Heat Engine Efficiency and Entropy

[UMath1]

In deriving the Carnot Efficiency, the assumption is made that theoretically most efficient engine will generate no net entropy, meaning that the entropy that enters the system during heat absorption must equal the entropy that leaves the engine during heat rejection. Why is the case? Why would the engine be less efficient if it gained more entropy than it lost, or vice versa?

 

[Chestermiller]

If entropy is generated within the working fluid, then $$\Delta S=\frac{Q_{in}}{T_{hot}}-\frac{Q_{out}}{T_{cold}}+\sigma$$where $\sigma$ is the entropy generated per cycle. But, if the engine is operating in a cycle, then $$\Delta S=0$$. Therefore, $$\frac{Q_{in}}{T_{hot}}-\frac{Q_{out}}{T_{cold}}+\sigma=0$$. The efficiency of the engine is $$\eta=\frac{Q_{in}-Q_{out}}{Q_{in}}$$where the numerator represents the amount of work done. Eliminating $Q_{out}$ between these two equations, we obtain: $$\eta=\left(1-\frac{T_{cold}}{T_{hot}}\right)-\frac{\sigma T_{cold}}{Q_{in}}$$The first term in parenthesis is the Carnot efficiency. Since the entropy generation $\sigma$ must always be positive, the efficiency is less than the Carnot efficiency.

 

[UMath1]

That makes sense. I still don't understand why ΔS must equal 0. What if the system continues to maintain the generated entropy? What would happen then?

 

[Chestermiller]

That makes sense. I still don't understand why ΔS must equal 0. What if the system continues to maintain the generated entropy? What would happen then?

By definition, in a cycle, the working fluid is returned to its initial state after the cycle is completed (so the change in its entropy is zero). Therefore, all the entropy generated in an engine cycle is transferred to the reservoirs. If that were not the case, the temperature of the working fluid would be changing from cycle to cycle.