Precipitate growth kinetics: Diffusion limited growth of a planar, incoherent precipitate-matrix boundary

Consider an alloy of composition c$_0$ which is cooled from a high temperature into a two phase region as shown in Figure. 22. At this temperature, the alloy phase separates into two phases, say, $\alpha$ and , with equilibrium compositions given by c$_e^{\beta}$ and c$_e^{\alpha}$. Let us consider the phase separation that takes place by the nucleation and growth mechanism. Suppose the phase has nucleated and is growing into the supersaturated $\alpha$ phase; we assume that the $\alpha$- boundary is planar and incoherent. Our aim, in this section, is to calculate the velocity of such a planar, incoherent boundary using an approximate method due to Zener (Theory of growth of spherical precipitates from solid solution, C. Zener, Journal of Applied Physics, Vol. 20, pp. 950-953, 1949).

Figure 22: Schematic of an incoherent precipitate with planar boundary growing into a supersaturated $\alpha$ phase.

[scale=0.4]Figures/GrowthSchematic.pdf

In Figure. 22 we also show, schematically, the microstructure; the two phases and $\alpha$ are separated by a planar interface. We also show the composition profile as we go from the precipitate into the superaturated matrix. The composition of the phase is c$_e^{\beta}$ as given by the phase diagram. However, on the $\alpha$ side the composition is not a constant. On the $\alpha$ side, at the interface, the composition of the $\alpha$ phase is c$_e^{\alpha}$ as given by the phase diagram. Far from the $\alpha$- interface, the composition of the $\alpha$ phase is the same as the initial alloy composition, namely, c$_0$. In the middle region, the composition changes from c$_e^{\alpha}$ to c$_0$ in a continuouos manner. This composition profile is schematically shown in the Figure. 22; this profile results due to the diffusion of B atoms in the $\alpha$ phase.

Consider a unit cross-section of the $\alpha$- interface which moves a small distance, say, $dx$ in a given time, say, $dt$. The number of B atoms needed for such a movement of the interface is given by $(c_e^{\beta} - c_e^{\alpha}) dx$. Since these B atoms are supplied to the interface by diffusion, using the Fick's first law, one can calculate the number of B atoms that diffuse in time $dt$ as $D \left( \frac{dc}{dx} \right) dt$ where $D$ is the interdiffusion coefficient. Equating these two quantities, one can obtain the velocity of movement $v$ of the interface as follows:

$$v = \frac{dx}{dt} = \frac{D}{( c_e^{\beta}-c_e^{\alpha})} \frac{dc}{dx}$$ (1)

Figure 23: the Zener approximation to calculate the growth velocity of a planar, incoherent precipitate-matrix boundary.

[scale=0.4]Figures/ZenerApprox.pdf

As the precipitate grows, $\frac{dc}{dx}$ keeps decreasing - and hence, more and more of the matrix regions participate in the diffusion process to supply B atoms to grow the precipitate. As noted earlier, Zener introduced a simplified profile to calculate the velocity approximately. This simplified profile is shown in Figure. 23. From the profile, clearly, $\frac{dc}{dx} = \frac{c_0-c_e^{\alpha}}{L} = \frac{\Delta C_0}{L}$. The value of L can be calculated using conservation of B atoms, which demands that the shaded rectangle on the left is equal in area to the shaded triangle on the right. Hence,

$$\frac{1}{2} \Delta c_0 L = x (c_e^{\beta} - c_0);$$ (2)

Thus,

$$L = \frac{2 x (c_e^{\beta} - c_0)}{\Delta c_0}.$$ (3)

Substituting Eq. 53 in Eq. 51, we obtain

$$v = \frac{dx}{dt} = \frac{D}{( c_e^{\beta}-c_e^{\alpha})} \f... ...ta c_0)^2}{2 x ( c_e^{\beta}-c_e^{\alpha})(c_e^{\beta} - c_0)}$$ (4)

Further simplifications are possible if we assume that the molar volumes of both the phases are the same - in which case, we can replace the concentrations by mole fractions (X = C V$_m$). Also, we can assume, for the sake of simplicity, $(c_e^{\beta} - c_0) = ( c_e^{\beta}-c_e^{\alpha})$. Thus, we obtain,

$$v = \frac{dx}{dt} = \frac{D (\Delta X_0)^2}{2 x ( X_e^{\beta}-X_e^{\alpha})^2}.$$ (5)

The expression in Eq. 55 can be integrated to obtain

$$x = \frac{\Delta X_0}{( X_e^{\beta}-X_e^{\alpha})} (\sqrt{Dt}).$$ (6)

Note that in the above equation (and for the rest of this module) we assume that at the initial time $t_0=0$.

Now, the expression in Eq. 56 can be differentiated to obtain the velocity as

$$v = \frac{\Delta X_0}{2 ( X_e^{\beta}-X_e^{\alpha})} \left( \sqrt{\frac{D}{t}} \right).$$ (7)