Math problem set Grade 12 Science
Sizing a Bioretention Cell at Redmarch Commons
A grade 12 science problem set that takes a redevelopment site from percent impervious to a treatment volume, a cell footprint, a drawdown time, and a solids load.
Objectives
- Students will compute a runoff coefficient and a water quality volume for a redevelopment site.
- Students will size a bioretention cell footprint from a required storage volume and a storage depth per unit area.
- Students will estimate the solids mass captured in one design storm using the pounds formula.
Vocabulary
- Total suspended solids (TSS)
- Total suspended solids is the amount of solid material floating or suspended in water, measured by filtering a sample and weighing what gets left behind. High TSS makes water look cloudy and can smother fish habitat and clog treatment equipment.
Also: water quality volume, runoff coefficient, impervious cover, bioretention, drawdown time
Teacher overview
Redmarch Commons is a fictional redevelopment on a 3.2-acre lot in the fictional town of Harrowden. The town requires every redevelopment to capture and treat the runoff from small, frequent storms on site, and the developer has proposed a single bioretention cell to do it. This problem set takes students from the site's impervious cover through the required treatment volume, the cell's footprint, its drawdown time, and the mass of solids one design storm delivers.
Students routinely confuse the water quality volume with a peak flow. Peak flow, the quantity the rational method produces, is a rate in cubic feet per second and it governs pipe and culvert sizing. The water quality volume is a volume in cubic feet, sized for the small storms that carry most of the year's pollutant load, and a cell sized correctly for that volume will still be overtopped by a large storm by design. The second frequent error is multiplying rainfall by the full site area and skipping the runoff coefficient, which overstates the volume by nearly half on a site like this one, as problem 4 shows.
Problems
All values below describe the fictional Redmarch Commons site and the fictional Harrowden stormwater manual. Show every step with units.
- Total site area: 3.2 acres; impervious area after redevelopment: 2.24 acres
- Harrowden runoff coefficient: Rv = 0.05 + 0.009 x I, where I is the percent impervious written as a whole number
- Harrowden water quality design rainfall: 1.2 inches
- Water quality volume in cubic feet = rainfall (inches) x Rv x drainage area (square feet) / 12
- Conversions: 1 acre = 43,560 square feet; 1 cubic foot = 7.48 gallons
- Cell design: 9 inches of ponding above the surface, over a 2.5-foot media layer with 30 percent porosity
- Media infiltration rate: 2 inches per hour
- Event mean total suspended solids in the runoff: 80 mg/L; the cell's design removal for solids: 80 percent
- Pounds formula: pounds = concentration (mg/L) x volume (million gallons) x 8.34
Calculate the percent of the site that is impervious after redevelopment.
Using the Harrowden formula and your answer to problem 1, calculate the runoff coefficient Rv. Round to two decimal places.
Convert the 3.2-acre site area to square feet.
Calculate the water quality volume in cubic feet, using the 1.2-inch design rainfall, your Rv from problem 2, and your area from problem 3. Round to the nearest cubic foot.
Convert the water quality volume from problem 4 to gallons. Round to the nearest hundred gallons.
Each square foot of cell holds the ponded depth plus the water stored in the pore space of the media. Calculate the storage the cell provides per square foot of surface, in cubic feet per square foot.
Using your answers to problems 4 and 6, calculate the surface area the bioretention cell must have, in square feet. Round to the nearest square foot.
The town requires the ponded water to drain within 24 hours. Using the 2-inch-per-hour media rate, calculate how long the 9 inches of ponding takes to drain, in hours.
Calculate the mass of total suspended solids delivered to the cell in one design storm, in pounds, using the volume from problem 5 and the 80 mg/L event mean concentration. Round to one decimal place.
Using your answer to problem 9 and the cell's 80 percent design removal, calculate the mass of solids the cell keeps out of the receiving stream in that storm, in pounds. Round to one decimal place.
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: 70 percent.
2.24 impervious acres / 3.2 total acres = 0.70. Written as a whole-number percent for the Harrowden formula, I = 70.Answer: 0.68.
Rv = 0.05 + 0.009 x 70 = 0.05 + 0.63 = 0.68. A little over two-thirds of the rain that falls on this site leaves it as runoff.Answer: 139,392 square feet.
3.2 acres x 43,560 square feet per acre = 139,392 square feet.Answer: 9,479 cubic feet.
1.2 inches x 0.68 = 0.816. Then 0.816 x 139,392 square feet = 113,743.87, and dividing by 12 to convert the rainfall depth from inches to feet gives 9,478.66 cubic feet, which rounds to 9,479 cubic feet. Skipping the 0.68 coefficient would have produced 13,939 cubic feet, about 47 percent too large.Answer: 70,900 gallons.
9,478.66 cubic feet x 7.48 gallons per cubic foot = 70,900.4 gallons, which rounds to 70,900 gallons.Answer: 1.5 cubic feet per square foot.
The ponding contributes 9 inches, which is 0.75 feet. The media contributes 2.5 feet x 0.30 porosity = 0.75 feet of water-filled depth. Together, 0.75 + 0.75 = 1.5 cubic feet of storage for every square foot of cell surface.Answer: 6,319 square feet.
9,478.66 cubic feet / 1.5 cubic feet per square foot = 6,319.11 square feet, which rounds to 6,319 square feet. A cell roughly 79 feet on a side would do it.Answer: 4.5 hours.
The media drains at 2 inches per hour, so 9 inches of ponding / 2 inches per hour = 4.5 hours. That is well inside the 24-hour requirement, and it means the surface is dry between storms rather than holding standing water.Answer: 47.3 pounds.
First convert the volume: 70,900 gallons is 0.0709 million gallons. Then 80 mg/L x 0.0709 MG x 8.34 = 80 x 0.5913 = 47.30 pounds, which rounds to 47.3 pounds of solids delivered in the design storm.Answer: 37.8 pounds.
47.30 pounds x 0.80 = 37.84, which rounds to 37.8 pounds captured. The remaining 9.5 pounds leaves through the underdrain, so the cell reduces the solids load without eliminating it.
Differentiation
Support
Give students a completed unit map for the fictional Redmarch Commons chain, showing acres to square feet to cubic feet to gallons to million gallons with the conversion factor written on each arrow. Work problems 1 through 3 together so the runoff coefficient and area are correct before the volume work begins. For problem 6, provide a cross-section sketch of the cell with the ponding depth and media depth labeled in feet rather than inches.
Extension
Have students redesign the fictional cell for a site with 90 percent impervious cover at the same 3.2 acres, recompute problems 2 through 7, and report how much larger the footprint has to be. Ask them to then find the ponding depth that would hold the footprint at 6,319 square feet, and to comment on why towns cap ponding depth even when deeper ponding would save land.
ELL supports
Provide a term card pairing each vocabulary word with the symbol or unit it uses, for example water quality volume with cubic feet and runoff coefficient with no units at all. Supply a sentence frame for the written part of problem 8, such as "The cell drains in ___ hours, which is ___ than the ___-hour requirement." Let students narrate their unit conversions aloud to a partner before writing them.
Extensions
Students can compare the fictional Redmarch Commons cell against a permeable-pavement alternative that stores the same volume in an aggregate base with 40 percent porosity, computing the footprint each option needs and listing two maintenance differences. A second option is to have students find their own municipality's stormwater manual online, locate its water quality design rainfall, and explain how a different design depth would change the volume they calculated in problem 4.