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.