Question about Reversible Engines and Carnot Efficiency

  • #1
Sum Guy
21
1

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?
dnhit4.png


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.
 

Answers and Replies

  • #2
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,809
1,670

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?
dnhit4.png


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.
 
  • #3
Sum Guy
21
1
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
 
  • #5
Sum Guy
21
1
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?
 
  • #7
Sum Guy
21
1

Suggested for: Question about Reversible Engines and Carnot Efficiency

Replies
1
Views
431
Replies
2
Views
654
Replies
2
Views
1K
Replies
8
Views
3K
Replies
16
Views
2K
Replies
1
Views
966
  • Last Post
Replies
8
Views
1K
Replies
2
Views
449
Replies
5
Views
815
Top