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

[jimmyw]

The pressure in the tank being filled is rising, there is adiabatic heating - even if the gas is cool to start with.

The pressure of the gas is what's important. That pressure is falling. Every molecule of gas in the scuba tank started off at higher pressure in the donor/fill tank and ended up at lower pressure in the scuba tank.

 

[randini]

I know this isn't the answer either. Charles law essentially says that at a constant pressure as absolute temperature increases, the volume of the gas also increases in proportion. We aren't looking at constant pressure and I'm trying to understand why some of the gas gets hot as it expands while other parts of the gas get colder. I'm pretty sure it has to do with the fact that this is an irreversible process in which entropy increases and the fact that the scuba tank is not thermally connected to the fill tank.
It's not Charles's Law, Dalton's Law, Boyle's Law, Henry's law or even the Ideal Gas Law. I'm becoming more confident that it has to do with the path the second 1cc parcel of air follows through P, V and T as it moves from the fill tank to the scuba tank.

---------- Post added February 21st, 2013 at 02:56 PM ----------


Well that's the whole point of my question. The gas in the scuba tank doesn't go from low pressure to high pressure. It starts at high pressure in the fill tank and room temperature. It ends up in a larger volume, at a lower pressure and a higher temperature. Just focus on the gas. It's at a high pressure in the fill tank and ends up at a lower pressure in the scuba tank. There's no way to avoid that and any explanation has to take that into account.

They are two separate tanks: the donor tank (the high pressure) and the recipient tank (the one being filled). Forget the donor tank for a second, the reason the temperature increases in the recipient tank is because gas went into it and increased it's (the recipient tank's) pressure. The interior of the recipient tank is increasing in pressure and, therefore, in temperature.

Likewise as the recipient tank fills, the interior of the donor tank is decreasing in pressure and cooling.

I don't know what you are talking about with the path of the second cc? Are you trying to use some sort of venturi effect to explain it?

 

[eponym]

Here's one question about the process, using a non-math version of Boyle's law . . .

First, consider the cooling that takes place when releasing air from a cylinder. Don't connect anything to it, just crack the valve open. The valve gets so cold that ice forms on it. Why? Because those air molecules that were crowded in the cylinder were moving very little. In order to move a lot faster when they expand into the surrounding one-atm field, they must speed up. They do this by drawing energy (heat) from their surroundings, especially the metal of the valve (and cylinder).

Now consider that process in reverse. Does it apply to a closed (two-cylinder) system? Is the above example a non-diabatic system?

 

[jimmyw]

When temperature changes, pressure changes
When pressure changes, temperature changes
expanding gasses cool. compressing gasses warm up.
That's pretty much the gist of it. It's not that hard.
R..

So why does the high pressure gas that starts in the donor fill tank get hot when it ends up at lower pressure and larger volume in the scuba tank? This is hard. It's very hard. Youi're missing something that's very fundamental in this whole process. It might help to think of the scuba tank as being welded to the side of the donor/fill tank and then punch a hole at the wall. The total amount of gas in the donor fill tank expands in volume and drops in pressure. By the simple rule it should all cool off.

Or think of the scuba tank as being a cylinder welded to the wall of the donor fill tank that's open to the donor fill tank and has a piston that's pushed against that wall. Then you slowly retract that piston until the cylinder has the volume of the scuba tank. Again, the total volume of gas in the donor/fill tank will have increased and the temperature should decrease. There's something very interesting going on here.

 

[lowviz]

...//...I'm pretty sure the answer lies in how the gas is transferred from the fill bottle to the scuba bottle, but I don't quite have it figured out. ...//...

Another wild a$$ guess.

Since I agree with your concerns from a simple energy conservation basis, try this: Starting with an evacuated tank, you get initial cooling in the lines and in the tank itself. The surrounding atmosphere will add energy to the tank and transfer lines. After the fill, and when everything equalizes, you get this energy back as an increase in temp. Now, if you could run your fill with both tanks in a vacuum chamber you may get the result that you expect...

(continuing eponym's post)

 

[jimmyw]

They are two separate tanks: the donor tank (the high pressure) and the recipient tank (the one being filled). Forget the donor tank for a second, the reason the temperature increases in the recipient tank is because gas went into it and increased it's (the recipient tank's) pressure. The interior of the recipient tank is increasing in pressure and, therefore, in temperature.

Gas laws apply to gas, not to tanks. It's what happens to the gas that's important. Imagine a tiny scuba tank that slowly expands in size as it fills. Like a balloon that holds a constant pressure of 3000 psi. Let that balloon fill at a constant pressure.

Likewise as the recipient tank fills, the interior of the donor tank is decreasing in pressure and cooling.
I don't know what you are talking about with the path of the second cc? Are you trying to use some sort of venturi effect to explain it?

With respect to the "path" of the second cc: By "path" I meant the graph of how the pressure temp volume changed as it went from the starting pressure temp volume to the final PVT.

---------- Post added February 21st, 2013 at 04:09 PM ----------

Here's one question about the process, using a non-math version of Boyle's law . . .

First, consider the cooling that takes place when releasing air from a cylinder. Don't connect anything to it, just crack the valve open. The valve gets so cold that ice forms on it. Why? Because those air molecules that were crowded in the cylinder were moving very little. In order to move a lot faster when they expand into the surrounding one-atm field, they must speed up. They do this by drawing energy (heat) from their surroundings, especially the metal of the valve (and cylinder).

