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

A 200 litre 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 vapour 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 if water that flowed out and the total heat transfer.
*I’ve seen others have posted this same question and that the answers are 8.8kg and 25MJ, respectively. However, those answers are apparently wrong (according to my professor).

Please help!

Solutions

Expert Solution

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


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