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

Step-by-step 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 * 4200 J/kg.K * (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 * 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 * 1800 J/kg.K * (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.