Daily Question 2016.7.5 (EIA Discussion): 3 Wealth points for replies, plus 5 Wealth points for correct answers (single-choice question)
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Question 9 from the 2016 Technical Methods test: The original moisture content of the sludge produced by a wastewater treatment plant is 96%. To reduce the amount of sludge to 10% of its original level, what percentage should the moisture content of the sludge be reduced to? A) 40% B) 50% C) 60% D) 70% Answer: C. Explanation: 1. Moisture content of sludge = (total mass of sludge – mass of dry matter) ÷ total mass of sludge × 100%. Since the mass of dry matter remains constant, it can be calculated as follows: 2. 96% = (total mass of sludge – mass of dry matter) ÷ total mass of sludge × 100%. 3. Mass of dry matter = total mass of sludge – total mass of sludge × 96%. When the amount of sludge becomes 10% of its original value: 4. Moisture content of sludge = (10% of total mass of sludge – mass of dry matter) ÷ total mass of sludge × 100%. 5. Substituting equation 3 into this formula gives: Moisture content of sludge = (10% of total mass of sludge – total mass of sludge + total mass of sludge × 96%) ÷ total mass of sludge × 100%. The resulting value is 60%, so the moisture content should be reduced to 60%
The original moisture content of the sludge produced by a wastewater treatment plant is 96%. To reduce the amount of sludge to 10% of its original level, the moisture content of the sludge needs to be reduced to (D). A) 40% B) 50% C) 60% D) 70%
The original moisture content of the sludge produced by a wastewater treatment plant was 96%. To reduce the amount of sludge to 10% of its original level, the moisture content of the sludge needs to be reduced to (C). A) 40% B) 50% C) 60% D) 70%