Consider 1100 degree F 2000 psia 5 psia Kankine cycle with 8
Solution
FROM MOLLIER DIAGRAM
ENTHALPIES AT DIFFEENT POINTS ARE AS FOLLOWS
h1=3600KJ/Kg at HP TURBINE INLET
h2=2773KJ/Kg at HP TURBINE OUT LET
STEAM IS REHEATED UN REHEATER UPTO1100 FAHREIGHNHEAT
AT THIS POINT h3=3650KJ/Kg
STEAM IS EXPANDED IN LP TURBINE UPTO 5psia=.34bar
at this point h4=2630KJ/Kg AT 5psia enthalpy of steam in liquid state is =300kj/kg
1)HEAT ADDED IN STEAM GENERATOR PER Kg OF STEAM= M(h1-h4)= 1*(3600-300 )=3300kW
2)ACTUAL WORK OF TURBINE=WORK DONE IN HP TURBINE+WORKDONE IN LP TURBINE
=(3600-2773)+(3650-2630)=1907KW
turbine is having .85 ejjicienvcy
turbine work=0.85*1907=1620 kw
3)HEAT REJECTED IN CONDENSOR PER Kg of steam =1*(2630-300)=2330 KW
4)THERMAL EFFICIENCY=WORK DONE/HEAT SUPPLIED
=1620/3300=0.491
=49.1%
5)WORK OF PUMP= M*(ENTHALPY PF STEAM LEAVING CONDENSOR) =1*300=300KW

