In this screencast we will show you how to use a degree of freedom analysis to make it
easier to solve material balances on a multiple unit process. So I am going to read the problem
statement and fill in our known variables on the schematic shown below as I go through
it. So we have a feed containing equimolar amounts of methanol and water it is mixed
with 10 moles of 40 mole percent aqueous methanol steam. Okay so our equimolar feed of methanol,
I will put 0.5 M and 0.5 W, is mixed with 10 moles. So we have a flow rate of 10 moles
that's 40 mole percent methanol. So 0.4 methanol and 0.6 water.
Now this mixture enters a separation
unit that creates two streams. A top stream exits that contains 70 mole percent methanol
and the rest water. So here we have 0.7 methanol and 0.3 water. The other stream, which is
70 moles, enters a second separation unit. So here we have 70 moles that is leaving unit
one and entering unit two. Now a top stream exits the second unit as a 50 percent methanol, 50
percent water mixture. So again we have 0.5 methanol and 0.5 water. The other stream is
unknown. Now if the fresh feed to the system is 100 moles. So 100 moles entering our system,
and the two top streams exiting the separation units have the same flow rate, we want to
find the molar flow rate and composition of the other stream exiting the second separation
unit. So we are trying to determine the composition and flow rate of this bottom stream from
unit two. So here is our schematic with our known variables for the flow rates or compositions.
So now we should label our unknowns. So the first stream entering the system has a flow
rate and composition, so we're good there. Our second stream, which now has a composition
that we had before and the flow rate is totally labeled, but our stream entering unit one
does not have a flow rate. So we will call that n1, and it does not have a composition
that we know of. So we will call that y1m for methanol. And we know that is is just methanol
and water, our binary system so the other variable that we don't have here is the composition
or mole fraction for water, but we do know that the two have to add up to one. So I will
just leave that off for now. So we have n1 and y1m entering unit one. We don't have a
flow rate for the top stream so I will call that n2. We have a flow rate for the bottom
stream, but we do not have a composition for methanol, so I will call that y3m. We don't
have a flow rate leaving unit 2, so I will call that n4. We don't have any information
for the last stream. So I will call that n5 and y5m for methanol. So now you see that
we have total of 7 variables, and we have to determine this last stream. So where
do we start? Now the reason that we do a degree of freedom analysis is that we can quickly
pick apart each process within our system in determining whether or not we could solve
those variables. So let's start by looking at the overall system, and doing a degree
of freedom analysis over the entire process. So I am going to draw my boundary to include
all streams that are entering or exiting the entire process. So recall in this situation
when we are doing a degree of freedom analysis, that we are basically taking the number of
unknowns and we are subtracting the number of relationships that we can write to calculate
those unknowns. And whatever is left is our number of degrees of freedom. Now if this is zero then
we know we can solve for the unknowns. If it is more then zero, we cannot solve for the
unknowns. And if it less then zero, if we get a negative number, then we have over specified
our system. So either the over specifications work and we still get what we need or it may
not match up depending on what we are calculating. So let's look at this overall balance.
Our unknowns involve nothing coming in on the left side, but it does involve n2 leaving
we also don't have n4, n5, or the composition of that last stream, so y5m. So we have 4
unknowns. Now in terms of relationships. Since we have two species we can write two independent
material balances. We have an overall balance, a methanol balance, and a water balance. We
can only write 2 of those 3 down. So 2 balances. We also have a relationship that says that n2 and n4
are the same. So I will say one material relationship, but that is all we have. So when we do this
we get 1 degree of freedom, and we cannot solve for these unknowns. So we can't start
with an overall balance. Now let's take a look at the first separation unit, and do
a degree of freedom analysis around this. For unknowns we now have n1 that we care about, and the composition,
y1m. We do not know n2
leaving nor do we know the compostion of the bottom stream y3m. So we have 4 unknowns.
Again we can write two balances, but we cannot write a relationship balance between n2 and
n4, since n4 is not one of the unknowns in this analysis. So this time we have 4 unknowns and 2 balances,
that leaves us with 2 degrees of freedom. So we can't start at unit 1. If we do a degree
of freedom analysis around unit 2, we have the unknowns y3m, n4, n5, and y5m.
Again, since we have two species we can write 2 balances and because we have 4 unknowns and two relationships
we have a degree of freedom for our second unit of 2. So again it is under-specified,
we cannot solve for these variables. The only thing left to look at is the mixing point. So if
we look at that mixing point, the two streams coming in at the entrance before unit 1.
The only unknowns we have are n1, and the composition of that stream or the mole fraction y1m.
The two relationships we can write are 2 material balances. This time since we have 2 unknowns
and 2 relationships we have a degree of freedom of zero. So we can start here, and by starting
here we get these two variables solved. So now we come up with our plan to solve for
these unknowns. We know that if we do a mixing point balance we are going to get n1 and y1m.
Knowing n1 and y1m we can then do a balance around unit 1, with 2 unknowns, n2 and y3m,
We can then solve for n2 and y3m. The last thing we are going to do is we are doing is an overall balance.
Our overall balance in the case that we don't know n4, n5 or y5m. So we have 3 unknowns. We can write
2 balances and we have the relationship between n2 and n4. So we right away we know n4. So
the two unknowns n5 and y5m are solved with our balances. So that's our plan. Now we
just have to execute it. So we are going to start with our mixing point. I am going to
write an overall balance, which says that 100 moles plus 10 moles coming in must be equal to n1.
Thus n1 is equal to 110 moles. Now a species balance on either methanol or water. I'll do methanol,
follows the same logic. We have 50 percent methanol of our 100 moles coming
in plus 40 percent methanol times our 10 moles coming in. That is going to be equal to our
mole fraction of methanol times our total. I can either put n1 here if I were
writing them both and they were unknown. Since we have already solved for n1, I will just plug it in.
This gives us a composition of methanol in the first stream of 0.49. So 49 mole percent methanol.
So I will plug these values into our process now. And the second part of our plan was to
perform material balances around unit number one. So again overall we have 110 coming out.
This must be equal to what is leaving. So this is n2 plus 70. So right away we can see
that n2 is 40 moles. We can write a methanol balance. 0.49 times 110 is equal to 0.7 times
n2 plus y3m times 70.
And we know n2 is 40. So you can see that we can solve for y3m.
Mole fraction of methanol on that bottom stream is equal to about 0.37.
So now we can plug that information into our schematic. And the last thing in our plan is to do the overall
material balances. So starting with an overall material balance, we have 100 moles entering
our process plus the 10 moles it is mixing with. This is equal to what's leaving our process.
This is 40 moles from unit one plus n4 plus n5. Methanol balance written in the same
fashion looks like the following. So in the past we were able to solve for the
2 variables at this point, but now we have 3 variables and 2 equations. Remember the
relationship that we are missing is that n4 was equal to the top stream of unit one.
So this is 40 moles. So now we know n4. We can solve for both n5 and the mole fraction of
methanol in that bottom stream. I get n5 is equal to 30 moles, and the mole fraction of methanol
in that bottom stream is 0.2. So these are the values we have calculated for this process.
Hopefully this screencast has given you an idea of why it's important to do a degree of
freedom analysis and how it can help you when we have multiple units involved in our process
and not necessarily knowing where we should start in approaching the problem.