C.V. Water. Energy
Eq. m(u2 − u1) = 1Q2 − 1W2
Entropy Eq : m(s2 − s1) = ∫ dQ/T Process: 1Q2 = 0 and reversible
State 1: (T, P) v1 = 0.5342, u1 = 2646.8, s1 = 7.1706 kJ/kg K
m = V1/v1 = 0.15/0.5342 = 0.2808 kg
With the assumed reversible process we have from entropy equation
s2 = s1 = 7.1706 kJ/kg K and from the energy equation
u2 = u1 − 1W2/m = 2646.8 - 30/0.2808 = 2540.0 kJ/kg
State 2 given by (u, s) check
sG (at uG = 2540) = 7.0259 < s1
⇒ State 2 must be in superheated vapor region.