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Carnot Cycle

  1. Nov 3, 2011 #1
    1. The problem statement, all variables and given/known data

    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.



    2. Relevant equations

    Qhot/Thot =Qcold/Tcold



    3. 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
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 4, 2011 #2

    rude man

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    Homework Helper
    Gold Member

    What about the first law of thermodynamics?
     
  4. Nov 6, 2011 #3
    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
     
  5. Nov 9, 2011 #4
    Thanks very much. Makes complete sense now
     
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