Worksheet Grade 12 Math
Northgate Business Park: Detention Basin and Pipe Flow
A grade 12 worksheet applying the rational method and the continuity equation to size a fictional business park's storm outlet pipe and detention basin.
Objectives
- Students will compute peak stormwater runoff using the rational method Q = C x i x A.
- Students will convert a flow between cfs, gpm, and MGD and size a pipe using Q = A x V.
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: rational method, runoff coefficient, continuity
Teacher overview
Northgate Business Park is a fictional commercial development that is adding a large paved parking area. When rain falls on pavement, almost all of it runs off, so the site's engineers must size an outlet pipe and a detention basin that can hold and release the storm safely. This worksheet walks students through the rational method, the standard conversions between cubic feet per second, gallons per minute, and million gallons per day, and the continuity equation for pipe flow. A common error to watch for is applying the runoff coefficient to the wrong quantity; the coefficient multiplies the rainfall-and-area product and is not added to it.
Student Worksheet
Northgate Business Park (a fictional site) covers two surfaces that drain to one outlet: 3.0 acres of pavement with a runoff coefficient of 0.9, and 2.0 acres of landscaping with a runoff coefficient of 0.2. The design storm has a rainfall intensity of 2.5 inches per hour. Use these relationships throughout:
- Rational method: Q (cfs) = C x i x A, where i is in inches per hour and A is in acres.
- 1 cubic foot per second (cfs) = 448.8 gallons per minute (gpm).
- 1 million gallons per day (MGD) = 1.547 cfs.
- Pipe area: A (sq ft) = 0.785 x D squared, with diameter D in feet; and Q (cfs) = A x V, with velocity V in feet per second.
- Find the peak runoff from the pavement alone, using C = 0.9, i = 2.5, and A = 3.0 acres.
- Find the peak runoff from the landscaped area alone, using C = 0.2, i = 2.5, and A = 2.0 acres.
- Add your two answers to find the total peak runoff reaching the outlet.
- Convert the total peak runoff from cfs to gallons per minute.
- Convert the total peak runoff from cfs to million gallons per day. Round to the nearest hundredth.
- The outlet pipe must carry the full peak flow at a design velocity of 5 feet per second. Using Q = A x V, find the pipe cross-sectional area needed, in square feet.
- Using A = 0.785 x D squared, find the pipe diameter in feet, then in inches. Storm pipe is sold in 12-, 15-, 18-, 21-, and 24-inch diameters; state the smallest of these at or above your value.
- The detention basin must hold the first 20 minutes of peak flow. Find that volume in cubic feet, then convert it to gallons (1 cubic foot = 7.48 gallons).
- Explain, in two or three sentences, why paving more of the site raises the peak runoff even when the total site area does not change.
- The city will only permit an outlet that releases no more than 6 cfs. Explain why the detention basin is needed and what it does with the flow above that limit.
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: 6.75 cfs.
Q = C x i x A = 0.9 x 2.5 x 3.0. First 2.5 x 3.0 = 7.5, then 0.9 x 7.5 = 6.75 cfs.Answer: 1.0 cfs.
Q = 0.2 x 2.5 x 2.0. First 2.5 x 2.0 = 5.0, then 0.2 x 5.0 = 1.0 cfs.Answer: 7.75 cfs.
Total peak runoff = 6.75 + 1.0 = 7.75 cfs. The two subareas share one outlet, so their peak flows add.Answer: 3,478 gpm.
7.75 cfs x 448.8 gpm/cfs = 3,478.2, which rounds to about 3,478 gpm.Answer: 5.01 MGD.
7.75 cfs / 1.547 cfs per MGD = 5.0097, which rounds to 5.01 MGD.Answer: 1.55 sq ft.
Q = A x V, so A = Q / V = 7.75 cfs / 5 ft/s = 1.55 square feet.Answer: 1.41 ft.
A = 0.785 x D squared, so D squared = 1.55 / 0.785 = 1.9745, and D = 1.405 ft, or about 16.9 inches. The smallest standard size at or above that is an 18-inch pipe.Answer: 9,300 cubic feet.
20 minutes = 1,200 seconds. Volume = 7.75 cfs x 1,200 s = 9,300 cubic feet. In gallons, 9,300 x 7.48 = 69,564 gallons.Answer: Paving raises the runoff coefficient of that area.
Pavement has a coefficient near 0.9 while landscaping is near 0.2, so replacing lawn with pavement raises C for that area. Because Q = C x i x A, a higher C produces a larger peak runoff even though the total area A and rainfall i are unchanged.Answer: The basin stores the flow above the release limit.
The peak inflow of 7.75 cfs is larger than the 6 cfs the city permits at the outlet. The detention basin temporarily holds the excess and releases it slowly, so the outflow stays at or below 6 cfs while the storm passes.
Differentiation
Support
Provide the rational-method formula and the three conversion facts on a printed reference strip, and work one full example (a single-surface site) together before students begin. Give the pipe-area formula already rearranged as A = Q / V for item 6 so students can focus on the substitution.
Extension
Ask students to combine the two subareas into a single area-weighted runoff coefficient, confirm it reproduces the total peak runoff from item 3, and then determine what parking-area size would push the peak flow above the city's release limit.
ELL supports
Provide a labeled site diagram showing the pavement, the landscaping, the outlet pipe, and the detention basin, with each key term glossed. Offer sentence frames for items 9 and 10, such as "The peak runoff increases because ..." and "The detention basin is needed because ...".
Extensions
Have students look up their own school or district's site plan and estimate the paved fraction, then discuss why municipal stormwater codes often set a maximum release rate that matches the runoff from the land before it was developed.