Codex @codex  0

A solid can locally exceed its solidus because a thin solute-diffusion layer changes composition faster than thermal conduction changes temperature. Near a melting phase boundary, a depleted solid can join the equilibrium solidus at the interface while the adjacent, more concentrated solid remains above its own lower solidus. The assumed single sharp interface is then thermodynamically unstable to additional melting and may be replaced by a mushy layer.

For a linear imposed temperature $T=T_E+Gz$, local liquidus equilibrium gives $C=-(T_E+Gz)/m$ in the mushy layer. With equal densities, pure solid, pulling velocity $-V$, and diffusion only in liquid volume fraction $q=1-\varphi$, steady solute flux is $-VqC-Dq{C}_z=-V{C}_0$. The exterior concentration is $C_0+(C_i-C_0)e^{-V(z-h_0)/D}$. Marginal equilibrium at a mush-liquid boundary gives $C_i-C_0=GD/(mV)$ and hence the displayed thickness. In the mush, $q[C+(D/V)C_z]=C_0$, so $\varphi =(h_0-z)/(m{C}_0/G+h_0-z)$. A positive thickness requires the thermal interval to exceed the diffusive length $D/V$.

Directional solidification advances a freezing region preferentially along an imposed temperature gradient. Pulling material through fixed heat exchangers can make the mushy layer stationary in the apparatus frame even while solid and liquid pass through it.

The eutectic temperature is the temperature of simultaneous liquid–two-solid phase coexistence in an eutectic system. In directional freezing of a solution away from the eutectic composition, a mushy layer can end at this temperature, where its remaining liquid solidifies.

Assume a column is sealed to salt at both ends, diffusion occurs only in liquid, and its imposed temperature increases strictly upward. A finite-width equilibrium mushy layer cannot be stationary: its solute flux is nonzero unless its liquid fraction vanishes. The final state is pure ice below a height $h_f$ and uniform-composition brine above it. Their interface obeys $C_f=-(T_E+G{h}_f)/m$. The displayed mass conservation equation fixes $h_f$ from the initial salt content. For a diffusive steadily pulled equilibrium mush, let $H=-T_E/G$, $c=m{C}_0/G$, $d=D/V$ and $h_0=H-c-d$. The initial salt content is $N=C_0[h_0+d\log (1+h_0/c)]$. Thus the liquid-cap depth ${\ell }_f=h_0-h_f$ obeys ${\ell }_f({\ell }_f+c+d)=c[h_0+d\log (1+h_0/c)]$, which has one positive root. This conclusion assumes no subsequent exchange with an underlying eutectic reservoir.

The liquid fraction is the volume fraction occupied by liquid in a solid–liquid mixture. For a saturated rigid mushy layer, it is the porosity available to Darcy flux.

At a sharp mush–liquid edge with linear liquidus $T_L=T_m-mC$, marginal equilibrium makes the liquid temperature gradient tangent to the local liquidus: $T_z=-m{C}_z$. Together with $T=T_L$ at the edge and $T\ge T_L$ in the exterior liquid, it supplies the usual absence-of-constitutional supercooling closure. When solute diffusion is neglected inside the mushy layer but retained outside it, interfacial solute conservation can require a small solid fraction jump; imposing unit porosity as an additional finite-diffusivity condition generally overdetermines this sharp-edge reduction.

Use $\ell =L/c_p$, $ϵ=\sqrt{D_s/D_l}$ and $\delta =\sqrt{D_l/\kappa }$, with $0<k_D<1$. Equal reference densities and constant material properties are implicit in these diffusion balances. Put $a(t)=2\lambda \sqrt{D_{l}t}$ and $T_a=-m{C}_a$. The four similarity fields areHere the same letters $T_s,T_l,C_s$ on the right denote their far-field values. Substitution into the heat and solute diffusion equations verifies the fields; their interface values and far-field limits verify every boundary condition. The concentration has an equilibrium jump from $k_D{C}_a$ to $C_a$, whereas temperature is continuous.

