Math and Calculations Practice Questions for Level 4 · Expert

1. Question 1

   Three identical 480-volt, three-phase motors each draw 80 amps at a power factor of 0.86 while operating. Other loads on the same distribution equipment total 120 kW, and its approved continuous limit is 300 kW. Using 1.732, what is the operating conclusion with all three motors running?

   1. Total demand is about 229 kW, leaving about 71 kW of capacity.
   2. Total demand is about 292 kW, leaving about 8 kW of capacity.
   3. Total demand is about 319 kW, exceeding the limit by about 19 kW.
   4. Total demand is about 358 kW, exceeding the limit by about 58 kW.
   Show the answer

   Answer: B. Total demand is about 292 kW, leaving about 8 kW of capacity.

   All three motors and the other loads total about 292 kW, leaving only about 8 kW below the approved continuous limit. Calculation: One motor = 1.732 × 480 V × 80 A × 0.86 ÷ 1,000 = 57.20 kW; Three motors = 171.61 kW; Total with other loads = 171.61 + 120 = 291.61 kW; Remaining capacity = 300 − 291.61 = 8.39 kW.

2. Question 2

   A pump upgrade costs $180,000. At the required duty point, it reduces demand from 150 kW to 110 kW for 16 hours per day and saves $8,000 per year in maintenance. Electricity costs $0.11/kWh. The plant requires a simple payback of no more than six years. Which conclusion is correct?

   1. Annual savings are about $25,700, so payback is about 7.0 years and misses the criterion.
   2. Annual savings are about $32,100, so payback is about 5.6 years and meets the criterion.
   3. Annual savings are about $33,700, so payback is about 5.3 years and meets the criterion.
   4. Annual savings are about $40,100, so payback is about 4.5 years and meets the criterion.
   Show the answer

   Answer: C. Annual savings are about $33,700, so payback is about 5.3 years and meets the criterion.

   Energy and maintenance savings total about $33,700 per year. The resulting 5.3-year simple payback meets the six-year criterion. Calculation: Demand reduction = 150 − 110 = 40 kW; Annual energy savings = 40 × 16 × 365 × $0.11 = $25,696; Total annual savings = $25,696 + $8,000 = $33,696; Simple payback = $180,000 ÷ $33,696 = 5.34 years.

3. Question 3

   A granular filter has 400 ft² of surface area and receives 3,600 gpm during backwash. The approved backwash rise-rate range is 13 to 15 in/min. Which conclusion is correct?

   1. The rise rate is about 9.0 in/min, below the approved range.
   2. The rise rate is about 12.0 in/min, just below the approved range.
   3. The rise rate is about 14.4 in/min, within the approved range.
   4. The rise rate is about 18.0 in/min, above the approved range.
   Show the answer

   Answer: C. The rise rate is about 14.4 in/min, within the approved range.

   Converting 9.0 gpm/ft² to vertical rise gives about 14.4 in/min, which is inside the stated 13–15 in/min range. Calculation: Backwash rate = 3,600 ÷ 400 = 9.0 gpm/ft²; Rise rate = 9.0 × 12 ÷ 7.48 = 14.44 in/min.

4. Question 4

   Three equal aeration basins each contain 6,000 lb of MLVSS. The plant receives 4,000 lb BOD/day, and the approved maximum F/M ratio during an outage is 0.25. One basin will be unavailable for one day. Equalization can defer up to 1,000 lb of that day's BOD load. Which plan keeps the operating basins at or below the approved F/M ratio?

   1. Defer 1,000 lb and treat 3,000 lb with the two available basins.
   2. Defer 500 lb and treat 3,500 lb with the two available basins.
   3. Treat all 4,000 lb with the two available basins.
   4. Keep all three basins in service and defer no load during the outage.
   Show the answer

   Answer: A. Defer 1,000 lb and treat 3,000 lb with the two available basins.

   Two basins contain 12,000 lb MLVSS, so treating 3,000 lb BOD produces an F/M of exactly 0.25. The full 1,000-lb equalization capability is required. Calculation: Available MLVSS = 2 × 6,000 = 12,000 lb; Maximum BOD at F/M 0.25 = 12,000 × 0.25 = 3,000 lb/day; Required deferral = 4,000 − 3,000 = 1,000 lb.

5. Question 5

   A nitrifying plant treats 5.0 MGD with 30 mg/L ammonia-nitrogen. Influent alkalinity is 180 mg/L as CaCO₃, and the operating plan requires at least 50 mg/L to remain. Use 7.14 mg/L alkalinity consumed per 1 mg/L ammonia-nitrogen oxidized. The alkalinity product is 50% active as CaCO₃, and the feed system can deliver 7,500 lb/day of product. Which conclusion is correct?

   1. No supplemental alkalinity is required because 180 mg/L exceeds the 50-mg/L residual target.
   2. About 7,020 lb/day of product is required, so the feed system has about 480 lb/day of capacity remaining.
   3. About 3,510 lb/day of product is required, so one-half of the feed-system capacity is unused.
   4. About 8,940 lb/day of product is required, so the feed system is short by about 1,440 lb/day.
   Show the answer

   Answer: B. About 7,020 lb/day of product is required, so the feed system has about 480 lb/day of capacity remaining.

   After accounting for nitrification consumption and the required residual, the plant needs 84.2 mg/L of active alkalinity as CaCO₃. At 50% active as CaCO₃, that requires about 7,020 lb/day of product, leaving about 480 lb/day of feed capacity. Calculation: Alkalinity consumed = 30 × 7.14 = 214.2 mg/L; Supplemental active alkalinity = 214.2 + 50 − 180 = 84.2 mg/L; Active pounds/day = 5.0 × 84.2 × 8.34 = 3,511.1 lb/day; Product pounds/day = 3,511.1 ÷ 0.50 = 7,022.1 lb/day; Remaining capacity = 7,500 − 7,022.1 = 477.9 lb/day.

Operator Master has 55 Level 4 · Expert math and calculations questions, free.

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