Calculating Maximum Cycle Temperature in Carnot Cycle | Homework Solution

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Discussion Overview

The discussion revolves around a homework problem involving the calculation of the maximum cycle temperature in a Carnot cycle engine. Participants explore the application of thermodynamic principles, specifically focusing on the Carnot efficiency and the relationships between heat transfer and temperatures in the cycle.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • The initial poster states the problem and expresses difficulty in applying the concept of entropy conservation in the context of the Carnot cycle.
  • One participant suggests considering the first law of thermodynamics as a relevant principle for the problem.
  • Another participant provides a calculation of the Carnot efficiency using the formula n=1-(Qc/Qh) and proposes that the heat supplied (Qh) can be determined from the power outputs and heat rejection rates.
  • This participant calculates the efficiency to be 62.5% and derives the maximum temperature (Th) to be approximately 66.67°C based on the relationship between temperatures and efficiency.
  • The initial poster expresses gratitude and indicates that the explanation clarified their understanding of the problem.

Areas of Agreement / Disagreement

While the initial poster finds the explanation helpful, there is no explicit consensus on the correctness of the calculations or methods used, as the discussion does not explore alternative approaches or validate the derived results.

Contextual Notes

Participants do not address potential limitations or assumptions in the calculations, such as the ideal nature of the Carnot cycle or the specific conditions under which the efficiency is calculated.

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Homework Statement



An ideal engine operating on the Carnot cycle with a minimum temperature of 25°C develops 20 kW and rejects heat a rate of 12 kW. Calculate the maximum cycle temperature and the required heat supply of the cycle.



Homework Equations



Qhot/Thot =Qcold/Tcold



The Attempt at a Solution



Since this is ideal entropy must be conserved. Δs hot =Δs cold is as far as I got.

Really stuck with this not sure how to go about it, any help would be greatly appreciated
 
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What about the first law of thermodynamics?
 
Carnot Efficiency is n=1-(Qc/Qh), Qc is heat rejected, Qh is heat supplied
also Carnot efficiency is n=1-(Tc/Th)

Drawing a control volume around the device, (20 KW + 12 leaving) means 32 input or Qh=32, thus you have an efficiency, and then you can use this efficiency to solve for Th

So n=1-(12/32)=62.5% efficient

.625=1-(tc/th)
Tc/.375=Th thus 25/.375=66.67 C

hope that helps
 
Thanks very much. Makes complete sense now
 

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