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Math problem set Grade 9 Math

Netherford Wastewater Plant: BOD Removal and Loading Problem Set

A grade 9 problem set applying removal efficiency and the pounds formula to BOD at a fictional wastewater plant, checked against the 30 mg/L secondary standard.

Objectives

  1. Students will calculate BOD removal efficiency across a treatment process.
  2. Students will use the pounds formula to convert concentration and flow into a daily pollutant load.

Vocabulary

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.

Also: BOD, removal efficiency, pounds formula

Teacher overview

The Netherford Water Resource Recovery Facility (a fictional wastewater plant used for this set) measures the strength of its sewage as biochemical oxygen demand, or BOD. Operators must lower the BOD before releasing water to a river, and federal secondary-treatment rules require the monthly-average effluent BOD to be 30 mg/L or less. This set walks students through removal efficiency and the pounds formula, which turns a concentration and a flow into a daily weight of pollutant. A common misconception is treating percent removal and load in pounds as the same thing; a plant can meet a percent target while still releasing many pounds per day when the flow is large.

Netherford Wastewater Plant: BOD Removal and Loading Problem Set — Grade 9 Math
Name: ____________________________ Date: ______________

Problems

Use these facts about Netherford (a fictional plant) for every problem. Round to one decimal place unless told otherwise.

  • Plant flow: 3.5 MGD (million gallons per day).
  • Influent (raw) BOD: 220 mg/L.
  • Final effluent BOD after treatment: 18 mg/L.
  • Removal efficiency = (influent - effluent) / influent x 100.
  • Pounds formula: load (lbs/day) = concentration (mg/L) x flow (MGD) x 8.34.
  1. Find the overall BOD removal efficiency, as a percent, from influent to final effluent.
  2. Does the 18 mg/L effluent meet the secondary standard of 30 mg/L? State how much room it has to spare.
  3. Find the influent BOD load in pounds per day using the pounds formula.
  4. Find the final effluent BOD load in pounds per day.
  5. Find the pounds of BOD removed per day (influent load minus effluent load).
  6. The primary clarifier alone removes 35 percent of the influent BOD. Find the BOD concentration, in mg/L, leaving the primary clarifier.
  7. The secondary process must lower the BOD from the item 6 value down to the final 18 mg/L. Find the removal efficiency the secondary process must reach.
  8. During a storm the flow rises to 5.0 MGD while the influent BOD drops to 160 mg/L. Find the new influent BOD load in pounds per day.
  9. The plant's design load limit is 7,000 lbs/day of BOD. Using item 8, how many pounds of headroom remain below the limit?
  10. Write and solve an equation for the maximum effluent BOD load, in pounds per day, that still meets the 30 mg/L standard at the storm flow of 5.0 MGD.

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: 91.8 percent. (220 - 18) / 220 x 100 = 202 / 220 x 100 = 91.8 percent.
  2. Answer: Yes, by 12 mg/L. The effluent is 18 mg/L, which is below the 30 mg/L limit, leaving 30 - 18 = 12 mg/L of room.
  3. Answer: 6,421.8 lbs/day. Load = 220 mg/L x 3.5 MGD x 8.34 = 770 x 8.34 = 6,421.8 pounds per day.
  4. Answer: 525.4 lbs/day. Load = 18 mg/L x 3.5 MGD x 8.34 = 63 x 8.34 = 525.4 pounds per day.
  5. Answer: 5,896.4 lbs/day. Removed load = 6,421.8 - 525.4 = 5,896.4 pounds per day. (Checking with the removed concentration: 202 mg/L x 3.5 x 8.34 = 5,896.4.)
  6. Answer: 143 mg/L. The clarifier removes 35 percent, leaving 65 percent: 220 x 0.65 = 143 mg/L.
  7. Answer: 87.4 percent. (143 - 18) / 143 x 100 = 125 / 143 x 100 = 87.4 percent.
  8. Answer: 6,672 lbs/day. Load = 160 mg/L x 5.0 MGD x 8.34 = 800 x 8.34 = 6,672 pounds per day.
  9. Answer: 328 lbs/day. Headroom = 7,000 - 6,672 = 328 pounds per day below the design limit.
  10. Answer: 1,251 lbs/day. Set load = 30 mg/L x 5.0 MGD x 8.34 = 150 x 8.34 = 1,251 pounds per day as the maximum effluent load allowed at that flow.

Differentiation

Support

Provide the removal-efficiency and pounds formulas on a printed card, and pre-fill the first multiplication step of item 3 (220 x 3.5 = 770) so students focus on the 8.34 step. Allow calculators, and let students check item 5 with either subtraction method.

Extension

Ask students to find the influent BOD concentration in mg/L that would exactly meet the 7,000 lbs/day design limit at the plant's normal flow, working backward through the pounds formula, and to explain why lower flows give the plant more room under a fixed pound limit.

ELL supports

Provide a simple flow diagram of the plant (influent, primary clarifier, secondary process, effluent) with the BOD value labeled at each stage. Pre-teach "influent," "effluent," and "load," and post the pounds formula with each variable's unit written beneath it.