Math problem set Grade 8 Engineering & Design

Sizing Corvale's Stormwater Detention Basin: A Cylinder Volume Set

A grade 8 problem set using the rational runoff method and cylinder volume formula to size an underground detention basin for a fictional paved business park.

Grade: Grade 8
Subject: Engineering & Design
Material: Math problem set
Time: 40 min
Prep: No prep
Materials: calculator

Standards

8.G.C.9 Common Core Math

Know the formulas for the volumes of cones, cylinders, and spheres and use them to solve real-world and mathematical problems.

How this material aligns: Problem 9 has students apply the cylinder volume formula to find the depth of an underground detention basin that must hold a required volume, the real-world volume application this standard targets.

All materials for 8.G.C.9 →

MS-ETS1-1 NGSS

Define the criteria and constraints of a design problem with sufficient precision to ensure a successful solution, taking into account relevant scientific principles and potential impacts on people and the natural environment that may limit possible solutions.

How this material aligns: Problems 1 through 3 and problem 9 have students define the runoff increase as a precise design constraint before sizing the basin, exactly the criteria-and-constraints work this standard describes.

All materials for MS-ETS1-1 →

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This material is in the public domain (CC0) and may be copied, edited, and shared freely, without attribution. Updated 2026-07-16.

Objectives

Students will calculate peak stormwater runoff before and after development using the rational method (Q = C x i x A).
Students will convert a flow rate and storm duration into a required storage volume in cubic feet, gallons, and acre-feet.
Students will apply the cylinder volume formula to find the depth a detention basin needs to store a required 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.
Acre-foot: The amount of water it takes to cover one acre of land, about the size of a football field, one foot deep. Western states use acre-feet to measure large amounts of water.
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, peak runoff

Teacher overview

The fictional Corvale Business Park is paving over 2 acres of grass to build a parking lot, and the town's engineering office requires the developer to install an underground detention basin that captures the extra stormwater runoff the pavement will create. This problem set has students use the rational method, a real formula engineers use to estimate peak runoff, to find how much additional water the paved lot will shed compared to the grass it replaces, then size a cylindrical underground basin to hold that difference.

A common misconception in this math is forgetting to convert the storm's one-hour duration into seconds before multiplying it by a flow rate measured in cubic feet per second. A student who multiplies cubic feet per second by "1" instead of by 3,600 seconds will get a volume that is far too small, off by a factor of 3,600, so check that conversion first whenever an answer to a volume question looks unexpectedly small.

Sizing Corvale's Stormwater Detention Basin: A Cylinder Volume Set — Grade 8 Engineering & Design

Name: ____________________________ Date: ______________

Problems

Corvale Business Park's fictional 2-acre lot is being paved. Use the rational method: Q = C × i × A. In that formula, C is the runoff coefficient, i is rainfall intensity in inches per hour, and A is area in acres. For this project, use a design storm intensity of i = 2 inches per hour. Use a runoff coefficient of C = 0.2 for the grass that used to cover the lot, and C = 0.9 for the new pavement. Assume the design storm lasts 1 hour, which is 3,600 seconds.

The lot used to be grass, with C = 0.2. Calculate the peak runoff from the 2-acre lot before development, in cubic feet per second (cfs).
Calculate the peak runoff, in cfs, from the same 2-acre lot after paving (C = 0.9).
Find the increase in peak runoff caused by paving the lot, in cfs.
Take the before-development flow rate from problem 1. The storm lasts 1 hour, or 3,600 seconds. Calculate the total runoff volume before development, in cubic feet.
Now take the after-development flow rate from problem 2. Use the same 3,600-second storm duration. Calculate the total runoff volume after development, in cubic feet.
Find the required detention volume, in cubic feet. It is the after-development volume from problem 5 minus the before-development volume from problem 4.
Convert the required detention volume from problem 6 into gallons. Use the conversion 1 cubic foot = 7.48 gallons.
Convert the required detention volume from problem 6 into acre-feet. Use the conversion 1 acre-foot = 43,560 cubic feet. Round to the nearest hundredth.
The engineering team proposes a cylindrical underground detention basin with a diameter of 36 feet. Use the formula Volume (cu ft) = 0.785 × Diameter² × depth. Find the minimum depth, in feet, the basin needs to hold the required volume from problem 6. Round to the nearest tenth of a foot.
Challenge. Suppose Corvale's engineers chose a basin with a 50-foot diameter instead. Would the required depth from problem 9 increase or decrease? Explain why, without working the full calculation again. Use the relationship between diameter and volume in the cylinder formula.

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: 0.8 cfs.
Q = C × i × A = 0.2 × 2 × 2 = 0.8 cfs.
Answer: 3.6 cfs.
Q = C × i × A = 0.9 × 2 × 2 = 3.6 cfs.
Answer: 2.8 cfs.
3.6 − 0.8 = 2.8 cfs.
Answer: 2,880 cubic feet.
0.8 cfs × 3,600 seconds = 2,880 cubic feet.
Answer: 12,960 cubic feet.
3.6 cfs × 3,600 seconds = 12,960 cubic feet.
Answer: 10,080 cubic feet.
12,960 − 2,880 = 10,080 cubic feet.
Answer: About 75,398 gallons.
10,080 × 7.48 = 75,398.4, which rounds to 75,398 gallons.
Answer: About 0.23 acre-feet.
10,080 ÷ 43,560 = 0.2314..., which rounds to 0.23 acre-feet.
Answer: About 9.9 feet.
0.785 × 36² = 0.785 × 1,296 = 1,017.36. Depth = 10,080 ÷ 1,017.36 = 9.909..., which rounds to 9.9 feet.
Answer: The required depth would decrease.
A 50-foot diameter is larger than 36 feet, so squaring it (50² = 2,500 versus 36² = 1,296) gives a much larger base area for the cylinder. Since the same 10,080 cubic feet must fill that larger base area, the basin does not need to be as deep to hold the same volume.

Differentiation

Support

Provide a filled-in unit-conversion reference card listing "1 cu ft = 7.48 gal" and "1 acre-ft = 43,560 cu ft," and give students the 3,600-seconds-per-hour conversion already written next to problems 4 and 5 so they only need to multiply.

Extension

Ask students to find the diameter a cylindrical basin would need, at a fixed depth of 6 feet, to hold the required 10,080 cubic feet, working backward through the volume formula to solve for diameter.

ELL supports

Provide a labeled diagram of the paved lot with arrows showing water flowing off the pavement into an underground basin, with "cubic feet," "gallons," and "acre-feet" labeled on a simple size-comparison scale. Allow problem 10's explanation to be given as a short phrase rather than a full paragraph.

Extensions

Have students investigate their own school's parking lot or roof area (measuring or estimating its size in square feet, then converting to acres) and calculate, using the same rational method and a 2-inch-per-hour design storm, how much peak runoff that surface alone produces compared to an equal area of grass.