In this screen cast we are going to work an example problem using a non dilute absorber.
The overall concept of this problem is we are taking a gas stream that contains CO2
in a nitrogen (N) carrier gas and we are trying to transfer a large amount of that CO2 into
a liquid stream, or a solvent stream and in this case we are using water.
We have been given a flow rate of the entering gas stream, the composition of the entering
gas stream and also a process specification in terms of how much of this CO2 we need to
remove from the gas stream.
And we are also given a process specification on the relative flow rates of the liquid and
the gas streams.
Now our overall approach to solving this problem is broken down as follows: first we are going
to look at this system and determine whether it is dilute or concentrated, the next step
would be to define the known flows and compositions for the system, then we are going to approach
this knowing that this is an equilibrium driven process and get the equilibrium data appropriate
for this problem, and then finally we are going to find the operating line and then
step off stages as appropriate.
Now the question of whether this is a dilute or concentrated system really comes down to
the idea of, are we having major change in the mass flow rates of the system?
When we do calculations in a dilute system we are assuming that the flow rates aren't
changing.
To determine whether or not that is valid assumption, all we have to do is determine
if there is a large change in flow rates of either of the streams.
Since we are pulling CO2 out of a gas stream, the simplest thing to do is to determine how
much CO2 is being pulled out and see how much that changes the mass flow rate of that gas
stream or the molar flow rate of that gas stream.
So we know we have 100 mol/hr of gas coming in the system total.
From that 100 mol/hr, we are removing 65% of the total CO2.
so we can calculate the amounts of CO2 removed from the stream just by multiplying the 100
moles/hr total that comes in times the fraction of that that is CO2 which is the 0.08 specification
we had in the problem statement.
These two numbers together determine the total amount of CO2 coming in and we are removing
65% of that.
So by multiplying these numbers together we have determined that we are removing 5.2 mol/hr
of CO2.
When we compare that to 100 mol/hr that is total coming in we see we have over a 5% change
in gas flow rate.
We would consider that a relatively large change in gas flow rate.
So as we move forward in this problem we are not going to assume that it is a dilute system.
So now let's look at our process schematic and start defining terms that are appropriate
for this system.
Since we are not working with a dilute system we are dealing with concentrated absorbers
we are going to look at gas flow rates, but these gas flow rates are going to be flow
rates of the carrier gas not the total gas flow rate.
And the flow rate of the liquid stream we are dealing with is just going to be moles/hr
of solvent.
Since we are only dealing with moles per hour of either carrier gas or solvent we know that
these flow rates aren't going to change throughout our system.
But in order to deal with flow rates in terms of carrier gas and solvents we need to redefine
our mole fractions in terms of mole ratios.
So in the gas stream we are going to have a mole ratio we are going to have a mole ratio
of moles of CO2 per mole of carrier gas and in the solvent stream we are going to be dealing
with a mole ratio of moles of CO2 per moles of solvent.
So now let's start to define some of these parameters.
For the flow rate of the carrier gas we know the flow rate of gas going in is 100 mol/hr
and we know the composition of that stream is 8 mole% CO2 and 92% N. So if we take the
92% N, multiply that by the flow rate, we can determine we have 92 moles/hr of N coming
in.
So that is going to be the flow rate of the N, both in the entering and exiting gas stream.
Now when we calculate the mole ratios, the mole ratios are easily determined by taking
the mole fraction and dividing by 1 minus the mole fraction.
Since we are given information about the mole fraction of the inlet gas stream we can quickly
now determine the mole ratio of the inlet gas stream.
So if we take this, look at the mole fraction coming in, of 8% CO2 per total moles of gas
coming in, and divide that by 1- this mole fraction, that mole fraction is now in terms
of N divided by total moles of gas stream.
Now we are going to notice when we do the division that the units are just going to
come down to moles of CO2 per moles of N. And that is what we refer to as a mole ratio.
Now that we have the mole ratio for the gas stream coming in, we would like to determine
the mole ratio of the gas stream exiting.
Well, we know the mole ratio of the gas stream exiting are just the moles of CO2 that are
in the outlet stream divided by the number of moles of carrier gas in the outlet stream.
Now we already know the molar flow rate of carrier gas exiting so we already know the
denominator of this in terms of moles per hour.
The next thing for us to figure out is how many moles of CO2 are going in that outlet
stream per hour.
So we can determine that by subtracting the moles of CO2 that are coming in per hour,
so the molar flow rate of CO2 in.
And subtracting from that the molar flow rate of CO2 that we removed.
We already calculated this term, moles of CO2 removed per hour, when we were determining
if this was a concentrated or dilute system.
So let's take the molar flow rate of CO2 in, we know that is 8 moles per hour, we are subtracting
from that the 5.2 moles per hour that is removed through the absorption process and we are
left with 2.8 moles of CO2 removed per hour.
Now when we plug that into this top equation we end up with 2.8 moles per hour of CO2 in
that exiting gas stream and we also have a molar flow rate of 92 moles per hour of N
in that gas stream.
So we have a molar ratio of 0.0304 in that exiting gas stream.
We also know that that solvent coming in is pure solvent so we know the mole ratio for
that solvent stream is 0.
The other parameters, the solvent flow rate we are going to figure out based on a minimum
solvent flow rate using the process specification and then the final composition is going to
be dependent upon that flow rate so we are going to have to figure these out later.
