Respuesta :
Answer:
(a) Volume of the tank is [tex]6.47\times 10^{- 3}\ m^{3}[/tex]
(b) Temperature is [tex]371^{\circ}C[/tex]
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 = [tex]200^{\circ}C[/tex]
Volume occupied by water, V = 25%[tex]V_{t}[/tex] = 0.25[tex]V_{t}[/tex]
Volume occupied by air, V' = 75%[tex]V_{t}[/tex]
where
[tex]V_{t}[/tex] = 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 [tex]200^{\circ}C[/tex], the specific volume, v = 0.001157[tex]m^{3}/kg[/tex]
At [tex]200^{\circ}C[/tex], the internal energy, u = 850.46 kJ/kg
Now, Volume of water, V = mv = [tex]1.4\times 0.001157 = 1.62\times 10^{- 3} m^{3}[/tex]
Thus
[tex]V = 0.25V_{t}[/tex]
[tex]V_{t} = \frac{1.62\times 10^{- 3}}{0.25} = 6.47\times 10^{- 3}\ m^{3}[/tex]
(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' = [tex]\frac{V_{t}}{m} = \frac{6.47\times 10^{- 3}}{1.4} = 4.63\times 10^{- 3}\ m^{3}/kg[/tex]
The corresponding temperature to this specific volume, T' = [tex]371^{\circ}C[/tex]
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:
[tex]\Delta U = u' - u = 2224 - 850.46 = 1373.54 kJ/kg[/tex]
