A 200 liter tank initially contains water at 100 kPa and a quality of 1%. Heat i

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 MJ

Explanation / 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

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