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

Neglecting energy losses, apply the the enthalpy of the fluid
From steam table
3 bar @ 375 C is 3220. 05 kJ/kg
25 deg C is 104.8 kJ/kg
80 deg C is 334.9 kJ/kg
Enthalpy balance
( 104.8 kJ/kg )( X kg ) + ( 3220.05 kJ/kg )( Y kg ) = ( 334.9 kJ/kg )( X kg + Y kg )

Initial amount of water X = 250 kg
Amount of superheated steam Y
Solve for Y kg

( 104.8 kJ/kg )( 250 kg ) + ( 3220.05 kJ/kg )( Y kg ) = ( 334.9 kJ/kg )( 250 kg + Y kg )

26200 kJ + 3220.05 Y kJ/kg = 83725 kJ + 334.9 Y kJ/kg
57525 kJ = 2885.15 Y kJ/Kg

Y = 19.94 Kg