PF simulation of Eutectic solidification
Re: PF simulation of Eutectic solidification
Dear Omid,
To be honest, I do not understand how you got the linearisation data from your sketch. At least, there are several inconsistencies:
 How did you get the solidus slopes? Why do you use so low values? Then, the solidus and liquidus will cross, and everything will be inpredictable!
 Why do you use the same reference compositions for phase 2 and phase 4 with liquid, respectively? This leads to no segregation, and I cannot see why you do that...
 You forgot to define phase 3 as stoichiometric (I guess from the solidus slopes).
My suggestion: As the solid compositions of all phases do not depend strongly on temperature, I would set all 4 phases as completely stoichiometric. Then you do not have to worry about solidus slopes.
PS: Composition limits do not have any effect if you do not use TQ coupling.
Bernd
To be honest, I do not understand how you got the linearisation data from your sketch. At least, there are several inconsistencies:
 How did you get the solidus slopes? Why do you use so low values? Then, the solidus and liquidus will cross, and everything will be inpredictable!
 Why do you use the same reference compositions for phase 2 and phase 4 with liquid, respectively? This leads to no segregation, and I cannot see why you do that...
 You forgot to define phase 3 as stoichiometric (I guess from the solidus slopes).
My suggestion: As the solid compositions of all phases do not depend strongly on temperature, I would set all 4 phases as completely stoichiometric. Then you do not have to worry about solidus slopes.
PS: Composition limits do not have any effect if you do not use TQ coupling.
Bernd
Re: PF simulation of Eutectic solidification
Dear Bernd,
Using the same value was due to collide of the liquidus surfaces at the adjusted point! Firstly I have defined all 4 phases as stoichiometric but since the phases 2 and 4 had a solubility range I switched them off! I just forgot to mention that those slopes were a result of playing with the numbers to find out the correlation! But I confirm that you are right I'm still idiot in linearisation job! Can I ask you for a favor? Just same as the previous guideline on a paper extract the data for i.e phase 0/2
Regards,
Omid.
Using the same value was due to collide of the liquidus surfaces at the adjusted point! Firstly I have defined all 4 phases as stoichiometric but since the phases 2 and 4 had a solubility range I switched them off! I just forgot to mention that those slopes were a result of playing with the numbers to find out the correlation! But I confirm that you are right I'm still idiot in linearisation job! Can I ask you for a favor? Just same as the previous guideline on a paper extract the data for i.e phase 0/2
Regards,
Omid.
Re: PF simulation of Eutectic solidification
Dear Omid,
The problem is that your liquidus sections do not provide any information which can be directly used for constructing linearized phase diagrams. The intersections of the liquidus planes do not have any meaning for the construction of a twophase linearisation, because you are interested in the interaction of each phase with liquid.
The only information which can be directly used is the eutectic composition (at the eutectic temperature) which gives you reference compositions of the three eutectic phases and of liquid at the eutectic temperature.
I don't know how the reference compositions for phases 0/4 can be determined.
If you do not assume all solid phases as stoichiometric, I believe, you have no chance. If you do, you still need to estimate the liquidus slopes and the reference composition of liquid in the 0/4 interface at the eutectic temperature. The liquidus slopes can be estimated by taking the individual liquidus planes and taking the derivative in the directions of the dissolved elements. For the reference composition of liquid in the 0/4 interface I don't know...
If you have access to Pandat and the database, of course, you could do calculations of the liquidus temperature for each solid phase at a given point (liquid composition) with variation of the liquid composition. This would allow you to get correct liquidus slopes at least at these points for each phase pair.
Bernd
The problem is that your liquidus sections do not provide any information which can be directly used for constructing linearized phase diagrams. The intersections of the liquidus planes do not have any meaning for the construction of a twophase linearisation, because you are interested in the interaction of each phase with liquid.
The only information which can be directly used is the eutectic composition (at the eutectic temperature) which gives you reference compositions of the three eutectic phases and of liquid at the eutectic temperature.
I don't know how the reference compositions for phases 0/4 can be determined.
If you do not assume all solid phases as stoichiometric, I believe, you have no chance. If you do, you still need to estimate the liquidus slopes and the reference composition of liquid in the 0/4 interface at the eutectic temperature. The liquidus slopes can be estimated by taking the individual liquidus planes and taking the derivative in the directions of the dissolved elements. For the reference composition of liquid in the 0/4 interface I don't know...
If you have access to Pandat and the database, of course, you could do calculations of the liquidus temperature for each solid phase at a given point (liquid composition) with variation of the liquid composition. This would allow you to get correct liquidus slopes at least at these points for each phase pair.
Bernd
Re: PF simulation of Eutectic solidification
Dear Bernd,
There is no chance to get database, whoever has it, does not want to share it or is not allowed to share it!
