In this video we are going to look at a binary distillation problem in which we have some
specific circumstances- one being that we have a total boil-up occurring so we have
no bottoms product.
We do have a distillate but we have no reflux.
So let's look at this problem and work through it together.
So we have acetone being separated from ethanol, we are given feed information and we are told
that the feed is at the fourth stage from the top- so this may or may not be the optimal
location.
We are told, instead of the reflux, we have a pure saturated liquid acetone entering the
column.
So this would be synonymous with maybe open steam heating, if we had a system that included
water and we didn't want to use a boiler.
But systems operating at steady state, bottoms shut off.
So we are told that everything that goes into the boiler is being boiled up into the column
and we are asked to determine the distillate composition, the composition of the liquid
sample if it was pulled from the reboiler, and the optimal feed stage.
So let's start with labeling our diagram with known information.
So I have labeled my column.
We are going to assume Constant Molar Overflow for our distillation.
So the first place to start would be an overall material balance for our column.
So we know whatever goes in must come out so we have a feed plus a pure stream of acetone
entering from the top and this is only coming out from our distillate product.
We are given information for these values.
So we could go ahead and plug that in and right away we are going to see our distillate
composition is going to be.
An acetone balance can be written as such, so we can go ahead and plug in information
we know for this.
And we see we have one unknown.
We can solve for our distillate composition and we get 0.87.
So let's go back to the problem statement.
We have determined the distillate composition is 0.87.
The second part of this problem asks, if we took a sample of liquid from the reboiler
what that composition would be.
So to do that we would want to determine what the operating conditions for our top and bottom
sections of our column would be.
So we can label this as our top section and this as our bottom section, or the rectifying
and stripping sections.
And we can start doing material balances on these sections to determine the operating
lines we would use in a McCabe-Theile analysis.
So let's determine an area that we are interested in for doing a material balance and we will
look at the top section first.
So draw a box around that area and write out our material balance for that top section.
Our overall balance would look like the following where we have some C entering plus we would
have some vapor coming up our column which we will designate as V.
We are going to have D leaving and some kind of liquid also going down our column and therefore
leaving our section.
So this would be our overall balance.
And we could also write a component balance.
And if I do this in terms of mole fractions, for our more volatile component, acetone,
I could write it as such.
Now we could rearrange the component balance into our more familiar form of our operating
lines that we use for our McCabe-Thiele analysis.
So when we do it in terms of y, we get the following equation.
And you will notice I made a mistake in the component balance.
Here I wrote xd times our distillate because I am so used to using a condenser that we
are usually talking about a liquid mole fraction . In this case we have vapor coming out of
the column and it's not being condensed.
So really this should be, y, and we should also use that in our component balance.
So this is the operating line for the top part of our column.
Now there are a couple things to distinguish about our specific operating conditions.
1) our component of acetone in our stream labeled C is saturated liquid and we are told
that it is pure acetone so our mole fraction is 1.
Furthermore since it is a saturated liquid we could say that C is just equal to L, that
being said we know that V must therefore be equal to D. So we can use this information
back in our equation and replace some of those variables.
So I have now written it in a simplified form, where we still have L/V as usual, and our
y intercept is the distillate composition minus our slope.
This means we have a solution at x=1 of y= our distillate composition.
Now we have an intercept on the right side of our McCabe Thiele diagram.
We also know what L and V are since we already did the material balance to solve for C and
D. So we said L was equal to 50 Kmol/hour and D we solved to be 150.
So we could plug this into our OL equation and get a line that we could then plot on
our diagram.
So this becomes our equation.
So now we have an equation that we could plot on our MT diagram which is shown here.
We know our slope and we know our intercept.
If we were looking for our y intercept at x=0 we have y=0.54, so this becomes our top
OL.
We could also plot our feed line.We are told we have a saturated vapor at composition of
0.8 so we go to the x=y line at 0.8 and draw a horizontal line for a saturated vapor all
the way across.
So now we have our top OL we have our feed OL so what is our bottom OL?
Well if we follow the same general rule where we would take the intercept of the feed in
our top section and draw down to our bottoms composition, it doesn't matter where it is
here it has to follow the x=y line.
Now another thing to keep in mind is for our bottoms section we had total reboil.
So everything is going back into the column and therefor the slope of that line is 1 and
follows that x=y line anyway.
So this is our bottom OL.
We will go ahead and show that just to double check.
But if we write this as Vbar and Lbar, standard nomenclature that we typically use, we could
do an overall balance for our bottom section- so say something like this.
And we know that, using CMO, Lbar here is going down and Vbar is leaving.Typically we
would have some bottoms product B but in this case we do NOT so in this case we say Lbar
equals Vbar.
Our component balance would just be Lbar(x) equals Vbar(y).
So in this case x must equal y and as we just showed that is the operating line on the MT
diagram.
So we start at our top distillate composition.
Which is going to be the endpoint of our operating line, so in this case we don't care about
that 45 degree line.
And we draw a stage over, and usually we would go down to the bottoms, because that is where
the switch is, but the feed stage isn't on 1 so we continue on our rectifying line or
the top OL until we count off 4 stages.
So that's 2, that'd be 3 and at our 4th stage we would go ahead and make that cross at our
fouth stage down to our bottoms.
And then continue from our bottom OL.
We are told we have six stages so that would be our sixth and our last stage would be right
there.
So again our first stage, we continued on our second, third, and on our fourth, since
that is where we are told that is where our feed is we made the transfer form the rectifying
line down to the stripping line.
We continue a fifth stage and a sixth stage.
Since we had no condenser and no reboiler, no partial reboiler, we do not include those
as our equilibrium stages.
So in this case we have our six stages as specified.
And using this 6 stage distillation column we get a bottoms of around 0.02.
Now another thing you will notice is our feed line intersects our lines at stage 1.
So our optimum feed stage, technically would be stage 1.
Remember, the definition for our optimal feed stage would be where we minimize our number
of stages in our system.
So you can imagine if we did that, what it would look like.
If feed one was our optimal feed location we would make that switch and start drawing
our stages down.
All the way to the bottom to get that same composition.
So here you see it would only take us 4 stages in our column to achieve the same composition
if we feed the vapor at stage one.
So hopefully this gives you an idea of how to write the balances for a column, regardless
of the type of conditions you are given.
And you can see that if a feed doesn't enter at its optimum position you may have to have
more stages than necessary to achieve the same separation.