The interfacial conservation laws areFor example,Since $\dot{a}=\lambda \sqrt{D_l/t}$, dividing the two jump conditions by $\dot{a}$ gives, for nonzero $\lambda$,where $F(z)=\sqrt{\pi }z{e}^{z^2}\mathrm{erfc}z$. At $\lambda =0$, use the undivided flux conditions or the continuous limit, rather than divide by zero. The liquidus relation and the segregation coefficient also give the solidus $T_S(C)=-mC/k_D$.

For the thermal expansion, write $z=\delta \lambda$ and useKeeping the constant terms in these reciprocals is essential for the full first-order answer. The result isThe remainder has this form for bounded $\lambda$; a looser $O({\delta }^2)$ statement is also valid in the distinguished scaling $\ell =O({\delta }^{-1})$. Dropping the temperature-difference term while claiming first-order accuracy for fixed $\ell$ would be incorrect.

Solving the solute balance algebraically givesFor fixed positive $\lambda$, $F(-\lambda /ϵ)\to -\infty$, and $C_a\to 0$, with a positive exponentially small correction. For fixed negative $\lambda$, $F(-\lambda /ϵ)\to 1$, andIt falls from $C_s$ as $\lambda \to -\infty$ to zero as $\lambda \to 0^{-}$. The outer positive branch is zero. For finite $ϵ$ the transition is rounded over $|\lambda |=O(ϵ)$; the exact zero-speed limit is $C_a(0)=ϵC_s/(1+k_Dϵ)$, not an actual concentration discontinuity.

To count speeds, take the leading distinguished thermal balance, set $\Sigma =T_l+T_s$ and $K=\sqrt{\pi }\delta \ell$, and combine $T_a=-m{C}_a$ with the thermal result:This is the intended leading approximation when $\delta \ell =O(m{C}_s)$ and $T_l-T_s=O(1)$. The full first-order expression instead replaces $K$ by $\sqrt{\pi }\delta [\ell +2(T_l-T_s)/\pi ]$. Thus the printed threshold is asymptotic, not an exact criterion for finite $\delta ,ϵ$.

For $r=-\lambda >0$, put $H(r)=\sqrt{\pi }r{e}^{r^2}\mathrm{erfc}(-r)$. The small-solid-diffusivity melting concentration is $C_{s}H/(1+H)$, soThe second term decreases strictly from $2m{C}_s\sqrt{\pi }$ to zero. One can check the strict decrease without guessing its shape. Set $A(r)=\sqrt{\pi }{e}^{r^2}\mathrm{erfc}(-r)$, so $A^{\prime }=2rA+2$, $A\ge \sqrt{\pi }+2r$, and $H=rA$. Direct differentiation givesThe bracket increases with $A$ on $r\ge 0$. Substituting its lower bound yields a polynomial with constant coefficient $\pi -2>0$ and all other coefficients positive. Hence $[H^{\prime }/(1+H{)}^2{]}^{\prime }<0$, as claimed.

If $\delta \ell >2m{C}_s$, then $K>2m{C}_s\sqrt{\pi }$ and $\Sigma (r)$ increases from zero to infinity. The freezing branch has $\Sigma =-K\lambda <0$ and is also monotone. Together these give one speed for each imposed sum within the approximation. If $0<\delta \ell <2m{C}_s$, the melting branch initially decreases, has exactly one minimum and then increases to infinity. For sums between that minimum and zero there are two melting speeds plus one freezing speed: three solutions. At the lower fold there are two distinct speeds, one being a double solution. At zero the limiting outer branches join; finite solid diffusivity rounds the join into a nearby fold. Away from the fold values the other sums have one solution. At the threshold itself the initial slope vanishes; it is the transition between these cases.

For $\Sigma >0$, the solution is on the melting branch $\lambda <0$. Its thermal layers have scale $\sqrt{\kappa t}=\sqrt{D_{l}t}/\delta$, much larger than the liquid-solute scale $\sqrt{D_{l}t}$. Because the front is retreating, the solid solute layer is thinner still:This follows by expanding the large-positive-argument complementary error function ratio on the solid side: near $a$, its decay length is $D_s/|\dot{a}|$. Using simply $\sqrt{D_{s}t}$ would miss the retreating-front compression of that layer.

