Ok this is a screen-cast for a distillation problem. We have a distillation with a partial
condenser and a total reboiler. So let's draw this. This is a little different from our
standard configuration. We have a feed going in and it goes up to this partial condenser
now. So now, the liquid from the partial condenser is going to go down whereas the distillate
will go up. So the distillate is now going to be yd, because instead of having a total
condenser, like we have had in the past, this time it is a partial condenser. So it acts
as an equilibrium contact and it separates the vapor and the liquid, and they will have
two different compositions. The Distillate and the Liquid streams will have different
compositions and the distillate will now be a vapor, not a liquid like it is in many of
the problems that we have done up until now. And we take the bottoms and we take it to
a total reboiler. So this total reboiler will then go off and go back up into the column.
So we will have Vbar going back up into the column and we will have out bottoms. And out
bottoms are now also a vapor. So the bottoms are going to be yb. So this free boilder is
no longer an equilibrium contact like it is whenever it is a partial reboiler. Now it
is a total reboiler. And so, both the bottoms and the vapor stream are going to be vapor.
Everything is boiled back up. There is no separation here. We only have a phase change,
there is no change in the mole fractions. So our feed is coming in. Our feed is 0.6
acetone so z=0.6, the feed is 200 kmol/hr. And it is superheated vapor, this is very
important. And our yd, yd= 0.9 and yb=0.05. We also know that L/D=3 and CMO is valid.
So let's start with part A. What is the equation for the feed line. I like to start with my
standard diagram where my feed is going into my feed stage and Vbar is going into the stage
and V is going out. And the liquid is going in and Lbar is going out. Now the feed consists
of a vapor portion in general and a liquid portion in general. In our problem because
it is superheated vapor, the whole thing is vapor. So Vf=F and Lf, there is no liquid
portion, so Lf=0. Let's do balances for both the liquid and the vapor. Let's do the vapor
first. So the vapor coming off of the feed stage equals the vapor going in, Vbar, plus
Vf, plus, some portion of liquid that condenses in order to cool down our superheated vapor.
So our liquid stream comes into contact with this superheated vapor and part of the liquid
becomes vapor. Part of it is vaporized. So 1/5F, 1 mole of liquid for every five moles
of feed, as it is given in the problem statement. So if we go back to our V mass balance, We
have V=Vbar+Vf+1/5F. Now let's do the liquid balance. Lbar=L+Lf. And here Lf equals 0.
And then it is minus 1/5 of the feed. This is because part of this liquid stream coming
down into the feed is vaporized because the feed is superheated vapor. So Lbar-L=-1/5F.
q=Lbar-L/F so it equals -1/5(F)/F= -1/5. So
our feed line is y=q/q-1 (x)-z/q-1. y=-1/5/(-1/5-1)(x)-z, z
is stated in the problem as 0.6/-1/5-1. So y=-1/6x-1/2. And that is our feed line. If
we want to plot some portions of the feed line we know that we can always plot (z,z)
you can always plot that for the feed line and here that is (0.6,0.6) And we can also
plot the point (0,0.5) when we sub in 0 for x. So that is it for the feed line. Now let's
move on to part b. Number from the top down, how many equilibrium stages are needed for
this separation. What are the total number of equilibrium contacts required? Now the
difference between these two is that equilibrium contacts include all of the equilibrium stages
plus a partial reboiler or a partial condenser. so equilibrium contacts equal the equilibrium
stages plus any partial condensers, condensers, or reboilers. If you have a total condenser
or a total reboiler this does not count as an equilibrium contact, because it does not
change the composition of any of the streams, all it does is change the phase of the streams.
If you have a partial condenser or a partial reboiler, this does count as an equilibrium
contact because you will be outputting a vapor and a liquid stream that both have different
compositions. So it acts as a stage, but it is different from a stage because equilibrium
stages are all part of the column. That is physically different than a furnace or a condenser
which will be separate, apart from the column. They will have different efficiencies and
they won't be housed as part of the column. So let's start this by starting as we always
do, with our top operating line. Top operating line (TOL). And how about I derive this equation
just to show you where they end up getting it from the book. We have the stuff going
in, we have V going in, equals D+L coming out. So V=D+L. Now our TOL species mass balance
is going to be different from the one you normally see. It is going to be Vy= Lx+Dyd,
not xd. So if we rearrange this we get y=L/Vx+D/Vyd. And our L/V = L/L+D, that equals L/D over
L/D+D/D and that equals R/R+1. Remember that R equals the reflux ratio equals L/D. Now
D/V equals D/L+D equals D/D oer L/D+D/D and that equals 1/R+1. So our TOL ends up being.
