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

a . ) Weight of water removed from 1000Kgs of wet material

= ( Water content in wet material ) - ( water content in dried material )

= (1000 x 10 /100) - ( 1000 x 5 /100 )

= 50.0 Kgs.

b.) Water removed per Kg of fresh air = ( water content perKg of recycled air ) - (water .................................................................content per kg of fresh air )

...............................................................= (0.08 - 0.008) = 0.072 Kgs.

So, for removing 50.0 Kgs of water the weight of fresh air required = (50.0 / 0.072)

....................................................................................................... = 694.44 Kgs.

& the weight of dry air present in 694.44 Kgs of fresh air = (weight of fresh air )- (weight of ...............................................................................................water in fresh air)

......................................................................................... = 694.44 - (694.44 x 0.008)

...........................................................................................= 688.89 Kgs.

c.) Similarly ,

The weight of dry air recycled = ( Weight of dry air required ) - ( weight of water )

................................................ = 688.89 - (688.89 x 0.08 )

................................................. = 633.78 Kgs.