I can think of two interpretations of this question. Did you give the exact problem statement?
One interpretation would be that, irrespective of the nature of the cyclic process, the efficiency is 50% over then entire path. So, $$Q_H=m_HC_H(T_{H0}-T_F)$$
$$Q_C=m_CC_C(T_F-T_{C0})$$
and $$\frac{Q_H-Q_C}{Q_H}=0.5$$
The other interpretation would be that $\Delta S$ is equal to zero up to the point that the temperatures are such that the efficiency is 50%, after which the efficiency is held constant at 50% (as in the first interpretation).
EDIT:
Whoops. I think I got the 2nd interpretation backwards. Start with the efficiency of 50%, and then switch to the $Delta S=0$ path after the absolute temperature of the cold reservoir reaches 50% of that of the hot reservoir.
EDIT 2
Try this and see if it works. Solve for intermediate temperatures $T_{H}$ and $T_C$ at which $T_C=T_H/2$ with 50% efficiency and with the instantaneous Carnot efficiency is also 50%: $$\frac{m_CC_C(0.5T_H-T_{C0})}{m_HC_H(1000-T_H)}=0.5$$
What values do you get for $T_{H}$ and $T_C$? Then, use these temperatures as the starting point for a second change in which you use your $\Delta S=0$ equation, with these values as the starting temperatures. What do you get for the final temperature?
I feel in the proof I have linked, there is no result used which is
only true for reversible engines, such as the change in entropy of the
system being 0.
If this is the basis of your concern for the proof, then you should know that the change in entropy of the system operating in a cycle is always zero, regardless of whether the cycle is reversible or irreversible.
When completing a cycle the system is returned to its initial equilibrium state. That means that all system properties, including entropy, are returned to their initial values. It is the change in entropy of the surroundings that differs. That change is zero for a reversible cycle and greater than zero for an irreversible cycle.
When a system cycle is irreversible it means that entropy is generated in the system. In order to return the system to its original entropy, the additional entropy generated due to irreversibility has to be transferred to the surroundings in the form of heat.
But in the actual proof, where did they use any result that is not
true for all engines in general? My concern was wrong I agree with
that, but my question still stands.
The key point in the proof is the following bullet:
The net result is that heat WA–WB is taken from sink and equal amount of work is produce. This violates second law of thermodynamics.
This means it violates the Kelvin-Plank statement of the second law which is
No heat engine can operate in a cycle while transferring heat with a single reservoir
Taking heat from a single reservoir (in this case the sink) and producing an equal amount of work violates the Kelvin-Plank statement.
The Corollary to the Kelvin Plank statement is
No heat engine can have a higher efficiency than a Carnot Cycle operating between the same reservoirs.
From the diagrams, it does not seem obvious to me that the work should
be any different for any other engine. I understand once you have
arrived to the final statement that the net work done = the heat taken
from the reservoir, there is a violation of the second law.
The cause of the confusion is the second diagram is wrong. For one thing, the input to A does not equal the sum of its outputs. For another, the output required by A to operated B is incorrect. The diagram should be as I show it below.
In order to run B in reverse A only has to supply work equal to $W_B$. Since for the super efficient A we have $W_{A}>W_{B}$ A now has a net work output of $W_{A}-W_{B}$ shown to the left that it can use for other purposes. Since the heat removed from the high temperature reservoir by the super efficient heat engine equals the heat put back by the reversible refrigerator, there is no net heat transfer to the high temperature reservoir, $T_1$. Therefore the overall result is A has taken heat from a single reservoir $T_2$ and produced a net work output of $W_{A}-W_{B}$, in violation of the Kelvin Plank statement of the second law.
If I am not mistaken, based on your answer if the refrigerator was
irreversible, some heat would be lost to the surroundings. Is this
correct?
Yes it is correct, but I would not necessarily use the term "lost". Better to say more heat must be transferred to the high temperature surroundings for the same desired heat extracted from the low temperature surroundings meaning more work input is needed than for the reversible refrigerator. There are various possible reasons for the irreversibility. One is mechanical friction, in which case the term heat "lost" comes to mind. But there are irreversible processes that do not involve mechanical friction as well.
An irreversible refrigerator requires more work input to extract the same heat from the low temperature reservoir resulting in lower efficiency (meaning, for a refrigerator, a lower coefficient of performance). That means more heat has to be rejected to the high temperature environment for the same amount of heat taken from the low temperature reservoir.
An irreversible heat engine has a lower work output for the same heat input for a lower efficiency. That means the irreversible heat engine transfers more heat to the low temperature reservoir than the reversible engine, leaving less heat to perform work than the reversible heat engine.
In both cases the reason for the lower efficiency is that entropy is generated in an irreversible cycle. In order to complete the cycle and return the system to its original entropy the generated entropy has to be transferred to the surroundings in the form of heat. @Chet Miller in his answer shows mathematically how that lowers efficiency.
The proof that is the subject of your post assumes the two devices are reversible, that is, they have the maximum possible efficiency for the cycle, regardless of the details of the cycle. It only seeks to prove that all reversible engines operating between two fixed temperatures have the same efficiency.
All of this relates to Carnot's theorem, which can be stated as follows:
"The efficiency of all reversible engines operating between the same two temperatures is the same, and no irreversible engine operating between these temperatures can have a greater efficiency than this"
The proof under discussion only relates to the first part of the theorem (which I show in italics). @Chet Miller addresses the second part of the theorem.
Hope this helps.
Best Answer
"I searched pretty extensively (using both Google and Google Scholar) and found no discussion of such a system." I expect you know that the set-up in your diagram is the one used in the traditional development of thermodynamics to show that the Second Law (Clausius version) implies that all reversible engines operating between the same two temperatures have the same efficiency, and that it is greater than for an irreversible engine. See Zemansky's Heat and Thermodynamics, Pippard's Classical Thermodynamics, Fermi's Thermodynamics...
I think that you would have a perpetual flow of energy in such a theoretical situation, but that it is theoretical in just as damning a way as the perpetual motion of a simple pendulum could be called 'theoretical'. It won't happen! This is for a number of reasons (which, arguably, have as much right to be called 'theoretical' in that we have a pretty good understanding of them!)