Explanation:
Using table A-3, we will obtain the properties of saturated water as follows.
Hence, pressure is given as p = 4 bar.
= 2553.6 kJ/kg
![v_{1} = v_{g} = 0.4625 m^{3}/kg](https://tex.z-dn.net/?f=v_%7B1%7D%20%3D%20v_%7Bg%7D%20%3D%200.4625%20m%5E%7B3%7D%2Fkg)
At state 2, we will obtain the properties. In a closed rigid container, the specific volume will remain constant.
Also, the specific volume saturated vapor at state 1 and 2 becomes equal. So, ![v_{2} = v_{g} = 0.4625 m^{3}/kg](https://tex.z-dn.net/?f=v_%7B2%7D%20%3D%20v_%7Bg%7D%20%3D%200.4625%20m%5E%7B3%7D%2Fkg)
According to the table A-4, properties of superheated water vapor will obtain the internal energy for state 2 at
and temperature
so that it will fall in between range of pressure p = 5.0 bar and p = 7.0 bar.
Now, using interpolation we will find the internal energy as follows.
![u_{2} = u_{\text{at 5 bar, 400^{o}C}} + (\frac{v_{2} - v_{\text{at 5 bar, 400^{o}C}}}{v_{\text{at 7 bar, 400^{o}C - v_{at 5 bar, 400^{o}C}}}})(u_{at 7 bar, 400^{o}C - u_{at 5 bar, 400^{o}C}})](https://tex.z-dn.net/?f=u_%7B2%7D%20%3D%20u_%7B%5Ctext%7Bat%205%20bar%2C%20400%5E%7Bo%7DC%7D%7D%20%2B%20%28%5Cfrac%7Bv_%7B2%7D%20-%20v_%7B%5Ctext%7Bat%205%20bar%2C%20400%5E%7Bo%7DC%7D%7D%7D%7Bv_%7B%5Ctext%7Bat%207%20bar%2C%20400%5E%7Bo%7DC%20-%20v_%7Bat%205%20bar%2C%20400%5E%7Bo%7DC%7D%7D%7D%7D%29%28u_%7Bat%207%20bar%2C%20400%5E%7Bo%7DC%20-%20u_%7Bat%205%20bar%2C%20400%5E%7Bo%7DC%7D%7D%29)
![u_{2} = 2963.2 + (\frac{0.4625 - 0.6173}{0.4397 - 0.6173})(2960.9 - 2963.2)](https://tex.z-dn.net/?f=u_%7B2%7D%20%3D%202963.2%20%2B%20%28%5Cfrac%7B0.4625%20-%200.6173%7D%7B0.4397%20-%200.6173%7D%29%282960.9%20-%202963.2%29)
= 2963.2 - 2.005
= 2961.195 kJ/kg
Now, we will calculate the heat transfer in the system by applying the equation of energy balance as follows.
Q - W =
......... (1)
Since, the container is rigid so work will be equal to zero and the effects of both kinetic energy and potential energy can be ignored.
= 0
Now, equation will be as follows.
Q - W =
Q - 0 =
Q = ![\Delta U](https://tex.z-dn.net/?f=%5CDelta%20U)
Now, we will obtain the heat transfer per unit mass as follows.
![\frac{Q}{m} = \Delta u](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7Bm%7D%20%3D%20%5CDelta%20u)
![\frac{Q}{m} = u_{2} - u_{1}](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7Bm%7D%20%3D%20u_%7B2%7D%20-%20u_%7B1%7D)
= (2961.195 - 2553.6)
= 407.595 kJ/kg
Thus, we can conclude that the heat transfer is 407.595 kJ/kg.