CHEMICAL ENGINEERING

HUMIDIFICATION AND WATER COOLING 319

1 kmol of vapour occupies 22.4m^3 at STP,

and: volume of water removedD 3. 933 ð 10 ^4 ð 22. 4 D 0 .00881 m^3 /s

Assuming no volume change on mixing, the gas flow after drying

D 0. 60  0. 00881 D 0 .591 m^3 /satSTP.

PROBLEM 13.3

Wet material, containing 70% moisture on a wet basis, is to be dried at the rate of 0.15 kg/s
in a counter-current dryer to give a product containing 5% moisture (both on a wet basis).
The drying medium consists of air heated to 373 K and containing water vapour with
a partial pressure of 1.0kN/m^2. The air leaves the dryer at 313 K and 70% saturated.
Calculate how much air will be required to remove the moisture. The vapour pressure of
water at 313 K may be taken as 7.4kN/m^2.

Solution

The feed is 0.15 kg/s wet material containing 0.70 kg water/kg feed.

Thus water in feedD 0. 15 ð 0. 70 D 0 .105 kg/s and dry solidsD 0. 15  0. 105 D
0 .045 kg/s.

The product contains 0.05 kg water/kg product. Thus, ifwkg/s is the amount of water
in the product, then:

w/wC 0. 045 D 0 .05 orwD 0 .00237 kg/s

and: water to be removedD 0. 105  0. 00237 D 0 .1026 kg/s.

The inlet air is at 373 K and the partial pressure of the water vapour is 1 kN/m^2.

Assuming a total pressure of 101.3kN/m^2 , the humidity is:

H 1 D[Pw/PPw]Mw/MA (equation 13.1)

D[1. 0 / 101. 3  1. 0 ] 18 / 29 D 0 .0062 kg/kg dry air

The outlet air is at 313 K and is 70% saturated. Thus, as in Example 13.1, Volume 1:

PwDPw 0 ðRH/ 100 D 7. 4 ð 70 / 100 D 5 .18 kN/m^2

and: H 2 D[5. 18 / 101. 3  5. 18 ] 18 / 29 D 0 .0335 kg/kg dry air

The increase in humidity is 0. 0335  0. 0062 D 0 .0273 kg/kg dry air and this must
correspond to the water removed, 0.1026 kg/s. Thus ifGkg/s is the mass flowrate of dry
air, then:
0. 0273 GD 0. 1026 and GD 3 .76 kg/s dry air

In the inlet air, this is associated with 0.0062 kg water vapour, or:

 0. 0062 ð 3. 76 D 0 .0233 kg/s