Gas Cooling on Expansion 2

[jhartis]

In calculating downstream temperatures in a 50 psig compressed air system, I cane across an apparent puzzle. I know the Joule Thomson coefficient for air at my conditions is roughly 0.25 °C/atm so an expansion from 3 atm -> 1 atm should cool a little less than 1°C (correct?). I did an experiment pumping a Coke bottle to 50 psig, let the bottle cool back to ambient, and release the pressure. I haven't found a way to measure the temperature, but I feel sure the air cools more than predicted by the JT effect. I would guess it is cooling 5-10°C and there is often visible condensation. The gas velocity inside is nearly zero throughout the process, so the kinetic energy change in the bottle should be negligible. Any thoughts?

 

[25362]

To denniskb.

I can understand that a gas expanding through a throttling valve does it isenthalpically. However, from the inlet to the "vena contracta" it appears the expansion follows a quasi-isentropic path with much deeper cooling to such a degree, that some gases may partly condense, CO₂ and ethylene, could serve as examples.

I think this is a fact that may happen even with expanding superheated steam, precipitously forming high-velocity water droplets within the valve body, or solid hydrates in the case of moist gas.

For these reasons the internals of throttling valves are constructed from materials that can withstand the erosive action of such streams, and measures are taken to abate the resulting noise.

Am I right ?

 

[sailoday28]

25362 (Chemical)states:
I can understand that a gas expanding through a throttling valve does it isenthalpically.

For an adiabatic process the energy equation,neglecting change in elevation is
enthalpy in + kinetic energy in = enthalpy out + kinetic energy out.

isenthalpic only if negligible change in KE

 

[denniskb]

Sorry, I had the blowdown times in the wrong order.

I have made some minor improvements to the calc so the result are slightly changed.

I have added the bottle discharge temp including JT cooling which you can see is quite small due to the small pressure drop.

- Orifice sizes 2, 3, 5, 10, 15 mm.
- Blowdown times 6.12, 2.55, 0.92, 0.22, 0.09 secs
- Gas in bottle temp 10.7, -0.12, -28.6, -61.5, -71.5 °C.
- Bottle temp 17.9, 19.6, 21.9, 24.0, 24.6 °C
- Gas outlet temp 10.5, -0.25, -28.7, -61.6, -71.6 °C

For 25362, yes the low temp can easily lead to condensation and since the velocity through the orifice is high (sonic) erosion can easily occur. Also I believe the JT cooling effect includes the reheating as the high velocity gas through the vena contracta slows down again to the downstream velocity so local temperatures at the vena contracta may be lower than indicated.

I have a pdf file of the above blowdown cases but am not sure how to make this easily available to readers.


Dennis Kirk Engineering

http://www.ozemail.com.au/~denniskb

 

[sailoday28]

denniskb (Mechanical)
You have stated that sonic velocity is reached. This represents a signilficant change in kinetic energy. I repeat again that this change does not represent an isenthalpic process. Only then is J-T coef valid.

 

[abeltio]

jarthis,
what if you use a bottle of pepsi?

saludos.
a.

 

[25362]

To sailoday28. You are right in regard to the kinetic energy across the valve restriction itself. I think, however, that no real throttling process is exactly isenthalpic.

We speak here of two consecutive fast processes: an isentropic expansion, with an enthalpy reduction, from P1 to about 0.5 P1, followed by an enthalpy recovery after a short distance downstream, where the kinetic energy has already been reduced to a level negligibly different from that upstream, at the final pressure.

The first step has a cooling effect with the gas crossing the vapor line into the two-phase region. As denniskb says, after a certain distance the gas becomes again superheated at a lower pressure and even a lower temperature, and at practically the same original enthalpy.

The overall result being an isenthalpic throttling process.

From an example from literature, CO₂ at 20 deg F and 300 psig is expanded through a throttling valve to atmospheric pressure.

The gas undergoes two temperature changes: first, to ~ -35 deg F at ~ 145 psia; then, after a short distance, the downstream gas reaches ~ -60 deg F and atmospheric pressure.

The duration of the whole process is so short that it is insufficient to cause formation of droplets.

The primary result is a pressure drop in the gas. In the absence of heat transfer, without an appreciable change in elevation or kinetic energy, and no shaft work produced, it reduces to [Δ]H = 0, or H₂ = H₁. Checking the data from the article, I found a very small overall drop (~1%) in enthalpy, which enables us to call it an isenthalpic process.

Any comments ?

 

[sailoday28]

If the adiabatic process just up and downstream of the orifice is bounded by piping, the pressure drop will result in a change in velocity up and downsteam of the orifice. With a known pressure drop and flow, and the STAGNATION ENTHALPY remaining constant, the conservation of mass and energy (stag. enth) should be solved. If change in velocity (and therefore KE) is negligible, the overall process may be considered isenthalpic.

It is further stated
"The duration of the whole process is so short that it is insufficient to cause formation of droplets."

I don't see what the short distance/time has to do with the process.--- Unless one is considering metastable equilibrium. Having the thermdynamic properties in the metastable state can be tricky.
I do know for ex that the ASME stm tables used to (probably still do) have steam properties in the metastable state.
Regards

 

[denniskb]

If anyone is still monitoring this thread I have added a new section to my website and put a pdf copy of the calculation outputs under "miscellaneous files".

Look under "design calculations" to find it.


Dennis Kirk Engineering

http://www.ozemail.com.au/~denniskb

 

[jhartis]

25362, et al

I'm struggling with much of this, but you say that for constant kinetic energy, delta H = 0. But there is a large velocity change, as denniskb mentions, the velocity is likely sonic at the restriction and perhaps again at the discharge on the expansion to atmosphere (for cases with higher pressure in the bottle and assuming there is an internal restriction or orifice at the entry to the valve stem).

This has always been an obstacle to my understanding of JT: Does the constant enthalpy necessary for the application of JT refer to no heat transfer from the surroundings or no change in enthalpy due to expansion (change in velocity & KE)? If you don't have a change in kinetic energy, is it only really applicable to a piston expansion where the velocity change isn't very significant? Am I mistaken that the change in kinetic energy for a gas acceleration comes from a change in enthalpy?