Engineering Initially evacuated chamber and maximum theoretical work

[Stephen Miller]

Homework Statement

Initially evacuated chamber and maximum theoretical work

Relevant Equations

Energy Balance Equation

Given: V_1 = 10 L, V_2 = 22 L

Not sure what I need to do here. We are told to ignore heat transfer, and the problem suggests that this is not a steady-state problem. The gage pressure is given, so I think that means we need to calculate for the absolute pressure, which would be 997 kPa. Cp and k are given, which yields Cv = 0.717.
I was thinking to use the fact that we have two volumes and the absolute pressure, so use the definition of work and compute that integral. However, the problem isn't that simple. Maybe we need to take a control volume about the entire system? Not sure, please help.

 

[collinsmark]

Homework Statement: Initially evacuated chamber and maximum theoretical work
Homework Equations: Energy Balance Equation

Given: V_1 = 10 L, V_2 = 22 L

Not sure what I need to do here. We are told to ignore heat transfer, and the problem suggests that this is not a steady-state problem. The gage pressure is given, so I think that means we need to calculate for the absolute pressure, which would be 997 kPa. Cp and k are given, which yields Cv = 0.717.
I was thinking to use the fact that we have two volumes and the absolute pressure, so use the definition of work and compute that integral. However, the problem isn't that simple. Maybe we need to take a control volume about the entire system? Not sure, please help.

The first step is to determine which type of process this is (eg., isobaric, adiabatic, isothermal, isometric, etc.)

(There may be an integral involved later, but first things first.)

 

[Stephen Miller]

It's adiabatic

 

[collinsmark]

It's adiabatic

Correct!

In an adiabatic process, PV \ne constant.

But there is a pressure and volume relationship that is equal to a constant (in an adiabatic process). Can you tell me what that relationship that is?

 

[Stephen Miller]

Is it pV^n = constant ?

 

[collinsmark]

Is it pV^n = constant ?

Correct again.

Except I think your coursework might use the variable k instead of n. I think. At least that's what I interpret in the problem statement when it says, "k = 1.4."

So

pV^k = C

where C is a constant.

Solve for p in terms of C and V^k.

Then you can plug that into W = \int_{V_1}^{V_{1+2}} pdV and voilà.

[Edit: Maybe "C" wasn't the best choice of variable name here. Call it K if you'd rather. Just don't confuse it with c_p or c_v. They're all different constants. ... No, K would be a bad choice too (your coursework uses k for something else). How about C_{\mathrm{konstant}}. ]

 

[Chestermiller]

It seems to me this is a pretty complicated problem. I've been thinking about it for a few hours, and I still don't see how to proceed. The main difficulty seems to me to be that tank that initially has vacuum in it. I would expect there to be some irreversibility associated with introducing the exit stream from the the turbine, and the mixing of the exit stream from the turbine with the existing contents of the tank at any time during the process.

It is clear that, in the end, the final pressures in the two tanks are going to be equal. But it also seems that the temperatures in the two tanks at the end of the process do not have to be equal.

I'll keep thinking about how to proceed.

 

[collinsmark]

It seems to me this is a pretty complicated problem. I've been thinking about it for a few hours, and I still don't see how to proceed. The main difficulty seems to me to be that tank that initially has vacuum in it. I would expect there to be some irreversibility associated with introducing the exit stream from the the turbine, and the mixing of the exit stream from the turbine with the existing contents of the tank at any time during the process.

It is clear that, in the end, the final pressures in the two tanks are going to be equal. But it also seems that the temperatures in the two tanks at the end of the process do not have to be equal.

I'll keep thinking about how to proceed.

My previously suggested approach does make a few assumptions. Primarily, it assumes that the turbine is somehow able to keep the gas in both tanks at equal temperatures [extracting any available work in the process]. Even if the temperature changes over time, the turbine is able to maintain equal temperatures in both tanks at any given time.

That assumption of mine is probably not very realistic. But hey, the creator if the problem did find a tank of ideal gas, so why not an ideal turbine? (Great yard-sale find! )

[Edit: This heat transfer is not wasted in this ideal turbine. Any temperature differential between the two tanks is utilized as part of the work extraction.]

