Answer:
η=0.19=19% for p=14.7psi
η=0.3=30% for p=1psi
Step-by-step 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=
![(h1-h2)/(h1-h3)](https://img.qammunity.org/2020/formulas/engineering/college/1nhq907livh2fues3cbkzdrzl5hvqakzi6.png)
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%