In this screen cast we are going to work an example problem using a non dilute absorber.
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
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
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.
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
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
gas stream and also a process specification in terms of how much of this CO2 we need to

remove from the gas stream.
remove from the gas stream.

And we are also given a process specification on the relative flow rates of the liquid and
And we are also given a process specification on the relative flow rates of the liquid and

the gas streams.
the gas streams.

Now our overall approach to solving this problem is broken down as follows: first we are going
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
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
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
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
for this problem, and then finally we are going to find the operating line and then

step off stages as appropriate.
step off stages as appropriate.

Now the question of whether this is a dilute or concentrated system really comes down to
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?
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
When we do calculations in a dilute system we are assuming that the flow rates aren't

changing.
changing.

To determine whether or not that is valid assumption, all we have to do is determine
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.
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
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
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.
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.
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.
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
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
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.
we had in the problem statement.

These two numbers together determine the total amount of CO2 coming in and we are removing
These two numbers together determine the total amount of CO2 coming in and we are removing

65% of that.
65% of that.

So by multiplying these numbers together we have determined that we are removing 5.2 mol/hr
So by multiplying these numbers together we have determined that we are removing 5.2 mol/hr

of CO2.
of CO2.

When we compare that to 100 mol/hr that is total coming in we see we have over a 5% change
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.
in gas flow rate.

We would consider that a relatively large 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 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
So now let's look at our process schematic and start defining terms that are appropriate

for this system.
for this system.

Since we are not working with a dilute system we are dealing with concentrated absorbers
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
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.
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
And the flow rate of the liquid stream we are dealing with is just going to be moles/hr

of solvent.
of solvent.

Since we are only dealing with moles per hour of either carrier gas or solvent we know that
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.
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
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.
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
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
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.
with a mole ratio of moles of CO2 per moles of solvent.

So now let's start to define some of these parameters.
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
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
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
92% N, multiply that by the flow rate, we can determine we have 92 moles/hr of N coming

in.
in.

So that is going to be the flow rate of the N, both in the entering and exiting gas stream.
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
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.
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
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.
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
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
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.
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
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.
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
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.
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
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.
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
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.
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
The next thing for us to figure out is how many moles of CO2 are going in that outlet

stream per hour.
stream per hour.

So we can determine that by subtracting the moles of CO2 that are coming in 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.
so the molar flow rate of CO2 in.

And subtracting from that the molar flow rate of CO2 that we removed.
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
We already calculated this term, moles of CO2 removed per hour, when we were determining

if this was a concentrated or dilute system.
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
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
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.
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
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
that exiting gas stream and we also have a molar flow rate of 92 moles per hour of N

in that gas stream.
in that gas stream.

So we have a molar ratio of 0.0304 in that exiting 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
We also know that that solvent coming in is pure solvent so we know the mole ratio for

that solvent stream is 0.
that solvent stream is 0.

The other parameters, the solvent flow rate we are going to figure out based on a minimum
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
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.
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
These absorbers are fundamentally equilibrium processes, we need to get equilibrium data

to design these.
to design these.

So to design these we are going to use equilibrium data derived from Henry's law.
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
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
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, divided by the total pressure and we are going to multiply that by the mole fraction

in the liquid stream.
in the liquid stream.

Now for our particular situation I looked up the Henry's law constant for this system
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
in terms of atmospheres per mole fraction and the total system pressure given in the

problem statement is 1 atm.
problem statement is 1 atm.

So now we know a way of relating mole fraction in the gas phase and the mole fraction in
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.
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
So now that we have this equilibrium data let's go ahead and plot that and start working

toward our equilibrium design.
toward our equilibrium design.

So I have set up an excel spreadsheet where we have mole fractions of both the liquid
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
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.
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
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 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
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
where we just take this gas phase fraction divide by 1640 and that gives us our liquid

phase fraction.
phase fraction.

We fill in those values just by taking the gas phase number, dividing by 1640 and now
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.
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
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 need to convert these into mole ratios so they agree with our operating line that

we are developing.
we are developing.

So the mole ratio, again, is just simply the mole fraction divided by 1 minus the mole
So the mole ratio, again, is just simply the mole fraction divided by 1 minus the mole

fraction.
fraction.

And that is true for both the liquid phase mole fractions and the gas phase mole fractions.
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.
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
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
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.
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
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.
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.
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
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
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
rate, multiplied by our mole ratio in the liquid and then we have some information on

an intercept.
an intercept.

But this is just the line defining the relationship between the mole ratios in the gas and the
But this is just the line defining the relationship between the mole ratios in the gas and the

liquid phase.
liquid phase.

Now since we know this is no more than a material balance we know the point x=0 and y= 0.0304
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.
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
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.
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
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
is knowing the process parameter that we are operating at 1.5 times the minimum S/G ratio

of flow rates.
of flow rates.

So the first step is to determine what that minimum ratio, S/G ratio is.
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
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.
slope of the operating line.

So we need to find a minimum slope that will get us from these conditions to our inlet
So we need to find a minimum slope that will get us from these conditions to our inlet

composition of 0.087.
composition of 0.087.

So that minimum slope is going to correspond to an infinite number of stages,which will
So that minimum slope is going to correspond to an infinite number of stages,which will

be a pinch point at 0.087.
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
So I have drawn this point here on the equilibrium line and we also have our exit composition

here.
here.

So this operating line represents the minimum ratio of S/G that will accomplish this separation.
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
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.
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
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
actually determined that the ratio of our flow rates is actually going to be this term

here 1.74X10^-3.
here 1.74X10^-3.

So now we want to define a new operating line using this actual S/G ratio of flow rates.
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.
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,
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.
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.
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
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.
times the change in x value.

And then we just fill in the rest of those cells using those same basic formulas.
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
Now all that is left to do is just to step off the stages and see how many equilibrium

stages we need.
stages we need.

Since we are using a graphical method I blew up this plot just so it is easier to see the
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.
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
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.
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 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
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.
in terms of the mole ratio of 0.087.

And from the operating line we can determine the composition of the exiting solvent stream.
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
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
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.
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
So if we are at an equilibrium stage we are going to be in contact with that equilibrium

data.
data.

So we are going to draw a line across to that equilibrium data and this is going to be the
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.
composition of our first equilibrium stage.

Now we are going to see what the composition is in between stages.
Now we are going to see what the composition is in between stages.

In between stages the compositions are diced by our operating line.
In between stages the compositions are diced by our operating line.

So we are going to draw a line up to the 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.
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.
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
And then we see when we do two full equilibrium stages we have actually overshot our target

compositions.
compositions.

So now to determine the number of fractional stages needed we are just going to see the
So now to determine the number of fractional stages needed we are just going to see the

fractional distance along this line.
fractional distance along this line.

This line indicates how much of that second stage we needed.
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
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.
we have approximately 1.9 equilibrium stages required for this system.