Why do tanks get hot when you fill them from higher pressure tanks?
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I offered to go through the mathematics and physics to show how and why the tanks heat/cool without any compression. No one asked to see that, but eventually someone may read this thread and wonder if I could show what's happening mathematically. I thought I'd at least lay out the basic process of following the gas from the time it starts out at room temp pressure, as it gets compressed into the donor tank, then as it expands into two tanks. Here it is:
1) We start with 80 cf of air at standard pressure, temp (P and T) and run it through a compressor. The compressor adds energy to the gas causing it to heat up as it squeezes it into the tank. We can solve for the new PVT as follows: The new volume V is 11 liters. The two equations PV=nRT and PV^gamma=constant can be solved for the new P and T. Gamma is 1.4 for air. The gas ends up at higher pressure and higher temperature. The equation with gamma is derived from the constant entropy assumption. It's the heat capacity ratio.
2) We let the donor tank cool to room temperature T. We lose heat energy as this is done, and the pressure P drops. Volume is unchanged at 11 liters. Using PV=nRT and the new temperature, we can solve for the new pressure P.
3) We are now ready to do the transfill operation from the room temp high pressure first tank into a vacuum empty second tank. We will consider two different volumes of gas in the donor tank, divided in half. The first half of the gas will stay in the donor tank and expand to fill that tank. The second half of the gas will leave the tank and freely expand in the receiving tank. The first half expands adiabatically at constant entropy and does work on the second half pushing it out of the tank. The first half of the gas cools as stored energy is released to the second half. The cooling due to expansion can be calculated using the two formulas in number one above. The second half of the gas expands adiabatically, but not at constant entropy, instead at constant enthalpy in the "free expansion" process and remains at constant temperature due to this process as it moves at high speed into the second tank. If you wonder why the second half of the gas expands without changing its temperature, see the "Joule expansion = free expansion" Wikipedia article linked above.
4) When the first half of the gas expands, it releases energy. That energy goes into accelerating the departing half of the gas, and initially appears as kinetic energy of motion of the mass of the second half. We can determine how much energy is released by the expansion from the equations in the Wikipedia article on energy storage in compressed air. Because the departing gas is initially moving rapidly and carries kinetic energy, but then stops when it arrives in the second tank we can calculate the heat released by that process. Since no energy was added to the system, the heat energy is equal to the energy given up by the expansion of the first half of the gas. It is this energy that heats the gas in the second tank.
We end up with two tanks of expanded gas, one hot and one cold. If we let them come to thermal equilibrium with each other, without adding/subtracting any heat from the environment, the heat in the second tank is just enough to heat the colder gas in the first tank to room temp. Once that happens, all of the gas can be said to have expanded according to the "Joule expansion" or "free expansion" process, which is a constant temperature gas expansion process. The existence of a constant temperature gas expansion process is what was new to me. Like others here (many others) I thought that gas always cools as it expands. It doesn't - it only cools when energy is extracted.
---------- Post added February 26th, 2013 at 04:26 PM ----------
1) We start with 80 cf of air at standard pressure, temp (P and T) and run it through a compressor. The compressor adds energy to the gas causing it to heat up as it squeezes it into the tank. We can solve for the new PVT as follows: The new volume V is 11 liters. The two equations PV=nRT and PV^gamma=constant can be solved for the new P and T. Gamma is 1.4 for air. The gas ends up at higher pressure and higher temperature. The equation with gamma is derived from the constant entropy assumption. It's the heat capacity ratio.
2) We let the donor tank cool to room temperature T. We lose heat energy as this is done, and the pressure P drops. Volume is unchanged at 11 liters. Using PV=nRT and the new temperature, we can solve for the new pressure P.
