The enthalpy in the quaternary example isn't (really) temperature dependent. The standard potentials depend on temperature because of the entropy term, but that's not at all the same functionality as the latent heat expression in the binary example. The quaternary example was just an illustration of how to flexibly deal with multiple components, not of how to deal with all of the possible physics of solidification.
To get the quaternary example to work like the binary example, you need to replace the simplistic standard potentials with ones based on the latent heats and melting points of the elements. If you don't have a particular system in mind, you can still "cook" the phase diagram by equating the solid and liquid chemical potentials for each element at some desired equilibrium temperature. On Nov 16, 2012, at 8:27 PM, Adam Stone wrote: > Thanks for the explanation, Jonathan. I think I see what you mean. I'm > now trying to extend the quaternary example to the dynamic case, as in > the binary example, such that the interface can move after > equilibration. However, in this case there is no definition of melting > temperature, and simply lowering the starting temperature does not cause > the interface to move. Some experimentation with temperature values > suggests that higher temperature creates a sharper, narrower interface > while lower temperature creates a broader interface, but it always > remains centered. > > I imagine this must be related to the different enthalpy terms between > the binary and quaternary cases? In the binary case the average enthalpy > is 0 for equilibration and < 0 in order to grow the solid phase > (interface moves right). Raising the temperature above Tm gives enthalpy >> 0 and shrinks the solid phase (interface moves left). This makes > sense as it is defined as proportional to the undercooling which is > giving the driving force for phase change. > > In contrast, the average enthalpy as defined in the quaternary case > seems to be > 0 for any positive temperature, yet the interface doesn't > move. So I suppose my next question is, why does the role of enthalpy > appear to change between the binary and quaternary cases, and how to go > about modifying the quaternary approach to account for melting temp, > undercooling, driving force, moving interface, etc? > > Thanks for being so responsive and patient with a phase-field newcomer. > > Adam > > > On 11/15/2012 09:00 AM, Jonathan Guyer wrote: >> On Nov 14, 2012, at 6:25 PM, Adam Stone wrote: >> >>> I'm going through the quaternary phase field example and having some >>> difficulty understanding the calculation of standard potentials. >> : >> : >>> But I don't understand how you "cook" these standard potentials: >>> >>> for Cj in interstitials + substitutionals + [solvent]: >>> ... Cj.standardPotential = R * T * (numerix.log(Cj.L/rhoL) >>> ... - numerix.log(Cj.S/rhoS)) >>> >>> Since this is the difference between the solid and liquid potentials, any >>> terms common to both would have disappeared and we should be left with the >>> remaining terms unique to each phase. But if these are indeed "standard" >>> potentials for the solid and liquid, it seems like none of them should >>> include R*T*ln(C/p) terms in the first place? So I don't see how this >>> expression is obtained (or why the solid term is being subtracted from the >>> liquid term rather than vice versa). >> This is just a statement of equality of chemical potentials in equilibrium. >> The difference between the standard potentials must be equal to the log of >> the ratio of the concentrations in equilibrium (for the ideal solution >> considered in that example, anyway). We start this example with the ansatz >> that an 0.3-0.4-0.2-0.1 solid is in equilibrium with a 0.4-0.3-0.1-0.2 >> liquid. >> >> The standard states of different phases are not in equilibrium; if they >> were, then vapor pressure would be a useless concept (pure bcc iron would be >> in equilibrium with 1 atm of Fe vapor under the same conditions that pure >> liquid helium was in equilibrium with 1 atm of He gas). We pick standard >> states because we need something to compare to, but they can't all have the >> same values or there would be no chemical specificity. >> >> It is also not true that the R*T*ln(C/p) terms don't appear in the standard >> state. It's just that, as a matter of convention, we usually take the >> standard state of condensed phases to have a unit activity and of gaseous >> phases to be at 1 atm and so the logs are zero. That's only because we chose >> to compare pressures in units of atmospheres, though; you don't have to. >> >> >> _______________________________________________ >> fipy mailing list >> [email protected] >> http://www.ctcms.nist.gov/fipy >> [ NIST internal ONLY: https://email.nist.gov/mailman/listinfo/fipy ] > > _______________________________________________ > fipy mailing list > [email protected] > http://www.ctcms.nist.gov/fipy > [ NIST internal ONLY: https://email.nist.gov/mailman/listinfo/fipy ] _______________________________________________ fipy mailing list [email protected] http://www.ctcms.nist.gov/fipy [ NIST internal ONLY: https://email.nist.gov/mailman/listinfo/fipy ]
