Modeling the kinetics of consecutive phase transitions in the solid state

Abstract

A theoretical approach for describing the kinetics of consecutive phase transformations ruled by nucleation and growth is reported. In the considered system, the mother phase (M) transforms to an intermediate phase (α) which, in turn, transforms to the final product (β). The classical Kolmogorov–Johnson–Mehl–Avrami theory is generalized to deal with a finite-size phase with moving boundary. To this end, the statistical method based on the differential critical region has been employed. The exact solution of the kinetics is computed in closed form for the transformation of a spherical α-nucleus growing into the mother phase. By resorting to an approximate expression for the probability function entering the differential critical region method, the consecutive transformation is studied in the case of nucleation and growth of the α-phase. The time dependence of the β/α volume fraction is found to be in very good agreement with the stretched exponential kinetics, and the dependence of Avrami’s exponent on both nucleation and growth rates of the two phases is investigated. Modeling of the non-isothermal kinetics at constant heating rate has also been performed which provides an insight into the shape of the differential scanning calorimetry curves for consecutive phase transitions.

Appendices

Appendix 1

In this Appendix, the computation of the volume of the critical region, ϖ(t, t ₁, t″), is reported. According to Fig. 2, the following cases are considered:

Case (i) 0 < t″ < t ₁ < t < t _f

For R _α (t″) + R _β (t, t″) < R _α (t ₁), the volume of the critical region is nil, ϖ = 0. For linear growth, this inequality implies,

$$ 0 < (1 - \rho )t^{\prime\prime} < t_{1} - \rho t , $$

(24)

where $\rho = \frac{{v_{\beta } }}{{v_{\alpha } }} < 1$. Therefore, for $\, \rho < \, \frac{{t_{1} }}{t} < 1 { }$ one defines the time $T_{1} = \frac{{t_{1} - \rho t}}{1 - \rho }$ and Eq. 24 provides

(a) $\, \rho < \, \frac{{t_{1} }}{t} < 1 { }$

$$ \varpi = 0 \, \leftrightarrow \, t'' < T_{1} < t_{1} $$

(25)

$$ \varpi = \omega (t,t_{1} ,t'') \, \leftrightarrow \, T_{1} < t'' < t_{1} $$

(26)

where ω(t, t ₁, t″) = ω[R _α (t″), R _β (t, t″); R _α (t ₁)], with ω[r ₁, r ₂; z] overlap volume of two spheres of radii r ₁ and r ₂ located at relative distance z.

For R _β (t, t″) > R _α (t ₁) + R _α (t″), the volume of the critical region is equal to ϖ(t″) = υ _α (t″) and its derivative, in Eq. 7, is nil. This condition implies

$$ (1 + \rho )t'' < \rho t - t_{1} . $$

(27)

For $\, \frac{{t_{1} }}{t} < \rho \,$ one defines the time $T_{0} = \frac{{\rho t - t_{1} }}{1 + \rho }$. Accordingly, for $\frac{\rho }{2 + \rho } < \frac{{t_{1} }}{t} < \rho,$ the inequalities 0 < T ₀ < t ₁ hold and the volume of the critical region becomes

b) $\frac{\rho }{2 + \rho } < \frac{{t_{1} }}{t} < \rho$

$$ \varpi = \upsilon_{\alpha } (t'') \leftrightarrow \, t'' < T_{0} < t_{1} $$

(28)

$$ \varpi = \omega (t,t_{1} ,t'') \, \leftrightarrow \, T_{0} < t'' < t_{1} . $$

(29)

On the other hand, when $\frac{{t_{1} }}{t}$ is lower than $\frac{\rho }{2 + \rho },$ the volume of the critical region is

c) $0 < \frac{{t_{1} }}{t} < \frac{\rho }{2 + \rho }$

$$ \varpi = \upsilon_{\alpha } \left( {t^{\prime\prime}} \right) \leftrightarrow t^{\prime\prime} < t_{ 1} $$

Case (ii) 0 < t ₁ < t″ < t < t _f

In this case, the volume of the critical region is always different from zero (Fig. 2).

For R _α (t″) > R _α (t ₁) + R _β (t, t″), the volume of the critical region is equal to ϖ(t, t″) = υ _β (t − t″). This inequality implies (1 + ρ)t″ > ρt + t ₁, namely $t'' > T_{2} = \frac{{t_{1} + \rho t}}{1 + \rho }$, where t ₁ < T ₂ < t. Furthermore, for R _β (t, t″) > R _α (t ₁) + R _α (t″), the volume of the critical region is equal to ϖ(t″) = υ _α (t″) and the derivative in Eq. 7 is nil. This condition leads to the inequality Eq. 27, (1 + ρ)t″ < ρt - t ₁ and, as above, for $\, \frac{{t_{1} }}{t} < \rho \,$ the time $T_{0} = \frac{{\rho t - t_{1} }}{1 + \rho }$ is defined. Consequently, for 0 < t″ < T ₀, the volume of the critical region is ϖ(t″) = υ _α (t″). Also, for $0 < \frac{{t_{1} }}{t} < \frac{\rho }{2 + \rho }$, t ₁ < T ₀ < T ₂ < t, while, for $\frac{\rho }{2 + \rho } < \frac{{t_{1} }}{t} < 1$, it follows that T ₀ < t ₁. All these conditions are summarized according to

a) $0 < \frac{{t_{1} }}{t} < \frac{\rho }{2 + \rho }$

$$ \varpi = \upsilon_{\alpha } (t'') \, \leftrightarrow \, t_{1} < t'' < T_{0} $$

$$ \varpi = \omega (t,t_{1} ,t'') \, \leftrightarrow \, T_{0} < \, t'' < T_{2} $$

$$ \varpi = \upsilon_{\beta } (t - t'') \, \leftrightarrow \, T_{2} < \, t'' < t $$

b) $\frac{\rho }{2 + \rho } < \frac{{t_{1} }}{t} < 1$

$$ \varpi = \omega (t,t_{1} ,t'') \, \leftrightarrow \, t_{1} < \, t'' < T_{2} $$

$$ \varpi = \upsilon_{\beta } (t - t^{\prime\prime} ) \leftrightarrow T_{ 2} < t^{\prime\prime} < t. $$

A graphical representation of the results here obtained for both cases (i) and (ii) is reported in Fig. 3.

Appendix 2

The overlap volume of two spheres of radii r ₁ and r ₂ whose centers are located at relative distance z is

$$ \omega (r_{1} ,r_{2} ;z) = \pi \left[ {\frac{2}{3}(r_{1}^{3} + r_{2}^{3} ) + \frac{1}{12}z^{3} - \frac{1}{2}z(r_{1}^{2} + r_{2}^{2} ) - \frac{1}{4z}[(r_{1}^{2} - r_{2}^{2} )^{2} ]} \right] , $$

(30)

which is further computed for r ₁ ≡ R _α (t″), r ₂ ≡ R _β (t, t″) and z ≡ R _α (t ₁). The derivative of Eq. 30 eventually gives

$$ \partial_{t} \omega (t,t_{1} ,t^{\prime\prime}) = \pi \left[ {2v_{\beta }^{3} (t - t^{\prime\prime})^{2} - v_{\beta }^{2} v_{\alpha } t_{1} (t - t^{\prime\prime}) + \frac{{v_{\beta }^{2} }}{{v_{\alpha } t_{1} }}(t - t^{\prime\prime})[v_{\alpha }^{2} t^{{\prime\prime}2} - v_{\beta }^{2} (t - t^{\prime\prime})^{2} ]} \right] $$

(31)

that is Eq. 11b.