The data which I sent is not useful?
For example for the phase 2 and 0: we have the junction points to liquid and solid, points a and g! And we have them at two different temperatures. Could we say that the reference temperature is i.e 2000 centigrade and reference point 1 is " a " and reference point 2 is " g " and we have the corresponding points at 2010 and consequently we can derive dT/dC!
Looks possible ?!
There is no chance to get database, whoever has it, does not want to share it or is not allowed to share it!
The data which I sent is not useful?
For example for the phase 2 and 0: we have the junction points to liquid and solid, points a and g! And we have them at two different temperatures. Could we say that the reference temperature is i.e 2000 centigrade and reference point 1 is " a " and reference point 2 is " g " and we have the corresponding points at 2010 and consequently we can derive dT/dC!
Looks possible ?!
Re: PF simulation of Eutectic solidification
Dear Omid,
can you explain which are the points "a" and "g"? Do you refer to some earlier diagrams?
Bernd
can you explain which are the points "a" and "g"? Do you refer to some earlier diagrams?
Bernd
Re: PF simulation of Eutectic solidification
Dear Bernd,
Sorry I forgot to name the points! I shown what I mean by A and G in the attached file!
Omid.
Sorry I forgot to name the points! I shown what I mean by A and G in the attached file!
Omid.
 Attachments

 diagram.pdf
 (248.51 KiB) Downloaded 105 times
Re: PF simulation of Eutectic solidification
Dear Omid,
yes, this is ok, but you need to calculate the derivative dT/dc for each dissolved element, i.e. you need to pick respective points on the liquidus intersection 0/2 for two temperatures which have an identical composition for one element. This is possible. And you can chose whichever point of the intersection as reference point (A is just one of the possible points) as long as all reference points are chosen at the same temperature.
Mathematically, you need 3 points for describing a liquidus plane. Thus, your liquidus surface with all derivatives and reference points can be perfectly determined. The critical point here is just that we do not know how the assumption of planarity (linearised phase diagram) affects your results, and where the reference points and slopes should best be determined to give you the best results.
For the solidus surfaces, you do not have sufficient information, even if you know point G at different temperatures: With 2 points you cannot construct a plane. Therfore, there is no other chance than to assume stoichiometric compositions for all 4 solid phases.
yes, this is ok, but you need to calculate the derivative dT/dc for each dissolved element, i.e. you need to pick respective points on the liquidus intersection 0/2 for two temperatures which have an identical composition for one element. This is possible. And you can chose whichever point of the intersection as reference point (A is just one of the possible points) as long as all reference points are chosen at the same temperature.
Mathematically, you need 3 points for describing a liquidus plane. Thus, your liquidus surface with all derivatives and reference points can be perfectly determined. The critical point here is just that we do not know how the assumption of planarity (linearised phase diagram) affects your results, and where the reference points and slopes should best be determined to give you the best results.
For the solidus surfaces, you do not have sufficient information, even if you know point G at different temperatures: With 2 points you cannot construct a plane. Therfore, there is no other chance than to assume stoichiometric compositions for all 4 solid phases.
Re: PF simulation of Eutectic solidification
Dear Bernd,
How can I tell the dri file that the phase 4 does not exist in the eutectic reaction?
Eutectic reaction occurs at 1991 and three solid phases meet each other there. and there is a peritectic reaction at 1993 and terminates at the specific composition, which leads to phase 4 but essentially when the liquid composition is adjusted to the eutectic composition and the top wall is fixed boundary condition to the Eutectic composition there should not be any nucleation of 4th phase! I want to have this phase in the phase diagram definition, so that it would be able to predict primary 4th phase when the composition of the liquid is not exactly the eutectic point but I have no idea how to tell to the setup to forget about this phase until the boundary and initial conditions are supposing so!
Omid.
How can I tell the dri file that the phase 4 does not exist in the eutectic reaction?
Eutectic reaction occurs at 1991 and three solid phases meet each other there. and there is a peritectic reaction at 1993 and terminates at the specific composition, which leads to phase 4 but essentially when the liquid composition is adjusted to the eutectic composition and the top wall is fixed boundary condition to the Eutectic composition there should not be any nucleation of 4th phase! I want to have this phase in the phase diagram definition, so that it would be able to predict primary 4th phase when the composition of the liquid is not exactly the eutectic point but I have no idea how to tell to the setup to forget about this phase until the boundary and initial conditions are supposing so!
Omid.
Re: PF simulation of Eutectic solidification
Dear Omid,
Let me first give you additional information how to calculate the liquidus slope at the example of the liquid/P2 interface above. I created a sketch which is attached.
