I Calculating Pressure Change Filling Gas Cylinders

[fonz]

I want to work out how to calculate the pressure change in a gas cylinder if it is used to fill another cylinder to a lower pressure.

For example, if a 50 litre gas cylinder initially at 200 bara is used to fill a 600 litre cylinder from atmospheric pressure to 1.5 Bara. What would the change in pressure be in the 50 litre cylinder?

My thoughts are that the temperature in both cylinders will change with pressure but that complicates the process. The number of moles of gas must be the conserved quantity in the process and assuming the process is adiabatic will presumably simplify the solution.

Any help would be very appreciated.

 

[russ_watters]

Wouldn't you let the temperatures equalize for a while?

 

[Chestermiller]

Are you familiar with the open system (control volume) version of the 1st law of thermodynamics?

 

[erobz]

assuming the process is adiabatic will presumably simplify the solution.

Assuming the process is isothermal will presumably simplify the solution. Are you sure you want adiabatic assumption (there are varying opinions of what some may consider simple)? Maybe an analysis for each?

 

[Chestermiller]

Are you willing to assume that we are dealing with an ideal gas, or do you want the solution for the case of a real non-ideal gas?

 

[fonz]

Thanks to everyone who has commented. I've responded to your comments below.

Wouldn't you let the temperatures equalize for a while?

I think that I can work out what the answer would be if the process was isothermal but I would like to compare the results for this specific example.

Are you familiar with the open system (control volume) version of the 1st law of thermodynamics?

I wasn't until now. I've searched for some examples and realised that it's more complicated than the typical closed system version. I noticed that the open system energy balance equations use the gas velocity as well which sounds even more complicated to have to factor in since the velocity will presumably change if the flow is throttled through a valve or regulator.

Assuming the process is isothermal will presumably simplify the solution. Are you sure you want adiabatic assumption (there are varying opinions of what some may consider simple)? Maybe an analysis for each?

I think that I have calculated the pressure change in the isothermal case and my result is shown below (critique would be appreciated). I'd like to compare it to the adiabatic case.

Assuming that mass is conserved and the process is isothermal, from Boyle's Law:

$$V_1 \Delta p_1 = V_2 \Delta p_2$$
Where 1 and 2 denote the 50L and 600L cylinders.

$$\Delta p_1 = \frac {V_2}{V_1} \Delta p_2$$
$$\Delta p_1 = \frac {600}{50} (1.5 - 1) = 6 Bar$$

Are you willing to assume that we are dealing with an ideal gas, or do you want the solution for the case of a real non-ideal gas?

I think that it would be valuable to recognise how the solution would need to be reformulated if it were a non-ideal gas.

 

[erobz]

Thanks to everyone who has commented. I've responded to your comments below.

I think that I can work out what the answer would be if the process was isothermal but I would like to compare the results for this specific example.I wasn't until now. I've searched for some examples and realised that it's more complicated than the typical closed system version. I noticed that the open system energy balance equations use the gas velocity as well which sounds even more complicated to have to factor in since the velocity will presumably change if the flow is throttled through a valve or regulator.I think that I have calculated the pressure change in the isothermal case and my result is shown below (critique would be appreciated). I'd like to compare it to the adiabatic case.

Assuming that mass is conserved and the process is isothermal, from Boyle's Law:

$$V_1 \Delta p_1 = V_2 \Delta p_2$$
Where 1 and 2 denote the 50L and 600L cylinders.

$$\Delta p_1 = \frac {V_2}{V_1} \Delta p_2$$
$$\Delta p_1 = \frac {600}{50} (1.5 - 1) = 6 Bar$$I think that it would be valuable to recognise how the solution would need to be reformulated if it were a non-ideal gas.

Oddly enough, I get the same solution for the adibatic case, which likely is an indication I don't know what I'm doing...

The equations of state for each tank:

$$ P_{50} = \frac{R}{{V\llap{-}}_{50} }m_{50}T_{50} \tag{50 L Tank} $$

$$ P_{600} = \frac{R}{{V\llap{-}}_{600} }m_{600}T_{600} \tag{600 L Tank} $$

Adiabatic process implies the internal energy of the system is conserved:

$$ \Delta u = 0 $$

Assuming both tanks are initially a state of equilibrium at common temp of surroundings $T_1$:

$$ u_f - u_o = c_v\left( m_{50}T_{50} + m_{600}T_{600} \right) - c_v\left( [m_{50}]_oT_1 + [m_{600}]_oT_1 \right) = 0 $$

$$\implies m_{50}T_{50} = k - m_{600}T_{600}$$

Where $k = [m_{50}]_oT_1 + [m_{600}]_oT_1$ (constant)

