In this screencast we will continue where we left off in our final review part 3. In
this previous screencast we used a degree of freedom analysis on an overall balance
to yield 4 unknowns. The 4 unknowns were n1, n2, zB, and xB. Now only 3 equations that we can use
were available. 2 species balances, 1 for benzene and 1 for toluene and our relationship
between the feed and the distillate stream where the problem states that 90% of the benzene
that entered the column exited in the distillate. So this meant we had 1 degree of freedom and
we could not solve for the overall balance. So here we are going to do a reboiler balance
and do the same type of analysis. So the reboiler is down here. So doing a degree of freedom
analysis yields 4 unknowns. We have the unknown n2, an unknown xB, we have yB, and we have
wB. Now we could write 2 species balances for toluene and for benzene and we could write
two Raolt's Law equations since we have vapor liquid equilibrium coming out of the reboiler.
Now recall that Raolt's Law states that the partial pressure of a species so here we have
the vapor mole fraction of benzene times the total pressure which is the partial pressure
of benzene is equal to the mole fraction of the species in the liquid phase. So designate
that xB, time the vapor fraction of the species so Psat of benzene at a given temperature.
In this case the reboiler temperature. So we could write the same thing for the other
species but I am going to write in terms of the benzene variables since we have written
the mole fractions as such. So it is just 1-yB times the total pressure is 1-xB times
Psat of toluene at the same temperature. So we need to look up the vapor pressure of toluene
and benzene at the 90 degrees that we have at the reboiler. So we would look these up
or use the Antoine equations and we get that Psat of benzene is equal to 1010 mmHg and
the vapor pressure of toluene at 90 degrees Celsius is approximately 405 mmHg. So now
we could plug in the vapor pressures into that equations. We now the total pressure
which is given to us as 500 mmHg and we could solve for yB and xB. So solving these two
equations I get yB is equal to 0.317 and xB is equal to 0.157. Again I will carry my numbers
through until the end before rounding them up. So what we have done here is we have solved
for 2 of our unknowns. We have yB and we have xB. So we should be able write our 2 species
balances and solve for n2 and wB so let's take a look at that. So remember we could
write 2 balances. We could us an overall or either of our species balances and the thing
to recognize is that if we do an overall balance we say that the incoming molar flow rate n2
is going to be equal to 0.4 n2 plus 0.6 n2.So what have we shown. That n2 is equal to n2.
So unfortunately given this information we do not have 2 unknowns left and 2 balances
we could write. We actually have 4 unknowns where we don't know the flow rate of this
stream and we don't know the flow rate of the vapor stream leaving so we have put them
into a relationship of our incoming stream but we didn't account for these unknowns.
So we cannot solve for these variable and at this point we are stuck. We do know that
we solve for some of the outlet conditions from the reboiler. We solved for xB and we
have yB. So maybe we could revisit our overall material balance. Recall that we had 4 unknowns
and 3 equations we could use. Well we have solved for 1 of the unknowns xB. Now we have
3 unknowns and 3 equations. So our degree of freedom analysis for an overall balance
gives us 0. So we could solve for the unknowns variables. let's take a look at an overall
balance. We write 100 moles in is equal to 0.5 n1 plus 0.6 n2. So we could either do
a benzene or a toluene species balance. So I will do the benzene. I have 0.6 as the mole
fraction of entering benzene. This is going to be equal to zB, the mole fraction of benzene
in the distillate stream, times 0.5 n1 plus xB, the mole fraction in the liquid stream
which is also known as the bottom stream, times 0.6 n2, but we known xB. So we have
2 equations and 3 unknowns. We need 1 more equation. That last crucial piece of information
is that 90% of incoming benzene comes out in the distillate stream. So how do we write
that? 90% of the incoming benzene so this is 0.6 times 100 moles per hour. Is going
to be equal to the benzene leaving in the top stream. In this case that's zB time 0.5
n1. So now you should see that we have 3 equations and 3 unknowns. We can solve for n1, n2 and
zB. Now when I do that I get n1 is equal to 123.6 moles per hour, n2 is equal to 63.7
moles per hour and zB is equal to 0.874. Again that's moles of benzene per mole. Now does
this makes sense. Benzene is more volatile than toluene so we would expect the benzene
mole fraction in the top of the column in the distillate to be higher than the mole
fraction in the bottom stream. In this case the zB is much higher than our xB. So at least
from a common sense perspective this does make sense. So now we have solved for zB.
We have n2 and we have n1. So the only thing left to solve for is our wB. We could do that
by revisiting our reboiler balances. So on the reboiler all we need is 1 balance since
we have 1 unknowns. I will write a benzene balance. The mole fraction of benzene times
the incoming stream is going to be equal to the mole fraction of benzene times the top
stream. In this case 0.4 n2 plus the mole fraction of benzene in the bottom stream which
is xB times 0.6 n2. Now using n2 of 63.7 and xB and yB values that our reported above we
can solve for wB. When we do we get an incoming mole fraction of benzene as 0.221. So now
we have solved for all of our unknowns and we can get an idea about the processing conditions
for this distillation setup. So hopefully you can see that doing a degree of freedom
analysis was important. It quickly helped us analyze certain parts of the system to
figure out where we could solve for the unknowns. In approaching this we may have to go back
and forth between systems once we can solve for some of those unknowns.