Worksheet Grade 12 Science
Rational Method Runoff for a Thorncliff Development
A grade 12 science worksheet applying the rational method to a fictional parcel, comparing peak runoff before and after paving and sizing the detention it needs.
Objectives
- Students will apply the rational method to calculate peak runoff before and after development.
- Students will compute a composite runoff coefficient from the mix of surfaces on a parcel.
- Students will convert peak flows between cubic feet per second and gallons per minute and size a detention volume.
Vocabulary
Key words: rational method, runoff coefficient, peak flow, detention, impervious
Teacher overview
Engineers estimate the peak runoff from a small drainage area with the rational method, Q = C x i x A, where C is a runoff coefficient, i is rainfall intensity in inches per hour, and A is the drainage area in acres. The result comes out in cubic feet per second. This worksheet uses a fictional 10-acre parcel in the town of Thorncliff, about to be developed near Kestrel Run, to compare runoff before and after paving and to size the detention needed to hold the difference.
A common misconception is that the runoff coefficient of a mixed parcel is just an average of the surface values. It is an area-weighted composite: a parcel that is mostly pavement behaves much more like pavement than like lawn. Item 2 asks students to compute that composite correctly before they use it.
Student Worksheet
A 10-acre parcel in Thorncliff (a fictional town used for this worksheet) drains toward Kestrel Run. The design storm has a rainfall intensity of 3.0 inches per hour. Before development the parcel is all meadow with a runoff coefficient of 0.2. After development it will be 6 acres of pavement and roof (coefficient 0.9) and 4 acres of lawn (coefficient 0.2). Use these formulas and conversions for every item.
- Rational method: Q (cfs) = C x i x A, with i in inches per hour and A in acres
- Composite C = (sum of each area x its coefficient) / total area
- Conversions: 1 cfs = 448.8 gpm; 1 cubic foot = 7.48 gallons
Calculate the peak runoff, in cfs, from the 10-acre parcel before development (all meadow, C = 0.2).
Calculate the composite runoff coefficient after development, combining the 6 acres of pavement (0.9) and the 4 acres of lawn (0.2).
Calculate the peak runoff, in cfs, after development, using the composite coefficient from item 2.
Calculate the increase in peak runoff, in cfs, caused by the development.
Convert the after-development peak runoff from item 3 to gallons per minute. Round to the nearest whole gpm.
Convert the increase in peak runoff from item 4 to gallons per minute. Round to the nearest whole gpm.
A detention pond must hold the extra runoff. Suppose it captures the increase from item 4 for 30 minutes at that rate. Calculate the detention volume needed, in gallons. Round to the nearest whole gallon.
Convert the detention volume from item 7 to cubic feet, using 1 cubic foot = 7.48 gallons. Round to the nearest whole cubic foot.
Explain why replacing meadow with pavement increases peak runoff even though the same amount of rain falls on the parcel.
Suppose the developer adds permeable pavement and a rain garden that lower the composite coefficient. Explain how that would change the after-development peak runoff and the detention volume the site needs.
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: 6.0 cfs.
Q = C x i x A = 0.2 x 3.0 x 10 = 6.0 cfs.Answer: 0.62.
Composite C = (6 x 0.9 + 4 x 0.2) / 10 = (5.4 + 0.8) / 10 = 6.2 / 10 = 0.62.Answer: 18.6 cfs.
Q = 0.62 x 3.0 x 10 = 18.6 cfs.Answer: 12.6 cfs.
Increase = 18.6 cfs - 6.0 cfs = 12.6 cfs.Answer: 8,348 gpm.
18.6 cfs x 448.8 gpm per cfs = 8,347.68 gpm, which rounds to 8,348 gpm.Answer: 5,655 gpm.
12.6 cfs x 448.8 gpm per cfs = 5,654.88 gpm, which rounds to 5,655 gpm.Answer: 169,646 gal.
5,654.88 gpm x 30 minutes = 169,646.4 gallons, which rounds to 169,646 gallons.Answer: 22,680 cu ft.
169,646.4 gallons / 7.48 gallons per cubic foot = 22,680 cubic feet (the same as 12.6 cfs x 1,800 seconds).Answer: Pavement lets almost no water soak in.
Meadow soil absorbs much of the rain, but pavement and roofs are nearly impervious, so a far larger share of the same rainfall runs off the surface quickly. That is why the runoff coefficient rises from 0.2 to a much higher composite value after development.Answer: Both the peak runoff and the detention volume would drop.
Permeable pavement and a rain garden let some rain infiltrate again, lowering the composite coefficient. A lower coefficient reduces the after-development peak runoff, which shrinks the increase over the pre-development flow and the detention volume needed to hold it.
Differentiation
Support (below grade level)
Provide the rational-method formula and both conversions on a reference card, and complete item 1 together as a worked example. Give students the composite-coefficient setup for item 2 with the two products written but not yet added.
Extension (above grade level)
Ask students to find the composite runoff coefficient that would hold the after-development peak runoff to no more than 12 cfs, then describe a mix of surfaces (pavement, lawn, permeable pavement) that could reach it.
ELL supports
Pre-teach "runoff," "impervious," "coefficient," and "detention" with a simple diagram of rain hitting a meadow versus a parking lot. Allow students to answer items 9 and 10 with a labeled sketch or a short sentence frame.
Extensions
Have students walk the school grounds and estimate the fraction that is roof, pavement, and lawn, then assign coefficients and compute a rough composite for their own site. Comparing that composite with Thorncliff's after-development value of 0.62 grounds the method in a place they know.