Classroom activity Grade 12 Math
Glenrowe Water Age Relay: Tracing Travel Time Through Mains
A 50-minute grade 12 team relay in which students convert flows to velocities, compute pipe travel times, and find which node holds the oldest water.
Objectives
- Students will convert a pipe flow in gallons per minute into a velocity in feet per second.
- Students will compute travel time through a pipe and sum travel times along a path.
- Students will compare two paths and explain why higher demand lowers water age at a node.
Vocabulary
Key words: water age, velocity, continuity, node, fire flow
Teacher overview
Water age is the time a parcel of water spends in the pipes between the treatment plant and a customer's tap, and it is the quantity operators reach for when chlorine residual sags at the end of a line. This relay puts teams of four in charge of the fictional Glenrowe distribution system, a five-pipe network running from the plant to two dead-end nodes. Each team member takes one pipe card, converts its flow into a velocity, and computes that pipe's travel time, and then the team assembles the legs into two path totals and decides which node holds the older water. A second round raises the flows to fire-flow levels so students can see the same network deliver much younger water.
Activity Directions
- Setup (6 min): Form teams of four. Give each team one printed map of the fictional Glenrowe network, drawn before class from the handout description, showing the plant, node B, and the branches running to nodes C, E, D, and F, plus the five pipe cards, one for each row of the handout table. Write the two working formulas on the board: area equals 0.785 times diameter squared with the diameter in feet, and flow equals area times velocity. Post the conversion 448.8 gallons per minute per cubic foot per second.
- Assign legs (2 min): Within each team, one student takes pipe P–B, one takes B–C and C–E, one takes B–D, and one takes D–F. Every student works only their own card until the handoff.
- Round 1, normal demand (14 min): Teams complete questions 1 through 11 on the handout using the round 1 flows. A leg is finished only when the student can state the velocity in feet per second and the travel time in minutes. Team members then hand their travel times to the recorder, who adds them along each path. Stop the round when the first three teams have both path totals, or at 14 minutes, whichever comes first.
- Check the handoff (5 min): Call out the correct travel time for pipe P–B, which is 24.5 minutes, and have every team confirm their own value before continuing. Teams whose figure is far off almost always divided by area in square inches rather than square feet.
- Round 2, fire flow (10 min): Announce that a hydrant near node F is flowing. Give the fire-flow figures on the handout: 2,100 gpm in pipe P–B, 1,400 gpm in B–D, and 850 gpm in D–F. Teams recompute the three legs on the path to node F and answer question 12. The round ends when every team has reported a fire-flow total.
- Scoring (4 min): Award each team one point per correct travel time and two points for each correct path total. Read out the round 2 totals and confirm which team is closest to 30.0 minutes.
- Debrief (9 min): Ask two questions and take responses from three teams each. First, why is node F the node where a chlorine residual problem would show up first in the fictional Glenrowe system. Second, why does opening a hydrant lower the water age at node F rather than raising it, when the hydrant is pulling more water out of the system.
Student Handout
The fictional Glenrowe water system runs from the treatment plant to node B, then splits. One branch runs to node C and on to node E. The other runs to node D and on to node F. Use area equals 0.785 times the diameter squared with the diameter in feet, and use 448.8 gallons per minute per cubic foot per second.
| Pipe | Diameter (in) | Length (ft) | Round 1 flow (gpm) |
|---|---|---|---|
| P–B | 12 | 5,000 | 1,200 |
| B–C | 8 | 3,000 | 500 |
| C–E | 6 | 2,000 | 180 |
| B–D | 10 | 4,000 | 700 |
| D–F | 6 | 2,500 | 150 |
- Find the cross-sectional area of the 12-inch pipe P–B, in square feet. Round to three decimal places.
- Convert the round 1 flow in pipe P–B to cubic feet per second. Round to three decimal places.
- Find the velocity in pipe P–B, in feet per second. Round to two decimal places.
- Find the travel time through pipe P–B, in minutes. Round to one decimal place.
- Find the travel time through pipe B–C, in minutes. Round to one decimal place.
- Find the travel time through pipe C–E, in minutes. Round to one decimal place.
- Find the travel time through pipe B–D, in minutes. Round to one decimal place.
- Find the travel time through pipe D–F, in minutes. Round to one decimal place.
- Add your rounded values to find the total travel time from the plant to node E, in minutes.
- Add your rounded values to find the total travel time from the plant to node F, in minutes.
