Heat engine and Heat pump in combination

  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. rude man

    rude man 6,085
    Homework Helper
    Gold Member

    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. 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. rude man

    rude man 6,085
    Homework Helper
    Gold Member

    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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