Carnot Engines Question -- Composite Configuration Efficiency

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Homework Help Overview

The discussion revolves around the efficiency of composite Carnot engines operating between three temperature reservoirs: Th, Tm, and Tc. Participants are exploring how adding an extra reservoir affects overall efficiency and the calculations involved in determining work done and heat exchanged in each engine.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to understand the implications of adding an extra reservoir on efficiency, with some suggesting it may not increase efficiency. Others are calculating individual efficiencies and questioning the rationale behind using the product of efficiencies for multiple processes.

Discussion Status

There is an ongoing exploration of calculations related to work done and heat exchanged for each engine. Some participants have provided guidance on calculating efficiencies and work, while others are questioning assumptions and seeking clarification on formulas. Multiple interpretations of the problem are being discussed without explicit consensus.

Contextual Notes

Participants are working within the constraints of the Carnot engine model and are discussing the relationships between heat input, heat output, and work done. There is a focus on ensuring calculations align with the principles of thermodynamics as they relate to Carnot engines.

lc99
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Homework Statement



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Homework Equations

The Attempt at a Solution


Trying to figure this out.

So, I'm thinking that adding a extra resevoir for another engine will not add efficiency. It will should split up the W done by the heat reservoir. It just seems to make sense that way. So would the answer be e?

I'm not sure how else to explain the answer.
 

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Calculate the efficiency for each engine, The overall efficiency for multiple processes is the product of the individual efficiencies. Compare to the original efficiency.
 
lc99 said:
So, I'm thinking that adding a extra resevoir for another engine will not add efficiency. It will should split up the W done by the heat reservoir. It just seems to make sense that way.
There is no need to guess or "feel" for what makes sense.
In detail, calculate the work done and heat exchanged in each part of the composite engine. You have all the formulas in front of you.
 
Dr Dr news said:
Calculate the efficiency for each engine, The overall efficiency for multiple processes is the product of the individual efficiencies. Compare to the original efficiency.
Why is it the product?
 
That can also be checked by completing the calculation as suggested by Dr Dr news
 
Chandra Prayaga said:
That can also be checked by completing the calculation as suggested by Dr Dr news
So far i have Engine A (between TH and TM) a e = 1-TM/TH.
For engine B (between TM and TC) , e =1-TC/TM
 
Start the way the original question posed it. Three temperatures Th, Tm and Tc. One carnot engine operating between Th and Tm, another between Tm and Tc. Now find heat input and output for each engine, work done in each case, as you would for a single engine. Then use the fact that the heat exhaust from the first is the heat input to the second
 
Just take it one engine at a time. Suppose you have an engine driving a propeller, The engine drives a gear box which in turn drives the propeller. If the engine outputs 250 horsepower and the gear box is 95% efficient then the input to the propeller is 0.95 x 250 HP and if the propeller is 80% efficient then the useful power driving the airplane is 0.95 x 250 HP x 0.80.
 
Chandra Prayaga said:
Start the way the original question posed it. Three temperatures Th, Tm and Tc. One carnot engine operating between Th and Tm, another between Tm and Tc. Now find heat input and output for each engine, work done in each case, as you would for a single engine. Then use the fact that the heat exhaust from the first is the heat input to the second
When i calculate the work for one engine, is it just the difference of Qh-Qc?
 
  • #10
Yes.
 
  • #11
Dr Dr news said:
Yes.
I think
Dr Dr news said:
Yes.
since here we are dealing with carnot, can i say W = Th-Tc?
 
  • #12
lc99 said:
I think

since here we are dealing with carnot, can i say W = Th-Tc?
Are there any other formulas i need for this. So far i am using W =Th-Tc and e = W/Qh

After writing out the individual engines and calculating the work based on the previous i have:

(Th-W1-Tc) /Th
 
  • #13
The work is not equal to the temperature difference. For a Carnot engine, Wk(out) = Q(absorbed) - Q(rejected) and the Carnot efficiency is η = Wk(out) / Q(absorbed) = [Q(absorbed) - Q(rejected)] / Q(absorbed) = (Qa - Qr) / Qa. Further, since this engine operates isentropically, we can relate Qa = Ta ΔS and Qr = Tr ΔS, and finally, η = (Ta ΔS - Tr ΔS) / Ta ΔS = (Ta - Tr) / Ta = 1 - (Tr / Ta). for each engine. For multiple engines η(overall) = η(1) η(2) η(3) ...
 
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