I have a problem that I need assistance with:
There is a well insulated rigid tank with 3 kg of saturated liquid-vapor mixture of water at 200 kPa. Initially, half of the mass is in the liquid phase. An electric resistor placed in the tank is connected to a 120-V source, and a current of 6 A flows throught the resistor when the switch is turned on. How long will it take to vaporize all the liquid in the tank?
The Attempt at a Solution
P = 200 kPa
x = .5
Therefore T = 120.21 C
v = vf +xvfg
vf = 0.001061 m3/kg
vfg = 0.88578 - 0.001061 m3/kg
Multiply v by m to obtain V = 1.33206 m3
Therefore V2 = 1.33206 and since the mass does not change, v2 = .4423595 m3/kg = vg
u1 = ufg @ 200 kPA = 2024.6 kJ/kg
We = Delta U => VI(Delta T) = m(u2 - u1)
Looking for the v2, it is assumed to be between 400 kPa and 450 kPa. Therefore, use the v values to interpolate the ug which is 2554.6669 kJ/kg.
((1 kJ/s)/(1000 VA))720 VA (Delta T) = 3 kg(2554.6669- 2024.6 kJ/kg)
Delta T = 2208.6 s or .6135 hours.
I just want to know if I made a mistake anywhere in this problem. I'm not the best with thermodynamics and usually wind up missing some minute detail that is vital to the rest of the problem.