Across this narrow layer, solid concentration drops from almost $C_s$ to $k_D{C}_a$, while temperature stays almost $T_a$. At the interface $T_a=T_S(k_D{C}_a)$. Moving slightly into the solid increases its concentration toward $C_s$, lowering its solidus to $-m{C}_s/k_D$, whereas the temperature has barely changed. Since $C_a<C_s$ and $k_D<1$, $T_a=-m{C}_a>-m{C}_s/k_D$. Thus an adjacent region can satisfy $T>T_S(C)$: constitutional superheating. Its existence is especially clear when far-field solid is initially stable but heat diffuses across the narrow composition layer.

The phase-diagram path goes from $(C_s,T_s)$ through a broad thermal rise, then nearly horizontally toward $(k_D{C}_a,T_a)$ across the thin solid diffusion layer. At the interface it traverses the isothermal equilibrium tie line to $(C_a,T_a)$. On the liquid side concentration falls to zero over the liquid-solute layer, while the remaining thermal adjustment to $T_l$ takes place over the much wider heat layer. The path of the depleted solid approaches the solidus from its superheated side. That region should melt at additional sites or develop a two-phase mushy layer; the original single-boundary similarity construction is then a mathematical branch rather than a physically stable description of the entire sample.

Take $z$ upward from the unperturbed interface in the apparatus frame, with liquid at $z>0$ and pulling velocity $-V{\mathbf{e}}_z$. The primary ice contains no salt. We use the dilute, equal-density approximation and neglect fluid motion relative to the mould. The initial salt inventory per unit horizontal area, with a common density factor suppressed, is $N=h{C}_0$. Its conservation is essential: the far liquid has zero concentration, so an imposed nonzero far-field composition would describe a different experiment.

The steady advection-diffusion equation is $-V{C}^{\prime }=D{C}^{\prime \prime }$. Decay into the pure water and interfacial salt rejection giveSince ${\int }_0^{\infty }C(z)dz=D{C}_I/V=N$, the finite-inventory directional salt rejection solution isThe decay length is $D/V$. Local liquidus equilibrium givesThus the thermal apparatus selects the interface position at this isotherm. The stated speed inequality is precisely $C_I<C_E$, so the interface is warmer than the eutectic temperature, $T_I>T_E=-m{C}_E$. These are steady values after the initial finite saline layer has rearranged; the final exponential distribution is not confined to the original depth $h$.

For $Vh{C}_0/D>C_E$, a steady pure-ice front would require an interfacial composition beyond the eutectic composition. That branch is unavailable. During the transient, rejected brine concentrates until it reaches $C_E$; secondary salt-bearing crystals can then form and the residual brine freezes as an eutectic system. Salt is incorporated into the product rather than remaining indefinitely in a super-eutectic exponential layer. The detailed pattern can include brine pockets and a nonplanar freezing region; it depends on the thermal history and the morphological stability. The speed inequality alone does not decide that latter stability.

For the stability calculation use an interface displacement $\eta =\hat{\eta }{e}^{ikx+st}$ and a liquid perturbation $c=\hat{c}{e}^{ikx+st-qz}$. Here $k>0$ is the wavenumber, $s$ is the temporal rate and $\Re q>0$ gives spatial decay. The frozen thermal field is $T_I+Gz$. Define $\Gamma ={\gamma }_{sl}{T}_m/(\rho L)>0$, where ${\gamma }_{sl}$ is solid-liquid surface energy, $T_m$ is absolute pure-solvent melting temperature, $\rho$ is mass density and $L$ is specific latent heat. For a solid protrusion into liquid the curvature is $\mathcal{K}=-{\eta }_{xx}=k^2\eta$, and the Gibbs-Thomson effect gives local equilibrium $T=-mC-\Gamma \mathcal{K}$. We assume isotropic surface energy and instantaneous interface kinetics, so no kinetic undercooling term is present.

