- [John] I'm gonna describe how an absorption column works
and then write down the mass balances that we use,
in order to determine the number of equilibrium stages.
So the general idea is we feed a vapor phase
that has a concentration of an impurity,
a salude, that we want to remove.
So we want the exit vapor phase,
to have much less of that salude.
Remove it by contacting the vapor,
with a liquid phase, in stages.
So the liquid phase is fed with a low concentration
of that salude and leaves with a higher concentration.
So, notice that the ratio of the liquid flow rate
to the vapor flow rate, for this particular example,
is a hundred.
So, of course, the liquid's much more dense
so we expect the molar flow rate,
to be much higher, for the this example.
And for this case,
we're gonna use these values as mole ratios.
So they're very close to the mole fraction
because the concentration of salude, is really quite low.
So if we look at a column, with five stages.
You can see that the concentration,
59 in the gas phase, decreases with each stage
whereas the concentration, 0.05 in the liquid phase,
increases with each stage.
Now notice, this corresponds to 59 mole per hour,
being fed of the salude, and we multiply by 10 to the six.
And this corresponds
because the flow rate is a hundred times higher,
54 moles per hour to salude.
So we're removing approximately 95% of the salude
that's in the vapor phase.
So let's look at the mass balances
that we use to analyze this type system.
So, on this diagram for mass balances,
we'll call this x zero
and then mole fraction or MO ratio and liquid phase,
the subscript corresponds to the stage that it's leaving.
And likewise ,for the vapor phase, the subscript.
So, we'll use Y vapor phase mole ratio.
The subscript corresponds to the phase that is leaving,
which means this is Y six.
So, if we look at the in stage
then the molar ratio is leaving XN and YN.
These values are in equilibrium.
That's what we're assuming, in this analysis
because such low concentrations
parts per million, in this case.
We will model this by Henry's Law
and say YN is equal to Henry's law of constant,
over pressure times XN.
So, Henry's law of constant,
in this case, has units of pressure.
There are different ways defining Henry's law of constant.
And so let's look at a mass balance,
for this part of the system.
And if we're gonna do now, a mass balance on the salude,
we're essentially saying, salute flow in,
equals salude flow out, flow in is,
x zero times liqs rate,
plus y four times the vapor flow rate.
And what's leaving then is,
X three liquid flow rate plus y one a vapor flow rate.
We can do a balance for any number of stages,
just one and two or all five.
And so the general form of this equation is,
YN plus one, 'cause remember this was stage three.
XN times L plus Y one times V.
So that just rearranged the equation.
And this is what we're gonna call, the operating line.
And so if we wanna determine the number of stages
we're gonna plot the operating line and the equilibrium line
and then we will step through, how we do that calculation,
determine number of stages.
So apply some example values.
This is the equilibrium line, Henry's law.
So it's linear.
This is the mass balance line, the operating line
and that's this equation here.
And I've picked a value for Y six of 21.
So that has to be on the mass balance line.
And the way we determine equilibrium stages,
is that if this value here is Y six,
this value then is going to be Y five and X five.
Because X five and Y five,
the values leaving stage five have to be an equilibrium.
So they need to be on equilibrium line.
Well then for the mass balance
this point is X four and Y five.
because X four and Y five, you can see,
here's where we do a mass balance.
We do, say, a mass balance around this.
They have to satisfy the equation.
And then we can step through, this is still X four
but now, it's Y four and of course X four and Y four,
leaving stage four, have to be an equilibrium
and we can continue.
And doing this by hand, of course, is not
as accurate, as if I was a little more careful.
But I can do this until here is X zero.
We don't have another stage.
Here is X zero
because that's what has to be on the mass balance line
and of corresponds then to Y one, right?
That's the feed and liquid phase coming in.
And we can count the stages
stage three, stage two, stage one.
So this is how we can step off stages
determine number of equilibrium stages required
for a certain separation and an absorption column.