Answer:
(a) Volume of the tank is ![6.47\times 10^{- 3}\ m^{3}](https://tex.z-dn.net/?f=6.47%5Ctimes%2010%5E%7B-%203%7D%5C%20m%5E%7B3%7D)
(b) Temperature is ![371^{\circ}C](https://tex.z-dn.net/?f=371%5E%7B%5Ccirc%7DC)
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](https://tex.z-dn.net/?f=200%5E%7B%5Ccirc%7DC)
Volume occupied by water, V = 25%
= 0.25![V_{t}](https://tex.z-dn.net/?f=V_%7Bt%7D)
Volume occupied by air, V' = 75%![V_{t}](https://tex.z-dn.net/?f=V_%7Bt%7D)
where
= 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
, the specific volume, v = 0.001157![m^{3}/kg](https://tex.z-dn.net/?f=m%5E%7B3%7D%2Fkg)
At
, 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}](https://tex.z-dn.net/?f=1.4%5Ctimes%200.001157%20%3D%201.62%5Ctimes%2010%5E%7B-%203%7D%20m%5E%7B3%7D)
Thus
![V = 0.25V_{t}](https://tex.z-dn.net/?f=V%20%3D%200.25V_%7Bt%7D)
![V_{t} = \frac{1.62\times 10^{- 3}}{0.25} = 6.47\times 10^{- 3}\ m^{3}](https://tex.z-dn.net/?f=V_%7Bt%7D%20%3D%20%5Cfrac%7B1.62%5Ctimes%2010%5E%7B-%203%7D%7D%7B0.25%7D%20%3D%206.47%5Ctimes%2010%5E%7B-%203%7D%5C%20m%5E%7B3%7D)
(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](https://tex.z-dn.net/?f=%5Cfrac%7BV_%7Bt%7D%7D%7Bm%7D%20%3D%20%5Cfrac%7B6.47%5Ctimes%2010%5E%7B-%203%7D%7D%7B1.4%7D%20%3D%204.63%5Ctimes%2010%5E%7B-%203%7D%5C%20m%5E%7B3%7D%2Fkg)
The corresponding temperature to this specific volume, T' = ![371^{\circ}C](https://tex.z-dn.net/?f=371%5E%7B%5Ccirc%7DC)
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](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%20u%27%20-%20u%20%3D%202224%20-%20850.46%20%3D%201373.54%20kJ%2Fkg)