solving Joule-Thompson pressure drop 4

[Davidowitz]

I need help solving a problem, probably with the Joules- Thompson effect. I know that if I have a pressure reducing valve (regulator), when I increase the inlet pressure by 300 psi, for the same outlet pressure, I get a temperature drop of 21 degrees F. I need the equation showing this. I'm changing the inlet from 600 psi to 900 psi, and the outlet pressure remains 350 psi, how would I demonstrate the associated temperature drop?

 

[BigInch]

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

need something else?

http://virtualpipeline.spaces.msn.com

"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain

 

[Davidowitz]

Yes- I couldn't solve the equation at the wikipedia site. First, going thru the regulator we have P1=600psi; P2=350; T1=60 degrees F; T2 unknown; V2=unknown and V1=unknown. then I have the same info but going from P1=900psi to P2=350psi. Too many variables.

 

[btrueblood]

J-T expansion is an isenthalpic process (enthalpy does not change), in the ideal world.

direct from wikipedia:

In an isenthalpic process:

h1 = h2
dh = 0
Isenthalpic processes on an ideal gas follow isotherms since dh = Cp(T2 ? T1)


(italics added for emphasis)

 

[Davidowitz]

Ok- so how would I use that in my case? I don't know enthalpy.

 

[vpl]

Can't you get the enthalpy from the steam tables? You know the initial pressure and temperature.

Patricia Lougheed

Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of the Eng-Tips Forums.

 

[Davidowitz]

I don't have he steam tables, and this is natural gas.

 

[vpl]

Try NIST:

http://webbook.nist.gov/chemistry/fluid/

While it doesn't list "natural gas", you probably can figure out which of the gases listed comes closest.

Patricia Lougheed

Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of the Eng-Tips Forums.

 

[BigInch]

Hint: methane

http://virtualpipeline.spaces.msn.com

"What gets us into trouble is not what we don't know, its what we know for sure" - Mark Twain

 

[zdas04]

I had a project last month where one of the options was to install a J-T valve and I find the assertion above that dh=cp(dT) to be wrong, possibly because Cp is a function of temperature and possibly just because it is wrong.

If you look at a p-h (Mollier) diagram for methane you'll see that none of the isotherms follow the isentrops (in fact the isotherms are shaped interestingly for methane)--i.e. THIS IS NOT AN ISOTHERMAL EVENT, it is close enough to isentropic for that to be a useful simplification.

In the NIST data referenced above there is a Joule-Thomson coefficient (units are F/psia). So if you start at 60F and 600 psia then the J-T coefficient is 0.054324 for methane. So dropping to 350 psia should give you a temp drop of about 13.6F. Changing to 900 psia (still at 60F) the J-T coefficient goes to 0.051265 and the temp drops by 28.2F.

Another way to approach this is using NIST's REFPROP.EXE you can put in starting conditions and on the next line put in your new pressure and the enthalpy from the line above and the program will calculate the temp. In this case it tells me that you should have 45.8F in the original case (temp drop of 14.2F, so the J-T method is off by 4%) and in the new case you get 28.3F (temp drop of 31.7F so the J-T method is off by 11%). The two methods are reasonably close.

Unless your gas is dehydrated I'd be really concerned about freezing in your new conditions.

David Simpson, PE
MuleShoe Engineering

http://www.muleshoe-eng.com

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

The main difference beteen humans and apes is that we have cooler tools

 

[btrueblood]

David,

It's not my assertion, it's how the thermo textbooks define a Joule-Thomson process. The assumption is that no work is done by the fluid in expanding, but heat transfer DOES occur (thus JT refrigeration processes). I'd guess that would take a certain length of line (or a hx) to allow the fluid to reach thermal equilibrium to its surroundings, and thus approximate a JT process.

You are correct that a typical orifice or nozzle expansion flow is better fit by an adiabatic or isentropic model (no heat transfer), and that may be what a JT valve looks like if you measure the u/s and d/s temperatures close to the valve. Dunno, never used one.

Finally, "my assertion" (the textbook's actually, via wikipedia to Sonntag & Wiley) included the italics (mine) that it was for an ideal gas; the OP never stated what he was using (natural gas) until after my post; you are correct that real gases, especially polyatomic ones, can/do have non-constant Cp.

