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Thermodynamics/Physics (P-v)

  1. Feb 27, 2009 #1
    1. The problem statement, all variables and given/known data

    The figure (http://session.masteringphysics.com/problemAsset/1001463/8/knight_Figure_19_54.jpg) shows the cycle for a heat engine that uses a gas having a specific heat ratio = 1.25 . The initial temperature is T1 = 300K ,and this engine operates at 20 cycles per second.

    a) What is the power output of the engine?
    b)What is the engine's thermal efficiency?

    2. Relevant equations

    P=W/t

    Not sure what else.

    3. The attempt at a solution

    T1=T2=300K

    P1*V1=P2*V2
    (1)(600)/200=3atm=P2=P3

    P1/T1=P3/T3
    (300)(3)/1=900K=T3

    Don't know where to go next.
     
  2. jcsd
  3. Feb 28, 2009 #2

    alphysicist

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    Hi j88k,

    What is the work done in one cycle of this engine?
     
  4. Feb 28, 2009 #3
    work done in the first cycle of this engine would be:

    Wb= P1V1 * ln (V2/V1)

    P1 = 101.325 Pa
    don't know how to get the specific volumes for V1 and V2

    need more help than this as soon as possible.
    Thank you.
     
  5. Feb 28, 2009 #4

    alphysicist

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    The PV diagram has V1 and V2 (as well as P1); isn't that all you need?
     
  6. Feb 28, 2009 #5
  7. Feb 28, 2009 #6

    Redbelly98

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    The equations for work, energy, etc. are in terms of actual volumes, not specific volumes. So just use the volumes given in the PV graph, as alphysicist said.

    Also, note that work done "in one cycle" means for one complete round-trip cycle, not just along path 1-2.
     
  8. Mar 1, 2009 #7
    For the first cycle I can use the formula - Wb= P1V1 * ln (V2/V1)
    For the second cycle I can use - Wb= P(V2-V1)
    For the second cycle Wb = 0

    is that correct ?
    and do I have to convert the pressure to Pa ?
     
  9. Mar 1, 2009 #8

    alphysicist

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    I'm not sure what you mean when you say first cycle and second cycle here. You need to calculate the total work done in one full cycle, which means the work done after it does a complete loop (from 1 to 2, then 2 to 3,and then 3 back to 1). Once you find the total work done in one cycle, you can use the fact that there are twenty cycles per second to find the power output.

    So your last post had the right idea. For one full cycle, the work done for the isothermal part will be your first equation, work done during the isobaric part will be your second equation, and no work is done during the isochoric part. (However, I think you have a sign error in one of your equations. What definition of work are you using?)

    Yes, you do have to convert the pressure to Pa if you want the work to be in Joules.
     
  10. Mar 1, 2009 #9
    i meant from 1 - 2, 2-3 and 3 to 1.

    can you point out which equation is wrong and correct it?

    after that i think i got it.
     
  11. Mar 1, 2009 #10

    Redbelly98

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    A good way to check is to remember that work is positive if the volume increases, and negative if the volume gets smaller. So you can tell whether work should be positive or negative along each path, 1-2 and 2-3.

    Then you can use the fact that V1 is larger than V2, and inspect the expressions you wrote in post #7, to see if they give a positive or negative result.
     
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