I Calculating Pressure Change Filling Gas Cylinders

[erobz]

correct. Let's see what you calculate for the final temperature, the initial number of moles, the final number of moles in the left chamber.

For the 50 L tank ( we are missing other factors related to the unknown gas for a numerical result):

$$ [m_{50}]_o = \frac{[P_{50}]_o {V\llap{-}}_{50}}{R T_1}$$

$$ [m_{50}]_f = \frac{[P_{50}]_o {V\llap{-}}_{50}}{R T_1}\left( \frac{[P_{50}]_f}{[P_{50}]_o}\right)^{\frac{1}{\gamma}}$$

$$ [T_{50}]_f = T_1 \left( \frac{[P_{50}]_f}{[P_{50}]_o}\right)^{\frac{\gamma - 1}{\gamma}}$$

Where the final pressure in the tank $[P_{50}]_f$ is given by:

$$[P_{50}]_f = [P_{50}]_o - \frac{ {V\llap{-}}_{600} }{ {V\llap{-}}_{50} } \left( [P_{600}]_f - [P_{600}]_o \right)$$

 

[Chestermiller]

For the 50 L tank ( we are missing other factors related to the unknown gas for a numerical result):

$$ [m_{50}]_o = \frac{[P_{50}]_o {V\llap{-}}_{50}}{R T_1}$$

$$ [m_{50}]_f = \frac{[P_{50}]_o {V\llap{-}}_{50}}{R T_1}\left( \frac{[P_{50}]_f}{[P_{50}]_o}\right)^{\frac{1}{\gamma}}$$

$$ [T_{50}]_f = T_1 \left( \frac{[P_{50}]_f}{[P_{50}]_o}\right)^{\frac{\gamma - 1}{\gamma}}$$

Where the final pressure in the tank $[P_{50}]_f$ is given by:

$$[P_{50}]_f = [P_{50}]_o - \frac{ {V\llap{-}}_{600} }{ {V\llap{-}}_{50} } \left( [P_{600}]_f - [P_{600}]_o \right)$$

Please evaluate the numbers.

 

[erobz]

Please evaluate the numbers.

What’s the gas? May I ask why it matters that I evaluate?

 

[Chestermiller]

What’s the gas? May I ask why it matters that I evaluate?

Assuming a diatomic gas, $\gamma=1.4$.

We specify $T_{1,i}=T_{2,i}=T_i=293$

$$n_{1,i}=\frac{P_{1,i}V}{RT_{1,i}}=\frac{(200)(50)}{(0.08314)(293)}=410.5\ moles$$

$$T_{1,f}=T_i\left(\frac{P_{1,f}}{P_{1,i}}\right)^{\frac{\gamma-1}{\gamma}}=293\left(\frac{194}{200}\right)^{\frac{0.4}{1.4}}=290.5$$

$$n_{1,f}=\frac{P_{1,f}V}{RT_{1,f}}=\frac{(194)(50)}{(0.08314)(290.5)}=401.7\ moles$$So the number of moles in cylinder 1 decreases by 8.8 moles, and the number of moles in cylinder 2 increases by this same amount.

$$n_{2,i}=\frac{(1)(600)}{0.08415)(293}=26.6\ moles$$

$$n_{2,f}=26.6+8.8=35.4\ moles$$

What is the final temperature In cylinder 2?

 

[erobz]

$$T_{2,f}=T_i\left(\frac{P_{2,f}}{P_{2,i}}\right)^{\frac{\gamma-1}{\gamma}}=293\left(\frac{1.5}{1}\right)^{\frac{0.4}{1.4}}=328.9~\rm{K}$$

 

[Chestermiller]

$$T_{2,f}=T_i\left(\frac{P_{2,f}}{P_{2,i}}\right)^{\frac{\gamma-1}{\gamma}}=293\left(\frac{1.5}{1}\right)^{\frac{0.4}{1.4}}=328.9~\rm{K}$$

This is not correct. The gas in cylinder 2 did not undergo an adiabatic reversible compression. There was irreversibility in the connecting tube and valve between cylinder 1 and cylinder 2, and it was mixed with cooler gas from cylinder 1. $$(401.7)(290.5s)+(35.4)T_{2,f}=(410.5 + 26.6)(293)$$
$$T_{2,f}=321.4\ K$$

 

[erobz]

This is not correct. The gas in cylinder 2 did not undergo an adiabatic reversible compression. There was irreversibility in the connecting tube and valve between cylinder 1 and cylinder 2, and it was mixed with cooler gas from cylinder 1. $$(401.7)(290.5s)+(35.4)T_{2,f}=(410.5 + 26.6)(293)$$
$$T_{2,f}=321.4\ K$$

Surely the irreversibility's can be ignored in our idealized tube connecting the tanks? You don't appear to account for them in the internal energy balance you just used to solve for the final tank 2 temp, so why mention it?

