Question

A 0.6-m3 rigid tank is filled with saturated liquid water at 135°C. A valve at the bottom of the tank is now opened, and one-half of the total mass is withdrawn from the tank in liquid form. Heat is transferred to water from a source of 210°C so that the temperature in the tank remains constant. Assume the surroundings to be at 25°C and 100 kPa.

Answer 1

The given question is incomplete. The complete question is as follows.

A 0.6-m3 rigid tank is filled with saturated liquid water at 135°C. A valve at the bottom of the tank is now opened, and one-half of the total mass is withdrawn from the tank in liquid form. Heat is transferred to water from a source of 210°C so that the temperature in the tank remains constant. Assume the surroundings to be at 25°C and 100 kPa. Determine the amount of heat transfer.

Explanation:

First, we will determine the initial mass from given volume and specific volume as follows.

              $m_{1} = \frac{V}{\alpha_{1}}$

                         = $\frac{0.6}{0.001075}kg$

                         = 558.14 kg

Hence, the final mass and mass that has left the tank are as follows.

             $m_{2} = m_{out} = \frac{1}{2}m_{1}$  

                          = $\frac{1}{2} \times 558.14 kg$

                          = 279.07 kg

Now, the final specific volume is as follows.

          $\alpha_{2} = \frac{V}{m_{2}}$

                        = $\frac{0.6}{279.07} m^{3}/kg$

                        = 0.00215 $m^{3}/kg$

Final quality of the mixture is determined actually from the total final specific volume and the specific volumes of the constituents for the given temperature are as follows.

           $q_{2} = \frac{\alpha_{2} - \alpha_{liq135}}{(\alpha_{vap} - \alpha_{liq})_{135}}$

                        = $\frac{0.00215 - 0.001075}{0.58179 - 0.001075}$

                        = $1.85 \times 10^{-3}$

Hence, the final internal energy will be calculated as follows.

          $u_{2} = u_{liq135} + q_{2}u_{vap135}$

                      = $(567.41 + 1.85 \times 10^{-3} \times 1977.3) kJ/kg$

                      = 571.06 kJ/kg

Now, we will calculate the heat transfer as follows.

            $\Delta U = Q - m_{out}h_{out}$

      $m_{2}u_{2} - m_{1}u_{1} = Q - m_{out}h_{out}$

                Q = $(279.07 \times 571.06 - 558.14 \times 567.41 + 279.07 \times 567.75) kJ$

                    = 1113.5 kJ

Thus, we can conclude that amount of heat transfer is 1113.5 kJ.