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

[jimmyw]

Close oh so close. . . so you've observed that filling a tank with O2 warms it more than filling it with air.

Actually, I didn't observe that. I learned it later. I did observe the heating, but didn't notice whether it got hotter or colder than air.

Ergo, something about air is different than pure O2.

I agree, there is a difference. It has to do with real world differences between ideal gases and real gases. However, air, and oxygen both heat up for the same basic reason, even though the temperature change is different.

If you were to observe a tank being filled with He, you'd observe slight cooling.

You're thinking of the fact that He and H both have a slightly negative Joule-Thomson coefficient at common pressures, while air has a slightly positive J-T coefficient. Ideal gases have a zero J-T coefficient. What this means is that at "constant enthalpy," air will heat slightly when expanded in a Joule-Thomson "free expansion" process, an ideal gas will not change temperature and He and H will slightly cool as they expand in the constant enthalpy process.

I know "enthalpy" is a pretty geeky term, but it just means that the internal energy of the gas in the system is unchanged. What this means is that if we filled the receiving tank with air from another higher pressure donor tank (as we've been discussing) and don't add or subtract any energy) the tank gets slightly hotter instead of remaining at constant temperature as described for the Joule expansion process, while He gets slightly cooler when it expands.

These are second order effects that deviate from the ideal gas law. By that I mean that air gets slightly cooler when it is compressed than an ideal gas, but that doesn't mean that compressing air with an air compressor will make the air colder. When you use a compressor to compress air, you are not engaged in a constant enthalpy process. You are increasing the "internal energy" of the gas and the J-T coefficient effect is just slight modification to the heating that occurs. The air gets slightly cooler than expected when it undergoes that compression even though the net effect is that the air gets hotter.

But wait ! This isn't accounted for by the Ideal Gas Law (or any of the equations you reference which are based on Ideal Gas(s) )

We can account for them if we need to, but it won't have a major effect.

Simple. Gas(s) are not ideal. Why is this such a problem for you?

It's not. It's just that the ideal gas is pretty close to the non-ideal gas and using the ideal gas laws will give the basic answers. Charles's Law still works pretty well.

Since you're getting twisted around by the equations (you've stated this many time), take a trip to your local technical LDS and ask to watch them fill a tank with He and the same pressure of O2. Measure the temperature change and get back to us. Right now you're just indulging in mental masturbation and counting (and re-counting) how many Angels are dancing on that pin.

I agree that He will cool. The He fill process is always a fill from an existing tank at higher pressure, which is a nearly constant enthalpy Joule expansion process.

Do you know what would happen if the room was filled with He and a compressor was used to take low pressure He at room temp and force it into the tank? That would be a constant entropy process, not constant enthalpy. The internal energy of the He would increase. The He would be heated and the amount of heating would be partly due to the increased internal energy (the same reason air is heated when compressed by a compressor) plus some additional heating due to the negative J-T coefficient of He. The second order effect is that He would get slightly hotter than expected, just as air gets slightly colder than expected when compressed by a compressor.

It's not angels on the head of a pin. It's physics.

---------- Post added February 24th, 2013 at 03:22 PM ----------

Are you ****ing kidding me? Of course the gas started at a higher pressure! That cannot be ignored. Why? Because there's no other way to get it out of the donor tank and into the receiving tank other than to have a pressure gradient sufficient to overcome the pressure inside the receiving tank.

At least we agree about this.

Please answer the two questions below:

After the tanks equalize, you close the valves and remove the fill whip. You put a reg and a BCD on the tank, open the valve, strap it to your back, and jump in the water. With the reg in your mouth, you attempt to draw a breath from the tank.

1.) What happens?
2.) Why?

Is this a trick question
1.) The gas passes through the regulator and expands into your lungs.
2.) The pressure differential and regulator act to ensure that the pressure in your lungs and the output of the regulator equalize.

I'm not sure where you're going with this.

---------- Post added February 24th, 2013 at 03:47 PM ----------

Well, I did ask for a one word answer.....

Anyhow, forget the freakin equations. So, you do agree that when gas gets compressed it warms up.

Next question, one word answer, yes or no. Do you accept the fact that the gas in the recipient tank is getting compressed?

I noticed you didn't answer mine, which scarcely seems fair.

If you tell me the starting temperature, pressure and volume of the gas you are talking about, I'll tell you whether the gas is getting compressed.

