Give the missing property of PvT and x for water at A 10 Mpa

Give the missing property of P,v,T and x for water at: A) 10 Mpa, 0.00104m^3/kg B) 400 degrees Celsius, 0.19 m^3/kg C) 10kpa, 10 degrees celcius

Solution

x = 0.02 - 0.001251 0.04887 = 0.38365
s = sf + x sfg = 2.7927 + 0.38365 × 3.2802 = 4.05 kJ/kg K

T = Tsat(P) = 64.97°C
x = (s – sf)/sfg = 7.70 - 0.893 6.9383 = 0.981
h = 271.9 + 0.981 × 2346.3 = 2573.8 kJ/kg

C.V. Water, which is a control mass. Adiabatic so: 1q2 = 0
Energy Eq.5.11: u2 u1 = 1q2 1w2 = -1w2
Entropy Eq.8.3: s2 - s1= dq/T = 0 (= since reversible)
State 1: Table B.1.3 u1 = 2945.21 kJ/kg; s1 = 7.127 kJ/kg K
State 2: (s, u): u2 = u1 - 1w2 = 2529.29 – 415.72 = 2529.49 kJ/kg
=> sat. vapor 200 kPa, T = 120.23°C
vPsT2112