Lesson plan Grade 11 Math

Wrenfield Development Runoff: Rational Method Design Lesson

A grade 11 lesson using the rational method to compare pre- and post-development peak runoff for a fictional site and design green infrastructure to a limit.

Objectives

  1. Students will compute peak runoff using the rational method Q equals C times i times A.
  2. Students will compare pre-development and post-development peak runoff for a site.
  3. Students will propose a green-infrastructure change that reduces post-development runoff toward a limit.

Vocabulary

Green infrastructure
Green infrastructure uses soil and plants, like rain gardens, green roofs, and street trees, to soak up rainwater where it falls instead of piping it away. It copies the way nature handles rain.

Also: rational method, runoff coefficient, peak runoff

Teacher overview

This lesson uses a fictional building site, the Wrenfield development, to show how covering land with pavement raises the peak rate of stormwater runoff, and how a design can bring that rate back down. Students apply the rational method, in which peak runoff equals a runoff coefficient times a rainfall intensity times a drainage area, and then compare a natural site to a paved one. A common misconception this lesson addresses is thinking that runoff depends only on how much rain falls, when the surface's runoff coefficient matters just as much. The lesson closes with a design task constrained by a common type of local rule that post-development runoff must not exceed the pre-development rate.

Procedure

Hook (5 min): Show two photos, a meadow and a parking lot, and ask students to predict which one sends more water to the storm drain during the same rainstorm, and why. Record predictions without resolving them yet.

Direct instruction (12 min): Introduce the rational method: peak runoff Q in cubic feet per second equals the runoff coefficient C, times the rainfall intensity i in inches per hour, times the drainage area A in acres. Explain that C is near 0.9 for pavement and near 0.2 for a lawn or meadow, so the same rain produces very different runoff depending on the surface. Introduce the fictional Wrenfield development, a 5-acre site, and a design storm intensity of 2.0 inches per hour.

Worked example (13 min): With the class, compute the meadow case first: Q equals 0.2 times 2.0 times 5, which is 2.0 cubic feet per second. Then compute the fully paved case: Q equals 0.9 times 2.0 times 5, which is 9.0 cubic feet per second. Highlight that the peak rate more than quadruples even though the rain is identical, and name that difference as the runoff the site must now manage.

Handout practice and design (22 min): Distribute the handout. Students confirm the two runoff values, then work a mixed-surface case and a green-infrastructure redesign, checking each result against the town's discharge limit. Circulate to confirm students multiply all three factors and keep the area in acres.

Closure and exit ticket (8 min): Poll the class on whether the permeable-pavement design in item 8 meets the town limit; if more than one-quarter answer incorrectly, review item 9 together. Collect the exit response naming one feature that reduces peak runoff.

Answer Key

  1. Answer: 2.0 cfs. Q = C × i × A = 0.2 × 2.0 × 5 = 2.0 cubic feet per second.
  2. Answer: 9.0 cfs. Q = 0.9 × 2.0 × 5 = 9.0 cubic feet per second.
  3. Answer: 7.0 cfs. Increase = 9.0 − 2.0 = 7.0 cubic feet per second.
  4. Answer: 6.2 cfs. Pavement: 0.9 × 2.0 × 3 = 5.4; lawn: 0.2 × 2.0 × 2 = 0.8; total = 5.4 + 0.8 = 6.2 cubic feet per second.
  5. Answer: 4.2 cfs. The mixed design's 6.2 exceeds the pre-development 2.0 by 6.2 − 2.0 = 4.2 cubic feet per second.
  6. Answer: 1,885 gpm. 4.2 cfs × 448.8 gallons per minute per cfs = 1,884.96, about 1,885 gallons per minute.
  7. Answer: A rain garden, bioswale, green roof, or permeable pavement. Any of these stores or infiltrates runoff so it leaves the site more slowly, which lowers the peak rate reaching the storm drain.
  8. Answer: 3.2 cfs. Paved area: 0.4 × 2.0 × 3 = 2.4; lawn: 0.2 × 2.0 × 2 = 0.8; new total = 2.4 + 0.8 = 3.2 cubic feet per second.
  9. Answer: no, over by 1.2 cfs. The 3.2 cubic feet per second still exceeds the 2.0 limit by 3.2 − 2.0 = 1.2, so more detention is needed.

Differentiation

Support

Provide the rational-method formula with the three factors labeled and the two runoff coefficients printed, and work item 1 and item 2 with students before releasing the mixed-surface items. Allow calculators throughout.

Extension

Ask students to show that any conventional pavement (C = 0.9) pushes the site above the pre-development 2.0 cubic feet per second, then find the largest area of permeable pavement (C = 0.4), in acres, the site could hold if the town allowed up to 2.5 cubic feet per second, with the rest left as meadow. (Answer: with x acres of permeable pavement, Q = 0.4(2.0)x + 0.2(2.0)(5 − x) = 2.0 + 0.4x, which is at most 2.5 when x is at most 1.25 acres.)

ELL supports

Pair "runoff coefficient" with images of a paved lot and a meadow labeled with 0.9 and 0.2, and provide a worked template showing the three factors multiplied in order. Give a sentence frame for item 7: "A ___ helps because it ___."