These absorbers are fundamentally equilibrium processes, we need to get equilibrium data
to design these.
So to design these we are going to use equilibrium data derived from Henry's law.
Henry's law is very simply taking the mole fraction in the gas phase and that is going
to be equal to the henry's law constant derived from the particular conditions you are working
in, divided by the total pressure and we are going to multiply that by the mole fraction
in the liquid stream.
Now for our particular situation I looked up the Henry's law constant for this system
in terms of atmospheres per mole fraction and the total system pressure given in the
problem statement is 1 atm.
So now we know a way of relating mole fraction in the gas phase and the mole fraction in
the liquid phase and relating those just by this number 1640.
So now that we have this equilibrium data let's go ahead and plot that and start working
toward our equilibrium design.
So I have set up an excel spreadsheet where we have mole fractions of both the liquid
phase and the gas phase and I have going ahead and filled in the mole fractions ranging from
0 all the way to the maximum mole fraction we have in the system which is 0.08 CO2.
Now for the mole fractions of the liquid phase that corresponds to these gas phase fractions
we need to fill that in using the equilibrium data from Henry's Law.
We know that they are related by 1640, so we are going to rearrange this equation above
where we just take this gas phase fraction divide by 1640 and that gives us our liquid
phase fraction.
We fill in those values just by taking the gas phase number, dividing by 1640 and now
we have a relationship between x and y.
Now since we are not dealing with a dilute system we are working with mole ratios so
we need to convert these into mole ratios so they agree with our operating line that
we are developing.
So the mole ratio, again, is just simply the mole fraction divided by 1 minus the mole
fraction.
And that is true for both the liquid phase mole fractions and the gas phase mole fractions.
So we can fill those in very quickly using excel.
Do those calculations and now we have plotted over here the relationship between the gas
phase mole fraction, which are moles of CO2 divided by moles of N. And the liquid phase
mole fraction which is the moles of CO2 divided by moles of water.
So now that we have our equilibrium data we are going to move forward and determine the
operating line that we are going to use for the design.
The operating line is nothing more than a material balance on CO2 in this system.
So we have a relationship that has been derived that is our gas phase mole ratio, is equal
to some slope which is defined on our solvent flow rate divided by our carrier gas flow
rate, multiplied by our mole ratio in the liquid and then we have some information on
an intercept.
But this is just the line defining the relationship between the mole ratios in the gas and the
liquid phase.
Now since we know this is no more than a material balance we know the point x=0 and y= 0.0304
absolutely has to be on that line since this is a set of conditions that exist in our system.
So when we come back to our excel spreadsheet I have entered the point (0,0.0304) and plotted
that on our graph as the first point on our operating line.
Now the other piece of information we can use to define our operating line for the system
is knowing the process parameter that we are operating at 1.5 times the minimum S/G ratio
of flow rates.
So the first step is to determine what that minimum ratio, S/G ratio is.
If we go back to the equation for the operating line we see the S/G is nothing more than the
slope of the operating line.
So we need to find a minimum slope that will get us from these conditions to our inlet
composition of 0.087.
So that minimum slope is going to correspond to an infinite number of stages,which will
be a pinch point at 0.087.
So I have drawn this point here on the equilibrium line and we also have our exit composition
here.
So this operating line represents the minimum ratio of S/G that will accomplish this separation.
We can calculate the slope of that line just by dividing the change in y over the change
in x and that is going to represent our S/G min.
We know that the actual S/G is going to be 1.5 times that S/G min, so with that we have
actually determined that the ratio of our flow rates is actually going to be this term
here 1.74X10^-3.
So now we want to define a new operating line using this actual S/G ratio of flow rates.
So I am going to define a new line.
I am going to use x values that are just the same as the x values for our mole ratios,
but the selection of these as our x values is relatively arbitrary.
I am going to find the y values using the equation for the operating line.
So for that cell I am just going to take our previous y value and add to that our slope
times the change in x value.
And then we just fill in the rest of those cells using those same basic formulas.
Now all that is left to do is just to step off the stages and see how many equilibrium
stages we need.
Since we are using a graphical method I blew up this plot just so it is easier to see the
stages and work with them.
Now my approach is going to be to start at one end of the column where we have a composition
of x is equal to 0, y is equal to 0.03.
And I am going to step off stages until I reach the other end of the column.
And the other end of the column we know is going to have a composition of gas exiting
in terms of the mole ratio of 0.087.
And from the operating line we can determine the composition of the exiting solvent stream.
Alright, so now we are going to step off the stages, so we are going to start off at this
end of the column and this is the composition of the exiting and entering streams and now
we need to go to that first equilibrium stage.
So if we are at an equilibrium stage we are going to be in contact with that equilibrium
data.
So we are going to draw a line across to that equilibrium data and this is going to be the
composition of our first equilibrium stage.
Now we are going to see what the composition is in between stages.
In between stages the compositions are diced by our operating line.
So we are going to draw a line up to the operating line.
Now were are going to move to our next equilibrium stage.
Our next equilibrium stage again is going to be in contact with the equilibrium data.
And then we see when we do two full equilibrium stages we have actually overshot our target
compositions.
So now to determine the number of fractional stages needed we are just going to see the
fractional distance along this line.
This line indicates how much of that second stage we needed.
So if we draw a line from our target final composition down to this line we can see that
we have approximately 1.9 equilibrium stages required for this system.