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I am reading through the text "Advanced Engineering Thermodynamics" by Bejan that I found hereand have encountered a conceptual issue that I'm having trouble resolving on my own.

On page 64, Bejan asks me to consider a "batch of fluid that expands adiabatically in a piston and cylinder apparatus." The piston rod is connected to some external device that can either create or consume work. He then presents a phase-space plot of the fluid volume vs. internal energy included in this post and then makes the claim that any point at $(V_2, U_2)$ which is adiabatically accessible from the state $(V_1,U_1)$ must lie above some coordinate point $(V_2, U_{2,\text{rev}})$ defined by the special property that the state may be reached by and return to $(V_1, U_1)$ adiabatically.

To see this, he says we can use a proof by contradiction as follows. Assume the point $(V_2, U^{'}_2)$ in which $U^{'}_2 < U_{2,\text{rev}}$ is adiabatically accessible from $(V_1, U_1)$. Then consider the counterclockwise path shown in the rightmost graph. Bejan states that the path from $(V_1, U_1$ to $(V_2, U^{'}_2)$ cannot be adiabatic because the only non-adiabatic part of the cycle, from $(V_2, U^{'}_2)$ to $(V_2, U_{2,\text{rev}})$, leads to an overall flow of heat into the fluid system which would violate the single thermal reservoir Kelvin-Planck statement. Further, Bejan indicates that such a positive heat flow is necessary by referring to the upward path from $(V_2, U^{'}_2)$ to $(V_2, U_{2,\text{rev}})$ as a "zero-work process."

However, couldn't we just let the "work device" that the piston rod is connected to rotate the piston rod to do rotational work on the fluid in order to bring the fluid from $(V_2, U^{'}_2)$ to $(V_2, U_{2,\text{rev}})$? Then, by making the path a "nonzero-work process", $(V_2, U^{'}_2)$ may still be reached adiabatically from $(V_1, U_1)$ and we would not be violating Kelvin-Planck's statement.

It seems that Bejan even suggests this possibility as, referring to the adiabatic accessibility of point $2^{''}$ from point 1 in the leftmost graph, he states:

the end-state energy $U_2$′′ can rise above $U_1$ if, following free expansion [from $U_1 \text{ to } U_2$], the external apparatus does work on the fluid. One way of achieving this effect is by rotating the piston rod around its axis while the fluid shear integrated over the piston face opposes the rotation. In this manner, the external apparatus does work on the system, from state 2' to $2^{''}$.

In summary, what forbids the same thing from happening in the rightmost diagram from point 2 to point $2_{\text{rev}}$?

enter image description here

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  • $\begingroup$ Until you made the edit, I had no way of understanding how the internal energy could possible increase in an adiabatic expansion involving only boundary work, or a free expansion involving no work. Now it appears the rotating piston in effect does stirrer work on the fluid. Are there any other "details" missing explaining the third and fourth graphs? $\endgroup$ – Bob D Sep 14 '20 at 17:00
  • $\begingroup$ Jacob, I have provided a new answer. Hope it finally helps. $\endgroup$ – Bob D Sep 15 '20 at 14:36
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I am providing a new answer in view of the additional details you have provided and our related discussions. I have also retracted my vote to close the post.

First of all, I believe we both agree that process $2$ to $2_{rev}$ in the third diagram and process $1$ to $2$ in the fourth diagram are impossible if adiabatic, based on the first law. The issue has to do the author's claim that process $1$ to $2$ in the fourth diagram would necessarily violate the Kelvin-Planck statement of the second law and why that is necessarily so, given the possibility of adiabatic stirrer work on the fluid going from state $2$ to $2_{rev}$ as you have suggested.

First, the author's argument-

To recall, the Kelvin-Planclk statement (from Wikipedia) is:

“It is impossible to devise a cyclically operating heat engine, the effect of which is to absorb energy in the form of heat from a single thermal reservoir and to deliver an equivalent amount of work”

If one assumes that process $1$ to $2$ in the fourth diagram were in fact possible, then one can device a heat engine cycle in which energy is absorbed from a single thermal reservoir and the cycle delivers an equivalent amount of net work, as follows:

After the process $1$ to $2$ expansion is completed the adiabatic boundary is removed. Then heat is transferred to (absorbed by) the fluid reversibly at constant volume from state $2$ to state $2_{rev}$. The adiabatic boundary is then replaced and the fluid is then compressed adiabatically and reversibly back to state $1$. We now have a heat engine cycle that produces net work by absorbing heat from a single reservoir during process $2$ to $2_{rev}$ in violation of the Kelvin-Planck statement.

