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Maximum efficiency of an engine taking heat from two hot reservoirs

  1. May 12, 2012 #1
    1. The problem statement, all variables and given/known data

    A heat engine is taking identical amounts of heat from two hot reservoirs at temperatures ##T_{H1}, T_{H2} ## doing work and then heat to a cold reservoir ##T_C ##

    What is the maximum efficiency of this heat engine?

    2. Relevant equations

    For a Carnot cyle it is ## \eta = 1 - T_C/T_H##

    3. The attempt at a solution

    First of all, the heat engine takes ##Q_1## from ##T_{H1}## and ##Q_2## from ##T_{H2} ## it then does work ##W## and heats ##T_C ##

    I thought that since the maximum efficiency was for a reversible process that does the same thing then I could make a new process that takes heat ##Q_1+Q_2## from a reservoir of temperature ## T_H = \frac{T_{H1}+T_{H2}}{2} ##

    Making the efficiency (max) ## \eta = 1 - T_C/T_H = 1 - \frac{2 T_C}{T_{H1}+T_{H2}} ##

    Am I able to do this?
     
  2. jcsd
  3. May 13, 2012 #2

    rude man

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    'Fraid not.

    Write down the equation for the 1st and 2nd laws and solve for η = W/2Q.
     
  4. May 13, 2012 #3
    Take it to be a reversible process

    ##\eta = |\frac{2Q-Q_C}{2Q}| ##

    Now the problem is to relate that to the temperatures ##T_{H1}## and ##T_{H2} ##. Going to say that they are Carnot engines. I'm not sure this right but my attempt:

    ##\eta = 1 - \frac{Q_C}{2Q}##

    For an ireversible process we have

    ## \oint \frac{\delta q}{T} = 0 ##

    ## \frac{Q}{T_{H1}} + \frac{Q}{T_{H2}} - \frac{Q_C}{T_C} = 0 ##

    ## \frac{Q(T_{H1}+T_{H2})}{{T_{H1}}{T_{H2}}} = \frac{Q_C}{T_C}##

    ## 2Q = \frac{2 T_{H1}T_{H_2}Q_C}{T_C(T_{H1}+T_{H2})} ##

    ##\eta = 1 - \frac{T_C(T_{H1}+T_{H2})}{2 T_{H1}T_{H_2}} ##
     
  5. May 13, 2012 #4

    rude man

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    You get an A+! Good shot!

    It's always best to go back to fundamentals instead of relying on formulas that may or may not apply. You did it the right way.

    PS - you said "irreversible process". You meant "reversible".
     
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