Question about Reversible Engines and Carnot Efficiency

[Sum Guy]

Homework Statement

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.

 

[SteamKing]

Homework Statement

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.

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.

 

[Sum Guy]

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.

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.

"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

 

[SteamKing]

"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

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

 

[Sum Guy]

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

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

 

[SteamKing]

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

No. In order for a process to be reversible, certain conditions must be met. The following article gives examples of certain things to avoid is one wishes to have a theoretically reversible process:

http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node34.html

 

[Sum Guy]

No. In order for a process to be reversible, certain conditions must be met. The following article gives examples of certain things to avoid is one wishes to have a theoretically reversible process:

http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node34.html

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