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

[rakpix]

No, I mean ... what does Queen have to do with kinetic theory?

 

[RJP]

I'm confused.

Anyone who isn't confused... hasn't been paying attention.

 

[rhwestfall]

This is why I became a Civil Engineer (rather than a Mechanical or Chemical)..........

one simple principal: "$h1t flows down hill".....

 

[windapp]

It is true that no work is being done on the system as a whole. However, work is being done by the gas in the bank to push the air through the hose. The expansion is why the air in the bank cools. If no work was being done during the expansion, there would be no cooling. Conversely, work is being done on the gas in the cylinder being filled when the gas that is already in there is compressed by the gas that is entering from the other cylinder. It is all a matter of point of view, and where exactly you draw your system boundaries to most easily solve the problem.

The easiest way to look at it is this:

The gas remaining in the bank is undergoing a reversible expansion as it pushes out gas into the tank being filled. It cools due to this expansion.

The gas flowing between the two tanks is undergoing a free expansion as it flows from the one tank to the other. Then Enthalpy remains unchanged. Because of this, depending on the pressure difference, it can either get a little bit warmer or a little bit cooler depending on the gas and pressure difference. If it was an ideal gas, the temperature would remain unchanged. It is not, so it changes a bit.

The gas in the tank being filled is compressed in a way that is somewhere between reversible and irreversible. It heats up substantially as a result of this compression.

The reason why this problem is confusing is that there is no simple equation to describe this event which is highly path dependent (exactly how the transfer occurs affects the final result), and is very far from steady state. Then you add the Wikipedia trained engineers who think that the JT effect somehow replaces the ideal gas law, and the problem gets that much more confusing.

 

[JohnN]

No, I mean ... what does Queen have to do with kinetic theory?

Another fish on the line. I'm sure Jimmyw is grinning his evil grin. . .

 

[rakpix]

It is true that no work is being done on the system as a whole. However, work is being done by the gas in the bank to push the air through the hose. The expansion is why the air in the bank cools. If no work was being done during the expansion, there would be no cooling. Conversely, work is being done on the gas in the cylinder being filled when the gas that is already in there is compressed by the gas that is entering from the other cylinder. It is all a matter of point of view, and where exactly you draw your system boundaries to most easily solve the problem.

The easiest way to look at it is this:

The gas remaining in the bank is undergoing a reversible expansion as it pushes out gas into the tank being filled. It cools due to this expansion.

The gas flowing between the two tanks is undergoing a free expansion as it flows from the one tank to the other. Then Enthalpy remains unchanged. Because of this, depending on the pressure difference, it can either get a little bit warmer or a little bit cooler depending on the gas and pressure difference. If it was an ideal gas, the temperature would remain unchanged. It is not, so it changes a bit.

The gas in the tank being filled is compressed in a way that is somewhere between reversible and irreversible. It heats up substantially as a result of this compression.

The reason why this problem is confusing is that there is no simple equation to describe this event which is highly path dependent (exactly how the transfer occurs affects the final result), and is very far from steady state. Then you add the Wikipedia trained engineers who think that the JT effect somehow replaces the ideal gas law, and the problem gets that much more confusing.

Correct. My point was that the system as a whole (high-pressure vessel, whip, and low-pressure vessel) does not experience a change in energy, if the system is adequately insulated from the atmosphere. Due to the increase in volume between the first state and final state of the gas, the final equilibrium temperature of the gas at intermediate pressure will be lower than the equilibrium temperature of either gas volume prior to opening the valve. However, the difference in temperatures of the metals in both vessels is only a transient phenomenon, and the metals of both vessels will eventually reach equilibrium through slow thermal conduction of the gasses if the system is adequately insulated from the atmosphere. So jimmyw's instincts are, partially, correct!

 

[windapp]

Correct. My point was that the system as a whole (high-pressure vessel, whip, and low-pressure vessel) does not experience a change in energy, if the system is adequately insulated from the atmosphere. Due to the increase in volume between the first state and final state of the gas, the final equilibrium temperature of the gas at intermediate pressure will be lower than the equilibrium temperature of either gas volume prior to opening the valve. However, the difference in temperatures of the metals in both vessels is only a transient phenomenon, and the metals of both vessels will eventually reach equilibrium through slow thermal conduction of the gasses if the system is adequately insulated from the atmosphere. So jimmyw's instincts are, partially, correct!