Plugging that into the equation of state for the 50 L tank:

$$P_{50} = \frac{R}{ {V\llap{-}}_{50} }\left( k - m_{600}T_{600} \right)$$

In $k$ we substitute:

$$ [m_{50}]_oT_1 = \frac{ [P_{50}]_o {V\llap{-}}_{50} }{R}$$

$$ [m_{600}]_oT_1 = \frac{ [P_{600}]_o {V\llap{-}}_{600} }{R}$$

and for $m_{600}T_{600}$ substitute:

$$ m_{600}T_{600} = \frac{ {V\llap{-}}_{600} P_{600}}{R}$$

Thus, the final pressure in the 50 liter tank is given by:

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

Maybe you can find the error in it? Otherwise, wait for @Chestermiller to arrive.

 

[Chestermiller]

Oddly enough, I get the same solution for the adibatic case, which likely is an indication I don't know what I'm doing...

The equations of state for each tank:

$$ P_{50} = \frac{R}{{V\llap{-}}_{50} }m_{50}T_{50} \tag{50 L Tank} $$

$$ P_{600} = \frac{R}{{V\llap{-}}_{600} }m_{600}T_{600} \tag{600 L Tank} $$

Adiabatic process implies the internal energy of the system is conserved:

$$ \Delta u = 0 $$

Assuming both tanks are initially a state of equilibrium at common temp of surroundings $T_1$:

$$ u_f - u_o = c_v\left( m_{50}T_{50} + m_{600}T_{600} \right) - c_v\left( [m_{50}]_oT_1 + [m_{600}]_oT_1 \right) = 0 $$

$$\implies m_{50}T_{50} = k - m_{600}T_{600}$$

Where $k = [m_{50}]_oT_1 + [m_{600}]_oT_1$ (constant)

Plugging that into the equation of state for the 50 L tank:

$$P_{50} = \frac{R}{ {V\llap{-}}_{50} }\left( k - m_{600}T_{600} \right)$$

In $k$ we substitute:

$$ [m_{50}]_oT_1 = \frac{ [P_{50}]_o {V\llap{-}}_{50} }{R}$$

$$ [m_{600}]_oT_1 = \frac{ [P_{600}]_o {V\llap{-}}_{600} }{R}$$

and for $m_{600}T_{600}$ substitute:

$$ m_{600}T_{600} = \frac{ {V\llap{-}}_{600} P_{600}}{R}$$

Thus, the final pressure in the 50 liter tank is given by:

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

Maybe you can find the error in it? Otherwise, wait for @Chestermiller to arrive.

Both these analyses are correct. But what if you wanted to determine the final temperatures and final numbers of moles in the two cylinders(adiabatic)? Or what if you wanted to allow the two chambers to equilibrate in pressure, and you had to determine the final states of the two cylinders?

 

[erobz]

Both these analyses are correct. But what if you wanted to determine the final temperatures and final numbers of moles in the two cylinders(adiabatic)? Or what if you wanted to allow the two chambers to equilibrate in pressure, and you had to determine the final states of the two cylinders?

Before I try that, could you offer some insight as to why the final pressure of the 50 L tank is the same for the isothermal and adiabatic process?

 

[Chestermiller]

Before I try that, could you offer some insight as to why the final pressure of the 50 L tank is the same for the isothermal and adiabatic process?

It’s an artifact of the ideal gas mathematics and with U being constant on both cases.

 

[Chestermiller]

Before I try that, could you offer some insight as to why the final pressure of the 50 L tank is the same for the isothermal and adiabatic process?

Here's a hint: Based on either the open system- or the closed system version of the 1st law of thermodynamics, the gas in the higher pressure cylinder at any time during the process has undergone an adiabatic reversible expansion.

 

[erobz]

Here's a hint: Based on either the open system- or the closed system version of the 1st law of thermodynamics, the gas in the higher pressure cylinder at any time during the process has undergone an adiabatic reversible expansion.

Is it because there is no entropy generated in either (reversible) process that the pressure (a property of the system) must be independent of the process?

 

[Chestermiller]

Is it because there is no entropy generated in either (reversible) process that the pressure (a property of the system) must be independent of the process?

I don’t understand this question.

 

[erobz]

I don’t understand this question.

Sorry, you must have been talking about a hint to the follow up questions. I haven't had a chance to think about them yet. Sun is shining here! Yard work to do.

