lahthffire
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Calculating pressure...
I'm still experimenting with the supercritical methanol biodiesel reaction, and I'm having trouble with a calculation.
What I would like to do is characterize the PρT relationship for methanol at high temperature and pressure.
In other words, if I completely fill a container with vegetable oil and methanol reactants, seal it off, and heat it up to a certain temperature, what
will the final pressure be?
I would be happy even to simplify the situation and envision the vegetable oil as not taking part in the action. So can anybody give me any leads on
how to calculate the pressure of a 1 liter container containing 24.71 moles of MeOH (791.8 g; 1 liter at STP) when it's raised to 300 deg. C.? I
imagine the ideal gas law probably isn't very ideal for this situation ivolving a supercritical fluid with a density equal to the density of liquid
methanol at STP.
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mick
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I do not think that the ideal gas laws would work. If the methanol dissolves in the oil under the pressure and the is no gas space then you need
hydraulic equations. I do not think the low temp. hydraulic equations would work. It sounds like a computer job.
mick
Typo
[Edited on 16-4-2006 by mick]
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leu
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Perry's lists the vapor pressure of methanol at 224 degrees Centrigrade to be 60 atmospheres, and the critical point to be 240 degrees Centrigrade and
78.7 atmospheres; while CRC lists these values to be 239.4 degrees Centrigrade and 79.2 
Chemistry is our Covalent Bond
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Magpie
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Supercritical pressures can be calculated from PV=ZnRT where Z is a compressibility factor. Z can be determined from the "reduced property"
correlations of Hougen, Watson & Ragatz. (See Introduction to Chemical Engineering Thermodynamics, 2nd ed, by Smith & Van Ness.)
I tried to determine the pressure using the parameters you gave but could not converge on a solution.
[Edited on 17-4-2006 by Magpie]
The single most important condition for a successful synthesis is good mixing - Nicodem
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lahthffire
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Hi guys, thanks for the replies.
I think I got it figured out. I used the virial equation of state and used the first three terms of the expansion to calculate a pressure of ~17,000
psi. This is exactly in line with what I was expecting, so I'm happy!
The virial equation can be found down a ways on this page:
http://en.wikipedia.org/wiki/Equation_of_state
Vc = critical volume
I found an online calculator for critical temperature, pressure, and volume, but I didn't trust it so I found the value for methanol elsewhere to be
Vc = .118 l/mol. It turns out the calculator was fairly close (.113).
Here's the calculator:
http://www.pirika.com/chem/TCPEE/CriP/ourCP.htm
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Magpie
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Lahthffire, I would be wary of using the virial equation in your situation. Your gaseous methanol will be very dense, as dense as a liquid, as you
have noted. Perry's Chemical Engineers' Handbook, 4th ed., says "The relative contributions of successive virial coefficients is a function
of the density, the higher virials becoming extremely significant at higher densities. Accordingly, the convergence of the series becomes poorer at
the higher densities and in fact diverges for densities approaching that of liquids..."
Using the "Generalized compressibility-factor diagram" from the Smith & Van Ness reference cited above I determined the pressure at the smallest
volume I could use without going off the chart. This yielded the following:
V= 1.43 liters
Pr = 28
Z= 2.71
P=PrPc = (28)(78.5 atm)(14.7 psi) = 32,311 psi
where Pc = critical pressure
and Pr = P/Pc = reduced pressure
and Z = the compressibility factor valid for PV=ZnRT equation of state
I would think that at V = 1 liter the pressure would be considerably higher than 32,311 psi.
[Edited on 18-4-2006 by Magpie]
The single most important condition for a successful synthesis is good mixing - Nicodem
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