Lesson plan Grade 10 Math
Thrapston School Flushing: Pipe Volume Before a Lead Sample
A 50-minute grade 10 lesson where students model school plumbing as cylinders to find how many gallons stand in the pipes and how long a flush must run.
Objectives
- Students will convert a pipe diameter in inches to feet and find its cross-sectional area.
- Students will calculate the volume of water standing in two runs of pipe, in cubic feet and gallons.
- Students will determine how long a fixture must run to clear a stated multiple of the standing volume.
Vocabulary
- Service line
- A service line is the small pipe that connects the water main under the street to the plumbing inside one home or building. It is the last leg of the trip water makes to your tap.
- 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.
Also: standing water, flushing, cross-sectional area
Teacher overview
Thrapston School is fictional, and its pipe lengths, diameters, and flow rate are set for this lesson. Sampling for lead in a school building depends on knowing how much water is standing in the pipes, because water that has sat overnight against old solder or an old service line carries a different lead concentration than water drawn straight from the main. The misconception this lesson addresses is treating pipe diameter in inches as though it can go directly into a volume formula that returns cubic feet; students who skip the inch-to-foot conversion get volumes that are wrong by a factor of 144. For context, the federal lead action level is 15 ppb and drops to 10 ppb in 2027, so how a sample is drawn matters as much as where.
Procedure
Hook (6 min): Ask students to estimate how many gallons of water sit inside the pipes between the street and a classroom drinking fountain. Record three guesses on the board and leave them there. Explain that this fictional district must decide which part of the water drawn from each fixture comes from the lead service line.
Direct instruction (12 min): Present the cylinder volume relationship in the form used for pipes: volume in cubic feet = 0.785 × diameter squared × length, with the diameter in feet. Model it on a teacher example that is not on the handout: a 1.5-inch pipe running 100 feet. The diameter is 1.5 ÷ 12 = 0.125 feet, the area is 0.785 × 0.125 × 0.125 = 0.01227 square feet, the volume is 1.227 cubic feet, and 1.227 × 7.48 = 9.18 gallons. Emphasize that the diameter must be converted before it is squared. Do not solve any handout value during this segment.
Guided practice (12 min): Distribute the handout. Work items 1 through 4 together, stopping after item 2 so pairs can compare their area to a neighbor's. Any pair with an area near 0.44 has squared inches instead of feet.
Independent practice (12 min): Students complete items 5 through 11 in pairs. Circulate and confirm that the 1-inch line is being treated as a separate cylinder rather than added to the copper run before the volume step.
Closure and formative check (8 min): Return to the three estimates from the hook and compare them with the item 6 total. As an exit check, collect item 11 and read a sample of responses. If more than one-third of the class cannot say why a 30-second flush might miss the service line water, reteach using item 9 as the anchor.
Student Handout
Thrapston School is fictional, and the plumbing described here was set for this lesson. Use these facts where needed: pipe volume in cubic feet = 0.785 × diameter squared × length, with the diameter in feet; 1 foot = 12 inches; 1 cubic foot holds 7.48 gallons.
A classroom drinking fountain at Thrapston School is fed by 60 feet of copper pipe with an inside diameter of 0.75 inches. That copper run connects to a 45-foot lead service line with an inside diameter of 1 inch, which reaches the water main in the street. The fountain flows at 2.0 gpm. Round each answer as the item directs.
Convert the 0.75-inch copper diameter to feet. Round to four decimal places.
Find the cross-sectional area of the copper pipe in square feet. Round to five decimal places.
Find the volume of water standing in the 60-foot copper run, in cubic feet. Round to three decimal places.
Convert your item 3 answer to gallons. Round to the nearest hundredth.
Find the volume of water standing in the 45-foot lead service line, in gallons. Round to the nearest hundredth.
Find the total standing volume in both runs, in gallons.
At 2.0 gpm, how many minutes does it take to move the item 6 volume out of the fixture? Round to the nearest tenth.
After the lead samples are collected, the district's plan calls for flushing three times the standing volume before the fountain returns to use. Find that flush volume in gallons.
At 2.0 gpm, how many minutes does the item 8 flush take? Round to the nearest tenth.
The building has 40 fixtures on identical pipe runs. If every fixture receives the item 8 flush, how many gallons does the whole building use?
Explain in two or three sentences why a 30-second flush might collect water that never touched the lead service line.
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: 0.0625 feet. 0.75 ÷ 12 = 0.0625 feet.
Answer: 0.00307 square feet. Area = 0.785 × 0.0625 × 0.0625 = 0.785 × 0.00390625 = 0.0030664, which rounds to 0.00307 square feet.
Answer: 0.184 cubic feet. Volume = 0.00307 × 60 = 0.1842 cubic feet.
Answer: 1.38 gallons. 0.184 × 7.48 = 1.376, which rounds to 1.38 gallons.
Answer: 1.83 gallons. The diameter is 1 ÷ 12 = 0.08333 feet, the area is 0.785 × 0.08333 × 0.08333 = 0.00545 square feet, the volume is 0.00545 × 45 = 0.2453 cubic feet, and 0.2453 × 7.48 = 1.8348, which rounds to 1.83 gallons.
Answer: 3.21 gallons. 1.38 + 1.83 = 3.21 gallons standing in the two runs together.
Answer: 1.6 minutes. 3.21 ÷ 2.0 = 1.605 minutes, which rounds to 1.6 minutes.
Answer: 9.63 gallons. Three times the standing volume is 3 × 3.21 = 9.63 gallons.
Answer: 4.8 minutes. 9.63 ÷ 2.0 = 4.815 minutes, which rounds to 4.8 minutes.
Answer: 385.2 gallons. 40 fixtures × 9.63 gallons = 385.2 gallons for the building.
Answer: At 2.0 gallons per minute, a 30-second flush draws only 1.0 gallon, which is less than the 1.38 gallons standing in the copper run alone. The sample would come entirely from pipe closest to the fountain and would never reach the lead service line 60 feet away. A result from that sample says nothing about the lead line.
Differentiation
Support
Give students item 1 and item 2 already completed so they begin at the volume step, and provide a printed diagram showing the two pipe runs end to end with their lengths and diameters labeled. A four-step conversion strip reading inches, feet, square feet, cubic feet, gallons helps students see which operation comes next.
Extension
Ask students to find the flush time for a fixture at the far end of the fictional building, fed by 110 feet of the same copper pipe instead of 60 feet, and compare it with the item 9 answer. A second challenge is to determine the fixture flow rate that would complete the three-volume flush in under three minutes.
ELL supports
Provide a labeled drawing of a pipe cross-section with the words diameter, area, and length placed on the figure, and a card that pairs each unit with what it measures. Allow students to answer item 11 with a labeled sketch showing where the 30-second flush water comes from, plus one short sentence.
Extensions
Have students obtain their own school's plumbing diagram or ask the facilities office for the length and diameter of the line feeding one fountain, then repeat items 1 through 9 with those numbers. The class can then compare its calculated flush time with the flushing practice the district actually follows.