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

by jimmyw1
Tags: pressure, tanks
Mentor
P: 9,655
 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?
Sure
 3) Would the recipient tank and donor tank equalize pressure regardless of whether we put a pneumatic motor/generator in the whip line?
Depends on details of the generator, but usually yes.
 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?
A part of it heats the low-pressure tank, another part is lost as friction in the whip line.

The pressure-drop happens in the donor tank only, in the other tank air gets compressed.

 If the two gases were allowed to freely mix or exchange heat energy, the total gas would come to the same original room temperature.
This could depend on details of the expansion process.
 P: 370 The donor tank got cold. That means the gas in the donor tank did some kind of work. The work was done on the gas that left the tank. So I propose following law of pressure tanks explains what happens: The gas that comes out first is hot, the gas that comes out last is cold. Oh I see, the OP says just that. Well, then I agree.
P: 3

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

 Quote by jartsa The donor tank got cold. That means the gas in the donor tank did some kind of work..... Oh I see, the OP says just that. Well, then I agree.
So how would you answer questions 5 and 6? Feel free to assume that there are no losses in the transfer process, i.e., that no energy is added or removed from the total gas in the system.
P: 370
 Quote by jimmyw1 So how would you answer questions 5 and 6? Feel free to assume that there are no losses in the transfer process, i.e., that no energy is added or removed from the total gas in the system.

Here's a qualitative description of an ideal energy extraction process:

At first when we let the smallest possible amount of gas (one molecule) to expand (through a pneumatic motor) into the vacuum of the receiving tank, the temperature of that gas becomes ridiculously low, because the expansion ratio is ridiculously high.

When the second molecule goes through the motor, the second molecule does a ridiculously small amount of work on the first molecule, because the first molecule causes an extremely small counter pressure, when the second molecule is cranking the pneumatic motor. That work becomes the heat energy of the first molecule. ... And so on ...

Now, if we put all the energy that we extracted to the gas in the receiving tank, we are putting back all the work that the gas in the receiving tank has done, plus the work that the gas that is still in the donor tank has done.

The gas in the donor tank indeed does do some work, so the temperature of the donor tank decreases, and the temperature of the receiving tank rises, in this scenario, which is equivalent to not doing any energy extraction at all.
 P: 50 Hi, Jimmy I used to dive myself but these days I mostly snorkel in the frigid waters of Ireland! I too recognised the phenomena of the donor tank rapidly cooling when decanting air into the scuba bottle. It is obviously acting like a heat exchanger. I have even at times seen a frost form on the neck of the donor bottle while the neck of the scuba bottle becomes to hot to handle! The air in the donor bottle is escaping via a convergent nozzle [the neck of the tank] causing the air to rapidly condense at a pressure higher than in the tank it came from, the air becomes a de-facto heat sink! When the air then meets the scuba tank, it encounters a divergent nozzle [again the neck of the tank] where the highly condensed air releases all of the energy bound up in it and transfers the heat from the donor tank to the scuba tank. If any of you are familiar with motorcycles [yes I'm a biker chick,GSX-R 750] you will often see the throat of the carburetor frosted up, here again we have the convergent/divergent throat, or look at a rocket launch and pay attention the the throat above the divergent rocket nozzle. I'm not really sure as to why a nozzle has this effect but it certainly is interesting.
 P: 3 I think that every parcel of gas - every definable volume of gas in the initial high pressure tank - ultimately expands from its initial high pressure state and small volume to a final larger volume and lower pressure. I think that's true regardless of whether the "defined parcel" of gas remains in the original tank or moves to the final tank. Is there anyone here who disagrees with that?
 HW Helper P: 6,779 My guess is that this issue is related to free expansion, wiki article: http://en.wikipedia.org/wiki/Free_expansion For an ideal gas in free expansion, the temperature doesn't drop because the expanding gas isn't performing work on some external (to the gas) component, such as a piston in a cylinder. So the initial temperature of each parcel of gas being transferred remains about the same as the temperature of the remaining gas in the high pressure tank, which is decreasing over time. The temperature of each parcel of gas in the once evacuated tank increases from it's initial temperature as the pressure increases.
P: 50
 Quote by jimmyw1 I think that every parcel of gas - every definable volume of gas in the initial high pressure tank - ultimately expands from its initial high pressure state and small volume to a final larger volume and lower pressure. I think that's true regardless of whether the "defined parcel" of gas remains in the original tank or moves to the final tank. Is there anyone here who disagrees with that?
The shape of the aperture plays a big part in the production of kinetic energy and heat even if you have two identical volumes of gas at identical pressure and temperature. All the energy of the first tank will be carried to the second tank in the form of kinetic energy and expended into it as a rapidly expanding hot gas. Condensing gases are cooler than expanding gases.