 

[Stephen Miller]

I don't know how to use LaTex, so here's an attached file instead :)
I hope that's right. I got a positive answer for W, which matches with W_out.

 

[collinsmark]

I don't know how to use LaTex, so here's an attached file instead :)
I hope that's right. I got a positive answer for W, which matches with W_out.

The final volume is not just the volume of tank-2, but rather the volume of tank-1 plus the volume of tank-2.

Is the volume of tank-2 12 liters or 22 liters? If it's 12 liters, then nevermind.

 

[collinsmark]

Also, in your final step you used the gauge pressure instead of absolute pressure for p_1.

 

[Stephen Miller]

The final volume is not just the volume of tank-2, but rather the volume of tank-1 plus the volume of tank-2.

Is the volume of tank-2 12 liters or 22 liters? If it's 12 liters, then nevermind.

$V_2 = 22 L$

 

[collinsmark]

$V_2 = 22 L$

Is that the volume of tank-2 or the combined volume of tank-1 and tank-2?

The final volume for this exercise needs to be the combined volume.

 

[Stephen Miller]

Is that the volume of tank-2 or the combined volume of tank-1 and tank-2?

The final volume for this exercise needs to be the combined volume.

That's the volume for tank 2

 

[collinsmark]

That's the volume for tank 2

OK, then you'll need to re-do the calculations such that the final volume is the combined volume. Also, don't forget to use absolute pressures (instead of gauge pressure).

That said, your approach (equations and whatnot) look good!

 

[Stephen Miller]

OK, then you'll need to re-do the calculations such that the final volume is the combined volume. Also, don't forget to use absolute pressures (instead of gauge pressure).

That said, your approach (equations and whatnot) look good!

So change the upper bound of my integral to $V_2 = 32$ and use $997 kPa$ instead of $896 kPa$

 

[collinsmark]

So change the upper bound of my integral to $V_2 = 32$ and use $997 kPa$ instead of $896 kPa$

Yep. I think that should do it.

 

[Stephen Miller]

So change the upper bound of my integral to $V_2 = 32$ and use $997 kPa$ instead of $896 kPa$

Yep. I think that should do it.

Thank you for your help, I'll have to swing by my teacher's office hours to check my work.

 

[Chestermiller]

In my judgment, you guys are significantly underestimating the complexity of this problem.

I don't see how the two tank temperatures can be at the same in the setup given (since the two tanks are physically separated). Any adiabatic reversible turbine is going to feature a decrease in both temperature and pressure.

I think I can solve this if I assume that, rather than a vacuum, there is some small initial pressure (and associated mass) within the 2nd tank. Maybe then I can examine the solution to see if it has a limit as the initial pressure approaches zero. It may take me some time to analyze this problem in this way.

 

[collinsmark]

In my judgment, you guys are significantly underestimating the complexity of this problem.

I agree that if the yard-sale turbine (great find, though it may be) does not have components that allow it to operate as a heat engine (capitalizing on temperature differences) in conjunction with it operating on the pressure differences, then this problem becomes much harder than we've treated it thus far.

 

[collinsmark]

I agree that if the yard-sale turbine (great find, though it may be) does not have components that allow it to operate as a heat engine (capitalizing on temperature differences) in conjugation with it operating on the pressure differences, then this problem becomes much harder than we've treated it thus far.

That does give me an idea though. If we can calculate the amount of work specifically obtained from keeping the two tanks at equal temperatures, we could then subtract that off the original answer, and that should be the answer for a turbine that does not attempt to keep the tanks at equal temperatures! (I'm not 100% sure on that, however.)

Hmm. Easier said than done though.

 

[Chestermiller]

Here is my analysis of the problem based on the open system (control volume) versions of the 1st and 2nd laws of thermodynamics applied to each of the two tanks and the turbine. It does not require the temperatures in the two tanks to be equal (as I definitely do not expect them to be). It also assumes that the mass holdup in the turbine is negligible, so that the turbine operates at quasi-steady-state, and all the mass of gas resides in the two tanks.

