Hunter-Nash Method: Liquid-Liquid Extraction

[@HellO-ly1bp]

oh my god i actually have a shot at passing my exam now thank you, you freakin life savers!

[@LearnChemE]

This screencast has been reviewed by faculty from other academic institutions.

[@xodn3300]

this is an immense help to my studies for the final. thank you.

[@LearnChemE]

Very welcome! Hope the videos help.

[@1man1sam]

Excellent video, but how do you know that the extract will be close to the solvent?

[@ghulamhabib948]

How do u plot tie line and curve inside the ternary diagram .kindly show it

[@jamesfx7227]

Can you explain me about 4:52 values? I did not understand where they came from. Thank you.

[@LearnChemE]

Thanks for your question. If you watch starting at 4:35, you'll see that a line is drawn between S and F, and a second line is drawn between E1 and RN. The intersection of these two lines is the mixing point, and it is marked with a red dot on the diagram. The numbers you see at 4:52 are the compositions that correspond to this point as read from the graph. Hope this helps!

[@jesussiller01]

Does the line R_N, M *always* intersect with E1 at the binodal curve? Following the S,F line, it seems to me that under arbitrary conditions (i.e. high IPA and high ether for the mixing point) the line R_N, M would lead to a point in the "raffinate side".

[@lamiabouzekri2074]

comment avoir les composition de Rn?

[@HoangNguyen-bj9qk]

I think, in order to perform extraction, the added solvent must completely dissolve the components to be separated right? It means that point R must be lower than point E, but on your diagram the point R is higher than E. Can you explain it to me? Thanks you P/s: Sorry, my English is not good

[@LearnChemE]

in response to KalifatS: The problem statement says that it is a cascade example.

[@alifhaikal6392]

how did you get 1.48? we dont know the value of S and F. We probably need to assume basis for F right but how to find S?

[@jahanzaibiqbal3211]

Use scale for measurement of line in cm🌸

[@sebrinaodaro1119]

Thanks so much for this wonderful explanation.

[@ahmadfadli1979]

thanks this video help me for my exam.

[@wanhamizan669]

may i ask, i want to do a liquid-liquid extraction to extract acetic acid from it's aquoues solution into ethyl acetate, i want to use 20 wt% of acetic acid to be extracted into ethyl acetate, my question is, how do i determine what is the concentration of my organic solvent to be used for the solvent extraction?

[@vitorhm1987]

Perfect! Thanks for teaching us

[@정하영-f1k]

i really appreciate your incredible explain! thank you

[@paulalima1769]

I have to make a program (python) to simulate a liquid-liquid extractor, however to calculate the number of stages, i can't use the graphical method on python, but adopt a transient system. Do you have any ideas of how can I do this? Thanks!

[@denizkurt517]

this is just great thank you for your effort sir !

[@LLHurt22]

I don't understand.. how find the R1?

[@LLHurt22]

Lídia Hurtado with Alders really???

[@amazingdeen6734]

How you found the value for R1, R2 and so on in the last phase of the video?

[@applecider3616]

Hello! I currently am cramming liquid-liquid extraction right now so do correct me if I am wrong. I believe you could do an overall material balance on stage 1 but wellllllll the problem didn't have flow rates so E2 is pretty much unknown I guess ? The internet is a haven for knowledge so you could search the experimental values of the tie line for E1. I do hope you passed your exam tho

[@Feanic]

Could you also explain how to find Pmin?

[@kalishalewa]

is this a counter current or cascade example?

[@RedDovah458]

Seems to be counter current if you look at the diagram from around 1:45

[@loveandcrychizita]

what are the equilibriu date`s isopropyl alcohol - eter?

[@LearnChemE]

Not sure what you are asking, but the equilibrium points form the equilibrium two phase envelope and contain all three components. Any point on the envelope or in the envelope form two phases with compositions that can be read using equilibrium tie lines.

[@zheli6676]

very clear and helpful! Thanks!

[@jesussiller01]

To expand my point: imagine this same problem. You are not given information about E1 but you have enouh information about the feed and the solvent to obtain M. Under certain circumstances (little water, high concentrations of ether and IPA) M could shift to the right side of the area where 2 phases are formed, and the line R_N, M would connect to a point in the "raffinate" side. What happens in that case? I think this image explains my point: i.imgur.com/WmJdWjE.jpg?1

[@tamaraandreatitovelarde9458]

What if the composition of the extract is not given ?

[@felixsubakti6907]

To be honest that's asking for trouble Or a chromatographic assessment who am I to judge

[@jmacmunn]

I can't seem to find the derivation video.

[@LearnChemE]

Sorry about that. It is linked now.

[@fahadalmuhandes1123]

From Iraq thank you so much

[@amalessid8278]

R1 how ?? THANKS

[@npepe097]

Hello man, I just have a correction, it's not 1.48, it's 1.35, but everything is great, greeting!

[@LearnChemE]

Thanks for the note! May I ask how you got the 1.35?

[@npepe097]

LearnChemE When you already know all the compositions of the mixture (for isopropyl alcohol in this case) you can use (S/F)=(XF-XM)/(XM-XS). This is (S/F)=(40-17)/(17-0) (S/F)=1.35

[@wBjw]

Thank you! nice work

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Thanks!