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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.

In the heat-exchanger frame the crystal velocity is ${\mathbf{v}}_s=-V{\mathbf{e}}_z$, while the liquid velocity is ${\mathbf{v}}_l={\mathbf{v}}_s+\mathbf{u}/(1-\phi )$. The steady vertical mass flux is constant and equals $-{\rho }_{s}V$ in the fully solid eutectic material. HenceThis gives the density-change flow in a pulled mushy layerThe laboratory mixture volume flux is $-V[r+(1-r)\phi ]{\mathbf{e}}_z$. In the pure liquid $\mathbf{u}=(1-r)V{\mathbf{e}}_z$, while in the fully solid region $\mathbf{u}=0$ and the material moves at $-V{\mathbf{e}}_z$. The liquid velocity formula applies wherever liquid exists; at solid fraction one there is no liquid phase to assign a velocity.

With pure solid crystals, negligible solute diffusion, and the density-change flow in a pulled mushy layer, steady solute balance gives $(1-\phi )C=C_0$. If $\theta =(T-T_0)/\Delta T$, $\mathcal{C}=m{C}_0/\Delta T$, and the liquidus is $T-T_0=-m(C-C_0)$, then