Question

Water in an open metal drum is to be heated from 27 °C to 80 °C by adding superheated steam slowly enough that all the superheated steam condenses. The drum initially contains 250 kg of water, and superheated steam is supplied at 5.0 bar and 375 °C via a 7.5 cm ID stainless steel pipe.

a.) How many kilograms of superheated steam should be added so that the final temperature of the water in the tank is exactly 80 °C?

b.) What is the velocity (m/Δt) of the superheated steam entering the open metal drum?

(Neglect all heat losses from the water in this calculation.)

Answer 1

Given Data:

Mass of water to be heated = 250 Kg

Temperature of cold water in = 27 deg C

Temperature of cold water out = 80 deg C

Superheated steam operating Pressure = 5

Tempearture of suoerheated steam = 375 deg C

Solution:

As the superheated steam is used to heat the water, it looses its temperatre in two steps:

1) superheated stem to condensed form

2) condensed vapour

The heat balance can be obtained as below:

heat required by water to raise temperature form 27 deg c to 80 deg c = heat realesed when superheated stem comes to its condensed form + heat released by the liquid.

thus,

m*cp*delta T = m*(steam)*cp  delta T+m(steam)* latent heat of condensation

250*4184*(80-27)= m(2.3289*(375-151.85) + (2095.9))

a) m(steam) = 2.119 Kg

Here, at 5 bar pressure, the superheated steam has been considered to have speccific heat = 2.3289

Boiling point = 151.85

Thus 2.119 Kg superheted steam should be added to reach final temperature of 80 deg C.