In this video we are going to look at a liquid-liquid extraction (LLE) problem for a multi-stage cascade. And to solve this problem we are

going to use the Hunter-Nash method. We won’t derive this method in this video we are just

going to demonstrate how it is used to solve these liquid-liquid extraction separations. Let’s take a look at the following problem in which we have a feed, 40 weight percent IPA, 55 weight percent ether,

5 weight percent water, that is mixed with pure water as the solvent to recover the ether.

Now we have our phase diagram shown here as a ternary phase plot. And we have collected

the data from a source and we have plotted that here with appropriate tie lines. The

goal is to produce a raffinate that has less than 3 wt% alcohol and an extract phase that

has at least 20 wt% alcohol. We are given a temperature and pressure, and we’re asked to calculate how many stages are required for this extraction. So let’s write down what we know, what we

are looking for, and a picture of our system. So we are given in this problem, three compositions

but no flow rates. So we are going to designate our compositions in mass fractions. We are

going to say xa is for our alcohol, xe for ether and xw for our water. We are not told

what the other compositions are, but we do know in this system that for our raffinate

and extract phase leaving, we are assuming that in each stage we reach equilibrium. So these

values right here will be on our equilibrium curve. I have drawn a schematic here to represent

our stages, our feed enters at stage 1 and then we will go into stage 2 and what leaves

stage 1 will be our raffinate from 1. We don’t know how many stages we have so we will have

little n and our final big N for the number of stages. So we know our raffinate leaving

will be from the last stage, so for our countercurrent system we are going to enter the solvent in here,

we will say the solvent is S. So we have our extract leaving from stage N, our extract leaving

stage 2 and our final extract leaving stage 1. So we are going to use this nomenclature

as we move forward in solving our problem. We are not given flow rates which means that

we are probably going to have to assume a basis. We are also going to have to assume

that a system operates isothermally, that any enthalpy associated with mixing is negligible.

Since we are given our composition of our raffinate and our extract, using the Hunter-Nash method, we need to determine our solvent flow rate. Once we have S we’re going to plot these points on our ternary

diagram. We are going to plot our feed composition, our solvent, our extract, our raffinate leaving,

and this all helps us calculate what is known as our mixture point. So let’s start with

step one and draw these compositions on our chart. Since we’re using a pure solvent this

is going to be all water. We could plot our feed by looking at 40 wt% alcohol and that

is going to fall somewhere along this line and then we could choose either of the other

two points. So in this case I am going to choose ether, we are going to go to 55% which

is right in between here. Where these two intersect gives us the composition of our feed. Now

our solvent is pretty easy. We are going to use 100% water so we are just going to draw

a point over here. Now our extract and raffinate phases have to be at equilibrium- not with

each other but leaving the appropriate stages. So our extract has alcohol of 20% so we know

that is going to be closer to our solvent side. We are going to look for where 20%, for

our alcohol, is on our equilibrium curve- that’s right here. We are going to do the same thing

for our Raffinate on the other side. We are looking for about 3%, somewhere around right

here. So we have our four points. Now we could calculate the mixture point if we had flow

rates for S and F. We know our feed enters our system with our solvent. So these two

combine to create our mixture. This mixture also must be equal to what leaves our system

which is our extract and our raffinate. So we know that these three are on a line and

these three are on a line. And we get that by using a material balance. And the derivation of why these points all fall on the same line is shown in another video. So

what we are going to do is draw a line from S to F and another line from our extract to

our raffinate. And where these intersect is our mixture point. So now we could read the

composition of our mixture point and use this in our material balances to solve for a ratio

of S to F. And so this would give us an idea of how much solvent we would need per amount

of feed to achieve this separation. Now that was not asked in the problem statement so

we don’t necessarily need to go down that road. But, just to point out, the one thing we can do is use the inverse lever rule as well since we know that we could write S/F equal to the length from M to F

over the length from M to S. And this gives us about 1.48. So we would need almost 1 and

a half times more solvent than our feed to achieve the given compositions that we are

proposing. Now we could also use the inverse lever rule for our extract and raffinate since we can then calculate the ratio between those two and using what we know from our feed and solvent,

forming our material balances gives us an idea of the absolute amounts of these two

phases. So step 2 in this process is to determine our operating point. We are going to say it

is called P here, and our operating lines. So what we are looking for is a point P, so

point P is a solution to the material balances and we know that these must be on a line as

well as those points. So we could draw these two lines going through our points. So we

will start with our line that goes from F through E1. We will draw another line that

goes between Rn and S. Where these intersect is going to be our point P. So the last step,

step 3 is to use our tie lines and equilibrium lines to determine the amount of stages that

we need. So we have our operating point. And we know that each line that goes through this

line is going to include the passing streams. And what I mean by that, let’s take a look

again at our schematic. We know the difference between this stream and this stream is the same as the difference between this stream and this stream and that difference is P. So that also means

that E2 and R1 fall on a line with P and EN and RN-1would fall on a line with P. So the

first thing we do from stage 1 at E1 is to follow the tie lines since we know that at

stage one we are going to get equilibrium. So we follow that tie line to the other side.

So I draw a point on the equilibrium curve on the other side for our raffinate and I

will label that R1. So we know that R1 and E2 are passing streams that fall on a line

with P. So we will draw a line from P to R1 and where it intersects on our equilibrium

curve is going to be where E2 is. So then from here we repeat the process of following the

tie line from E2 to the other side where we think R2 is going to be. Now I had to interpolate here, so that’s probably my best guess without doing our interpolation method to draw more tie lines. So again we know R2 and E3 fall on the same line with P so we draw a line to R2 and we could figure out

where E3 is. We are going to get rid of these lines just to make it a little easier to see,

again we follow the tie line from E3 and this time we get to our RN. So based on this we

have drawn three tie lines, or three equilibrium lines for our stages, and so we would have three

stages. Now it is possible you would want to zoom in on this, be a little bit more precise.

The first time I did this I actually got around 2.8 so you would have a fractional stage, so it is important to try to do your best to draw appropriate tie lines and not guess. Hopefully this gives

you a better idea of how to use the Hunter-Nash method to solve for parameters in a liquid-liquid extraction separation.

Very welcome! Hope the videos help.

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

Could you also explain how to find Pmin?

God Bless you son

<3

thanks!

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

Thanks!

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

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

like >>> 1000 like

I can't seem to find the derivation video.

is this a counter current or cascade example?

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

Does the line R_N, M

alwaysintersect 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".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: http://i.imgur.com/WmJdWjE.jpg?1

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

Thank you! nice work

this is just great thank you for your effort sir !

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

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!

how to determine new V minimum? which is finding new M

very clear and helpful! Thanks!

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

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?

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?

بفرونسي

i really appreciate your incredible explain! thank you

From Iraq thank you so much

R1 how ?? THANKS

What if the composition of the extract is not given ?

Thank you.

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

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