3) We are now ready to do the transfill operation from the room temp high pressure first tank into a vacuum empty second tank. We will consider two different volumes of gas in the donor tank, divided in half. The first half of the gas will stay in the donor tank and expand to fill that tank. The second half of the gas will leave the tank and freely expand in the receiving tank. The first half expands adiabatically at constant entropy and does work on the second half pushing it out of the tank. The first half of the gas cools as stored energy is released to the second half. The cooling due to expansion can be calculated using the two formulas in number one above. The second half of the gas expands adiabatically, but not at constant entropy, instead at constant enthalpy in the "free expansion" process and remains at constant temperature due to this process as it moves at high speed into the second tank. If you wonder why the second half of the gas expands without changing its temperature, see the "Joule expansion = free expansion" Wikipedia article linked above.
4) When the first half of the gas expands, it releases energy. That energy goes into accelerating the departing half of the gas, and initially appears as kinetic energy of motion of the mass of the second half. We can determine how much energy is released by the expansion from the equations in the Wikipedia article on energy storage in compressed air. Because the departing gas is initially moving rapidly and carries kinetic energy, but then stops when it arrives in the second tank we can calculate the heat released by that process. Since no energy was added to the system, the heat energy is equal to the energy given up by the expansion of the first half of the gas. It is this energy that heats the gas in the second tank.
We end up with two tanks of expanded gas, one hot and one cold. If we let them come to thermal equilibrium with each other, without adding/subtracting any heat from the environment, the heat in the second tank is just enough to heat the colder gas in the first tank to room temp. Once that happens, all of the gas can be said to have expanded according to the "Joule expansion" or "free expansion" process, which is a constant temperature gas expansion process. The existence of a constant temperature gas expansion process is what was new to me. Like others here (many others) I thought that gas always cools as it expands. It doesn't - it only cools when energy is extracted.
---------- Post added February 26th, 2013 at 04:26 PM ----------
Here is the physical evidence of what happened when Joule did this test (from the Joule expansion page):
I learned here that helium has a slightly negative index as compared to an ideal gas and He and hydrogen would heat up slightly if allowed to expand. Those are second order effects as compared to the underlying process explained by assuming the gas is an ideal gas.
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Irrelevant, since there IS compression.
You're like my roommate who developed a formula that proved that bats can't fly. Of course, there was only one problem with his calculations...
---------- Post added February 26th, 2013 at 07:30 PM ----------
Because it's irrelevant. Per that article (I can't believe I actually just read the thing!) the second container begins as an evacuated container...
"The Joule expansion is an irreversible process in thermodynamics in which a volume of gas is kept in one side of a thermally isolated container (via a small partition), with the other side of the container being evacuated."
If this example applied to our scuba tanks you would have one tank holding 10 liters of gas (pressurized donor tank) and a receiving tank holding ZERO liters of gas (because it is evacuated there is no volume of gas in the tank) so your starting volume of gas is 10 liters (10 + 0 = 10) When the partition/valve opens you do in fact end up with expansion of the gas from donor into receiving tank. If you closed the valve/partition as soon as the receiving tank pressure was 14psi/1atm you would then have 20 liters of gas - 10 liters at 2986psi and 10 liters at 14psi. And - you are correct AT THAT POINT IN TIME, AND ONLY IN A NONEXISTANT HYPOTHETICAL SITUATION
- there are now 20 liters of gas, so there has been expansion from 10+0 to 10+10
- the 14psi of gas in the receiving tank would be colder
However, I'm assuming you want to scuba dive in the real word, not a nonexistant hypothetical world. In the real world you wouldn't have a very good dive with 14psi of breathing gas, so you open the valve again and let the tanks equalize. At the end of the transfill process the receiving tank will end up with 1507psi of gas having been compressed into that tank... and that gas will be warmer than it was when it was in the tank it came from.
In the REAL WORLD there is NO expansion in this situation. As has been stated repeatedly, in the real world you START with 20 lites of gas (10 liters at 3,000psi and 10 liters at 14psi) and END with 20 liters of gas (10 liters at 1507psi in the donor tankl and 10 liters at 1507psi in the receiving tank.) If any of those 20 liters of gas had expanded, you would end up with more than 20 liters. But you don't; you end up with the same 20 liters of gas that you started with. And, has been observed millions and millions of times - by everyone from Cambridge PhDs to uneducated tank-monkees - the air in the receiving tank ends up warmer than it was when it was in the donor tank.
