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
