In this screencast I'll describe the Joule-Thomson expansion process and what the Joule-Thomson
coefficient is. The idea is we have a high pressure gas and we go through a porous plug.
Something that causes the pressure to drop to low pressure and in the process, typically
what happens is, we also lower the temperature. So this can be a gas at low temperature or
it can be a gas liquid mixture if the temperature is lowered enough to get into the two phase
region. The way that we write the first law for Joule-Thomson expansion is that delta H
is equal to zero. We are assuming that the process is fast enough that the rate of heat
transfer is zero. There's no moving parts so the shaft work is zero. And we're going
to ignore velocity differences so we're ignoring the differences in kinetic energy. Because
it's a non-ideal gas there will be this temperature change. This Joule-Thomson coefficient is
just the derivative of temperature with respect to pressure. This is a partial derivative.
This is at constant enthalpy. And then this turns out that it can be written as the temperature
times the partial derivative of volume with respect to temperature at constant pressure
minus the specific volume, the molar volume, divided by the heat capacity. This means that
we could use the equation of state to determine the Joule-Thomson coefficient. Now this can
be greater than zero or this can be less than zero. We're most interested in as we lower
the pressure we lower the temperature. But this depends on which pressure and which temperature
regime the high pressure gas entering responds to. Then we'll do an example here of using
Joule-Thomson expansion to liquefy gas and get a mixture of liquid and vapor. We have
oxygen at minus 100 degrees C and 110 bar. We are going to undergo Joule-Thomson expansion
to a final pressure of 2 bar. The question is, What is the final temperature after this
expansion? And if two phases are present, what fraction is in the liquid phase? So we're
going to use the Peng-Robinson Equation of State, use the spreadsheet from the thermodynamics
textbook by Elliot and Lira. That's available online, the details to use the spreadsheet
are given in other screencasts. For oxygen we have to look up the critical, the heat
capacity values, the temperature, pressure, and eccentric factor from the spreadsheet.
And I put in the initial conditions. You can see in yellow, coming in at 173 kelvin which
is -100 degrees C. And 11 mega pascals which corresponds to 110 bar. And we see that the
incoming enthalpy then is - 7285 essentially. So this of course means that H2, or the enthalpy
leaving, is also - 7285 J/mol. We have expansion, H1 and leaving H2. So we
want to use now the same spreadsheet and put in the new pressure and try and determine
the temperature. So here's the spreadsheet where we put in the pressure, 2 bar or 2 tenths
of a MPa and I guess the temperature of 100. Now we are in the three solution region
but the correct solution in the equation of state is the one with the lower fugacity.
Fugacity is lower here, that says we're in the vapor phase and this is the enthalpy.
Of course that's higher than our value of -7285 so this means we need to go to a lower
temperature. So here's the Peng-Robinson Equation of State for a lower temperature, 95 kelvin.
And three solutions. Lower fugacity is the liquid phase in this case and this number
is much lower than our enthalpy H2. That means we must be in the two phase region at a temperature
between these two because we go from vapor at 100 to liquid at 95. So what we can do,
certainly use solver, or it doesn't take much trial and error. Here's the spreadsheet where
I've adjusted the temperature such that the fugacity in the vapor phase is identical,
in of course significant figures, to the fugacity in liquid phase. So this is the saturation
temperature at this pressure for oxygen. And now we can read off enthalpy for the vapor
phase, enthalpy for the liquid phase, and we're in between these two so we can calculate
the quality by saying that H2 which is -7285 J/mol, is equal to some fraction
that's vapor. So this is quality, fraction that is vapor, times the enthalpy of the vapor
phase and then the fraction remaining is liquid and then enthalpy of the liquid phase. We
don't know x but we can substitute in the values from the equation of state spreadsheet
for the vapor liquid phase enthalpies. So I've made the substitution and done the math
and calculated that the quality is 0.78 which means that 78% is vapor and 22% is liquid.
We expand oxygen, high pressure, higher temperature, the Joule-Thomson expansion we convert some
fraction this creates 22% liquid.