Math problem set Grade 9 Math
Cravenshaw Meadows Detention Basin: Storage and Release
A grade 9 problem set applying the rational method to a fictional subdivision, then testing whether the detention basin fits the site under a depth limit.
Objectives
- Students will apply the rational method to find peak runoff before and after development.
- Students will convert a detained flow and a storm duration into a required storage volume.
- Students will test a basin design against a depth constraint and judge whether it is viable.
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.
Also: runoff coefficient, peak flow, rational method, viable solution
Teacher overview
When meadow becomes rooftops and streets, the same storm produces a much larger peak flow, and towns require builders to hold the extra water in a detention basin and release it slowly. The rational method, Q = C x i x A, gives the peak flow from a runoff coefficient, a rainfall intensity in inches per hour, and a drainage area in acres. This set sizes the basin for one fictional subdivision and then checks the design against the parcel the builder actually owns. The error it targets is storing the whole post-development peak flow instead of only the increase over the pre-development peak, which produces a basin more than half again as large as the town requires.
Problems
Cravenshaw Meadows is a fictional subdivision proposed on 22 acres of meadow. Use these values throughout.
- Drainage area: 22 acres
- Design storm intensity: 2.4 inches per hour, lasting 45 minutes
- Runoff coefficient before development: C = 0.20
- Runoff coefficient after development: C = 0.55
- Rational method: Q = C x i x A, with Q in cubic feet per second when i is in inches per hour and A is in acres
- Town rule: the basin must hold the increase in peak flow for the full storm, and standing water may be no deeper than 4 feet
- Parcel set aside for the basin: 130 feet by 95 feet
- Find the peak runoff leaving the site before development.
- Find the peak runoff leaving the site after development.
- Find the flow the basin must detain, in cubic feet per second.
- Convert the 45-minute storm to seconds and find the storage volume the basin must provide, in cubic feet.
- Find the footprint, in square feet, that volume would need at the 4-foot depth limit.
- Find the area of the parcel set aside for the basin, in square feet.
- Find the depth the basin would need to hold your item 4 volume on the parcel from item 6. State whether that design is viable under the town rule.
- The builder adds pervious pavement, which lowers the after-development runoff coefficient to 0.50. Find the new storage volume required, in cubic feet.
- Find the depth the pervious-pavement design needs on the same parcel, and state whether it is viable.
- The basin outlet releases water at 3.0 cubic feet per second. Find how many hours the basin from item 8 takes to drain. Round to the nearest hundredth of an hour.
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: 10.56 cfs. Q = 0.20 x 2.4 x 22 = 0.48 x 22 = 10.56 cubic feet per second.
- Answer: 29.04 cfs. Q = 0.55 x 2.4 x 22 = 1.32 x 22 = 29.04 cubic feet per second.
- Answer: 18.48 cfs. The basin holds only the increase: 29.04 - 10.56 = 18.48 cubic feet per second.
- Answer: 49,896 cubic feet. The storm lasts 45 x 60 = 2,700 seconds, so the volume is 18.48 cubic feet per second x 2,700 seconds = 49,896 cubic feet.
- Answer: 12,474 square feet. Footprint = volume / depth = 49,896 / 4 = 12,474 square feet.
- Answer: 12,350 square feet. Area = 130 x 95 = 12,350 square feet.
- Answer: 4.04 feet. Depth = 49,896 / 12,350 = 4.0402 feet. The town caps standing water at 4 feet, so this design is not viable; the parcel is about 124 square feet too small.
- Answer: 42,768 cubic feet. The new after-development peak is 0.50 x 2.4 x 22 = 26.4 cubic feet per second, so the detained flow is 26.4 - 10.56 = 15.84, and 15.84 x 2,700 = 42,768 cubic feet.
- Answer: 3.46 feet. Depth = 42,768 / 12,350 = 3.4630 feet, which is below the 4-foot cap, so this design is viable and leaves about half a foot of freeboard.
- Answer: 3.96 hours. Drain time = 42,768 cubic feet / 3.0 cubic feet per second = 14,256 seconds, and 14,256 / 3,600 = 3.96 hours.
Differentiation
Support
Give students a unit-labeled version of the rational method for this fictional site, with the runoff coefficient, the intensity in inches per hour, and the area in acres each written in its own box, and work problems 1 and 2 together as a class. Provide the 45-minute-to-seconds conversion already completed so problem 4 is a single multiplication.
Extension
Ask students to solve for the runoff coefficient that would let the fictional basin fit the parcel at exactly the 4-foot depth limit, then describe what mix of pavement, roof, and lawn could produce that value. Have them explain why lowering the coefficient is usually cheaper than buying more land.
ELL supports
Pre-teach "detention," "peak flow," "runoff coefficient," and "viable" with a two-panel sketch of the same hillside as meadow and as a paved subdivision, with a wide arrow and a narrow arrow for the flows. Provide a sentence frame for problems 7 and 9: "The basin needs ___ feet of depth, which is ___ the 4-foot limit, so the design ___."