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Worksheet Grade 11 Math

Thackery Mills Reclamation Plant: Loading and Removal Efficiency

A grade 11 worksheet applying the pounds formula and removal-efficiency percentages to a fictional water reclamation facility's BOD and TSS data.

Objectives

  1. Students will compute pollutant loading in pounds per day from concentration and flow using the pounds formula.
  2. Students will calculate removal efficiency as a percentage from influent and effluent concentrations.
  3. Students will choose an appropriate quantity for reporting treatment performance.

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, TSS, loading, removal efficiency

Teacher overview

This worksheet works through one fictional plant, the Thackery Mills Water Reclamation Facility, using the same pounds formula that appears on operator licensing exams. Students convert concentrations and flow into daily pollutant loads, then measure how well the plant removes organic material and solids. The federal secondary-treatment standard, an effluent BOD and TSS of 30 mg/L or less, gives students a real benchmark to check the plant against. A common misconception this worksheet corrects is treating a low effluent concentration and a high removal percentage as the same claim, when they answer different questions.

Thackery Mills Reclamation Plant: Loading and Removal Efficiency — Grade 11 Math
Name: ____________________________ Date: ______________

Student Worksheet

The Thackery Mills Water Reclamation Facility (a fictional plant) treats a flow of 4.0 million gallons per day. Its influent biochemical oxygen demand, or BOD, is 220 mg/L, and its effluent BOD is 18 mg/L. Its influent total suspended solids, or TSS, is 240 mg/L, and its effluent TSS is 22 mg/L. Use the pounds formula, loading in pounds per day = concentration in mg/L × flow in million gallons per day × 8.34, and the removal-efficiency formula, removal percent = (influent − effluent) ÷ influent × 100.

  1. Calculate the influent BOD loading in pounds per day.
  2. Calculate the effluent BOD loading in pounds per day.
  3. Calculate the plant's BOD removal efficiency as a percentage. Round to the nearest tenth.
  4. The federal secondary standard sets effluent BOD at 30 mg/L or less. Does the plant's 18 mg/L effluent meet that standard, and by what margin in mg/L?
  5. Calculate the plant's TSS removal efficiency as a percentage. Round to the nearest tenth.
  6. Calculate the effluent TSS loading in pounds per day.
  7. Using your answers from problems 1 and 2, find the pounds of BOD removed per day.
  8. The plant manager wants to describe the plant's success to the town council. Decide which single quantity communicates that success more clearly, the effluent concentration in mg/L or the removal efficiency as a percent, and explain your choice in one or two sentences.
  9. Suppose the flow rises to 6.0 million gallons per day while the influent BOD stays at 220 mg/L. Calculate the new influent BOD loading in pounds per day.
  10. The secondary standard corresponds to roughly 85% removal. Does the plant's BOD removal from problem 3 clear that mark, and by how many percentage points?

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: 7,339.2 lb/day. Loading = 220 mg/L × 4.0 MGD × 8.34; 220 × 4.0 = 880, then 880 × 8.34 = 7,339.2 pounds per day.
  2. Answer: 600.48 lb/day. Loading = 18 mg/L × 4.0 MGD × 8.34 = 72 × 8.34 = 600.48 pounds per day.
  3. Answer: 91.8%. Removal = (220 − 18) ÷ 220 × 100 = 202 ÷ 220 × 100 = 91.818, which rounds to 91.8 percent.
  4. Answer: yes, by 12 mg/L. The effluent BOD of 18 mg/L is at or below the 30 mg/L standard, clearing it by 30 − 18 = 12 mg/L.
  5. Answer: 90.8%. Removal = (240 − 22) ÷ 240 × 100 = 218 ÷ 240 × 100 = 90.833, which rounds to 90.8 percent.
  6. Answer: 733.92 lb/day. Loading = 22 mg/L × 4.0 MGD × 8.34 = 88 × 8.34 = 733.92 pounds per day.
  7. Answer: 6,738.72 lb/day. BOD removed = influent load − effluent load = 7,339.2 − 600.48 = 6,738.72 pounds per day.
  8. Answer: Removal efficiency. The percent removed shows how much work the plant does relative to what came in, so 91.8% communicates performance clearly; a single effluent number of 18 mg/L is better for a compliance check than for describing overall success. Also accept the effluent concentration if the student ties the 18 mg/L result to the 30 mg/L federal standard.
  9. Answer: 11,008.8 lb/day. Loading = 220 mg/L × 6.0 MGD × 8.34 = 1,320 × 8.34 = 11,008.8 pounds per day.
  10. Answer: yes, by 6.8 percentage points. The plant removes 91.8% of BOD, which clears the roughly 85% mark by 91.8 − 85 = 6.8 percentage points.

Differentiation

Support

Provide the pounds formula and the removal formula printed at the top of the page with one fully worked example using different numbers, so students can pattern-match the setup before substituting the plant's own values. Allow a calculator for every item.

Extension

Ask students to find the highest influent BOD the plant could receive at 4.0 million gallons per day and still meet the 30 mg/L effluent standard, if its removal efficiency holds steady at 91.8 percent.

ELL supports

Pre-teach "influent" and "effluent" with a simple flow diagram showing water entering and leaving the plant, and label BOD and TSS as two things the plant is measured on. Provide a sentence frame for item 8: "The plant's success is clearer as ___ because ___."