(1 point) An insulated rigid tank contains a saturated liquid-vapor mix of water
ID: 703815 • Letter: #
Question
(1 point) An insulated rigid tank contains a saturated liquid-vapor mix of water initially at a pressure of 100 kPa. The mass of the mixture is 5 kg, but only 75.6 % of the total mass is liquid. An electric resistance heater is turned on within the tank until all the water has ust vaporized. The heater power is a constant 2.6 kW. Hint: you will not need to interpolate to find the solution to part (b) Note: quantities shown in the figure are not necessarily to scale )LxeNd s Niel Crews, 2013 a) What is the volume of the container? 2.0707 b) How long was the heater on? 56.89 minutesExplanation / Answer
At 100 kPa from the steam table
Saturated steam
Specific volume of fluid vf = 0.001043 m3/kg
Specific volume of gas vg = 1.6941 m3/kg
Steam quality x = 1 - 0.756 = 0.244
Specific volume balance
v1 = vf + x(vg - vf)
= 0.001043 + 0.244 x (1.6941 - 0.001043)
= 0.414148908 m3/kg
Volume of container = 0.414148908 m3/kg x 5kg
= 2.0707 m3
Internal energy U1 = Uf + x(Ug - Uf)
= 417.4 + 0.244 x (2505.6 - 417.4)
= 926.9208 kJ/kg
At final state
v2 = v1 = 0.414148908 m3/kg
U2 = Ug (at v2) = 2557.1 kJ/kg
Energy balance
2.6 kJ/s x time = 5 kg x (2557.1 kJ/kg - 926.9208 kJ/kg)
Time = 3134.96 s x 1min/60s
= 52.25 min
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