Nomenclature
$n_1(t)=$ moles of gas in tank 1
$n_2(t)=$ moles of gas in tank 2
$T_1(t)=$ temperature of gas in tank 1 = temperature of gas entering turbine
$T_2(t)=$ temperature of gas in tank 2 (assumed well-mixed)
$P_1(t)=$ pressure of gas in tank 1 = pressure of gas entering turbine
$P_2(t)=$ pressure of gas in tank 2 = pressure of gas exiting turbine
$T_f(t)=$ temperature of gas exiting turbine and entering tank 2
$u_1(t)$ = internal energy per mole of gas in tank 1 = internal energy per mole of gas exiting tank 1 and entering turbine
$h_1(t)$ = enthalpy per mole of gas in tank 1 = enthalpy per mole of gas exiting tank 1 and entering turbine
$h_f(t)$ = enthalpy per mole of gas exiting turbine and entering tank 2
$u_2(t)$ = internal energy per mole of gas in tank 2

TANK 1
Application of the open system version of the 1st law to Tank 1 yields:
$$\frac{d(n_1u_1)}{dt}=\frac{dn_1}{dt}h_1=\frac{dn_1}{dt}(u_1+RT_1)$$
Differentiating the left hand side by parts then yields:
$$n_1\frac{du_1}{dt}=RT_1\frac{dn_1}{dt}$$or equivalently $$n_1C_v\frac{dT_1}{dn_1}=RT_1$$ This equation integrates immediately to $$\frac{T_1}{T_{10}}=\left(\frac{n_1}{n_{10}}\right)^{(\gamma-1)}\tag{1}$$
The corresponding pressure is $$P_1=\frac{n_1RT_1}{V_1}=\frac{n_1RT_{10}}{V_1}\left(\frac{n_1}{n_{10}}\right)^{(\gamma-1)}=\frac{n_{10}RT_{10}}{V_1}\left(\frac{n_1}{n_{10}}\right)^{\gamma}=P_{10}\left(\frac{n_1}{n_{10}}\right)^{\gamma}\tag{2}$$
Eqns. 1 and 2 express the temperature in Tank 1 parametrically in terms of the number of moles of gas $n_1(t)$ remaining in the tank at any time. As expected, both the temperature and pressure decrease as gas leaves the adiabatic tank during the process.

TURBINE
Application of the open system version of the 2nd law of thermodynamics to the Turbine operating at steady state, adiabatically, and reversibly yields: $$-\frac{dn_1}{dt}\Delta s=0$$ where $\Delta s$ is the change in entropy per mole of gas passing through the turbine, given by:
$$\Delta s=C_p\ln{(T_f/T_1)}-R\ln{(P_2/P_1)}=0$$or equivalently$$\frac{T_f}{T_1}=\left(\frac{P_2}{P_1}\right)^{\frac{(\gamma-1)}{\gamma}}\tag{3}$$where $P_2$ is the exit pressure from the turbine (also equal to the pressure in Tank 2)

TANK 2
Application of the open system version of the 1st law to Tank 2 yields:
$$\frac{d(n_2u_2)}{dt}=\frac{dn_2}{dt}h_f=\frac{dn_2}{dt}(u_f+RT_f)$$
Differentiating the left hand side by parts then yields:
$$n_2\frac{du_2}{dt}=(u_f-u_2+RT_f)\frac{dn_2}{dt}$$or equivalently
$$n_2C_v\frac{dT_2}{dt}=(C_pT_f-C_vT_2)\frac{dn_2}{dt}$$or equivalently
$$\frac{d{(n_2T_2)}}{dn_2}=\gamma T_f$$

Next, from the ideal gas law, we have that $n_2T_2=\frac{P_2V_2}{R}$. Substitution of this into the previous equation gives:$$\frac{V_2}{R}\frac{dP_2}{dn_2}=\gamma T_f\tag{4}$$
If we next substitute Eqns 1- 3 into Eqn. 4, we obtain
$$\frac{V_2}{R}\frac{dP_2}{dn_2}=\gamma T_{10}\left(\frac{P_2}{P_{10}}\right)^{\frac{(\gamma-1)}{\gamma}}$$Further manipulation using the ideal gas law and integration leads to the astonishingly and unexpectedly simple result that $$\frac{P_2}{P_{10}}=\left(\frac{n_2V_1}{n_{10}V_2}\right)^\gamma\tag{5}$$

It will complete this analysis in a subsequent post.

