Math problem set Grade 7 Engineering & Design
Thistledown Constructed Wetland: Nutrient Removal Design Set
Students size a fictional constructed treatment wetland, compute nutrient removal percentages, and check the design against a budget and a water-quality goal.
Objectives
- Students will calculate nutrient removal efficiency as a percentage from influent and effluent concentrations.
- Students will scale a wetland's required area in proportion to flow and check the design against a budget.
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: constructed wetland, removal efficiency, nutrient
Teacher overview
The town of Thistledown, a fictional community, is building a constructed treatment wetland. The wetland will use plants, soil, and gravel to pull extra nutrients out of the town's treated wastewater before that water flows into a small creek. Nutrients like phosphorus help plants grow, yet too much phosphorus in a creek can trigger heavy algae growth that harms fish. This set walks students through sizing the wetland, finding how much nutrient it removes, and checking the design against the town's budget.
A common misconception in this math is treating a removal percentage as if it were the leftover amount. Removal efficiency is the share taken out, not the share that stays in the water. If a wetland lowers phosphorus from 6.0 to 1.5 mg/L, the removal is the drop divided by the starting amount, not the final reading. Watch for students who report the effluent value as the "percent removed."
Problems
All values below describe the fictional Thistledown design. Use them for every problem.
- Wastewater flow into the wetland: 40,000 gallons per day.
- Incoming (influent) phosphorus: 6.0 mg/L; target outgoing (effluent) phosphorus: 1.5 mg/L.
- Incoming nitrogen: 20 mg/L; target outgoing nitrogen: 8 mg/L.
- Design rule: the wetland lowers phosphorus by 0.5 mg/L for every 200 square feet of surface.
- Build cost: 12 dollars per square foot. Town budget for the wetland: 25,000 dollars.
- Removal efficiency (%) = (influent − effluent) ÷ influent × 100.
Find the phosphorus removal efficiency for the wetland, using the influent of 6.0 mg/L and the target effluent of 1.5 mg/L. Round to the nearest whole percent.
Find the nitrogen removal efficiency, using the influent of 20 mg/L and the target effluent of 8 mg/L. Round to the nearest whole percent.
The design must drop phosphorus by 4.5 mg/L (from 6.0 to 1.5). Using the rule of 0.5 mg/L removed per 200 square feet, how many square feet of wetland surface does that require?
Suppose the town grows and the flow into the wetland doubles to 80,000 gallons per day. Engineers say the required area grows in direct proportion to the flow. What surface area, in square feet, would the larger flow need to reach the same phosphorus drop?
Thistledown's rule says the wetland must remove at least 70 percent of the incoming phosphorus to protect the creek. Does the design from problem 1 meet that goal? By how many percentage points does it clear or miss the goal?
The build cost is 12 dollars per square foot. Using the area from problem 3, what is the total build cost of the wetland?
The town budgeted 25,000 dollars for the wetland. What percent of that budget does the build cost from problem 6 use? Round to the nearest whole percent.
In one or two sentences, explain why removing nutrients before the water reaches the creek matters, even though nutrients like phosphorus are helpful to plants in a garden.
From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.
Answer Key
Answer: 75%.
Removal = (6.0 − 1.5) ÷ 6.0 × 100 = 4.5 ÷ 6.0 × 100 = 0.75 × 100 = 75%.Answer: 60%.
Removal = (20 − 8) ÷ 20 × 100 = 12 ÷ 20 × 100 = 0.60 × 100 = 60%.Answer: 1,800 square feet.
The phosphorus must drop 4.5 mg/L. Number of 0.5 mg/L steps = 4.5 ÷ 0.5 = 9 steps. Each step needs 200 square feet, so 9 × 200 = 1,800 square feet.Answer: 3,600 square feet.
Area is proportional to flow. The flow doubles from 40,000 to 80,000 gallons per day, so the area doubles: 1,800 × 2 = 3,600 square feet.Answer: Yes, by 5 percentage points.
The design removes 75 percent (problem 1), and the goal is at least 70 percent. 75 − 70 = 5, so it clears the goal by 5 percentage points.Answer: 21,600 dollars.
Build cost = 1,800 square feet × 12 dollars per square foot = 21,600 dollars.Answer: 86%.
21,600 ÷ 25,000 = 0.864, which rounds to about 86 percent of the budget.Answer: Extra nutrients in a creek cause problems.
Accept any answer explaining that too much phosphorus or nitrogen in a creek feeds heavy algae growth, which can lower oxygen and harm fish, so the same nutrients that help garden plants become a pollutant when they reach open water in large amounts.
Differentiation
Support
Give students the removal-efficiency formula already set up for problem 1 as "(6.0 − 1.5) ÷ 6.0 × 100 = ___" and allow a calculator throughout. Provide a printed reminder that removal is the drop divided by the starting amount.
Extension
Ask students to find the total build cost for the larger design in problem 4 (3,600 square feet) at the same 12 dollars per square foot, then decide whether the town would need to raise its budget and by how much.
ELL supports
Provide a simple labeled diagram showing water entering the wetland (influent) on one side and leaving (effluent) on the other, with the nutrient values marked. Allow the problem 8 response to use a sentence frame: "Too much phosphorus in a creek can ___, so the wetland removes it before the water leaves."