Carnot Cycle

  • Thread starter db725
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  • #1
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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

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
rude man
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What about the first law of thermodynamics?
 
  • #3
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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
 
  • #4
35
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Thanks very much. Makes complete sense now
 

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