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Proof that in Brayton reverse cycle heat is negative

  1. Nov 30, 2015 #1
    1. The problem statement, all variables and given/known data
    I was given the Brayton reverse cycle and asked to prove that the total heat is negative (hence the heat pump cools the system).
    I was to assume that all the steps are reversible.
    2. Relevant equations
    An ideal diatomic gas.
    1st step: adiabatic compression.
    2nd step: isobaric heat.
    3rd step: adiabatic expansion.
    4th step: isobaric heat absorption.
    I managed to get to the equation:
    Qtot = Cp*((T3-T2) + (T1 - T4)

    3. The attempt at a solution
    Since I have two adiabtes there are only two places in the cycle where Q can change - the two isobars I summed the two heata calculated from the Energy differential and managed to get to the equation above for total heat but didn't manage to prove that Qtot<0
     
  2. jcsd
  3. Dec 5, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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