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Worksheet Grade 8 Engineering & Design

Hartley Mills Wastewater: Basin or Constructed Wetland

A grade 8 worksheet solving a break-even cost equation for two wastewater treatment options and calculating BOD removal efficiency against a compliance standard.

Objectives

  1. Students will write and solve a linear equation to find the break-even treated volume between two wastewater treatment options.
  2. Students will calculate each option's total cost at a stated and a projected treatment volume.
  3. Students will calculate BOD removal efficiency and check it against the real secondary treatment standard.

Vocabulary

Key words: break-even point, constructed wetland, removal efficiency, BOD, ecosystem service

Teacher overview

The fictional town of Hartley Mills must upgrade its wastewater treatment to meet the real federal secondary treatment standard, and its engineering team is weighing two options: expanding the town's mechanical aeration basin, or building a constructed wetland that treats wastewater using plants and soil instead of mechanical equipment. This worksheet has students compare the two options' costs using a linear equation, then check whether the chosen system's treated water meets the real BOD standard that municipal wastewater plants must meet.

Hartley Mills Wastewater: Basin or Constructed Wetland — Grade 8 Engineering & Design
Name: ____________________________ Date: ______________

Student Worksheet

Hartley Mills' fictional engineering team is comparing two ways to treat wastewater. One is a mechanical aeration basin. The other is a constructed wetland. A constructed wetland cleans wastewater with plants and soil instead of machines. Let x stand for the number of thousands of gallons treated per year.

  • Mechanical aeration basin: capital cost of $250,000, plus an operating cost of $0.80 for every 1,000 gallons treated.
  • Constructed wetland: capital cost of $110,000, plus an operating cost of $1.50 for every 1,000 gallons treated.
  1. Write a total-cost equation, C(x), for the mechanical aeration basin.
  2. Write a total-cost equation, C(x), for the constructed wetland.
  3. Set the two equations equal to each other. Solve the linear equation for x. That value of x is the break-even treated volume.
  4. Find the total cost of each option at the break-even volume from item 3. Confirm that the two costs match.
  5. Hartley Mills now treats about 120,000 thousand gallons of wastewater per year. Find each option's total cost at this volume. State which option costs less, and by how much.
  6. Town planners expect treated volume to reach 260,000 thousand gallons per year within ten years. Find each option's total cost at this volume. State which option costs less, and by how much.
  7. A constructed wetland does more than treat water. It also provides an ecosystem service: it creates wildlife habitat and filters pollutants naturally. The mechanical basin does neither. Weigh that against your answers from items 5 and 6. Then write two or three sentences recommending which option Hartley Mills should choose, and justify your recommendation.
  8. BOD stands for biochemical oxygen demand. Real secondary treatment standards require finished wastewater to have a BOD of 30 mg/L or less. Suppose Hartley Mills' raw sewage has a BOD of 240 mg/L. The new system cuts that to 25 mg/L. Influent is the water going in. Effluent is the water coming out. Calculate the removal efficiency. Use Removal efficiency (%) = (influent − effluent) ÷ influent × 100.
  9. Look at your answer to item 8. Does Hartley Mills' new system meet the real 30 mg/L secondary treatment standard? Explain, using the effluent value given.
  10. Challenge. Suppose the team instead demanded effluent BOD no higher than 20 mg/L. Would the required removal efficiency be higher or lower than your answer in item 8? Calculate the new required removal efficiency. Assume the same 240 mg/L raw sewage BOD.

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: C(x) = 250,000 + 0.8x.
    The $250,000 capital cost is the fixed starting value, and $0.80 is charged for every one of the x thousands of gallons treated.

  2. Answer: C(x) = 110,000 + 1.5x.
    The $110,000 capital cost is the fixed starting value, and $1.50 is charged for every one of the x thousands of gallons treated.

  3. Answer: x = 200,000 thousand gallons.
    250,000 + 0.8x = 110,000 + 1.5x. Subtract 0.8x from both sides: 250,000 = 110,000 + 0.7x. Subtract 110,000 from both sides: 140,000 = 0.7x. Divide both sides by 0.7: x = 200,000.

  4. Answer: $410,000 for both options.
    Mechanical: 250,000 + 0.8(200,000) = 250,000 + 160,000 = 410,000. Wetland: 110,000 + 1.5(200,000) = 110,000 + 300,000 = 410,000.

  5. Answer: The wetland is cheaper, by $56,000.
    Mechanical: 250,000 + 0.8(120,000) = 250,000 + 96,000 = 346,000. Wetland: 110,000 + 1.5(120,000) = 110,000 + 180,000 = 290,000. 346,000 − 290,000 = 56,000.

  6. Answer: The mechanical basin is cheaper, by $42,000.
    Mechanical: 250,000 + 0.8(260,000) = 250,000 + 208,000 = 458,000. Wetland: 110,000 + 1.5(260,000) = 110,000 + 390,000 = 500,000. 500,000 − 458,000 = 42,000.

  7. Answer: Accept any well-reasoned response. A strong answer notes that the wetland is the cheaper choice at Hartley Mills' current volume and provides ecosystem services the mechanical basin does not, but the mechanical basin becomes the cheaper choice once treated volume grows past 200,000 thousand gallons, so the recommendation should weigh how quickly Hartley Mills expects that growth to happen against how much the town values the wetland's added habitat and natural filtering benefits.

  8. Answer: About 89.6 percent.
    (240 − 25) ÷ 240 × 100 = 215 ÷ 240 × 100 = 89.58..., which rounds to about 89.6 percent.

  9. Answer: Yes, it meets the standard.
    The system's effluent BOD of 25 mg/L is below the real secondary treatment standard of 30 mg/L or less.

  10. Answer: The required removal efficiency would need to be higher, about 91.7 percent.
    (240 − 20) ÷ 240 × 100 = 220 ÷ 240 × 100 = 91.66..., which rounds to about 91.7 percent, higher than the 89.6 percent found in item 8, since removing more BOD (down to 20 mg/L instead of 25 mg/L) requires taking out a larger share of the original 240 mg/L.

Differentiation

Support

Provide both cost equations from items 1 and 2 already written, and give students a printed removal-efficiency formula card with a worked example using different numbers (150 mg/L influent, 15 mg/L effluent) before they attempt item 8.

Extension

Ask students to find the treated volume at which the mechanical basin costs exactly $100,000 more than the wetland, setting up and solving a new equation based on the two cost expressions from items 1 and 2.

ELL supports

Provide a labeled diagram comparing the two systems: a mechanical basin with pipes and tanks, and a constructed wetland with plants and open water, with "capital cost" and "operating cost" labeled on each. Allow item 7's recommendation to be written as two or three short bullet points instead of full sentences.

Every figure in this scenario except the 30 mg/L federal standard is fictional, chosen so students work with realistic engineering numbers.

Extensions

Have students investigate (using only the figures given in this worksheet) how the break-even volume from item 3 would change if the wetland's operating cost dropped from $1.50 to $1.20 per 1,000 gallons, and have them explain whether that change makes the wetland the better choice for a larger or smaller range of treated volumes than before.