Distillation of a Two Component Mixture (Part - 1) Video Lecture | Mass Transfer - Chemical Engineering

Here we have a distillation column. So here
is our distillation column and the feed to
the distillation column which is right here
is 45 mol% n-pentane and 55 mol% n-hexane. It
is a liquid. The vapor stream leaving the
column which contains 98 mol% of the pentane
and the balance hexane goes to a total condenser.
This vapor stream goes to a total condenser.
Half of the liquid condensate is returned
to the column and the rest is withdrawn as
a product at a rate of 85.0 kmol/hr. The over
head product contains 95% of the pentane fed
to the column. The liquid stream leaving the
bottom of the column, right here, goes to
a reboiler. Part of this stream is vaporized
and recycled to the bottom of the column and
the liquid is taken off as a product as well.
Let's label what we know. Both of these are
85 kmol/hr which means this is 170 kmol/hr. This is some n sub L and this is some n sub
vapor. We know that this stream is 98% pentane
and 0.02 hexane. Why do we put that in even
though it is the composition of the vapor
stream leaving the column? Well when it totally
condenses you are are not going to change
any of the mol fractions so whatever it came
in at it is going to come out at. Here we
have some x sub P which is the mole fraction
in the liquid of the pentane and 1 minus x
sub P. Here we have a y sub P and 1- y sub
P. Those are the mole fractions in the liquid
and in the vapor respectively. One of the
things that we are told is that the overhead product contains (and we will call this n
sub 1) 95% of the pentane fed to the column.
Our overhead product which is 85 kmol/hr.
The amount of pentane in that is 98% of it
and that is going to equal 95% of the pentane
which is 0.45 times n1 fed to the column.
That piece of information allows us to solve
our n1 which is 195 kmol/hr. That is the
molar flowrate of the feed stream. Now the
molar flowrate and composition of the bottom
liquid stream. We can do an overall balance
to find n sub L. 195 equals 95 plus n sub
L. Remember when you are doing an overall
balance you do not include the streams that
are not part of what is going in and out of
the process. We only have 3 streams and n1,
85 kmol/hr and nL that are in or out of the
process. Now we have the molar flow rate of
the bottom liquid stream. That equals 110
kmol/hr. To find the x sub P we could either
do a pentane balance or a hexane balance but
let's do a pentane balance since we have x
sub P there. So 0.45 times n1 which 195 kmol/hr
equals 0.98 times 85 kmol/hr, that is what
is leaving up at the top, plus our x sub P
times our n sub L which is 110 kmol/hr. When
we solve for x sub P we find that it is 0.0405.
The next thing that we are asked to do is
find the temperature of the vapor as it enters
the condenser assuming it is saturated. (as soon as you see those words "it is saturated"
it should tell you that you should be using
Raoult's law) at an absolute pressure of 1
atm and then we need to find the volumetric
flowrates of the vapor leaving the column
and the top liquid product. Let's start with
the temperature of the vapor as it enters
the condenser. That is at 170 kmol. Again
it is in equilibrium because it is saturated.
What that means is that the y sub P of the
pentane times the total pressure equals the
mol fraction in the liquid times the vapor
pressure at that temperature which is what
we are trying to find. We set up the same
relationship for toluene. This is P star of
toluene at the temperature. We know that y
sub P equals 0.98. y sub T equals 0.02. We
also know and this is how we are going to
rewrite it that x sub P plus x sub T equals
1. Now let's rewrite both of our equations
in terms of our x's. Our x sub P is y sub
P times the total pressure over our vapor
pressure at the temperature. That is our x
sub P that we got from here. We add that to
y sub T times P divided by the vapor pressure
of toluene at that temperature and that we
got from here. That has to equal 1. We write
it as 0.98, that is our y sub P, times 760
mmHg, that is our pressure of 1 atm, divided
by now we have to write out the vapor pressure
of pentane which is 6.85221 minus 1064.63,
again this is antoines equations, divided
by T plus 232. That is our first term. plus
now we are going to look at the toluene. These
have to equal 1. Using some kind of solver
or mathcad or some program like that we find
that the temperature equals 37.3 degrees C.
Now we are asked to find the volume or the
volumetric flow rate of the stream that is
coming out of the distillation column. This
vapor stream that is 170 kmol/hr. Because
it is a vapor we are going to assume the ideal
gas law. V equals nRT over P which is 170
kmol/hr times 0.08206 meter cubed atm per
(kmol*K) and our temperature that we just
found which is 310.5 K divided by 1 atm. However,
our volumetric flow rate of liquid product
coming out because it is a liquid we can't
use the ideal gas law. Instead we have to
use a combination of the density and the molecular
weight. Let's start with our pentane. That
is 0.98 times 85 kmol/hr and the molecular
weight is 72.15 kg/kmol. Now our density is
0.621 kg of pentane per 1 liter. That is just
the volumetric flow rate of the pentane. Now
we have to do the same for the hexane. 0.02
times 85 kmol/hr. Again we are going to have
different molar masses or molecular weights.
86.17 kg/kmol and a different density of
0.659 kg of hexane per liter. When we add
this all together our volumetric flow rate
of the liquid product is 9.9*10^3 L/hr.