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A mushy layer contains solid crystals and liquid in local phase equilibrium, rather than a single sharp phase boundary. Its solid fraction changes as heat and solute are transported. A concentration-dependent liquidus couples temperature to liquid composition; a connected crystal framework makes the residual liquid behave as a porous-media flow.

Let a representative fixed volume have mixture mass density $\bar{\rho }={\rho }_s\phi +{\rho }_l(1-\phi )$. Since the crystal framework is rigid and stationary, the liquid mass flux is ${\rho }_l\mathbf{u}$, where $\mathbf{u}$ is Darcy velocity. Mass conservation givesThe constant phase densities therefore imply the phase-change volume source in a rigid mushFor ice less dense than brine, increasing solid fraction creates an expansion flow; when the phase densities agree, this source vanishes.

Use $\zeta =Vz/\kappa$, $H=Vh/\kappa$, and $S=L/(c_p\Delta T)$, the latent-to-sensible heat ratio. Transforming the heat equation to the fixed apparatus frame and substituting Darcy flux gives the exact steady mush equationFor $\mathcal{C}\gg 1$, the solid fraction is $\phi =-\theta /\mathcal{C}+O({\mathcal{C}}^{-2})$. Since $S/\mathcal{C}=O(1)$, the latent term remains leading order, whereas the correction $(1-r)\phi {\theta }_{\zeta }$ is small. Define $\Omega =1+S/(r\mathcal{C})$. The large-concentration thermal profile of a pulled mush then obeysWith $\theta (H)=0$ and $\theta \to {\theta }_{\infty }$ in the liquid,The thermal conductivity agrees on both sides and the solid fraction vanishes at the mush–liquid interface, so there is no jump in latent production there: continuity of heat flux gives ${\theta }_m^{\prime }(H)={\theta }_l^{\prime }(H)=r{\theta }_{\infty }$. ThusThe boundary value ${\theta }_m(0)=-1$ determines $H$, as calculated next. The crystal fraction is obtained from the previous part's formula, with the temperature field understood to this leading asymptotic accuracy.

For a steady field in the apparatus frame, the time derivative in the translating crystal frame is $-V{\partial }_z$. The given solute equation therefore becomesSubstituting the density-change flow in a pulled mushy layer reduces this to $(1-\phi )C_z=C{\phi }_z$. Thus the solute conservation in a steadily pulled mush integrates to $(1-\phi )C=C_0$, using the zero solid fraction at the top boundary.

Set $\mathcal{C}=m{C}_0/\Delta T$. The liquidus gives $C=C_0(1-\theta /\mathcal{C})$, soIn particular, $\phi =0$ at $\theta =0$, and immediately above the eutectic temperature, where $\theta =-1$, the mush has $\phi =1/(\mathcal{C}+1)$. It becomes fully solid across the eutectic front below.