I hesitate to waste more time on this, but... there are some screaming fundamental errors being made. One cannot "do" math without sorting some of this out.
I'm game - let's do it
Regarding these tanks you speak of: 11 litre cylinders. These have a wet volume of 11 litres... about. Let's keep the arithmetic simple and say that we substitute 10 litre tanks instead. Is that OK?
Great - yes 10 litre tanks.
Empty one will contain about 10 litres of gas at one bar. The charged one will contain 10 litres multiplied by its actual fill pressure... let's say 200 bar. So, there are 10 litres in one and 2,000 litres of gas in the other.
If you insist on keeping air in the second tank, OK. If you can do it with a vacuum in there, the math is easier, but I'll do it your way.
Now, let's also assume that the gas and the transfill system itself (ALL components) have reached a state of thermal equilibrium... a fancy was of saying they are the same temperature... normally we might cite room temperature, let's say 20 degrees (68 F)... but we are trying to do maths around gas which is basic chemistry so we should work in Standard Temperature and Pressure (STP), and that should be 273 K (0° Celsius) and 100 kPa or one bar. And we can also work out the quantity of gas we have... in MOLES. This is somewhat pedantic but hey, read this thread.
OK
At STP one mole of gas has a volume of around 22.4 litres. Therefore we have approximately 89.33 MOLES OF GAS... The 2000 litres in one cylinder and the one litre in the other... plus a few ccs in the fill whip. Cool.
Remember, the "2000 litres in one cylinder" are at a totally different pressure from the receiving tank, and you said "one litre in the other" when I think you meant 10 litres in the other.
I'm going to work out the numbers this way -
We have 10 liters of air at 200 bar and temp 273 K (0° Celsius) .... I got to this point and then realized you didn't actually do the math. I have done the math. Repeatedly. Is it worth going through it? To do that, I have to state whether we are adding energy to the gas or not. That's why I asked these 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?
Are they that hard?
Let me give the answers I think are pretty clear. If my answers are wrong, tell me which ones are wrong. At least give me some hints on where you think this is wrong:
1) Did it take energy to compress the gas into the donor tank?
A: Yes. We have to do work on the gas to push it into the donor tank. The compressor does that work. As a side note, this is called a reversible adiabatc process of constant entropy. The internal energy of the gas is raised as it goes from 80cf at STP to a higher temperature pressure and smaller volume. Because this process is "reversible" (another way of saying the entropy is unchanged), we could extract the stored energy by letting the gas expand and do mechanical work on a turbine/generator. The gas would cool if we did that and return to STP. The cooling would exactly counteract the heating from the original compression. The extracted energy could be used to reverse the process again and push the gas back into the donor tank, which would heat it up. In theory, we can go back and forth - compress hot - expand cool. Below I will say we could let the expanding gas out into a tall vertical cylinder with a weight on a piston. We could let it push the cylinder of mass m upwards to a height h and store mgh potential energy. We could then push the cylinder and weight downward to compress the gas and make it hotter. We can go back and forth (neglecting friction).
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?
A: Yes. As described above. We don't have to use a pneumatic motor and generator. We could let the gas expand into a cylinder and push a weight on a piston upwards and use mgh where m is the mass of the rising piston and h is the height the gas pushes that weight up. It doesn't matter how we extract the energy or how we store it. What is important, however, is that if we *do* extract the energy, all the gas in the donor tank, including the gas we let out of the donor tank into the cylinder not only expands, it also cools. Again this is a reversible constant entropy process. I'm sorry I'm using technical terms, but they accurately describe what's happening.
3) Would the recipient tank and donor tank equalize pressure regardless of whether we put a pneumatic motor/generator in the whip line?
A: Yes. If we let the gas push a mass upwards, it's a reversible process, and we extract energy, but if we were to just let the gas into an empty cylinder, with the weight already at the top of the cylinder, it would be an irreversible, process. If I can be excused for using another technical term, this type of expansion is called constant enthalpy. The constant enthalpy process is not reversible, but has constant internal energy of the gas. Either way, the final pressure would be the same in both tanks, but in one case, the temperature of the gas would decrease and in the other (called "free expansion") the temperature would be unchanged in the "Joule expansion = free expansion" process - provided we let the gas freely mix. If we don't let it freely mix, some gas is hotter and some gas is colder.
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: The energy that could have been extracted is released as heat into the gas in the receiving tank. Instead of the expansion cooling the gas in the receiving tank, as happened in the reversible process, it is released as heat, and there's just enough energy to counteract the cooling of all the gas from the expansion that would have occurred if we'd let the expansion raise the weight in a reversible process.
I can do the math on all of this. I can start with 80cf of air at STP and calculate how hot it gets when we compress it reversibly. If preferred, I can let the excess heat escape and figure out the pressure (PV is constant since T is constant) I can then assume no energy (no work no heat) is added to the gas and determine the final temp during free expansion. It will be room temp if all the gas is allowed to mix and equalize temp. I can also calculate how hot the gas in the receiving tank gets if we assume that the gas in the donor tank expanded reversibly, and all the released energy is transferred to the gas in the receiving tank. I can even calculate what happens if there is residual gas in the receiving tank. And if necessary, I can even add in the final details of assuming that the gas is not ideal so that the J-T coefficient is non-zero.
It's long and boring, and I suspect that it won't convince anyone, but it convinced me.
If anyone wants to understand the physics, they have to figure out the difference between extracting energy from the compressed high pressure gas in the donor cylinder versus not extracting/storing that energy and letting it appear as heat in the receiving tank gas. All of this works for the ideal gas assumptions, and for real world gases.
Here's another example - we let the gas from the expanding donor tank push a bullet through the whip hose. The bullet has mass and picks up energy, then slams into the receiving tank and it's kinetic energy converts to heat. It's the same as letting the expanding gas of the donor tank push and accelerate the gas leaving the donor tank, but it might be easier to see it that way.
