\sim T_m$, and $\lambda \sim r/(\sqrt{\pi }S)$. ConsequentlyHere heat removal through the poorly conducting substrate is rate limiting, and only a small temperature drop is needed across the much more conducting ice.

Measure $z$ upward from the heat exchanger and let the steady ice front be at $z=h>0$. In the exchanger frame, salt in the liquid satisfies the advection-diffusion equationThe decaying solution and the prescribed total salt mass aresoThe linear liquidus condition fixes the interface temperature asA solid layer between the exchanger and the interface can therefore exist only if

Write $\vartheta =T-T_{\infty }$. Heat advection, conduction, and environmental loss giveIts characteristic exponents areBelow the exchanger and above the ice front, boundedness givesIn $0<z<h$, the ice temperature iswhereSubstitution of these fields into the Stefan conditiongives one scalar equation for the steady height $h$, which can be solved numerically. The heat flux may jump at $z=0$ because the exchanger supplies the required localized cooling.

The local equilibrium freezing temperature ahead of the front isBecause $T_l(h)=T_L(h)$, constitutional supercooling begins when the actual liquid-temperature gradient at the interface is smaller than the liquidus gradient:Using the solutions above, this is

At criticality the inequality is an equality. If $F/k\ll V^2/{\kappa }^2$, then $q\sim V/\kappa$ and, for $\kappa >D$,The critical curve is proportional to $S^{-1}$. If $F/k\gg V^2/{\kappa }^2$, put $q\sim \sqrt{F/k}$ to obtainThis curve behaves as $S^{-1/2}$ for small $S$ and approaches $D\sqrt{F/k}$ for large $S$. Supercooling occurs above the corresponding critical curve, where solute rejection steepens the liquidus faster than heat transport raises the actual temperature.