An insulated rigid tank initially contains 1.4-kg saturated liquid water and water vapor at 200°C. At this state, 25 percent of the volume is occupied by liquid water and the rest by vapor. Now an electric resistor placed in the tank is turned on, and the tank is observed to contain saturated water vapor after 20 min. Determine: (a) the volume of the tank

(b) the final temperature

(c) the electric power rating of the resistor

Solution:

Mass of liquid water and water vapor in the insulated tank initially = 1.4 kg

Temperature = 200 °C

And 25% of the volume by liquid water is steam.

State 1

      (taking the value of and at 200°C  )

Now quality of vapor

Internal energy at state 1 can be found out by

    = 856.54 kJ/kg

After heating with the resistor for 20 minutes, at state 2, the tank contains saturated water vapor

Tank is rigid, so volume of tank is constant.

Now interpolate the value to get temperature at state 2 with specific volume value to get final temperature

   = 369.11° C

Internal energy at state 2

Now power rating of the resistor

  = 1.51 kW

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