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Lesson plan Grade 12 Science

The Sablefield Polishing Wetland: Loading and Nitrogen Removal

A grade 12 science lesson on treatment wetlands, with handout problems on hydraulic loading rate, areal nitrogen loading, detention time, and removal efficiency.

Objectives

  1. Students will explain how microbial processes rather than filtration remove nitrogen in a treatment wetland.
  2. Students will calculate hydraulic loading rate, areal nitrogen loading, and detention time for a wetland.
  3. Students will predict how a change in flow affects detention time and nitrogen removal.

Vocabulary

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: treatment wetland, hydraulic loading rate, areal loading rate, denitrification, removal efficiency

Teacher overview

The fictional Sablefield Water Reclamation Facility already meets its permit, yet the lake it discharges to keeps producing summer algae blooms, so the district built a twelve-acre treatment wetland to polish the effluent before it leaves the site. This lesson works through the three numbers that govern how such a wetland performs: how much water it receives per unit of area, how much nitrogen it receives per acre, and how long the water stays inside.

The misconception worth confronting is that a wetland strains nitrogen out of the water the way a filter strains particles. Nitrogen is dissolved, and it leaves mainly because bacteria in the sediment convert nitrate to nitrogen gas, a process that needs time, carbon, and low oxygen. That is why the handout keeps returning to detention time, and why a wetland pushed to twice its design flow removes a smaller share of the nitrogen it receives.

Procedure

Hook (5 min): Ask students how a marsh could clean water without pumps, chemicals, or moving parts. Collect two or three answers on the board, then name the fictional Sablefield wetland and state the lesson's question: what determines how much nitrogen a wetland can remove.

Direct instruction (15 min): Explain the mechanism. Nitrate arrives dissolved in the effluent, and in the low-oxygen sediment beneath the plants, bacteria use nitrate instead of oxygen to break down organic carbon, releasing nitrogen gas to the atmosphere. Plants take up some nitrogen too, though most of that returns to the water when the plants die back, so the permanent removal pathway is the microbial one. Introduce the three design quantities: hydraulic loading rate, which is flow divided by wetland area; areal loading rate, which is pounds of nitrogen per acre per day; and detention time, which is wetland water volume divided by flow. Emphasize that all three describe the same wetland from different directions.

Worked example (10 min): Model the calculations with a different wetland so students learn the method rather than the answers. For a 5-acre wetland receiving 0.8 MGD, the area is 5 x 43,560 = 217,800 square feet and the flow is 800,000 / 7.48 = 106,952 cubic feet per day, so the hydraulic loading rate is 106,952 / 217,800 = 0.49 feet per day. Walk through the units aloud, showing why gallons must become cubic feet before the division makes sense.

Independent handout (18 min): Students complete the handout individually. Circulate and check question 3 early, since a student who divides gallons by square feet will carry that error through the rest of the page.

Exit ticket and formative check (7 min): Pose one question: if the fictional plant doubled its flow to the wetland without enlarging it, what happens to detention time and what happens to nitrogen removal? Collect one-sentence answers. The check is met when at least 80 percent state that detention time halves and removal falls, rather than saying removal is unaffected because the wetland is the same size.

The Sablefield Polishing Wetland: Loading and Nitrogen Removal — Grade 12 Science
Name: ____________________________ Date: ______________

Student Handout

Use these values for the fictional Sablefield treatment wetland. Show your work with units.

  • Wetland surface area: 12 acres
  • Average water depth in the wetland: 1.2 feet
  • Flow to the wetland: 1.5 MGD
  • Total nitrogen entering the wetland: 14 mg/L
  • Total nitrogen leaving the wetland: 5 mg/L
  • Conversions: 1 acre = 43,560 square feet; 1 cubic foot = 7.48 gallons
  • Pounds formula: pounds per day = concentration (mg/L) x flow (MGD) x 8.34
  • Hydraulic loading rate = flow (cubic feet per day) / wetland area (square feet)
  • Detention time (days) = wetland water volume (gallons) / flow (gallons per day)
  • Removal efficiency (percent) = (in - out) / in x 100

  1. Convert the 12-acre wetland surface to square feet.

  2. Convert the 1.5 MGD flow to cubic feet per day. Round to the nearest cubic foot.

  3. Calculate the hydraulic loading rate in feet per day. Round to two decimal places.

  4. Calculate the total nitrogen load entering the wetland, in pounds per day. Round to one decimal place.

  5. Calculate the areal nitrogen loading rate, in pounds per acre per day. Round to one decimal place.

  6. Calculate the wetland's nitrogen removal efficiency as a percent. Round to one decimal place.

  7. Calculate the mass of nitrogen the wetland removes each day, in pounds per day. Round to one decimal place.

  8. Calculate the volume of water standing in the wetland, in gallons, using the surface area and the 1.2-foot average depth. Round to the nearest gallon.

