So in this video we are going to talk about internal material balances on a distillation
column for a special circumstance. One in which we have some kind of side stream that
we are going to pull off with a different kind of composition than our distillate and
our bottoms. So let's create an example problem and we will work through this. So here we
have a problem where we have methanol and water being fed into a distillation column.
You can read the details of the problem statement and go along as we go through the problem,
but the main point is that we are trying to figure out where the optimal stage is to pull
off a side stream with a certain composition that is different from the distillate and
the bottoms. So first things first, we should draw a diagram and label it appropriately.
So here is our information for our feed, 500 kmol/hr, it's a 50:50 mixture and 50% of it
enters as a vapor and 50% of it enters as a liquid. Now i will go ahead and write our
duties for our condenser and our reboiler but it is not part, necessarily, for this
problem but it is good to label it. We also have our distillate which has 0.95 and our
bottoms at 0.05. And although it is not told to you you need to figure out what component
that is talking about. Since we are talking about methanol water, methanol being the more
volatile component, both of these components are for methanol. We are operating at 1 atm.
And we have a side stream that is a compositions 0.75. Last thing we know is that the reflux
is 2. But there are some other things that we could draw in here that would help us solve
our problem. First, we know this is vapor going into our condenser so we will label
that as V. We also know we have liquid going back into our column since we are operating
a total condenser and that is what is going to happen. We have some liquid entering our
boiler and that is going to be at a different flow rate since we have a feed and a side
stream that will alter our internal material balances. We also have a vapor which we will
designate as Vbar that is leaving the boiler and entering the column. So now we know that
we have Vbar and Lbar at the bottom of the column. Somewhere at the feed we are going
to have a change in those which we will say we have V now and since our side stream is
only taking off liquid we will not be changing the flow rate of the vapor given that we make
the appropriate assumptions. However this L that comes in does change so we will label
that L' and then L' will change to Lbar since we have a V that is contributing to that flow
rate. So after drawing the picture it is good to write down what we are looking for. We
are looking for a feed stage, we are also looking for where our side stream stage will
be and we are looking for the total number of stages. I am going to make our assumptions
up front and in this case for our distillation column it is a good assumption to make the
constant molar overflow assumption also known as CMO. So how do we start this? Since we
are using vapor liquid equilibrium information to separate two components it is probably
important that we get that data first. So we would have to look up vapor liquid equilibrium
data so that we can plot all this on our McCabe-Theile diagram. Once place to do this would be Perry's
Chemical Engineering Handbook and that is where we get this data for methanol and water.
So I will show you that. So we have built our x-y diagram with our 45 degree line and
now we can start trying to determine our operating lines for the column. So a great place to
start is in our rectifying section. And we are not going to look at the side stream right
now. So we will just look at this area that I have highlighted in a box and write a material
balance for what is going out and coming into that region. So we will call this area right
here that involves our reflux and above our side stream pullout, section 1. So as it is
drawn, the only thing we have coming out is D, we also have V entering and L leaving.
And we are saying that L0 is going to be the same thing as L based on our assumptions for
CMO. So we write two material balances, one where V=L+D. And Then we are going to write
one for methanol, do a material balance. Then we are going to combine these and solve for
y so that we could get that relationship between y and x and plot that on our y-x diagram.
So as it is shown this is our operating line for our rectifying section above the side
stream pull-off. Now, although we have not done material balances on the entire column
to solve to try and get some of these values itself, one thing that we can do is relate
our L, D and V to our reflux ratio which we are given. So now that we have reorganized
this into terms of our reflux ratio, and that is something that we are given, we have a
nice mx+b and we could plug in our reflux ratio to get a line that has a slope of 2/3
plus an intercept of around 0.3. So we could plot this on our x-y diagram. So we start
at our x=y intercept of 0.95, and go to our y-intercept around 0.32 and that slope of
that line should be 2/3 and that is something that you could check. Remember one of the
solutions to this line is x=y=the distillate composition, xD. So this is our top rectifying
line. So one down, three more to go. And let me go ahead and write our feed line equation
which is explained in other screen-casts for distillation. And this is given in terms of
a q value and our composition of the feed. So what is our q value here? q is defined
as the amount of feed that crosses the feed plate. So that has to be in the liquid state.
So if we designate Lbar as what goes down and L as what is coming in, then that change,
or feed, is going to be a ratio which we are going to say is q. So we could re-write this
in terms of the fraction that is vaporized and we know that this value is 0.5 from the
problem statement. So our q line or our feed line, operating, is just going to be y=-x+1.