What case are we talking about? There is no gas in the recipient tank if it's evacuated, so I answer NO - it's not getting compressed. That's a simpler case to deal with and would answer my question if anyone could explain .

If the recipient tank was not evacuated, and had some air at room temperature, then I answer YES. The air that started in there is getting compressed. That gas gets hotter. That heat is not enough to explain the amount of heat produced, but it does have a small effect. There is also a small effect from the non-zero J-T coefficient of air.

It really is as simple as that. Consider this one of those glorious times in the universe when there's a super-easy-peezy answer.

Why don't you answer these questions for me.
1) Do you agree it took energy to compress the gas into the donor tank?
2) Do you agree that we could 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) Do you agree that the recipient tank and donor tank would equalize pressure regardless of whether we put a pneumatic motor/generator in the whip line?
4) What happened to all the energy we could have extracted from the pneumatic motor/generator if we didn't put the pneumatic motor/generator in the whip line?

Now put down the equations and go outside, have some fun!

Good idea.

Particular thanks to JohnN. His J-T coefficient comments took me the most effort to figure out.

RJP's and your thoughts on compressing the preexisting air in the tank were also good, but didn't deal with the evacuated tank problem. I could show mathematically that the heating of the preexisting air isn't enough, but that should be apparent from the evacuated tank problem.

I forget who raised the possibility of heat leaking into the system at the cold valve, but thanks to him, Until I understood Joule expansion (free expansion) and the difference between constant entropy and constant enthalpy expansion, I thought that was the answer.

 

[redacted]

Let me ask one, yes or no: Do you accept the fact that air does not always get cold when allowed to expand - specifically, the temperature is unchanged when allowed to "freely expand" (Joule expansion) as described in the wiki article on Joule expansion?

I'll answer that. It always get cooler when pressure decreases and warmer when pressure increases, even when the source or recipient of the thermal energy is other molecules. Of course, it is a zero sum game so overall, the temperature (thermal energy) remains constant.

I wouldn't describe it quite like this. Even if the donor tank and all the valves are well insulated, the gas remaining in the donor tank itself will cool as it expands. I'd also say that the reason it cools is that it is doing work. What work is it doing? It's pushing the gas going into the receiving tank and getting it up to speed. We could have extracted the energy that was given to that gas by running the gas through a turbine generator or a pneumatic motor. If we had, the gas in both tanks would have cooled. Because we didn't, the gas in the receiving tank got more of the energy than the gas in the donor tank, even though they both expand and both are at the same final pressure.

In practice, we always extract some of the energy. That is why the walls of the tank and the valve get cooler. That is why a 1st stage can freeze (form ice) in cooler waters. Gas in the receiving tank does not expand. Gas in the receiving tank is compressed and always gives off thermal energy in the process - and the tank absorbs some of the from the gas and gets warmer.

Again, I don't agree with this description. Thermodynamics laws don't allow the gas to extract energy from the room temperature tank and valve, so the gas never has any excess energy. The total energy is unchanged. However, the compressed gas of the donor tank has internal energy (in the form of both pressure which is derived from the XYZ motion of the gas and its internal vibrations and rotations). That internal energy can be released as kinetic energy (speeding up the molecules leaving the tank) and that kinetic energy can be transformed into heat as the molecules slam into the walls of the receiving tank.

Back to the freezing 1st stage. The total thermal energy is unchanged but the distribution of it sure changes until it again equalizes. Just last week I managed to ice up a 1st stage using a tank as a compressed gas source for operating power tools.

BTW, it really does not matter whether the recipient tank is evacuated or not. It just matters that the recipient tank is at a lower pressure than the donor tank. If it is evacuated, there will be a first molecule that arrives. It is going to bring with it the thermal energy it gained as it mover towards lower pressure. But it will not have the tank to itself for long because another molecule will be right behind it and will put it under increased pressure. That will cause it to give up some thermal energy. And so on and so on..... So whether the recipient tank is at 0 atm of 1 atm, the effect is the same until the recipient tank and the donor tank are at the same pressure (and temperature).

 

[halocline]

There is no gas in the recipient tank if it's evacuated, so I answer NO - it's not getting compressed.

