The purpose of this screencast is to show an example of setting up and solving material
balances on a multi-unit system where we have a liquid-liquid extraction column in series
with a distillation column. We'll touch on the ideas of fully labeling a process diagram,
degrees of freedom, and setting up a system of material balances. In this problem we have
acetic acid and water that is being fed to the process. These two species are difficult
to separate by distillation. So the purpose of the liquid-liquid extraction column is
to transfer acetic
acid from water into hexanol. And the acetic acid-hexanol mixture is then fed to the distillation column,
where the acetic acid can be separated. The schematic of this process is shown below.
However we don't have any information that is provided in the problem labeled on the flow
chart. At this point it is important to take a moment and look at what has been given in
the problem statement and see how we can apply this information to fully label the process
diagram below.
After reading the problem statement it's clear that we don't have any
flow rates that have been given. It's important that the total flow rate of every stream is
either known or labeled with an unknown variable. So I am going to go through the process of
labeling each of the six streams in this process. All the compositions that have been given
are in weight percent. So I've given a variable of m1 though m6 to represent the mass flow
rate of the six streams in the process. The easiest way to fully label the diagram is
to go line by line through the problem statement and apply the information that has been given
to the appropriate stream. For example we have some information about the feed to the
liquid-liquid extraction column. We know that we have a mixture of 18 weight percent acetic acid,
and the balance water that's being fed to the extraction vessel. So we can apply that information
to stream 1. Here we're given that the mass fraction of acetic acid is 0.18, and if the
balance is water, we know that the mass fractions need to sum up to one. So the mass fraction of
water must be 0.82. The problem states that the other feed to the liquid-liquid extraction column
is pure hexanol, which corresponds to stream 2. The water rich stream that's leaving the
extraction column is 99.5 weight percent water. Keep in mind it's the acetic acid and hexanol
that's going to the distillation column. So this is the stream m4, that's not being
fed to the distillation column. In this stream the mass fraction of water is 0.995, and we
know that hexanol and water do not mix. So without hexanol in the stream the mass fraction
of acetic acid must be 0.005 so that the mass fractions sum to one. The hexanol rich extract
leaving the extraction column is fed to the distillation column. Here it does not provide
any information about the composition so we're going to have to label m3 with an unknown
composition variable to represent that the mass fraction of hexanol and acetic acid.
So here I've labeled the mass fraction of hexanol arbitrarily with the variable x. And
again, we know that there is no water in this stream so the mass fraction of acetic acid
must be 1 minus x. We do have some information about the composition of both the distillate
and the bottom streams leaving the distillation column. We know the composition of the distillate
is 96 weight percent acetic acid and the balance hexanol. The bottom stream is 97.2 weight
percent hexanol and the balance acetic acid. So that information has been applied to m5
and m6 respectively. We have one more piece of information that has been given to us.
The problem states that 95% of the hexanol that's fed to the extraction vessel is recovered
in the bottoms stream. So that can't be applied directly to the information on any particular
stream, but we can represent that information mathematically. 95% of the feed should be
0.95 times m2 and that's equal to the flow rate of the hexanol in the bottoms stream,
which is 0.972 times m6. So at this point we've accounted for all the information that's
been provided in the problem statement. Let's take a look at what's being asked in the problem.
We need to calculate the percentage of acetic acid in the process feed that's recovered
in the distillate. So we want to know how much of the acetic acid going in is being
recovered in the distillate here. Before we start writing and solving equations kind of
haphazardly, the next thing to do is do a degree of freedom analysis on every system
in this particular process. That will illustrate to us the easiest way to proceed to calculate
what's being asked for. Alright, I've scrolled down a little bit to make some room do the
degrees of freedom analysis on this process. We can apply this analysis to every system
in the process. In this case, we have three. So we have the liquid-liquid extraction column,
we have the distillation column, and we have the overall process. So the feed to the process
and the output to the process. So we'll go through these one by one to see where we should
start setting up mass balances to solve for what's asked for. So let's start with the
liquid-liquid extraction column and count up how many unknowns we have. We have m1,
m2, m3, the composition variable x, as well as m4. So we have 5 unknowns. We can relate
these unknowns with material balances. However, the balances need to be independent and the
number of independent mass balances that we can write will always be equal to the number
of species that we have. We have A, we have W, water, and we also have hexanol. So we
have 3 species in this particular system. We can write 3 species balances that are independent.
Keep in mind we can also write a total balance, however, the total balance is going to be
the sum of each species balance. So it is not independent from the 3 species balances.
So in this case we have 3 independent mass balances that we can write. We also don't
have any other information that relates any of the variables in the extraction column.
So, without any other information, 5 minus 3 gives us 2 degrees of freedom. We're short
on equations that describe these variables, hopefully we can find another system where
we have more information specified. Next, let's take a look at the distillation column.
