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: 276 j/kg

Explanation:

The given data is as follows.

      = (83 + 273) K = 356 K

      = (100 + 273) K = 373 K

         m = 3.9 kg/s

Relation between heat energy and specific heat is as follows.

             Q =

Putting the given values into the above formula as follows.

              Q =

                =

                = 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 is as follows.

            = 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 or (100 + 273)K = 373 K. Therefore, final temperature will  be calculated as follows.

              Q =

         8282460 J/s =

              = 1552.83 K

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