Joules Thompson Effect Calculation - Help required 1

[GasProcess]

Will some body please help in calculating the temerature of the natural gas of known composition when flashed from a known pressure and temperature to a lower pressure.

There are condensable hydrocarbon fractions in the gas that will condense as the pressure is reduced.

Regards

 

[Homayun]

Do you have access to a simulation tool like HYSYS or Aspen? If you do, why don't you use that?

 

[jsummerfield]

Work.

If you flash across a control valve or choke valve there is no work. If you flash across an expander that drives a compressor, the work increases the refrigeration. Now let the chemical or mechanical engineers clarify this thing work but it likely does not apply of you are dropping across a valve.

John

 

[mbeychok]

When a liquid is at its boiling point at a given pressure and that liquid undergoes a pressure drop across a valve, some of the liquid will immediately vaporize (i.e., "flash" into a vapor). Both the residual liquid and the flashed vapor will decrease in temperature to the boiling point of the liquid at the lower pressure.

When a real gas undergoes a pressure drop across a valve, there is no "flashing". However, when a real gas expands at constant enthalpy (i.e., no heat is transfered to or from the gas, and no external work is extracted), the real gas will be either cooled or heated by the expansion. The change in real gas temperature with the change in pressure is called the Joule-Thomson coefficient and is denoted by mu. Mathematically:

mu = (dT/dP) at constant enthalpy

The value of mu depends on the specific gas, as well as the temperature and pressure of the gas before expansion. For all real gases, mu will equal zero at some point called the "inversion point".

If the gas is below its inversion point temperature, mu will be positive when the isenthalpic gas expansion occurs (i.e., the gas will be cooled because when mu is positive, dT must be negative since dP is negative for a gas expansion).

If the gas is above its inversion point, mu will be negative when the isenthalpic gas expansion occurs (i.e., the gas will be heated because when mu is negative, dT must be positive since dP is negative for a gas expansion).

You can look up tabulations of mu versus temperature and pressure in various references. Perry's Chemical Engineers' Handbook has some such tabulations.

It should be noted that mu is always equal to zero for perfect ideal gases (i.e., they will neither heat nor cool upon being expanded).

I suppose that if the Joule-Thomson effect cools a real gas mixture (such as raw natural gas) sufficiently, then some of the heavier molecular weight components of the gas could condense.

As for how the Joule-Thomson effect is achieved in practice:

(1) The real gas is allowed to expand through a throttling device (usually a valve) and that throttling device must be very well insulated to prevent any heat transfer to or from the gas.

(2) There must be no external work extracted from the gas during the expansion (i.e., the gas must not be expanded through a turbine, for example).


Milton Beychok
mbeychok@xxx.net (replace xxx with cox)
(Visit me at www.air-dispersion.com)