I want to calculate the phase diagram for the Al-In system, which has an immiscible phase. As the solid elements don't have any mutual solubility, there is no excess Gibbs description for the solid mixing phase.

With the excess Gibbs parameters from here, which are given on page 67 for pure Al as $G^{liq}_{Al} - G^{fcc}_{Al} = 10711 - 11.473 T$, for pure In as $G^{liq}_{In} - G^{tet}_{In} = 3283 - 7.639 T$, and for the excess Gibbs energy as a Redlich-Kister polynomial $G_E = x_{In} x_{Al} \sum^2_{v = 0} (x_{Al} - x_{In})^v (A_v + B_v T)$ with the following parameters: | $v$ | $A_v$ | $B_v$ | | --- | ----- | ----- | | $0$ | $21259.6$ | $-0.48737$ | | $1$ | $3850.3$ | $-1.20564$ | | $2$ | $5479.2$ | $-3.16805$ |

I get a correct liquid-liquid phase transition (by finding the common tangent for the liquid $G(x)$), but I can not get the solid↔liquid transition below $x_{In} = 0.05$, nor at above $x_{In} = 0.85$, as there is no excess Gibbs for the solid phase.

And when I set the excess Gibbs to zero, with then only the ideal mixing ($R T (x_{Al} \ln x_{Al} + x_{In} \ln x_{In})$, I get a too high equilibrium point. (finding the equilibrium by finding the common tangent between $G^{liq}(x_1)$ and $G^{sol}(x_1)$

What am I missing? How do I calculate the equilibria of the other parts than the $L_{Al} + L_{In}$ phase?

My calculated points over the plot from the source, low concentration.

My calculated points over the plot from the source

  • $\begingroup$ How are you treating the solid phases, particularly since they are immiscible in each other? $\endgroup$
    – Jon Custer
    Jan 17, 2023 at 15:46

1 Answer 1


Two suggestions up front - first, the accepted compilation of pure elements is AT Dinsdale, Calphad 15(4) 317-425 (1991). That gives slightly different evaluations for the liquids vs the solid phases for Al and In. Second, there are several good book on the Calphad process and either Saunders and Miodownik or Lukas, Fries and Sundman are fine.

Here, since fcc Al is considered insoluble in tetragonal In, and vice versa, the question comes down to how you want to handle those (separate) solid phases. One could either (1) consider them line compounds or (2) use a very large heat of mixing (and don't forget to use tabulated or estimated free energies for tet-Al or fcc-In).

I implemented the Al-In system using I. Ansara et al., Calphad 18(2) 177-222 (1994) (which covers many III-V and references Coughanowr's thesis for Al-In). I get:

enter image description here

This was obtained by setting the tetragonal phase of Al to being 15kJ above fcc (not particularly critical), and the fcc phase of In to the Dinsdale value (tet-Al is not in Dinsdale). Then I added an ad hoc mixing term of 100,000 J/mol to both the tetragonal and fcc phases. Now your mutual tangents will do the right things.

Python code for the 3 phases (fcc, tet, liquid) is:

def AlIn_tet(x): # x is Al concentration throughout
    return(Gideal(x,lambda: 15000.,In_tet)+x*(1-x)*100000.)
def AlIn_fcc(x):
def AlIn_liq(x):

where the ideal contribution (Gideal) is the linear combination of the elemental free energies plus the entropy of mixing term. Note that all free energy equations have to use the same concentration variables, and the one you choose may lead to changing the sign of the mixing terms in odd powers of (2*x-1).

  • $\begingroup$ Thank you for the reply! Now, I must admit that I'm still a bit confused, and hope you could help me a bit further. I implemented the fcc-In from Dinsdale, and the guesstimate tet-Al, as you recommended (G0Al_tet = G0Al_fcc + 1.5e4). I get something close to the correct curve for the high-Al case, but it still does not give a valid solution for the high-In case. So my Gibbs for the tet/fcc phase (p) is G = xIn * G0In_p + xAl * G0Al_p + RT (xIn * log(xIn) + xAl * log(xAl)) + 1e5*xAl*xIn. Could you maybe share the Matlab(?) code for your plot? $\endgroup$
    – asbjos
    Jan 19, 2023 at 14:51
  • $\begingroup$ Thank you very much, I have now gotten the correct solution for the entire range! To anyone else attempting the same thing in the future: beware that the result for the x2 value (where the mutual tangent from the liquid and solid phases crosses the solid phase) is very small (~1e-7). So the initial guess for the tangent finding algorithm should also be very small. Using scipy.optimize.least_squares, I had to use values lower than 0.05, which I didn't use at first, explaining why I did not get the correct result. $\endgroup$
    – asbjos
    Jan 20, 2023 at 15:58
  • $\begingroup$ Yes, the perils of essentially immiscible systems. I chose to go about the calculation differently to trade off speed for exact answers. $\endgroup$
    – Jon Custer
    Jan 20, 2023 at 16:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.