Now, your argument-

To the above, you have raised the following question in regard to the fourth diagram:

However, couldn't we just let the "work device" that the piston rod is connected to rotate the piston rod to do rotational work on the fluid in order to bring the fluid from $(V_2, U^{'}_2)$ to $(V_2, > U_{2,\text{rev}})$? Then, by making the path a "nonzero-work process", $(V_2, U^{'}_2)$ may still be reached adiabatically from $(V_1, U_1)$ and we would not be violating Kelvin-Planck's statement. In summary, what forbids the same thing from happening in the rightmost diagram from point 2 to point $2_{rev}$?

You are correct this would not violate the Kelvin-Planck statement since no heat transfer occurs. However, the difference between this and the author's argument is that there is also no net work done in the cycle you propose. If no net work is done, the Kelvin-Planck statement does not apply.

Applying the first law to the complete cycle, we have

$$\Delta U_{cycle}=\sum Q_{i}-\sum W_{i}=0$$

Since the cycle is adiabatic, $\sum Q_{i}=0$ and

$$\sum W_{i}=0$$

Or

$$W_{Net}=0$$

$$W_{Net}=W_{1-2}+W_{2-2r}+W_{2r-1}=0$$

The first term is positive since the fluid does work in expanding against resistance of the apparatus. The second term is negative since the stirrer work is done on the fluid. The third term is also negative since the reversible adiabatic compression is work done on the fluid. Since the overall net work has to be zero, the work done by the fluid has to equal the work done on the fluid for a net work of zero.

In spite of no violation of the second law, process $1$ to $2$ in the last diagram still violates the first law.

Hope this helps.

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  • $\begingroup$ The update you provided is certainly helpful. I had not thought about applying the first law to the problem. The total work must be zero since the cycle is adiabatic as you point out, though I don't understand why the sum of $W_{1-2}$, $W_{2-2r}$, and $W_{2r-1}$ can't be zero. As shown in the rightmost diagram, if stirring work is done on the fluid, it seems that the change in internal energy is zero as required. $\endgroup$ – Jacob Wilson Sep 15 '20 at 23:34
  • $\begingroup$ I’m not following you. The sum of the works is zero as I said. I’m sorry but I’ve already spent a lot of time on this have no more time. Either you can accept my answer or I will move on and wish you luck! $\endgroup$ – Bob D Sep 15 '20 at 23:40
  • $\begingroup$ I understand that the Kelvin-Planck statement does not apply if no net work is done (as in the case of my replacement of heat transfer with stirring work). So something other than the Kelvin-Planck statement must be used to explain why the stirring work cannot complete the cycle, right? $\endgroup$ – Jacob Wilson Sep 15 '20 at 23:56
  • $\begingroup$ Thank you so much for spending so much time talking with me about this. If you have to move on, I definitely understand :) $\endgroup$ – Jacob Wilson Sep 15 '20 at 23:58
  • $\begingroup$ Just to be clear, I never said the Kelvin-Planck statement says the stirrer work cannot complete the cycle. It can. But even given that there is no net work done exchanging heat with single reservoir, which is what the Kelvin-Planck statement says cannot occur. $\endgroup$ – Bob D Sep 16 '20 at 0:51
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Bob D pointed out that stirring work is irreversible. With that insight, perhaps the following point is why my claim in the question is false:

Referring to the rightmost graph of the figure, suppose point $2$ may be reached adiabatically from point $1$ and suppose point $2_{\text{rev}}$ can reach point $1$ adiabatically. Consider the cycle $1\rightarrow 2 \rightarrow 2_{\text{rev}}\rightarrow 1$. If stirring work is used to move from point $2$ to $2_{\text{rev}}$ in the rightmost graph, there is no entropy loss in the cycle. Thus, the counterclockwise cycle is reversible.

However, because stirring work is irreversible, the counterclockwise cycle is not reversible. So we have a contradiction.

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