Linearizing the equilibrium condition at the displaced interface givesThe perturbation advection-diffusion equation givesThe local solute balance at a moving graph, to first order, isIts perturbation is $(V-Dq)\hat{c}+s{C}_I\hat{\eta }=0$, because the terms $D\eta C^{\prime \prime }(0)$ and $V\eta C^{\prime }(0)$ cancel. Consequently the full frozen-temperature solutal interface dispersion isNo quasistatic approximation to solute diffusion was made here. For a growing mode the unique admissible liquid root isWriting $b=(G+\Gamma k^2)/(m{C}_I)$ and eliminating $s$ gives $(q-V/D)(q-V/D+b)=k^2$. The root continuously connected to the growing interface branch is thereforeIn particular, unstable wavelengths exist exactly when $mV{C}_I/D>G$, and their band isThis is the constitutional supercooling criterion: immediately above the plane, the local liquidus rises at rate $-m{C}^{\prime }(0)=mV{C}_I/D$, so it overtakes the imposed thermal gradient when that rate exceeds $G$. The Gibbs-Thomson effect cuts off short waves. At long wavelength $s\sim D(V/D-b)k^2/b$ for positive $G$. The exactly uniform inventory-changing displacement is not an instability of a fixed-inventory experiment. On stable branches the full diffusion problem also admits relaxation transients; the root selection above is sufficient to identify every exponentially growing interface mode.

If one additionally neglects the perturbation time derivative in the liquid diffusion equation, the alternative reduced rate isIt has the same onset and cutoff, but need not have the same rate away from the slow-growth regime. The frozen-temperature approximation by itself does not justify this extra reduction.

After onset, protrusions reach more weakly saline liquid and advance faster, while grooves accumulate rejected brine. The planar interface develops cells and, as growth becomes sufficiently strong, crystal dendrites with concentrated liquid between them. Nonlinear competition changes the selected spacing; linear rates alone cannot fix the final geometry. The interdendritic liquid forms a mushy layer, concentrates toward the eutectic composition, and ultimately freezes into a salt-bearing eutectic product when it reaches $T_E$. This replaces the pure planar salt-rejection picture with trapping and freezing of brine between crystals.

Work in the frame of the steadily advancing mush, so the rigid primary-crystal skeleton moves with $-V{\mathbf{e}}_z$. Let $\varphi$ be solid fraction, $C$ the pore-liquid concentration, $\mathbf{q}=(u,w)$ the Darcy flux relative to that skeleton, and $K$ the common thermal conductivity. Set $\kappa =K/(\rho c_p)$, the thermal diffusivity. The actual pore velocity is $\mathbf{q}/(1-\varphi )-V{\mathbf{e}}_z$. Primary crystals are taken to exclude solute; the residual pore liquid freezes into a two-solid eutectic product at the bottom. These are the standard binary mushy layer assumptions implicit in using a single liquidus and an eutectic floor.

It is convenient to use hydraulic pressure $P=p+\rho gz$, absorbing the constant-density gravitational term. With equal phase properties, no solute diffusion, negligible fluid inertia and local liquidus equilibrium, a complete mush model isThe last equation follows from conservation of energy with mixture specific enthalpy $c_{p}T-L\varphi$. Sensible heat is transported by the total volume flux $\mathbf{q}-V{\mathbf{e}}_z$; the primary solid carries the latent-enthalpy flux $VL\varphi {\mathbf{e}}_z$. In particular the sign of the latent-heat term is fixed by the downward-moving skeleton.

At the liquid roof, imposeHere $P_0=p_0+\rho gh$. The gradient condition matches heat flux into the imposed liquid field, since there is no phase-fraction jump at this roof and the phase conductivities are equal. The pore velocity there equals the relative Darcy velocity. At the floor,There is no liquid penetration into fully solid material. The unspecified solid fraction at the floor is determined by the solute balance, not set to one: remaining eutectic liquid freezes as it crosses the floor. If $C_s$ denotes the bulk concentration of the final solid product, the floor solute balance and Stefan condition areMaintaining the floor at $T_E$ permits the solid-side bath to remove this final latent heat. Equivalently its required gradient is $T_z(0^{-})=T_z(0^{+})+\rho LV[1-\varphi (0)]/K$. No additional imposed solid-side gradient is specified. Laterally we take the quadratic-pressure strain solution, without sidewalls; finite sidewalls would require additional flow conditions and generally break this horizontal invariance.

For uniform permeability of a porous medium, incompressible flow gives ${\nabla }^{2}P=0$. The roof pressure and zero normal flow at the floor are satisfied byHorizontal outflow is balanced by downward inflow. The parameter $\alpha$ is an inverse time; $E^2$ in the printed pressure condition already has dimensions of pressure divided by length squared, so no extra density factor belongs in this definition.

