# Entropy generation in an endoreversible engine

An endoreversible engine is one that is internally reversible (i.e., a perfect Carnot engine) but the heat transfers occur across imperfect conductors (resistances), thus rendering the engine externally irreversible. It's a nice step forward for more realistic modelling of power plants.

Here's a diagram with a control volume that I'll use:

I'm trying to distinguish between maximum and minimum entropy generation by this system, and ran into a conundrum.

The total entropy generation is found by applying an entropy balance to the red control volume in the figure:

$$S_{gen} = \frac{Q_L}{T_L} - \frac{Q_H}{T_H}$$

Here, it looks like $$S_{gen}$$ is maximum when $$Q_H = 0$$.

Now consider another entropy balance on a control volume around the top "hot" resistance:

$$(S_{gen})_H = \frac{Q_H}{T_{H'}} - \frac{Q_H}{T_H}$$

where $$T_{H'}$$ is the temperature at the end of the resistance.

This implies that $$(S_{gen})_H = 0$$ if $$Q_H = 0$$.

This is a major conundrum of our previous result with the larger control volume, that says $$(S_{gen})_H$$ is maximum when $$Q_H = 0$$.

What is the source of my confusion?

• You are aware that, once you set $Q_H$ to zero, you no longer have an engine operating in a cycle, right? – Chet Miller Nov 27 '18 at 2:14
• Yeah that's true, I should have said as $Q_H$ approaches zero. – Drew Nov 27 '18 at 2:34
• In order for that to happen, $Q_C$ will have to decrease in proportion. – Chet Miller Nov 27 '18 at 2:56
• Oh yeah... Duh this seems simple now. I might delete this question. – Drew Nov 27 '18 at 3:14

You correctly determined that the rate of entropy generated in the hot resistance is given by $$(\dot{S}_{gen})_H=\left(\frac{1}{T_{H'}}-\frac{1}{T_H}\right)\dot{Q}_H=\frac{T_H-T_{H'}}{T_HT_{H'}}\dot{Q}_H\tag{1}$$And in the cold resistance, it is given by$$(\dot{S}_{gen})_L=\left(\frac{1}{T_{L}}-\frac{1}{T_{L'}}\right)\dot{Q}_L=\frac{T_{L'}-T_{L}}{T_LT_{L'}}\dot{Q}_L\tag{2}$$The total rate of entropy generation in the universe is obtained by summing these equations to yield:$$\dot{S}_{gen}=(\dot{S}_{gen})_H+(\dot{S}_{gen})_L=\frac{\dot{Q}_L}{T_L}-\frac{\dot{Q}_H}{T_H}\tag{3}$$
From the heat conduction equation, we also know that $$\dot{Q}_H=kA\frac{T_H-T_{H'}}{\Delta L}\tag{4}$$where $$\Delta L$$ is the thickness of the resistance, A is the area for heat flow, and k is the thermal conductivity. If we substitute this into Eqn. 1, we obtain:$$(\dot{S}_{gen})_H=kA\Delta L\frac{[(T_H-T_{H'})/\Delta L]^2}{T_HT_{H'}}\tag{5}$$This shows that the rate of entropy generation is roughly proportional to the square of the temperature gradient within the resistance. A similar relation can be written for the cold resistance.
• That is interesting insight. It clearly answers my question though; if $\dot{Q}_H$ is zero, then $(\dot{S}_{gen})_H = 0$ due to entropy balance on hot resistance, and $\dot{Q}_L=0$ due to the Clausius condition on the perfect engine. – Drew Nov 27 '18 at 19:17