## Driving force for stoichiometric phase

issues about thermodynamics and their coupling to MICRESS
zhubq
Posts: 84
Joined: Mon Jun 22, 2009 7:33 pm

### Driving force for stoichiometric phase

Hi, Bernd
I have a problem that i am not sure about the answer. For stoichiometric phase, is it assumed that the concentration of solute is constant, i.e. both within the interface and inside the phase? So the extropolated driving force is DG=DS_ab*DT, ringht?

Bernd
Posts: 997
Joined: Mon Jun 23, 2008 9:29 pm

### Re: Driving force for stoichiometric phase

Hi zhubq,

Your question is a bit unprecise. The driving force is a property of an interface between two phases - I guess you mean one phase which is stoichiometric for all alloying elements which is in contact to another phase which is not stoichiometric...
In general, whether stoichiometric or not, the composition dependency of the driving force is determined by the compositions in both phases inside the interface. As we assume local (quasi-)equilibrium, there should be no difference if you use the composition in one or the other phase for the calculation. In MICRESS, for pure numerical and symmetry reasons, we take the average of both:

+ +

But in case that for example is stoichiometric, the corresponding slopes would be infinite and the concentration differences 0, so the terms for this phase cannot be calculated numerically. In those cases, only the term of phase are used without factor 1/2:

+

If both phases are stoichiometric for a given cNew publication on MICRESS phase-field modeomponent k, no driving force can be calculated and the phase interaction is not allowed.

PS: This thread was split from its original position in the thread New publication on MICRESS phase-field model because it seems not to be thematically linked.

zhubq
Posts: 84
Joined: Mon Jun 22, 2009 7:33 pm

### Re: Driving force for stoichiometric phase

Hi, Bernd.

I am thinking how to partition the solute between two phases e.g. ferrite and carbide? I think Eq. (68) cannot not be used
since Kab is infinite or zero. Instead, after get total C and because we know C_theta., we can get the concentration in ferrite phase by C_a=(C-C_theta*phi_theta)/phi_a. But what about three coexisting phases, e.g. ferrite, austenite and carbide?
I suppose we use {C-C_theta*phi_theta} as the total concentration to partition only between ferrite and austenite[Eq. (68)].

Bernd
Posts: 997
Joined: Mon Jun 23, 2008 9:29 pm

### Re: Driving force for stoichiometric phase

Dear zhubq,

is is much easier than you say - if you just rearrange phases in Eqn. 68 (Phys. Ref. E 73, 066122 (2006)) such that

for stoichiometric phases, then everything is correct, even for multiple phases. The trick is just that alpha should not be the stoichiometric phase!

If all phases are stoichiometric for the same component, then partitioning is impossible - of course.

Bernd

zhubq
Posts: 84
Joined: Mon Jun 22, 2009 7:33 pm

### Re: Driving force for stoichiometric phase

Hi, Bernd.
I am confused by the derivation of the partition coefficient K_ab.
I am suspecting the validity of Equation (59)!

Eq. (58) is contradictory to Eq.(59).
In Eq. (59)
I think it should be

rather than

Then it should be like
(look at eq.(67))

If you compare Eq. (43) with Eq. (53). The expression for K is also confusing.

Could you please clarify this?

Thank you.

Ben

Bernd
Posts: 997
Joined: Mon Jun 23, 2008 9:29 pm

### Re: Driving force for stoichiometric phase

Dear zubq,

you are right, obviously there is some confusion in there!
I think that in eqns. (53) and (59) the definition of K is wrong- I will discuss this with Janin as soon as she is back to her office!

Thanks for the comment!

Bernd

Bernd
Posts: 997
Joined: Mon Jun 23, 2008 9:29 pm

### Re: Driving force for stoichiometric phase

Dear zubq,

I discussed the issue with Janin, and she agreed that the right part of equation (53) as well as eqn (59) are wrong, i.e. the phase subindices are reversed!

We are sorry for that!

Bernd