R or y=R/R+1 x, that is the same as usual, plus yd/R+1. Now we are given in the problem
statement that R=3 so y=3/3+1x+yd is given as 0.9/3+1. So y equals 0.75x+0.225. And we
will plot the points (0 and 0.225) and then the other point is (yd,yd) which is (0.9,0.9).
And why can we plot the second point. Let's plug it into our second equation. When x=yd,
the value of yd, We have y= to R/R+1 yd + yd over R+1. So y= R+1 yd over R+1. y=yd.
So when you are at physically on your chart at the point x=yd, the y value is also going
to be yd. And so we have 0.9,0.9. So let's plot this. That 0.9,0.9 and we have got 0
and 0.225. And let's just draw this line. Now we are going to plot the feed line. The
feed line. We got that previously in part A and we said that we could plot 0 and 0.5.
(0 and 0.5) and (0.6 and 0.6). 0.6,0.6 is just the feed composition and then 0 and 0.5
and let's plot this. And let's check this slope. For a superheated vapor our slope should
be less than 1 but positive and that is exactly what we have. And then finally our bottom
operating line (BOL). Operating line goes trough (yb,yb) and the intersection of the
top and the feed lines. And it is yb instead of xb because we have a total reboiler so
everything coming out of this reboiler is vapor. Ok. So (yb,yb) is (0.05 and 0.05).
So we will plot that here, (0.05 and the intersection). Now all we have left to do is step off out
stages. We will start at the top and we will go across, this is supposed to be a straight
line. We stay on the TOL, we go across and we are still on the TOL and we go across and
look we have moved passed where the feed and the top and bottom operating lines all intersect
and so now it is time to move to the bottom operating line. We will go across one more
time and we get an exact number. Now how do we number these feed stages.We are going to
start at the top and it says that we have a partial condenser so that acts as an equilibrium
contact. So we have PC at the top. Now at the very bottom we would often put R our partial
reboiler, so PR here, but we don't have a partial reboiler this time, we have a total
reboiler. So it does not act as an equilibrium contact. So we can just step off our stages
to the very bottom one. We have out partial condenser first, then our stage 1, then number
2, then our third stage. And what is the optimal feed location, it is stage 2. So let's go
back to this. What are the number of stages? The stages: 3. The contacts, the equilibrium
contacts: we have 4. Three stages and our partial condenser. Now we just move on to
part d. Calculate Lbar and Vbar in the bottom part of the column. Let's just do a mass balance
for this. We know we have a our feed coming in. We have got our distillate and our bottoms
coming out. Feed equals the Bottoms plus the Distillate and the feed times z equals the
distillate times yd plus the bottoms times yb. So if we combine these two we have 200kmol/hr
*(0.6) = D* (0.9) + (200-D)* (0.05). This will give us a distillate of 129.4 kmol/hr.
And a bottoms of 70.6. All of these flow rates are going to be kmol/hr. Now to find the liquid
flow rate in the top part of the column. We have the liquid times the reflux ratio, L/D
times D that equals 3*(129.4) and that equals 388.2. equals L. Lbar we found equals L-1/5(F)
we found that in part A when we did a balance on the feed stage so that equals 388.2-1/5(200)
and that equals 348.2=Lbar. Now let's do a balance on the bottom portion of the column.
Here we have Lbar going out to our total reboiler and let's say that is Lbar and then Vbar coming
out and B coming out. So Lbar equals Vbar+B so that means that 348.2=Vbar+70.6 and Vbar
eqals 277.6. We have a couple different ways that we can check. We can check number 1 and
that is V=L+D. This is just a mass balance over the top of the column. That equals 388.2
+ 129.4. So V= 517.6. Now in part A we did a balance on the vapor coming off from the
feed stage and we found that V=F+Vbar plus (1/5)F. So let's just show this, becuase Vf
equals F. The entire feed is vapor. So we have 517.6 just plugging everything in equals
200+ Vbar+(1/5) of 200 and Vbar equals 277.6. That checks. So check number 2. We will use
the graph. From the plot, Lbar/Vbar equals 0.58-0.05 over 0.475-0.05. So Lbar over Vbar
equals 1.25. Let me show you this 0.58 will be about here, 0.58 and about here, 0.475.
And then this point is of course 0.05 and 0.05. So that is the slope of the BOL. And
if Lbar equals 348.2, then Vbar equals 348.2 over 1.25 and that equals 278.6. And that
is close enough if we take in to account any error introduced in just reading off of the
plot. And so we have Lbar and Vbar and we have checked them out and they look great.