State 1:
From steam tables;
T1 = 100°C
P1 = 1.01325 Bar
S1=Sg = 7.355 kJ/kg.K
State 2:
P= 1.5 Bar
T2 = ?
This point is super-heated but entropies should be equal.
S1=Sg=S2 = 7.355 kJ/kg.K
This point lies between saturation point at 1.5 Bar and 150 °C super-heat state. By interpolation,
T2 = 111.4+ (7.355-7.223)/(7.420-7.223)*(150-111.2) = 137.26 °C
Work done:
W=ΔU = m(u2-u1)
u1= 2506 kJ/kg
u2 = 2519 + (7.355-7.223)/(7.420-7.223)(2580-2519) = 2559.87 kJ/kg
m= volume/specific volume, where specific volume = 1.159+(7.355-7.223)/(7.420-7.223)(1.286-1.159) = 1.244 m^3/kg
Then, m=0.25/1.244 = 0.2 kg
Therefore,
W= 0.2*(2559.87-2506) = 10.77 kJ