http://en.wikipedia.org/wiki/De_Laval_nozzle
P: 50
 Quote by rcgldr My guess is that this issue is related to free expansion, wiki article: http://en.wikipedia.org/wiki/Free_expansion For an ideal gas in free expansion, the temperature doesn't drop because the expanding gas isn't performing work on some external (to the gas) component, such as a piston in a cylinder. So the initial temperature of each parcel of gas being transferred remains about the same as the temperature of the remaining gas in the high pressure tank, which is decreasing over time. The temperature of each parcel of gas in the once evacuated tank increases from it's initial temperature as the pressure increases.
Strange I always considered a scuba tank to be rather like a cylinder. And the pressure of the decanting cylinder is acting like a piston.
Don't forget the scuba tank is at lower pressure than the decanting tank. You are filling a 10 ltr tank with 2000 ltrs of air at a minimum pressure of 200 Bar!
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P: 6,779
 Quote by rcgldr For an ideal gas in free expansion, the temperature doesn't drop because the expanding gas isn't performing work on some external (to the gas) component, such as a piston in a cylinder.
 Quote by Velikovsky Strange I always considered a scuba tank to be rather like a cylinder.
Yes, but a cylinder without a piston. The work done on the piston is the integral of force x distance. Without the piston, you have force, but zero distance, so zero work done.

Without any work done, the total energy of the gas remains the same, but this isn't enough to explain why temperature remains constant in a free expansion. The temperature is related to the average speed of the molecules, which for an ideal gas remains constant in a free expansion because without a receding surface (like a piston) for the molecules to bounce off of, which would reduce speed (transfer of energy and momentum), the molecules average speed remains constant as they travel through a vacuum and eventually bounce off non-moving surfaces.
 P: 827 The correct way to calculate this is going through the energy per particle (the chemical potential), because the main change in the target bottle is the change of the particle number. It should go something like this: There are two ways to express the energy per particle. One is $$e:=E/N=k_B T$$ We know that the pressure is an intensive quantity, while the volume is extensive. Therefore the change of energy per particle is the pressure times the change of Volume per particle. $$\mathrm{d}e=- p\,\mathrm{d}V/N= - p\, \mathrm{d}v$$ Therefore it should be possible to calculate the temperature change as $$\mathrm{d}T = \frac{1}{k_B} \mathrm{d}e = -\frac{1}{k_B} p(N,T) \mathrm{d}v$$ The two terms on the right can be expressed by the ideal gas equation. $$p(N,T)=\frac{N k_B T}{V_B}$$ $$v=V_B/N \implies \mathrm{d}v= - \frac{V_B}{N^2} \mathrm{d}N$$ Where $V_B$ is the bottle volume. Therefore $$\frac{\mathrm{d}T}{T} = \frac{\mathrm{dN}}{N}$$ $$\int_{T_0}^{T_1}\frac{\mathrm{d}T}{T} = \int_{N_0}^{N_1} \frac{\mathrm{d}N}{N}$$ So expressed using the particle number $$\ln \frac{T_1}{T_0} = \ln \frac{N_1}{N_0} + c^\star \implies \frac{T_1}{T_0} = c\frac{N_1}{N_0}$$ Now I am too lazy to do the complete adiabatic treatment. But roughly if you double the pressure, you multiply the number of particles by $\sqrt 2$ and you multiply the temperature by $\sqrt 2$. I am not entirely sure if I should have considered the chemical potential of the particles in the source bottle, because with the way how I am calculating right now I think you are adding particles at the same temperature as the target bottles temperature. But you get the gist. More pressure $\implies$ less Volume per particle $\implies$ more energy per particle due to $\mathrm{d}e = - p\, \mathrm{d}v$ $\implies$ higher temperature due to $e=k_b T$ The kinetic energy of the source gas is probably not very important, except for the pressure it represents. Of course you could extract energy from the flow of gas. The pressures would still equalize. The necessary energy would presumably come from the change in Entropy.
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P: 6,779
 Quote by 0xDEADBEEF Therefore it should be possible to calculate the temperature change as ...
But in a free expansion of an ideal gas, the (average) temperature doesn't change, so aren't a different set of equations needed?
P: 827
 Quote by rcgldr But in a free expansion of an ideal gas, the (average) temperature doesn't change, so aren't a different set of equations needed?
Then $V_B \rightarrow \infty$ and $N \rightarrow \infty$ therefore $\frac{\mathrm{d}v}{\mathrm{d}N} \rightarrow 0$ and $\mathrm{d}e = - p\, \mathrm{d}v \rightarrow 0$ because adding a few particles to something like the atmosphere will not noticeably change the volume per particle and no energy is expended.

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