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Reversible work of compression

  1. Jun 21, 2017 #1
    1. The problem statement, all variables and given/known data
    Air enters a compressor at 100 [Kpa] and 10 [c], the air leave the compressor at 12[Mpa]
    the isentropic efficiency η is 87%, calculate:
    • work done
    • reversible work
    • irreversibility

    2. Relevant equations

    T2=T1(P2/P1)(1.4-1)/1.4
    W = (Cp(T2-T1))/η
    Wrev=Δh - T0Δs
    Δs = cpln(T2/T1) - Rln(P2/P1)
    3. The attempt at a solution

    Work done can be calculated with:
    T2=T1(P2/P1)(1.4-1)/1.4
    W = (Cp(T2-T1))/η
    T1 , P1 , P2 are given.
    Δs = cpln(T2/T1) - Rln(P2/P1)
    I tried to find Wrev with Δh - T0Δs
    and Δs = cpln(T2/T1) - Rln(P2/P1)

    the correct answers are :
    • work : -338.2 kj/kg
    • reversibility : -317.4 kj/kg
    • irreversibility : 20.8 kj/kg
     
  2. jcsd
  3. Jun 26, 2017 #2
    Thanks for the thread! 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? The more details the better.
     
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