This screencast is going to go over a phase diagram for a ternary system. In the following example
we have a stream that is 50 % mass acetone and 50% by mass water. This is mixed with
a pure stream of methyl isobutyl ketone (MIBK). The mass feed rate of the acetone/water stream
is half that of the MIBK stream. We are asked to determine what the outlet mass of the water-rich
phase is per mass of water in the feed. With any material balance the first place we should
start with is a diagram. Here is my diagram where we have two feed streams. One which
is 50% by mass acetone and 50% by mass water. As well as a second stream that is 100% MIBK.
These are mixed in the tank and then they equilibrate and separate into two streams.
A water rich phase and an MIBK rich phase. You can see here that I have written out the
three unknown compositions since each phase will contain parts of each of those species. The only
other information that we are given that we should write here is that the mass feed rate
of the acetone/water stream so m1 is half of that of the MIBK stream. This can be confusing
the way to write this but just double check it. Basically we are saying that the MIBK
stream is twice the first stream. If that makes sense now we can write it the same way
where m1 would equal 1/2 m2. I just rewrote it so that it was 2 m1 equals m2. The last
thing to do with any diagram if it hasn't been given is to choose a basis. We are going
to go ahead and choose 1 g/s of acetone in the feed stream. It just keeps it simple that
way. Since the question is asking for the amount of water, per mass of water, I could
have written it that way too but because it is a 50/50 stream it is easy to see that if
you have 1 g/s of acetone and 1 g/s of water and that is in m1. That is all of the information
that we have been given. Now we go basically write our material balances. Our material
balance for the total is m1 plus m2 and that is what is going to go into the system. What
comes out of the system has to also be the total mass and we have written it as mW and
mM. I have in equals out. In this case m1 plus m2 equals mW plus mM. We have been given
a relationship for m2 to m1 so we will go ahead and rewrite that. If you recall m2 equaled
2 m1 so we have 3m1 equals mW plus mM. That is our total balance. Now we can write an
acetone or a water or a MIBK balance. I will go ahead and write a water balance since that
is what we are interested in here. We know that 50% of stream 1 is water and that is
going into the system. That has to equal some mass fraction for the water rich phase and
some mass fraction in the MIBK-rich phase. We will go ahead and put the unknown times mW plus some unknown
times the mass flow rate in the MIBK-rich phase. We first need to find where we lie on this
diagram going into the mixer. If we use the basis of 1 g/s of water and 1 g/s acetone
then we know there must be 4 g/s MIBK based on that relationship of 2:1. If we do this
we need to determine the mass fraction of each. As you can see the total is going to
be 6 g. So 1 g and 6 g is going to give us 0.167, 0.167, and 0.667. I will round up a
little bit on this diagram since we are reading it on a chart anyways and follow you through
this. Let's use a highlighter here. We are going for a mass fraction of around 17% acetone.
We are going up the chart so it is going to be between the 15 and the 20 so somewhere
in there. For the water we are going to the left corner so we start on the right side
and it is going to be somewhere between 15 and 20 so some where in here. Then 67% is
going to be somewhere between 65 and 70 which puts us somewhere right in here. That should
read 60. At this point you can see we are in the envelope so we are going to get phase
separation. We are going to go ahead and draw our tie line. If we follow over to the right
side we should be right there and if we follow down to the left side it looks like it is
about half way between those two tie lines so I will draw it right there. Now we have
our MIBK rich phase and our water rich phase. We can start filling those in. Since we are
doing our water balance that is what we are interested in. our water percentage looks
like it is between 85 and 90 so we will go ahead and say it is 87. This is going to be
0.87 and in our MIBK phase water looks like it is just less than 5 so we will say it is
4. Those don't have to add up to 1. Remember that. They are just the fractions in the two
streams of water. Now we have our balance there and our balance for the total. It looks
like we have 3 unknowns and 2 equations but we do have a basis for m1. We know it is 2
g/s. We will fill that in for m1. We have 2 equations and 2 unknowns so we can go ahead
and solve this. You can see I have simplified the water balance to get 1 equals 0.87 mW
plus 0.04 mM. The total simplifies to mW equals 6 minus mM. We plug the second equation into
the first and I will do that on the original page. You can see at the top right corner
I have rewritten the balance and simplified down to solve for the mass flow rate of both
MIBK and water. I get 5.08 g/s of the MIBK rich phase and 0.92 g/s of the water rich
phase. You have to ask yourself if this makes sense. Recall that we did have 4 g/s of MIBK
coming in and based on the equilibrium of the tie line we have very little MIBK in the
water rich phase. The majority of MIBK is going to be in that MIBK phase and we are
taking some of the acetone and water with it so it should be higher. You can double
check those answer but the question asks what is the outlet mass of the water rich phase
per mass of water in the feed. Since we used mass of water in the feed as a basis of 1
we are just going to say that it is 0.92 over 1. I will just highlight this answer since
that is what we are looking for. Hopefully this gives you an idea of using a ternary
phase diagram in conjunction with a material balance and solving what the equilibrium concentrations
are going to be in the two phases if there are two phases.