This problem we are given a size for a propane tank and we are going to try to put 3.5 kg
of propane.
And we are going to keep the temperature 25 C. And what we want to do is use the Peng-Robinson
EOS to model the propane at these conditions and we want to calculate, what's the pressure
in the tank.
So the spreadsheet that has been developed for the Peng-Robinson EOS is from the Elliot
and Lira textbook and I've provided a link here to that spreadsheet.
We are going to go to that spreadsheet but first let's look at the conditions that we
are going to use.
The pressure we need to determine, the temperature, 25 degrees C so 298 K. And then the volume,
so we have 80 L and keep in mind this is the specific volume, so we want L/mol, so we want
80 L and 3.5 kg.
So I'm going to do some unit conversion to put the volume in the units that are appropriate
for the spreadsheet.
And the spreadsheet uses cubic centimeters per mole.
So that means 1 kg, 10^3 grams, 10^3 cm^3/L and then the molecular weight, 44g/mol.
I do this calculation, I get 1,006 cm^3/mol.
So I have temperature and volume and I want to determine pressure.
So what the spreadsheet is going to use is the critical pressure, critical temperature,
and the eccentric factor.
This information will be available in the spreadsheet.
So now let's go look at the spreadsheet and see how we determine the pressure.
So here we have the Peng-Robinson EOS spreadsheet.
And I have done a couple things- I've made some of the fonts bigger and highlighted them
with the parts that we are most interested in.
Now what I've done is I've put in the critical temperature and pressure and eccentric factor.
I just went to the critical properties here and copied and pasted them here.
I put in the temperature of the system, and I just guessed the pressure.
Because the spreadsheet is setup to enter the pressure and temperature and calculate
the volume.
Since we are given the volume we have to work a little differently.
One other thing I have to be able to do to work on this, is I'm going to go into review
and I'm going to un-protect the spreadsheet.
That allows me to change cells that otherwise are restricted.
So I guessed 0.1 mPa.
The volume of my system, 1006, well you can see that this condition the vapor volume is
much larger than that.
This is the lower fugacity of the two.
Here's the fugacity of the vapor, here's the fugacity of the liquid.
Since this is lower that means that at these conditions we would be at the vapor phase,
much too large a volume.
So let's increase the pressure.
Well, still in the vapor phase, still much too large.
Still in the vapor phase, too large, let's double it again.
And now notice the lower fugacity is for the liquid.
It is much too small a volume and so the vapor of course, too large a volume.
What does this mean?
We went to liquid phase much too small, we are in the vapor phase, too large.
This means that at these conditions we must be in vapor liquid equilibrium.
So that means we need to determine the saturation pressure and then we should be able to see
if indeed that is the condition where we have equilibrium for the system at 298 K.
So what I'm going to do, here's the fugacity ratio.
That's the ratio of the vapor to the liquid fugacity.
At saturation this number is 1 so I'm going to go into data, into solver, and I'm going
to change this fugacity ratio, that's my objective, I want the fugacity ratio to equal 1.
I'm going to change the cells for pressure, namely B8.
So if I tell it to solve, comes up with the solution.
Gives the pressure 0.95 essentially mPa.
Well now the fugacities are equal so we are in vapor liquid equilibrium.
Here's the liquid volume per mole, here's the vapor volume.
1000 is between these two and so this means that roughly, it says, that half the volume
is liquid and half is vapor.
So we could calculate the exact amount but what we are interested in here is what is
the pressure.
So when we fill the tank at these conditions, with this many kg, we are going to be at saturation
pressure which is 0.95 mPa or 9.5 bar pressure.