Heat engine and Heat pump in combination

  1. Nov 22, 2011 #1
    Question is attached

    I have spent a great deal of time on this problem and I am hoping someone can help me out.
    My attempt at problem.

    first for the heat engine
    QL/Qw1 = TL/Twaste
    so efficiency = 1- Tl/Twaste
    efficiency = 0.93808
    and I know that efficiency = Wout/ Qw1


    This is as far as i got for the heat engine...

    For Heat pump
    i found cop = TH/(TH-TL) = 4.23
     

    Attached Files:

  2. jcsd
  3. Nov 23, 2011 #2

    rude man

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    Never mind efficiency. The problem doesn't ask for efficiency. It only wants you to compute dQh/dt.

    ANYWAY ...

    write the equations for the 1st and 2nd laws. That's 4 equations.

    Then write the relationship between Qw1, Qw2 and 5 MW.

    5 equations, 5 unknowns. Away you go. Solve for Qh.

    (Note - the 1st and 2nd laws are usually written in terms of energy rather than power. Don't let that slow you down. Just pretend it's energy for 1 second, which of course is 1 J/s = 1 W.
     
  4. Nov 23, 2011 #3
    I am still a little confused. There are 2 equations for both first and second law?
    i only have delta U = Q-W
    and Q = integral of TodeltaS
     
  5. Nov 23, 2011 #4

    rude man

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    Yes, but you have two mechanisms, the heat engine and the heat pump. Each has its own
    1st & 2nd law equations.

    BTW dQ = TdS is not an expression of the second law. It's just a formula to calculate entropy. What you need is an expression reflective of the fact that entropy is unchanged in a reversible cycle.

    As for the 1st law, you are right, but what can you say about Qw1, QL, Qw2, QH and W? Remember that U is a state function so over a cycle it doesn't change either, reversible or not.
     
    Last edited: Nov 23, 2011
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