Soft impingement in diffusion-controlled growth of binary alloys: moving boundary effect in one-dimensional system

Abstract

The impact of soft impingement on the kinetics of diffusion-controlled growth of binary alloys is investigated. An analytical approach is developed which takes into account the process of island growth, that is the time dependence of the position of the nucleus/parent phase interface. The concentration profile, the growth law, and the kinetics of the fraction of transformed phase are computed and compared with those attained for point islands. At odd with the point island approach the local kinetics of growth depends on initial supersaturation. On the other hand, the whole transformation kinetics is in good agreement with that of the point island model with an Avrami exponent close to the theoretical value n = 0.5. The concentration profile is well described by a polynomial function in the whole spatial domain, with an exception for the initial stage of the phase separation. The effect of the spatial distribution of the nuclei on the kinetics is also studied in the model case of hard-core correlation among nuclei.

Appendices

Appendix 1

The computation of Eq. 1 requires the use of the following integrals:

$$ \int\limits_{ - \xi }^{\xi } {\sin \frac{\pi nx}{\xi }} \sin \frac{\pi mx}{\xi }{\text{d}}x = \xi \delta_{n,m} $$

(14)

$$ \int\limits_{ - \xi }^{\xi } {x\sin \frac{\pi mx}{\xi }} \cos \frac{\pi nx}{\xi }{\text{d}}x = - \frac{{\xi^{2} }}{\pi }\left( {\frac{{( - )^{m + n} }}{m + n} + \frac{{( - )^{m - n} }}{m - n}} \right) = - \frac{{\xi^{2} }}{\pi }\frac{2m}{{m^{2} - n^{2} }}\;\left( {m \ne n} \right) $$

(15)

$$ \int\limits_{ - \xi }^{\xi } {x\sin \frac{\pi mx}{\xi }} \cos \frac{\pi mx}{\xi }{\text{d}}x = - \frac{{\xi^{2} }}{2\pi m}, $$

(16)

where the terms n + m and n − m are both even numbers for n and m are odd integers.

Appendix 2

The mass balance in the region 0 < x < L reads $\int\limits_{l(t)}^{L - l(t)} {c(x,t){\text{d}}x + 2l(t)c^{(\alpha )} = c^{(0)} L}$. The derivative of this expression is $\int\limits_{l(t)}^{L - l(t)} {\frac{\partial c}{\partial t}} {\text{d}}x + 2\dot{l}(t)(c^{(\alpha )} - c^{(\beta )} ) = 0$ that is, using Eq. 1, $\int\limits_{l(t)}^{L - l(t)} {\frac{\partial c}{\partial t}} {\text{d}}x + 2D\left. {\frac{\partial c}{\partial x}} \right|_{x ={l}({t}) } = 0$. In terms of Fourier coefficients, this equation becomes

$$ \sum\limits_{n} {\left( {\dot{c}_{n} \frac{\xi }{\pi n} + c_{n} \frac{{\dot{\xi }}}{\pi n} + c_{n} \frac{\pi nD}{\xi }} \right)} = 0. $$

(17)

It is worth noting that one is not allowed to set each term in the brackets of Eq. 17 equal to zero. In fact, in this case, Eqn. 3a reduces to $c_{m} = 4\sum\limits_{n \ne m} {\frac{mn}{{m^{2} - n^{2} }}c_{n} }$ which implies, in general, c _n  = 0. On the other hand, retaining the sums in Eq. 17, compatibility between Eqs. 3a and 17 implies

$$ 4\sum\limits_{n,m \ne n} {\frac{{nc_{n} }}{{m^{2} - n^{2} }}} = \sum\limits_{m} {\frac{{c_{m} }}{m}} $$

(18)

that is $nc_{n} \sum\limits_{m \ne n} {\frac{1}{{m^{2} - n^{2} }}} = \frac{{c_{n} }}{4n}$ and

$$ n\sum\limits_{m \ne n} {\frac{1}{{m^{2} - n^{2} }}} = \frac{1}{4n}, $$

(19)

where m and n are odd integers. In fact for n = 1 Eq. 19 is satisfied as $\sum\limits_{m > 1}^{{}} {\frac{1}{{m^{2} - 1}}} = \frac{1}{2}\sum\limits_{m > 1}^{{}} {\left( {\frac{1}{m - 1} - \frac{1}{m + 1}} \right) = \frac{1}{4}} \sum\limits_{k = 1}^{\infty } {\left( {\frac{1}{k} - \frac{1}{k + 1}} \right)} = \frac{1}{4}$. For n > 1 Eq. 19 becomes $\frac{1}{4}\sum\limits_{s = 1}^{\infty } {\left( {\frac{1}{s} - \frac{1}{s + n}} \right)} + \sum\limits_{m < n} {\frac{n}{{m^{2} - n^{2} }} = } \frac{1}{4n}$ which leads to $\frac{1}{4}\sum\limits_{s = 1}^{n} {\left( {\frac{1}{s} - \frac{1}{n}} \right)} = \sum\limits_{m < n} {\frac{n}{{n^{2} - m^{2} }}}$ and, eventually, to

$$ \frac{1}{4}\sum\limits_{s = 1}^{n - 1} \frac{1}{s} = \sum\limits_{k = 0}^{{k_{n} }} {\frac{n}{{n^{2} - (2k + 1)^{2} }}} , $$

