Question

A rigid tank initially contains 1.4 kg saturated liquid water at 200°C. At this state, 25 percent of the volume is occupied by water and the rest by air. Now heat is supplied to the water until the tank contains saturated vapor only. Determine (a) the volume of the tank, (b) the final temperature and pressure, and (c) the internal energy change of the water.

Answer 1

Given:

• Mass of liquid (m) = 1.4 kg
• The temperature of the liquid water (T) = 200°C
• The volume occupied by water ($V_1$) = 25%V"
• The volume occupied by air ($V_2$) = 75%V"

To Find:

• The volume of the tank (V).
• Final temperature and pressure.
• Internal energy change with the water.

Solution:

• In order to find volume of the tank, we are using the steam table for specific volume at a particular temperature.
• At 200°C, the specific volume, v = 0.001157$m^{3}/kg$
• At 200°C. the internal energy, u = 850.46$kJ/Kg$
• So, volume of water V = mv = 1.4*0.001157 = $1.6198*10^{-3} m^{3}$
• $V_1$ = 0.25V
• V" = $\frac{V_1}{0.25}$
• V" = $\frac{1.6198*10^{-3} }{0.25}$ = $6.47*10^{-3} m^{3}$
• To find temperature and pressure, we first find out specific volume v' and the using the specific volume we find out the corresponding temperature and pressure from the steam table.
• v' = V/m
• v' = $\frac{6.47*10^{-3} }{1.4}$ =  $4.62*10^{-3} m^{3}$
• The corresponding temperature to this specific volume, T' = 371°C
• The corresponding pressure to this specific volume, P' = 21.3kPa
• The corresponding internal energy to this specific volume, u' = $2224kJ/Kg$
• The change in internal energy ΔU = u'-u = 2224-850.46 = $1373.54kJ/Kg$

(a) Volume of the tank = $6.47*10^{-3} m^{3}$

(b) Final temperature = 371°C and Pressure = 21.3kPa

(c) The change in internal energy = $1373.54kJ/Kg$