 

[fonz]

Both these analyses are correct. But what if you wanted to determine the final temperatures and final numbers of moles in the two cylinders(adiabatic)? Or what if you wanted to allow the two chambers to equilibrate in pressure, and you had to determine the final states of the two cylinders?

I'm stuck with this for a couple of reasons:

1. I'm trying to work out the final pressure in the 50 litre cylinder but I don't know the number of moles of gas that leaves the cylinder or the final temperature of the cylinder, so the ideal gas equation alone can't help.
2. To work out the number of moles of gas that leaves the 50 litre cylinder I tried to work out the number of moles of gas that must exist in the 600 litre cylinder after the compression (using the ideal gas equation), but that requires knowing the temperature, which I can't figure out how to calculate.

 

[erobz]

I'm stuck with this for a couple of reasons:

1. I'm trying to work out the final pressure in the 50 litre cylinder but I don't know the number of moles of gas that leaves the cylinder or the final temperature of the cylinder, so the ideal gas equation alone can't help.
2. To work out the number of moles of gas that leaves the 50 litre cylinder I tried to work out the number of moles of gas that must exist in the 600 litre cylinder after the compression (using the ideal gas equation), but that requires knowing the temperature, which I can't figure out how to calculate

We have equations of state( ideal gas laws), conservation of energy, conservation of mass, and that pressure relationship to work with simultaneously. From what I’ve looked at so far, it’s messy.

 

[Chestermiller]

We have equations of state( ideal gas laws), conservation of energy, conservation of mass, and that pressure relationship to work with simultaneously. From what I’ve looked at so far, it’s messy.

You know the final pressure in the 50 liter cylinder, right? 194 bars,

 

[erobz]

You know the final pressure in the 50 liter cylinder, right? 194 bars,

I am using that fact, but perhaps not well. I'm real close to waiving the white flag . Every time I think I could be getting somewhere all the variables I'm working with disappear and I'm left with a 1 = 1 result

I might sleep on it and try it again tomorrow.

 

[Chestermiller]

I am using that fact, but perhaps not well. I'm real close to waiving the white flag . Every time I think I could be getting somewhere all the variables I'm working with disappear and I'm left with a 1 = 1 result

I might sleep on it and try it again tomorrow.

Envision an imaginary balloon that surrounds the air that will eventually fill the high pressure cylinder at the final pressure. During the process, this air expands to push the air that surrounds it ahead of it out of the exit valve. The air in the imaginary balloon is at the same temperature and pressure as the air that surrounds it, and so the air within the imaginary balloon experiences an adiabatic reversible expansion. The initial specific volume of the air within the balloon is $v_{1.i}=V/m_{1,i}$ and its final specific volume is $v_{1,f}=V/m_{1,f}$.

 

[erobz]

Here is where I run out of useful equations:

$$ [m_{600}]_f [T_{600}]_f = \frac{1}{R}\left( [P_{600}]_o {V\llap{-}}_{600} + {V\llap{-}}_{50} \left( [P_{50}]_o - [P_{50}]_f \right) \right) $$

The entire RHS is known.

On the LHS two variables remain, and I appear to have run out of useful equations. We don't know $T_1$ ( the initial temperature common to both tanks, nor do we know anything about the initial masses of the tanks, I can't understand why we should be able to solve for these when we are unable to fix the initial state for either(both) tank(s)?

I'm waiving the white flag.

That is derived from combining (1), which by itself has 7 unknows:

$$ [m_{600}]_f [T_{600}]_f+[m_{50}]_f [T_{50}]_f = T_1 \left( [m_{50}]_o + [m_{600}]_o \right) \tag{1}$$

Then to eliminate variables on the RHS by subbing 2 and 3:

$$ [m_{600}]_o T_1 = \frac{[P_{600}]_o {V\llap{-}}_{600} }{R} \tag{2}$$

$$ [m_{50}]_o T_1 = \frac{[P_{50}]_o {V\llap{-}}_{50} }{R} \tag{3}$$

Giving 1 equation 4 unknows:

$$ [m_{600}]_f [T_{600}]_f+[m_{50}]_f [T_{50}]_f = \frac{1}{R} \left( [P_{50}]_o {V\llap{-}}_{50} + [P_{600}]_o {V\llap{-}}_{600} \right) \tag{4}$$

Then, since we know the final pressure $[P_{50}]_f$ in the 50 L tank, eliminate $[m_{50}]_f [T_{50}]_f$ by subbing 5:

$$ [m_{50}]_f [T_{50}]_f = \frac{ [P_{50}]_f {V\llap{-}}_{50} }{R} \tag{5} $$

$$ \boxed{ [m_{600}]_f [T_{600}]_f = \frac{1}{R}\left( [P_{600}]_o {V\llap{-}}_{600} + {V\llap{-}}_{50} \left( [P_{50}]_o - [P_{50}]_f \right) \right) }$$

 

[Chestermiller]

You can solve for T and n algebraically, but, say instead, we specify $T_{1,i}=293\ K$. Then you have enough information to determine $n_{1,i}$. Same for cylinder 2.

