# Question about Reversible Engines and Carnot Efficiency

• Sum Guy
In summary, the conversation discussed a question regarding heat engines and the confusion of the individual regarding the efficiency and entropy of their calculation. The summary also includes a link to review the Carnot cycle and its efficiency, as well as a clarification that not all processes are reversible in theory. The individual is then asked to explain which processes in their calculation are irreversible and why.
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.

Sum Guy said:

## 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.

SteamKing said:
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.
SteamKing said:
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

Sum Guy said:
"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.

SteamKing said:
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 said:
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?

## 1. What is a reversible engine?

A reversible engine is a theoretical engine that operates in a way that all of its processes can be reversed without any loss of energy. This means that the engine can convert all of the work done into an equal amount of useful work without any waste.

## 2. How does a reversible engine work?

A reversible engine works by using a series of reversible processes, such as isothermal and adiabatic processes, to convert heat energy into useful work. These processes are carefully controlled to minimize any energy losses.

## 3. What is Carnot efficiency?

Carnot efficiency is the maximum efficiency that a reversible engine can achieve. It is calculated by dividing the temperature difference between the hot and cold reservoirs by the temperature of the hot reservoir.

## 4. How is Carnot efficiency related to the second law of thermodynamics?

Carnot efficiency is related to the second law of thermodynamics because it is the maximum efficiency that can be achieved by any heat engine. This law states that it is impossible to convert all of the heat energy into work in a cyclic process without any energy loss.

## 5. Can any real engine achieve Carnot efficiency?

No, it is impossible for any real engine to achieve Carnot efficiency. This is because real engines experience energy losses due to friction, heat transfer, and other factors. However, the closer an engine's efficiency is to Carnot efficiency, the more efficient it is considered to be.

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