So we can go ahead and plot this on our yx diagram which will look like the following
where we start at x=y=composition of our feed, 0.5. And we have a slope of -1 up to a intercept
of 1. So in the past if we didn't have the side stream, we would just take that intercept
and draw it down to our composition in our bottoms to get our stripping line, but we
have a side stream. So that intersection here actually doesn't mean anything because we
have an operating line that goes in between that feed and our top rectifying section.
So we need to figure that out. So we need another section that includes our side stream
here that includes to our feed. So this is still part of our rectifying section but we
will call it section 2. So I will blow up this balance so you can see what it is going
to look like. So this is the region of interest that we are doing this around that includes
our side stream. So we know that based on the vapor flow rate going into our condenser,
that above our side stream will have the same vapor flow rate, using our assumptions, and
will have the same liquid flow rate that is coming in. However, now with the side stream
being pulled off we have a different liquid flow rate than before and we will designate
that will a prime sign. Thing is that the side stream is pulling up a liquid. So the
stream going up as a vapor can stay as a vapor. So now we know we have some terminology that
we designated for this system and we can write the same material balances that we did before.
So our overall balance is going to look like this following and then we are going to do
a component balance and I have simplified this already and dropped some of the tray
nomenclature since we are assuming CMO. And just like before we could solve for y, so
this becomes our operating equation for that middle section designated as 2 in the diagram.
So obviously the trick to plotting this would be to know some information about those variables.
And we know xD and xS but we don't really know about the flow rates. So a good place
to go from here would be to do the material balances that we can to solve for the flow
rates. So the overall balance on our distillation column gives us the following where the feed
has to be equal to the distillate bottoms and the side stream flow rates. The component
balance for our methanol or any component for that matter will give us the following.
And as you can see we have a lot of unknowns and only two equations. But we were told that
our side stream was 100 kmol/hr. and our feed was 500 kmol/hr. We were also told compositions.
So plugging these values into our material balance shows us that we have two equations
and two unknowns and we could solve for D and B. So after solving I get a distillate
flow rate of 172 kmol/hr and a bottoms flow rate of 227.8 kmol/hr. So that is great, that
gives us ideas on the flow rates leaving and entering our system however what we are really
interested in is those operating lines and as you saw before we have to figure out something
about L, V, Vbar, Lbar, L', and so forth. So we need some relationships for those. First
place to start is our reflux ratio which we know is the amount that is sent back into
the column over our distillate. As you can see that is going to give us our liquid flow
rate into the column. That should give us 344 kmol/hr. The next thing is we can figure
out what L' is and the reason we can do that, if we think back to this diagram, we have
L coming down, we have a side stream being pulled off and then we have L'. So L- L' is
going to be S. So this should be lower since we are pulling something out. We have 244.4
kmol/hr of our L' flowing down. The last thing we need is our vapor flow rate V and we know
that is equal to L'+D+S from one of our overall balances on that section. Plugging those values
in and we get V=516.6. So if we go back to our section of interest for our operating
line we plug in values for L' over V as well as our other flow rates and get an equation
for this section. So we have a slope a y intercept, we know it is going to intersect with our
rectifying line since it is a saturated liquid and it is at 0.75 we can draw that. So the
side stream is kind of like the opposite of a feed. So we would start at x=y=0.75, draw
a straight line up for a saturated liquid. At this point this intersection matters, the
rest of the line does not. And now we have an operating line for our middle section between
our feed line and that side stream. We know the slope is 0.47 and our y-intercept is 0.46.
Just use the y-intercept and that becomes our middle operating line. And here again
all we care about is that intersection and not necessarily the rest of the line, same
thing with our feed. And then the last line to draw would be our stripping line and we
can do the same thing as we do for other cases and go from that intercept to our composition
on the x=y line 0.05. And we have drawn all our operating lines for this column. And now
the real question is, where are we pulling the side stream, and how many stages are there.
So the first thing that we would do, would be draw a stage that precisely ends at our
side stream because we are pulling off a liquid, it has to have that composition so it has
to be at that stage. And then we would just continue filling in stages throughout the
diagram to count how many stages we have and see where our side stream stage enters. So
remember when we cross over those intersection points we drop down to the other line. Since
I have cut those other lines it is easier to see that. So we start counting them off.
Since we have a total condenser we start stage 1, stage 2, and that would be our stage for
our side stream. And then we have stage 4, stage 5, stage 6. So our feed stage is stages
6 because the intersection occurs between stage 5 and 6. And then we have stage 7 and
then we have our partial reboiler. So we have 7 stages plus a partial reboiler. We have
a side stream being pulled off at stage 3 and our feed being feed at stage 6.