Calculating temp change releasing a compressed gas

• matt621
In summary, the gas released from a compressed gas container will be cooler than the surrounding air. This is because it takes a long time for the pressure to build up and the gas will heat up as it is released. If the gas is released immediately, the temperature change will be smaller. Finally, if the gas is released into a vortex tube (aka Maxwell's Demon), the released air will be colder on the cold side than what comes out of the container.
matt621
Hello,

I understand when you release a compressed gas generally it comes out cooler than atmosphere.

But here are my questions:

When you fill the container, the gas warms up as the pressure builds. Generally it takes a long time to compress and I believe the temperature of the gas when it's released will vary by the rate of release. Is there anyway to calculate what the temp will be based on Total vol. of the container, pressure at the time of release and the rate of release?

Also If you compress it and immediately release it, what is the temp change compared to if you compress it today and let the container cool to ambient tomorrow? Or if you put the container in the freezer overnight? How much cooler would the gas coming out be?

(assume just general air for the gas.)

Now what if the air being released it then feed into a vortex tube (aka Maxwell's Demon.) Will the vortex tube separate the released air into even colder air on the cold side then what comes out of the container? Ie, the air released would be concentrated into the warmest parts and coldest parts, so the resulting air would be even lower than what it comes out of the container? (Since what comes out of the container is an average of all the air molecules.)

And if this true, then is there some way to calculate how much (% of the original containers vol.) going to the "hot" and "cold" sides?

Sorry for such a long question, but it's a question I have been trying to understand now for a long time. Trying to quantify is as much as possible.Thank you.

As long as the pressure isn't ridiculously high, you can just ratio the pressures and temperatures using the ideal gas law. That answers most of your questions: not sure about the relevance of Maxwell's demon here.

russ_watters said:
As long as the pressure isn't ridiculously high, you can just ratio the pressures and temperatures using the ideal gas law. That answers most of your questions: not sure about the relevance of Maxwell's demon here.

Pressure is 4500psi.

Not sure if you consider that "ridiculously high" or not.

Sorry boss, but I'm not getting it.

I have 2 transformations.

P1V1/T1=P2V2/T2

I can't solve this because I don't know V1.

I only know the pressure I'm staring with 1ATM, Temp I'm starting with 100F, or 310K

So T2=P2V2T1/P1V1

But at this point I don't know T2 because I don't know V1.

Then I have a 2nd transformation:

P2V2/T2=P3*V3/T3

T3 is what I'm looking for.

So T3=P3V3T2/P2V2

But using the above I found T2 is P2V2T1/P1V1 (as a function of V1)So T3=(P3V3)(P2V2T1)/P1V1)(P2V2) or P3V3P2V2T1/P1V1P2V2) or P3V3T1/P1V1 (P2's and V2's cancel) But V1=V3, so I'm left with P3T1/P1 but P3 and P1 are the same 1ATM so this shows T3=T1, Which can't be right because we all know the gas goes in ambient temp, compresses, heats up and then comes out colder, usually much colder. Right?

So where did I go wrong?

Also nothing in here takes into consideration TIME. The first transformation takes a long, long time. The 2nd takes just seconds. That has to have some effect on this.

But it still does not answer the question: If I compress the gas, it heats up in the container to some higher temp than what I started with, to what I call T2. But then over time, T2 eventually ( I think) goes back to ambient or T1.(maybe I'm wrong, but that heat is going to escape that container no matter how well I insulate it.) So then we I start the 2nd transformation, T2 is not really T2, it could less (if I put it in the freezer) or more (if I warm it) II still need to know what the ultimate temp of the escaping gas will be. And how much cooling the container at the end of the first step will effect the final temp of escaping gas.

Or is the problem my expectations? Do we only get a temp drop of it involves a phase change? Compressing "air" will not get a phase change. :-(

Yes, 4500 psi is "ridiculously high" and the ideal gas law won't work. What happens in this case will be highly dependent on the shape of the nozzle, but the potential exists for liquefying the air: that's how industrial gas plants do it.
http://en.wikipedia.org/wiki/Hampson–Linde_cycle

I was thinking more in terms of what you might find when you pump up and/or let the air out of your bike/car tires.

1. What factors affect the temperature change when releasing a compressed gas?

The temperature change when releasing a compressed gas is affected by the initial pressure of the gas, the volume of the container, and the type of gas being released. The specific heat capacity of the gas and the surrounding environment can also play a role.

2. How do you calculate the temperature change when releasing a compressed gas?

The temperature change can be calculated using the ideal gas law: PV = nRT. By rearranging the equation to solve for T, you can find the change in temperature by dividing the change in pressure by the initial pressure and multiplying by the initial temperature.

3. Is the temperature change the same for all compressed gases?

No, the temperature change will vary depending on the specific heat capacity of the gas being released. Some gases, like helium, have a lower specific heat capacity and will experience a larger temperature change compared to gases with a higher specific heat capacity, like carbon dioxide.

4. Can the temperature change be negative when releasing a compressed gas?

Yes, the temperature change can be negative if the gas is expanding into a larger volume and the surrounding environment is colder than the initial temperature of the gas. This can occur when releasing gas into a colder room or environment.

5. How does the volume of the container affect the temperature change when releasing a compressed gas?

The volume of the container can have a significant impact on the temperature change. As the volume increases, the gas molecules have more space to spread out and the temperature change will be smaller. Conversely, releasing gas into a smaller volume will result in a larger temperature change.

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