= Solution
Use a planar <saline Stefan problem> with upward coordinate $z$, the initial contact at $z=0$, and <phase boundaries> $z=a(t)$ and $z=b(t)$ enclosing pure <ice>, where $a<b$. Fresh liquid occupies $z>b$ and brine occupies $z<a$. We assume negligible bulk flow, equal constant <mass density> $\rho$, <specific heat capacity> $c_p$ and <thermal diffusivity> $\kappa$ in all regions, and zero salt content and salt transport in the <ice>. Equal densities remove phase-change volume flow; suppressing convection isolates the molecular-transport mechanism. Different material properties would change the numerical coefficients. Set $K=\rho c_p\kappa$ for the common <thermal conductivity> and $\Delta T=T_m-T_\infty>0$. Denote the brine salt <diffusion coefficient> by $D$.
Use a linear <ice> <liquidus>, $T_L(C)=T_m-mC$, with $m>0$ and $C$ the salt mass fraction. Assume the initial brine is a stable liquid, $T_\infty>T_L(C_0)$, and all concentrations considered are below the <eutectic composition>. Thus $R=mC_0/\Delta T>1$. These assumptions exclude independent nucleation in a supercooled brine or an <eutectic system> event, neither of which is specified by the initial data alone. Start with a negligible seed of <ice> and use local phase equilibrium without interfacial kinetics or curvature. <Temperature> differences may be in Celsius, since only differences enter this calculation.
Write
$$
S=\frac{L}{c_p\Delta T},\qquad \epsilon=\sqrt{D/\kappa}.
$$
The paper calls $S$ a <Stefan number>; it is the <latent-to-sensible heat ratio>, reciprocal to another common <Stefan number> convention. At the lower <phase boundary>, define $T_i=T_L(C_i)$ and let
$$
\vartheta=\frac{T_m-T_i}{\Delta T},\qquad c=\frac{C_i}{C_0}.
$$
The <liquidus> relation is $\vartheta=Rc$.
A <similarity solution> gives an explicit calculation of the two positions. Put
$$
a(t)=2\delta\sqrt{\kappa t}=2A\sqrt{Dt},\qquad
b(t)=2B\sqrt{\kappa t},\qquad
\delta=\epsilon A,
$$
and use $u=z/(2\sqrt{\kappa t})$, $v=z/(2\sqrt{Dt})$. The <temperatures> and salinities are
$$
T_f=T_m,\quad C_f=0\qquad(z>b),
$$
$$
T_s=T_m-\Delta T\,\vartheta
\frac{\operatorname{erf}B-\operatorname{erf}u}
{\operatorname{erf}B-\operatorname{erf}\delta},
\quad C_s=0\qquad(a<z<b),
$$
$$
T_l=T_\infty+\Delta T(1-\vartheta)
\frac{\operatorname{erfc}(-u)}{\operatorname{erfc}(-\delta)},
\quad
C_l=C_0+(C_i-C_0)
\frac{\operatorname{erfc}(-v)}{\operatorname{erfc}(-A)}
\qquad(z<a).
$$
These <error function> and <complementary error function> profiles satisfy the <heat equation> and the salt <diffusion equation>, their far-field conditions, and the interfacial <temperature> conditions. The fresh-water <temperature> is constant because the interface and the far field are both at $T_m$.
Salt conservation at the moving lower <phase boundary> requires
$$
D\,C_{l,z}(a,t)=-\dot a\,C_i.
$$
For upward motion this is dilution by <melting> salt-free <ice>, rather than <salt rejection> by <freezing> brine. Substitution gives the <dilution function for a melting saline Stefan front>:
$$
\boxed{c=\frac1{1+\sqrt\pi A e^{A^2}\operatorname{erfc}(-A)}}.
$$
For $A>0$, $0<c<1$, so the interfacial brine is fresher than the remote liquid.
The <Stefan condition> at the upper boundary is $\rho L\dot b=K T_{s,z}(b)$, since the fresh liquid has no <temperature> gradient. At the lower boundary it is
$$
\rho L\dot a=K\bigl(T_{s,z}(a)-T_{l,z}(a)\bigr).
$$
These signs follow from the jump in enthalpy: the upper front freezes liquid while a positive $\dot a$ melts solid. With $E=\operatorname{erf}B-\operatorname{erf}\delta$, they reduce to
$$
\boxed{
SB=\frac{\vartheta e^{-B^2}}{\sqrt\pi E},\qquad
S\delta=\frac{\vartheta e^{-\delta^2}}{\sqrt\pi E}
-\frac{(1-\vartheta)e^{-\delta^2}}
{\sqrt\pi\,\operatorname{erfc}(-\delta)},\qquad
\vartheta=\frac{R}{1+\sqrt\pi A e^{A^2}\operatorname{erfc}(-A)}.}
$$
Together with $\delta=\epsilon A$, these equations determine $A,B,\vartheta$, and hence both positions, without discarding the salt-diffusion correction.
