Math problem set Grade 9 Math

Kettleridge Wastewater Plant: BOD Loading and the Pounds Formula

A grade 9 problem set using the pounds formula to track BOD loading and removal at Kettleridge's wastewater plant, then writing an equation for load from concentration.

Objectives

  1. Students will apply the pounds formula to find daily loading from concentration and flow.
  2. Students will calculate removal efficiency as a percent.
  3. Students will write and solve an equation relating load to concentration at a fixed flow.

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.

Also: pounds formula, BOD, removal efficiency

Teacher overview

The fictional Kettleridge Water Resource Recovery Facility cleans sewage before returning it to a river. Operators use the pounds formula, where loading in pounds per day equals concentration in mg/L times flow in MGD times 8.34, to track how much material moves through the plant each day. This set follows one day's flow from dirty water to treated water, then turns the pounds formula into an equation students solve for a missing concentration. A common misconception is putting flow in gallons per minute into the pounds formula instead of MGD, or dropping the 8.34 factor.

Answer Key

  1. Answer: 9,174 lb/day. 220 mg/L x 5 MGD x 8.34 = 9,174 pounds per day. (220 x 5 = 1,100; 1,100 x 8.34 = 9,174.)
  2. Answer: 1,251 lb/day. 30 mg/L x 5 MGD x 8.34 = 1,251 pounds per day. (30 x 5 = 150; 150 x 8.34 = 1,251.)
  3. Answer: 7,923 lb/day. 9,174 - 1,251 = 7,923 pounds per day of BOD removed.
  4. Answer: 86 percent. (220 - 30) / 220 x 100 = 190 / 220 x 100 = 86.36%, about 86%.
  5. Answer: P = 41.7c. Since 5 MGD x 8.34 = 41.7, the effluent load is P = c x 41.7 = 41.7c.
  6. Answer: 20 mg/L. 834 = 41.7c, so c = 834 / 41.7 = 20 mg/L.
  7. Answer: 83.4 lb/day. 2.0 mg/L x 5 MGD x 8.34 = 83.4 pounds per day of chlorine. (2.0 x 5 = 10; 10 x 8.34 = 83.4.)
  8. Answer: 333.6 lb/day. 8 mg/L x 5 MGD x 8.34 = 333.6 pounds per day of nitrogen. (8 x 5 = 40; 40 x 8.34 = 333.6.)

Differentiation

Support

Give students the pounds formula on a card with its three slots labeled (concentration, flow, and the 8.34 factor) and one filled example that is not in this set. Provide the product 5 times 8.34 equal to 41.7 already computed for problems 5 and 6, and allow a calculator.

Extension

Ask students to find the plant's BOD load if the flow rises to 6 million gallons per day while the influent concentration stays the same, then explain in one sentence why the load grows even though the concentration did not change.

ELL supports

Pre-teach "influent," "effluent," "load," and "removal" with arrows on a simple plant diagram showing dirty water going in and treated water coming out. Provide a sentence frame for problem 4: "The removal efficiency is ___ percent because ___."