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Thermodynamics- Work done- Adiabatic process

  1. Jan 15, 2014 #1
    I have a small doubt regarding the workdone derivation in adiabatic process. Here is the derivation
    W= (P1V1-P2V2)
    = mR(T1-T2)/n-1

    AS we know R/n-1 = Cv and Cp/Cv=1 also considering m=1

    W=Cv(T1-T2)

    but in textbook it is quoted as

    W=Cp(T1-T2)


    please can u help me in solving this.( i found this during deriving Brayton cycle)
     
  2. jcsd
  3. Jan 15, 2014 #2
    The adiabatic compression step and the adiabatic expansion step in the Brayton cycle takes place in an open system operating at steady state, with gas flowing through the system. For such circumstances, you need to use the flow form of the first law, which focuses on each parcel of mass flowing through the system. The form of the first law that applies here is Δh=-ws, where Δh is the enthalpy change per unit mass passing through the system, and ws is the shaft work per unit mass. The enthalpy change per unit mass is Cp(T2-T1). You need to go back and review the derivation of the form of the first law that applies to flow systems operating at steady state.

    Chet
     
  4. Jan 16, 2014 #3
    thank you
     
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