Compressed AirTank

[bhasker2]

I would appreciate if someone could correct me in this understanding.

If a compressed air tank with 3000 psi and an internal saturation pressure much higher than ambient temperature is exposed to the ambient, the vapor pressure within the tank would decrease releasing heat and consequently decreasing internal temperature. Additionally, condensate would accumulate as long as vapor pressure is present.

Am I correct on that? I'm trying to quantify the condensate accumulated in the tank if I let a compressed tank cool by using ambient temperature, without dropping absolute pressure.

 

[racookpe1978]

Theoretically -> Yes. Practically? Yes.

My air compressor tank collects (condenses from the mild but humid air sucked into the compressor) a considerable mount of water in the bottom of the tank every hour that it runs.

How much will depend on your inlet conditions (relative humidity) and amount of air that is being used.

 

[bhasker2]

Racookpe1978,

Thanks for your reply. I understand the thermodynamics within the compressor and am able to calculate the accumulation of condensate after each compressor cycle.

What I need to know is when I've compressed the air into the tank. If I let enough time pass by, how much additional condensation will occur, or if any. The case remains that the saturation temperature is much higher than the ambient temperature in the tank, which means the dew point is much less than the ambient temperature, so condensation should occur.

I'm just not certain what happens within - since there's no volume change and pressure remains at compressed pressure in the tank, how would temperature change?

 

[bhasker2]

Sorry, I didn't mean "dew point is LESS" I meant MORE.

 

[fegenbush]

bhasker2,

If you used an ideal gas, the temperature would remain the same for a given pressure. Air, however, is not an ideal gas. The condensation of the vapor is a product of the cooling of the air.

If you look at a psychrometric chart to determine the difference between the air at elevated temperature with known relative humidity and the air at ambient temperature with 100% relative humidity, you can calculate the amount of condensation present.

 

[bhasker2]

fegenbush,
I can't use the psychrometric chart, as the temperatures I'm dealing with are much higher. And, to be honest, I'm still unsure if that's the correct method for my problem.

Perhaps if I provide some values and known information:

I have already calculated the condensation occurring after the compressor and its intercooling. So, after the compressor has done its service, I'm left, in the tank, with 3000 psi and temperature at 120 deg C. Since the saturation temperature is a function of vapor pressure, that's at 160 deg C. Now, saturation temp. > tank temperature, so I will get condensation. I have this quantified.

Now, I guess you can think of the next process as a cooling stage. So, if I leave that closed compressed tank with its temperature at 120 deg C and pressure at 3000 psi in a room at 25 deg C, what will happen?

My assumption is that cooling will occur within the tank by the decrease of the tank's vapor partial pressure, which in turn will result in additional condensation. This is inline with what you said, "condensation of the vapor is a product of the cooling of the air"

If correct, how do I go about quantifying that?

 

[racookpe1978]

OK. Well, think about it, how would "more humid air" get into the compressed air tank other than through the compressor itself?

So, cool but humid air goes into the compressor inlet, comes out hot (due to the compression) and still humid. Your tank now gets 8, 10, 15, 25 times as much humid air into its belly as was outside. When all that air cools back down to ambient, the relatively rises, the moisture condenses and runs down to the drain. IF you don't get that water out of the tank right away, it re-saturates the compressed air, rusts the tank bottom, and will get through your system into the application (paint, tools, instruments, motors, valve actuators, etc. If that moisture is trapped in some line downstream of the tank, it can freeze there and plug the line (cutting off instruments or causing valves to freeze or just rust out and plug the line.

Once that condensed water is drained, (by definition) there should only be humid air - but no physical water - in the tank at the tank's new ambient temperature. By volume, the humidity is the same as original since the tank volume did not change. By mass, the relative humidity inside is much less, since the water vapor remaining is now divided up among all the extra air crammed into the tank .

But, other than air trapped in some depressurized line away from the tank, there is no other way to get minsture into the system tank.

 

[racookpe1978]

So, if I leave that closed compressed tank with its temperature at 120 deg C and pressure at 3000 psi in a room at 25 deg C, what will happen?

Within an hour or so, the tank and air will cool from 120 C (393 K) to 25 C (298 K).

(Are you sure the air will go in at over the boiling point of water?

PV = n RT, (and no extra air has gone in the tank), and the tank volume (realistically) has not changes, so pressure will drop by 3000 x (298/393) = 2275 psig.
So your new water condensation calc needs to be for 2275 and 25 C, not 3000 psig and 25C.

Until you turn the compressor on again to regain your 3000 psig set point. 8<) So, eventually, you will be back at 3000 psig AND 25 C, but with MORE air in the tank than you expected from your first calc.

 

[bhasker2]

racookpe1978,

Firstly, thank you so much for your time.

Are you sure the air will go in at over the boiling point of water? - The outlet temperature of the compressor is given at 250F (120C). So, yes the temperature in question is correct, but am I missing something here?

And, that's what I wasn't sure of - the proportionality between pressure drop and temperature drop. (3000 x (298/393) = 2275 psig)

What I had done was: assumed the tank temperature was going to cool all the way to 25C. Then, take the saturation pressure at that point as vapor pressure of the tank within, since it is 100% humidity after compression. Obtain the humidity ratio at that point in the tank and subtract from the humidity ratio right after compression. But, I now realize that if I assume vapor pressure was going to decrease, so must the total pressure, since total pressure = partial pressure of vapor + partial pressure of dry air.

So, all in all, I must re-adjust my humidity ratio(0.622*Pv/(P-Pv)) at 25C from [0.622*Pv/3000-Pv] to [0.622*Pv/2275-Pv], correct?

 

[tr1ntx]

Compressed air initially fills the tank. You close it. As presented, you're saying the tank contents are 100% gas phase at that instant. Air temp drops, some water vapor changes to liquid, thus mass of gas phase contents drops. Liquid occupies space at bottom of tank, thus tank internal free volume declines. Less mass, less volume, same pressure as initially. I can't do the math any longer without significant refreshing, so I'll leave it to you. Is this the answer for your issue

 

[bhasker2]

tr1ntx,

That's what I've gathered after piecing information. My need is to quantify the condensate collected, so that I can accurately size a moisture separator and a heat exchanger.

I can share with you my computation, if desired. I've almost convinced myself that it's correct, but I can definitely use another pair of eyes. Let me know, and I can share an excel file.

I do want to, once again, thank everyone for their input. It's absolutely remarkable how much support there is on this forum. Hopefully, as I progress in my Engineering career, I can embark onto a similar capacity as many of you.

 

[Compositepro]

All you need to know is the humidity of the air entering the compressor, and the vapor pressure of water versus temperature. Water will condense in your tank until the partial pressure of water equals the vapor pressure at the tank temperature. Don't confuse yourself with "humidity ratio".

 

[bhasker2]

Compositepro,

That's the scaffold I used. Good to know that it's in the right direction.

Humidity ratio, aka specific humidity, allows me to quantify the vapor present after the whole cooling scenario. Once obtained, I can subtract from the initial vapor content, which then gives me the total condensate in the tank.