Calculate the quantity of dry air required to dry 50 metric tons of
food from 19 to 11% moisture. The air enters with a moisture content
of 0.003 kg/water/kg dry air and exits with a moisture content of
0.009 kg water/kg dry air?

ANSWER:
GIVEN:     FOOD is 50metric tonnes
           19% moisture means =(19/100)*50=9.5 metric tonnes(MT)
           11% moisture means=(11/100)*50=5.5 metric tonnes
           moisture to be removed=9.5-5.5=4 metric tonnes
             1kg of dry air enter with 0.003kg water
             1kg of dry air exit with 0.009kg water
        moisture removed by 1kg of water is 0.006kg
        to remove 4metric tonnes of moisture required dry air
=(4/0.006)*1=666.6667 MT
                     
                                           reference -chemical process
calculation by G.K.ROY
                                       (Any chemical engineering 
faculty)
 
 
                                                                          
                   Answered by
                                                                          
                    k.a.velavan