Worksheet Grade 10 Math
Marloch Commerce Park: Stormwater Runoff and Detention Design
A grade 10 worksheet using the rational method and rectangular-volume geometry to size a detention basin for a fictional town's commerce park development.
Objectives
- Students will calculate peak stormwater runoff using the rational method.
- Students will find a weighted runoff coefficient for a mixed-cover site.
- Students will size a rectangular detention basin to hold a required storage volume.
Vocabulary
- Detention basin
- A detention basin is a dug-out area that fills with stormwater during a storm and then slowly drains empty over hours or days. It holds the water back so streets and streams downstream do not flood.
- Impervious surface
- An impervious surface is any hard surface, like a road, roof, or parking lot, that rain cannot soak through. The more impervious surface a town has, the more stormwater rushes off it.
Also: runoff coefficient, rational method, peak flow
Teacher overview
The town of Marloch, a fictional town invented for this worksheet, is reviewing plans for a new commerce park. Paving over open ground raises the peak rate of stormwater runoff, so the town requires a detention basin to hold the extra water and release it slowly. Students use the rational method to compare runoff before and after development, then use rectangular-volume geometry to size a basin that meets a fixed storage requirement. The scenario keeps every quantity tied to a real design decision the town must make.
Student Worksheet
The town of Marloch is reviewing a commerce park proposed on an 8-acre site. After construction the site will be 5 acres of pavement and roof and 3 acres of lawn. Before construction the whole 8 acres was lawn.
Use these facts for all problems:
- Rational method: peak runoff Q (cubic feet per second) = C × i × A, where C is the runoff coefficient, i is rainfall intensity in inches per hour, and A is area in acres.
- Runoff coefficient: pavement and roof ≈ 0.9; lawn ≈ 0.2.
- Design storm intensity for this site: i = 2.5 inches per hour.
- 1 cubic foot holds 7.48 gallons.
Find the peak runoff, in cubic feet per second, from the 5 acres of pavement and roof.
Find the peak runoff, in cubic feet per second, from the 3 acres of lawn.
Find the total peak runoff from the developed site by adding your answers to items 1 and 2.
Find the weighted (composite) runoff coefficient for the whole 8-acre developed site. Round to the nearest hundredth.
Before construction the whole 8 acres was lawn. Find the site's peak runoff before development.
Find the increase in peak runoff caused by the development by subtracting item 5 from item 3.
The town requires a rectangular detention basin that stores 20,000 cubic feet of stormwater. Suppose the basin is 100 feet long and 50 feet wide. Using volume = length × width × depth, find the required depth in feet.
Convert the 20,000 cubic feet of basin storage to gallons.
For safety the developer wants the basin only 3 feet deep while still storing 20,000 cubic feet, keeping the length at 100 feet. Find the width the basin now needs. Round to the nearest hundredth.
Explain, using the runoff coefficients, why replacing lawn with pavement raises the peak rate of runoff from the same rainstorm.
The basin's outlet is sized so downstream flow never exceeds the predevelopment peak from item 5. Explain how the basin protects the downstream creek even though the total volume of rain that falls on the site does not change.
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: 11.25 cfs.
Q = C × i × A = 0.9 × 2.5 × 5 = 11.25 cubic feet per second.Answer: 1.5 cfs.
Q = 0.2 × 2.5 × 3 = 1.5 cubic feet per second.Answer: 12.75 cfs.
Total = 11.25 + 1.5 = 12.75 cubic feet per second.Answer: 0.64.
Weighted C = (0.9 × 5 + 0.2 × 3) ÷ 8 = (4.5 + 0.6) ÷ 8 = 5.1 ÷ 8 = 0.6375, which rounds to 0.64.Answer: 4.0 cfs.
Before development the site was all lawn: Q = 0.2 × 2.5 × 8 = 4.0 cubic feet per second.Answer: 8.75 cfs.
Increase = developed − predevelopment = 12.75 − 4.0 = 8.75 cubic feet per second.Answer: 4 ft.
Depth = volume ÷ (length × width) = 20,000 ÷ (100 × 50) = 20,000 ÷ 5,000 = 4 feet.Answer: 149,600 gal.
20,000 cubic feet × 7.48 gallons per cubic foot = 149,600 gallons.Answer: 66.67 ft.
Width = volume ÷ (length × depth) = 20,000 ÷ (100 × 3) = 20,000 ÷ 300 = 66.667, which rounds to 66.67 feet.Answer: Pavement lets far less water soak in.
Lawn has a low runoff coefficient (about 0.2) because much of the rain soaks into the soil, while pavement and roof have a high coefficient (about 0.9) because almost all the rain runs off. A higher coefficient in the rational method produces a larger peak flow from the same rainfall intensity and area.Answer: The basin slows the release rate.
The basin catches the extra runoff and lets it out through a small outlet no faster than the predevelopment peak of 4.0 cubic feet per second. The same total volume still leaves the site, but it leaves over a longer time, so the downstream creek never sees the higher peak flow that would erode its banks.
Differentiation
Support
Provide the rational-method formula and the two runoff coefficients on a printed reference card, and work item 1 through together so students see one full C × i × A calculation before continuing. For item 7, remind students that depth is the unknown and that volume, length, and width are given.
Extension
Ask students to find the single runoff coefficient the whole 8-acre site would need to keep its developed peak runoff at the predevelopment 4.0 cubic feet per second, then explain why no realistic mix of pavement and lawn could reach that value.
ELL supports
Provide a labeled site diagram showing the pavement area, the lawn area, and the detention basin with an arrow for the slow outlet flow. Pre-teach "runoff," "coefficient," and "detention" with the diagram, and allow students to answer items 10 and 11 with a labeled sketch plus a short phrase.