Math problem set Grade 12 Careers & CTE

Netherstowe Stormwater District: Rational Method Technician Math

Grade 12 problem set applying the rational method, area-weighted runoff coefficients, and pipe capacity to a fictional stormwater district facing new development.

Objectives

  1. Students will calculate an area-weighted runoff coefficient for a drainage area with mixed surfaces.
  2. Students will apply the rational method to find peak stormwater flow and convert it among cfs, gpm, and MGD.
  3. Students will size a detention basin footprint from a required storage volume and a fixed basin depth.

Vocabulary

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.
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, peak flow, continuity equation

Teacher overview

A stormwater technician in the fictional town of Netherstowe reviews every development plan submitted to the district and checks whether the existing pipe network can carry the runoff the project will generate. This set follows one such review from an area-weighted runoff coefficient through peak flow, pipe velocity, and finally the detention storage the district would require from the developer. The misconception the set is designed to correct is the habit of applying a single runoff coefficient to a whole drainage area, because students commonly use the pavement coefficient of 0.9 for a site that is more than half lawn and then overstate peak flow by roughly eighty percent. Problem 1 forces the area weighting first, and every later answer depends on getting it right.

Answer Key

  1. Answer: 0.50. Weighted C = (3.6 acres x 0.9 + 4.8 acres x 0.2) / 8.4 acres. The pavement contributes 3.6 x 0.9 = 3.24 and the lawn contributes 4.8 x 0.2 = 0.96, for a total of 4.20. Dividing, 4.20 / 8.4 = 0.50.
  2. Answer: 10.5 cfs. Q = 0.50 x 2.5 in/hr x 8.4 acres. First, 0.50 x 2.5 = 1.25. Then 1.25 x 8.4 = 10.5 cubic feet per second.
  3. Answer: 4,712.4 gpm. Flow = 10.5 cfs x 448.8 gpm per cfs = 4,712.4 gallons per minute.
  4. Answer: 6.79 MGD. Flow = 10.5 cfs / 1.547 cfs per MGD = 6.787, which rounds to 6.79 million gallons per day.
  5. Answer: 2.1 ft/s. A 30-inch pipe is 2.5 feet in diameter, so its area = 0.785 x 2.5 x 2.5 = 4.90625 square feet. Rearranging Q = A x V gives V = 10.5 cfs / 4.90625 sq ft = 2.140, which rounds to 2.1 feet per second.
  6. Answer: 0.70. Weighted C = (6.0 acres x 0.9 + 2.4 acres x 0.2) / 8.4 acres. The pavement contributes 6.0 x 0.9 = 5.40 and the lawn contributes 2.4 x 0.2 = 0.48, for a total of 5.88. Dividing, 5.88 / 8.4 = 0.70.
  7. Answer: 14.7 cfs. Q = 0.70 x 2.5 in/hr x 8.4 acres. First, 0.70 x 2.5 = 1.75. Then 1.75 x 8.4 = 14.7 cubic feet per second.
  8. Answer: 4.2 cfs. The increase = 14.7 cfs minus 10.5 cfs = 4.2 cubic feet per second, an increase of 40 percent over the pre-development peak.
  9. Answer: 7,560 cubic feet. Storage = 4.2 cubic feet per second x 1,800 seconds = 7,560 cubic feet. Students who use 30 rather than 1,800 will report 126 cubic feet, which is a signal that minutes were never converted to seconds.
  10. Answer: 2,520 square feet. Footprint = 7,560 cubic feet / 3 feet of depth = 2,520 square feet, roughly the area of a rectangle 60 feet by 42 feet.

Differentiation

Support

Give students a partially completed weighting table for problem 1 with the pavement row filled in, so the structure of the calculation is visible before they attempt problem 6 on their own. Provide the diameter conversion from 30 inches to 2.5 feet directly in problem 5, and supply a worked example of the continuity equation rearranged for velocity. Students who finish problems 1 through 5 have met the core objective even without the redevelopment sequence.

Extension

Ask students to determine the pipe diameter that would carry the post-development peak flow at the same velocity found in problem 5, then discuss why a district would prefer detention storage over replacing a pipe. Students can also recalculate the whole sequence at a higher design rainfall intensity of their choosing and describe how sensitive the required basin footprint is to that single input.

ELL supports

Post the rational method equation with each variable labeled in words as well as symbols, and pre-teach impervious, runoff coefficient, peak flow, and detention. Provide a diagram of the drainage area with the pavement and lawn portions shaded and labeled with their acreages so the weighting in problems 1 and 6 has a visual anchor. Allow calculations to stand as the answer without a written explanation.

Extensions

Have students obtain a real development plan set from a local planning department, identify the drainage area boundaries and surface types on the site plan, and repeat this sequence with the plan's own numbers. Students can also compare how two nearby municipalities define their design storm and write a short memo explaining what a different design intensity would change about the basin they sized in problem 10.