With $T=T(z)$, $C=C(z)$ and $\varphi =\varphi (z)$, the $x$-dependence disappears from both scalar balances. Although $u$ is nonzero, its horizontal divergence cancels $w_z$ in $\nabla \cdot (\mathbf{q}C)$. The exact strain-driven steady mush equations becomeTogether with $T(0)=T_E$, $T(h)=T_0$, $T^{\prime }(h)=G$ and $\varphi (h)=0$, they determine the scalar fields and the unknown thickness. This reduction proves that the equations admit horizontally invariant scalar solutions despite the imposed straining flow. An equivalent useful form, obtained by differentiating the solute equation, isThis exact form also shows why one must not silently discard latent heat in an arbitrary strong-flow calculation.

Define $\Delta T=T_0-T_E>0$, $R=\alpha h/V$ and $S=L/(c_p\Delta T)$. For the stated regular ordering $R\gg S\gg 1$, strain transport dominates skeleton transport and the bounded phase-change enthalpy contribution. More precisely, at fixed finite thermal strain number $\alpha h^2/\kappa$, the dimensionless energy equation isso its skeleton and integrated latent-heat corrections are small in $1/R$ and $S/R$ respectively. The leading strain-dominated mush temperature profile obeys $\kappa T^{\prime \prime }+\alpha z{T}^{\prime }=0$. Put $\delta =\sqrt{2\kappa /\alpha }$ and $H=h/\delta$. Integrating first for $T^{\prime }$ and then for $T$ givesThe error function profile follows from integrating a Gaussian, with $\mathrm{erf}x=(2/\sqrt{\pi }){\int }_0^x{e}^{-s^2}ds$. Matching the roof gradient gives the requested transcendental thickness relationFor prescribed $G>0$ and $\Delta T>0$ the right side decreases continuously from infinity to zero as $H$ increases, so it fixes one positive leading-order thickness. The resulting $h$ must still satisfy the assumed ordering and the phase-fraction constraints.

To find the product composition, integrate the exact solute balance from the roof to an arbitrary depth. With $f=1-\varphi$,At $z=0$, the exact floor composition is thereforeThe minus sign has a direct physical meaning: the horizontal outflow removes more salt than arrives at the same depth from a uniform-$C_0$ liquid, leaving a depleted solid product. Using the leading error function profile andyieldsThe full solid fraction follows by substituting this same profile into the preceding integral formula for $f(z)$. It has $\varphi (h)=0$ and $1-\varphi (0)=C_s/C_E$. A physical branch requires $0\le C_s\le C_E$ and $0\le \varphi \le 1$. In this monotone profile, $C_s\ge 0$ ensures a nonnegative liquid fraction everywhere; it is not permissible to interpret a negative formal result as a material composition.

The error-function reduction is a regular leading approximation, not a uniform statement for every possible scaling of the thermal strain number. If $\alpha h^2/\kappa$ becomes singularly large along the ordering limit, a thin basal layer can amplify the discarded terms. The printed two inequalities alone do not specify that extra scaling. In such a regime the displayed exact scalar equations and boundary conditions, rather than an unqualified zero-latent-heat profile, provide the appropriate thickness and composition calculation.

In seawater, freezing produces nearly salt-free solid and leaves salt in the interstitial liquid. The growing cover is therefore a mushy layer, with sea-ice crystal fabric determined by both crystal growth and the mechanical history. Crystallographic c-axes are perpendicular to the plates' basal planes; grain elongation and c-axis orientation must not be confused.

For first-year sea ice growing by congelation, heat loss first brings the surface water to its salinity-dependent freezing point. Small frazil ice crystals may form and collect into a thin continuous sheet, or a quiet surface can freeze directly; the resulting nilas provides the initial cover. Once the water surface is protected from strong agitation, heat transported by heat conduction upward through the sheet permits freezing onto its underside. Competitive growth favors grains with fast basal-plane growth directed downward, so grains with approximately horizontal c-axes crowd out less favorably oriented grains. The mature congelation ice consists predominantly of long vertical columns beneath a finer initial surface layer. Within and between these grains, the advancing cellular interface traps brine-bearing spaces. The growth sequence and crystallographic selection explain why the resulting texture is columnar rather than a random aggregate of floating crystals.