Okay, then we are not talking about the same thing at all. I thought you were talking about filling a tank from another tank, hence the title of this rather amazingly nutty thread. The tank you are filling is the 'recipient' tank. The recipient tank starts at ambient pressure, or even completely empty of all gas as you want to prefer, right? And as you fill it, per the title of the thread, there's an increasing amount of gas under an increasing amount of pressure, right? That's why the little dial on the gauge moves.

So, if you do not agree that the gas in the recipient tank is being compressed, then all the equations in the world won't help you. You're just not perceiving basic reality.

Edit; I doubt this will help because it must have been mentioned several times, but I guess you're confused because you think since the air in the donor tank is more compressed than the air in the recipient tank even after it's filled, that the air in the donor tank is not getting compressed? Is that it? You think it just goes straight from 3000 PSI to 1500? (for example)

Maybe I'm not perceiving basic reality.....

 

[RJP]

Maybe I'm not perceiving basic reality.....

I'm working on a formula for basic reality for jimmyw currently (R/F=WT) where R is reality, F is the force of gravity, W is the weight of the earth, and T is the amount of time I've spent on this asinine thread.

Since the force of gravity (F) is a constant, we can multiply both sides of the equation by "F" to show that for jimmyw... Reality = WTF!

If anyone needs me, I'll be in the basement evacuating a few bottles of beer.

 

[jimmyw]

Let me ask one, yes or no: Do you accept the fact that air does not always get cold when allowed to expand - specifically, the temperature is unchanged when allowed to "freely expand" (Joule expansion) as described in the wiki article on Joule expansion?

I'll answer that. It always get cooler when pressure decreases and warmer when pressure increases, even when the source or recipient of the thermal energy is other molecules. Of course, it is a zero sum game so overall, the temperature (thermal energy) remains constant.

I think you agree with me then. During "free expansion" the temperature drop from expansion is compensated for by the energy release from that expansion and the temperature remains constant. All one needs to do then, is to understand that when crossfilling a tank, the donor tank is not undergoing "free expansion" and the heat released is transferred to the receiving tank so that if the two gases were mixed together, the net temperature change would be zero as described in Wikipedia and the entire gas (all gas in both tanks) would be undergoing "free expansion" and remain at constant temperature, despite being at lower pressure.

In practice, we always extract some of the energy. That is why the walls of the tank and the valve get cooler.

I agree - the energy is extracted by accelerating the molecules of the escaping gas to a high speed. It should be possible to see that the receiving tank will fill regardless of whether we capture the energy released or not.

That is why a 1st stage can freeze (form ice) in cooler waters. Gas in the receiving tank does not expand.

Do you still think that if the receiving tank started out empty? What gas are we talking about?

BTW, it really does not matter whether the recipient tank is evacuated or not. It just matters that the recipient tank is at a lower pressure than the donor tank. If it is evacuated, there will be a first molecule that arrives. It is going to bring with it the thermal energy it gained as it mover towards lower pressure. But it will not have the tank to itself for long because another molecule will be right behind it and will put it under increased pressure.

I can show you the math that when the gas leaves the donor tank and expands, then gets recompressed and ends up at a lower pressure, the result is expansion. That's the source of my original question. I can show that if the gas cools from the first expansion, that the later heating from subsequent recompression is always less since it always ends up at lower pressure.

You seem pretty reasonable. If anyone can find my error, it's likely to be you. Why don't you try these questions:
Why don't you answer these questions for me.
1) Do you agree it took energy to compress the gas into the donor tank?
2) Do you agree that we could 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) Do you agree that the recipient tank and donor tank would equalize pressure regardless of whether we put a pneumatic motor/generator in the whip line?
4) What happened to all the energy we could have extracted from the pneumatic motor/generator if we didn't put the pneumatic motor/generator in the whip line?

---------- Post added February 25th, 2013 at 09:34 AM ----------

Okay, then we are not talking about the same thing at all. I thought you were talking about filling a tank from another tank, hence the title of this rather amazingly nutty thread. The tank you are filling is the 'recipient' tank. The recipient tank starts at ambient pressure, or even completely empty of all gas as you want to prefer, right? And as you fill it, per the title of the thread, there's an increasing amount of gas under an increasing amount of pressure, right? That's why the little dial on the gauge moves.

We are talking about the same thing - filling a tank from another higher pressure tank.

So, if you do not agree that the gas in the recipient tank is being compressed, then all the equations in the world won't help you. You're just not perceiving basic reality.