 

[mfig]

I am not sure why one couldn't simply use the change in entropy formula considering the gas as the system undergoing a process from state 1 to 2 - with the turbine volume negligible. If so, then we have

$\Delta S = \int_{1}^{2} \frac{\delta Q}{T} + \sigma$

Since there is no heat transfer the integral is zero. If we are looking for the maximum theoretical work then the entropy generation term is also zero. So

$\Delta S = 0$

If this is correct, then we can use, for an ideal gas,

$\Delta S = 0 = c_v ln(\frac{T_2}{T_1}) -R Ln(\frac{V_1}{V_2})$

This means that we can use the relation

$\frac{T_2}{T_1} = (\frac{V_1}{V_2})^{(k-1)}$

So if we plug in numbers we find $T_2 = 187.2K$. Then we can use the 1st Law for a system

$\Delta U = Q-W = 0 - W \rightarrow \Delta U= -W$

Now use the ideal gas relation for the change in internal energy,

$\Delta U = c_v \Delta T$

Finally, this results in

$c_v(T_2 - T_1) = -79.6 kJ = -W$

so the maximum theoretical work is $W_{max} = 79.6 kJ$.

 

[Chestermiller]

I am not sure why one couldn't simply use the change in entropy formula considering the gas as the system undergoing a process from state 1 to 2. If so, then we have

$\Delta S = \int_{1}^{2} \frac{\delta Q}{T} + \sigma$

Since there is no heat transfer the integral is zero. If we are looking for the maximum theoretical work then the entropy generation term is also zero.

The entropy generation is zero only if the process is reversible. And the discharge and mixing of the expanded gas from the turbine with the gas already in tank 2 during the process is going to be irreversible.

 

[mfig]

The entropy generation is zero only if the process is reversible. And the discharge and mixing of the expanded gas from the turbine with the gas already in tank 2 during the process is going to be irreversible.

Right. But if we are looking for the maximum theoretical work, we should consider all processes reversible. Of course this is never achievable, but this would establish an upper limit. So it seems to me.

 

[Chestermiller]

Right. But if we are looking for the maximum theoretical work, we should consider all processes reversible. Of course this is never achievable, but this would establish an upper limit. So it seems to me.

The problem statement clearly says that the only equipment available are the 2 tanks and the turbine. We must come as close as possible to reversible within this constraint.

 

[mfig]

The problem statement clearly says that the only equipment available are the 2 tanks and the turbine. We must come as close as possible to reversible within this constraint.

I assumed no other equipment than the tanks and turbine, so I am confused about why you would point this out. "As close as possible to reversible" is very near reversible. I assumed exactly reversible, which puts an upper limit on the "very near" part.

What am I missing?

 

[Chestermiller]

This is a continuation of the analysis presented in post #22. At the final state of this system, we must have that $P_2$ determined from Eqn. 5 is equal to $P_1$ determined from Eqn. 2. From this, it follows that $$\frac{n_2V_1}{n_{10}V_2}=\frac{n_1}{n_{10}}$$or equivalently: $$\frac{n_2}{n_1}=\frac{V_2}{V_1}$$So, amazingly, the final numbers of moles in each of the two tanks is just proportional to the tank volume. From this (and previous equations), it follows that, in the final state of the system, $$n_1=n_{10}\frac{V_1}{(V_1+V_2)}$$
$$n_2=n_{10}\frac{V_2}{(V_1+V_2)}$$
$$T_1=T_{10}\left(\frac{V_1}{V_1+V_2}\right)^{(\gamma-1)}$$
$$T_2=T_1=T_{10}\left(\frac{V_1}{V_1+V_2}\right)^{(\gamma-1)}$$

Now, for the parameters in this problem, $$n_{10}=\frac{(997000)(0.010)}{(8.314)(298)}=4.024\ moles$$
So, in the final state of the system, $$n_1=4.024(10/32)=1.258\ moles$$
$$n_2=4.024(22/32)=2.767\ moles$$
$$T_1=298\left(\frac{10}{32}\right)^{0.4}=187.1K$$
$$T_2=298\left(\frac{10}{32}\right)^{0.4}=187.1K$$
The change in internal energy for the system between the initial and final states is given by:
$$\Delta U=n_{10}C_v(T_1-T_{10})=(4.024)(20.79)(187.1-298)=-9278\ J$$
This is minus the maximum amount of work, so the maximum amount of work is +9278 J.

