In: Other
Let make the diagram
Initial conditions
Pressure P1 = 100 kPa
Quality of steam x = 1% = 0.010
Specific volume of fluid vf = 0.001043 m3/kg
vfg = 1.69296 m3/kg
Specific volume at initial conditions
v1 = vf + x*vfg
= 0.001043 + 0.01*1.69296
= 0.0179726 m3/kg
Internal energy
U1 = Uf + x*Ufg
= 417.33 + 0.01*2088.72
= 438.2172 kJ/kg
Initial mass
m1 = total volume / Specific volume
=( 200 L x 1m3/1000L)/0.0179726 m3/kg
= 11.128 kg
State 2 after opening the valve
Pressure P2 = 2 MPa
Quality of steam x2 = 0.90
Specific volume
v2 = vf + x*vfg
= 0.001177 + 0.90*0.09845
= 0.089782 m3/kg
Internal energy
U2 = Uf + x*Ufg
= 906.42 + 0.90*1693.84
= 2430.88 kJ/kg
Initial mass
m1 = total volume / Specific volume
=( 200 L x 1m3/1000L)/0.089782 m3/kg
= 2.227 kg
At the final conditions at 2 MPa
Enthalpy at exit = enthalpy of gas = he = 2799.51 kJ/kg
Mass balance
Mass in = mass out
me = m1 - m2 = 11.128 - 2.227 = 8.901 kg
Energy balance
Energy in = energy out
m1*U1 + Q = m2*U2 + me*he
11.128 * 438.2172 + Q = 2.227 * 2430.88 + 8.901 * 2799.51
Q = 25455.5 kJ x 1MJ/1000 kJ
= 25.45 MJ