Answer:
a) 440° C
b) 808.354 kJ/kg
Explanation:
Properties of 2.0 MPa and 300° C
[tex]h_1[/tex] = 3024.2 kJ/kg
[tex]s_1[/tex] = 6.7684 kJ/kg.K
Properties at 1 MPa and [tex]h_1 = h_2[/tex] = 3024.2 kJ/kg
Using properties at 1MPa and [tex]h_2[/tex] from steam Table
[tex]T_2[/tex] = 440° C
At the turbine exit
the steam is at 8 KPa which is equivalent ⇒ 0.08 bar
Properties at 0.08 bar
[tex]h_{fg}[/tex] = 173.84 kJ/kg
[tex]h_{fg3}[/tex] = 2402.36 kJ/kg
[tex]x_3=[/tex] 85% = 0.85
[tex]h_3}= h_{f_3}} + x * h_{fg_3}[/tex]
= (173.84+(0.85*2402.36)
= 2215.846 kJ/kg
P = (3024.2 - 2215.846) kJ/kg
P = 808.354 kJ/kg