Question 7 of 8: Classical Nucleation Theory and Solidification Growth Morphology

Question 7: Classical Nucleation Theory and Solidification Growth Morphology (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

7.1 — (a) Deriving $r^*$ and $\Delta G^*$ for a spherical nucleus

Given. A spherical solid nucleus of radius $r$ forming from the liquid, with volumetric free-energy change on transformation $\Delta G_v$ (negative below the melting point) and solid/liquid interfacial energy $\gamma_{SL}$.

Find. $r^*$ and $\Delta G^*$ in terms of $\gamma_{SL}$ and $\Delta G_v$.

Approach. Write the total free-energy change of forming a nucleus of radius $r$ as the sum of the volume (driving-force) and surface (interfacial-penalty) contributions, then find the stationary point.

1. Total free-energy change. $$\Delta G(r)=\dfrac{4}{3}\pi r^3\Delta G_v+4\pi r^2\gamma_{SL}$$ The first term is negative (favourable, driving the transformation) and grows as $r^3$; the second is always positive (unfavourable, the energy cost of creating new solid/liquid interface) and grows only as $r^2$ — so the surface term dominates at small $r$ and the volume term dominates at large $r$, producing a maximum in $\Delta G(r)$ at intermediate $r$.
2. Find the critical radius. Setting $d(\Delta G)/dr=0$: $$4\pi r^2\Delta G_v+8\pi r\gamma_{SL}=0\ \Longrightarrow\ \boxed{r^*=-\dfrac{2\gamma_{SL}}{\Delta G_v}}$$ (positive, since $\Delta G_v<0$ below the melting point).
3. Find the activation barrier. Substituting $r^*$ back into $\Delta G(r)$: $$\Delta G^*=\Delta G(r^*)=\dfrac{4}{3}\pi\left(-\dfrac{2\gamma_{SL}}{\Delta G_v}\right)^3\Delta G_v+4\pi\left(-\dfrac{2\gamma_{SL}}{\Delta G_v}\right)^2\gamma_{SL}=\boxed{\dfrac{16\pi\gamma_{SL}^3}{3\Delta G_v^2}}$$

7.2 — (b) Why undercooling and the $2\gamma/r^*$ term matter

The volumetric driving force scales with undercooling as $\Delta G_v\approx-\dfrac{L\,\Delta T}{T_m}$ (Turnbull approximation, $L$ = latent heat of fusion, $\Delta T=T_m-T$), i.e. $\Delta G_v$ is exactly ZERO at the melting point and grows in magnitude linearly with undercooling below it. Since $r^*\propto1/|\Delta G_v|$ and $\Delta G^*\propto1/\Delta G_v^2$, at $T=T_m$ ($\Delta T=0$) $r^*\rightarrow\infty$ and $\Delta G^*\rightarrow\infty$ — nucleation is IMPOSSIBLE exactly at the melting point, no matter how long one waits, because an infinitely large (and hence infinitely improbable) fluctuation would be required. Undercooling is therefore not merely helpful but strictly necessary: it is what makes $r^*$ and $\Delta G^*$ finite (and, at sufficient undercooling, small enough to be reached by ordinary thermal fluctuations on laboratory timescales).

The term $2\gamma_{SL}/r^*$ (which appears directly from $r^*=-2\gamma_{SL}/\Delta G_v\Rightarrow\Delta G_v=-2\gamma_{SL}/r^*$) is the extra pressure/energy-density a curved solid/liquid interface of radius $r$ exerts on the solid, from the Gibbs-Thomson (capillarity) effect — physically, it is the amount by which a curved solid nucleus's effective melting point is DEPRESSED below $T_m$, exactly balancing the chemical driving force at $r=r^*$. It is the quantitative statement of why small particles are less stable (require more driving force to survive) than large ones, which is the same physics that governs coarsening/ripening of any curved interface (precipitates, grains) once nucleation is complete.

