Calculating Pressure Change Filling Gas Cylinders

[erobz]

Nobody is saying that the pipe has no effect. @Chestermiller is saying that the change in enthalpy of the gas going through the connecting pipe is zero. But that does not mean nothing about the gas changes as it goes through the pipe. Obviously that is false, since the problem specifies that the outlet pressure into the second tank is 1.5 bar, which is a huge pressure drop from the inlet pressure from the first tank.

It is true that the OP did not specify an exact length for the connecting pipe, but it seems to me that an assumption of a few feet (or meters, if you are using SI units) is much more reasonable than contemplating a pipe that circumnavigates the globe.

My point was that outside of adiabatic fantasy land the pluming has direct consequence for generated entropy. However, inside of adiabatic fantasy land a pipe that circumnavigates the globe is no different than an imaginary partition separating the gases in the same tank. The plumbing has no effect on the final states of the "two tanks" inside of adiabatic fantasy land. So when it is said that entropy generation from the "plumbing" has been accounted for (although not technically incorrect in adiabatic fantasy land) I simply ask what change has been accounted for in the pipes? We are literally arguing about "nothing" as far as I can tell.

 

[PeterDonis]

the pluming has direct consequence for generated entropy

Yes, and that is how @Chestermiller has been treating the connecting pipe. He has been saying all along that the process of the gas expanding through the connecting pipe increases its entropy.

 

[PeterDonis]

inside of adiabatic fantasy land a pipe that circumnavigates the globe is no different than an imaginary partition separating the gases in the same tank

Only in the much more improbable "fantasy land" in which all such arrangements impose the same outlet pressure in tank 2.

 

[PeterDonis]

I simply ask what change has been accounted for in the pipes?

That's easy: the pressure drop between tank 1 and tank 2.

 

[erobz]

That's easy: the pressure drop between tank 1 and tank 2.

I meant to say “what change in what pipes” The imaginary partition achieves the same thing as any “pipes” being imagined for this problem. Chestermiller is saying the entropy generation is in the "plumbing" (a presently unmodeled component), and my text book confirms (to me at least ) it's in the mixing of the gasses a two different temperatures ( a presently modeled component). They derive the same entropy as Chester quotes as for the "pipes" as for an imaginary partition between the gases that is removed.

 

[Chestermiller]

I think we've pretty much said our opinions regarding where in the system the entropy generation/irreversibility occurs in this process. I am about to close this thread. But before I do, I would like to present the calculated entropy generated in the connecting tube and in the low pressure cylinder. The overall entropy change is determined as follows:

1. All the gas remaining in the high pressure cylinder has undergone a reversible adiabatic expansion, and the entropy generated in tis 401.7 moles of gas is zero.

2. 8.8 moles of gas pass through the connecting tube, starting at 200 bars in cylinder 1 at 293 K, while the 26.6 moles of gas initially in cylinder 2 remain in that cylinder, starting out at 1 bar and 293 K.

3. The final state of the gas in cylinder 2 is 8.8+26.6=35.4 moles gas at 321.4 K and 1.5 bars.

Therefore, the overall change of the entropy for this adiabatic process is $$\Delta S=8.8R\left[\frac{7}{2}\ln{(321.4/293)}-\ln{(1.5/200})\right]$$$$+26.6R\left[\frac{7}{2}\ln{(321.4/293)}-\ln{(1.5/1)}\right]=363.6\ J/K$$

We can get a lower bound estimate of the entropy generated in the connecting tube by recognizing that there is no temperature change between inlet and outlet (zero enthalpy change), the minimum that the pressure is at the inlet is 194 bars and the maximum that the pressure at the outlet is 1.5 bars. Therefore, the lower bound to the amount of entropy generated in the connecting tube is $$\Delta S_{lower\ bound\ in\ connector}=8.8R[-\ln{(1.5/194})]=355.7\ J/K$$Comparing this lower bound estimate of the entropy generated in the connecting tube to the total amount of entropy generated in the process shows that the vast majority of the entropy generated takes place in the connecting tube.

 

[PeterDonis]

The imaginary partition achieves the same thing as any “pipes” being imagined for this problem.

No, it doesn't, because an imaginary partition cannot enforce a 1.5 bar outlet pressure into tank 2.

 

[PeterDonis]

"plumbing" (a presently unmodeled component)

I have no idea what you mean by this. If you mean that the OP didn't explicitly specify that he was connecting tank 1 to tank 2 with a pipe of a specific length, you're quibbling. The OP did specify the outlet pressure into tank 2 and the starting pressure in tank 1. That information is already sufficient to know that some kind of device that enforces a huge pressure drop from tank 1 to tank 2 is necessary to connect the tanks. So any valid model has to include such a device.

 

[erobz]

I have no idea what you mean by this. If you mean that the OP didn't explicitly specify that he was connecting tank 1 to tank 2 with a pipe of a specific length, you're quibbling. The OP did specify the outlet pressure into tank 2 and the starting pressure in tank 1. That information is already sufficient to know that some kind of device that enforces a huge pressure drop from tank 1 to tank 2 is necessary to connect the tanks. So any valid model has to include such a device.

@Chestermiller
@PeterDonis

 

[fonz]

Sorry to bump this old thread but despite all the help I still have some issue with this problem.

1. How can the final pressure in the 50 litre cylinder be calculated without knowing the final temperature?
2. Similarly, how can the final temperature in the 50 litre cylinder be calculated without knowing the final pressure?
3. How is it possible to work out the number of moles of gas that initially fills the hypothetical balloon inside the 50 litre cylinder, that eventually expands to fill the whole cylinder at the final pressure and temperature, if it’s initial volume is unknown?

It seems like there are too many unknowns to use the adiabatic equation.

In the examples in this thread, I took it for granted that the final pressure I calculated was correct. However, my calculation for the final pressure assumed that the initial and final temperature of gas in both cylinders was the same i.e. a completely isothermal process, which isn’t the case.