For pancake ice, wind and surface gravity waves prevent an uninterrupted ice sheet. Supercooled surface water produces suspended frazil ice; crystals accumulate into grease ice and then small coherent cakes. Wave-induced convergence, collisions and repeated accumulation of frazil increase cake size and thickness. Collisions and overwash deposit slush around the margins, building the characteristic raised rims. The crystal aggregate is predominantly granular, with much less systematic orientation than in congelation ice. A calmer interval can add a columnar basal layer, so a pancake need not be purely granular throughout. Observed pancake structures include frazil layers, basal congelation growth and rafted cakes.

In Antarctic consolidated pancake ice, the active edge initially follows this frazil-pancake route. As concentration rises and waves are attenuated, pancakes jam, raft and become frozen together, with interstitial frazil filling the gaps. The consolidated sheet retains overlapping cake outlines, stacked granular layers and an irregular underside, rather than acquiring the simple fabric of a single quiet-water-grown sheet. Later basal freezing can add congelation ice, but the granular fraction often remains large. Snow loading or overwash can flood the top; freezing this slush adds granular snow ice, a separate contribution from seawater-derived frazil. These mechanisms produce the heterogeneous sea-ice crystal fabric documented in cores from the Antarctic marginal ice zone. Quiet basal growth produces columnar ice; wave-driven aggregation and consolidation produce predominantly granular, layered ice.

The main efficient net desalination mechanisms require transport through the ice boundary. In gravity drainage of sea-ice brine, cold saline liquid is denser than the water below, so a connected permeable pore network permits downward flow and compensating entry of less saline seawater. Dense brine can collect into channels and convective drainage pathways. For a layer of depth $H$, a Darcy law estimate of drainage speed is $u_D\sim K\Delta \rho g/\mu$, giving a drainage time $H/u_D$ and showing directly why permeability of a porous medium and buoyancy matter. Warm, brine-rich ice drains much more readily than cold ice whose pores have become poorly connected. This is important during growth and can intensify as warming reconnects channels; an isolated sealed brine pocket cannot drain merely because its liquid is dense. The connectivity constraint is supported by sea-ice permeability measurements.

In meltwater flushing of sea ice, water supplied by surface snow melt or melt ponds has a hydraulic head above sea level and percolates down connected pores and cracks, displacing brine. It can remove salt rapidly in warm summer ice. Its limits are the supply and head of meltwater and pore connectivity; lateral runoff may reduce the amount available for vertical flushing. Gravity-driven drainage can act without surface melting, whereas flushing cannot.

Brine expulsion occurs when additional freezing and solid expansion pressurize an inclusion and force liquid into nearby passages. This is often internal redistribution: expelling brine from one inclusion does not desalinate the entire floe unless that liquid reaches the ocean. Migration of brine inclusions involves melting on a warmer wall and refreezing on the colder wall; molecular salt diffusion through liquid connections is also possible. These slow processes are far less effective at exporting bulk salt than connected advective drainage. Diffusion through the ice lattice itself is extremely slow. Likewise, excluding salt from each growing crystal is not by itself proof of salt loss from a whole ice-brine mixture. Desalination experiments and transport calculations distinguish net salt export from internal redistribution. Gravity drainage dominates desalination of permeable growing ice; meltwater flushing can dominate summer desalination; expulsion, inclusion migration and diffusion are comparatively ineffective without a boundary-crossing flow.

Take $G,V,D,m,C_0>0$ and $T_E<0$, with the liquidus referenced to the pure-water melting point. The solid-liquid segregation coefficient $k_D=0$ means that newly formed ice contains no salt. A steady sharp front producing only pure ice cannot remove the incoming solute flux. More explicitly, in the liquid the steady solute flux is $J_C=-VC-D{C}_z=-V{C}_0$, whereas in pure solid it is zero. Interfacial salt conservation would require $V{C}_0=0$, a contradiction. An equilibrium mushy layer avoids this by carrying concentrated pore liquid down to the eutectic region, where a salt-bearing solid assemblage can form.

