Ternary Interaction Parameters in Calphad Solution Models

Abstract

For random, diluted, multicomponent solutions, the excess chemical potentials can be expanded in power series of the composition, with coefficients that are pressure- and temperature-dependent. For a binary system, this approach is equivalent to using polynomial truncated expansions, such as the Redlich-Kister series for describing integral thermodynamic quantities. For ternary systems, an equivalent expansion of the excess chemical potentials clearly justifies the inclusion of ternary interaction parameters, which arise naturally in the form of correction terms in higher-order power expansions. To demonstrate this, we carry out truncated polynomial expansions of the excess chemical potential up to the sixth power of the composition variables.

Appendix: Proof of (5)

Equation 5 is but a generalization of the expansion proposed by [1], who presented only the first few terms. The proof of the general expression, using the Gibbs-Duhem equation, may be of some interest. To that end, we write (3) in the following form

$$ x_{B}\frac{d\mu^{ex}_{B}}{d\, x_{A}}= x_{A}\frac{d\mu^{ex}_{A}}{d\,x_{B}}, $$

(A1)

which is valid for a binary system A−B. More compactly, (4) can be rewritten as follows:

$$ \mu^{ex}_{A}=\sum_{\eta=1}^{\eta_{max}} \frac{{{\Omega}}_{A,\eta}}{\eta} {x_{B}}^{\eta},\qquad \mu^{ex}_{B}=\sum_{\eta=1}^{\eta_{max}} \frac{{{\Omega}}_{B,\eta}}{\eta} {x_{A}}^{\eta}. $$

(A2)

Substitution of the right-hand sides in (A2) for the excess chemical potentials in (A1) yields the equality

$$ x_{B} \left({{\Omega}}_{B,1}+ \sum_{\eta=2}^{\eta_{max}} {{\Omega}}_{B,\eta} {x_{A}}^{\eta-1} \right)= x_{A} \left({{\Omega}}_{A,1}+ \sum_{\eta=2}^{\eta_{max}} {{\Omega}}_{A,\eta} {x_{B}}^{\eta-1} \right). $$

(A3)

From (A3), we deduce immediately that Ω _A,1= Ω _B,1=0, as pointed out by Darken [34]. We also have to equate the sums within the parentheses on both sides of (A3). Division of both sides by the product x _A x _B then yields the expression

$$ \sum_{\eta=2}^{\eta_{max}} {{\Omega}}_{B,\eta} {x_{A}}^{\eta-2} = \sum_{\eta=2}^{\eta_{max}} {{\Omega}}_{A,\eta} {x_{B}}^{\eta-2} $$

(A4)

We now recall that x _B =1−x _A and use the binomial theorem to expand the resulting powers of 1−x _A , which leads to the equality

$$ \sum_{\eta=2}^{\eta_{max}} {{\Omega}}_{B,\eta} {x_{A}}^{\eta-2} = \sum_{\eta=2}^{\eta_{max}} \sum_{\lambda=0}^{\eta-2} {{\Omega}}_{A,\eta} \binom{\eta-2}{\lambda}(-1)^{\lambda}{{x_{A}}}^{\lambda} $$

(A5)

Since the right-hand side of (A5) is a polynomial in x _A , we can easily reverse the order of the double sum, to find that

$$ \sum_{\eta=2}^{\eta_{max}} {{\Omega}}_{B,\eta} {x_{A}}^{\eta-2} = \sum_{\lambda=0}^{\eta_{max}-2} \sum_{\eta=\lambda+2}^{\eta_{max}} {{\Omega}}_{A,\eta} \binom{\eta-2}{\lambda}(-1)^{\lambda}{{x_{A}}}^{\lambda} $$

(A6)

The formal transformations λ→η−2 and η→λ adjusts the power of x _A in the summand on the right-hand side of (A6) to make it identical to the power in the summand on the left-hand side:

$$ \sum_{\eta=2}^{\eta_{max}} {{\Omega}}_{B,\eta}\, {x_{A}}^{\eta-2} = \sum_{\eta=2}^{\eta_{max}} \sum_{\lambda=\eta}^{\eta_{max}} {{\Omega}}_{A,\lambda} \binom{\lambda-2}{\eta-2}(-1)^{\eta}\,{x_{A}}^{\eta-2} $$

(A7)

since (−1)^η−2=(−1)^η.

Comparison between the coefficients of $x_{A}^{\eta -2}$ in the summands on both sides of (A7) then yields the final result

$$ {{\Omega}}_{B,\eta} = \sum_{\lambda=\eta}^{\eta_{max}} {{\Omega}}_{A,\lambda} \binom{\lambda-2}{\eta-2}(-1)^{\eta}. $$

(A8)