How many tons of sludge will be dehydrated when the moisture content is reduced from 80% to 60%?
In 10 tons of sludge, the content of dry sludge is: 10x (100% -80%)=2 tons, which is equivalent to 60% of the sludge. The proportion of water accounts for 60%, and the content of dry sludge accounts for 40%. The actual weight of dry sludge remains unchanged at 2 tons. According to the ratio: 40%: 60%=2: X, X=(60% x2) ÷ 40%=3 tons, the total weight of sludge with a moisture content of 60% is 2+3=5 tons. When 10 tons of sludge are reduced from 80% to 60%, the reduced weight, i.e. the amount of desalinated water, is 10-5=5 tons. Assuming: 10 tons of mud, initial moisture content A, final moisture content B, the weight of the dry mud is 10 (100% - A), the moisture content in the dried mud is 10 (100% - A) * B ÷ (100% - B), and the weight of the dried mud is 10 (100% - A)+10 (100% - A) * B ÷ (100% - B). The total amount of water removed is 10-10 (100% - A) -10 (100% - A) * B ÷ (100% - B)
