Past exam of the topics in fluid mechanics course of the University of Oxford / 2010 / 3

Write $\Delta T=T_0-T_B$, $\Delta C=C_0-C_B$ and assume these scales are nonzero. The two reference liquidus relations imply $\Delta T=-m\Delta C$. For the similarity variable $\eta =z/(2\sqrt{\kappa t})$ and front $h=2\lambda \sqrt{\kappa t}$,Thus a temperature profile $T=T_{\ast }+\Delta T\theta (\eta )$ has $T_t=-\Delta T\eta {\theta }^{\prime }/(2t)$ and $\kappa T_{zz}=\Delta T{\theta }^{\prime \prime }/(4t)$. Each heat equation becomes ${\theta }^{\prime \prime }+2\eta {\theta }^{\prime }=0$. The solute diffusion equation becomes ${\mathcal{C}}^{l\prime \prime }+2Le\eta {\mathcal{C}}^{l\prime }=0$, withThis Lewis number compares thermal and solutal diffusivities. Inside the mushy layer, $T^s-T_B=-m(C^s-C_B)$ gives $\boxed{{\mathcal{C}}^s={\theta }^s}$. The wall and far-field conditions become ${\theta }^s(0)=0$ and ${\theta }^l,{\mathcal{C}}^l\to 0$ as $\eta \to \infty$.

At the moving phase boundary, equality of temperatures gives ${\theta }^s={\theta }^l+(T_{\infty }-T_B)/\Delta T$. Equality of concentrations gives ${\mathcal{C}}^s={\mathcal{C}}^l+1$. Therefore the dimensionless parameters with the convention used here areNeither concentration nor temperature offsets should be dropped when enforcing continuity. In particular ${\mathcal{R}}_{\theta }$ can be negative. If $k=\rho c_p\kappa$, then $S=L/[c_p\Delta T]$: this is the inverse Stefan number, although the paper calls $S$ its Stefan number.

To check every factor of two in the flux conditions, divide the thermal Stefan condition by $k\Delta T/(2\sqrt{\kappa t})$, and the solute balance by $D\Delta C/(2\sqrt{\kappa t})$. This givesFinally $T_z^l=-m{C}_z^l$ and $-m\Delta C=\Delta T$ imply $\boxed{{\theta }^{l\prime }={\mathcal{C}}^{l\prime }}$. This is marginal equilibrium at a mush-liquid boundary. Together these calculations produce all the rescaled differential equations and all three interface balances.

Integrating $Y^{\prime \prime }+2a\eta Y^{\prime }=0$ once gives $Y^{\prime }=K{e}^{-a{\eta }^2}$. A second integration expresses the solution using the error function when the value at zero is prescribed, and the complementary error function when decay at infinity is prescribed. Write ${\theta }_{\lambda }={\theta }^s(\lambda )$ and take $\lambda >0$, $Le>0$. The wall condition and the interface value giveFor the exterior liquid, continuity at the interface fixes ${\theta }^l(\lambda )={\theta }_{\lambda }-1+{\mathcal{R}}_{\theta }$ and ${\mathcal{C}}^l(\lambda )={\theta }_{\lambda }-1$. Thus the two decaying similarity solutions areThe factor $\sqrt{Le}$ is required because solute and heat diffuse on different length scales. Both denominators are positive for a finite positive $\lambda$. These formulas use the differential equations, wall/far-field data and continuity; the remaining three flux conditions determine their unknown constants.

Introduce three positive functions of the unknown front parameter:Differentiating the error function profiles gives ${\theta }^{s\prime }=2P{\theta }_{\lambda }$, ${\theta }^{l\prime }=-2Q({\theta }_{\lambda }-1+{\mathcal{R}}_{\theta })$ and ${\mathcal{C}}^{l\prime }=-2R({\theta }_{\lambda }-1)$ at the interface. The three flux conditions are therefore exactlyThese undivided equations are the safest description of the constant-fraction mushy-layer similarity solution, including degenerate parameter choices. In the nondegenerate case the third equation first givesThe second then gives the solid fractionUse the third equation to replace the $Q$ term in the first, and eliminate $\varphi$. The front parameter is a positive root of the single transcendental equationThen substitute back for ${\theta }_{\lambda },\varphi$ and $h(t)=2\lambda \sqrt{\kappa t}$. Since the physical parameters have not been assigned numbers, an implicit scalar equation is the appropriate determination of $\lambda$.

The divided formulas require $R\ne Q$ and ${\theta }_{\lambda }-1+{\mathcal{R}}_C\ne 0$. If either vanishes, return to the boxed three-equation system. For example $Le=1$ gives $R=Q$, so the last equation demands ${\mathcal{R}}_{\theta }=0$; otherwise this proposed similarity solution cannot exist. A physical branch must also have $\lambda >0$ and $0\le \varphi \le 1$. The algebra alone does not establish existence or uniqueness of such a branch for arbitrary parameters.