Level II sample questions
5 free sample questions on math and calculations for the Washington Level II wastewater treatment operator exam. Try each one, then open the answer to read the explanation.
A rectangular tank is 40 feet long, 20 feet wide, and contains 12 feet of water. Using 7.48 gallons per cubic foot, approximately how many gallons does it contain?
Answer: C. 71,808 gallons
Multiply length, width, and water depth to obtain cubic feet, then convert cubic feet to gallons. Calculation: 40 × 20 × 12 × 7.48 = 71,808 gallons.
A plant treats 2.0 MGD with a chemical dose of 3.0 mg/L. Using 8.34, approximately how many pounds of chemical are applied per day?
Answer: C. 50.0 lb/day
Pounds per day equals flow in MGD multiplied by dose in mg/L and 8.34. Calculation: 2.0 × 3.0 × 8.34 = 50.04 ≈ 50.0 lb/day.
A 30-minute settleometer result is 240 mL/L, and the MLSS concentration is 3,000 mg/L. What is the sludge volume index?
Answer: B. 80 mL/g
SVI is the settled-sludge volume in mL/L multiplied by 1,000 and divided by MLSS in mg/L. Calculation: (240 × 1,000) ÷ 3,000 = 80 mL/g.
A clarifier receives 1.8 MGD and has 600 feet of total effluent-weir length. What is its weir overflow rate?
Answer: C. 3,000 gpd/ft
Weir overflow rate equals total daily flow divided by the total effective weir length. Calculation: 1,800,000 gpd ÷ 600 ft = 3,000 gpd/ft.
An aeration basin contains 2.0 MG at an MLSS concentration of 2,500 mg/L. The plant removes 4,000 lb/day in waste sludge and loses 200 lb/day in the effluent. Using 8.34, what is the approximate mean cell residence time?
Answer: B. 9.9 days
MCRT equals the solids inventory in the treatment system divided by the daily solids leaving through wasting and effluent. Calculation: 2.0 × 2,500 × 8.34 = 41,700 lb`; `41,700 ÷ (4,000 + 200) = 9.93 ≈ 9.9 days.
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Operator Master has 21 Level II math and calculations questions, free, with an explanation after every answer.
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