Why do tanks get hot when you fill them from higher pressure tanks?

[rjack321]

He is right. Ignore the source. Pretend you just have a valve on the wall the gas comes out of. Could be from a compressor or could be from the bank. You don't know. Now you would expect it to heat the tank, wouldn't you? As the gas in the tank goes from low pressure to higher pressure it gives off heat by raising its temperature.

Close.
Gas will only ever flow from high pressure to a lower pressure. Btw, water will only flow downhill too. Well except for tides but that's just extraterrestial gravity at work.

 

[randini]

this is wrong

---------- Post added February 21st, 2013 at 06:03 PM ----------

It has to do with the rate of collisions between the individual molecules of gas. As they become more compressed they have more impacts, more "vibration" which we understand to be temperature. It is adiabatic cooling process in reverse.

Care to tell me what about my post is wrong?

 

[dumpsterDiver]

Care to tell me what about my post is wrong?

P versus T is the equation? .... Do a few example problems for me will you?

How how hot will the tank get when it goes from 50 to 3000 psi?

after you do a few examples, then you can tell me where you went wrong.

 

[randini]

No, P versus T is not the equation. More precisely P is proportional to T is the formula, but the version I posted earlier allows the comparison of a gas in two situations. Go do your own problems.

Since you claimed that I was wrong, provided a definition of temperature and proceeded to claim that adiabatic "cooling in reverse" is the answer, explain to me how adiabatic warming (or if you prefer go ahead and explain it using your adiabatic cooling in reverse) works without G-L's law.

 

[Ghanguss]

If you have two identical tanks, one full (t1) and one empty (t2) (absolute pressure) joined by a hose, and open the valves on both tanks what happens?

1. The volume of the gas has doubled; the pressure halved-therefore the temperature remains constant. This is known as an isothermal process.

What actually happens is t1 gets colder while t2 gets warmer. This is due to the location where the air is expanding (t1 valve only) and where the air is compressing (t1 valve through to t2).

Now you only need to consider the effect of closing the valve before the pressure equalizes between the two vessels, but I think you will find the same logic applies.

 

[RJP]

This is sort of a geeky physics/gas law question, but I thought someone here might be able to answer it for me or at least point me in the right direction. I've been diving for over 40 years, but I've never really understood why tanks get hot when you fill them from other tanks. I signed up here just to ask.

Do you believe that the air going into the tank being filled somehow KNOWS it's being filled from a compressor vs being transfilled from another tank... and should somehow behave differently vis-a-vis heating? Or do you think that the laws of physics apply in either case?

:d

 

[Ghanguss]

Do you believe that the air going into the tank being filled somehow KNOWS it's being filled from a compressor vs being transfilled from another tank... and should somehow behave differently vis-a-vis heating? Or do you think that the laws of physics apply in either case?

:d

I think the point of the OP is that overall the net pressure is dropping from the full tank to the empty one, therefore why is there not a net cooling effect.

 

[RJP]

I think the point of the OP is that overall the net pressure is dropping from the full tank to the empty one, therefore why is there not a net cooling effect.

The "overall net pressure" is not dropping. The overall net pressure IN THE SYSTEM is the same. This is why we say that tanks "equalize" - pressure in one is dropping, pressure in the other is rising. The one dropping cools, the one rising heats up.

 

[dumpsterDiver]

No, P versus T is not the equation. More precisely P is proportional to T is the formula, but the version I posted earlier allows the comparison of a gas in two situations. Go do your own problems.

Since you claimed that I was wrong, provided a definition of temperature and proceeded to claim that adiabatic "cooling in reverse" is the answer, explain to me how adiabatic warming (or if you prefer go ahead and explain it using your adiabatic cooling in reverse) works without G-L's law.

You are wrong, if you will not do a few example problems, then I will not use any more time to try to explain it to you. Are you really a scientists?

 

[halocline]

I think the missing part of the equation is time. The OP is wondering why gas in a tank at say 3000PSI at room temp (say 70F) ends up in another tank at 1500 PSI, but is now warmer, say 90F. I think he's wondering where the heat energy is produced. It 'should' be such that the net cooling and heating is zero, or less than zero, rise due to the net loss in pressure.

And actually, if you were able to measure the actual temp of each air molecule and track it through time, you'd probably find that was the case. But simply feeling the tank, you're feeling radiant heating/cooling from contact with the air, and the metal in the tank absorbs and retains this heat relatively slowly. And, it's probably a different rate of radiance depending on the ambient temperature and maybe even the properties of the metal, I really don't know. But certainly if tank one cools off as air goes from 3000 PSI to near ambient, that 'could' be a measurable amount of heat release, then tank two heats as the air goes from near ambient to 1500 PSI, and that 'could' be measurable, and if so, the net heat release would exceed the net heat increase. And if you could somehow measure these amounts and display real-time numbers representing them, you'd see the delay in radiance pass from one tank to the next, and to the ambient temp, eventually equalizing. And if you could measure actual total heat loss/gain, you'd find a net loss, but only after the radiant effect on the tank metal had passed. At least that's my guess!