Math problem set Grade 9 Math

Willowvane Constructed Wetland: Nitrogen Loading and Sizing

A grade 9 problem set on a fictional polishing wetland, using the pounds formula to find nitrogen load, removal rate, detention time, and acres needed.

Grade: Grade 9
Subject: Math
Material: Math problem set
Time: 45 min
Prep: No prep
Materials: Calculator

Standards

HSN-Q.A.1 Common Core Math

Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays.

How this material aligns: Problems 1 through 8 carry units through square feet, acres, milligrams per liter, million gallons per day, pounds per day, and days of detention.

All materials for HSN-Q.A.1 →

HSA-CED.A.1 Common Core Math

Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions.

How this material aligns: Problem 9 has students write and solve a one-variable equation for the effluent concentration once the wetland reaches its removal capacity.

All materials for HSA-CED.A.1 →

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This material is in the public domain (CC0) and may be copied, edited, and shared freely, without attribution. Updated 2026-07-23.

Objectives

Students will apply the pounds formula to convert a concentration and a flow into a daily nitrogen load.
Students will compute removal efficiency, areal removal rate, and detention time for a treatment wetland.
Students will write and solve an equation predicting effluent quality after a flow increase.

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.
Detention time: Detention time is how long water stays inside a tank or basin before flowing out. Water must stay long enough for each treatment step, like settling or disinfecting, to finish its job.

Also: constructed wetland, loading

Teacher overview

A constructed wetland is a shallow basin planted with cattails and rushes, where plant roots and the bacteria growing on them take up nitrogen from treated wastewater before it reaches a stream. Operators size these wetlands by comparing the pounds of nitrogen arriving each day against the pounds each acre can remove. This set follows one fictional village wetland from its current flow through a growth scenario that pushes it past its capacity. The error the set is built to correct is reporting a removal in milligrams per liter as though it were a removal in pounds per day; a concentration difference only becomes a mass once it is multiplied by the flow and by 8.34.

Willowvane Constructed Wetland: Nitrogen Loading and Sizing — Grade 9 Math

Name: ____________________________ Date: ______________

Problems

The Willowvane Water Reclamation Facility is a fictional village plant. Its treated effluent flows into a constructed wetland before it reaches Willowvane Brook. Use these values throughout.

Current flow through the wetland: 0.45 MGD
Nitrogen entering the wetland: 18 mg/L
Target nitrogen leaving the wetland: 8 mg/L
Wetland cell: a rectangle 660 feet by 264 feet, with water 1.5 feet deep
Loading (lbs/day) = concentration (mg/L) x flow (MGD) x 8.34
1 acre = 43,560 square feet; 1 cubic foot = 7.48 gallons

Find the surface area of the wetland cell in acres.
Find the pounds of nitrogen entering the wetland each day at the current flow.
Find the pounds of nitrogen leaving the wetland each day when the effluent holds 8 mg/L.
Find the pounds of nitrogen the wetland removes each day.
Find the removal efficiency as a percent. Round to the nearest tenth.
Divide your item 4 answer by the wetland area to find the areal removal rate in pounds per acre per day. Round to the nearest hundredth.
Find the volume of water standing in the wetland, in gallons.
Find the detention time in days, using detention time = volume divided by flow. Round to the nearest hundredth.
Willowvane grows and the flow rises to 0.60 MGD, while the nitrogen entering stays at 18 mg/L and the wetland still removes the same pounds per day you found in item 4. Write an equation for the effluent concentration c in milligrams per liter, then solve it.
The village wants the effluent held at 8 mg/L even at the higher 0.60 MGD flow. Using the areal removal rate from item 6, find how many additional acres of wetland that would take. Round to the nearest hundredth of an acre.

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: 4.00 acres. Area = 660 x 264 = 174,240 square feet. Then 174,240 / 43,560 = 4.00 acres.
Answer: 67.55 lbs/day. Load = 18 mg/L x 0.45 MGD x 8.34 = 8.1 x 8.34 = 67.554, which rounds to 67.55 pounds per day.
Answer: 30.02 lbs/day. Load = 8 x 0.45 x 8.34 = 3.6 x 8.34 = 30.024, which rounds to 30.02 pounds per day.
Answer: 37.53 lbs/day. Removal = 67.554 - 30.024 = 37.53 pounds per day. The same result comes from the 10 mg/L difference: 10 x 0.45 x 8.34 = 37.53.
Answer: 55.6 percent. Efficiency = (18 - 8) / 18 x 100 = 0.5556 x 100 = 55.56, which rounds to 55.6 percent. Note that this percent matches the concentrations because the flow is the same on both sides.
Answer: 9.38 lbs per acre per day. Rate = 37.53 pounds per day / 4.00 acres = 9.3825, which rounds to 9.38 pounds per acre per day.
Answer: 1,954,972.8 gallons. Volume = 174,240 sq ft x 1.5 ft = 261,360 cubic feet. Then 261,360 x 7.48 = 1,954,972.8 gallons.
Answer: 4.34 days. A flow of 0.45 MGD is 450,000 gallons per day, so 1,954,972.8 / 450,000 = 4.3444, which rounds to 4.34 days. That is about 104 hours.
Answer: 10.50 mg/L. The load arriving is 18 x 0.60 x 8.34 = 90.072 pounds per day. The wetland still removes 37.53 pounds per day, so the load leaving is 90.072 - 37.53 = 52.542 pounds per day. The equation is c x 0.60 x 8.34 = 52.542, or 5.004c = 52.542, so c = 10.50 milligrams per liter. The effluent would rise above the 8 mg/L target.
Answer: 1.33 acres. Holding the effluent at 8 mg/L means removing (18 - 8) x 0.60 x 8.34 = 50.04 pounds per day. At 9.3825 pounds per acre per day, that needs 50.04 / 9.3825 = 5.3333 acres. The wetland already covers 4.00 acres, so the village must add 5.3333 - 4.00 = 1.33 acres.

Differentiation

Support

Give students a filled-in unit map for the pounds formula in this fictional scenario, showing that milligrams per liter times million gallons per day times 8.34 produces pounds per day, with the units cancelling written out. Work problems 2 and 3 together, and let students carry forward the class values for problems 4 through 6.

Extension

Ask students to build a table of effluent concentration against flow for the fictional wetland at 0.45, 0.50, 0.55, 0.60, and 0.65 million gallons per day, holding removal fixed at the item 4 value. Have them describe how the effluent rises and identify the flow at which the wetland can no longer meet the target.

ELL supports

Pre-teach "influent," "effluent," "load," and "detention time" with a labeled cross-section of a wetland showing water entering on one side and leaving on the other. Provide a sentence frame for problem 9: "When the flow rises, the same pounds removed spread over ___ water, so the concentration leaving ___."