CT-Rich
Contributor
hate to rain on your parade, but in the close system you are describing, there are two tanks at 10 liters each (10+10=20)
one tank has 3,000psi, the other has 0 psi. You can't magically have a tank appear in a closed system. Because it is a closed system. The tank is there whether it has 0 psi or 14 psi, or 3,000 psi. PhD's from cambridge love to bitch slap people who make foolish assertions. Most (perhaps not all) tank-monkees would bitch slap you for calling them tank monkees. But the air fill personell would certainly have noticed that the donor tanks would get cooler and the receiving tank would warm up. The cambridge professors while contendedly enjoy the sting on their hands after administering a good bitch slapping, would understand that the total energy in the system would remain constant. A smaller bottle would feel warmer than the larger tank, only because the change in energy would be spread out over a larger mass.
There is no free ride in thermodynamics, unless it involves a good bitch slapping....
---------- Post added February 26th, 2013 at 10:14 PM ----------
I thought this thread was in advanced. Please forgive, the above. It is primarily in jest (although at its core I was making a valid observation about an obvious flaw in logic that undermined his entire arguement). Sorry if I was too snarky...
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the second tank is "there" of course, and the exterior of the tank displaces 10 liters of volume/air in the room it is in. however, because in this hypothetical situation the second tank is a perfect vacuum, it has zero volume INSIDE the tank. I can show the the formula, if you like...
:d
PS - as a part time tank monkey myself (+/- education) I used that term with all due respect and affection.
PfcAJ
Contributor
In response to "I don't agree that tanks being filled always get hotter" I propose a simple experiment:
Acquire empty scuba tank. Quickly fill scuba tank. Put tongue on tank. Video. Post video to SB.
Acquire full scuba tank. Open tank valve. Put tongue to tank valve. Wait. Get friend to take pics. Post pics to SB.
Acquire empty scuba tank. Quickly fill scuba tank. Put tongue on tank. Video. Post video to SB.
Acquire full scuba tank. Open tank valve. Put tongue to tank valve. Wait. Get friend to take pics. Post pics to SB.
CT-Rich
Contributor
Unless you are flying a TARDIS, I can't see the logic of the arguement. Saying it has zero volume in a perfect vacuum and then popping into existance after opening the valve. Following your arguement, the addition of a single molecule of gas would magically make the 10 liters pop into existence. The mass of gas in the empty cylinder would be zero, the energy in the empty cylinder would be zero, but the volume of the cylinder would be zero? what you are describing is the big bang, not a tank fill. Add a little matter and BOOM a 10 liter universe? A quantum SCUBA tank?
Of course all this is relative. From the perspective of the vacuum tank, you add ANY volume of gas, you are increasing the pressure and energy. From the perspective of the donor tank the pressure is always going down. From the perspective of the outside observer the you are shifting gas with in a system with no change in the overall volume or energy of the system... With no external force acting on the components, the system remains the same. If you want to prove your point, draw and post a graph of what is going on in the system, and each tank respectively. You will immediately see why your zero volume arguement doesn't hold up.
---------- Post added February 27th, 2013 at 06:55 AM ----------
There is no brief moment of expansion. That would assume that the outside effect of the atmosphere has any effect at all, which it wouldn't. The pressure inside the tank is always going up.
I feel like I'm in one of those movies where there's a plot twist at the end where it turns out that the PROTAGONIST in the story is actually the one who is insane.
I probably shouldn't have gotten "sucked into" the OP's hypothetical vacuum argument. (See what I did there? :d)
Yeah, I got sloppy with my thinking/wording. What I was trying to say is that there is ZERO volume of gas in the hypothetical "perfect vacuum" receiving tank, not that the space inside the tank doesn't exist. (Though, if you had a hypothetical "perfectly collapsible" scuba tank for the "perfect vacuum" receiving tank, there would in fact be 0liters of space to go along with the 0liters of gas at 0psi.)
