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Math problem set Grade 12 Math

Ostwick Recovery Facility: Nutrient Loading and Phosphorus Recovery

A grade 12 problem set on nutrient loads at a recovery facility: nitrogen and phosphorus loading, removal efficiency, and phosphorus recovered as fertilizer.

Objectives

  1. Students will convert nutrient concentrations to daily loads in pounds using the pounds formula.
  2. Students will calculate nutrient removal efficiency and the mass of phosphorus recovered as a resource.

Vocabulary

Key words: nutrient loading, phosphorus, nitrogen, removal efficiency, resource recovery

Teacher overview

Nitrogen and phosphorus are nutrients that plants need, yet in a river they feed algae blooms that strip oxygen from the water. Modern plants remove them and, increasingly, recover the phosphorus as struvite, a slow-release fertilizer. This set follows the fictional Ostwick Water Resource Recovery Facility through the nutrient math an operator and a process engineer run together. A common misconception to correct is comparing influent and effluent concentrations in mg/L and stopping there; the amount of nutrient actually kept out of the river is a daily load in pounds, which depends on flow as well as concentration.

Ostwick Recovery Facility: Nutrient Loading and Phosphorus Recovery — Grade 12 Math
Name: ____________________________ Date: ______________

Problems

Ostwick Water Resource Recovery Facility is a fictional plant that treats 4.0 MGD. Use the pounds formula (pounds per day equals concentration in mg/L times flow in MGD times 8.34) and these readings for every problem.

  • Influent total phosphorus: 6.5 mg/L
  • Effluent total phosphorus: 1.2 mg/L
  • Influent total nitrogen: 35 mg/L
  • Effluent total nitrogen: 8 mg/L
  • Struvite recovery captures 25 percent of the phosphorus the plant removes each day
  • The recovered phosphorus product sells for $3.20 per pound of phosphorus
  1. Calculate the influent phosphorus load in pounds per day using the pounds formula.
  2. Calculate the effluent phosphorus load in pounds per day.
  3. How many pounds of phosphorus per day does the plant remove from the wastewater?
  4. Calculate the phosphorus removal efficiency as a percent. Round to the nearest tenth.
  5. The struvite process recovers 25 percent of the phosphorus removed in problem 3. How many pounds of phosphorus per day are recovered?
  6. Calculate the influent nitrogen load in pounds per day.
  7. Calculate the effluent nitrogen load in pounds per day, then find how many pounds of nitrogen per day the plant removes.
  8. Calculate the nitrogen removal efficiency as a percent. Round to the nearest tenth.
  9. Using the recovered phosphorus from problem 5 and the $3.20-per-pound price, find the daily revenue from the recovered phosphorus product. Round to the nearest dollar.
  10. Phosphorus is described as both a pollutant and a resource in this scenario. Explain what makes it each one, referring to the river downstream and to the recovered product.

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: 216.8 pounds per day. Phosphorus load equals 6.5 times 4.0 times 8.34: 6.5 times 4.0 is 26; 26 times 8.34 is 216.84, or about 216.8 pounds per day.
  2. Answer: 40.0 pounds per day. Effluent load equals 1.2 times 4.0 times 8.34: 1.2 times 4.0 is 4.8; 4.8 times 8.34 is 40.03, or about 40.0 pounds per day.
  3. Answer: 176.8 pounds per day. The plant removes 216.84 minus 40.03, which is 176.81, or about 176.8 pounds per day.
  4. Answer: 81.5 percent. Removal efficiency equals (6.5 minus 1.2) divided by 6.5, times 100: 5.3 divided by 6.5 is 0.8154, or 81.5 percent.
  5. Answer: 44.2 pounds per day. Recovery is 0.25 times 176.81, which is 44.20, or about 44.2 pounds of phosphorus per day.
  6. Answer: 1,167.6 pounds per day. Nitrogen load equals 35 times 4.0 times 8.34: 35 times 4.0 is 140; 140 times 8.34 is 1,167.6 pounds per day.
  7. Answer: 900.7 pounds per day. Effluent nitrogen load is 8 times 4.0 times 8.34, which is 266.88 pounds per day; removed nitrogen is 1,167.6 minus 266.88, which is 900.72, or about 900.7 pounds per day.
  8. Answer: 77.1 percent. Removal efficiency equals (35 minus 8) divided by 35, times 100: 27 divided by 35 is 0.7714, or 77.1 percent.
  9. Answer: $141 per day. Daily revenue is 44.20 pounds times $3.20 per pound, which is $141.44, or about $141 per day.
  10. Answer: Phosphorus is a pollutant in the river and a resource once recovered. Released downstream, it feeds algae that use up dissolved oxygen and harm fish; captured as struvite, the same phosphorus becomes a slow-release fertilizer the plant can sell, turning a pollutant into a product.

Differentiation

Support

Provide the pounds formula on a reference card with one problem fully worked, and give students the flow value already substituted so they focus on the concentration and the 8.34 factor. Let them complete problem 2 by mirroring problem 1.

Extension

Ask students to compute the annual revenue from recovered phosphorus at the fictional plant, then estimate how many years of that revenue would offset a one-time $250,000 cost to install the recovery equipment, stating any assumption they make.

ELL supports

Pre-teach "nutrient," "load," "removal," and "recovery" with a diagram tracing phosphorus from influent to effluent and to the recovered product. Provide the sentence frame "The load is ___ pounds per day because ___" for problems 1 and 6.

Extensions

Have students research one real water resource recovery facility that recovers struvite and compare its reported recovery rate with the fictional 25 percent used here, discussing why a real plant's rate might be higher or lower.