Question

Liquid water at 83 C and at 1 atm flows through a heated pipe at a flow rate of 3.9 kg/s. It then leaves the pipe as steam. The water receives 12378600 J/s from the pipe. Calculate the temperature of the steam leaving the pipe. The water boiling point at the pressure of the system is 100 C. Thermal properties: Co of liquid water: 4200 J/kg.K Cp of steam: 1800 J/kg.K Enthalpy of evaporation of water: 2760000 J/kg

Answer 1

Explanation:

The given data is as follows.

      $T_{1} = 83^{o}C$ = (83 + 273) K = 356 K

      $T_{2} = 100^{o}C$ = (100 + 273) K = 373 K

         m = 3.9 kg/s

Relation between heat energy and specific heat is as follows.

             Q = $mC_{p} \Delta T$

Putting the given values into the above formula as follows.

              Q = $mC_{p} \Delta T$

                = $3.9 kg/s \times 4200 J/kg.K \times (373 - 356)K$

                = 278460 J/s

As it is given that enthalpy of evaporation of water is 2760000 J/kg.

Hence, energy left is as follows.

                    (2760000 - 278460) J/s

                  = 2481540 J/s

So, enthalpy of vapor energy at $100^{o}C$ is as follows.

              $3.9 kg/s \times 2760000 J/kg$

            = 10764000 J/s

Hence, energy left for steam is as follows.

            (10764000 J/s - 2481540 J/s)

            = 8282460 J/s

Now, steam is formed at $100^{o}C$ or (100 + 273)K = 373 K. Therefore, final temperature will  be calculated as follows.

              Q = $mC_{p} \Delta T$

         8282460 J/s = $3.9 kg/s \times 1800 J/kg.K \times (T_{2} - 373)K$

              $T_{2}$ = 1552.83 K

Thus, we can conclude that the temperature of steam leaving the pipe is 1552.83 K.