temerature drop due to pressure drop

[usamoh]

how much temperature drop for a pressure drop
i have to answers
1- 12 celisus for 6.78 bar
2- 1 celisus for 3 bar

 

[Feric]

Hi there:

Depends upon what kind of process your are dealing with.

Thanks,

Gordan

http://engware.i-dentity.com

 

[usamoh]

it is natural transmission gas pipe line

 

[dcasto]

natural gas as a JT coefficient of 6 deg F per 100 psi. In SI 3 deg C per 6 bar is a close value.

This is for pressures near 50 barg and 20 C.

 

[mbeychok]

dcasto:

Unless the gas undergoes a sudden pressure drop by passing through a valve, an orifice plate or a similar device, I don't think that there will be any J-T effect at all. Any temperature drop due to the pressure drop would simply be due to heat transfer ... and that would depend on the pipeline length, the type and amount of insulation, the pipe material, and the outside temperature.

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

 

[25362]

Concerning the J-T coefficient.

Assuming no heat transfer from or to the surroundings, and assuming a J-T effect takes place due to the expansion in the line, from Perry VI (table 3-150) there is an inversion of the J-T coefficient at about 60 Mpa (reduced pressure[≥]12).

Meaning that the gas could heat up on expansion at all temperatures at pressures this high.

At T_R[≈]1.5, the inversion would take place at reduced pressures higher than 10.

 

[dcasto]

mbeychock, the gas could get even colder too. the JT effect is not dependant on sudden change.

25362, yep, thats why I stated near 50 barg and 20C.

 

[mbeychok]

dcasto:

Please read thread69-37069 (in the Petroleum production engineering forum) to see just how much disagreement there is as to whether or not J-T effect depends on having a flow obstruction device (valve, orifice etc) in the pipeline.

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

 

[vzeos]

The gas does not need to undergo a sudden pressure drop by passing through a valve or an orifice plate. The JT effect applies to pipe flow. In a pipeline the pressure loss is due to friction, therefore the process is completely irreversible and the enthalpy is constant. Since the flow is isenthalpic, you can follow a constant enthalpy line on a Mollier diagram (for any fluid) from any P₁, T₁ to P₂ and you can read T₂ without considering the pressure reduction element (valve or an orifice plate). Also see the attached paper, Physics of Pipeline Flow.

http://www.energy-solutions.com/pdf/tech_paper_Modisette_Physics_of_Pipeline_Flow.pdf

 

[dcasto]

and I've seen the effects on an ethylene line. It's just that the heat transfer from thr ground into the pipe offset the temperature effect so it isn't noticable. How would a molecule know the difference between knocking around a mountain on the side of a pipe or smashing around a sharp curve in some valve?