I Calculating Pressure Change Filling Gas Cylinders

[erobz]

Obvious to you, not to me. I guess we are just going to have to consider this a disagreement between experts.

I respect your opinion, and consider you the expert (not I). But thank you for offering the extension of "no hard feelings". I reciprocate.

But seriously though, if I don't get off of this computer today my wife (mother of my children) is probably going to divorce me...and I wouldn't blame her

 

[Chestermiller]

I respect your opinion, and consider you the expert (not I). But thank you for offering the extension of "no hard feelings". I reciprocate.

But seriously though, if I don't get off of this computer today my wife (mother of my children) is probably going to divorce me...and I wouldn't blame her

Here is Problem 6.91 in the book Fundamentals of Engineering Thermodynamics by Moran et al:

Air at 200 kPa, 52C, and velocity of 355 m/s enters an insulated duct of varying cross-sectional area. The air exits at 100 kPa, 82C. At the inlet, the cross sectional area is 6.57 cm^2. Assuming the ideal gas model for the air, determine
(a) the exit velocity, m/s
(b) the rate of entropy production within the duct, in kW/K

Do you think that this problem is solvable, given that there are no details provided regarding the duct internal geometry or its length? If it is solvable, what is your solution to the problem. Do you see any relationship between this system and the duct joining our two cylinders in our problem?

 

[erobz]

Here is Problem 6.91 in the book Fundamentals of Engineering Thermodynamics by Moran et al:

Air at 200 kPa, 52C, and velocity of 355 m/s enters an insulated duct of varying cross-sectional area. The air exits at 100 kPa, 82C. At the inlet, the cross sectional area is 6.57 cm^2. Assuming the ideal gas model for the air, determine
(a) the exit velocity, m/s
(b) the rate of entropy production within the duct, in kW/K

Do you think that this problem is solvable, given that there are no details provided regarding the duct internal geometry or its length? If it is solvable, what is your solution to the problem. Do you see any relationship between this system and the duct joining our two cylinders in our problem?

Different problem. The states are fixed at the inlet and outlet of a pipe of fixed geometry. Someone is in effect measuring it and telling us "that is what it is for this pipe, can you infer X,Y,Z ... I think this illustrates my point perfectly. If we changed something about the duct "in this experiment" the state at the outlet will no longer be 100 kPa, 82C.

This problem is not the same as me going to two pipes that start in a room ( one terminating across the hall, and the other in China), noting the inlet temp, pressure, velocity is the same in each and exclaiming the temperature, pressure, velocity at the outlets are both X,Y,Z.

 

[Chestermiller]

Different problem. The states are fixed at the inlet and outlet of a pipe of fixed geometry. Someone is in effect measuring it and telling us "that is what it is for this pipe, can you infer X,Y,Z ... I think this illustrates my point perfectly. If we changed something about the duct "in this experiment" the state at the outlet will no longer be 100 kPa, 82C.

So you don't accept the result from the open system version of the 1st law of thermodynamics that, in the absence of significant kinetic energy and potential energy effects, in the adiabatic flow of a gas through a duct, the change in enthalpy of the gas between inlet and outlet is zero.

 

[erobz]

So you don't accept the result from the open system version of the 1st law of thermodynamics that, in the absence of significant kinetic energy and potential energy effects, in the adiabatic flow of a gas through a duct, the change in enthalpy of the gas between inlet and outlet is zero.

I probably allowed myself to walk in and out of reality a bit on that "analysis" of the two pipes. It's easy to forget we are living in "adiabatic fantasy land" in this discussion.

Fundamentally I feel this is an application of Occam's razor. If the pipe has no effect, the there is "no pipe" is also just as valid as a 1 foot long pipe, or a pipe that circumnavigates the universe and returns for the "completely theoretical" adiabatic flow through it. We don't need the pipe, the entropy shows up whether it's there or not. Why should I believe that something that has no effect has an effect? I think the entropy is from the mixing of the gases (the only entities that truly have a proper model in this exercise) at two different temps.

In my Thermo Text, (same one you reference Moran,Shapiro 6th), they interpret statistical entropy by examining an ideal gas in a partially evacuated chamber. They remove the partition( allowing it to expand into the other half of the chamber) They don't mention anything about the "entropy of the partition", then they state that the entropy change for the process is yada yada - what you quoted earlier (bottom of #46 arranged to work with volume change). The focus is on the "mixing/expansion of the gas", not on the partition.

 

[Chestermiller]

I probably allowed myself to walk in and out of reality a bit on that "analysis" of the two pipes. It's easy to forget we are living in "adiabatic fantasy land" in this discussion.

