Lesson plan Grade 10 Engineering & Design

Osprey Flats Secondary Treatment: A Wastewater Design Lesson

A 55-minute grade 10 engineering lesson: students split a wastewater plant into stages and find the removal efficiency needed to meet the secondary standard.

Objectives

  1. Students will identify the stages of a wastewater treatment train and the job each stage performs.
  2. Students will calculate removal efficiency and use it to check a design against the secondary standard.
  3. Students will explain how the biological stage relies on microbes to remove dissolved organic matter.

Vocabulary

Aeration
Aeration means mixing air into water. At a wastewater plant, blowers bubble huge amounts of air through tanks so the helpful microbes have the oxygen they need to eat the waste.

Also: treatment train, BOD, removal efficiency, secondary standard

Teacher overview

A wastewater plant does not clean sewage in one step; it chains several stages, each solving one part of the problem. In this lesson, students study the fictional Osprey Flats Water Pollution Control Plant and break its process into stages, then focus on the biological (secondary) stage, where microbes in an aeration tank consume dissolved organic matter measured as biochemical oxygen demand (BOD). Students calculate how much BOD each stage must remove for the plant to meet the federal secondary standard of 30 mg/L. A common misconception this lesson addresses is that removal percentages simply add up across stages: because each stage acts on what is left after the previous one, students must apply each removal to the running concentration, not to the original raw value.

Procedure

Hook (5 min): Ask students what happens to everything that goes down a drain in their town. Show a simple diagram of a treatment train with the stage labels hidden, and ask pairs to guess, in order, what each of the four boxes might do before dirty water can be returned to a river.

Direct instruction (12 min): Reveal the stages and the job each performs: screening removes rags and grit; primary settling lets heavy solids sink; the biological stage uses microbes in an aerated tank to eat dissolved organic matter (the environmental-biotechnology step); and the secondary clarifier settles the microbes back out. Introduce BOD as a measure of that dissolved organic load, and state the secondary standard: effluent BOD averaging 30 mg/L or less over 30 days, and removal of at least 85 percent of the incoming BOD, starting from raw sewage that typically runs about 100 to 300 mg/L.

Model a worked example (10 min): Using Osprey Flats (a fictional plant) with a raw influent BOD of 220 mg/L, model the running-concentration idea. If primary settling removes 35 percent of the BOD, the concentration after primary is 220 × (1 − 0.35) = 220 × 0.65 = 143 mg/L. Then the secondary stage must bring 143 down to the 30 mg/L standard, a removal of (143 − 30) ÷ 143 × 100 = about 79 percent. Finally show the overall removal from raw to final: (220 − 30) ÷ 220 × 100 = about 86 percent, which meets the 85 percent minimum removal.

Guided practice (13 min): In pairs, students work a second plant from the handout with a raw influent BOD of 260 mg/L and a final effluent of 24 mg/L. They calculate the overall removal efficiency and decide whether the 24 mg/L effluent meets the 30 mg/L secondary standard. Circulate and confirm pairs subtract effluent from influent before dividing.

Independent practice (10 min): Each student solves the exit-style setup on the handout: a plant that achieves exactly 85 percent removal on an influent BOD of 200 mg/L. They find the effluent BOD and state whether it meets the standard.

Closure and exit ticket (5 min): Students write one sentence naming which stage does the most to lower BOD and why, then hand it in. Scan for correct identification of the biological (aeration) stage before dismissal.

Answer Key

  1. Answer: 90.8 percent.
    Guided-practice overall removal: (260 − 24) ÷ 260 × 100 = 236 ÷ 260 × 100 = 90.77, which rounds to about 90.8 percent.

  2. Answer: Yes.
    The 24 mg/L effluent is below the 30 mg/L secondary standard, so this plant meets the requirement.

  3. Answer: 30 mg/L.
    Independent practice: effluent = 200 × (1 − 0.85) = 200 × 0.15 = 30 mg/L, which exactly meets the secondary standard.

  4. Answer: The biological (aeration) stage.
    Accept any answer identifying the aeration or secondary biological stage, since the microbes there consume the dissolved organic matter that makes up most of the remaining BOD after primary settling.

Differentiation

Support (below grade level)

Provide the treatment-train diagram with the stages already labeled and the removal-efficiency formula written with labeled blanks, and complete the primary-settling step of the worked example as a class so students begin the secondary calculation from the after-primary value.

Extension (above grade level)

Ask students to determine what removal efficiency the secondary stage alone would need if primary settling were skipped entirely at Osprey Flats, then explain why plants use primary settling first even though the microbes could, in principle, do more of the work.

ELL supports

Pre-teach treatment train, BOD, aeration, and removal efficiency with a labeled diagram and simple icons for each stage. Provide sentence frames for the closure ticket: "The ___ stage removes the most BOD because ___," and allow calculations to be shown as numbered steps.