I do agree that the gas in the receiving gas is being compressed! But I also know that before it gets compressed, it started at higher pressure in the donor tank, and it has to expand to get into the lower pressure receiving tank. What is wrong with that statement?

If it expands, it cools. If it gets compressed later, it gets hot. If the recompression is less than the expansion, then the heating is less than the cooling. What is wrong with that statement?

That was my original formulation of this problem, but it's enough to show the problem - later I realized that there were two types of expansion - one where we don't extract energy in "free expansion" and one where we do extract energy.

Edit; I doubt this will help because it must have been mentioned several times, but I guess you're confused because you think since the air in the donor tank is more compressed than the air in the recipient tank even after it's filled, that the air in the donor tank is not getting compressed? Is that it? You think it just goes straight from 3000 PSI to 1500? (for example)

Maybe I'm not perceiving basic reality.....

I think you are perceiving reality. I don't think that the gas instantly jumps pressure and I do think the gas in the receiving tank gets compressed - but only after it expanded. It starts in one tank and ends up in two tanks. How can that not be expansion????????? Is there anyone who wants to argue it's not?

I doubt it will help, but I'll confirm that I fully understand that the gas is going through an expansion and recompression cycle before ending up at lower pressure. The filling process, when filling from a donor tank is a process that doesn't add any energy (heat or work). It's unlike the direct compressor fill process where the tank is filled by adding energy derived by powering the compressor. Since no energy is added to the gas (the total gas of both bottles) the total "internal energy" of the gas remains constant - see Wikipedia on "internal energy"

Since the internal energy of the total gas is unchanged we know this, quoted from Wikipedia on the internal energy page:
"The internal energy is a state function of a system, because its value depends only on the current state of the system and not on the path taken or process undergone to arrive at this state." The intervening process of expansion and compression does not matter as long as we look at all the gas.

This is the result of more work done over the weekend in trying to understand this.

If anyone will try to respond to these questions, I think we'll get somewhere:

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 pneumatic motor/generator if we don't put the pneumatic motor/generator in the whip line?

 

[RJP]

I can show you the math that when the gas leaves the donor tank and expands, then gets recompressed and ends up at a lower pressure, the result is expansion.

The gas does not and cannot expand into the other tank. The tanks are rigid. The tanks do not expand.

I had a bit of an epiphany this morning. I think you are confusing yourself when you refer to volume. (Never mind what's happening in your head with all the other nonsense you're dragging into this discussion.) I believe you are thinking that the volume of the donor tank is 80cf and the volume of the receiving tank is rougly 0cf.

You need to think "actual tank volume" vs "amount of gas molecules stuffed into that volume." (And for god's sake, please don't bring up a hypothetical "vacuum tank" again.)

So here's the situation:

You have two tanks that we (in the US) call 80cf tanks, but in fact that is merely a naming convention, and has no bearing on the actual volume of the tank. Each tank's actual volume is 11 liters, based on how many cc's of distilled water would fit in them. This is the VOLUME with which you need to concern yourself when doing your fancy calculations. The US naming convention of "80cf" notwithstanding.

At the beginning of the process the gas in those two tanks occupies 22 liters of space. 22 liters is the volume in the equation. (not 80cf)
- Donor tank has 11 liters of gas at 3,000psi. Correct?
- Receiving tank has 11 liters of gas at 14psi Correct?
- Total gas volume is 22 liters. Correct? (11x2=22, for those who like formulas)

When you open the tank valves, the gas equalizes from one tank to another. Correct?

At the end of the equalization process, the two tanks each contain 1,507psi. Correct?
- Donor tank contains 11 liters of gas at 1,507psi.
- Receiving tank contains 11 liters of gas at 1,507psi.
- Total gas volume is 22 liters at 1,507psi. Correct?

If the volume that the gas occupies at the END of the process is the same volume that the gas occupied at the BEGINNING of the process, the volume of the gas remains unchanged (each tank individually and the total system as a whole). There has been no "expansion of gas" in the system. Nor has there been an expansion of gas into the receiving tank.

Roughly half of the gas molecules have migrated into the receiving tank due to the pressure gradient. But the total volume occupied by the gas - in the system as a whole and each tank individually - remains unchanged.

No change in volume means no expansion.

res ipsa loquitur

 

[jimmyw]

The gas does not and cannot expand into the other tank. The tanks are rigid. The tanks do not expand.