In order to get the liquidus slope for this interface, based on the eutectic point E at 1991°C (which is part of the liquidus surface) and the intersection of the liquidus surface at 2000°C (line between A and A^{'}), one has to draw isocomposition lines through the eutectic point E along the direction c(B)=const. (red dashed line) and c(A)=constant (green dashed line). Now you can calculate the slopes by evaluation of the coordinates of the intersects of these dashed lines with the liquidus line (AA^{'}):
dT/dc(A)= (2000°C1991°C)/(c^{E}(A)c^{A*}(A))
dT/dc(B)= (2000°C1991°C)/(c^{E}(B)c^{A+}(B))
Point E could be used as reference point for the liquidus plane and G for the (stoichiometric, i.e. infinitely steep) solidus plane.
For phase 4, the procedure must be different because E is not part of the liquidus surface 0/4. Therefore, a third temperature (e.g. 2010°C) must be included. One point could be chosen on the intersection of 0/4 at 2000°C, and A^{*} and A^{+} can be constructed on the 2010°C line in the same way as shown above. Then, as reference point at 1991°C, one would have to backextrapolate to this temperature (all reference points have to be at the same temperature).
In order not to form phase 4 at the eutectic composition, the liquidus slopes for interface 0/4 has to be so steep that it lies below the eutectic temperature at this point. If not, formation of P4 cannot be avoided (which means that the phase diagram would not be eutectic and badly constructed). And this is also the answer to your latest question: You can tell the .dri file not to form phase 4 by providing a correct phase diagram description  whether this is possible at all with a single linearized phase diagram description you will have to find out...
Bernd
Let me first give you additional information how to calculate the liquidus slope at the example of the liquid/P2 interface above. I created a sketch which is attached.
In order to get the liquidus slope for this interface, based on the eutectic point E at 1991°C (which is part of the liquidus surface) and the intersection of the liquidus surface at 2000°C (line between A and A^{'}), one has to draw isocomposition lines through the eutectic point E along the direction c(B)=const. (red dashed line) and c(A)=constant (green dashed line). Now you can calculate the slopes by evaluation of the coordinates of the intersects of these dashed lines with the liquidus line (AA^{'}):
dT/dc(A)= (2000°C1991°C)/(c^{E}(A)c^{A*}(A))
dT/dc(B)= (2000°C1991°C)/(c^{E}(B)c^{A+}(B))
Point E could be used as reference point for the liquidus plane and G for the (stoichiometric, i.e. infinitely steep) solidus plane.
For phase 4, the procedure must be different because E is not part of the liquidus surface 0/4. Therefore, a third temperature (e.g. 2010°C) must be included. One point could be chosen on the intersection of 0/4 at 2000°C, and A^{*} and A^{+} can be constructed on the 2010°C line in the same way as shown above. Then, as reference point at 1991°C, one would have to backextrapolate to this temperature (all reference points have to be at the same temperature).
In order not to form phase 4 at the eutectic composition, the liquidus slopes for interface 0/4 has to be so steep that it lies below the eutectic temperature at this point. If not, formation of P4 cannot be avoided (which means that the phase diagram would not be eutectic and badly constructed). And this is also the answer to your latest question: You can tell the .dri file not to form phase 4 by providing a correct phase diagram description  whether this is possible at all with a single linearized phase diagram description you will have to find out...
Bernd
 Attachments

 WP_20160602_001.jpg (112.83 KiB) Viewed 1064 times
Re: PF simulation of Eutectic solidification
Dear Bernd,
Thanks for the helpful guides. You gave me good glasses
I still have a problem for the linearization. for the phase 3/0 interaction in which the corresponding line is AB in the attached photo, if we draw the parallel lines to Boron and silicon fraction lines, it does not cut the line! what should I do in this case? There are two possibilities for me, first instead of Si and B, choosing i.e Mo and B, which cut the AB line and then extract the Si out of these data,
Or draw the line till it cuts the line at an imaginary point "I" and applying the specifications of I in the phase diagram definition.
Which one or any other alternative looks wisely?
Omid.
Thanks for the helpful guides. You gave me good glasses
I still have a problem for the linearization. for the phase 3/0 interaction in which the corresponding line is AB in the attached photo, if we draw the parallel lines to Boron and silicon fraction lines, it does not cut the line! what should I do in this case? There are two possibilities for me, first instead of Si and B, choosing i.e Mo and B, which cut the AB line and then extract the Si out of these data,
Or draw the line till it cuts the line at an imaginary point "I" and applying the specifications of I in the phase diagram definition.
Which one or any other alternative looks wisely?
Omid.
 Attachments

 New Microsoft WordDokument.pdf
 (194.31 KiB) Downloaded 100 times