Directional solidification

Directional solidification advances a freezing region preferentially along an imposed temperature gradient. Pulling material through fixed heat exchangers can make the mushy layer stationary in the apparatus frame even while solid and liquid pass through it.

A restricted passage in a casting mould that progressively excludes competing crystal grains so a single surviving orientation can grow into the component.

The frozen-temperature approximation prescribes the external thermal field during interface perturbations, rather than solving a perturbation heat equation. An interface displacement samples that prescribed field at its new location. Local liquidus and Gibbs-Thomson effect conditions still apply, and the solute field may retain full time-dependent diffusion.

For pure solid rejecting all solute, pulling speed $V>0$, solutal diffusivity $D$ and finite solute inventory per unit area $N$, a stationary planar front has $C(z)=C_I{e}^{-Vz/D}$ in its liquid half-space. The advection-diffusion equation gives this profile, and ${\int }_0^{\infty }Cdz=N$ fixes $C_I=VN/D$. A liquidus $T_I=-m{C}_I$ fixes its position in an imposed temperature gradient. The pure-solid branch requires $C_I<C_E$, where $C_E$ is the eutectic composition.

For an interface perturbation $\eta e^{ikx+st}$ with instantaneous interface kinetics, take the liquid concentration perturbation proportional to $e^{-qz}$. Here $\Gamma >0$ is the coefficient of curvature-induced melting-temperature depression. The advection-diffusion equation and interfacial mass conservation giveFor a growing normal mode the unique decaying liquid branch has $q=(V+\sqrt{V^2+4D(s+D{k}^2)})/(2D)$. Writing $b=(G+\Gamma k^2)/(m{C}_I)$, its rate is $s=(D/2)(V/D-b)(\sqrt{b^2+4k^2}-b)$. Instability requires $G+\Gamma k^2<mV{C}_I/D$. The Gibbs-Thomson effect suppresses sufficiently short waves; a quasistatic solute diffusion to solute diffusion changes the rate and must be stated separately.

A quasistatic approximation to the solute field neglects its perturbation time derivative while retaining the interface motion in the solute balance. For steady pulling, a lateral wavenumber $k$ then has decaying exponent $q=(V+\sqrt{V^2+4D^2{k}^2})/(2D)$. It is justified when the interface growth time is long compared with the relevant solute diffusion time. Fixing the thermal field does not by itself impose this solutal timescale ordering.

For a linear imposed temperature $T=T_E+Gz$, local liquidus equilibrium gives $C=-(T_E+Gz)/m$ in the mushy layer. With equal densities, pure solid, pulling velocity $-V$, and diffusion only in liquid volume fraction $q=1-\varphi$, steady solute flux is $-VqC-Dq{C}_z=-V{C}_0$. The exterior concentration is $C_0+(C_i-C_0)e^{-V(z-h_0)/D}$. Marginal equilibrium at a mush-liquid boundary gives $C_i-C_0=GD/(mV)$ and hence the displayed thickness. In the mush, $q[C+(D/V)C_z]=C_0$, so $\varphi =(h_0-z)/(m{C}_0/G+h_0-z)$. A positive thickness requires the thermal interval to exceed the diffusive length $D/V$.

If the solid skeleton translates with velocity $-V{\mathbf{e}}_z$ and the fully solid material has density ${\rho }_s$, steady mass conservation fixes Darcy flux toThe liquid's laboratory velocity is therefore ${\mathbf{v}}_l=-rV{\mathbf{e}}_z$ wherever liquid exists. This upward relative flux for $r<1$ supplies the expansion needed when liquid becomes a less dense solid.

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

For $\mathcal{C}\gg 1$ and latent-to-sensible heat ratio $S\gg 1$ 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 ${\theta }^{\prime \prime }+r\Omega {\theta }^{\prime }=0$ in the mush and ${\theta }^{\prime \prime }+r{\theta }^{\prime }=0$ in the liquid. Matching temperature and heat flux at $\zeta =H$, with $\theta (0)=-1$ and liquid temperature ${\theta }_{\infty }>0$ far above, givesThe latent heat modifies the mush's effective thermal decay rate and therefore its thickness.