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Math problem set Grade 10 Math

Foxtail Commons Runoff: A Grade 10 Rational-Method Detention Set

Grade 10 stormwater math: students use the rational method to find peak runoff, size a detention basin, and find a composite runoff coefficient.

Objectives

  1. Students will apply the rational method to calculate peak stormwater runoff from paved and landscaped drainage areas.
  2. Students will size a detention basin's storage volume and calculate a composite runoff coefficient for a mixed site.

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, rainfall intensity, peak flow

Teacher overview

When a developer paves over open ground, more rain runs off the surface and less soaks in, so engineers size a detention basin to hold the extra water and release it slowly. This set uses the rational method, the standard tool for estimating peak runoff, to design a basin for the fictional Foxtail Commons redevelopment. A common misconception in this math is adding the runoff coefficients of two surfaces together; the coefficient is a ratio between 0 and 1, so a mixed site's composite coefficient is a weighted average of its parts, never a sum that climbs above 1.

Foxtail Commons Runoff: A Grade 10 Rational-Method Detention Set — Grade 10 Math
Name: ____________________________ Date: ______________

Problems

Foxtail Commons (a fictional site built for this set) covers 5.0 acres: 3.0 acres of new pavement and 2.0 acres of landscaped lawn. The design storm has a rainfall intensity of 4.0 inches per hour. Use the rational method, Q = C × i × A, where Q is peak runoff in cubic feet per second (cfs), C is the runoff coefficient, i is intensity in inches per hour, and A is area in acres. Use a runoff coefficient of 0.9 for pavement and 0.2 for lawn. The detention basin is a rectangle 120 feet long, 80 feet wide, and 4 feet deep.

  1. Find the peak runoff from the 3.0 acres of pavement in cfs, using C = 0.9 and i = 4.0 inches per hour.

  2. Find the peak runoff from the 2.0 acres of lawn in cfs, using C = 0.2 and i = 4.0 inches per hour.

  3. Find the total peak runoff reaching the basin by adding your answers from problems 1 and 2.

  4. Convert the total peak runoff from problem 3 to gallons per minute, using the conversion that 1 cfs equals 448.8 gpm.

  5. Convert the total peak runoff from problem 3 to million gallons per day, using the conversion that 1 MGD equals 1.547 cfs.

  6. Find the full storage volume of the detention basin in cubic feet, using Volume = Length × Width × Depth.

  7. Convert the basin volume from problem 6 to gallons, using the conversion that one cubic foot holds 7.48 gallons.

  8. The basin releases water through an outlet at a controlled 2.0 cfs. If the total peak runoff from problem 3 flows in while 2.0 cfs flows out, the basin fills at the net rate. Find how many minutes the empty basin would take to fill, dividing the basin volume in cubic feet (problem 6) by the net inflow in cubic feet per minute (net cfs × 60).

  9. Find the composite runoff coefficient for the whole 5.0-acre site by dividing the total peak runoff (problem 3) by (i × total area), that is, by 4.0 × 5.0.

  10. Explain why replacing lawn with pavement raises the site's peak runoff, and how the detention basin protects the storm drain downstream.

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key

  1. Answer: 10.8 cfs.
    Q = 0.9 × 4.0 × 3.0 = 10.8 cfs.

  2. Answer: 1.6 cfs.
    Q = 0.2 × 4.0 × 2.0 = 1.6 cfs.

  3. Answer: 12.4 cfs.
    Total peak runoff = 10.8 cfs + 1.6 cfs = 12.4 cfs.

  4. Answer: about 5,565 gpm.
    12.4 cfs × 448.8 gpm/cfs = 5,565.12, which rounds to about 5,565 gpm.

  5. Answer: about 8.0 MGD.
    12.4 cfs ÷ 1.547 cfs/MGD = 8.016, which rounds to about 8.0 MGD.

  6. Answer: 38,400 cubic feet.
    Volume = 120 × 80 × 4 = 38,400 cubic feet.

  7. Answer: 287,232 gallons.
    38,400 cubic feet × 7.48 gallons/cubic foot = 287,232 gallons.

  8. Answer: about 61.5 minutes.
    Net inflow = 12.4 − 2.0 = 10.4 cfs; 10.4 × 60 = 624 cubic feet per minute. Fill time = 38,400 ÷ 624 = 61.54, which rounds to about 61.5 minutes.

  9. Answer: 0.62.
    Composite C = 12.4 ÷ (4.0 × 5.0) = 12.4 ÷ 20 = 0.62. This matches the area-weighted value (0.9 × 3 + 0.2 × 2) ÷ 5 = 3.1 ÷ 5 = 0.62.

  10. Answer: Pavement raises peak runoff.
    Accept any answer explaining that pavement has a far higher runoff coefficient (0.9) than lawn (0.2), so the same storm sheds much more water off the paved area instead of soaking in. The detention basin catches that extra water and lets it out slowly through a small outlet, so the storm drain downstream never has to carry the full peak flow at once.

Differentiation

Support (below grade level)

Provide the rational-method formula and the two flow conversions on a reference card, and complete problem 1 together so students see how the coefficient, intensity, and area multiply before they work problem 2 on their own.

Extension (above grade level)

Ask students to find the smallest outlet release rate, in cfs, that would keep the basin from overflowing during a 90-minute storm at the problem 3 peak flow, working from the basin volume in problem 6, and to explain the trade-off between a faster release and a smaller required basin.

ELL supports

Provide a labeled site diagram showing the paved area, the lawn area, and the basin, with the terms runoff coefficient, intensity, and peak flow marked beside a short definition. Allow students to show problems 1 through 9 as numbered calculations and to answer problem 10 with a labeled sketch plus one or two sentences.