For the stated large <latent-to-sensible heat ratio>, the layer is thin relative to the thermal <diffusion length>. Expanding the thermal equations for $B,\delta\ll1$ gives
$$
S(B-\delta)\simeq\frac{1-\vartheta}{\sqrt\pi},\qquad
\vartheta\simeq2SB(B-\delta).
$$
Consequently a useful leading calculation is
$$
\boxed{
h(t)=b-a\simeq\frac{2\sqrt{\kappa t}}{S\sqrt\pi},\quad
a(t)=2A\sqrt{Dt},\quad
b(t)\simeq2\left(\epsilon A+\frac1{S\sqrt\pi}\right)\sqrt{\kappa t},}
$$
where the positive $A$ is obtained from
$$
\frac{R}{1+\sqrt\pi A e^{A^2}\operatorname{erfc}(-A)}
\simeq\frac2{\pi S}+\frac{2\epsilon A}{\sqrt\pi}.
$$
The left side decreases with $A\ge0$ while the right side increases, and their values at zero and infinity guarantee a unique positive root for large $S$. Thus this leading calculation includes the translation of the lower boundary as well as the increasing thickness.
If the scale separation also obeys $\epsilon SA\ll1$, the familiar simpler result is
$$
B\simeq\frac1{S\sqrt\pi},\qquad
\vartheta\simeq\frac2{\pi S},\qquad
\frac{C_i}{C_0}\simeq\frac2{\pi RS},
$$
with
$$
A e^{A^2}\simeq\frac{\sqrt\pi RS}{4},
\qquad A=O\!\left(\sqrt{\log(RS)}\right).
$$
This last simplification needs the logarithmic refinement $\epsilon S\sqrt{\log(RS)}\ll1$. The algebraic ordering $\epsilon\ll S^{-1}$ alone should not be used to discard an arbitrarily large logarithmic correction; the preceding coupled equations remain the appropriate calculation when that refinement is unavailable. This distinction is captured by <large latent heat in a freezing and melting ice layer>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-71-fields.png]
{title=Fresh-water freezing and basal ice melting with the temperature and salinity profiles on their distinct diffusion scales}
{height=650}
The illustration uses $S=40$, $\epsilon=2.5\times10^{-4}$ and $R=2$. Solving the full equations gives $A=1.73707$, $B=0.0143053$ and $\vartheta=0.0158766$. It shows the <ice layer between freshwater and cold brine>, with both fronts advancing upward and a strongly diluted lower liquid boundary.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-71-phase-trajectory.png]
{title=Temperature-salinity trajectory from cold bulk brine to the ice liquidus and through salt-free ice to fresh water}
{height=620}
The <phase diagram> shows the spatial path from remote brine to the lower interface: <temperature> first changes over $\sqrt{\kappa t}$ at almost unchanged <salinity>, then <salinity> changes over its much smaller diffusion scale at nearly constant <temperature>. The liquid path ends on the <liquidus> at $(C_i,T_i)$; the solid has $C=0$ and its <temperature> rises from $T_i$ to $T_m$. The eutectic position is schematic, with $C_E=2C_0$ in the illustration, and is not used in the calculation. A point initially in the fresh layer freezes when $b$ reaches it, cools in the <ice>, and later melts when $a$ reaches it. The inset indicates this temporal path in the opposite direction through the solid branch and into the liquid.
The mechanism is \b[simultaneous upper <freezing> and lower <melting>, with net growth of the <ice> thickness]. Cold brine accepts most of the <latent heat> released by upper <freezing>. Salt reaching the lower contact lowers its equilibrium <melting> <temperature>, and <melting> freshens that liquid until its <liquidus> is close to $T_m$. The heat delivered through the <ice> supplies the much smaller <melting> demand there. The remote brine remains liquid even though it is below the pure-water <melting> <temperature>, because it lies above its own saline <liquidus>. Thus “cold” does not by itself decide which phase is stable. These conclusions apply while the layers remain effectively deep and the stated no-convection, pure-<ice> and phase-equilibrium assumptions hold.
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