adrian bejan advanced engineering thermodynamics pdf

The quantity in the square brackets is positive because k >1 and V 10, directed into the system Q12 = 0, adiabatic U2− U1 = m c (T 2− T1), incompressible substance In conclusion, T2 ( Wi  f /m 1 )pa rt (b (b ) ( Wi  f /m f ) pa r t ( a ) The second group of terms on the right-hand side is the work output during the reversible and adiabatic expansion (path: PV k = constant). M1 = P1V1/RT1, based on the solution for mf given in part (a), and

The final ideal-gas mass admitted is m f  M f  Noting that V 2 = Mf Vf and and dividing everything by M f yields yields cvTf  P1vf  cvT1  Pv1 Integrating in time, fĪnd recognizing that U i = 0 and M i = 0, the first law reduces to fįor the “ideal gas” working fluid, we write U f  M f cv (Tf  T0 ) h1  cv (T1  T0 )  Pv1Ĭv (T  T0 )  Mf  P1V2

U are the mass and energy inventories of the system (the “system” “s ystem” is the cylinder volume).

Is the instantaneous flow rate into the cylinder and M and Next, to calculate T f , we note that from state Ch ap t er 1 THE FIRST LAW Prob Pro b lem 1.1 1.1 

Adrian bejan advanced engineering thermodynamics pdf