But in any case... in the bizzarro world example, you start with gas occupying 10 liters of space (10l in donor tank at 3000psi and 0l in receiving tank at 0psi) and end with gas occupying 20 liters of space. Therefore, in bizzarro world, expansion has occurred. In the real world, however, you would start with gas occupying 20 liters of space (10l in donor tank at 3000psi and 10l in receiving tank at 14psi) and end with gas occupying 20 liters of space. (10l in donor tank at 1507psi and 10l in receiving tank at 1507psi) 20=20, therefore no expansion has occurred.
All of this marginalia aside, let's get back to the original question:
Q: "Why do tanks get hot when you fill them from higher pressure tanks?"
A: "Because the gas in that tank has been compressed*, and the simple, immutable laws of physics tell us that when a gas is compressed its temperature rises."
*whether the energy that caused that compression came directly from a compressor or from the stored energy of a 3,000psi donor scuba tank (which came from a compressor) is irrelevant.
It is what's happening if you start with a vacuum empty receiving tank. I've done that with an oxygen tank fill and the expanding oxygen still gets hot. If you start with a tank of residual room temp air, then you have to add in the heat from the residual air that is compressed. I don't disagree with anyone that wants to say that residual air gets compressed and that makes it hot. I agree with that, but it's just some additional heat that gets added in. It doesn't explain why the expanding gas from the donor tank also gets hot.
---------- Post added February 27th, 2013 at 09:53 AM ----------
If the receiving tank started out empty, there is no gas to be compressed. If the receiving tank gets its gas from the full tank, then according to your argument, shouldn't that gas have gotten colder due to expansion before getting hotter due to compression? By your logic, shouldn't the fact that it expanded first more than it was compressed second mean that it should end up colder, not hotter? Edit:For those interested, that's the answer to question #5 below - extracting energy causes the gas expanding into the second tank to cool.)
OK, I said I'd leave once I was sure I knew the answer. It's not fair to stay here and challenge the viewpoints of others. That wasn't my purpose. I did want to see if I could figure it out, and I think I have. The most thoughtful posts here pointed me to some good links. I'd read many Wikipedia links on gas thermodynamics, but the one on energy storage in compressed air was particularly interesting and dovetailed nicely with my understanding. Gas in two tanks at half the pressure has less air pressure energy stored because, as the Joule expansion article explains, the Joule expansion process results in increasing entropy and as others have posted, when it comes to entropy, you can't win and you can't break even.
Those who think that gas expanding from one tank to two is somehow being compressed are just fooling themselves, but that's their right and it's not my job to convince them. I would like to have seen some decent arguments in favor of or against my understanding of the process. I almost left once, but by staying, I got the link to the article on energy storage in compressed air, and that article helped me put the whole process into perspective.
I'll check back occasionally (I know -- don't bother) but I would like to see a decent attack on the physics of the process as I outlined it from someone who understands this (if I'm wrong), or a recognition from that person that I did understand it correctly. To be credible, such an attack would have to answer the four questions I asked early on (relating to two tanks of equal volume, the receiving tank being vacuum empty, the donor tank being at high pressure and the gas being ruled by ideal gas laws, with no energy added or subtracted except by the optional pneumatic motor/generator in a whip line):
1) Did it take energy to compress the gas into the donor tank?
2) Could we get some of that energy back by putting a pneumatic motor/generator in the whip line between the tanks and driving the generator by the pressure differential during the fill?
3) Would the recipient tank and donor tank equalize pressure regardless of whether we put a pneumatic motor/generator in the whip line?
4) What happens to the energy we could have extracted from the gas with the pneumatic motor/generator if we don't put the pneumatic motor/generator in the whip line?
They would have to explain why Doppler's answers to those simple questions are wrong. They might also like to answer these:
5) What would happen to the temperature of the gas in the receiving tank if we do extract energy by putting a pneumatic motor/generator in the whip line?
6) What would happen to the temperature of the gas in the receiving tank if we don't extract energy by putting a pneumatic motor/generator in the whip line?
As far as I can tell the answers to those questions lead inexorably to the conclusion I reached. It's odd to find myself alone on this (or mostly alone), but so be it. The physics is what it is.
Good luck to all, and dive safe out there - even those who disagree with me
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