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Worksheet Grade 11 Engineering & Design

Bristow Reclamation: BOD and TSS Removal-Efficiency Worksheet

A grade 11 worksheet where students compute BOD and TSS removal efficiency and loading for the fictional Bristow plant and check it against the secondary standard.

Objectives

  1. Students will calculate BOD and TSS removal efficiency and compare each against the secondary treatment standard.
  2. Students will use the pounds formula to find influent and effluent loading and the mass of BOD removed.

Vocabulary

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, TSS, removal efficiency, loading

Teacher overview

A water reclamation plant is judged on how much of the incoming pollution it removes before returning water to a river. Two of the most-watched numbers are BOD, a measure of the oxygen-demanding organic load, and TSS, the total suspended solids. Secondary treatment must bring both down to a 30-day average of 30 mg/L or less and remove at least 85 percent of each. This worksheet uses a fictional plant, Bristow Water Reclamation, to practice removal-efficiency and loading calculations, then checks the results against that standard. A common mistake is confusing concentration with load; a concentration in mg/L only becomes a mass in pounds per day once you multiply by the flow and the 8.34 factor.

Bristow Reclamation: BOD and TSS Removal-Efficiency Worksheet — Grade 11 Engineering & Design
Name: ____________________________ Date: ______________

Student Worksheet

Use these facts about the fictional Bristow plant for every item:

  • The plant treats a flow of 4 MGD (million gallons per day).
  • Influent BOD is 240 mg/L; effluent BOD is 20 mg/L.
  • Influent TSS is 260 mg/L; effluent TSS is 18 mg/L.
  • Removal efficiency = (influent − effluent) ÷ influent × 100.
  • Loading in pounds per day = concentration (mg/L) × flow (MGD) × 8.34.
  • Secondary treatment requires effluent BOD and TSS of 30 mg/L or less and at least 85 percent removal of each. Raw sewage BOD usually runs 100 to 300 mg/L. Most fish need dissolved oxygen above 5 mg/L.
  1. Find the BOD removal efficiency as a percent.
  2. Does the 20 mg/L effluent BOD meet the 30 mg/L secondary standard?
  3. Does the BOD removal efficiency meet the 85 percent minimum removal requirement?
  4. Find the influent BOD loading in pounds per day, using the pounds formula.
  5. Find the effluent BOD loading in pounds per day.
  6. Find the mass of BOD removed per day, using item 4 minus item 5.
  7. Find the TSS removal efficiency as a percent.
  8. Does the 18 mg/L effluent TSS meet the 30 mg/L secondary standard?
  9. Raw sewage BOD usually runs 100 to 300 mg/L. Is Bristow's 240 mg/L influent BOD inside that range?
  10. Name the treatment step that removes most of the BOD, and explain why the 30 mg/L limit protects the river the plant discharges into.

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: 92 percent. BOD removal = (240 − 20) ÷ 240 × 100 = 220 ÷ 240 × 100 = 91.7, about 92 percent.
  2. Answer: Yes. Effluent BOD of 20 mg/L is below the 30 mg/L secondary standard.
  3. Answer: Yes. At about 92 percent, the removal exceeds the 85 percent minimum removal required for secondary treatment.
  4. Answer: 8,006.4 lbs/day. Influent loading = 240 × 4 × 8.34 = 8,006.4 pounds per day.
  5. Answer: 667.2 lbs/day. Effluent loading = 20 × 4 × 8.34 = 667.2 pounds per day.
  6. Answer: 7,339.2 lbs/day. BOD removed = 8,006.4 − 667.2 = 7,339.2 pounds per day.
  7. Answer: 93 percent. TSS removal = (260 − 18) ÷ 260 × 100 = 242 ÷ 260 × 100 = 93.1, about 93 percent.
  8. Answer: Yes. Effluent TSS of 18 mg/L is below the 30 mg/L secondary standard.
  9. Answer: Yes. The 240 mg/L influent BOD falls inside the usual 100-to-300 mg/L range for raw sewage.
  10. Answer: The biological stage. The aeration or activated-sludge step, where bacteria consume the organic matter, removes most of the BOD. The 30 mg/L limit protects the river because organic load consumes dissolved oxygen as it breaks down; holding effluent BOD low keeps river oxygen above the roughly 5 mg/L most fish need.

Differentiation

Support

Provide the removal-efficiency and pounds formulas on a reference strip, and pre-label each item with the unit its answer should carry (percent, pounds per day) so students can check that their result matches the expected unit.

Extension

Ask students to find the highest influent BOD, in mg/L, that Bristow could receive and still meet the 30 mg/L effluent standard if its BOD removal efficiency held at 92 percent, and to state what the plant would do if incoming load rose past that point.

ELL supports

Pre-teach "influent," "effluent," "load," and "removal efficiency" with a simple before-and-after diagram showing water entering dirty and leaving cleaner. Allow yes-or-no items to be answered with the word plus a short reason rather than a full sentence.