Question

1. You are to analyze a simple steam power plant. At the boiler exit, the pressure is 8 MPa and the temperature is 1000 °C. The turbine isentropic efficiency is 85%. The condenser pressure is 15 kPa and the water is a saturated liquid at the condenser exit. The pump isentropic efficiency is 80%. Answer the following:

a) What is the quality at the exit in %, if the turbine is assumed isentropic?

b) What is the work output of the actual turbine?

c) What is the work input required by the actual pump?

d) How much heat is added in the boiler?

e) What is the actual thermal efficiency of this steam power plant in %?

Answer 1

8 MPa NT=0.85 The WT mp=0.80 15 kea N4 Turbine inlet P,=8 MPa Ti = 1000°c from superheated steam table I hi = 4619, 11 kilig si=7.gul kolka dal Turbine outlet : P = 15kpa hfa 225.94 kilkg hg= 2598.303 kJ/kg verso Sf = 0.755 killigk Sy=8.007 kulkg k

Tsay = 327.12 k l = 0.00 m² kg, If expansion process in turbine is Isentropic then = > Si=S2 Si= Sf f d (Sg-ff) F.ghl= 0.75 & 2218.007 - 6. AST) z= 0.991. I has hftalha-hf) - 225-94 + Ooggi (@598.3 - 225.947 = 2576. ar kiflg. Now Nr = hi ha hi ha 0.85 = 4619.11- h₂ 4619.11-2576.95 h = 2883.27. kolkg, - At Condensor outlet: P = 15 kpg sut liquid h₂=225.94 kilkg

Isendoopic work done of pump Wp = Vf. de = 0.001 (8000-15) Wps 7.gg kulkg. - Actual work Actual wor done of pump ! Wp = Wp = 9.98 kJ/kg.. Wp = ha-hz hü= 235.92 killeg. 9 of turbine if expansion is Quality at exit isentropic : x2 = 99.1 % work output of turbine : Wy = nCh, - ha) = 0.85 (4619.11-2576.95) Wy = 1735.84 kilka

Work input by actual pump! wp= 9.98 kulkg) I d) Heat added in boiler i Qin = hi ha = 4619-11-235.92 Qin = 4383.19 kilkol e thermal efficiency ! NT = Wy- We 1735.86-9-99 4383.19 = 39.37 % n