Let's forget about the temperature of the metal for a minute, assuming that all of the rigid parts are perfect insulators. Then, the temperature of the gas is our only concern. Allow the expansion, and then allow the gasses in the two vessels to reach thermal equalibrium. I know this cannot be done in practice, but at this point, we are totally into the theoretical. Because this is a free expansion, and therefore 100% irreversible, the overall enthalpy of the gas will remain the same. For an ideal gas, this means the temperature will be the same as when the process started. For a real gas, it will be a little different depending on exactly what the gas is, and the proportion of gas that was originally high pressure, and low pressure in the system. This is the Joule-Thompson effect that JohnN has been obsessing over.

Essentially what has happened here is the potential energy in the bank has been converted first to kinetic energy in pushing the gas through the valve, and then back to thermal energy when the gas stops.

The transient result, however is that the bank gets cooler, and the cylinder being filled gets warmer during the actual expansion.

 

[robertarak]

23 pages to explain why a tank gets hot when filled? This sums up scubaboard perfectly!

 

[lowviz]

...//... what does Queen have to do with kinetic theory?

I think I got that one. Search on Queen and bicycle...

 

[jimmyw]

It is true that no work is being done on the system as a whole. However, work is being done by the gas in the bank to push the air through the hose. The expansion is why the air in the bank cools. If no work was being done during the expansion, there would be no cooling. Conversely, work is being done on the gas in the cylinder being filled when the gas that is already in there is compressed by the gas that is entering from the other cylinder. It is all a matter of point of view, and where exactly you draw your system boundaries to most easily solve the problem.

The easiest way to look at it is this:

The gas remaining in the bank is undergoing a reversible expansion as it pushes out gas into the tank being filled. It cools due to this expansion.

The gas flowing between the two tanks is undergoing a free expansion as it flows from the one tank to the other. Then Enthalpy remains unchanged. Because of this, depending on the pressure difference, it can either get a little bit warmer or a little bit cooler depending on the gas and pressure difference. If it was an ideal gas, the temperature would remain unchanged. It is not, so it changes a bit.

The gas in the tank being filled is compressed in a way that is somewhere between reversible and irreversible. It heats up substantially as a result of this compression.

The reason why this problem is confusing is that there is no simple equation to describe this event which is highly path dependent (exactly how the transfer occurs affects the final result), and is very far from steady state. Then you add the Wikipedia trained engineers who think that the JT effect somehow replaces the ideal gas law, and the problem gets that much more confusing.

I dropped back to see where this thread had gone. I'm not here to restart a flame war, but I did ask the original question, and I worked hard to understand the answers given. Windapp's answer quoted above and his other answers are the best I saw. I agree with all he wrote. The only difference I saw between what he wrote, and my personal conclusions were:
1) He discussed non-ideal gas effects during constant enthalpy free-expansion. It's a second order effect and one can understand the whole process without considering that effect.
2) He broke the process down into three parts that were slightly different from the way I analyzed it. In the explanation quoted above, the gas immediately free expands in the whip, and moves with relatively low velocity through the whip to the destination tank. I analyzed it as though it did not expand in the whip, but moved at high speed through the whip, carrying excess energy in the form of kinetic energy of motion (delivered by the reversible expansion in the source tank). The gas then free expands in the destination tank, and stops immediately, converting the excess kinetic energy into heat. His analysis is probably the better one, although I found mine to be easier for me to grasp.

Either analysis gives the same answer if you correctly treat the gas in the destination tank and both are idealized approximations. I think his is a better explanation for those who want to analyze it from the point of view that the gas is heated due to being compressed in the destination tank. I liked mine because it let me understand what happened to the energy delivered by the reversible expansion process in the form of kinetic energy. I realize that my analysis requires one to imagine that each parcel of gas free expands only to its final destination volume in the destination tank, without first expanding to fill the entire tank volume, then be compressed to its final volume. Either way, the temperature of that parcel is the same. It either is heated by conversion to heat of the kinetic energy it carried (my analysis) as it travels fast through the fill whip, or by compression after arriving at a slow speed (his analysis), but both produce the same results.

Rakpix's answers also seemed basically correct, where he emphasized the isolated nature of the entire system, but without discussing free expansion and reversible/irreversible processes, it's hard to put his comments into the context I learned about the ideal gas laws.

Be safe out there.