Again, we need to know how many unknowns we have. So the feed to the distillation column
is m3, and again we have the unknown composition variable x. We also have an unknown m5 and
an unknown m6. So in this case we have 4 total unknowns. The number of material balances
that we have or independent material balances is equal to the number of species. However,
in this case we only have two, we have hexanol and acetic acid. Water is not being introduced
to the distillation column, so in this case we only have 2 independent balances that we
can write. And again in this case, we don't have any other information that's relating
the unknowns. So with out any further information, we also have 2 degrees of freedom for the
distillation column. So let's take a look at the overall system. The system boundaries
for the two unit operations here, the extraction column and the distillation column, are pretty
straightforward. For the overall system, we are concerned with the inputs and the outputs
to the process. So to help us visualize the system boundary for the overall process, I'm
going to draw a line that intersects the inputs and the outputs to the system. So when we
do a degrees of freedom analysis on the overall system we are only concerned with the streams
that are intersecting the system boundary for the overall process. That includes the
two inputs to the extraction column, the two outputs to the distillation column, but it
also includes one of the outputs from the extraction column that does not go into the
distillation column. So let's see what variables we have. We have m1 going into the extraction
column. We have m2. We have m4 leaving the extraction column. We are not worried about
m3 because that is internal to the system. We are concerned with m5 and m6. So it looks
like we have 5 unknowns. For the overall system, we have 3 species in acetic acid, water, and
hexanol. So we have 3 independent balances we can write. We also have another piece of
information that relates the unknowns in another equation different from a mass balance. We
were given some information about how much of the hexanol going in ends up in the bottoms
stream. So we have one other equation that we can use. So we're left with one degree
of freedom. However, if you look at the flow chart, you'll notice that there is not one
flow rate that is specified. We always need to have at least one flow rate specified.
In other words, we need a basis. So if we assume a basis in one of the streams in the
overall process, we can go from one degree of freedom down to zero and start solving
for the unknowns. Which flow rate we assume for a basis is arbitrary, and I'm going to
use m2 and assume that we have 100. So you can see that if we assume a basis of 100 for
m2 that essentially eliminates one of the unknowns and therefore we'll have one less
degree of freedom and give us a zero degree of freedom system. I simplified our flow chart
here so that we can start calculating unknowns related to the overall system. Keep in mind
what we're trying to calculate, we want to know what the recovery of acetic acid is in
the distillate, which is here, relative to what goes into the process which is here.
So minimally, we need to calculate values for m1 and m5. We can do that using mass balances
and the other equation that has been given to us that relates the amount of hexanol in
the feed to that in the bottoms stream. So we wrote that mathematically using this equation.
Since we assumed a basis of 100 for m2, we can quickly calculate m6. So substituting
in 100 for m2, gives m6 is equal to 97.7. And here I'm not assigning units with these
flow rates, the units are really arbitrary for calculating a percentage recovery or a
ratio of any two flow rates. If we want to calculate values for m1 and m5, we're going
to need to relate them by material balances. We can write a material balance on the overall
mass. We can also write a species balance for the 3 species that we have. We'll assume
that we're operating at steady-state, so all of our material balances are going to have
the form of in equals out. If we apply that equation to the total balance, then we have
m1 plus m2 which we assumed was 100 is equal to m4 plus m5 plus m6. If we want to write
the balance on acetic acid, on the inlet side we only have acetic acid in one of the inlets
to the extraction column. So we have a mass fraction of 0.18 times the total flow rate
of m1 and that's equal to the sum of the acetic acid in the 3 outlet streams. I'll quickly
write the balances for both water and hexanol as well. I also substituted the value for
m6 that we calculated into the acetic acid balance. With all the material balances written
out, we can come up with a strategy that is easiest to solve for the remaining unknowns.
So looking at this system of equations, we can see that our H balance here, just has
one unknown in m5. So we can calculate that first. And with m5 calculated, we can solve
any two of the three remaining balances simultaneously. So I'll pick the acetic acid and the water
balance. So solving the H balance first to solve for m5, gives a value of m5 is equal
to 125. Once we known m5, we can calculate m1 and m4 by solving acetic acid and the water
balance simultaneously. So I won't go through that algebra but will just provide the answers.
With all of our remaining unknowns calculated from the mass balances, we can finally calculate
what we ultimately want, which is the recovery of acetic acid in the distillate. So mathematically
we can write that the percent recovery is equal to the concentration in the distillate
times the flow rate which was m5 divided by what went in to the feed. The composition
that was given originally was 0.18 and our flow rate for m1 is 697.9. So if we calculate
that out, the recovery is 0.956 or in percentile, we have 95.6% of the acetic acid is recovered
in the distillate.