1000 points 1 out of 5 attempts Assume the sumoundings to be
Solution
From a 1st law energy balance on the system, by denoting entry condition as state 1 and exit as state 2 we get
m(h1 + V12 /2) = Wout + m(h2 + V22 /2)
Wout = m*[h1 - h2 + (V12 - V22) /2]
From steam properties at P1 = 6 MPa and T1 = 600 deg C, we get specific enthalpy h1 = 3660 kJ/kg and specific entropy s1 = 7.17 kJ/kg-K
From steam properties at P2 = 50 kPa and T2 = 100 deg C, we get specific enthalpy h2 = 2680 kJ/kg and specific entropy s2 = 7.7 kJ/kg-K
We are given that V1 = 86 m/s and V2 = 154 m/s and Wout = 5.4 MW
Thus, 5.4*106 = m*[3660*103 - 2680*103 + (862 - 1542) /2 ]
Mass flow rate m = 5.556 kg/s
For max. power output, Wrev,out = m*[(h1 - h2) + (KE1 - KE2) - T0*(s1 - s2) ].......where KE is kinetic energy, T0 = Ambient temp = 21 deg C = 294 K
Wrev,out = 5.556*[3660*103 - 2680*103 + (862 - 1542) /2 - 294*(7.17*103 - 7.7*103) ]
Wrev,out = 6.2658 MW = 6266 kW
