Question

Steam enters a turbine in a Rankine cycle power plant at 200 psia and 500 °F. a) Calculate the isentropic thermal efficiency if the condenser pressure is 14.7 psia. b) How much will the isentropic thermal efficiency change if the condenser pressure is 1.0 psia.

Answer 1

Answer:

η=0.19=19% for p=14.7psi

η=0.3=30% for p=1psi

Explanation:

enthalpy before the turbine, state: superheated steam

h1(p=200psi,t=500F)=2951.9KJ/kg

s1=6.8kJ/kgK

Entalpy after the turbine

h2(p=14.7psia, s=6.8)=2469KJ/Kg

Entalpy  before the boiler

h3=(p=14.7psia,x=0)=419KJ/Kg

Learn to pronounce

the efficiency for a simple rankine cycle is

η=$\frac{h1-h2}{h1-h3}$

η=(2951.9KJ/kg-2469KJ/Kg)/(2951.9KJ/kg-419KJ/Kg)

η=0.19=19%

second part

h2(p=1psia, s=6.8)=2110

h3(p=1psia, x=0)=162.1

η=(2951.9KJ/kg-2110KJ/Kg)/(2951.9KJ/kg-162.1KJ/Kg)

η=0.3=30%