# Homework Help: Question about Reversible Engines and Carnot Efficiency

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1. Apr 11, 2016

### Sum Guy

1. The problem statement, all variables and given/known data
I have a question regarding heat engines that cropped up whilst I was doing a practice question. I will summarise the results I obtained for the previous parts of the question so as to save your time. The highlighted parts of the image are where I am having some issues.

I confused because:
-I thought all reversible heat engines operate exactly at the carnot efficiency $\frac{T_H - T_C}{T_H}$
-I thought that *in theory*, the engine cycle below is reversible and so should operate at this efficiency
-From my workings, I found this not to be the case and I also found that there is an increase in entropy in the universe after each engine cycle (a reversible carnot engine would have no net change in entropy)

So my question is - are my workings wrong or can my results be explained as I misunderstand something?

The attempt at a solution (Important Results)

$$\gamma = \frac{5}{3}$$ $$T_2 = 2T_1$$ $$T_3 = \frac{p_3}{p_1}2T_1$$ $$T_3 = \left(\frac{1}{2}\right)^{2/3}T_1 = 0.63T_1$$ $$W_{31} = p_{1}V_{1}\left( \frac{2^{1-\gamma} - 1}{1-\gamma}\right) = 0.56p_{1}V_{1}$$ $$Q_{23} = \frac{3}{2}\left( (1/2)^{2/3} - 2\right)p_{1}V_{1} = -2.06p_{1}V_{1}$$
$$Efficiency = \frac{W_{net}}{Q_{Hot}} = \frac{W_{net}}{Q_{Cold} + W_{net}}$$
$W_{net} =$area enclosed by loop $= p_{1}V_{1}\left(1 + \frac{1-2^{1-\gamma}}{1-\gamma}\right)$ $$Q_{Cold} = Q_{23}$$ $$Efficiency = 0.18$$ $$\Delta S_{HotRes} = \frac{Q_{in}}{T_H} = \frac{-p_{1}V_{1}}{T_{1}}$$ $$\Delta S_{universe} = \Delta S_{HotRes} + \Delta S_{ColdRes}$$
$$\Delta S_{ColdRes} = -\frac{Q_{23}}{0.5T_1} = 3(2-(1/2)^{2/3})\frac{p_{1}V_{1}}{T_{1}}$$
$$\Delta S_{universe} = 3.11\frac{p_{1}V_{1}}{T_{1}}$$
$$Carnot \quad efficiency = \frac{T_{H} - T_{C}}{T_{H}} = 0.8$$

So in short - why does the efficiency I've calculated differ from the Carnot efficiency and why do I find there to be a net increase in entropy.

2. Apr 11, 2016

### SteamKing

Staff Emeritus
You probably should review what the Carnot cycle consists of:

https://en.wikipedia.org/wiki/Carnot_cycle

There are P-V and T-S diagrams of said cycle included in the article.

If your cycle doesn't match exactly the Carnot cycle, it won't be capable of working at a Carnot efficiency.

3. Apr 11, 2016

### Sum Guy

"All Reversible Heat Engines have same efficiency when operating between the same two temperature reservoirs."

See: http://aether.lbl.gov/www/classes/p10/heat-engine.html

4. Apr 11, 2016

### SteamKing

Staff Emeritus
You are assuming that the cycle in the OP is reversible. That claim is not made in the problem description.

5. Apr 12, 2016

### Sum Guy

All of the processes are reversible in theory though, no?

6. Apr 12, 2016

### SteamKing

Staff Emeritus
7. Apr 12, 2016

### Sum Guy

Please could you explain which of the processes above are irreversible and why?