Answer:
a) 440° C
b) 808.354 kJ/kg
Step-by-step explanation:
Properties of 2.0 MPa and 300° C
= 3024.2 kJ/kg
= 6.7684 kJ/kg.K
Properties at 1 MPa and
= 3024.2 kJ/kg
Using properties at 1MPa and
from steam Table
= 440° C
At the turbine exit
the steam is at 8 KPa which is equivalent ⇒ 0.08 bar
Properties at 0.08 bar
= 173.84 kJ/kg
= 2402.36 kJ/kg
85% = 0.85
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= (173.84+(0.85*2402.36)
= 2215.846 kJ/kg
P = (3024.2 - 2215.846) kJ/kg
P = 808.354 kJ/kg