So in this example we have a binary distillation column and we are going to get rid of our
reboiler. And so now we have nothing to send our vapor stream back to our column. However
if we decided to use the direct application of steam and pull our bottom stream from the
bottom of the tower. Now we don't need a reboiler to send the vapor since we are are directly
applying it. This would be applicable in cases where maybe we are working with water anyway.
So let's take a look at an example. An equimolar water and methanol mixture that is entering
our distillation column in which 50% of it is vaporized. and we are given information
about the column in terms of operating conditions the compositions we want coming out of the
top and bottom. But as we mentioned instead of a reboiler we will use saturated steam
directly which is also known as open steam heating or open steam heating. We want to
determine the optimum feed location and the number of stages needed. So like any problem
at this point we get to start with our diagram, labeling it properly. So taking our column,
and filling it in with relevant information from our problem statement. We know that with
our steam it is pure, so xm or x steam is going to be 0. And that is important and we
will see why later on. So another thing I have designated is we have our stripping section
which is below our feed, including feed. And we have our rectifying section which is above
the feed. Now these flow rates that I have designated for our vapor flow rate and our
liquid flow rate are those for the rectifying section. We know that, if we make certain
assumptions, those values will be similar to the ones leaving the column and going into
the condenser. However, our feed will change those flow rates for our stripping section,
so we designate those with a bar to show they are different. But, what happens toward the
bottom of the column since we are using direct steam? How does that change it? So let's figure
out what we want. We want our location of our feed stage, our optimal. We also want
the number of stages. We are going to make a couple of assumptions for our distillation
column we are using CMO which is constant molar overflow. The main thing with this is
it allows us to say that the flow rates are constant. There are a number of things that
go into the CMO assumption but one thing we can do to check is look at the enthalpy associated
with vaporization of the two components and see how similar they are. And if they aren't
similar then we probably should take this to a simulation software program to do this
kind of analysis or at least verify the work we do here. So where do we go from here? What
is going to be our approach to solve this? Well first things first we need to get phase
equilibrium data. A good place to look would be Perry's Chemical Engineering Handbook or
some online source. So we will look this up for 1atm pressure. So here you see our xy
diagram of phase equilibrium data between methanol and water at one atmosphere. So as
we solve our distillation column we are looking for operating conditions for these conditions.
And we can use internal material energy balances for certain sections to figure out those operating
conditions. So typically a good place to start would be a rectifying section which would
include that top section of our feed. So we write an overall material balance where we
have our vapor flow rate equaling our liquid flow rate from the column plus whatever leaves
as distillate. We also write a component balance for methanol. Now we can solve for y because
we want to be able to plot it on our MT diagram our xy diagram. And we get the following for
our OL. Now we are given these four rates, we could probably solve using material balances.
The other way is to use the reflux ratio and relate our reflux to the amount we are sending
back to the column over the amount we are taking out. If we use that ratio then we can
rewrite our OL in terms of the reflux ratio and then use the reflux ratio given in the
statement as well as our top composition and you should get 2/3x +0.32. We will plot that
on our xy diagram. So we start at our x=y=xd and we go down to what our intercept was,
which was about 0.32. So this is our rectifying line. And at the point xd, xd. And that is
a solution to the OL. So you plug it in and verify. Now another condition we know is our
feed line. That again will be discussed in another screen-cast. But we can relate that
to what is known as a q value. And our OL looks like the following. I will discuss q
here in a second. q gives us an idea of the fraction that comes in and crosses our feed
stage. So for it to cross our feed stage it needs to be liquid, some form of liquid. So
the fraction of the feed that is liquid gives us an idea of the q value. It is also related
to the energy that comes in with our feed. Like I said that is discussed in another screen-cast.
We can write it Lbar minus L, remember those are the flow rates of our liquid in the rectifying
and the stripping sections. We can also change this, we are given this value, that is the
fraction of the feed that is vaporized. So we can plot our feed line as -x+1 plot this.
So since we have an equimolar feed it comes in at 0.5 on the x=y line and goes to an intercept
of y=1. So that is our rectifying line and our feed line. Now the last thing is to determine
our stripping line. Now the thing about the stripping line is, in the past, we have drawn
from the intersection down to the x=y line. However since we don't have a partial reboiler
that is not the case, and I will show you here in a second. So let's do a balance over
this bottom section, where the steam enters into our column. So in terms of what is coming
in we have Lbar coming in as well as our steam so we have vapor leaving plus our bottoms.
Now if we do a methanol balance we have designated ys as our mole fraction of methanol in the
steam which right away we should see is 0. Again we can solve in terms of y. You can
see I have that second term still there just to show that if it did come in as some methanol
for some reason that we could account for that term, but in this case our ys is zero.
so our OL simplifies into the following form. Now a very important thing to note is that
we are using saturated steam so our V is just equal to S which also means that our L is
equal to B and the reason is CMO and saturated steam feed. So we can rewrite the OL equaiton
with our Lbar and we can see now that when y is equal to zero, x must be equal to xb.
This is in contradiction to a reboiler in which we are working at y=x=xb. So, now we
have our intercept and we know that it is supposed to intercept the feed. We go back
to our plot, y=0, x=0.05, draw that up to our intersection. I like to get rid of the
excess parts of our line. Counting our stages, since we are using a total condenser that
is going to be our first stage, that is going to be our second stage, keep going down the
plot until we get to this changeover point and you want to go to the OL the 45 is just
a tool to see we are not quiet there so we do one more at that point which takes us to
a pretty low composition. And granted this is a a graph and just an interpretation it
is a nice way to see what we are looking at for our column. So here I count 6 stages,
we don't have a partial reboiler so we probably have to use our 7th stage. And our feed here
enters here between 4 and 5 so our feed is at stage 5. And we have 7 stages. So a big
difference with open steam heating is this point down here which our intercept is at
y=0, x= the bottom composition. So hopefully that gives you an idea of how to use open
steam heating in column distillation.