Steam enters an adiabatic turbine steadily at 3 mpa and 400°c and leaves at 50 kpa and 100°c. if the power output of the turbine is 2 mw, determine (a) the isentropic efficiency of the turbine and (b) the mass flow rate of the steam flowing through the turbine.

P1 = 3 MPa , T1 = 400°C.
 p2 = 50 kPa, T2 = 100°C.
 Wout=2MW.
 The enthalpies at various states are
 h1=3231.7 KJ/Kg, s1=6.9235 KJ/Kg.K
 h2a=2682.4 KJ/Kg.
 The exit enthalpy of the steam for the isentropic process h2S is determined from the requirement that the entropy of the steam remain constant (s2s = s1)
State 2s: P2s=50Kpa, s2s=s1.
 sf=1.0912 KJ/Kg.K, sg=7.5931 KJ/Kg.K
 Obviously, at the end of the isentropic process steam exists as a saturated mixture since sf < s2S < sg. Thus we need to find the quality at state 2s first:
 x2s=((s2s-sf)/(sfg))=((6.9235-1.0912)/(6.5019))= 0.897.
 h2s=hf +x2s*hfg= 340.54 + 0.897*2304.7 = 2407.9 KJ/Kg. 
 By substituting these enthalpy values into, the isentropic efficiency of this turbine is determined to be:
 wa = h1 – h2 = 3231.7– 2682.4 kJ/kg =547.3 kJ/kg
 ws = h1 - h2S = 3231.7 – 2407.9 kJ/kg = kJ/kg = 823.8 kJ/kg 
 ηT = wa / ws = 547.3 / 823.8 = 0.6644
 ηT %= 0.6644 ×100% = 66.44 %.
 (b) The mass flow rate of steam through this turbine is determined from the energy balance for steady-flow systems:
 Wout=m*Wa
 m=Wout/Wa= 2000 kW / 547.3 kJ/kg = 3.654 kg/s