Is a Two-Stage Carnot Engine More Efficient Than a Single-Stage Engine?

Click For Summary

Homework Help Overview

The discussion revolves around the efficiency comparison between a two-stage Carnot engine and a single-stage Carnot engine, focusing on their operational temperatures and the implications of their efficiencies. Participants explore the calculations of efficiency based on temperature variables and question the validity of their setups.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants attempt to derive the efficiencies of both the two-stage and single-stage Carnot engines using temperature relationships. Questions arise regarding the correctness of their calculations and the assumptions made about the temperatures involved.

Discussion Status

Several participants are actively engaging in recalculating efficiencies and clarifying their understanding of the relationships between work and heat in the context of Carnot engines. There is an ongoing exploration of the implications of their findings, with some participants expressing uncertainty about their previous conclusions.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can use or the assumptions they can make about the engines' operations. There is a noted concern about potential violations of thermodynamic laws in their reasoning.

sapiental
Messages
110
Reaction score
0
Consider a Carnot engine operating between temperatures T_h_ and T_c_, where T_c_ is above the ambient temperature T_o_. A second engine operates between the temperatures T_c_ and T_o_.

Calculate (in terms of temperatures) the overall efficiency of the two-stage engine operating this way, and compare with the efficiency of a single Carnot engine operating between the high and low temperatures T_h_ and T_o_.
Which has the higher efficiency, the two-stage or the single engine?

e = 1 - (T_c_/T_h_)

e_2stage_ = 1 - (T_c_/T_h_)(T_o_/T_c_) = 1 - (T_h_-T_o_)/(T_h_)

If I set this up correctly, the 2tage carnot engine will be by far more efficent since it uses the discarded heat to create more work.

I have a funny feeling as if this is violating some law of thermodynamics however..

any advice and feedback is much appreciated.

Thanks.
 
Physics news on Phys.org
Maybe this image helps :)
 

Attachments

  • untitleds.JPG
    untitleds.JPG
    4.2 KB · Views: 754
What are the hot and cold temperatures for the single stage engine?
 
hot temp = T_h_ and cold temp = T_c_ for single stage disregarding the second engine, which is T_o_ for the cold temp and T_c_ (above ambient temp) for the hot temp.
 
sapiental said:
Consider a Carnot engine operating between temperatures T_h_ and T_c_, where T_c_ is above the ambient temperature T_o_. A second engine operates between the temperatures T_c_ and T_o_.

Calculate (in terms of temperatures) the overall efficiency of the two-stage engine operating this way, and compare with the efficiency of a single Carnot engine operating between the high and low temperatures T_h_ and T_o_.
Which has the higher efficiency, the two-stage or the single engine?

e = 1 - (T_c_/T_h_)

e_2stage_ = 1 - (T_c_/T_h_)(T_o_/T_c_) = 1 - (T_h_-T_o_)/(T_h_)

If I set this up correctly, the 2tage carnot engine will be by far more efficent since it uses the discarded heat to create more work.

I have a funny feeling as if this is violating some law of thermodynamics however..

any advice and feedback is much appreciated.

Thanks.
I think that what you are calling e_2stage is in fact the efficiency of a single stage operating between T_h and T_o. I can't see the equality between the two expressions you wrote. How did you come up with e_2stage?

Which is more efficient, an engine operating between T_h and T_c or one operating between T_h and T_o?
 
Last edited:
hmm, I see my mistake, let me try it this waye = W/Q
e2stage = (W1 + W2)/(Q1+Q2)
e2stage = W1/Q1 + W2/Q1 + W1/Q2 + W2/Q2
e2stage = (Q1 - Q2)/Q1 + (Q1 - Q2)/Q2 + (Q2 - Q3)/Q1 + (Q2 - Q3)/Q2QH/QC = TH/TC

e2stage = (T_h_ - T_c_)/T_h_ + (T_h_ - T_c_)/T_c_ + (T_c_ - T_o_)/T_h_ + (T_c_- T_o_)/T_c_

is this it? I can't believe how far off I was earlier. Sorry for confusing you guys as much as I did.
 
Last edited:
sapiental said:
hmm, I see my mistake, let me try it this way


e = W/Q = (W1 + W2)/Q <-- added by OlderDan
e2stage = (W1 + W2)/(Q1+Q2)
e2stage = W1/Q1 + W2/Q1 + W1/Q2 + W2/Q2
e2stage = (Q1 - Q2)/Q1 + (Q1 - Q2)/Q2 + (Q2 - Q3)/Q1 + (Q2 - Q3)/Q2


QH/QC = TH/TC

e2stage = (T_h_ - T_c_)/T_h_ + (T_h_ - T_c_)/T_c_ + (T_c_ - T_o_)/T_h_ + (T_c_- T_o_)/T_c_

is this it? I can't believe how far off I was earlier. Sorry for confusing you guys as much as I did.
The first two lines are good, but the third violates some basic rules of Algebra. The line after that I don't get what you are doing.

The way I interpret your first two lines is this
e is the efficiency of the single stage engine
W is the work done by the single stage engine
Q is the energy input to the single stage engine
W1 and W2 are the work done by the individual stages of the two-stage engine, and Q1 and Q2 are the energy inputs to the two stages.

To compare the efficiencies, e to e2stage, I would set the work outputs equal (as I did by adding the blue terms to yoru equation), and compare the energy input Q to the sum of the stage inputs Q1 + Q2. The known efficiencies for the carnot engine can be incorporated into the second equation to begin the analysis. The relative magnitude of Q and (Q1 + Q2) can be interpreted to tell which efficiency is greater. You need to state your conclusion, not just come up with an equation for e2stage
 

Similar threads

Replies
14
Views
2K
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 25 ·
Replies
25
Views
7K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 8 ·
Replies
8
Views
15K
Replies
1
Views
2K