For the final pressure, I get 197 kPa

 

[Chestermiller]

I assumed no other equipment than the tanks and turbine, so I am confused about why you would point this out. "As close as possible to reversible" is very near reversible. I assumed exactly reversible, which puts an upper limit on the "very near" part.

What am I missing?

I don't know. Have you looked over my analysis? Does anything look incorrect about that?

 

[mfig]

I don't know. Have you looked over my analysis? Does anything look incorrect about that?

No, for your analysis is very detailed! One thing I notice is that the pressures aren't equal at the end, unless I have misapplied your results? (Perhaps this is round-off error.)

$P_1=P_{10}\left(\frac{n_1}{n_{10}}\right)^{\gamma}=997 \left(\frac{1.258 }{4.024 }\right)^{1.4}=195.7kPa$

$P_2=P_{10}\left(\frac{n_2V_1}{n_{10}V_2}\right)^\gamma = 997\left(\frac{2.676 *10}{4.024 *22}\right)^{1.4}=186.8kPa$

Also, the ideal gas law is not satisfied in tank 2, if the pressure above is correct (but it is satisfied in tank 1 given the above pressure).

For tank 1:

$P_1 = n_1*R*T_1/V_1 =1.258*8.314*187.1/.01 = 195.7kPa$

This is as above.

But for tank 2:

$P_2 = n_2*R*T_2/V_2 = 2.676*8.314*256.5/0.022 = 259.4kPa$

This differs significantly from above.

 

[Chestermiller]

No, for your analysis is very detailed! One thing I notice is that the pressures aren't equal at the end, unless I have misapplied your results? (Perhaps this is round-off error.)$P_1=P_{10}\left(\frac{n_1}{n_{10}}\right)^{\gamma}=997 \left(\frac{1.258 }{4.024 }\right)^{1.4}=195.7kPa$

$P_2=P_{10}\left(\frac{n_2V_1}{n_{10}V_2}\right)^\gamma = 997\left(\frac{2.676 *10}{4.024 *22}\right)^{1.4}=186.8kPa$

Also, the ideal gas law is not satisfied in tank 2, if the pressure above is correct (but it is satisfied in tank 1 given the above pressure).

For tank 1:

$P_1 = n_1*R*T_1/V_1 =1.258*8.314*187.1/.01 = 195.7kPa$, as above.

But for tank 2:

$P_2 = n_2*R*T_2/V_2 = 2.676*8.314*256.5/0.022 = 259.4kPa$, which differs from above.

That's because I made a mistake in arithmetic or a typo. n2 should be 2.767. I have corrected it in all the subsequent equations.

 

[mfig]

That's because I made a mistake in arithmetic or a typo. n2 should be 2.767

Ah, that explains the pressure discrepancy. The second problem remains, however.

 

[Chestermiller]

Ah, that explains the pressure discrepancy. The second problem remains, however.

Hmmm. I'll have to recheck my algebra.

 

[Chestermiller]

Ah, that explains the pressure discrepancy. The second problem remains, however.

Yes. I made an algebra error. Amazingly, T2 comes out to be equal to T1. I've gone back and corrected that too. Thanks for pointing these out.

 

[mfig]

Yes. I made an algebra error. Amazingly, T2 comes out to be equal to T1. I've gone back and corrected that too. Thanks for pointing these out.

That's great. But now I noticed something about my solution in post 23. I left out the mass in the final step of my isentropic solution. When I include the mass (0.1165 kg), calculated from the ideal gas law on the initial contents of tank 1, our solutions closely match. Unfortunately, I can't go back and edit my work. I think I understand why they match. What a neat exercise!

