Irreversibility of thermodynamic cycles

Question

For any heat engine operating between two sources of different temperatures, would not the process it goes through be irreversible? Since the heat is transferred to the system from a source with a finite temperature difference from the system, the system loses equilibrium and thus loses reversibility. What is wrong with this thinking?

Answer 1

The system is usually taken to be the working substance. When at a temperature, $T_{hi}$, this receives heat from a source at a higher temperature. Having been cooled by doing adiabatic work, the working substance 'excretes' heat at a lower temperature, $T_{lo}$, to a sink at a lower temperature. The working substance is then taken through the rest of the reversible 'Carnot' cycle.

I assume that you are concerned with the heat flow from the source to the working substance at temperature $T_{hi}$, and from the working substance at temperature $T_{lo}$, to the sink. Surely these flows won't take place unless there are temperature differences in each case? Yes, but, in this thought-experiment (which is really what a Carnot cycle is) we can make these temperature differences as small as we like (and operate the Carnot cycle very slowly)!

Answer 2

Since the heat is transferred to the system from a source with a finite temperature difference from the system, the system loses equilibrium and thus loses reversibility. What is wrong with this thinking?

You are absolutely correct, if you think that's how the engine works. Only that it doesn't.

In between the processes of heat exchange with the sources, there are processes which reversibly bring the temperature of the system down/up to the temperature of the source with which it is going to interact. As you might have guessed, these intermediate processes occur without exchange of heat (as we have the heat sources for that purpose) - they are adiabatic processes.

Now that these intermediate processes are reversible and have made the heat exchange processes reversible too, there is no process which remains irreversible.

Answer 3

Nothing. In real life, that's what basically happens - no heat engine is truly reversible. The Carnot engine is just an idealisation; it tells us that in theory, if you operated an engine like that extremely slowly and carefully, with no losses to piston friction, vortices in the fluid, etc. you could get a reversible engine. "Reversible" meaning it can be operated in reverse: of course, since the engine draws heat $Q_1$ from a source and dumps $Q_2$ into a sink, producing work $W$, we have that

$$ Q_1 = Q_2 + W $$

so if you truly wanted to reverse the engine, you would have to compress the fluid (acting with a work $W$) and that would allow you to drain $Q_2$ from a colder source and move it to a hotter one, now with the addition of $W$ on top. This is nothing extraordinary: moving heat from cold to hot by paying a price in energy is what a fridge does. The 2nd principle of thermodynamics only states that $\Delta S \geq 0$, and since for reversed operation

$$ \Delta S = - \frac{Q_2}{T_2} + \frac{Q_1}{T_1} $$

it's fine as long as $Q_1$ is big enough to compensate for the fact that $T_1 > T_2$. It being just the right amount is the condition for a reversible Carnot cycle.

Answer 4

That's right. A heat engine operating between just two reservoirs at a finite different temperature undergoes an irreversible process. Thermodynamic cycles are idealized processes in which the heat engine is at thermodynamic equilibrium with the reservoirs at all stages. This can only occur if heat is transferred between two reservoirs with infinitesimally different temperatures. In practice, when a thermodynamic cycle is depicted putting a heat engine between two reservoirs at different temperatures we effectively mean that the engine exchanges heat with infinitely many reservoirs at infinitesimally decreasing temperatures from the hot to the cold one. In this case, since the system has been at thermodynamic equilibrium at all stages, the process is reversible as desired.