Question

PROCESS MASS and ENERGY BALANCES A solid material with 15% water by weight is to be dried to 7% water. Fresh air is mixed with recycled air and blown over the solid. Fresh air contains 0.01 kg moisture per kg of dry air and recycled air, which is part of the air leaving the drier, contains 0.1 kg moisture per kg of dry air. Mixed air entering the drier contains 0.03 kg moisture per kg of dry air. Determine the following: 1) (a) The amount of water removed per 100 kg of wet material fed to the drier. (b) The amount of dry air in fresh air per 100 kg of wet material. (c) The amount of dry air in recycled air per 100 kg of wet material.

Answer 1

a) Water removed = 8.6 kg

b) Dry air in the fresh air = 95.6 kg

c) Dry air in the recycled air = 27.3 kg

Step-by-step explanation:

To solve this problem we have to make mass balances of the different streams.

1) Material balance for the dry solid

For every 100 kg of feed, we have 85 kg of dry solid and 15kg of water.

If the exit material has 7% of moisture content, the total dry solid represents 93% of the mass exiting the drier.

If the dry solid is 85 kg and represents 93% of the total exit material, the total amount of exit material is 85/0.93=91.4 kg. The difference (7%) is water, weighting (91.4-85)=6.4 kg.

The water removed for every 100 kg of feed is (15-6.4)=8.6 kg.

2) Material balance for the water

The water entering the system has to be the same that exit the system.

Let da be the amount of dry air. Then the water entering the drier is (15+0.01*da) and the water exiting the drier is (6.4+0.1*da). We can calculate the amount of dry air:

`15+0.01*da=6.4+0.1*da\\(15-6.4)=(0.1-0.01)*da\\da=8.6/0.09=95.6`

For every 100 kg of feed, 95.6 kg of dry air is entering the drier.

3) Recycled air

Let rda be the amount of dry air in the recycled stream. We can balance the water content like:

water in the fresh air + water in the recycled air = water in the air entering the drier

`0.01*da+0.1*rda=0.03*(da+rda)\\\\0.1*rda-0.03*rda=0.03*da-0.01*da\\\\0.07*rda=0.02*da\\\\rda=(0.02/0.07)*da=0.286*da=0.286*95.6=27.3 kg`

The amount of dry air in the recycled stream is 27.3 kg.