Question

Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 360°C. For the water, determine the heat transfer, in kJ per kg of water. Kinetic and potential energy effects can be ignored.

Answer 1

Explanation:

Using table A-3, we will obtain the properties of saturated water as follows.

Hence, pressure is given as p = 4 bar.

$u_{1} = u_{g}$ = 2553.6 kJ/kg

$v_{1} = v_{g} = 0.4625 m^{3}/kg$

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, $v_{2} = v_{g} = 0.4625 m^{3}/kg$

According to the table A-4, properties of superheated water vapor will obtain the internal energy for state 2 at $v_{2} = v_{g} = 0.4625 m^{3}/kg$ and temperature $T_{2} = 360^{o}C$ 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.

 $u_{2} = u_{\text{at 5 bar, 400^{o}C}} + (\frac{v_{2} - v_{\text{at 5 bar, 400^{o}C}}}{v_{\text{at 7 bar, 400^{o}C - v_{at 5 bar, 400^{o}C}}}})(u_{at 7 bar, 400^{o}C - u_{at 5 bar, 400^{o}C}})$

     $u_{2} = 2963.2 + (\frac{0.4625 - 0.6173}{0.4397 - 0.6173})(2960.9 - 2963.2)$

                   = 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 = $\Delta U + \Delta K.E + \Delta P.E$ ......... (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.

            $\Delta K.E = \Delta P.E$ = 0

Now, equation will be as follows.

           Q - W = $\Delta U + \Delta K.E + \Delta P.E$

           Q - 0 = $\Delta U + 0 + 0$

           Q = $\Delta U$

Now, we will obtain the heat transfer per unit mass as follows.

          $\frac{Q}{m} = \Delta u$

         $\frac{Q}{m} = u_{2} - u_{1}$

                      = (2961.195 - 2553.6)

                      = 407.595 kJ/kg

Thus, we can conclude that the heat transfer is 407.595 kJ/kg.