# Engineering Initially evacuated chamber and maximum theoretical work

#### Chestermiller

Mentor
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

Mentor
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

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#### Chestermiller

Mentor
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.

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#### Chestermiller

Mentor
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

Mentor
Ah, that explains the pressure discrepancy. The second problem remains, however.
Hmmm. I'll have to recheck my algebra.

#### Chestermiller

Mentor
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!

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• Chestermiller

#### Chestermiller

Mentor
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?

• collinsmark

#### 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

#### collinsmark

Homework Helper
Gold Member
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. Last edited:
• mfig

#### collinsmark

Homework Helper
Gold Member
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

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Gold Member

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)

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#### Chestermiller

Mentor
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.

• collinsmark

#### Chestermiller

Mentor
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

Homework Helper
Gold Member
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. )

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[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.)]

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• Chestermiller

#### Chestermiller

Mentor
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.

• collinsmark