Gas laws apply to gas, not tanks. It's the volume of the gas that's important, not the volume of some nearby tank that doesn't have any gas in it. Tanks are only relevant if we let the gas expand into them or we force the gas to compress into them and then only the volume of the gas is relevant, but since teh volume of the gas is the same as the volume of the tank, we can just use the volume of the tank in the equations. Surely there are some others who understand this point.

I had a bit of an epiphany this morning. I think you are confusing yourself when you refer to volume. (Never mind what's happening in your head with all the other nonsense you're dragging into this discussion.) I believe you are thinking that the volume of the donor tank is 80cf and the volume of the receiving tank is rougly 0cf.

No, I think the volume of the gas in the donor tank is 80cf and the volume of the gas in the receiving tank is 0cf

So here's the situation:

You have two tanks that we (in the US) call 80cf tanks, but in fact that is merely a naming convention, and has no bearing on the actual volume of the tank. Each tank's actual volume is 11 liters, based on how many cc's of distilled water would fit in them. This is the VOLUME with which you need to concern yourself when doing your fancy calculations. The US naming convention of "80cf" notwithstanding.

I agree to here.

At the beginning of the process the gas in those two tanks occupies 22 liters of space.

The volume of the high pressure gas is 11 liters. If the receiving tank is empty, that's all the gas. If It's not, we have to do two sets of equations - one for what happens to the 11 liters of high pressure gaas and one for what happens to the other 11 liters of room temp/pressure gas in the not vacuum empty tank. The two volumes of gas need to be treated separately since they aren't at the same pressure.

22 liters is the volume in the equation. (not 80cf)

Yes, in two different tanks at different pressures.
One equation has 11 liters of gas at 3,000psi. and one either has no gas and we can ignore it, or it has 11 liters of room temp press gas and we have to analyze what happens to it when it gets compressed. We can do both sets of equations, but at some point we have to know what happens to the 11 liters of high pressure gas.

- Donor tank has 11 liters of gas at 3,000psi. Correct?

Correct.

- Receiving tank has 11 liters of gas at 14psi Correct?

Agreed - if it's not vacuum empty.

- Total gas volume is 22 liters. Correct? (11x2=22, for those who like formulas)

Yes, but we can't use that volume in any equations. All gas equations assume all the gas is at the same PVT.

When you open the tank valves, the gas equalizes from one tank to another. Correct?

Agreed.

At the end of the equalization process, the two tanks each contain 1,507psi. Correct?

You have to define the "equalization process." Did we let the hot tank cool off the cold tank? If so, then the two tanks are at 1500 psi and room temp. If not, then they aren't equalized in temp.

If the volume that the gas occupies at the END of the process is the same volume that the gas occupied at the BEGINNING of the process, the volume of the gas remains unchanged (each tank individually and the total system as a whole). There has been no "expansion of gas" in the system. Nor has there been an expansion of gas into the receiving tank.

One volume of gas expanded and one volume was compressed. We have to analyze them separately to figure out what happens if we let the gas mix. I know it drives you crazy to start with the final tank vacuum empty, but unless we do that we have to analyze what happens to both volumes of gas separately and then see what happens. I've done that. The analysis for the high pressure gas is exactly the same as the analysis for that gas expanding into a vacuum empty tank. A vacuum empty receiving tank still gets hot. I've seen it. The math and physics explains it. There's not much point analyzing the 1 atm room temp gas in the receiving tank if the analysis of the expanding gas from the donor tank shows it is also getting hot.


Again:
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?

Surely someone wants to tackle these questions? They are very close to what has convinced me. The more I study this, the more I get the same answer. I could replace the "pneumatic motor generator" with a weight being raised, or a lifting bag coming up from depth and study the energy stored in raising sea level as the bag rises and expands, but it's the same question.

 

[redacted]

Gas laws apply to gas, not tanks. It's the volume of the gas that's important, not the volume of some nearby tank that doesn't have any gas in it. Tanks are only relevant if we let the gas expand into them or we force the gas to compress into them and then only the volume of the gas is relevant, but since teh volume of the gas is the same as the volume of the tank, we can just use the volume of the tank in the equations. Surely there are some others who understand this point.

Hopefully not. Gas does not have volume as such. Gas has mass. The notion of gas volume only comes into play when the gas is in a container which has volume. The tank is important. Boyle's law - Wikipedia, the free encyclopedia

 

[cool_hardware52]

The gas in the higher pressure tank will expand and cool *primarily* at the point of restriction between the two cylinders. This is typically some where on the transfill whip, i.e. the needle valve that controls the transfer rate.

