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Problem 1.10 The system is closed, the process 1−2 is at constant temperature, T, and the volume change is quasistatic, W = PdV. The quantity in the square brackets is positive because k >1 and V10, directed into the system Q12 = 0, adiabatic U2− U1 = m c (T2− T1), incompressible substance In conclusion, T2  T1  (Wif /m 1 )part ( b ) (Wif /m f )part (a ) Finally, the goodness ratio is Wif  m1Ĭv (P1V1  P2 V2 )  c R  RT1 1  v P1V1 / (RT1 )  R The second group of terms on the right-hand side is the work output during the reversible and adiabatic expansion (path: PVk = constant). M1 = P1V1/RT1, based on the solution for mf given in part (a), and Hence the goodness ratio Wif P1V2   RT1 mf P1V2 / (RT1 ) The final ideal-gas mass admitted is mf  Mf  Noting that V2 = MfVf and dividing everything by Mf yields cvTf  P1vf  cvT1  Pv1 (1') becomes M f cv (T  T0 )  M f  P1V2 Integrating in time, fĪnd recognizing that Ui = 0 and Mi = 0, the first law reduces to U f  M f h1  P1V2įor the “ideal gas” working fluid, we write U f  M f c v (Tf  T0 ) h1  c v (T1  T0 )  Pv1 Where m is the instantaneous flow rate into the cylinder and M and U are the mass and energy inventories of the system (the “system” is the cylinder volume).

(i) to state (f), we have dM m dt dU  W  mh 1 dt Next, to calculate Tf, we note that from state i