Help with a gas compression problem

[drew_goring]

I am just wondering if anyone has any idea how to calculate the increase of temperature that results from compressing a gas to a certain pressure over a certain time. Is this even possible to do simply?

I'm just wondering how hot a volume of air gets when compressed to 5000 psi over a minute or so.

[FiveFreshFish]

http://www.chemicool.com/idealgas.html

[drew_goring]

Originally posted by FiveFreshFish
http://www.chemicool.com/idealgas.html

The ideal gas law doesn't take into account time, which I am assuming would involve thermodynamics to some degree.

[ExtraSlow]

The amount of time you take isn't important unless you want to take into account how fast you are radiating energy into the environment. If that's the case, you'd need to know the temperature of the outside environment, and the thermal conductivity of the container.

If this is for your thermodynamics class, try enthalpy. That's probably the ticket.

[drew_goring]

Originally posted by ExtraSlow
The amount of time you take isn't important unless you want to take into account how fast you are radiating energy into the environment. If that's the case, you'd need to know the temperature of the outside environment, and the thermal conductivity of the container.

If this is for your thermodynamics class, try enthalpy. That's probably the ticket.

Not school related. Personal interest in a situation that I came across actually.

[ExtraSlow]

well, I used to work daily with natural gas compression problems, and at some point I passed all those university thermodynamics classes. So I may be able to help if you can give more details about the specific situation.

In theory, the speed doesn't matter if you aren't losing energy to the outside environment.

[97'Scort]

In the real world, not a simple problem. The air will heat up an amount equal to the total electrical energy put into the compressor, minus friction losses, minus electrical losses, minus the heat dissipated from the system over the time period given. I believe it works out to a second order differential once you're all done.

If you want to approximate within 15% or so, you can idealize it with simple thermodynamics. It will depend on what kind of compressor you're using, which will most likely be a piston or screw type.

[drew_goring]

Originally posted by 97'Scort
In the real world, not a simple problem. The air will heat up an amount equal to the total electrical energy put into the compressor, minus friction losses, minus electrical losses, minus the heat dissipated from the system over the time period given. I believe it works out to a second order differential once you're all done.

If you want to approximate within 15% or so, you can idealize it with simple thermodynamics. It will depend on what kind of compressor you're using, which will most likely be a piston or screw type.

A compressor is not involved. I should have been more clear. Basically a large volume of water is compressed at 5000psi and it is slowly equalized with a much smaller sealed volume of air at atmospheric pressure. The water which is already at 5kpsi is compressing the air.

[nobb]

How much time is involved here? If it's just a few seconds and the container is relatively well insulated, then you can ignore time. Otherwise, to take into account time, you need to calculate the heat lost to the environment using thermal conductivity, area, difference in temperature.

[97'Scort]

I would need this in some context other than the description given. Can you tell me what's actually going on, as in, why you need to do this in the first place?

Edit: I'll add this in case it helps:

From what I gather, you are just equalizing water pressure with pressurized air as you are returning it to atmospheric pressure. I very much doubt that the heating of the air due to the compression will have a large effect on your process for two reasons: 1) If the volume of water is exponentially larger than the volume of air, the water will absorb the heat generated in the process, and 2) if you are doing this REALLY slowly, then you could neglect the heat from compression, since the system will have adequate time to dissipate the heat.

What I don't understand is how you can equalize the water with air at atmospheric pressure. You would need to expose it to air at 5 kpsi and gradually reduce the air pressure, otherwise you run the possibility of the water simply vaporizing when immediately exposed to atmospheric pressure.