# Shape factors

Calculations of driving forces, interfacial compositions and growth rates were previously shown for spherical precipitates, in which the radius of curvature is constant across the surface. For non-spherical precipitates, the curvature varies.

While it is possible to model a single precipitate to investigate the thermodynamic and kinetic behavior across the surface, this is impractical for the KWN model. Rather, a correction factor is applied to the thermodynamic and kinetic terms.

In defining the precipitate shape, three terms will be used: the long axis ($l$), the short axis ($r$) and the equivalent spherical radius ($R$). Additionally, for ellipsoids, the eccentricity can be defined (which simplifies some of the following equations). The equivalent spherical radius is defined as the radius of a sphere that gives the same volume as the non-spherical precipitate. Defining the precipitate size by the equivalent radius minimizes the amount of changes needed to the KWN model (such as calculating volume fraction and composition).

$$\alpha = \frac{l}{r}$$

$$e = \sqrt{1 - \frac{1}{\alpha^2}}$$

The changes made to the growth rate and Gibbs-Thomson effect are then:

$$\frac{dR}{dt} = f(\alpha) \frac{dR}{dt} \biggr\rvert_{sphere}$$

$$\mu_A^\alpha = \mu_A^\beta + \left(g(\alpha) \frac{2\gamma}{R} + \Delta G_{el} \right) V_M^\beta$$

### Sphere #

The equations for a spherical precipitate reduces the KWN model to its original equations.

$$R_{eq} = r$$

$$f(\alpha) = 1$$

$$g(\alpha) = 1$$

### Needle #

$$R_{eq} = r \sqrt{\alpha}$$

$$f(\alpha) = \frac{2 e \sqrt{\alpha^2}}{\ln{(1+e)} - \ln{(1-e)}}$$

$$g(\alpha) = \frac{1}{2 \sqrt{\alpha^2}} \left(1 + \frac{\alpha}{e} \sin^{-1}{e} \right)$$

### Plate #

$$R_{eq} = r \sqrt{\alpha^2}$$

$$f(\alpha) = \frac{e \sqrt{\alpha^2}}{\pi/2 - \cos^{-1}{e}}$$

$$g(\alpha) = \frac{1}{2 \sqrt{\alpha^4}} \left(\alpha^2 + \frac{1}{2 e} \ln{ \left(\frac{1+e}{1-e} \right)} \right)$$

$$R_{eq} = \sqrt{\frac{3\alpha}{4\pi}} l$$
$$f(\alpha) = 0.1 \exp{ \left(-0.091 (\alpha-1)\right)} + \frac{1.736 \sqrt{\alpha^2 - 1}}{\sqrt{\alpha} \ln{ \left( 2\alpha^2 + 2\alpha \sqrt{\alpha^2 - 1} - 1 \right)}}$$
$$g(\alpha) = \frac{2\alpha + 1}{2\pi} \left( \frac{4 \pi}{3 \alpha} \right)^{2/3}$$