Math problem set Grade 10 Science

Greywillow Plant: Detention Time and Filter Loading Rates

A grade 10 problem set at the fictional Greywillow treatment plant, computing basin volume, detention time, and filter loading rate and comparing each to design ranges.

Grade: Grade 10
Subject: Science
Material: Math problem set
Time: 45 min
Prep: No prep
Materials: Calculator

Standards

HSN-Q.A.1 Common Core Math

Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays.

How this material aligns: Students carry cubic feet, gallons, gpm, and hours through the multistep detention-time and loading-rate calculations in problems 3, 4, and 6, keeping units consistent throughout.

All materials for HSN-Q.A.1 →

HSG-MG.A.1 Common Core Math

Use geometric shapes, their measures, and their properties to describe objects (e.g., modeling a tree trunk or a human torso as a cylinder).

How this material aligns: In problem 9 students model a round clarifier as a cylinder and use the 0.785 x diameter squared x depth formula to estimate its volume, describing a real tank with a geometric shape.

All materials for HSG-MG.A.1 →

HS-ETS1-2 NGSS

Design a solution to a complex real-world problem by breaking it down into smaller, more manageable problems that can be solved through engineering.

How this material aligns: The set breaks the plant into separately solvable stages, with problems 2 through 5 sizing the sedimentation basin against its detention-time range, problems 6 through 8 sizing the filters against their loading-rate range, and problem 9 sizing the clarifier.

All materials for HS-ETS1-2 →

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This material is in the public domain (CC0) and may be copied, edited, and shared freely, without attribution. Updated 2026-07-23.

Objectives

Students will calculate basin volume in cubic feet and convert it to gallons.
Students will calculate detention time and compare it to the standard design range.
Students will calculate a filter loading rate and judge it against the typical range.

Vocabulary

Detention time: Detention time is how long water stays inside a tank or basin before flowing out. Water must stay long enough for each treatment step, like settling or disinfecting, to finish its job.
Clarifier: A clarifier is a big round or rectangular tank where water sits calmly so solids can sink to the bottom and be scraped away. Treatment plants use clarifiers at more than one stage of cleaning.
Flow rate: Flow rate measures how much water moves past a point in a certain amount of time, like gallons per minute through a garden hose. Bigger pipes and stronger pushes give bigger flow rates.

Also: sedimentation basin, filter loading rate

Teacher overview

The Greywillow Water Treatment Plant is a fictional plant that removes particles from river water before disinfection. Two numbers tell operators whether the plant is sized correctly: the detention time, or how long water sits in the settling basin, and the filter loading rate, or how fast water passes through the filters. This set walks students through both, using the plant's stated flow and tank dimensions. A common misconception is dividing a volume in cubic feet by a flow measured in gallons; the volume must be converted to gallons first, and problem 3 forces that step before problem 4.

Greywillow Plant: Detention Time and Filter Loading Rates — Grade 10 Science

Name: ____________________________ Date: ______________

Problems

The Greywillow plant (a fictional plant) treats a flow of 2.0 MGD. Its rectangular sedimentation basin is 80 ft long and 30 ft wide, with the water 12 ft deep. Use these conversions and facts: 1 MGD = 694 gpm; 1 cu ft = 7.48 gal; 2.0 MGD = 2,000,000 gallons per day; sedimentation basins are typically designed for 2 to 4 hours of detention; conventional filters typically run 2 to 6 gpm per sq ft.

Convert the plant's flow of 2.0 MGD to gallons per minute, using 1 MGD = 694 gpm.
Find the volume of the rectangular sedimentation basin in cubic feet, using volume = length x width x depth.
Convert the basin volume from problem 2 to gallons, using 1 cu ft = 7.48 gal.
Find the detention time in the basin in hours. Use detention time = volume (gal) divided by flow, with the flow of 2,000,000 gallons per day, then multiply the result in days by 24 to get hours. Round to the nearest tenth of an hour.
Sedimentation basins are typically designed for 2 to 4 hours of detention. Does Greywillow's basin fall within that range?
The plant's filters have a total surface area of 250 sq ft. Find the filter loading rate in gpm per sq ft, using loading rate = flow (gpm) divided by filter area, with the gpm from problem 1.
Conventional filters typically run 2 to 6 gpm per sq ft. Is Greywillow's loading rate within that range, and is it near the low or the high end?
To lower the loading rate, the plant adds filter area, bringing the total surface area to 350 sq ft. Find the new loading rate in gpm per sq ft. Round to the nearest hundredth.
A separate round clarifier is 40 ft in diameter with the water 10 ft deep. Find its volume in cubic feet, using volume = 0.785 x diameter squared x depth.
Using your detention-time answer, explain why running the plant at a higher flow than 2.0 MGD would shorten the detention time and could let more particles pass through without settling.

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: 1,388 gpm. 2.0 MGD x 694 gpm per MGD = 1,388 gallons per minute.
Answer: 28,800 cu ft. Volume = 80 ft x 30 ft x 12 ft = 28,800 cubic feet.
Answer: 215,424 gal. 28,800 cu ft x 7.48 gal per cu ft = 215,424 gallons.
Answer: 2.6 hours. Detention time = 215,424 gal divided by 2,000,000 gal per day = 0.10771 days, and 0.10771 x 24 = 2.585, which rounds to 2.6 hours.
Answer: Yes. The detention time of about 2.6 hours falls inside the typical 2-to-4-hour design range.
Answer: 5.55 gpm per sq ft. Loading rate = 1,388 gpm divided by 250 sq ft = 5.552, which rounds to 5.55 gpm per square foot.
Answer: Within the range, near the high end. 5.55 gpm per sq ft is inside the 2-to-6 range but close to the top of 6, so the filters are working near their design limit.
Answer: 3.97 gpm per sq ft. Loading rate = 1,388 gpm divided by 350 sq ft = 3.966, which rounds to 3.97 gpm per square foot.
Answer: 12,560 cu ft. Volume = 0.785 x 40 squared x 10 = 0.785 x 1,600 x 10 = 12,560 cubic feet.
Answer: A higher flow means the same basin volume is divided by a larger flow, so the detention time gets shorter. With less time in the basin, particles have less time to settle, so more of them can carry through to the filters and beyond.

Differentiation

Support

Provide a printed conversion card with "1 MGD = 694 gpm," "1 cu ft = 7.48 gal," and the detention-time and loading-rate formulas. Work problem 1 with students before they continue, and allow calculators throughout.

Extension

Ask students to find the maximum flow, in MGD, at which Greywillow's basin would still hold water for the full 2-hour minimum detention time. They should rearrange the detention-time formula, solve for the flow, and convert the result to MGD.

ELL supports

Provide a labeled diagram of the plant showing the sedimentation basin and the filters, with the dimensions marked. Allow students to show their work using the diagram and numbers rather than full sentences, and pair the words "detention time" and "loading rate" with simple definitions.