  9. Calculate the detention time in the wetland, in days. Round to one decimal place.

  10. The plant doubles its flow to the wetland to 3.0 MGD without enlarging it. Calculate the new detention time, in days. Round to one decimal place.

  11. Explain why the shorter detention time in question 10 lowers the wetland's nitrogen removal efficiency, using the microbial mechanism from the lesson.

  12. In late autumn the wetland sometimes releases more nitrogen than it receives for a short period. Explain how that is possible in a system built to remove nitrogen.

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: 522,720 square feet.
    12 acres x 43,560 square feet per acre = 522,720 square feet.

  2. Answer: 200,535 cubic feet per day.
    1.5 MGD is 1,500,000 gallons per day. Dividing by 7.48 gallons per cubic foot: 1,500,000 / 7.48 = 200,534.76, which rounds to 200,535 cubic feet per day.

  3. Answer: 0.38 feet per day.
    200,534.76 cubic feet per day / 522,720 square feet = 0.3836 feet per day, which rounds to 0.38. Expressed another way, the wetland receives about 4.6 inches of water spread over its whole surface each day.

  4. Answer: 175.1 pounds per day.
    14 mg/L x 1.5 MGD x 8.34 = 21 x 8.34 = 175.14, which rounds to 175.1 pounds of nitrogen per day.

  5. Answer: 14.6 pounds per acre per day.
    175.14 pounds per day / 12 acres = 14.595, which rounds to 14.6 pounds per acre per day. Designers compare this figure against the loading a wetland of this type can handle, since a wetland that receives more nitrogen per acre than its sediment can process will pass the excess through.

  6. Answer: 64.3 percent.
    (14 - 5) / 14 = 9 / 14 = 0.642857, which is 64.3 percent removal.

  7. Answer: 112.6 pounds per day.
    The concentration drop is 14 - 5 = 9 mg/L. Then 9 mg/L x 1.5 MGD x 8.34 = 13.5 x 8.34 = 112.59, which rounds to 112.6 pounds per day. The same result comes from taking 64.3 percent of the 175.1 pound load.

  8. Answer: 4,691,935 gallons.
    522,720 square feet x 1.2 feet = 627,264 cubic feet. Then 627,264 x 7.48 gallons per cubic foot = 4,691,934.7, which rounds to 4,691,935 gallons.

  9. Answer: 3.1 days.
    4,691,934.7 gallons / 1,500,000 gallons per day = 3.128 days, which rounds to 3.1 days.

  10. Answer: 1.6 days.
    4,691,934.7 gallons / 3,000,000 gallons per day = 1.564 days, which rounds to 1.6 days. Doubling the flow halves the time each parcel of water spends in the wetland.

  11. Answer: The bacteria need contact time, and halving detention time cuts the reaction short.
    Nitrogen leaves the water when bacteria in the sediment convert nitrate to nitrogen gas, and that conversion proceeds at a limited rate. In 3.1 days a given parcel of water sits over the sediment long enough for much of its nitrate to be converted, while in 1.6 days it is carried out of the wetland before the reaction finishes. The wetland also receives twice the nitrogen load per acre at the higher flow, so the same sediment has to process more nitrogen in less time. Removal efficiency drops for both reasons at once.

  12. Answer: Plants die back and decompose, releasing the nitrogen they took up during the growing season.
    Plant uptake stores nitrogen temporarily rather than removing it. When the vegetation senesces in autumn and decomposes, that nitrogen returns to the water as ammonia and organic nitrogen, and cold water slows the microbial nitrification and denitrification that would otherwise remove it. For a short period the wetland can export more nitrogen than arrives in the effluent, which is why annual averages rather than single samples are used to judge performance.

Differentiation

Support

Give students a unit map for the fictional Sablefield chain, showing acres to square feet and gallons to cubic feet with the conversion factor on each arrow, and complete questions 1 and 2 together as a class. Provide a labeled cross-section of the wetland with the depth marked so question 8 reads as length times width times depth. Allow questions 11 and 12 to be answered in three short sentences.

Extension

Ask students to calculate the wetland area the fictional plant would need at 3.0 MGD to hold the original 3.1-day detention time, and to express that area in acres. Have them then estimate the areal loading rate at that size and comment on whether land or loading is the tighter constraint on expanding a wetland.

ELL supports

Provide a diagram of the fictional wetland with arrows for water in, water out, plant uptake, and nitrogen gas leaving, and place the six vocabulary terms on the diagram where each applies. Give a sentence frame for questions 11 and 12, such as "Removal falls because the bacteria need ___, and at 3.0 MGD the water only stays ___ days." Pair students for questions 3 and 5 so the phrase "per unit of area" can be talked through before it is written.

Extensions

Students can compare the fictional Sablefield wetland with a mechanical denitrification filter sized for the same 1.5 MGD, listing the land, energy, and maintenance differences between the two approaches and stating which they would recommend for a district with abundant land. A second option is to have students find a published account of a treatment wetland operating in their region and describe what wildlife habitat value the operators report alongside the water quality results.