3 replies to this topic

[alfirin]

Hi everyone,

I have the following problem in one of my assignments and although I have studied some literature I still fail to understand how the right triangular diagram works.

In the diagram attached, A is the solute, B is the Diluent and S is the solvent. We have a 100kg A-B mixture with compositions at point Xf =0.29 to be extracted by 80kg pure solvent S in a single process. Question asks to identify the mass fractions of A in the extract and raffinate and the total masses of the extract and the raffinate.

any help will be appreciated!

Attached Files

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[USR]

alfirin,
Why don't you tell us what are the specific issues about the rectangular coordinate diagram that is bothering you?
But inspite of that i'll give you a few hints about how to approach the problem:
What you show us is actually a rectangular coordinate diagram with the wt fraction of the solute ploted along the vertical line and the weight fraction of solvent along the horizontal line.
Pure solvent will be located at y=0,i hope you are aware of the terminologies we use in liquid liquid extraction.This is the point where weight fraction of solvent is 1 i.e. at point S.
Next you do a material balance to calculate the wt fraction of solute in the mixture.This will be located inside your solubility curve on the line joining xf and y=0 line.
You have been given the tie lines from which you can generate the distribution curve which will help you locate x1 and y1 i.e. concentrations of solute in extract and raffinate phases once they reach equillibrium.This calls for a trial and error.
Please show us till what stage have you solved your assignment question?Good luck solving..

[alfirin]

USR:

Thanks a lot for the quick answer and the hints.

I collected a couple of books from the library and I'm going to work on it during the weekend. I will post my progress here in the meantime.

Cheers

[alfirin]

It took me a while to figure out how right triangular diagrams work, but I managed to solve the question. I found point M and then I tried to do trial-and-error to calculate E and R. The problem is that I didn't work good on it and in the end the raffinate and extract masses added up to 186kg instead of 180kg.

I hope this won't be a major reduction in my mark!