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?

 

[quark]

This is not a steady state process and can't be a JT expansion. Why can't you just use ideal gas equation (of state) to get the final temperture? What you should know is the initial and final mass of the air inside the bottle. (can a coke bottle withstand 50psig?)

Regards,


Believe it or not : Eratosthenes, a 3rd century BC true philologist, calculated circumference of earth with the help of a stick and it's shadow. The error was just 4% to the present day calcuated value.

 

[25362]

The J-T data mentioned by jhartis is for air free of moisture and carbon dioxide.

 

[Productmanager]

For joul Tomson effect to take place, the gas has to expand from high pressure to low pressure. The temperature of expanded gas can be found from the gas tables distributed by NIST. You take the difference between the enthropy at two state and using C calculate the change in tepmerature. Fast expansion is an adiabatic process and there is no loss of energy to the surrounding.

 

[25362]

Just for the sake of exactitude the change in temperature on a isenthalpic-adiabatic drop in pressure (a free expansion) is called the Joule-Thomson coefficient.

Quark is correct in the sense that in a "bottle" experiment the upstream pressure drops as the expansion takes place and the pressure difference is not constant.

Although the cooling effect is practically independent of the pressure at high temperatures, it increases markedly at lower temperatures.

Thus a difference of 200 atm lowers the temperature of air at 0^oC by 45 degrees, whilst at -90^oC the drop is of 100 degrees.

Hydrogen, for example, that heats up on expansion at ordinary temperatures, when precooled with liquid air, can be further cooled and eventually liquified by its own free expansion.

 

[1969grad]

The JT coefficient of 0.25 C/atm may or may not be valid as the coefficient does change with pressure.

To get a good JT coefficient check a pressure-enthalpy diagram assuming the expansion process to be a constant enthalpy process. Plot the start gas pressure and temperature and then draw a straight vertical line to the end pressure and read the temperature at that point.

 

[25362]

The J-T process is a throttling process without any appreciable change in kinetic energy of the downstream outgoing gas. Perry VI Table 3-151a gives J-T coefficients for air for such processes.

A quick (high flow rate) adiabatic expansion on the other hand, would result in T₁/T₂=(p₁/p₂)^(k-1)/k.

The experiment carried out by jhartis seems to be more a quick adiabatic expansion than a throttling process where the pressure ratio is coming from a starting 4.4 down to a final 1.0 .

 

[25362]

To make myself clear, even when the gas comes out through a true throttling process -with little cooling as the J-T coefficient brought by jhartis indicates- the gas left behind in the bottle undergoes cooling by expansion in a non-flow process. This expansion can be approximated by the adiabatic (reversible) expansion formula given above, although the "bottle" experiment doesn't seem to be adiabatic and the bottle would exchange heat with the surroundings.

So, the outgoing gas is cooler than expected by the J-T free expansion, because the "upstream" gas expanding inside the bottle is cooling down. This effect doesn't happen when the system (upstream) pressure is constant, and the only process playing a role is the isenthalpic expansion of the air coming out.

 

[iainuts]

Expansion of a gas through a valve or orifice is isenthalpic, ie: enthalpy remains constant.

Expansion of a gas in a cylinder where pressure is released is approximated by an isentropic process, ie: entropy remains constant.

In the case of the isenthalpic process, no work is done by the fluid.

In the case of the isentropic process, work is done, and the gas will be significantly cooler because of this.

 

[quark]

Actually this process is adiabatic and lowest possible temperature of the gas left in the bottle can be checked by the equation provided by 25362. But I never perceived such lower temperatures when evacuating gas cylinders, for example. This is because we can't vent out the bottle so fast that any heat input into the bottle can be neglected.

Still, I doubt the practicality of conducting this experiment in a coke bottle and original poster didn't turn up again to excuse my ignorance.

Regards,


Believe it or not : Eratosthenes, a 3rd century BC true philologist, calculated circumference of earth with the help of a stick and it's shadow. The error was just 4% to the present day calcuated value.

 

[iainuts]

Adiabatic means only lack of heat transfer. The first law is applicable here. Change of internal energy = enthalpy in - enthalpy out plus etc ...

For expansion through a valve, the adiabatic process is isenthalpic.

For expansion of gas in a cylinder, work is done by the gas. Yes, this is an adiabatic process also, but easiest to calculate as isentropic assuming you have the NIST tables or other thermodynamic data as mentioned earlier.

Hope that clarifies just a little.

 

[jhartis]

Thanks to all. While some of these posts help, I'm still a little confused. To answer Quarks question about the pressure capability of a Coke bottle, I pumped a standard 2 liter PET Coke bottle to nearly 150 psi, before I chickened out and quit pumping.

I see now that I really have 2 processes, the expansion of the gas in the bottle and the gas that blasts out though the valve. In spite of the differences, the bottle gets almost as cold as the nozzle. And Quark, I know the experiment isn't very practical, but it was some fun and piqued my curiosity, not a bad bargain.

I don't believe work is being done by either the expanded gas or the throttled discharge gas. Or am I wrong and energy lost somehow by accelerating the discharge gas? A JT expansion occurs though a nozzle. What happens to the kinetic energy in the accelerated gas in that case?

If the eqn suggested by 25362 & Quark (T1/T2=(p1/p2)(k-1)/k) applies, the gas should get quite cold as mentioned. Admittedly the process is not entirely adiabatic, but it only takes a few seconds to release the pressure in the bottle and the plastic is not very conductive nor is it a large heat sink, it should be nearly adiabatic. How is this different from a JT expansion?

 

[25362]

You are absolutely right. For an adiabatic expansion the temperature T₂' of the remaining air inside the bottle would be T1/1.5 or about 300/1.5=200 K=-73^oC !!

