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Thermodynamics fun!

  1. Apr 10, 2005 #1
    For a brayton cycle on a pressure and temperature diagram, using helium gas, is it reasonable to have the net work for the cycle and the change in Q for the cycle to come out to be equal to each other?
     
  2. jcsd
  3. Apr 10, 2005 #2
    i thought the first law of thermodynamics said "change in work is equal to change in heat"

    so yeah it is reasonable. and i just worked some brayton cycle problems for homework, and yeah that's what i had
     
  4. Apr 11, 2005 #3
    First law of Thermodynamics

    [tex]\Delta U = q + w[/tex]

    Or in differential form

    [tex]dU = dq + dw[/tex]

    In an ISOLATED system, [tex]\Delta U = 0[/tex]. A Brayton cycle, however, is not isolated. Actually its not even closed. So offhand I would not expect the net work to be the same as the net heat evolved. I would however, calculate the work done and heat evolved through all 4 steps of the cycle for an ideal gas. You have two isobars and two isoentropes (constant entropy).
     
    Last edited: Apr 11, 2005
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