Water for Schools
Menu

Lesson plan Grade 11 Engineering & Design

Aldermoor Corrosion Control: Choosing a Lead-Reduction Treatment

A grade 11 engineering lesson where students weigh corrosion-control options for a lead exceedance and compute an orthophosphate feed rate with the pounds formula.

Objectives

  1. Students will compare corrosion-control options for a lead exceedance using stated criteria and trade-offs.
  2. Students will calculate an orthophosphate feed rate in pounds per day using the pounds formula.

Vocabulary

Corrosion control
Treatment that keeps water from slowly eating away at the inside of pipes. It helps stop metals like lead from getting into drinking water.
Action level
A trigger number in lead and copper testing. If too many homes test above this level, the utility must take steps like improving treatment and telling the public.
Orthophosphate
A chemical that utilities add to water in very small amounts. It coats the inside of pipes so lead cannot easily get into the water.
Lead service line
A lead service line is an old water pipe made of lead metal that connects a house to the water main. Lead can get into the drinking water, so these pipes are being dug up and replaced across the country.

Also: pounds formula

Teacher overview

The fictional Aldermoor water system exceeded the federal lead action level, and its engineers must decide how to reduce the lead reaching customers' taps. This lesson frames that decision as an engineering trade-off between two corrosion-control approaches, adding orthophosphate or adjusting pH and alkalinity, and has students compute the chemical feed rate one option would require. A common misconception in the pounds-formula math is treating the dose in milligrams per liter as pounds directly; the dose must be multiplied by the flow in million gallons per day and by 8.34 pounds per gallon to reach a feed rate in pounds per day.

Procedure

Hook (5 min): Project the fictional Aldermoor system's most recent 90th-percentile lead result of 21 parts per billion beside the federal action level of 15 parts per billion. Ask students what "exceeding the action level" should require the utility to do, and record their first ideas without comment.

Direct instruction (15 min): Explain how lead enters drinking water through the corrosion of lead service lines and old solder, not from the source water itself. Introduce the two corrosion-control strategies: raising pH and alkalinity so a protective scale forms inside pipes, and adding orthophosphate so a phosphate film coats the pipe wall. Note that the action level is 15 parts per billion, dropping to 10 parts per billion in 2027, and that the Lead and Copper Rule Improvements require replacing nearly all lead service lines within about 10 years.

Guided practice (15 min): Model the pounds formula with a worked example that is not on the handout: a plant treating 1.0 million gallons per day at an orthophosphate dose of 2.0 milligrams per liter needs 2.0 times 1.0 times 8.34, which is 16.7 pounds per day. Walk through how each unit cancels so the answer lands in pounds per day.

Independent practice (15 min): Students complete the Student Handout, which uses the fictional Aldermoor system's own numbers. Circulate and check that students multiply the dose, the flow, and 8.34 in the right order on items 3 and 4.

Closure / exit ticket (5 min): On an index card, each student answers: "State one advantage and one drawback of choosing orthophosphate over pH adjustment for the fictional Aldermoor system." Collect and scan for a genuine trade-off before dismissal.

Aldermoor Corrosion Control: Choosing a Lead-Reduction Treatment — Grade 11 Engineering & Design
Name: ____________________________ Date: ______________

Student Handout

For the fictional Aldermoor water system, use these values: the plant treats 2.0 million gallons per day, the proposed orthophosphate dose is 1.5 milligrams per liter, the 90th-percentile lead result is 21 parts per billion, and the system has 4,000 lead service lines.

  1. Aldermoor's 90th-percentile lead result is 21 ppb and the action level is 15 ppb. By how many ppb does the system exceed the current action level?
  2. Under the 2027 rule the action level drops to 10 ppb. By how many ppb would the same 21 ppb result exceed the lower level?
  3. The plant treats 2.0 MGD and will dose orthophosphate at 1.5 mg/L. Using pounds per day = milligrams per liter x million gallons per day x 8.34, find the orthophosphate feed rate in pounds per day.
  4. If the plant flow rises to 3.0 MGD at the same 1.5 mg/L dose, find the new feed rate in pounds per day.
  5. The engineers can add orthophosphate or raise pH and alkalinity. Orthophosphate reliably coats pipes but adds phosphorus that reaches the town's wastewater plant and its discharge. Judging the orthophosphate option on how reliably it lowers lead and on its effect on the receiving water, state one advantage and one drawback, then say which of the two criteria you would treat as more important and why.
  6. The rule requires replacing nearly all 4,000 lead service lines within about 10 years. On average, how many lines must Aldermoor replace per year to finish in 10 years?

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: 6 ppb. The result of 21 ppb minus the action level of 15 ppb equals 6 ppb over the limit.
  2. Answer: 11 ppb. The result of 21 ppb minus the lower 10 ppb level equals 11 ppb over the future limit.
  3. Answer: 25.0 lbs/day. Feed rate equals 1.5 milligrams per liter times 2.0 million gallons per day times 8.34, which is 25.02, or about 25.0 pounds per day.
  4. Answer: 37.5 lbs/day. At 3.0 million gallons per day, the feed rate is 1.5 times 3.0 times 8.34, which is 37.53, or about 37.5 pounds per day.
  5. Answer: Accept any answer naming a real trade-off and ranking the two criteria, such as: an advantage is that orthophosphate forms a protective film and reliably lowers lead, while a drawback is that it raises phosphorus in the wastewater plant's discharge, which can worsen nutrient loading downstream. Either ranking earns credit when it is defended, though most students place reliable lead reduction first because lead exposure is a direct health risk while the added phosphorus can be managed at the wastewater plant.
  6. Answer: 400 lines/year. Replacing 4,000 lines over 10 years is 4,000 divided by 10, which equals 400 service lines per year.

Differentiation

Support

Provide the pounds formula written with labeled blanks for dose, flow, and the 8.34 constant, so students only substitute and multiply on items 3 and 4. Give the worked guided-practice example on paper for students to reference.

Extension

Ask students to work out how the required orthophosphate feed rate would change if the plant blended two sources and the combined flow reached 4.5 million gallons per day, and to explain why the feed rate rises in proportion to flow while the dose stays fixed.

ELL supports

Pre-teach "corrosion," "dose," and "feed rate" with a labeled cross-section of a pipe showing a protective film on the inside wall. Provide a sentence frame for item 5: "One advantage of orthophosphate is ___, and one drawback is ___."