Air may function as an ideal gas and the process was quick and thus adiabatic, but definitely irreversible. There is indeed no shaft work done. However, because of friction effects and the energy spent in "pushing" the outcoming air molecules the efficiency of the expansion would be just, say, 30%.
Taking Cp=1.0 kJ/kg-K for air, the total "available" energy of expansion would be = Cp(T1-T₂') equal to

1.0*(300-200)=100 kJ/kg air.

The final "actual" air temperature T₂ inside the bottle after expanding it down to atmospheric pressure can be estimated as follows.

Assume 30% efficiency, the "lost" work would be 70 kJ/kg.

The entropy change related to this lost work would be

S₂-S₁=70/200=0.35 kJ/kg-K

And since S₂-S₁=Cp ln(T₂/T₂')

0.35=1.0*ln(T₂/200), thus T₂=283 K=10^oC.

Which is a reasonable result confirming your findings, and may serve to explain quark's own experience when emptying gas cylinders.

The J-T expansion didn't contribute much to this cooling, especially considering it was not a "steady" process been effected with a small (decaying) delta P.

 

[unclesyd]

What would an Infra-Red camera shown in the space occupied by "jhartis's" bottle if it had burst?

 

[quark]

Jhartis!

Are you still there? I am very sorry that I was very quick in deciding the practicality of the problem. I saw closed coke bottles (empty) bulging when exposed to Sun for a while. That is why I thought it was dangerous to pressurize them (ofcourse, I forgot junkyard wars). I did a lot of reading after my second post and didn't find a single clue to prove JT expansion for the gas which remains in the bottle. Some of the books stated it as adiabatic expansion.

One more thing you have to mind is that, 1lb of moisture requires 1000 btu/lb and if you measure the quantity condensed, you can have some estimation about the actual temperature of gas in the botle. I will have some discussions meanwhile and I will let you know if anything comes up further.

PS:I think 25362 is very close to the answer.

Regards,

 

[owg]

I would like to know if the thermodynamic "daemon" available for online calculations at the link shown below, can help in solving this problem. I input P1 and T1 conditions for a real gas (air) of 3 atm and 60 deg F, and final conditions of 1 atm, and the initial enthalpy condition as calculated by the daemon for the P1, T1 conditions. The answer is shown below and results in a drop of about 1 deg F for the expanded gas.

State-1: Air > Superheated Vapor;
Given: p1= 3.0 atm; T1= 60.0 deg-F; Vel1= 0.0 m/s;
z1= 0.0 m;
Calculated: p_r1= 0.080629975 UnitLess; T_r1= 2.1789098 UnitLess; Z1= 0.99847066 UnitLess; v1= 0.27215365 m^3/kg; u1= -92.871346 kJ/kg; h1= -10.143444 kJ/kg;
s1= 6.507626 kJ/kg.K; e1= -92.871346 kJ/kg; j1= -10.143444 kJ/kg;

State-2: Air > Superheated Vapor; Given: p2= 1.0 atm; h2= -10.14344 kJ/kg; Vel2= 0.0 m/s;
z2= 0.0 m;
Calculated: p_r2= 0.026876656 UnitLess; T2= 59.10502 deg-F; T_r2= 2.1751573 UnitLess; Z2= 0.9994693 UnitLess; v2= 0.81587 m^3/kg; u2= -92.81147 kJ/kg;
s2= 6.8225365 kJ/kg.K; e2= -92.81147 kJ/kg; j2= -10.14344 kJ/kg;

http://eng.sdsu.edu/testcenter/testhome/indexdaemons.html

HAZOP at

http://www.curryhydrocarbons.ca

 

[25362]

To owg, why have you assumed a constant enthalpy expansion for the air in the bottle ?

 

[owg]

I was running the original problem of letting down 3 atm air to 1 atm in a steady state operation. I believe that would be enthalpic. I also ran the isentropic case for air saturated with water vapour but I did not save the results. Try this link for the thermodynamic daemon. It seems to have more features than the testcenter version referenced in my previous post.

http://www.me.udel.edu/testcenter/indexdaemons.html

HAZOP at

http://www.curryhydrocarbons.ca

 

[iainuts]

Saying "Adiabatic" is very missleading. Adiabatic only indicates an absence of heat transfer.

Adiabatic expansion through a flow restriction is very different from adiabatic expansion of a gas through an expansion engine. In the first case, the gas that is expanding does no work. In the case of the gas expanding in the cylinder work is performed by the gas.

Using the first law of thermodynamics is the proper way to solve either of these problems. In the case of a gas being released from the bottle, the change in internal energy of the gas inside the bottle is equal to the enthalpy of the gas exiting the bottle.

In this case, you must perform the analysis iteratively since as the pressure decays, the gas gets colder, and the enthalpy of the gas venting from the bottle drops.

If done iteratively, you will eventually get the correct answer.

Alternatively, you can draw a control volume around the gas that never leaves the bottle. At first, it occupies a very small amount of room at the bottom of the bottle, and as it expands, it eventually displaces all the other gas and fills the bottle.

In this case, one can immediately conclude the gas that has expanded has undergone an isentropic expansion, just like the piston in a cylinder.

Note that I've done this analysis both ways and both methods give identical answers.

The result is about -190 F. I doubt it actually gets that cold though, since the process is not adiabatic. Heat transfer from the thermal mass of the bottle is still very significant. I've had to do numerous problems similar to this one at cryogenic temperatures, and I've always found that the heat transfer from the vessel is a very significant factor, so "adiabatic" it is not.

 

[25362]

If the bottle were well insulated and the expansion quick enough, an adiabatic process could be safely assumed, not so for reversibility, thus it wouldn't be isentropic, and the thermophysical properties of the leaving gas would differ from those of the gas left behind in the bottle. Iainuts please comment.