A 200 liter tank initially contains water at 100 kPa and a quality of 1%. Heat i
ID: 707571 • Letter: A
Question
A 200 liter tank initially contains water at 100 kPa and a quality of 1%. Heat is transferred to the water thereby raising its pressure and temperature. At a pressure of 2 MPa a safety valve opens and saturated vapor at 2 MPa flows out. The process continues, maintaining 2 MPa inside until the quality in the tank is 90% then stops. Determine the total mass of water that flowed out and the total heat transfer sat vap cv O m 8.90 kg -125.46 kJ m 8.90 kg Q 25.46 MJ C m. 18.90 kg Q 25.46 MJ m 88.90 kg Q 215.46 MJExplanation / Answer
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
Option B is the correct answer
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.