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Lesson plan Grade 12 Engineering & Design

Dunmore Reclamation Facility: BOD Removal Efficiency and Loading

A 50-minute grade 12 engineering lesson where students calculate BOD removal efficiency and pounds-per-day loading for a fictional secondary wastewater plant.

Objectives

  1. Students will calculate BOD removal efficiency for a wastewater plant from influent and effluent concentrations.
  2. Students will calculate the pounds per day of BOD a plant removes using the pounds formula.

Vocabulary

Influent
Influent is the incoming dirty water that arrives at a treatment plant, ready to be cleaned. Operators test it to see how much work the plant has to do.
Effluent
Effluent is the cleaned water that leaves a treatment plant at the end of the process. It is tested carefully before being released to a river, lake, or the ocean.
Secondary treatment
Secondary treatment uses billions of helpful microbes that eat the dissolved waste in the water. The plant keeps the microbes supplied with air, then settles them out, leaving much cleaner water.

Also: BOD, removal efficiency

Teacher overview

A wastewater plant is judged largely on how much of the incoming pollutant load it removes before discharging to a river. Biochemical oxygen demand (BOD) is the standard measure of that load, and secondary treatment must bring the monthly-average effluent BOD to 30 mg/L or less and remove at least 85 percent of the incoming BOD, from raw sewage that typically runs about 100 to 300 mg/L. This lesson has students calculate removal efficiency and pounds-per-day loading for the fictional Dunmore Reclamation Facility, then decide whether it meets the standard. A common misconception is that removal efficiency and the pounds of BOD removed measure the same thing; one is a percent, and the other is a mass that also depends on the plant's flow.

Procedure

Hook (5 min): Ask students how they would prove a treatment plant is actually working. Steer the discussion toward two questions: what fraction of the pollution is removed, and how many pounds of it are removed each day. Both matter, and this lesson computes each.

Direct instruction (12 min): Introduce the two formulas. Removal efficiency (percent) = (influent - effluent) / influent x 100. Pounds loading (pounds per day) = concentration (mg/L) x flow (MGD) x 8.34. Explain that efficiency is a percent that says how good the process is, while the pounds figure is a mass that says how much material is handled, and that a plant treating more flow removes more pounds even at the same efficiency.

Worked example (13 min): Work the fictional Dunmore facility, which treats a flow of 2.5 MGD. Its influent BOD is 220 mg/L and its effluent BOD is 25 mg/L. Removal efficiency = (220 - 25) / 220 x 100 = 195 / 220 x 100 = 88.6 percent. Because 88.6 percent is above 85 percent and the effluent of 25 mg/L is below the 30 mg/L limit, the plant meets the standard. Pounds of BOD removed = 195 mg/L x 2.5 MGD x 8.34 = 4,065.75 pounds per day. Pounds of BOD still leaving in the effluent = 25 mg/L x 2.5 MGD x 8.34 = 521.25 pounds per day.

Guided practice (12 min): In pairs, students evaluate a second month at Dunmore: influent 240 mg/L, effluent 30 mg/L, flow 2.5 MGD. They answer two numbered questions in order, then state whether the plant meets the effluent limit.

  1. What is the plant's BOD removal efficiency this month?
  2. How many pounds of BOD does the plant remove per day?

Closure / exit ticket (8 min): On an index card, students answer question 3: "A plant lowers BOD from 200 mg/L to 30 mg/L. What is its removal efficiency, and does it meet the 30 mg/L limit?" Collect and scan for the correct setup before dismissal.

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: 87.5 percent (guided practice, question 1). Removal efficiency = (240 - 30) / 240 x 100 = 210 / 240 x 100 = 87.5 percent. Because 87.5 percent is above 85 percent and the effluent of 30 mg/L is at the 30 mg/L limit, the plant meets the standard that month, with no margin to spare.
  2. Answer: 4,378.5 pounds per day (guided practice, question 2). Pounds of BOD removed = (240 - 30) mg/L x 2.5 MGD x 8.34 = 210 x 2.5 x 8.34 = 4,378.5 pounds per day.
  3. Answer: 85 percent (exit ticket, question 3). Removal efficiency = (200 - 30) / 200 x 100 = 170 / 200 x 100 = 85 percent. The effluent of 30 mg/L is at the 30 mg/L limit, so the plant meets the standard.

Differentiation

Support

Provide the two formulas on a printed card and work the removal-efficiency step of the guided-practice item as a class before students finish the pounds calculation. Allow a calculator throughout.

Extension

Ask students to find, for the worked Dunmore example, what effluent BOD would drop the plant below 85 percent removal at the same influent, and to explain why a plant operator watches both the percent and the pounds figures rather than either one alone.

ELL supports

Pre-teach "influent," "effluent," and "removal efficiency" with a labeled diagram of water entering and leaving a plant. Provide a sentence frame for the exit ticket: "The removal efficiency is ___ percent, so the plant ___ the limit."

Extensions

Have students compare the pounds of BOD removed by a small plant and a large plant that operate at the same removal efficiency but different flows, and write a short paragraph explaining why the two plants remove very different masses of pollutant despite equal efficiency.