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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:

enter image description here

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?

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  • $\begingroup$ You are aware that, once you set $Q_H$ to zero, you no longer have an engine operating in a cycle, right? $\endgroup$ – Chet Miller Nov 27 '18 at 2:14
  • $\begingroup$ Yeah that's true, I should have said as $Q_H$ approaches zero. $\endgroup$ – Drew Nov 27 '18 at 2:34
  • $\begingroup$ In order for that to happen, $Q_C$ will have to decrease in proportion. $\endgroup$ – Chet Miller Nov 27 '18 at 2:56
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    $\begingroup$ Oh yeah... Duh this seems simple now. I might delete this question. $\endgroup$ – Drew Nov 27 '18 at 3:14
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I think that this is an interesting set-up. For ideal infinite reservoirs and an ideal engine, at least at steady state, all the entropy generation occurs in the two resistances (although the generated entropy in the resistances is then continually transferred to the working fluid and reservoirs).

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.

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  • $\begingroup$ 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. $\endgroup$ – Drew Nov 27 '18 at 19:17

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