solution-manual-advanced-engineering-thermodynamics-4th-edition-adrian-bejan

Chapter 1 THE FIRST LAW Problem 1.1 f (a) Wi f i PdV PV 1 2 . Next, to calculate Tf, we note that from state (i) to state (f), we have dM  m dt dU  W mh1 dt 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). Integrating in time, f Mf  Mi i U f  U i P V1 2 h (M1 f M )i (1) and recognizing that Ui = 0 and Mi = 0, the first law reduces to Uf M hf 1 PV1 2 (1′) For the “ideal gas” working fluid, we write Uf M c (Tf v f T )0 h1  c (Tv 1 T )0 Pv1 hence, eq. (1') becomes M c (Tf v  T )0 M [c (Tf v 1  T )0 Pv ] PV1  1 2 Noting that V2 = MfVf and dividing everything by Mf yields c Tv f Pv1 f c Tv 1 Pv1 or c Tv f RTf c Tv 1 RT1 in other words, Tf = T1. The final ideal-gas mass admitted is PV1 2 m Mf  f  RT1 hence the goodness ratio Wi f  P V1 2  RT1 mf P V /(RT )1 2 1 (b) m1 = P1V1/RT1, based on the solution for mf given in part (a), and V1 V2 Wi f  0 PdVV1 PdV cv (P V1 1 P V )2 2  P V1 1  R The second group of terms on the right-hand side is the work output during the reversible and adiabatic expansion (path: PVk = constant). Finally, the goodness ratio is  cv (P V1 1 P V )2 2  cv 1 P V2 2  Wi f P V1 1  R  RT 11     m1 P V /(RT )1 1 1  R  P V1 1  (c) The relative goodness is (Wi f /m )1 part(b)  1 cv 1 P V2 2  (Wi f /m )f part(a) R  P V1 1  c   V k 1   1 v 1 1   R   V2   The quantity in the square brackets is positive because k >1 and V1< V2; therefore,  Wi f   Wi f       m1 part(b)  mf part(a) Problem 1.2 (a) Given are m = 1 kg, T1 = 100°C, and x1 = 0.5. The path is constant volume. (b) To pinpoint state (2), we must determine two properties at the final state. The first one is the volume v2  v1 vf,T1  x v1 fg,T1   0.0010440.5(1.67290.001044)  0.837 m / kg3 The second property is the internal energy: this comes from the first law Q1 2 W1 2 m(u2 u )1 (1) where W1−2 = 0 and u1  uf,T1 x u1 fg,T1  418.94(0.5)(2087.6) 1462.74 kJ/kg Equation (1) yields 1 u2  u1  Q1 2 3662 kJ/kg m (c) To find T2 and P2, we must first locate state (2) on the P(v, t) surface (or tables). At state (2), we know u2 and v2; therefore, one way to proceed is to look at the table of superheated steam properties and find the u values of order 3662 kJ/kg. This is the equivalent of traveling along the u = u2 line and looking for the v value that comes closest to v2. This search leads to this portion of the table: T P = 0.5 MPa P = 0.6 MPa v u v u 800°C 0.9896 3662.1 0.8245 3661.8 Fitting v2 between 0.9896 and 0.8245, we interpolate linearly for pressure and find P2 0.592 MPa The final temperature is T2 800°C. (d) At state (2), the system is superheated steam. This particular fluid approaches ideal gas behavior if near state (2) the following two conditions are met: (i) u = u(T)