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-   -   Why do tanks get hot when you fill them from higher pressure tanks? (http://www.physicsforums.com/showthread.php?t=675162)

jimmyw1 Feb28-13 09:18 AM

Why do tanks get hot when you fill them from higher pressure tanks?
 
I'm a scuba diver - diving since the 1960's. For more years than I'd like to admit, I've wondered why tanks get hot when you fill them from other high pressure tanks. I've seen it happen with "empty" scuba tanks (really at atmospheric pressure, temp). And with vacuum evacuated oxygen tanks.

I'd like to set up a simple scenario: A starting tank of air (assume ideal gas laws if you wish) at 3000 psi and 11 liters of volume and room temp is connected with a hose (called a transfill whip) to an identical tank that is vacuum empty. The valves are opened and the donor tank gets cold and the receiving tank gets hot.

Many times I've asked others - more experienced divers - those teaching diving classes, etc. all who are teaching ideal gas laws, Henry's law, Dalton's law, etc. why the receiving tank gets hot. I've always gotten one of two answers. The first is that the gas going into the empty tank is getting compressed and is heated by compression. The second is that the tank has residual air and that residual air is heated by compression. The first answer made no sense to me. The gas expands into two tanks and the pressure drops. By any reckoning that makes sense to me, that's expansion, not compression. The second doesn't explain why vacuum empty tanks get hot.

I asked the question in an advanced diving forum. After lots of research in Wikipedia and linked references there I concluded that the key was "free expansion" also called "Joule expansion." I found myself in a minority of one - being repeatedly ridiculed that i didn't understand what was happening. I laid out my analysis as best I could - referring to formulas for each stage and asking some simple questions:

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?

The answers to those questions led me to the answers to 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 took me a lot of research to figure this out and perhaps I made some fatal errors. I just can't be sure, so I'm here.

My ultimate conclusion was essentially that the gas in the donor tank is expanding roughly adiabatically and isentropically. It is doing work on the departing gas, resulting in a temperature drop and a transfer of energy to the departing gas. The entire gas system has no energy added or subtracted. The departing gas expands roughly isothermally via "free expansion" or "Joule expansion" at constant enthalpy without change to its internal energy (at least not until its kinetic energy from the donor gas tank is converted into heat). The energy released by the isentropically expanding gas in the donor tank appears as heat energy in the gas in the receiving tank, resulting in heating above room temperature. I'm not totally sure about how the energy is carried or transferred and converted to heat, but my simple answer is basically that it's carried as kinetic energy and when the departing gas impacts the walls of the receiving tank, that energy is released as heat.

If the two gases were allowed to freely mix or exchange heat energy, the total gas would come to the same original room temperature. (with some minor second order effects from the non-ideal nature of air as compared to an ideal gas.)

I really do want to know the answer to this. However, I'm the only one who thinks that the answer above is right. Despite the fact I think I can show it mathematically, and I think the physics above is fundamentally right, not one expert in the advanced diving forum agreed with me. I stood alone, and that was a worrisome sign that I'm doing something seriously wrong. I did notice that only one person would try to answer the questions above (the first four) and those answers seemed to match mine, so if I am doing something seriously wrong, no one there could tell me what - at least not in a way that made sense to me.

I would try to give you more info on what the competing answers were, but generally they boiled down to increasing pressure in the second tank means the gas is being compressed there and compressed gas always gets hot. Another answer was that there is residual air in the second tank in the real world and the compression of that gas produces the heat so we should ignore the "free expansion" idea. Some said I had the initial gas volume wrong. They pointed to the two tanks and said it was the volume of the full tank plus the volume of the empty tank that was the correct volume to start with. Not one person would agree with me that gas that expands from one high pressure tank into twice the volume (two tanks) is undergoing expansion.

If you want to see the thread, I'll link it here.

When I asked for the best answer from those who disagreed with me, it was post #2 at the link above.

I did put some basic equations and links in the argument/explanation together in some later posts, but I suspect people here can do it better than I did. Here are my best answers to the first four questions.

Here is my best summary of how I would explain the heating and the physics mathematically.