Now consider that process in reverse. Does it apply to a closed (two-cylinder) system? Is the above example a non-diabatic system?

Thank you for the response. I agree that the valve gets very cold and the molecules escaping speed up. I don't agree that the reason they speed up is because the molecules draw heat from the surrounding. The reason they speed up is the pressure difference is accelerating them and the reason they cool is that they are expanding adiabatically.

I've thought about the process in reverse. In reverse, we'd have to add energy to push those molecules in against the pressure. The gas would heat adiabatically. If the gas started cold (as it came out of the nozzle, then the heating would be just enough to bring the gas back to room temp.

---------- Post added February 21st, 2013 at 04:17 PM ----------

Another wild a$$ guess.

Since I agree with your concerns from a simple energy conservation basis, try this: Starting with an evacuated tank, you get initial cooling in the lines and in the tank itself. The surrounding atmosphere will add energy to the tank and transfer lines. After the fill, and when everything equalizes, you get this energy back as an increase in temp. Now, if you could run your fill with both tanks in a vacuum chamber you may get the result that you expect...

Thank you for this response. It's excellent. I also thought this could be the answer. However I ran some numbers and it can't be the complete answer. You can get some heating from the fact that there's cooling at the valve and heat leaks into the system there. However, that heat isn't enough. Even running in a vacuum chamber with perfect insulation, one tank would cool and one would heat up.

 

[rjack321]

The pressure of the gas is what's important. That pressure is falling. Every molecule of gas in the scuba tank started off at higher pressure in the donor/fill tank and ended up at lower pressure in the scuba tank.

Whatever pressure the gas used to be at is not relevant. The pressure could be ambient, go through a compressor, come out of the backpressure reg at 2000psi and go into a tank at 500psi. Or it can start out in a 4000psi bank and go into a near empty 500psi tank. Does not matter.

You are filling a tank, from any source, the pressure is RISING!

Edit: To add a little bit. An individual molecule has no "pressure". Pressure is the sum of all the collisions between the individual gas molecules and the wall of the tank. More collisions per second= higher pressure.

Adiabatic cooling by its very definition involves zero transfer of heat or energy. Its caused simply the change in pressure. The supply tank is losing pressure (molecular collisions) and the gas (+tank) gets colder. The receiving tank is gaining pressure and the gas inside is getting warmer, it transfers some of this energy to the tank. The tank and valve on the supply are colder than ambient and they gain some energy from their surroundings. The receiving tank is warmer so it loses some energy. The amount of heat/energy lost by the receiver and the amount gained by the supply are equal.

Has absolutely nothing to do with nozzles and venturi's etc. They only serve to slow down the process slightly and limit the ability for the gains and loses to equalize via the gas through the hose.

 

[dumpsterDiver]

Guy-Lussac's law is closer : P1/T1 = P2/T2

@OP: The physics doesn't care if you are filling the tank from a compressor or from higher pressure tank. In your original post you were on the right track when you talked about how the energy put into a tank from a compressor could be recovered by opening the tank's valve. That's what you're doing to the donor tank to fill the recipient tank: you open it's valve! Moreover you should notice that that the donor tank cools as the air leaves it (pressure decreases).

From your ideal gas law discussions, the volume does not significantly change because the tank's walls are rigid (you'd need a hydrostatic test station to measure the tank's expansion). What you are adding to the tank is molecules of gas (i.e. 'n' in the IGL), thereby increasing the density of the air inside the tank (hence why it gets heavier and negatively buoyant as you fill it) by compressing it (adding pressure).

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.

 

[BRT]

I know this isn't the answer either. Charles law essentially says that at a constant pressure as absolute temperature increases, the volume of the gas also increases in proportion. We aren't looking at constant pressure and I'm trying to understand why some of the gas gets hot as it expands while other parts of the gas get colder. I'm pretty sure it has to do with the fact that this is an irreversible process in which entropy increases and the fact that the scuba tank is not thermally connected to the fill tank.
It's not Charles's Law, Dalton's Law, Boyle's Law, Henry's law or even the Ideal Gas Law. I'm becoming more confident that it has to do with the path the second 1cc parcel of air follows through P, V and T as it moves from the fill tank to the scuba tank.

---------- Post added February 21st, 2013 at 02:56 PM ----------


Well that's the whole point of my question. The gas in the scuba tank doesn't go from low pressure to high pressure. It starts at high pressure in the fill tank and room temperature. It ends up in a larger volume, at a lower pressure and a higher temperature. Just focus on the gas. It's at a high pressure in the fill tank and ends up at a lower pressure in the scuba tank. There's no way to avoid that and any explanation has to take that into account.

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.

 

[Diver0001]

So why does the high pressure gas that starts in the donor fill tank get hot when it ends up at lower pressure and larger volume in the scuba tank? This is hard. It's very hard. Youi're missing something that's very fundamental in this whole process. It might help to think of the scuba tank as being welded to the side of the donor/fill tank and then punch a hole at the wall. The total amount of gas in the donor fill tank expands in volume and drops in pressure. By the simple rule it should all cool off.

Or think of the scuba tank as being a cylinder welded to the wall of the donor fill tank that's open to the donor fill tank and has a piston that's pushed against that wall. Then you slowly retract that piston until the cylinder has the volume of the scuba tank. Again, the total volume of gas in the donor/fill tank will have increased and the temperature should decrease. There's something very interesting going on here.

Can you explain in more detail exactly what's happening?

R..