Lesson plan Grade 12 Science

Phosphorus Recovery via Struvite at Ashgrove Reclamation

A grade 12 lesson on struvite precipitation, where students calculate phosphorus loading and recovery at a fictional plant and treat a nutrient as a resource.

Objectives

  1. Students will explain how struvite precipitation recovers phosphorus from wastewater as a marketable fertilizer product.
  2. Students will calculate phosphorus loading and the mass recovered using the pounds formula and a stated recovery efficiency.
  3. Students will evaluate why treating a nutrient as both a pollutant and a resource changes how a utility manages it.

Vocabulary

Eutrophication
Eutrophication is what happens when a lake or bay gets overfed with nutrients: algae grows out of control, and when it dies and rots, the water loses the oxygen that fish need.
Biosolids
Biosolids are the nutrient-rich solid material left over after sewage sludge has been treated and made safe. Many farms use treated biosolids as fertilizer for their fields.

Also: struvite, phosphorus recovery, nutrient loading

Teacher overview

Ashgrove Water Reclamation Facility, a fictional plant used for this lesson, is installing a struvite precipitation process to pull phosphorus out of its wastewater stream before it reaches a receiving lake. Phosphorus is a nutrient in two senses at once: left in the effluent, it fuels the algae growth that can crash dissolved oxygen below the roughly 5 mg/L most fish need to survive, while recovered as solid struvite crystals, it becomes a slow-release fertilizer a utility can sell. This lesson uses the pounds formula to size how much phosphorus Ashgrove's process handles and recovers each day.

A common misconception in this math is assuming that a nutrient-recovery process is purely an environmental cost with no offsetting value. The guided and independent practice steps are built so students calculate a specific recovered mass, in pounds per day, that Ashgrove's plant could market, making the "resource" half of the topic as concrete as the "pollutant" half.

Procedure

Hook (5 min): Ask students what they think happens to the phosphorus in wastewater after it leaves a home, and whether they consider phosphorus a pollutant, a resource, or both. Record initial responses without comment.

Direct instruction (15 min): Explain struvite precipitation: operators add magnesium to phosphorus-rich wastewater, causing magnesium ammonium phosphate crystals, called struvite, to form and settle out, where they can be harvested and sold as fertilizer. Model the loading calculation using the pounds formula: at Ashgrove's flow of 5 MGD and an influent phosphorus concentration of 8 mg/L, Loading (lbs/day) = Concentration (mg/L) x Flow (MGD) x 8.34 = 8 x 5 x 8.34 = 333.6 lbs/day of phosphorus entering the process. Real plants run this process on the small, phosphorus-rich liquid stream left after digested sludge is dewatered, where concentrations are far higher than in raw influent. This lesson simplifies by applying one recovery rate to the whole influent load.

Guided practice (15 min): In pairs, students calculate how much of that phosphorus Ashgrove's process actually recovers, given a stated recovery efficiency of 85 percent: Recovered = 333.6 lbs/day x 0.85 = 283.56 lbs/day, which rounds to about 283.6 lbs/day. They then subtract to find how much phosphorus remains in the effluent stream: 333.6 - 283.56 = 50.04 lbs/day, which rounds to about 50.0 lbs/day, and discuss how that remaining amount compares to the original 333.6 lbs/day.

Independent practice (10 min): Students solve two related problems. First, they recalculate phosphorus loading if Ashgrove's flow rises to 6 MGD during wet weather while influent concentration stays at 8 mg/L. Second, they compare Ashgrove's 283.6 lbs/day recovered to a fictional neighboring facility, Dunmore Reclamation, which recovers phosphorus at only 60 percent efficiency from the same 333.6 lbs/day loading.

Closure / exit ticket (5 min): Students respond in two or three sentences to this prompt: why might a wastewater plant invest in phosphorus recovery technology even though it costs money to install, considering both the pollutant and resource sides of the nutrient? Collect responses and scan for whether students reference both sides before dismissal.

Answer Key

  1. Answer: 400.3 lbs/day.
    Loading = 8 mg/L x 6 MGD x 8.34 = 400.32 lbs/day, which rounds to about 400.3 lbs/day.

  2. Answer: 83.4 lbs/day less.
    Dunmore's recovered mass = 333.6 lbs/day x 0.60 = 200.16 lbs/day. Compared to Ashgrove's 283.56 lbs/day recovered at 85 percent efficiency, Dunmore recovers about 283.56 - 200.16 = 83.4 lbs/day less.

  3. Exit ticket answer: Answers will vary but should reference both sides of the nutrient.
    A strong response notes that recovering phosphorus reduces the amount discharged to the receiving lake, lowering the risk of algae blooms and oxygen crashes that harm fish, while the recovered struvite can be sold as fertilizer, which can offset some or all of the technology's installation and operating costs over time.

Differentiation

Support (below grade level)

Provide the pounds formula pre-filled with Ashgrove's flow and concentration values so students only need to complete the multiplication, and work through the 85 percent recovery calculation together as a class before releasing pairs to the independent problems.

Extension (above grade level)

Ask students to determine what recovery efficiency Ashgrove's process would need to reach in order to recover exactly 300 lbs/day of phosphorus from the same 333.6 lbs/day loading, setting up and solving an equation for the unknown efficiency.

ELL supports

Pre-teach "struvite," "loading," and "recovery efficiency" with a simple diagram showing wastewater entering a tank, crystals settling out, and two labeled output arrows, one for treated water and one for harvested struvite. Allow the exit ticket to be answered as a two-column list (pollutant side / resource side) instead of full sentences.