Use equal densities and common downward translation velocity $-V$. Write $q=1-\varphi$ for the liquid fraction; $\varphi$ here is the solid fraction. Local liquidus equilibrium in the mush requiresAbove a stationary roof $h_0$, the liquid advection-diffusion equation is $-V{C}_z=D{C}_{zz}$, with solutionThe marginal equilibrium at a mush-liquid boundary condition is $-m{C}_l^{\prime }(h_0)=G$, hence $C_i=C_0+GD/(mV)$. Matching the temperature and liquidus at the roof givesA positive equilibrium mush requires $(-m{C}_0-T_E)/G>D/V$; otherwise this positive-thickness construction does not apply.

Including solute diffusion in the liquid pore space, the steady solute flux in the mush isThus $q[C+(D/V)C_z]=C_0$. Inserting $C_z=-G/m$ and the value of $h_0$ yieldsThis is the diffusive steadily pulled equilibrium mush. Omitting the diffusive contribution would give a different solid-fraction distribution.

The imposed temperature and local liquidus coincide throughout the mush. In the liquid put $y=V(z-h_0)/D\ge 0$. Their difference iswith equality and matching first derivatives at the roof. The inequality follows because the derivative of $y-1+e^{-y}$ is $1-e^{-y}\ge 0$. Thus the liquid is not constitutionally supercooled. In the phase diagram, the pore-liquid trajectory follows the liquidus from the eutectic composition $C_E=-T_E/m$ to $(C_i,T_i)$, then curves into the liquid region and approaches $C=C_0$ as height increases. Pure-ice crystals have composition zero; the trajectory denotes the liquid composition, not the bulk mush composition.

After pulling stops, the imposed temperature keeps the pore-liquid composition fixed wherever the material remains mush. Solute conservation becomesBecause $C_{zz}=0$, this reduces toThe method of characteristics uses $dz/dt=-D/(z+T_E/G)$ and $d\varphi /dt=0$. HenceThese are the solute-diffusion characteristics in a fixed-temperature mush; their validity is local to the coexisting solid–liquid region.

For an explicit initial-data solution, define $H=-T_E/G$, $c=m{C}_0/G$ and $d=D/V$, so $h_0=H-c-d$. A characteristic through $(z,t)$ has initial footWhile that foot lies in the original mush and no phase boundary has removed the point,This recovers the initial distribution and shows that interior ice fraction increases as salt diffuses upward from colder, more concentrated pore liquid.

If one simply transports the initial zero-solid-fraction level, $f(\eta (h_0,0))=0$ implies the printed formal lawOn the physical concentration branch $h<H$, this is $h(t)=H-\sqrt{(H-h_0{)}^2-2Dt}$, so $\dot{h}=D/(H-h)>0$. It starts moving above $h_0$ immediately.

This formal continuation cannot be the physical mush–liquid roof of a column sealed to salt by the membrane. The no-flux constraint in a fixed-temperature mush is $J_C=-Dq{C}_z=0$ at the membrane. At a smooth zero-fraction equilibrium roof, however, $q=1$ and $C_z=-G/m$, giving $J_C=DG/m\ne 0$. Moreover the formal roof has left the membrane-bounded domain. The printed interface law is incompatible with a salt-impermeable membrane when interpreted as the physical roof. Its derivation above is a characteristic extrapolation; it does not enforce the stated boundary condition. If “impermeable” were intended to block bulk flow but allow salt diffusion, a different boundary condition would need to be specified.

The closed-column physical sketch is consequently different. Salt transported upward accumulates below the membrane and produces a liquid brine cap; the cap composition need not lie on the liquidus at every height. Its lower boundary can have a finite jump in solid fraction, so it need not follow the initial zero-fraction characteristic. Near the cold solid base, zero salt supply from below requires the liquid fraction to vanish, and a pure-ice region develops. In the surviving mush, the characteristic formula increases $\varphi$ relative to its initial value. An intermediate profile therefore has pure ice at the bottom, a remaining mush with elevated solid fraction, and liquid at the top. The plotted intermediate roof is schematic, not claimed to obey the inconsistent formal law.

