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

Answer:

(a) Volume of the tank is $6.47\times 10^{- 3}\ m^{3}$

(b) Temperature is $371^{\circ}C$

Pressure is 21.3 kPa

(c) The change in internal energy is 1373.54 kJ/kg

Solution:

As per the question:

Mass of liquid in the tank, m = 1.4 kg

Temperature, T = $200^{\circ}C$

Volume occupied by water, V = 25%$V_{t}$ = 0.25$V_{t}$

Volume occupied by air, V' = 75%$V_{t}$

where

$V_{t}$ = Volume of tank

Now,

(a) In order to calculate the volume of the tank, we make use of the steam table for specific volume at  as given temperature:

At $200^{\circ}C$, the specific volume, v = 0.001157$m^{3}/kg$

At $200^{\circ}C$, the internal energy, u = 850.46 kJ/kg

Now,  Volume of water, V = mv = $1.4\times 0.001157 = 1.62\times 10^{- 3} m^{3}$

Thus

$V = 0.25V_{t}$

$V_{t} = \frac{1.62\times 10^{- 3}}{0.25} = 6.47\times 10^{- 3}\ m^{3}$

(b) For the final temperature and pressure, we calculate the specific volume, v' and then find the corresponding temperature and pressure from the steam table:

v' = $\frac{V_{t}}{m} = \frac{6.47\times 10^{- 3}}{1.4} = 4.63\times 10^{- 3}\ m^{3}/kg$

The corresponding temperature to this specific volume, T' = $371^{\circ}C$

The corresponding pressure to this specific volume, P' = 21.3 kPa

The corresponding internal energy to this specific volume, u' = 2224 kJ/kg

(c) The change in the internal energy of water is given by:

$\Delta U = u' - u = 2224 - 850.46 = 1373.54 kJ/kg$