Question 13.3: Wet material, containing 70% moisture on a wet basis, is to ...
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 1.0 kN/m² . 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 7.4 kN/m².
The feed is 0.15 kg/s wet material containing 0.70 kg water/kg feed.
Thus water in feed = (0.15 × 0.70) = 0.105 kg/s and dry solids = (0.15 – 0.105) = 0.045 kg/s.
The product contains 0.05 kg water/kg product. Thus, if w kg/s is the amount of water in the product, then:
w/(w + 0.045) = 0.05 or w = 0.00237 kg/s
and: water to be removed = (0.105 – 0.00237) = 0.1026 kg/s.
The inlet air is at 373 K and the partial pressure of the water vapour is 1 kN/m².
Assuming a total pressure of 101.3 kN/m², the humidity is:
= [)](/) (equation 13.1)
= [1.0/(101.3 – 1.0)](18/29) = 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:
× RH/100 = (7.4 × 70/100) = 5.18 kN/m²
and: = [5.18/(101.3 – 5.18)](18/29) = 0.0335 kg/kg dry air
The increase in humidity is (0.0335 – 0.0062) = 0.0273 kg/kg dry air and this must correspond to the water removed, 0.1026 kg/s. Thus if G kg/s is the mass flowrate of dry air, then:
0.0273G = 0.1026 and G = 3.76 kg/s dry air
In the inlet air, this is associated with 0.0062 kg water vapour, or:
(0.0062 × 3.76) = 0.0233 kg/s
Hence, the mass of moist air required at the inlet conditions
= (3.76 + 0.0233) = 3.783 kg/s