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Math problem set Grade 10 Math

Cattail Slough Phosphorus Load: A Grade 10 Nutrient Removal Set

Grade 10 nutrient-control math: students use the pounds formula to find phosphorus loading, the removal needed, and the fertilizer recovered at a recovery facility.

Objectives

  1. Students will apply the pounds formula to calculate phosphorus loading and required removal at a water resource recovery facility.
  2. Students will calculate the mass of phosphorus recovered as fertilizer and interpret nutrients as both a pollutant and a resource.

Vocabulary

Key words: pounds formula, phosphorus, loading, removal efficiency, nutrient recovery

Teacher overview

Phosphorus is a nutrient that plants need, yet too much of it in a river feeds algae blooms that strip the water of oxygen, so a water resource recovery facility both removes phosphorus and, increasingly, captures it as a fertilizer product. This set has students size the phosphorus job at the fictional Cattail Slough Water Resource Recovery Facility using the pounds formula. A common misconception in this math is confusing a concentration with a load: the same 6.0 mg/L of phosphorus is a much larger daily mass at a high flow than at a low flow, which is why engineers convert concentration to pounds per day before they design a removal or recovery step.

Cattail Slough Phosphorus Load: A Grade 10 Nutrient Removal Set — Grade 10 Math
Name: ____________________________ Date: ______________

Problems

Cattail Slough (a fictional facility built for this set) treats a steady 4.0 MGD. Its incoming (influent) water carries 6.0 mg/L of total phosphorus. The facility's discharge permit allows an effluent of 1.0 mg/L of total phosphorus, and after an upgrade the plant reaches an effluent of 0.8 mg/L. Use the pounds formula, Loading (lbs/day) = Concentration (mg/L) × Flow (MGD) × 8.34, and removal efficiency = (influent − effluent) ÷ influent × 100.

  1. Find the influent phosphorus load in pounds per day, using 6.0 mg/L, the flow of 4.0 MGD, and the 8.34 factor.

  2. Find the phosphorus load the permit allows in the effluent, using the 1.0 mg/L permit limit and the flow of 4.0 MGD.

  3. Find how many pounds of phosphorus per day the facility must remove to bring the influent load (problem 1) down to the permit-allowed load (problem 2).

  4. Find the removal efficiency, as a percentage, needed to go from the 6.0 mg/L influent to the 1.0 mg/L permit limit.

  5. Find the removal efficiency the upgrade actually reaches, going from 6.0 mg/L influent to the 0.8 mg/L effluent.

  6. Find the effluent phosphorus load in pounds per day at the upgraded effluent of 0.8 mg/L and the flow of 4.0 MGD.

  7. Find how many pounds of phosphorus per day the upgraded plant removes, subtracting the effluent load (problem 6) from the influent load (problem 1).

  8. The facility recovers 40 percent of the phosphorus it removes as a struvite fertilizer. Find the pounds of phosphorus recovered per day, using your answer from problem 7.

  9. During a wet-weather peak the flow rises to 5.5 MGD while the influent stays at 6.0 mg/L. Find the new influent load in pounds per day.

  10. Explain how the same phosphorus can be both a pollutant the facility must remove and a resource the facility can sell.

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key

  1. Answer: 200.16 lbs/day.
    Loading = 6.0 mg/L × 4.0 MGD × 8.34 = 24 × 8.34 = 200.16 pounds per day.

  2. Answer: 33.36 lbs/day.
    Loading = 1.0 mg/L × 4.0 MGD × 8.34 = 4 × 8.34 = 33.36 pounds per day.

  3. Answer: 166.8 lbs/day.
    Required removal = 200.16 − 33.36 = 166.8 pounds per day.

  4. Answer: about 83.3 percent.
    Removal efficiency = (6.0 − 1.0) ÷ 6.0 × 100 = 5 ÷ 6 × 100 = 83.33, which rounds to about 83.3 percent.

  5. Answer: about 86.7 percent.
    Removal efficiency = (6.0 − 0.8) ÷ 6.0 × 100 = 5.2 ÷ 6 × 100 = 86.67, which rounds to about 86.7 percent.

  6. Answer: about 26.69 lbs/day.
    Loading = 0.8 mg/L × 4.0 MGD × 8.34 = 3.2 × 8.34 = 26.688, which rounds to about 26.69 pounds per day.

  7. Answer: about 173.47 lbs/day.
    Removed = 200.16 − 26.69 = 173.47 pounds per day.

  8. Answer: about 69.39 lbs/day.
    Recovered = 0.40 × 173.47 = 69.388, which rounds to about 69.39 pounds per day.

  9. Answer: 275.22 lbs/day.
    Loading = 6.0 mg/L × 5.5 MGD × 8.34 = 33 × 8.34 = 275.22 pounds per day.

  10. Answer: Phosphorus is both.
    Accept any answer explaining that phosphorus released to the river is a pollutant, because it overfeeds algae and lowers the oxygen fish need, while the same phosphorus captured as struvite becomes a slow-release fertilizer the facility can sell, turning a waste it must control into a product with value.

Differentiation

Support (below grade level)

Provide the pounds formula and the removal-efficiency formula on a reference card with the 8.34 factor already written in, and work problem 1 together so students see how concentration, flow, and the factor multiply before they reuse the pattern in problems 2, 6, and 9.

Extension (above grade level)

Ask students to find what influent phosphorus concentration, in mg/L, would put the upgraded plant's effluent right at the 1.0 mg/L permit limit if the removal efficiency from problem 5 held steady, and to explain why a rising influent concentration threatens permit compliance even when the plant's percentage removal does not change.

ELL supports

Provide a labeled diagram of the treatment train marking where phosphorus enters, where it is removed, and where struvite is recovered, with the terms load, removal efficiency, and recovery defined beside it. Allow students to show problems 1 through 9 as numbered calculations and to answer problem 10 with a labeled sketch plus one sentence.