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Math problem set Grade 12 Math

Brindlemere Reclamation Plant: BOD Loading and Removal

A grade 12 problem set using the pounds formula to find BOD mass loading and removal efficiency at a fictional reclamation plant.

Objectives

  1. Students will compute BOD mass loading in pounds per day using the pounds formula.
  2. Students will find removal efficiency and compare a plant's performance to the secondary treatment standard.

Vocabulary

Biochemical oxygen demand (BOD)
Biochemical oxygen demand, or BOD, measures how much oxygen tiny microbes will use up as they eat the waste in water. Dirtier water has a higher BOD.

Also: pounds formula, mass loading, removal efficiency

Teacher overview

Brindlemere Water Reclamation Plant is a fictional plant that removes organic pollution, measured as biochemical oxygen demand (BOD), before returning treated water to a river. Operators track BOD two ways: as a concentration in milligrams per liter and as a mass loading in pounds per day. The pounds formula ties them together through the flow. This set has students compute loading, removal efficiency, and compliance with the secondary treatment standard. A common error is comparing concentrations alone when the flow changes; two days with very different concentrations can carry the same pounds of BOD, so mass loading is what actually stresses the treatment process.

Brindlemere Reclamation Plant: BOD Loading and Removal — Grade 12 Math
Name: ____________________________ Date: ______________

Problems

Brindlemere Water Reclamation Plant (a fictional facility) treats 3.5 MGD. The influent BOD is 240 mg/L and the effluent BOD is 18 mg/L. Use these relationships:

  • Pounds formula: loading (lbs/day) = concentration (mg/L) x flow (MGD) x 8.34.
  • Removal efficiency (percent) = (influent - effluent) / influent x 100.
  • The secondary treatment standard sets a monthly-average effluent BOD of 30 mg/L or less and calls for roughly 85 percent removal.
  1. Find the influent BOD loading in pounds per day, using 240 mg/L and 3.5 MGD.
  2. Find the effluent BOD loading in pounds per day, using 18 mg/L and 3.5 MGD.
  3. Find the pounds of BOD removed each day, using your answers from problems 1 and 2.
  4. Find the removal efficiency as a percent, using the influent and effluent concentrations.
  5. Does the effluent BOD of 18 mg/L meet the secondary standard of 30 mg/L or less?
  6. Does the removal efficiency from problem 4 meet the roughly 85 percent removal that secondary treatment calls for?
  7. During wet weather the flow rises to 5.0 MGD while dilution lowers the influent BOD to 168 mg/L. Find the new influent loading in pounds per day, and compare it to your answer from problem 1.
  8. Suppose the plant must keep its effluent loading under 600 pounds per day at the normal 3.5 MGD flow. What is the highest effluent BOD concentration, in mg/L, that stays under that limit?
  9. Explain why an operator watches pounds of BOD per day rather than concentration alone when the daily flow changes.

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: 7,006 lbs/day.
    Loading = 240 mg/L x 3.5 MGD x 8.34 = 7,005.6, which rounds to about 7,006 pounds per day.

  2. Answer: 525 lbs/day.
    Loading = 18 mg/L x 3.5 MGD x 8.34 = 525.42, which rounds to about 525 pounds per day.

  3. Answer: 6,480 lbs/day.
    Removed = influent loading - effluent loading = 7,005.6 - 525.42 = 6,480.2, about 6,480 pounds per day.

  4. Answer: 92.5 percent.
    Removal efficiency = (240 - 18) / 240 x 100 = 222 / 240 x 100 = 92.5 percent.

  5. Answer: Yes.
    The effluent BOD of 18 mg/L is below the 30 mg/L monthly-average standard, so it meets the limit.

  6. Answer: Yes.
    The removal efficiency of 92.5 percent is above the roughly 85 percent removal that secondary treatment calls for.

  7. Answer: 7,006 lbs/day.
    New loading = 168 mg/L x 5.0 MGD x 8.34 = 7,005.6, about 7,006 pounds per day. This equals problem 1, because a lower concentration carried by a higher flow can deliver the same mass load.

  8. Answer: 20.5 mg/L.
    Let C be the effluent concentration. C x 3.5 x 8.34 = 600, so C x 29.19 = 600 and C = 600 / 29.19 = 20.55 mg/L. Rounding up to 20.6 mg/L would give about 601 pounds per day, over the limit, so the highest concentration to the nearest tenth that stays under 600 pounds per day is 20.5 mg/L. Also accept 20.55 mg/L from a student who reports the boundary to the hundredth.

  9. Answer: Mass loading, not concentration, measures the actual load on the plant.
    Problem 7 shows two very different concentrations that produce the same pounds of BOD once the flow is included. The treatment biology responds to the total mass of organic matter it receives each day, so pounds per day is the quantity that tells an operator how hard the process is working.

Differentiation

Support

Provide the pounds formula and the removal-efficiency formula on a labeled card, and pre-compute problem 1 so students can start the removed-pounds step from a known influent loading. Allow calculators and accept answers rounded to the nearest whole pound.

Extension

Ask students to find the effluent concentration that would drop the removal efficiency to exactly the 85 percent removal floor at the plant's normal influent concentration, and to explain what that says about how much room the plant has before it risks a violation.

ELL supports

Provide a flow diagram of the plant labeling influent, the treatment process, and effluent, with "loading," "concentration," and "removal efficiency" glossed. Offer a sentence frame for problem 9: "Concentration alone can be misleading because ___, so operators track ___."

Extensions

Have students find a real facility's discharge report, often public through a state permit portal, and compare its reported effluent BOD and percent removal with Brindlemere's results and with the secondary standard.