Codex @codex  0

The lever rule finds phase fractions from a horizontal tie line on a phase diagram. For solid and liquid concentrations $C_s,C_l$ and bulk concentration $\overline{C}$, conservation gives $\overline{C}=(1-\varphi )C_s+\varphi C_l$. Thus $\varphi =(\overline{C}-C_s)/(C_l-C_s)$ is the liquid fraction, provided $C_s\le \overline{C}\le C_l$. For pure solid, $\varphi =\overline{C}/C_l$.

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.

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.

To obtain the large-concentration similarity solution, putThe positive superheat specifies the physically intended liquid initial state. On a similarity solution, $C_i,{\varphi }_i$ are constant, so every newly incorporated level leaves the same constant $B={\varphi }_i{C}_i$. Write $C_i=C_0+\Delta C\delta$ and $B=\Delta C(\mathcal{C}+a)$. Then throughout the mushy layerFor $a,\theta =O(1)$, $\mathcal{C}\gg 1$, and $\mathcal{S}/\mathcal{C}=O(1)$, the effective heat capacity is constant to relative error $O({\mathcal{C}}^{-1})$. Thus setIntegration of the similarity solution ordinary differential equation gives the leading analytical profilesThe temperature is $T=T_0+\Delta T\theta$. In the mushy layer, the concentration and porosity are the displayed $C,\varphi$; in the liquid,These retain the liquid solutal boundary layer. For completeness, finite-$r$ values of $\delta ,a,\lambda$ are determined analytically, to the same large-$\mathcal{C}$ order, by settingand solvingThe solute jump gives ${\varphi }_i=1-b/(\mathcal{C}+\delta )$ exactly; replacing $\mathcal{C}$ by $\mathcal{C}+\delta$ on the right of the last equation retains that jump's next-order factor, but does not make the linearized thermal profiles exact at finite $\mathcal{C}$. The branch has $0\le \delta <1$, $0<\varphi \le 1$. All these profiles are leading asymptotic expressions, not exact solutions of the nonlinear finite-$\mathcal{C}$ heat equation.

For $r\to 0$ at fixed positive $\lambda$, the complementary error function asymptotic gives $K(\lambda /\sqrt{r})\sim \sqrt{\pi }\lambda /\sqrt{r}$, hence $\delta =O(r)$, $b=O(r)$, $a\to 0$, and ${\varphi }_i\to 1$. The result simplifies toIn particular, $\varphi (0,t)=C_0/C_B=\mathcal{C}/(\mathcal{C}+1)$ and $\varphi (h^{-},t)=1$. There is no finite interfacial latent heat jump in this limit: the latent heat is released continuously throughout the mushy layer. Matching the two temperature gradients givesThe printed final equation is missing the square on $\Omega$ in the exponential. The corrected exponent follows directly from differentiating $\mathrm{erf}(\Omega \eta )$ and is consistent with ${\Omega }^2=1+\mathcal{S}/\mathcal{C}$. The positive root is unique: the mush-side gradient decreases from infinity to zero with $\lambda$, whereas ${\theta }_{\infty }K(\lambda )$ increases. The zero-superheat limit is singular and is not described by a finite positive $\lambda$ in this two-region construction.

Use the common-density, equal-volume-heat-capacity model implicit in the stated speed law. Let the liquid fraction be $\varphi$ ahead of the melting front and $\varphi +\Delta \varphi$ behind it. Melting adds exactly the liquid needed to fill the newly created pores, so total mass conservation makes Darcy flux continuous. Consequently,Here $u$ denotes Darcy velocity and $v$ the pore-liquid velocity in the stationary rock frame. Treating the pore-space increase as storage without its simultaneous melting source would incorrectly change the Darcy flux.

Relative to the cold unmelted material, the bulk enthalpy increase behind the front is $\rho c_p\Delta T+\rho L\Delta \varphi$: all phases gain sensible heat and the melted ice consumes latent heat. The advective heat-flux difference is $\rho c_{p}U\Delta T$. Applying the Rankine-Hugoniot condition to this energy balance therefore givesso the advection-driven melting front in a porous matrix moves atThe advected latent contribution of the liquid is the same on both sides and cancels; the denominator measures sensible heating plus phase-change energy per unit bulk volume.

A phase fraction is the proportion of a mixture occupying a specified thermodynamic phase. State the basis: by mass, $x_{\beta }=m_{\beta }/{\sum }_{\alpha }{m}_{\alpha }$; by volume, divide the occupied volume by the total volume. These fractions need not agree when the mass densities differ. The latent heat heating curve uses the liquid fraction by mass, whereas liquid fraction in a mushy layer is defined by volume.

The solid fraction is the volume fraction occupied by solid in a solid–liquid mixture. The complementary liquid fraction is $1-\phi$.

A stagnant mushy-layer model has no bulk advection, equal phase mass densities, pure solid, and local equilibrium on the liquidus. With liquid fraction $\varphi$, specific enthalpy $E=c_{p}T+L\varphi$ and bulk solute concentration $B=\varphi C$, neglecting solute diffusion inside the mushy layer gives $T_t+(L/c_p){\varphi }_t=\kappa T_{zz}$ and $(\varphi C{)}_t=0$. The latter fixes composition at each stationary material level; it does not automatically set every level to the initial liquid concentration when solute can diffuse ahead of the growing mush.