Problem 2 Recycle in a Dryer. (Geankoplis, Pb. 1.5-8) A solid material containin
ID: 700803 • Letter: P
Question
Problem 2 Recycle in a Dryer. (Geankoplis, Pb. 1.5-8) A solid material containing 15.0wt % moisture is dried so that it contains 7.0 wt % water by blowing fresh warm air mixed with recycled air over the solid in the dryer. The inlet fresh air has a humidity of 0.01 kg water/kg dry air, the air from the drier that is recycled has a humidity of 0.1 kg water/kg dry air, and the mixed air to the dryer, 0.03kg water/kg dry air For a feed of 100kg solid/h fed to the dryer, calculate the kg dry air/h in the fresh air, the kg dry air/h in the recycled air, and the kg/h of "dried" product. Answer. 95.6kg/h dry air in fresh air, 27.3kg/h dry air in recycled air, and 91.4kg/h "dried" productExplanation / Answer
Input stream
Dry solid in wet material = 85%
Water = 15%
Dry air in fresh air = 1 kg
Water in fresh air = 0.01 kg
output stream
Dry solid = 93%
Water = 7%
Recycled air
Dry air = 1 kg
Water = 0.1 kg
Feed rate of wet solid = 100 kg/h
Dry solid in feed = 0.85*100 = 85 kg/h
Water in feed = 0.15*100 = 15 kg/h
Mass Balance of the solid
Dry solid will remain same in the output
Dry solid in output stream = 85 kg/h
Output stream rate = 85*100/93 = 91.4 kg/h
water in the output stream = 91.4 - 85 = 6.4 kg/h
Mass balance of air
Let rate of input dry air = A kg/h
Water in fresh air inlet = 0.01A kg/h
moist air in the output stream = A kg/h dry air + 0.10A kg/h water
Mass balance of water
Water in Input stream = 15 kg/h in wet solid + 0.01A kg/h in fresh air
Water in Output stream = 6.4 kg/h in the product + 0.10A kg/h in the moist air
Basic mass balance equation
Total input of water = total output of water
15+0.01A=6.4+ 0.10A
8.6=0.09A
A= 8.6/0.09
A = 95.6 kg/h = dry air in fresh air
Water in fresh input air = 0.01 A = 0.01*95.6
= 0.956 kg/h
Do the balance on recycled air
Let the rate of dry air in recycled air = B kg/h
Water in recycled air = 0.1B kg/h
It mix with fresh air & produces the mixture with
0.03 kg water/ kg dry air
Mass balance on this mixing point
Total input water in air = 0.03 x total input dry air
0.1B + 0.956 = 0.03*(B+95.6)
0.07B=1.91
B = 1.91/0.07 = 27.3 kg/h = dry air in recycled air
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