Today's screen cast is for packed column distillation and we have a bunch of information given to
us here.
I will let you read through the problem and I will start writing.
So we have a column the has a feed of 50 mol/hr and it is a saturated liquid.
And z equals 0.5.
We are told that it is a total condenser and a partial reboiler and we know what the distillate
composition is.
The desired one is 0.95, and the desired bottoms are 0.05.
We have a certain packing.
It is run at a reflex ratio that is 1.5 times the minimum.
So L/D equals 1.5 (L/D)min. and we know that Hg is 1.3, we are given that down here.
And we are also given kxA/kyA values in the stripping and enriching sections.
And what we are looking for is the height of the column using the HTU, NTU graphical
method.
Let's review what this method utilizes.
So the first stage to to determine nG or the NTU's.
And so first we will plot the top and bottom operating lines.
And then we will determine the limits of integration from these lines.
Step c is we will pick points on the top and the bottom operating lines to determine xi,
yi.
We will calculate 1/(yi-y) for each set of points.
Then we will plot 1/(yi-y) and finally we will calculate the areas under these curves
to plug that into our height equation.
And step two, with the height equaling (Hg*nG)enriching plus (Hg*nG)stripping.
So I want to point out something that often Hg in enriching and Hg in stripping are not
the same.
But for us they are the same.
V equals Vbar and this is because the feed is a saturated liquid and for this type of
feed, V equals Vbar.
So our Hg values are going to be the same in the enriching and the stripping sections.
Well lets do the problem.
Part a: plot the top and bottom operating lines (TOLs and BOLs).
So, by the way, I am just going to go through each of the steps that we have just gone through
right here on this slide.
So we will start with 1a) which is plotting the top and bottom operating lines.
So what is L over D?
We need to find it from the minimum L over D that we use with the pinch point.
So first we need to figure out what kind of feed we have.
Feed is a saturated liquid.
So it is going to be a horizontal line at the point (z,z).
So the feed line will just look like this.
So this is the feed line.
Now this is a standard distillation column so we can use the point xd, or yeah, (xd,xd)
for the top operating point for the TOL.
So that is here.
And the pinch point, we use my ruler, the pinch point will be here.
And so this will be a slope of (L/V)min.
And if you don't understand why we are doing that look at the screencast for distillation-reflux.
How we determine minimum reflux.
So (L/V)min equals 0.95-0.84 over 0.95-0.5.
So this is this point here.
That equals 0.24.
Now (L/V)min equals (L/D)min over L/D min plus 1.
So that equals 0.24 and L/D min equals 0.316.
So our L/D value, L/D equals 1.3*(L/D)min equals 0.474.
And L/V equals L/D over L/D plus 1, equals 0.322.
So our top operating line is y= 0.322x + 0.644.
This value is just the y intercept of xd over L/D +1 and that equals 0.644.
So let's plot this TOL.
Our intercept is 0.644 and plot this, and then we will plot the BOL.
It goes through (xb,xb) and the intersection of the TOL and the feed.
It looks like that.
And let's go to the next page where we have this already plotted.
This is our TOL and this is our BOL.
So our limits of integration, yOUT enriching equals 0.95, so that's up here.
yIN enriching equals yOUT stripping equals 0.802, here.
And then yIN stripping equals, because it is a partial reboiler 0.15.
Because for a partial reboiler we have xOUT which is xb is in equilibrium with the vapor
coming off the partial reboiler so we go up here to get 0.15.
And this 0.802 was right here and the 0.95 was right here.
Now we are going to move on to our c: pick points on the TOL and BOL (x,y) and determine
(xi, yi).
And so just pick random points.
Start here, my equation for this is yi=-kxa/kya xi + kxa/kya x+y.
So I pick a point on my operating line and I plug in the values for x and y.
And then I have got that point and then I usually just plug in a value of 0 for x and
then I calculate what yi is.
So I get that point.
And then I plot that in excel and I will get something like this and I continue and do
that for all these different points.
Just plug in different values for x and y and then I just get another point with which
to plot the line.
So I make xi equal to 0 so I can determine what yi is and then I plot that line, it goes
way up.
And it will look like this.
And then for the bottoms I do the same thing.
They start here and I will get lines that look more like this.
And for each one of these I find out what the intersection is with he equilibrium data
and that is my value of xi, yi.
So I have done this on the next page, so for these first values, let's see, x=0.11.
So I started with that and I found out what y was based on the BOL.
y=0.15 and then I plotted my line using the equation that we have on the other pages.
y=-kxa/kya xi + kxa/kya x+y.
So I plot my line and then I see where it intersects with the equilibrium data and here
we have xi =0.075 and y=0.24.
Then I do that for the next point 2.
We have go 0.2 and the corresponding way is 0.3.
So I put that 0.3 in here, I draw my line and I see where it intersects the equilibrium
data.
So xi = 0.15 and yi = 0.54, whoops I think that is supposed to be .45.
And then we do that also for here, so that is going to be the 0.3 here And then we do
it up until we get to the intersection of the BOL and the TOL.
And at that point we switch to the slope of the TOL.
So we have values of -kxa over kya for the stripping and enriching sections.
And those we are given here.
So we have a stripping slope and we have got a slope for the enriching section.
Note that they are different.
So after I have tabulated all of these values and I calculate 1/(yi-y) for each then I plot
1/(yi-y) over y and I get this plot as seen here.
Now I want to determine the area is under this plot.
So we need to determine our limits of integration.
We have got that is the bottom, the bottoms going into the stripping.
And then we switch from the bottoms to the enriching at 0.8 and then the output from
the, from the enriching section is 0.95, 0.8 and 0.15.
So this section is the stripping and this section is the enriching.
To determine the are under this I use the trapezoidal rule and I use a while bunch of
areas.
Area 1, area 2, area 3, area 4 , area 5, area 6.
And let me show you how I got the area in each one of these.
So for the trapezoidal rule, you could also use Simpson's rule, the area equals h*((b1+b2)/2).
So for example for area 1 I have the height equals 0.35 - 0.15 and that equals 0.2, b1=11,
moving across here we get 11, b2 = 4, so A1 equals 1.4.
Let's do one other area, we will do area 2.
The height equals 0.72-0.35.
So this is 0.72. that equals 0.37.
b1=4, b2=18.
So this is approximately 18 and A2 equals 4.07.
So I did this for each of these areas and got A enriching equal to 6.01 and this equals
nG enriching.
And H stripping, this is the area under the curve so this is the area under the 1/yi-y
curve.
So A in the stripping equals 7.57 equals nG stripping.
Finally we will calculate the overall height.
The height equals 1.3 which is our Hg that was given to us at the beginning in the problem
statement.
Hg is 1.3 ft times 6.01 plus 1.3 ft times 7.57.
And so the overall height equals 17.7 ft.
That is the total height required for this column.