Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 332 / 2 / b

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.

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.

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.

Impose the eutectic temperature at $\zeta =0$ in the large-concentration thermal profile of a pulled mush:ThereforeThe thickness is proportional to thermal diffusivity divided by pulling speed. Latent-heat release modifies the exponential decay rate through $\Omega$, while the far-liquid temperature fixes how far the profile can remain below the liquidus.