 

[Chestermiller]

That's great. But now I noticed something about my solution in post 23. I left out the mass in the final step of my isentropic solution. When I include the mass, calculated from the ideal gas law on the initial contents of tank 1, our solutions closely match. Unfortunately, I can't go back and edit my work. I think I understand why they match. What a neat exercise!

I think the interesting part is explaining mechanistically why they match. I had an idea about this, and I'd like to try it out on you. My thought is that the gas that had already entered tank 2 (earlier in the process) is being adiabatically and reversibly compressed by the new gas that enters tank 2 later in the process. So even in tank 2, the process is proceeding isentropically. So at all times through the process, the change in entropy of the total system contents is zero. Is this what you had in mind?

 

[mfig]

I think the interesting part is explaining mechanistically why they match. I had an idea about this, and I'd like to try it out on you. My thought is that the gas that had already entered tank 2 (earlier in the process) is being adiabatically and reversibly compressed by the new gas that enters tank 2 later in the process. So even in tank 2, the process is proceeding isentropically. So at all times through the process, the change in entropy of the total system contents is zero. Is this what you had in mind?

Yes, that's what I imagine happens in the limit. It also explains why the temperatures are the same even though the tanks aren't thermally connected. When we consider a reversible and adiabatic process it is in the limit as a quasi-static, or infinitely slow, process. Thus, the gas temperature during the entire process equalizes by the normal kinetic means because the gas is "self connected" so to speak.

I think this is what happens explicitly in my solution and implicitly in yours. When you used only the 1st law for open systems on the tanks, there is no consideration of entropy so there is no accounting for any change. Does that sound right?

 

[collinsmark]

our solutions closely match.

I think the interesting part is explaining mechanistically why they match.

By golly, my solution also matches, and thus I assume that after correcting for a couple of minor mistakes (that were known to be mistakes), that @Stephen Miller's (the OP's) final answer matches too!

So, I suppose, in the end we were all correct. (Although I agree that @Chestermiller was much more rigorous in the analysis.)

Thermodynamics is neat.

 

[collinsmark]

I think the reason all our different solutions match is we each ultimately modeled isentropic processes, albeit in different ways.

@mfig's isentropic approach, invoking entropy from the get-go, is arguably a more applicably approach than mine, since the volume of the system is not changing gradually.

However, in my approach, I assumed reversibility, then integrated over the volume. Because the system being modeled was assumed to be completely reversible, the ideal turbine (in this model, using the assumptions I used) is equivalent to an ideal piston in a cylinder, in terms of total work done.

@Chestermiller's approach using separate processes (individually analyzing each tank in partial-isolation), seems to have also modeled a system which is ultimately isentropic.

 

[collinsmark]

I've been thinking about this for the last couple of hours.

So the crux of the question is, why does tank1 and tank2 ultimately end up in thermal equilibrium in the final condition? Why did @Chestermiller's result lead this this conclusion, even though each tank was analyzed separately?

Firstly, I think @mfig's idea about quasi-static equilibrium is crucial here.

This problem varies between (a) happening over centuries [possibly longer] or (b) happening instantaneously with throttling*. But to maximize the problem, we're going to have to go with (a). The gas isn't being throttled through the turbine, but it's presesd rather slowly, -- possibly very slowly --, being fed/forces through it, generating work in the process, for all it's worth. (I guess what I'm saying is it's important is that the turbine is actually able to extract work from the transfer. All the work. Even it it takes awhile).

Scenario (a) captures all the work where (b) merely increases entropy* (in which several approches we've taken are assumed zero). So if we're maximizing work output, we just must be patient.

So the solution that we all came up with (everybody involved [solution (a)]) is the maximum work that can be obtained, it might take awhile to achieve it. But it's not out of reach.

-----

I've learned a lot with this problem. And there might be more to learn. But right now, I need to rest. Maybe more to learn tomorrow.

*Constant enthalpy with throttling! Yes. But enthalpy be damned! (kidding)

 

[Chestermiller]

Yes, that's what I imagine happens in the limit. It also explains why the temperatures are the same even though the tanks aren't thermally connected. When we consider a reversible and adiabatic process it is in the limit as a quasi-static, or infinitely slow, process. Thus, the gas temperature during the entire process equalizes by the normal kinetic means because the gas is "self connected" so to speak.

