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