Question

A dryer is operated with a material containing 10% H2O (by mass) to a final dried material containing 5% H2O (by mass). The fresh air is mixed with recycled air and is blown over the solid. The fresh air contains 0.008 kg H2O/kg dry air and the recycled air contains 0.08 kg H2O/kg dry air. For 1000 kg of wet material fed to the dryer. Calculate (a) weight of H2O removed, (b) weight of dry air in the fresh air feed and (c) weight of dry air recycled.

Answer 1

The initial material containg 10% by mass of water

The dried material contains 5% by mass of water

If we take 1000 Kg of wet material it will contain 100 Kg water so the dry material is 900 Kg

when it is dried the material will contain 5% by mass of water to find the toal weight after drying

900 Kg + 0.05 x total weight = total weight

900 Kg = 0.95 total weight

total weight = 900/0.95

total weight = 947.4 Kg

a) So weight of water removed is 1000 - 947.4 = 52.6 Kg

The fresh air contains 0.008 kg H2O/kg dry air and the recycled air contains 0.08 kg H2O/kg dry air

Since the ratio of dry air and recycled air is not given I will assume them to be 1:1

52.6 Kg water has to be removed

so 2 Kg dry air will remove 0.088 Kg of water after it goes over the solid it will contain 0.16 Kg of H2O

so 2 Kg of air removes 0.072 Kg of water /2 Kg of dry air

so to remove 52.6 Kg of water we will need 52.6/0.072 = 730.5 Kg of dry air as fresh air

730.5 Kg of the air will be recycled air