For the adiabatic reversible expansion in cylinder 1, you have $$\frac{n_{1,f}}{n_{1,i}}=\left(\frac{P_{1,f}}{P_{1,i}}\right)^{\frac{1}{\gamma}}$$and$$\frac{T_{1,f}}{T_{1,i}}=\left(\frac{P_{1,f}}{P_{1,i}}\right)^{\frac{\gamma-1}{\gamma}}$$

 

[erobz]

You can solve for T and n algebraically, but, say instead, we specify $T_{1,i}=293\ K$. Then you have enough information to determine $n_{1,i}$. Same for cylinder 2.

I suspect you meant to say "You can't solve for T and n algebraically"?

 

[Chestermiller]

I suspect you meant to say "You can't solve for T and n algebraically"?

I meant “express the results algebraically in terms of the unspecified initial temperature .”

 

[erobz]

Envision an imaginary balloon that surrounds the air that will eventually fill the high pressure cylinder at the final pressure. During the process, this air expands to push the air that surrounds it ahead of it out of the exit valve. The air in the imaginary balloon is at the same temperature and pressure as the air that surrounds it, and so the air within the imaginary balloon experiences an adiabatic reversible expansion. The initial specific volume of the air within the balloon is $v_{1.i}=V/m_{1,i}$ and its final specific volume is $v_{1,f}=V/m_{1,f}$.

I don't think I've fully considered how the adiabtic expansion works before. The air is going from the high pressure cylinder filling the low pressure cylinder.

What do you mean by:

Envision an imaginary balloon that surrounds the air that will eventually fill the high pressure cylinder at the final pressure.

It seems like the balloons (represented by the dashed boundary) are initially:

and in the final state they are like:

 

[Chestermiller]

I don't think I've fully considered how the adiabtic expansion works before. The air is going from the high pressure cylinder filling the low pressure cylinder.

What do you mean by:It seems like the balloons (represented by the dashed boundary) are initially:

View attachment 326432

and in the final state they are like:

View attachment 326433

No. The red ballon initially fills only part of the left chamber. In the final state, it fills the entire left chamber.

 

[erobz]

No. The red ballon initially fills only part of the left chamber. In the final state, it fills the entire left chamber.

I read(skimmed) this Adiabatic Process and I see that to get to:

$$\frac{T_{1,f}}{T_{1,i}}=\left(\frac{P_{1,f}}{P_{1,i}}\right)^{\frac{\gamma-1}{\gamma}}$$

The number of moles ( or mass) is assumed to be constant. These tanks are losing/gaining mass?

 

[Chestermiller]

I read this Adiabatic Process and I see that to get to:

The number of moles ( or mass) is assumed to be constant. These tanks are losing/gaining mass?

The number of moles inside the balloon is constant.

 

[erobz]

The number of moles inside the balloon is constant.

Ok, "the balloon" contains the final mass in the cylinder. I think I get it now. Thanks!

 

[Chestermiller]

Ok, "the balloon" contains the final mass in the cylinder. I think I get it now. Thanks!

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.

 

[Chestermiller]

Here is the development obtained by applying the open system (control volume) version of the 1st law of thermodynamics to the left (higher pressure) cylinder. The starting equation is $$dU=hdn$$where n is the number of moles of gas in this tank during the process, U is the internal energy of the gas contained in the cylinder, and h is the enthalpy per mole of the exit stream from the cylinder, with $$h=u+Pv=u+RT$$where P is the pressure and v=V/n is the volume per mole of the exit stream and cylinder contents. In addition, the internal energy of the gas in the cylinder is related to the internal energy per mole and number of moles in the tank by $$U=nU$$If we combine these equations, we obtain: $$ndu=nC_VdT=RTdn$$Dividing this equation by nT then yields:
$$C_vd\ln{T}=Rd\ln{n}$$This leads to $$\frac{T_{1,i}}{T_{1,f}}=\left(\frac{n_{1,i}}{n_{1,f}}\right)^{\gamma-1}=\left(\frac{v_{1,f}}{v_{1,i}}\right)^{1-\gamma}$$

 

[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.