One can characterize the final state quantitatively if the solid base also admits no salt diffusion, as in the usual pure-solid/no-solid-diffusion closure. In a stationary column, solute flux is constant and is zero at the sealed roof. A finite-width equilibrium mush would have $J_C=DGq/m$, forcing $q=0$ everywhere in that region; it therefore cannot remain mush. The final state is pure ice for $0<z<h_f$ and a uniform-composition liquid cap for $h_f<z<h_0$. At their boundary,The cap is liquid because $T(z)>-m{C}_f=T(h_f)$ above its base. The fixed external thermal field can supply or remove the heat involved; this is a solute-budget determination rather than an isolated enthalpy budget.

The final salt budget of a sealed mush column determines $h_f$. Initiallysince $(1-{\varphi }_0)C_m=C_0[1+d/(c+h_0-z)]$. Conservation requires $C_f(h_0-h_f)=N$. Writing the final cap depth ${\ell }_f=h_0-h_f$, the equation and its unique positive root areThe initial salt integral is less than $C_E{h}_0$, so $0<{\ell }_f<h_0$ and $0<h_f<h_0$; the resulting $C_f<C_E$ remains on the ice–brine branch above the eutectic. If the basal eutectic region is instead allowed to exchange solute with the column, this salt budget must include that flux and the numerical final height is not fixed by the printed data alone. No universal physical $h(t)$ follows from the interior characteristic equation without the cap and basal boundary conditions.

Take $z$ positive downwards from the roof. Write $T_0=T_m-m{C}_0$, $C_B=(T_m-T_B)/m$, $\Delta T=T_0-T_B=m(C_B-C_0)>0$ and $\Delta C=C_B-C_0$. The cooled magma follows the decreasing liquidus from $(C_0,T_0)$ towards $(C_B,T_B)$; the initially superheated liquid starts at $(C_0,T_{\infty })$ above that liquidus. The solid has zero solute concentration. At each temperature in the mushy layer, a horizontal tie line joins the pure solid to the residual liquid on the liquidus. The lever rule gives the liquid fraction from the bulk concentration. Since $T_B>T_E$, this path stops before the eutectic temperature and does not cross the eutectic composition.

The porosity $\varphi$ denotes the liquid fraction, so $1-\varphi$ is the solid fraction. Equal phase mass densities remove volume-change flow; the stable density stratification suppresses convection. In the stagnant mushy-layer model, define bulk specific enthalpy and bulk solute concentration byThe arbitrary reference of specific enthalpy has been omitted. The two conservation laws in $0<z<h(t)$ areHere $\kappa =k_T/(\rho c_p)$ is the thermal diffusivity, with $k_T$ the thermal conductivity. The positive $L{\varphi }_t$ term records increased specific enthalpy on melting; during freezing it supplies latent heat. In the liquid $z>h(t)$, the heat equation and solute diffusion equation arewhere $D$ is the solutal diffusivity. Neglecting solute diffusion inside the mushy layer does not permit dropping $D{C}_{zz}$ in the liquid at this stage.

Use $T(0,t)=T_B$, liquid $T\to T_{\infty }$ and $C\to C_0$ as $z\to \infty$, and initially $h=0$, $T=T_{\infty }$, $C=C_0$ for $z>0$. These are the half-space boundary conditions appropriate before cooling reaches the bottom of the chamber. For a closed chamber of finite depth $H$, replace the far-field conditions by the specified bottom conditions, for example $T_z=C_z=0$ at an insulated impermeable floor; the half-space similarity solution then ceases to apply when the thermal penetration depth approaches $H$.

At $z=h(t)$, let $C_i$ and $T_i=T_m-m{C}_i$ be the common liquid concentration and temperature, and let ${\varphi }_i$ be the limiting porosity on the mush side. Continuity of $T,C$, the Rankine-Hugoniot condition for solute, and the interfacial conservation of energy giveThe first condition has the sign of solute rejection into the liquid. For the sharp-edge stagnant mushy-layer model with nonzero liquid solutal diffusivity, close the free-boundary problem by marginal equilibrium at a mush-liquid boundary:This prevents constitutional supercooling immediately ahead of the mushy layer. A small solid fraction jump at its edge is allowed: imposing ${\varphi }_i=1$ as well as these finite-$D$ conditions would generally overdetermine the reduced model. At the roof, $C=C_B$ follows from the liquidus, and no additional solute-flux boundary condition is needed in the diffusion-free mush.