I think this is what happens explicitly in my solution and implicitly in yours. When you used only the 1st law for open systems on the tanks, there is no consideration of entropy so there is no accounting for any change. Does that sound right?

Yes. I might add however that, even though, in the end, the temperatures in the two tanks are equal, during the process path that I considered (essentially the simplest quasi-static path that one would use for this system), the temperature in the two tanks are not equal except at the end.

 

[Chestermiller]

This problem varies between (a) happening over centuries [possibly longer] or (b) happening instantaneously with throttling*.

*Constant enthalpy with throttling! Yes. But enthalpy be damned! (kidding)

Please be careful with wording. When I think of "throttling," I'm immediately thinking of dissipation in a valve or porous plug (along with entropy generation). I'm sure that is not what you meant here, but less experienced members might get confused.

 

[collinsmark]

Please be careful with wording. When I think of "throttling," I'm immediately thinking of dissipation in a valve or porous plug (along with entropy generation). I'm sure that is not what you meant here, but less experienced members might get confused.

Actually, that is precisely what I meant.*

If instead of asking "what is the maximum theoretical work," the problem statement had asked, "what is the minimum theoretical work," we would model the turbine (not such a great yard-sale find after all here) as a simple valve that does no work at all. The final result in the tanks is that they would contain equal pressure but might contain a temperature difference. Entropy would certainly increase.

The other extreme -- the one we did -- where the work is maximized, leads to zero entropy change. Combine that with adiabatic process and all our answers match.

Real-world turbines will fall somewhere in-between these two extremes.

But the fact that we were finding the maximum work in this ideal turbine (great yard sale find indeed) is my explanation as to why both tanks have equal temperatures in the end, even when analyzed individually.

*(Btw, please forgive my earlier "enthalpy be damned" outburst. Sometimes thermodynamics makes me very excited. ♥)

-------------

[Edit: By the way, I do acknowledge your observation that at any given point in time the temperatures in both tanks might not be equal; they're only equal in the end. I redact my earlier comments about the ideal turbine requiring special elements specifically to keep the temperatures equal. While such mechanisms would not change the final outcome, they are not necessary. The process is reversible so long as the turbine can be treated as quasi-static. (At least that's what I seem to have recently convinced myself of.)]

 

[Chestermiller]

I would like to share the following additional thoughts. Part of my rationale for analyzing this problem the way I did was that, if the temperature of the gas exiting the turbine and entering tank 2 were not exactly equal to the temperature of the gas already contained inside tank 2 (from previous discharge into the tank) at all instants of time during the process, we would be mixing gas at one temperature with gas at a different temperature. This mixing would involve irreversibility and entropy generation, and the process could therefore not be considered isentropic. So I have gone back and checked my equations to compare the gas temperature $T_f(t)$ coming out of the turbine with the temperature $T_2(t)$ of the gas already in the tank at time t. I found that, amazingly, the two temperatures are indeed exactly equal at all times during the process, and given by:
$$T_f(t)=T_2(t)=T_{10}\left(\frac{n_2(t)V_1}{n_{10}V_2}\right)^{(\gamma-1)}$$This confirms the validity of assuming an isentropic process.

 

[Stephen Miller]

Thank you all for the help. The correct answer is in the attached file. That's the teacher's solution, so just plug and chug from there. My answer came out to be $9.4367 𝑘𝐽$

 

[Chestermiller]

With all due respect to your teacher, I stand by the solution I obtained in post #28: 9.278 kJ.

 

[mfig]

Thank you all for the help. The correct answer is in the attached file. That's the teacher's solution, so just plug and chug from there. My answer came out to be $9.4367 𝑘𝐽$

Yes, it seems the teacher made an error in writing:

$\Delta U = m c_p (T_2 - T_1)$

This should be written (for an ideal gas) as:

$\Delta U = m c_v (T_2 - T_1)$

Thus the work should be written, in terms only of the given parameters, as:$W = \frac{P_1 V_1 c_p}{R_{air} k} \left(1-\left(\frac{V_1}{V_1 + V_2}\right) ^{k-1}\right)$