(20)

where 2k _n  + 1 = n − 2. It is straightforward to show that this equation holds for n = 3. The validity of Eq. 20 for all n can easily be verified by induction. In fact, if Eq. 20 is verified for n = 2r − 1 (i.e., $\frac{1}{4}\sum\limits_{s = 1}^{2r - 2} \frac{1}{s} = (2r - 1)\sum\limits_{k = 0}^{r - 2} {\frac{1}{{(2r - 1)^{2} - (2k + 1)^{2} }}}$), then for n = 2r + 1 (r is integer) the relation holds $\frac{1}{4}\left( {\sum\limits_{s = 1}^{2r - 2} \frac{1}{s} + \frac{1}{(2r - 1)} + \frac{1}{(2r)}} \right) = (2r + 1)\sum\limits_{k = 0}^{r - 1} {\frac{1}{{(2r + 1)^{2} - (2k + 1)^{2} }}}$

that is

$$ (2r - 1)\sum\limits_{k = 0}^{r - 2} {\frac{1}{{(2r - 1)^{2} - (2k + 1)^{2} }}} + \frac{1}{4(2r - 1)} + \frac{1}{4(2r)} = (2r + 1)\sum\limits_{k = 0}^{r - 1} {\frac{1}{{(2r + 1)^{2} - (2k + 1)^{2} }}} . $$

(21)

Eq. 21 is equivalent to the equality $\sum\limits_{k = 0}^{r - 2} {\left\{ {\frac{1}{r - k - 1} + \frac{1}{r + k}} \right\}} + \frac{1}{2r - 1} + \frac{1}{2r} = \sum\limits_{k = 0}^{r - 1} {\left\{ {\frac{1}{r + k + 1} + \frac{1}{r - k}} \right\}}$ in which validity can be checked by changing the sum indexes.

Appendix 3

By equating the derivative of $\tilde{\xi }^{2}$ as given by Eqs. 5 and 7 one obtains

$$ 2\varepsilon \rho_{1} \cong \frac{\text{d}}{{{\text{d}}\tau }}\frac{1}{{\dot{\rho }_{1} }}\left\{ {\rho_{1} - \varepsilon \frac{1}{2}\rho_{1}^{2} } \right\} $$

(22)

where terms of the order of $\rho_{m}^{{}}$, with m ≥ 3, have been neglected when compared to $\rho_{1}$. Equation 22 reads,

$$ 2\varepsilon \rho_{1} \cong - \frac{1}{{\dot{\rho }_{1}^{2} }}\ddot{\rho }_{1} \left\{ {\rho_{1} - \varepsilon \frac{1}{2}\rho_{1}^{2} } \right\} + 1 - \varepsilon \rho_{1}^{{}} $$

(23)

that is

$$ \frac{{{\text{d}}\ln \left| {\dot{\rho }_{1} } \right|}}{{{\text{d}}\rho_{1} }} = \frac{{1 - 3\varepsilon \rho_{1}^{{}} }}{{\rho_{1} - \frac{\varepsilon }{2}\rho_{1}^{2} }}. $$

(24)

Equation 24 leads to Eq. 9,

$$ \ln \left| {\dot{\rho }_{1} } \right| = 2\int {\frac{{1 - 3\varepsilon \rho_{1}^{{}} }}{{2\rho_{1} - \varepsilon \rho_{1}^{2} }}} {\text{d}}\rho_{1} + C, $$

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where the integration constant, C, has to be determined from the initial condition $\dot{\rho }_{1} (1) \cong - (1 - \varepsilon /2)$.

The approximation here employed is not expected to hold in the initial stage of the growth, owing to the contribution of the series. The uncertainty brings about by this approximation can be estimated by retaining the sum of the series in the short time limit (with τ ≠ 0). For τ ≪ 1 the derivative of $\rho_{1}$ reads $\dot{\rho }_{1} \approx - [1 - \varepsilon (1 + s_{1} (\tau ))(s_{2} (\tau ) + 1/2)]$ where $s_{1} (\tau ) \cong \sum\limits_{n > 1} {{\text{e}}^{{ - n^{2} \tau }} }$ and $s_{1} (\tau ) \cong \sum\limits_{n > 1} {\frac{2}{{1 - n^{2} }}{\text{e}}^{{ - n^{2} \tau }} }$ (with n odd integers). For ε = 0.5 and τ = 0.01 one obtains $\dot{\rho }_{1} = - 0.64$ to be compared with the value $\dot{\rho }_{1} = - 0.75$ used in Eq. 24. Similarly, for $\dot{\rho }_{3}$ the computation gives $\dot{\rho }_{3} \cong - (3.46)^{2}$ to be compared with the figure $\dot{\rho }_{1} = - (3)^{2}$.