7.3 — (c) Why heterogeneous nucleation dominates despite unchanged $r^*$

Heterogeneous nucleation on a foreign surface (the vessel wall) does not change $r^*$, because $r^*$ depends only on $\gamma_{SL}$ and $\Delta G_v$, both bulk/interfacial properties of the solid-liquid system itself — the geometry of nucleation does not enter that ratio. What DOES change is the SHAPE (and hence volume and surface area) of the critical nucleus: on a flat wall, the energetically most favourable nucleus shape is a spherical cap (wetting angle $\theta$ set by the relative surface energies $\gamma_{SL}$, $\gamma_{\text{wall-S}}$, $\gamma_{\text{wall-L}}$) rather than a full sphere. A spherical cap of the SAME radius of curvature $r^*$ has a much smaller volume and a much smaller solid/liquid interfacial area than a full sphere of that radius (the wall itself supplies part of the required "surface" at no interfacial-energy cost, since the wall/solid interface simply replaces part of the wall/liquid interface that already existed). The activation barrier for heterogeneous nucleation is therefore

$$\Delta G^*_{\text{het}}=\Delta G^*_{\text{hom}}\cdot S(\theta),\qquad S(\theta)=\dfrac{(2+\cos\theta)(1-\cos\theta)^2}{4}$$

where the shape factor $S(\theta)\le1$ for any wetting angle $0<\theta<180^{\circ}$, and $S(\theta)\rightarrow0$ as $\theta\rightarrow0$ (perfect wetting). Because $r^*$ is unchanged but the VOLUME of material that must be assembled to reach that same critical radius is drastically smaller, $\Delta G^*_{\text{het}}\ll\Delta G^*_{\text{hom}}$, and since nucleation rate depends exponentially on $\Delta G^*$ ($I\propto\exp(-\Delta G^*/kT)$), even a modest reduction in the barrier produces an enormous increase in the practical nucleation rate — which is why heterogeneous nucleation on the container wall (or on any available foreign surface) overwhelmingly dominates over homogeneous nucleation in real castings.

7.4 — (d) Planar vs. cellular growth: temperature gradient and constitutional supercooling

For a binary alloy solidifying with a partition coefficient $k<1$ (solute rejected into the liquid at the interface, the normal case), solute builds up ahead of the advancing solid/liquid interface, forming a boundary layer whose composition (and hence local equilibrium LIQUIDUS temperature, from the binary phase diagram) varies with distance $x$ ahead of the interface: $T_L(x)$ rises from the interface value toward the bulk liquid value over the diffusion boundary-layer thickness.

1. (i) Planar growth. If the ACTUAL temperature gradient imposed in the liquid, $G=dT_{\text{actual}}/dx$, is everywhere STEEPER than the liquidus-temperature gradient $dT_L/dx$ set by the solute pile-up, then $T_{\text{actual}}(x)>T_L(x)$ everywhere ahead of the interface — the liquid everywhere ahead of the interface is hotter than its own local equilibrium freezing point, so any small protrusion of solid that pokes ahead of a flat interface finds itself in liquid that is ABOVE its local liquidus and would remelt. The interface therefore remains stable and PLANAR.
2. (ii) Cellular (and, at greater supercooling, dendritic) growth. If $G$ is shallow enough (or the solidification rate high enough that solute rejection outpaces diffusion) that $T_{\text{actual}}(x)<T_L(x)$ over some region ahead of the interface, that region is constitutionally supercooled: the liquid there is below its own local equilibrium freezing point even though it is further from the (colder) interface. Any small protrusion that pokes into this supercooled liquid finds itself in liquid that wants to freeze, so the protrusion is thermodynamically favoured to grow FASTER than the flat interface around it, breaking the interface up into a cellular (mild supercooling) or, with progressively more supercooling, dendritic (branching, most severe supercooling) morphology.

Both the temperature gradient $G$ and constitutional supercooling matter because interface stability is set by their COMPARISON, not by either alone: the standard criterion is $G/R>m\,C_0(1-k)/(kD)$ for a planar interface to remain stable, where $R$ is the growth rate, $m$ the liquidus slope, $C_0$ the bulk alloy composition, $k$ the partition coefficient and $D$ the liquid diffusivity — a shallow gradient (small $G$), a fast growth rate (large $R$), a steep liquidus slope, a high bulk solute content, or a small partition coefficient all independently push the alloy toward cellular/dendritic breakdown of the interface.