Otherwise, the mixing of the gasses at different temps is something I obviously didn't consider. Good trap you set up there.

 

[erobz]

Is @fonz (the OP) finding good use in my public lashings?

 

[Chestermiller]

Surely the irreversibility's can be ignored in our idealized tube connecting the tanks? You don't appear to account for them in the internal energy balance you just used to solve for the final tank 2 temp, so why mention it?

Who says they're not?

This can be settled easily. The entropy change of the gas finally remaining in cylinder 1 is zero. What do you get for the entropy change of the 8.8 moles that started in cylinder 1 and ended up in cylinder 2 ? What do you get for the entropy change of the 26.6 moles that started and ended in chamber 2?

 

[erobz]

Who says they're not?

This can be settled easily. The entropy change of the gas finally remaining in cylinder 1 is zero. What do you get for the entropy change of the 8.8 moles that started in cylinder 1 and ended up in cylinder 2 ? What do you get for the entropy change of the 26.6 moles that started and ended in chamber 2?

Tank 1 is in California and tank 2 is in Japan. If the claim that the entropy generated isn't dependent on the geometry of the pipe joining the tanks in a process where irreversibility's are present, how is it possible that it occurred there?

 

[erobz]

The equations that are used are in complete ignorance of a pipe or a valve even existing, all they know is that two thermal reservoirs are in thermal contact. If irreversibility's occurred in this hypothetical pipe, heat was generated. With irreversibility's present in the purely hypothetical pipe and valve, assuming no heat was lost to the surroundings in going from tank 1 to tank 2, (like you did ) that implies that the temperature of the gas exiting the pipe was necessarily greater than the temperature at which it entered the pipe.

Therefore, in an answer as to what entropy that was generated by the gas in flowing from tank 1 to tank 2 is indeterminate. We simply don't know the outlet temperature (inlet of tank 2). We only have the entropy generated in that gas after it was completely mixed with the gas in tank 2. The mixing of the gases generated the entropy, not the arbitrary chosen purely theoretical pipes and valves connecting the two.

 

[Chestermiller]

The equations that are used are in complete ignorance of a pipe or a valve even existing, all they know is that two thermal reservoirs are in thermal contact. If irreversibility's occurred in this hypothetical pipe, heat was generated. With irreversibility's present in the purely hypothetical pipe and valve, assuming no heat was lost to the surroundings in going from tank 1 to tank 2, (like you did ) that implies that the temperature of the gas exiting the pipe was necessarily greater than the temperature at which it entered the pipe.

It sounds that you are not familiar with the irreversible process of steady Joule-Thomson adiabatic flow through a porous plug, a valve, or a pipe in which the change in enthalpy per unit mass of the fluid is zero and, for an ideal gas, the temperature is constant.

Therefore, in an answer as to what entropy that was generated by the gas in flowing from tank 1 to tank 2 is indeterminate. We simply don't know the outlet temperature (inlet of tank 2). We only have the entropy generated in that gas after it was completely mixed with the gas in tank 2. The mixing of the gases generated the entropy, not the arbitrary chosen purely theoretical pipes and valves connecting the two.

Oh yeah? The initial pressure of the 8.8 moles of gas that entered the tank was 200 bars and its initial temperature was 293 K; its final pressure was 1..5 bars and its final temperature was 321.4 K.

And, for the 26..6 moles of gas that were inside tank 2 to begin with. its initial pressure was 1 bar and its initial temperature was293 k; its final pressure was 1.5 bars and its final temperature was 321.4 K.

Are you saying that you can't determine the entropy changes for these?

 

[erobz]

It sounds that you are not familiar with the irreversible process of steady Joule-Thomson adiabatic flow through a porous plug, a valve, or a pipe in which the change in enthalpy per unit mass of the fluid is zero and, for an ideal gas, the temperature is constant.

Irrelevant here. You are evoking losses of a fictious valve and arbitrary length of pipe/diameter connecting them. You are in effect trying to sell me the two scenarios are equivalent (non-zero) in terms of entropy generation in each completely fictious connecting pipe. Neither of which are represented in any way shape or form in the model.

 

[Chestermiller]

You have a system in an initial state and a final state. What is the entropy change?

State 1: 8.8 moles of a diatomic ideal gas at 293 K and 200 bars, and 26.6 moles of the same gas at 293 K and 1 bar

State 2: 35,4 moles of the same gas at 321.4 K and 1.5 bars.

What do you calculate for the entropy change for this system?

 

[erobz]

You have a system in an initial state and a final state. What is the entropy change?

State 1: 8.8 moles of a diatomic ideal gas at 293 K and 200 bars, and 26.6 moles of the same gas at 293 K and 1 bar

State 2: 35,4 moles of the same gas at 321.4 K and 1.5 bars.