 

[btrueblood]

Wups, sorry. The other part of the difference between JT and isentropic expansion is the recovery of the work done (acceleration of the fluid) in isentropic expansion by fluid-dynamic drag, converting the kinetic energy of the gas to turbulence and thus to heat. This too takes some time and distance to occur in real gas flows, thus temperatures close to the valve can/do approximate isentropic process values, and farther away and with heat transfer, they approximate the JT numbers.

Clear as mud now?

 

[zdas04]

OK, you got me going back to my graduate school texts.

Change in enthalpy is the definite integral from T1 to T2 of (cp)(dT) which is cp*(T2-T1) only if cp is a constant [sorry about the crappy representation, eng-tips.com really needs an equation editor].

In real gases, cp is defined as T*((partial s)/(partial T)) at constant pressure. This means that for a truly isentropic process (i.e., both adiabatic and reversible; which requires dS=0) then cp must be zero, but its not ever zero.

I need to ponder this somemore.

David

 

[insult2injury]

The assumption is that no work is done by the fluid in expanding, but heat transfer DOES occur (thus JT refrigeration processes)

Heat transfer does not occur. It's conversion of energy between two different forms that results in the temperature change.

h = u + pv

In a J-T expansion, internal energy (molecular motion, vibration, etc.) is converted to pv (pressure-volume) energy. The decrease in internal energy results in a change in temperature while not affecting the overall enthalpy as the decrease in u is equal to the increase in pv (no losses).

If heat transfer were to occur, then there definitely would be a change in enthalpy.

I2I

 

[mbeychok]

Davidowitz:

Ideal gases do not undergo the J-T effect. The J-T coefficient for ideal gases is zero.

The response from "insulttoinjury" is correct. The J-T effect is an isenthalpic process (constant enthalpy) and no work is done. The J-T valve should be heavily insulated to prevent any heat transfer.

If a gas expands through a turbo-expander or if steam expands through a steam turbine, mechanical work is done. That is an isentropic process (constant entropy) and also results in a large temperature drop (much more than in a J-T expansion).

Read the Citizendium article at

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

The article at Wikipedia has been messed up by too many fingers in the pie.

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

 

[mbeychok]

btrueblood:

Just to clear up what you wrote earlier in this thread:

The assumption is that no work is done by the fluid in expanding, but heat transfer DOES occur (thus JT refrigeration processes).

The J-T valve itself is heavily insulated to prevent any heat transfer to the ambient amosphere during the gas expansion. The expansion results in lowering the gas temperature (i.e., cooling the gas) and that cooled gas is then used to cool whatever process needs cooling. In other words, heat transfer does not occur during the expansion itself ... any heat transfer (from the cooled gas to whatever process needs cooling) occurs after the J-T expansion has cooled the gas.

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

 

[sailoday28]

While this may seem minor, adiabatic flow thru a valve is not a constant enthalpy process. Stagnation enthalpy is constant. Neglecting elevation effects, h+v^2/2= constant.
KE will change thru a valve. If KE effects are negligible, then J-T is appropriate.
Regards

 

[mbeychok]

sailoday28:

Hi, sailoday28. Now that you've chipped in , I'm outta here!! Bye-bye.

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

 

[LeSabre]

Davidowitz:

Getting back to your problem, the Joules- Thompson effect applies to natural gas regulation. To approximate the outlet temperature, you can use the basic definition of Mu, the Joules- Thompson coefficient,

Mu=(T2-T1)/(P2-P1)

Note that the J-T effect is independent of V1 and V2, so you can calculate the regulator outlet temperature as,

T2 = T1 + Mu * (P2 - P1)

The J-T coefficient is a function of both P and T and you can get good values of this from the NIST website given above. However, if you note that within a small range of temperatures typical of pipeline problems, the relation between Mu and P is weak compared to the relation between Mu and T, so you can get good approximate values of Mu if you interpolate between Mu= 0.07375 @ 460 R and Mu=0.049753 @ 560 R.

 

[insult2injury]

If you have enthalpy data, then you are better off setting h(P1,T1) = h(P2,T2) and solving, or following a constant enthalpy line on a diagram. Your results will be more accurate than using an estimated/interpolated value for the J-T coefficient (mu).

If you do use mu to determine the result, I would recommend looking up the enthalpy value at the calculated exit condition and comparing to the inlet enthalpy to ensure reasonable accuracy of the result.

I2I