I am struggling to understand ternary phase diagrams for reactor engineering. The course slides offer little to no help so I'm hoping to find an answer to this question

How does the slope of the tie-lines affect on which side of the equilibrium line the extract/raffinate is?

From what I understand is that if the lines slope down, the extract is on the right and the raffinate is on the left. If they slope up, the sides are flipped. Is this correct?

Sidenote: does anyone know a good resource where ternary diagrams for chemical/reactor engineering are well explained?

  • $\begingroup$ I can't really comment on the extract/raffinate aspects - that is more chemical engineering that I never did. However, in metallurgical phase diagrams (binary, ternary, ...) the only point is that the line (plane in ternary) touches the axes at the free energy value that corresponds to the equilibrium value. BUT - you are free to reference that free energy to anything you want to. Using absolute (SGTE), relative phases, or whatever, you will end up with the exact same diagram, but different tie line slopes. $\endgroup$
    – Jon Custer
    Commented Apr 3, 2015 at 13:38

1 Answer 1


It has been 7 years for this question, but it may help others with the same doubt.

First of all, ternary phase diagrams are not related to reactor design, they are related to the field of liquid-liquid extraction. This is, we have a liquid flow of two components, but only one is of interest, so we try to obtain an exiting flow that is enriched in the component of interest. We do this by adding a third component, the solvent, with the hope that the resulting liquid-liquid equilibria behaviour will help us.

The tie-line is the line (it may be a curve) that represents the equilibrium compositions of the two liquid phases, that given enough time, will be formed by an an initial solution that lies along the tie line. The nature of the tie-line solely depends on the behaviour of the equilibrium established by the three componenents. Its "slope" is a result of this behaviour, and it has nothing to do with respect to on which side you are going to end.

Lets give an example. Say we have a mixture of components $A$ and $C$, but we want $A$. Mixing both with $B$, the solvent, results in the emergence of liquid-liquid equilibria, delimitated by the following dome:

example image

If we mix an adequate molar flow of $F$ (containing $A$ and $C$) and of $S$ (containing the three components, but very rich in the extracting solvent $B$), the result is the mixture $M_1$. Given enough time, we will end up with two molar flows, in equilibrium with each other and of the following characteristics:

  1. An extract $E_1$, that we generally don't care, with a high composition of $B$ and then of $C$.
  2. A raffinate $R_1$, with a high composition of $A$, that we really do care.

The slope of the tie-line depends, as I said, in the equilibrium established. I will on purpose increase the slope of that tie-line:

example image

All the tie-lines should change in reality, but for illustrating purposes, I just changed the one that we are interested in. As we can see, the equilibrium is much more favorable to us now. The raffinate $ R_1 $ has a much more pure composition in $A$ than before. Did the slope send $ A $ to the right and $ B$ to the left? No, they stay where they were, they don't get flipped.

For more details in LL extraction, you can check Equilibrium-Stage Separation Operations in Chemical Engineer by Henley & Seader. However, Mass Transfer by Treybal is also fine, and a classic.


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