Problem 2 (20 points): A vertical piston-cylinder device contains 20 kg of satur
ID: 717773 • Letter: P
Question
Problem 2 (20 points): A vertical piston-cylinder device contains 20 kg of saturated liquid water. The weight of the piston is such that a constant pressure of 200 kPa is maintained in the cylinder. Heat is added to the water by a resistance heater until the quality is 0.3, as shown in figure. A 12-V battery supplies a current of 2 A for 200 minutes. Determine: ttft P Vapor 12 V Liquid (a) Initial and final volumes of the water in the cylinder (b) Total work done (mechanical and electrical) in this process (10) (10)Explanation / Answer
Part a
Mass of saturated liquid water
m = 20 kg
Initial pressure P1 = 200 kPa
Saturation temperature Ts = 120.20°C
Initial steam quality at 200 kPa
x1 = 0
At 200 kPa
Specific volume of fluid vf = 0.001061 m3/kg
Total volume of system = specific volume x mass
= 0.001061 m3/kg x 20 kg
= 0.02122 m3
At final state of saturated mixture
Steam quality x2 = 0.30
At 200 kPa
Specific volume of gas vg = 0.8857 m3/kg
Specific volume of mixture
v2 = vf + x2*(vg - vf)
= 0.001061 + 0.30*(0.8857 - 0.001061)
= 0.2665 m3/kg
Final volume V2 = mass x v2
= 20 kg x 0.2665 m3/kg
= 5.329 m3
Part b
Work done in mechanical process
W1 = P(V2 - V1)
= 200*1000 Pa x (5.329 - 0.2122)m3
= 1023360 J
Work done in electrical process
W2 = voltage x current x time
= (12 x 2) J/s x (200 x 60)
= 288000 J
Net work done
W = W1 - W2
= 1023360 - 288000
= 735360 J x 1kJ/1000 J
= 735.36 kJ