We're going to do a flash calculation for an ideal solution so we're going to use Raoult's
Law.
So we have a feed of liquid.
We're given the liquid composition, and then we're given the outlet temperature and pressure.
So measurements were made if you like, otherwise we would have to do energy calculations.
And we're given Antoine's calculations for saturation pressures.
And the question is what fraction of the effluent from this flash is liquid and what are the
compositions of the two phases that are in equilibrium?
So the first thing we want to do is draw a diagram representing the system.
So let me pause and make that diagram.
So here is a representation.
We have the feed composition assuming moles of feed and what we're going to do is make
this one mole as a basic.
We calculate the fraction of vapor and the fraction of liquid.
The liquid feeds at a temperature and a pressure that we don't know for this problem because
we're given the outlet temperature and pressure, those are what's important.
Clearly it's a high enough pressure to make it a liquid.
We're going to solve for the mole fractions in the vapor phase and the liquid phase, so
this should be X2.
And then L is the number of moles of liquid and V is the number of moles of vapor, and
they're going to have to add to one because we're feeding just one mole.
So we have Raoult's Law that is going to relate the mole fractions and mass balances that
will hep us determine the amount of liquid and vapor.
And so to apply Raoult's Law we're going to have to use saturation pressures.
So I'm going to rewrite these equations for saturation pressures and substitute in the
temperature in degrees Centigrade, 150 degrees C. Notice these are natural log, often Antoine's
equation can be log base 10.
In this case they're natural log.
So I substitute in 150 degrees C and I get the log of the saturation pressure, 7.475.
Which means if I take the exponent of both sides, I can calculate P1-sat and that's 1763
kilo-pascals.
Do the same thing for saturation pressure two.
Now that we have saturation pressures, we can substitute into Raoult's Law which relates
the mole fraction in the liquid phase, saturation pressure of that component to its mole fraction
in the vapor phase times the total pressure.
Let me write them for both components.
If I had these together, I'm eliminating the mole fractions in the vapor phase and Y1 plus
Y2 times pressure, so this is pressure.
We know the saturation pressures now, we know the total pressure, we can solve for X1.
So note that when I made the substitution, this was X2.
So now I have an equation just in X1 that I can rearrange and solve for X1.
And so I calculated X1 and of course X2 is such that X1 plus X2 adds to one.
We can now go back to Raoult's Law and solve for Y1, which is just X1 P1-sat over the total
pressure.
We just calculated it, X1.
We calculated P1-sat already - 1763 - and total pressure in the same units so that Y1
is equal to 0.765 and of course then Y2 again so the Ys add up to one, so it will be 0.235.
So we're almost finished, what we need to do now is go back to the mass balance to determine
what is the fraction that is liquid.
The mass balance, the overall mole fraction Z1 times the total molar flow rate X1 times
the liquid molar flow rate, Y1 times the vapor.
Or you can look at is as the fraction of liquid and the fraction of vapor where I'm going
to pick a basis of one.
So mole fraction in the feed, pick a basis of one mole.
I just calculate X1, I don't know L, this is going to be the fraction now because we
picked a basis of one, the fraction of the effluent that is liquid.
And then I calculated Y1 times V, and V of course is one minus L because I picked a basis
of one mole.
V plus L equals the feed, one mole.
So remember, this equation is just a mass balance.
It says the amount of component one entering is equal to the amount of component one leaving.
So I can rearrange this equation to solve for L. So I've rearrange and I've solved for
L. So this says we have 69% liquid, the effluent from the flash, 31% vapor and we calculated
for the liquid that X1 to two significant figures is 0.53, Y1 is 0.77.
So we have the compositions and we have the fractions for this flash calculation.