I should say that I'd never heard of "enthalpy" when I started this, nor "free expansion" or "Joule expansion" nor "internal energy" of gas. My understanding of gas thermodynamics is all recent and shaky. I'd heard of entropy, but only vaguely - as in it always increases and the universe dies in some sort of maximum entropy heat death. I knew what degrees of freedom meant, but had no idea it related to internal energy of gas or heat capacity ratios (constant pressure to constant volume). I had never even fully understood why gas was heated when compressed with a compressor until I read about internal energy and heat capacity ratios, so I won't be upset if you tell me I'm full of it, as long as you tell me why.

Am I way off base or do I have the physics right? If I'm wrong, where did I go wrong? Is the gas in the second tank expanding and getting hotter or is it being compressed as everyone else says?

I'd appreciate it if you would answer the above questions for me, or if they aren't relevant, let me know why. Thank you very much for any assistance you can give me in understanding this. BTW, if the answers are significantly different when the receiving tank starts at "empty" 1 bar pressure, but not vacuum empty, I'd appreciate comments on that.

mfb Feb28-13 11:59 AM

Re: Why do tanks get hot when you fill them from higher pressure tanks
 
Quote:

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
Quote:

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.
Quote:

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.

Quote:

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.

jartsa Feb28-13 04:54 PM

Re: Why do tanks get hot when you fill them from higher pressure tanks
 
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.

jimmyw1 Mar1-13 09:51 AM

Re: Why do tanks get hot when you fill them from higher pressure tanks
 
Quote:

Quote by jartsa (Post 4289927)
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.

jartsa Mar1-13 08:24 PM

Re: Why do tanks get hot when you fill them from higher pressure tanks
 
Quote:

Quote by jimmyw1 (Post 4290942)
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.

Velikovsky Mar2-13 09:15 PM

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.

jimmyw1 Mar6-13 10:56 AM

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?

rcgldr Mar6-13 11:34 AM

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.

Velikovsky Mar6-13 03:58 PM

Quote:

Quote by jimmyw1 (Post 4297415)
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

Velikovsky Mar6-13 04:21 PM

Quote:

Quote by rcgldr (Post 4297452)
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!

rcgldr Mar6-13 07:27 PM

Quote:

Quote by rcgldr (Post 4297452)
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:

Quote by Velikovsky (Post 4297761)
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.

0xDEADBEEF Mar7-13 01:57 PM

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 [tex]e:=E/N=k_B T[/tex]
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.
[tex]\mathrm{d}e=- p\,\mathrm{d}V/N= - p\, \mathrm{d}v[/tex]
Therefore it should be possible to calculate the temperature change as
[tex]\mathrm{d}T = \frac{1}{k_B} \mathrm{d}e = -\frac{1}{k_B} p(N,T) \mathrm{d}v[/tex]
The two terms on the right can be expressed by the ideal gas equation.
[tex]p(N,T)=\frac{N k_B T}{V_B}[/tex]
[tex]v=V_B/N \implies \mathrm{d}v= - \frac{V_B}{N^2} \mathrm{d}N[/tex]
Where [itex]V_B[/itex] is the bottle volume. Therefore
[tex]\frac{\mathrm{d}T}{T} = \frac{\mathrm{dN}}{N}[/tex]
[tex]\int_{T_0}^{T_1}\frac{\mathrm{d}T}{T} = \int_{N_0}^{N_1} \frac{\mathrm{d}N}{N} [/tex]
So expressed using the particle number
[tex]\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}[/tex]
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 [itex]\sqrt 2[/itex] and you multiply the temperature by [itex]\sqrt 2[/itex]. 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 [itex]\implies[/itex] less Volume per particle [itex]\implies[/itex] more energy per particle due to [itex]\mathrm{d}e
= - p\, \mathrm{d}v[/itex] [itex]\implies[/itex] higher temperature due to [itex]e=k_b T[/itex] 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.

rcgldr Mar7-13 04:05 PM

Quote:

Quote by 0xDEADBEEF (Post 4298883)
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?

0xDEADBEEF Mar7-13 05:46 PM

Quote:

Quote by rcgldr (Post 4299053)
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 [itex]V_B \rightarrow \infty[/itex] and [itex]N \rightarrow \infty[/itex] therefore [itex]\frac{\mathrm{d}v}{\mathrm{d}N} \rightarrow 0[/itex] and [itex]\mathrm{d}e = - p\, \mathrm{d}v \rightarrow 0[/itex] 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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