Simple Joule Thomson Cooling Question

[spikedturtle]

Very simple question but its doing my head in.

How do you use the joule thomson tables in Perrys engineering handbook?

I have a table of pressures and temperatures. If I know I am at a certain temperature and pressure and I know the pressure drop how do I get the resultant temperature. The JT coefficient is the rate of change of temperature with pressure, so it makes sense that the coefficient times the drop should be the temperature change.

So are the pressure given to be taken as drops? And I just multiply that by the JT value and take it away from my temperature?

Any direction given would be much appreciated.

Table reference is 2-151

 

[dcasto]

The JT effect can be found in tables and typically the tables show change in temperature per pressure change. Like & degrees per 100 psi. SO a 500 psi drop would have 5 times 7 of a 35F drop.

Another way is to calculate the enthaphy at the beginning pressure and temp, then iterate the cahnge in temp to get the same enthalpy at the new pressure.

 

[mbeychok]

spikedturtle:

You might find it useful to read this article in the online Citizendium encyclopedia:

Click here ==>

http://en.citizendium.org/wiki/Joule-Thomson_effect

Milton Beychok
(Visit me at www.air-dispersion.com)
.

 

[PaoloPemi]

In my edition (sixth) of Perry there are several tables with Joule Thomson values specified for substance and ranges of temperatures and pressures, for example for carbon Dioxide at 60 atm and 0 C you get a value of 0.0370 C / atm which means that dt (0.0370 C) per atm
The tables in Perry are limited to a limited number of pure substances (with the exception of air) a more preferable solution is to adopt an equation of state,
for the Joule Thomson (dT/dP)H = (dH/dP)T / (dH/dT)P
so to calculate the value you need a software with the enthalpy derivatives vs. temperature and pressure, tools as Prode Properties or Nist Refprop have that and can calculate all the values which you need.