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Carnot Engine Question

  1. Jul 26, 2016 #1
    1. The problem statement, all variables and given/known data
    A Carnot engine moves a piston which applies 5N of force to a rotating disc by moving a piston 0.25cm for
    a total of 5W of power. If the high temperature reservoir is 100K, what must the temperature of the low
    temperature reservoir be?

    2. Relevant equations
    The efficiency of a Carnot engine is given by the formula
    e = 1-TLo/THi.


    3. The attempt at a solution
    I know you can rearrange the efficiency equation and solve for Tlo which is TLo = (1-e)*THi. After this this step I am drawing a blank on what to do next. I know that it gives you the work done by the piston and the power of the engine, but I dont know what to do next.
     
  2. jcsd
  3. Jul 27, 2016 #2

    Simon Bridge

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    Do you know anything about the carnot cycle besides the efficiency equation?
     
  4. Jul 28, 2016 #3

    rude man

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    Hmm - is there a solution?
    The work per carnot cycle is presumably 5N x 0.0025m = 0.0125J.
    There are an infinite number of combinations of T2 and Q1 that can provide this work (T2 = low temp, Q1 = high heat) while still satisfying carnot efficiency.
    The power of 5W just indicates how many carnot cycles have to be performed per second.
    Comments/critiques welcome!
     
  5. Jul 28, 2016 #4
    Rude man/JoeSmith1013, I feel your pain on this one. To let you know where I'm at on this one, I even had to search to find out what a carnot engine was. Maybe the efficiency equation has nothing to do with it. Maybe it has to do with the fact that the gas temperature has to be reduced enough so that when it is warmed up, the gas expansion is sufficient to produce the 5N and 0.25cm. It is possible for a situation where the temperature difference is so small that the gas expansion cannot even produce the 5 N, 0.25 cm movement. So it seems to me that there has to be some minimum temperature delta to accomplish that. Just my $0.02.
     
  6. Jul 28, 2016 #5

    rude man

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    There is always some minimum temperature delta required in any engine.
    I did notice that the "high" temperature is pretty da**ed low (almost to frozen Nitrogen) which seems odd but I guess there it is.
    Th efficiency of a carnot engine is determined by the fact that the entropy loss of the "high" reservoir = entropy gained by the low (lower!) reservoir. The possible combinations of Q1 and T2 I referred to previously is based on this fact.
     
  7. Jul 29, 2016 #6

    Simon Bridge

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    I think there is context missing.
    Its pretty specific about how the carnot cycles is being implimented so there may be an example in course notes that can help.
     
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