Fundamentally I feel this is an application of Occam's razor. If the pipe has no effect, the there is "no pipe" is also just as valid as a 1 foot long pipe, or a pipe that circumnavigates the universe and returns for the "completely theoretical" adiabatic flow through it. We don't need the pipe, the entropy shows up whether it's there or not. Why should I believe that something that has no effect has an effect? I think the entropy is from the mixing of the gases (the only entities that truly have a proper model in this exercise) at two different temps.

In my Thermo Text, (same one you reference Moran,Shapiro 6th), they interpret statistical entropy by examining an ideal gas in a partially evacuated chamber. They remove the partition( allowing it to expand into the other half of the chamber) They don't mention anything about the "entropy of the partition", then they state that the entropy change for the process is yada yada - what you quoted earlier (bottom of #46 arranged to work with volume change). The focus is on the "mixing/expansion of the gas", not on the partition.

This is very different from removing a partition between the two cylinders since, in such a case, the two chambers would rapidly equilibrate, and the combination would end up homogeneous in temperature and pressure. In our case, the pipe and valve allow gas to bleed gradually from the high pressure cylinder to the low pressure cylinder, shutting off the flow after the pressure in the low pressure side only rises by 0.5 bars. And there is no thermal interaction between the gases the two cylinders.

So, again do you accept the result from the open system version of the 1st law of thermodynamics that, in the absence of significant kinetic energy and potential energy effects, in the adiabatic flow of a gas through a duct, the change in enthalpy of the gas between inlet and outlet is zero (or not)? This does not mean that the pipe has no effect. The outlet pressure is much lower than the inlet pressure, even though the enthalpy change is zero.

See section 4.10 Throttling Devices, Fundamentals of Engineering Thermodynamics, Moran et al.

Do you not know how to analyze the problem of gas flow through a duct at constant temperature in terms of the relationship between pressure drop and flow rate?

 

[erobz]

So, again do you accept the result from the open system version of the 1st law of thermodynamics that, in the absence of significant kinetic energy and potential energy effects, in the adiabatic flow of a gas through a duct, the change in enthalpy of the gas between inlet and outlet is zero (or not)? This does not mean that the pipe has no effect. The outlet pressure is much lower than the inlet pressure, even though the enthalpy change is zero.

As I said before I accept that in "adiabatic fantasy land" its true. I was guilty of drifting into "reality" a bit.

This is very different from removing a partition between the two cylinders since, in such a case, the two chambers would rapidly equilibrate, and the combination would end up homogeneous in temperature and pressure. In our case, the pipe and valve allow gas to bleed gradually from the high pressure cylinder to the low pressure cylinder, shutting off the flow after the pressure in the low pressure side only rises by 0.5 bars. And there is no thermal interaction between the gases the two cylinders.

I think you are allowing yourself to also drift in and out of "fantasy land" when it's convenient. In the limit of this model there is absolutely no difference between a "valve" and a "partition". What size is the pipe. What size is the valve. That can't be answered. You are just as ignorant of the "dynamics" as I and everybody else surrounding these completely made up entities here. The proof is in the pudding...the same formula for entropy change is derived for the partition as you are exclaiming is the pipe. 1+ 0 = 1

Do you not know how to analyze the problem of gas flow through a duct at constant temperature in terms of the relationship between pressure drop and flow rate?

Again, with the bullying. This whole fiasco is a direct result of you bullying me because I dared to answer a question from the OP on an internet forum. You've done this to me several times now. I answer a question, and you challenge me to solve another. So much for agreeing to disagree. Bye now.

 

[Chestermiller]

As I said before I accept that in "adiabatic fantasy land" its true. I was guilty of drifting into "reality" a bit.I think you are allowing yourself to also drift in and out of "fantasy land" when it's convenient. In the limit of this model there is absolutely no difference between a "valve" and a "partition". What size is the pipe. What size is the valve. That can't be answered. You are just as ignorant of the "dynamics" as I and everybody else surrounding these completely made up entities here. The proof is in the pudding...the same formula for entropy change is derived for the partition as you are exclaiming is the pipe. 1+ 0 = 1Again, with the bullying. This whole fiasco is a direct result of you bullying me because I dared to answer a question from the OP on an internet forum. You've done this to me several times now. I answer a question, and you challenge me to solve another. So much for agreeing to disagree. Bye now.

Do you want me to specify a pipe design and its analysis, because o know how to do this.

 

[erobz]

Do you want me to specify a pipe design and its analysis, because o know how to do this.

No, I want you to leave me alone, as was suggested by yourself in post #50. I agreed with that ( noting that you were overly generous in considering me an expert), and you rescinded this morning.

 

[PeterDonis]

If the pipe has no effect

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.

 

[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.