Respuesta :
Explanation:
Using table A-3, we will obtain the properties of saturated water as follows.
Hence, pressure is given as p = 4 bar.
[tex]u_{1} = u_{g}[/tex] = 2553.6 kJ/kg
[tex]v_{1} = v_{g} = 0.4625 m^{3}/kg[/tex]
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, [tex]v_{2} = v_{g} = 0.4625 m^{3}/kg[/tex]
According to the table A-4, properties of superheated water vapor will obtain the internal energy for state 2 at [tex]v_{2} = v_{g} = 0.4625 m^{3}/kg[/tex] and temperature [tex]T_{2} = 360^{o}C[/tex] 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.
 [tex]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}})[/tex]
   [tex]u_{2} = 2963.2 + (\frac{0.4625 - 0.6173}{0.4397 - 0.6173})(2960.9 - 2963.2)[/tex]
          = 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 = [tex]\Delta U + \Delta K.E + \Delta P.E[/tex] ......... (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.
      [tex]\Delta K.E = \Delta P.E[/tex] = 0
Now, equation will be as follows.
      Q - W = [tex]\Delta U + \Delta K.E + \Delta P.E[/tex]
      Q - 0 = [tex]\Delta U + 0 + 0[/tex]
      Q = [tex]\Delta U[/tex]
Now, we will obtain the heat transfer per unit mass as follows.
     [tex]\frac{Q}{m} = \Delta u[/tex]
     [tex]\frac{Q}{m} = u_{2} - u_{1}[/tex]
           = (2961.195 - 2553.6)
           = 407.595 kJ/kg
Thus, we can conclude that the heat transfer is 407.595 kJ/kg.