What do you calculate for the entropy change for this system?

You told me that a fixed amount entropy was generated by the addition of a fictious valve and connecting pipe of arbitrary dimension. A pipe which is completely unmodeled is generating entropy. Someone better call Bernoulli and let him know. I'm sorry, but it's a ludicrous proposition... I'm no longer entertaining this idea. The entropy in this model can only come from a part of the system that is modeled. There is no pipe, or valve, it's a figment of your imagination in this exercise.

The mixing of the gases at two temperatures in tank 2 is what is generating the entropy you wish me to calculate.

 

[Chestermiller]

You told me that a fixed amount entropy was generated by the addition of a fictious valve and connecting pipe of arbitrary dimension. A pipe which is completely unmodeled is generating entropy. Someone better call Bernoulli and let him know. I'm sorry, but it's a ludicrous proposition... I'm no longer entertaining this idea. The entropy in this model can only come from a part of the system that is modeled. There is no pipe, or valve, it's a figment of your imagination in this exercise.

The mixing of the gases at two temperatures in tank 2 is what is generating the entropy you wish me to calculate.

I think our point of contention is that I am maintaining that most of the irreversible entropy generation takes place in the connecting duct between the two cylinders and you are maintaining that most of it is generated by mixing of the cool gas from cylinder 1 with the hotter gas in cylinder 2. Is that your understanding?

I am envisioning the connecting pipe and valve between the cylinders as being of short length and high fluid flow resistance, having negligible volume and negligible mass and energy holdup. During the process, the gas flowing through this pipe starts out at 194-200 bars at the inlet end and exits at 1.0 to 1.5 bars at the exit end. Bernoulli equation does not describe this high frictional drag flow with significant pressure drop.

In first semester Thermodynamics, we study the application of the open system version of the 1st law of thermodynamics to the steady adiabatic flow of an ideal gas through a short length of pipe containing a porous plug or valve (or just a small diameter pipe having very significant frictional flow resistance). We conclude that the change in enthalpy per unit mass of the gas is zero, as is its temperature change in passing through the pipe from the high pressure end to the low pressure end. The deformation that the gas experiences in the pipe is highly irreversible, and significant entropy generation occurs in the gas between entrance and exit from the pipe. The change in entropy per mole of gas passing through the pipe is given by $$\Delta s=-R\ln{(P_{out}/P_{in})}$$So the pipe is not "completely unmodeled," and, on the contrary, the entropy generation in the pipe is accurately accounted for.

 

[erobz]

I think our point of contention is that I am maintaining that most of the irreversible entropy generation takes place in the connecting duct between the two cylinders and you are maintaining that most of it is generated by mixing of the cool gas from cylinder 1 with the hotter gas in cylinder 2. Is that your understanding?

Yes, that is our point of contention.

I am envisioning the connecting pipe and valve between the cylinders as being of short length and high fluid flow resistance, having negligible volume and negligible mass and energy holdup.

And to make a point (that you keep flat out ignoring), I'm envisioning a pipe joining tank1 and tank2 that circumnavigates the globe. Neither connection type (or anything in-between) has been specified, nor would they generate the same amount of entropy.

 

[Chestermiller]

Yes, that is our point of contention.And to make a point (that you keep flat out ignoring), I'm envisioning a pipe joining tank1 and tank2 that circumnavigates the globe. Neither connection type (or anything in-between) has been specified, nor would they generate the same amount of entropy.

Well, I'm not sure I can design a tube 25000 miles long with a pressure difference between the ends of 200 bars to 1 bar having low enough holdup (say << 1 liter) so that the holdup is << the the tank volumes. But I am sure I can design one with a pipe of length anywhere from 6 inches to 5 feet, all of which have negligible holdup and have the same amount of entropy generated for 8 moles of diatomic gas flowing between the two tanks. How does that grab you?

 

[erobz]

Well, I'm not sure I can design a tube 25000 miles long with a pressure difference between the ends of 200 bars to 1 bar having low enough holdup (say << 1 liter) so that the holdup is << the the tank volumes. But I am sure I can design one with a pipe of length anywhere from 6 inches to 5 feet, all of which have negligible holdup and have the same amount of entropy generated for 8 moles of diatomic gas flowing between the two tanks. How does that grab you?

It doesn't grab me at all. It's immaterial...same as the unspecified pipe joining the tanks. Its apparent to me that you are incapable of conceding this minuscule (but completely obvious ) point I'm making. I'm walking away now. It Mothers Day. Take care.

 

[Chestermiller]

It doesn't grab me at all. It's immaterial...same as the unspecified pipe joining the tanks. Its apparent to me that you are incapable of conceding this minuscule (but completely obvious ) point I'm making. I'm walking away now. It Mothers Day. Take care.

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

 

[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.