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Heat Engine

  1. Aug 23, 2014 #1
    1. The problem statement, all variables and given/known data
    We have two different set ups:
    a) A 200K brick in a 300K environment
    b) A 300K brick in a 200K environment
    From which one of these can you extract more work?
    Assume equal mass and heat capacity.

    2. Relevant equations
    Efficiency - η=[itex]\frac{T_H - T_L}{T_H}[/itex]
    and:
    W = ΔQ => W≤[itex] Q_H * \eta_C[/itex]


    3. The attempt at a solution
    I regard both as cases as infinitesimal carnot engines.
    a) [itex]T_L[/itex] is varying, and after integration I get m*c*16.67 (m and c are the mass and heat capacity)
    b) [itex]T_H[/itex] is varying, after integration the result is m*c*19.
    therefore, I conclude that i can get more work from b.
    The problem is that the correct answer is marked a.
    Anyone has an idea what i am doing wrong?
     
    Last edited: Aug 23, 2014
  2. jcsd
  3. Aug 23, 2014 #2

    rude man

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    You have the right approach (infinire sum of infinitesimal reversible cycles) but you have not shown your work so we can't help yoiu until you do.

    Careful with the signs of dQ and dT in (a) and (b). I would go with dQ > 0 always but dT < 0 in (b).
     
  4. Aug 23, 2014 #3
    Sorry, guess I was kind of lazy with copying it.
    a)
    [itex] \int_{200}^{300} {\frac{300 - T}{300}}{dT} = 100 - 250/3 = 16.67[/itex]
    b)
    [itex] \int_{300}^{200} {\frac{T - 200}{T}}{dT} = -100 + 200 * \log(3/2) = 19[/itex]
     
  5. Aug 23, 2014 #4

    rude man

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    You'll have to derive the integral in (a), for starters. it's close but not what I got. Check the denominator! I got > 19.


    (b) I got the same as you.
     
  6. Aug 23, 2014 #5

    haruspex

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    Maybe I'm missing something but I get your answers.
    In fact, it seemed to me straight away that the answer is (b). At each point in time, the temperature difference between brick and environment is the same, but in (a) the overall temperature is higher. So you must be getting more work from (b).
     
  7. Aug 23, 2014 #6

    rude man

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    Well, you saw my post. Do you want to compare notes privately?
     
  8. Aug 24, 2014 #7

    Andrew Mason

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    I don't think you have to do any calculation here.

    The real issue is how to extract work with the least amount of change in temperature difference per cycle. The first Carnot cycle will produce the same amount of work in either case. But in a) (the brick is receiving heat flow Qc) the heat flow to the brick (bringing the brick's temperature closer to surroundings by ΔT = Qc/Cv) is less than the heat flow from the brick in b) (bringing the brick's temperature closer to the surroundings by ΔT = Qh/Cv). So the next cycle will produce a bit more work in a) than in b) and so on.

    AM
     
    Last edited: Aug 24, 2014
  9. Aug 24, 2014 #8

    TSny

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    In part (a) heat is added to the brick.

    I'm getting that the work in part (a) is greater than in (b), in line with rude man.
     
  10. Aug 24, 2014 #9

    Andrew Mason

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    Yeah. Sorry about that. The reasoning in my earlier post was correct but I had in my mind a and b reversed. I have corrected it.

    AM
     
  11. Aug 24, 2014 #10

    haruspex

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    Yes, I agree. I had allowed myself to be misled by the efficiency calculation in posts 1 and 3. My argument in favour of (b) only leads to the conclusion that the efficiency is greater.
    Reading the whole question, I see it asks for the work extracted, not the efficiency in doing so.
    There's no contradiction, since the heat flow is greater in (a).
     
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