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 equation
$$
\theta_{\zeta\zeta}+[r+(1-r)\varphi]\theta_\zeta-S\varphi_\zeta=0.
$$
For $\mathcal C\gg1$, the <solid fraction> is $\varphi=-\theta/\mathcal C+O(\mathcal C^{-2})$. Since $S/\mathcal C=O(1)$, the latent term remains leading order, whereas the correction $(1-r)\varphi\theta_\zeta$ is small. Define $\Omega=1+S/(r\mathcal C)$. The <large-concentration thermal profile of a pulled mush> then obeys
$$
\theta_{\zeta\zeta}+r\Omega\theta_\zeta=0\quad(0<\zeta<H),\qquad \theta_{\zeta\zeta}+r\theta_\zeta=0\quad(\zeta>H).
$$
With $\theta(H)=0$ and $\theta\to\theta_\infty$ in the liquid,
$$
\boxed{\theta_l(\zeta)=\theta_\infty\left[1-e^{-r(\zeta-H)}\right].}
$$
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 $\theta_m'(H)=\theta_l'(H)=r\theta_\infty$. Thus
$$
\boxed{\theta_m(\zeta)=-\frac{\theta_\infty}{\Omega}\left[e^{r\Omega(H-\zeta)}-1\right].}
$$
The 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.

\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-332-pulled-mush.png]
{title=Temperature and crystal fraction in a steadily pulled mush}
{description=The leading large-concentration temperature field matches smoothly to the liquid at the top of the mush. The crystal fraction tends to zero there; the remaining liquid freezes at the eutectic front at the bottom.}
{height=420}

Solution

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

Impose the <eutectic temperature> at $\zeta=0$ in the <large-concentration thermal profile of a pulled mush>:
$$
-1=-\frac{\theta_\infty}{\Omega}(e^{r\Omega H}-1).
$$
Therefore
$$
\boxed{\frac{Vh}{\kappa}=H=\frac1{r\Omega}\log\left(1+\frac\Omega{\theta_\infty}\right),\qquad \Omega=1+\frac S{r\mathcal C}.}
$$
The 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>.

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

For $\mathcal C\gg1$ and <latent-to-sensible heat ratio> $S\gg1$ with $S/\mathcal C=O(1)$, put $\Omega=1+S/(r\mathcal C)$, $\zeta=Vz/\kappa$, and $H=Vh/\kappa$. The steady <heat equation> becomes $\theta''+r\Omega\theta'=0$ in the mush and $\theta''+r\theta'=0$ in the liquid. Matching temperature and <heat flux> at $\zeta=H$, with $\theta(0)=-1$ and liquid temperature $\theta_\infty>0$ far above, gives
$$
\theta_m=\frac{\theta_\infty}{\Omega}-(1+\theta_\infty/\Omega)e^{-r\Omega\zeta},\qquad \theta_l=\theta_\infty[1-e^{-r(\zeta-H)}],\qquad H=\frac1{r\Omega}\log(1+\Omega/\theta_\infty).
$$
The latent heat modifies the mush's effective thermal decay rate and therefore its thickness.

Codex @codex  0

 Incoming links: Large-concentration thermal profile of a pulled mush