- Which node, E or F, holds the older water under round 1 flows?
- During the fire-flow round, pipe P–B carries 2,100 gpm, pipe B–D carries 1,400 gpm, and pipe D–F carries 850 gpm. Add your three rounded leg times to find the new total travel time from the plant to node F, in minutes.
- Explain in two or three sentences why the water age at node F drops when a hydrant near it is flowing.
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: 0.785 sq ft.
A 12-inch pipe is 1.000 foot across, so the area is 0.785 times 1.000 squared, or 0.785 square feet.Answer: 2.674 cfs.
Dividing 1,200 gpm by 448.8 gallons per minute per cfs gives about 2.6738 cubic feet per second.Answer: 3.41 ft/s.
Velocity is flow divided by area, so 2.6738 divided by 0.785 is about 3.406, which rounds to 3.41 feet per second.Answer: 24.5 minutes.
Dividing 5,000 feet by 3.406 feet per second gives about 1,468 seconds, and 1,468 divided by 60 is about 24.5 minutes.Answer: 15.7 minutes.
An 8-inch pipe has an area of 0.349 square feet. The flow is 500 divided by 448.8, or 1.114 cfs, giving 3.19 feet per second, and 3,000 feet at that speed takes about 940 seconds, or 15.7 minutes.Answer: 16.3 minutes.
A 6-inch pipe has an area of 0.196 square feet. The flow of 180 gpm is 0.401 cfs, giving 2.04 feet per second, and 2,000 feet takes about 979 seconds, or 16.3 minutes.Answer: 23.3 minutes.
A 10-inch pipe has an area of 0.545 square feet. The flow of 700 gpm is 1.560 cfs, giving 2.86 feet per second, and 4,000 feet takes about 1,398 seconds, or 23.3 minutes.Answer: 24.5 minutes.
The 6-inch pipe has an area of 0.196 square feet. The flow of 150 gpm is 0.334 cfs, giving 1.70 feet per second, and 2,500 feet takes about 1,468 seconds, or 24.5 minutes.Answer: 56.5 minutes.
Adding the rounded legs 24.5, 15.7, and 16.3 gives 56.5 minutes from the plant to node E.Answer: 72.3 minutes.
Adding the rounded legs 24.5, 23.3, and 24.5 gives 72.3 minutes from the plant to node F.Answer: Node F.
Its path total of 72.3 minutes is longer than node E's 56.5 minutes, so water reaching node F has spent more time in the pipes.Answer: 30.0 minutes.
At 2,100 gpm the P–B leg runs 5.96 feet per second and takes 14.0 minutes. At 1,400 gpm the B–D leg runs 5.72 feet per second and takes 11.7 minutes. At 850 gpm the D–F leg runs 9.65 feet per second and takes 4.3 minutes. The three rounded legs sum to 30.0 minutes.Answer: Higher flow raises velocity, so the same pipe length is crossed in less time.
The pipe diameters do not change, so pulling more water through them raises the velocity in every leg on the path. Water age is length divided by velocity, and a larger denominator over the same distance means the water reaching node F left the plant more recently.
Differentiation
Support
Give these students the area column already computed for all five pipes in the fictional Glenrowe table, so their relay leg begins at the flow conversion. Provide a three-box conversion strip labeled gpm, then cfs, then feet per second, with the operation written under each arrow, and allow them to work in pairs on a single card rather than one card each.
Extension
Have students decide whether any flow in pipe D–F alone could bring node F's total travel time under 45 minutes in the fictional Glenrowe network while the other two legs stay at their round 1 values, and explain what their answer shows about which leg controls the water age at node F. Then ask them what a looped connection between nodes E and F would do to the water age at both, and why a utility cannot simply run higher flows all the time to keep water young.
ELL supports
Pre-teach "water age," "velocity," "node," and "fire flow" with a labeled map of the fictional Glenrowe network showing an arrow along each pipe and a clock symbol at each node. Provide sentence frames for the debrief and for item 13, such as "When the flow increases, the velocity ___, so the travel time ___," and allow students to answer item 11 by circling a node on the map before writing.
Extensions
Teams can redraw the fictional Glenrowe network with a new pipe joining nodes E and F, then argue from their travel times which direction water would move in the new pipe and how each node's water age would change. A second follow-up gives each team a chlorine residual that falls by 0.1 mg/L for every 20 minutes of travel and asks them to predict the residual at both nodes under round 1 flows.