A solid material containing 15.0 wt % moisture is dried so that it contains 7.0

Question

A solid material containing 15.0 wt % 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, 0.03 kg water/kg dry air. For a feed of 100 kg 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

Explanation / Answer

The amounts of each component (dry solid, dry air, and water) going and coming out of the dryer per hour must balance. Here, the inputs are, moist product (dry solid 85%, water 15%) and fresh air (dry air1 kg, water 0.01 kg) and the outputs are dried product (dry solid 93%, water 7%) and moist air (dry air 1 kg, water 0.1 kg).

Dry solid

100kg of the input product contains water 15kg and dry solid 85kg. Therefore, the output must contain dry solid 85 kg (which is 93%)

Therefore, mass of the product output = (85x100) / 93 = 91.4 kg.

Mass of the water content of the output = 91.4 Kg – 85 Kg = 6.4 kg.

Dry air

Suppose the input dry air be A kg/h. Then the fresh air inlet contains 0.01A kg/h of water. Therefore, the output moist air from the dryer contains A kg / h dry air + 0.10A kg/h water

Water

Here, input is 15 kg/h (moist product) + 0.01A kg/h (fresh air). Also, output is 6.4 kg/h (product) + 0.10A kg/h (moist air).

But, total input = Total output

Therefore,

15 + 0.01 A = 6.4 + 0.10 A

0.09A = 8.6

A = 8.6 / 0.09 = 95.6 kg

The dry air input is 95.6 kg/h.

Recycled air

Suppose the rate of dry air in the recycled air is B kg/h. The water it holds is 0.1B kg/h. It is mixed with the fresh air of composition 95.6 kg/h dry air and 0.956 kg/h water to a mixture with 0.03 kg water per kg of the dry air.

Therefore, airborne water input = 0.03 x dry air input

0.1B + 0.956 = 0.03 (B + 95.6)

1.91 = 0.07 B

B =1.91/0.07= 27.3

The mass rate of the dry air in the recycled stream = 27.3 kg/h

Therefore, the answer is,

95.6 kg/h dry air in the fresh air

27.3 kg/h dry air in recycled air and 91.4 kg/h dried product

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