What is wrong with this proof that an irreversible process does not always entail an increase in entropy?

Question

In the paper, Clausius and Darwin Can Both Be Right, the author claims to show an apparent paradox in the status of the heat that is not converted into work in a modified version of the Carnot Cycle. In this modified version, after one reversible cycle (at which point the transfer of heat to the cold bath is reversible since the work done by the cycle is available to reverse the heat flow), the available work is then dissipated into heat that flows into the cold bath, e.g, via driving a paddle in the cold bath causing friction.

The author claims that the entire change in entropy in this overall process is due to the increase in entropy caused by the dissipation of the work into heat, which is irreversible. But if so, this would apparently show that the heat that was not converted to work caused no increase in entropy. But since this latter heat flow can no longer be reversed since the work done has been dissipated, the author claims that this shows an irreversible heat flow with no increase in entropy.

This seems wrong to me, since presumably someone would have shown that there are irreversible processes that do not increase entropy long ago if the reasoning were correct. But I can't quite put my finger on where his argument breaks down in the paper.

Is it simply that the status of the heat that was not converted to work incrementally changes from reversible to irrevesible as the work is dissipated? If so, this feels a little strange, almost like entanglement's spooky action at a distance. The dissipation of the work produced by a local cycle in some remote system (e.g., in the lifting of a weight) can change the status of the heat that flowed into cold reservoir.

My apologies for not including the equations in my question, I'm not proficient in creating such equations.

The relevant discussion takes place in section 3.1. I am not concerned with any other part of the paper or any of the other claims made in other sections, eg claims about evolution vs entropy. Also, note that the flow of the text on p.56 is a little strange. It seems that Fig.2 should have been a page break. The right column above Fig.2 should be read after the left column above Fig.2. Then the right column below the figure should be read after the left.

Answer 1

The author never explains why in the graph on the right $Q_{rev}$ is now $Q_{irr}$. That is not the case in a Carnot engine and it is the origin of his statement that $S_{rev}=S_{irr}$. Just ignore the work path, which is not adding anything to the argument.

An easy way to see this is:

1) If you do not change $Q_{rev}$ into $Q_{irr}$ the conclusion no longer holds.

2) If you make W=0, you still get the same result.

Now let us do the same argument but with a different system. We have two separate containers that do not interact with each other. They are both connected to two heat sources. Let us say that the first subsystem is a Carnot engine, so it absorbs $Q_{hot}$ from the hot reservoir and releases $Q_{cold}$ to the cold reservoir. This entire process being reversible. At the same time, the second subsystem also absorbs heat from the hot reservoir and releases some or all to the cold reservoir in an irreversible way. Now, the system as a whole suffered an irreversible transformation. Would you say that because of this fact, the Carnot cycle had become irreversible? I am sure you will not, and for the same reason you cannot rename $Q_{rev}$ to $Q_{irr}$ in the article's diagram.

Answer 2

I see a different flaw in the conversion that was made. The heat generated is dumped directly and completely into the cold bath. It does not flow from a hotter source temperature. How can such a step be called irreversible? Spontaneous, irreversible heat flow occurs only with a temperature difference, ostensibly across a boundary between a system and a surroundings.

Otherwise, conversion of work goes first to make a change in the internal energy in a system. We never convert work to heat directly. These are flows in/out of system from/to its surroundings. Internal energy has no basis to leave a system as (irreversible) heat flow unless the surroundings to the system are at a colder temperature.

In short, we cannot convert work to irreversible heat flow at a given temperature. We can convert work to internal energy change at a given temperature. Then, we convert internal energy an irreversible heat flow by having it leave the system a hot temperature compared to a colder temperature in the surroundings.

The picture for the proposition must appear as below. The second step through an intermediate temperature $T_i$ is required, and that is a second system in the larger box.

The heat $q_i$ is along an irreversible path. The existence of the flow of heat from $T_i$ as an irreversible step has no impact that can permit an unprincipled change in the designation of the reversibility for the engine to the left of it.

The above is where the analysis fails for me.

In addition, the discussion is propagating a second fatal flaw. A reversible process is one where the system and surroundings are in exact thermal and mechanical equilibrium at all points during the process. A reversible cycle is one that can be reversed through the entire cycle and experience no change (either direction has a net zero internal energy, entropy, enthalpy ... change). A reversible cycle is by definition a combination of reversible paths (processes). The minute that a forward cycle is complete, you cannot remove part of the cycle, claim the cycle fails because you have removed part of it, and thereby state that it is now an irreversible cycle. In fact, it is NOT EVEN A CYCLE anymore. You also cannot claim that a reversible cycle becomes irreversible because the heat or work that enters/leaves the cycle goes somewhere/comes from somewhere special.

So, we cannot say that because the work is dissipated, that work is no longer able to cause the cycle to make heat flow back to the hot reservoir, and therefore the cycle that was reversible going clockwise is now suddenly irreversible in all cases even going forward. The minute you take away the work in the first system, you have no cycle that can allow you to make or change labels!

Answer 3

Heat and work are path dependant quantities. Saying "an amount of heat $Q$ has been transfered from the hot to the cold resovoir" does not define the final state of the system, and so I can find many values for its final entropy. So you can follow the path laid out in the paper and generate an entropy of $\frac{Q}{T_c}$ or I could find something monstrously inefficent and generate $\frac{10,000 Q}{T_c}$ or I could set up a Carnot engine and run it until an amount of heat $Q$ has been deposited in the cold resovoir while depositing the produced work back in the hot resovoir (as reversible work) and generate no entropy at all.

Really the paper should be talking about the amount of energy transfered, since that is the relavent state function, and the lower limit on entropy generatin to transfer the energy is 0.

The second law is an inquallity; you can always find a way to generate more entropy than nessarsery (and this paper has found one). You cannot generate less.

Answer 4

I think I figured out an answer that seems intuitive to me. It is related to @BobD's observation that the apparent flow of heat (Qc) directly from the hot bath to the cold bath via Path A in Fig.1 seems fishy. I think the mistake in logic begins with equation:

dSirr,engine = dSA,irr + dSD, irr = dSA,rev + dSB,rev + dSC, irr = 0 + 0 + W/Tc =Q/Tc (7)

dSA,rev and dSB,rev do not equal zero; they equal Qc/Tc and -Q/Th, respectively. The author's justification for asserting they are both zero is equation (4):

dSA,rev + dSB,rev = 0 (4)

While this is true, it does not imply that dSA,rev and dSB,rev are individually zero. At the end of one reversible cycle of a Carnot engine, the change in entropy of the working medium is 0 (the right side of equation (4)). The decrease in entropy in the system that work is done on is -Q/Th and the increase in entropy of the cold bath is Qc/Tc, such that they balance one another. A more accurate rendition of (4) would be something like this:

Qc/Tc + Q/Th = 0 (4')

So equation (7) should be written as:

dSirr,engine = dSA,irr + dSD, irr = dSA,rev + dSB,rev + dSC, irr = Qc/Tc - Q/Th + W/Tc = Q/Tc (7')

This means that equations (8)-(11) are also all incorrect because they rely on eliminating dSA,rev (or equivalently, dSA,irr) because it is supposedly equal to 0. But it is not; dSA,rev is equal to Qc/Tc.

To summarize: the increase of entropy of the cold bath, Qc/Tc (via Path A) is an increase of entropy whether or not such increase is reversible. If the work produced by the cycle is still available, the entropy is reversible; if the work has been dissipated, it is irreversible.