The gas in the lower pressure tank will be compressed, and it will heat up.

What's being lost is the effects of the transfill whip. The needle valve is cooled, and drops well below ambient. Often the needle valve will ice up. That means that cooled expanded gas is in fact gaining heat, extracting it from the ambient air, and or the latent heat of fusion, i.e. turning condensed water vapor in to ice, phase changes require large energy inputs.

The reason why the recipient tank does not gain the same amount of heat as the donor tank looses is because a great deal of the cooling from expansion is lost via the transfill whip.

If there was no restriction in the transfill whip, meaning the pressures equalized instantly, and the transfill whip was perfectly insulated than the heat loss in the donor tank and heat gain in the recipient would be equal.

Once again if one understands the refrigeration cycle all this is easily understood.


Tobin

 

[jimmyw]

Hopefully not. Gas does not have volume as such. Gas has mass. The notion of gas volume only comes into play when the gas is in a container which has volume. The tank is important. Boyle's law - Wikipedia, the free encyclopedia

I agree - the tank is important, but when trying to figure out the pressure volume and temperature of some gas, the tank's importance lies in the fact that the volume of the tank defines the volume of the gas it contains.

---------- Post added February 25th, 2013 at 11:42 AM ----------

The gas in the higher pressure tank will expand and cool *primarily* at the point of restriction between the two cylinders. This is typically some where on the transfill whip, i.e. the needle valve that controls the transfer rate.

Some of the cooling occurs in the needle valve. Some occurs in the gas in the donor tank, which everyone here agrees gets colder during the fill process. I don't think we disagree on this.

The gas in the lower pressure tank will be compressed, and it will heat up.

I agree with this. I did the math - this heating is secondary. It's not enough to produce all the heat released. We still need to understand what happens if the low pressure tank is vacuum empty. I think it still gets hot. I'm not sure what you think.

The needle valve is cooled, and drops well below ambient. Often the needle valve will ice up.

Agreed.

That means that cooled expanded gas is in fact gaining heat, extracting it from the ambient air, and or the latent heat of fusion, i.e. turning condensed water vapor in to ice, phase changes require large energy inputs.

You make a good argument. You've made two of the three good arguments made here (the other is the J-T coefficient of air - the first is the compression of residual air and the second is the gain of environmental heat at the valve.) However, we can isolate the gas and the needle valve from the environment by using insulation. Or we can put the needle valve inside the donor tank (or the receiving tank) so the energy gain/loss to the gas is zero. What happens then?

You have two good arguments, but you still have to address the case where we don't add any energy to the system (heat added at a cold fill valve) and the case of vacuum in the receiving tank (no residual air compression heating). Both are very realistic experiments. They've been done. The transfilled tank still gets hot! That heat comes from the energy released by the donor tank when the air inside escapes to a lower pressure. That energy was originally put into the donor tank by a compressor that did a lot of work to fill the donor tank. (The donor tank still gets cold - but you know that.)

The reason why the recipient tank does not gain the same amount of heat as the donor tank looses is because a great deal of the cooling from expansion is lost via the transfill whip.

Agreed - any heat added or lost has to be accounted for.

If there was no restriction in the transfill whip, meaning the pressures equalized instantly, and the transfill whip was perfectly insulated than the heat loss in the donor tank and heat gain in the recipient would be equal.

I agree. The donor tank will cool and the receiving tank will heat up. If no heat energy leaks in or out, the two balance out. All that is left is to explain why the one tank got hot and the other got cold. that lies at the root of my question. Why didn't they both cool or heat the same amount? The gas pressure ends up the same in both. I understand now, but didn't when I came here.
(edit: Well actually the root of my problem was that I thought expanding gas always cools. I didn't understand that freely expanding gas stays at the same temperature. I had trouble relating a description of constant temp free gas expansion to what was happening with a transfill operation where one tank cools and the other heats up).

Once again if one understands the refrigeration cycle all this is easily understood.
Tobin

I think of the refrigeration cycle as relating to the transfer of heat energy to/from the environment, but I won't disagree that it would be understood if you understood the refrigeration cycle. Whether that understanding is "easily" obtained is certainly debatable in light of this thread.
For others:
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?