Math problem set Grade 10 Math

Scarrow Bend Clarifier: Circle Geometry and Loading Rates

A grade 10 problem set on a fictional circular clarifier where students use area, volume, and circumference to find overflow rates, detention time, and a new tank size.

Grade: Grade 10
Subject: Math
Material: Math problem set
Time: 50 min
Prep: No prep
Materials: Calculator

Standards

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: Problems 1, 3, and 5 model the clarifier as a cylinder and a circle to obtain its settling area, storage volume, and weir circumference.

All materials for HSG-MG.A.1 →

HSG-MG.A.2 Common Core Math

Apply concepts of density based on area and volume in modeling situations (e.g., persons per square mile, BTUs per cubic foot).

How this material aligns: Problems 2, 6, and 8 express plant flow as gallons per day per square foot of surface and per foot of weir, applying density concepts based on area and length.

All materials for HSG-MG.A.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 the surface area, volume, and weir length of a circular clarifier.
Students will express plant flow as a surface overflow rate and a weir overflow rate.
Students will resize a clarifier so a peak flow stays within a stated loading limit.

Vocabulary

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.
Weir: A weir is a low wall that water flows over, used to control water levels and to measure flow. The deeper the water pouring over the weir, the more water is flowing.
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.

Also: surface overflow rate, circumference

Teacher overview

The Scarrow Bend Water Resource Recovery Facility is fictional, and its tank dimensions, flows, and design limits are set for this problem set. A circular clarifier lets solids settle out while clarified water spills over a weir at the rim, so its performance depends on how much flow each square foot of surface and each foot of weir has to handle. The misconception this set targets is expecting settling area to grow at the same rate as diameter; problems 9 and 10 show that a 15.7 percent wider tank gains 33.9 percent more area, because area depends on the square of the diameter. Students also tend to use the diameter in place of the radius, so the 0.785 form of the area formula is given to keep the focus on the modeling.

Scarrow Bend Clarifier: Circle Geometry and Loading Rates — Grade 10 Math

Name: ____________________________ Date: ______________

Problems

The Scarrow Bend facility is fictional, and all values here are set for this set of problems. Its secondary clarifier is a circular tank 70 feet across with a side water depth of 12 feet. Average flow through the tank is 2.4 MGD, which is 2,400,000 gallons per day. Clarified water leaves over a weir that runs around the full rim. The plant's design limits are 700 gallons per day per square foot of surface and 15,000 gallons per day per foot of weir. Use these formulas and conversions: circle area = 0.785 × diameter squared; circumference = 3.14 × diameter; 1 cubic foot holds 7.48 gallons; 1 day = 24 hours.

Find the water surface area of the clarifier in square feet.
Find the surface overflow rate at 2.4 MGD, in gallons per day per square foot. Round to the nearest tenth.
Find the tank volume in gallons at the 12-foot side water depth. Round to the nearest gallon.
Find the detention time in hours at 2.4 MGD. Round to the nearest hundredth.
Find the weir length in feet, using the circumference of the 70-foot circle. Round to the nearest tenth.
Find the weir overflow rate at 2.4 MGD, in gallons per day per foot of weir. Round to the nearest whole number.
Does the problem 6 result meet the plant's limit of 15,000 gallons per day per foot? Answer yes or no.
During a wet-weather peak the flow rises to 3.6 MGD, or 3,600,000 gallons per day. Find the surface overflow rate at that flow. Round to the nearest tenth.
The plant wants a larger replacement clarifier that, on its own, holds the surface overflow rate at 700 gallons per day per square foot while passing the full 3.6 MGD. Find the diameter it needs, rounded up to the nearest whole foot.
Compare the surface area of the 70-foot tank with the surface area of the tank you found in problem 9. By what percent is the new area larger? Round to the nearest tenth.

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: 3,846.5 square feet. Area = 0.785 × 70 × 70 = 0.785 × 4,900 = 3,846.5 square feet.
Answer: 623.9 gallons per day per square foot. 2,400,000 ÷ 3,846.5 = 623.94, which rounds to 623.9. This is below the 700 limit, so the tank is within its design range at average flow.
Answer: 345,262 gallons. Volume = 3,846.5 × 12 = 46,158 cubic feet, and 46,158 × 7.48 = 345,261.8, which rounds to 345,262 gallons.
Answer: 3.45 hours. Detention time = 345,262 ÷ 2,400,000 = 0.14386 of a day, and 0.14386 × 24 = 3.45 hours.
Answer: 219.8 feet. Circumference = 3.14 × 70 = 219.8 feet of weir around the rim.
Answer: 10,919 gallons per day per foot. 2,400,000 ÷ 219.8 = 10,919.0 gallons per day for each foot of weir.
Answer: Yes. The rate of 10,919 gallons per day per foot is below the 15,000 limit, so the weir is not overloaded at average flow.
Answer: 935.9 gallons per day per square foot. 3,600,000 ÷ 3,846.5 = 935.92, which rounds to 935.9 and exceeds the 700 limit.
Answer: 81 feet. The area needed is 3,600,000 ÷ 700 = 5,142.9 square feet. Then diameter squared = 5,142.9 ÷ 0.785 = 6,551.4, and the square root is 80.94, which rounds up to 81 feet. Checking, 0.785 × 81 × 81 = 5,150.4 square feet, and 3,600,000 ÷ 5,150.4 = 699.0, just under the limit.
Answer: 33.9 percent. The increase is 5,150.4 − 3,846.5 = 1,303.9 square feet, and 1,303.9 ÷ 3,846.5 = 0.339, or 33.9 percent. The diameter grew only 15.7 percent, because area depends on the square of the diameter.

Differentiation

Support

Provide a top view of the tank with the diameter drawn and the weir traced around the rim, so students can see which measurement each formula uses. Give the area from problem 1 already computed for use in problems 2, 3, and 8, and supply the square-root step in problem 9 as a guided sequence of three labeled blanks.

Extension

Ask students to find the side water depth that would give the fictional 70-foot tank a detention time of exactly 4.00 hours at average flow, then say whether that depth seems reasonable for a settling tank. A second challenge is to determine how many 70-foot clarifiers running in parallel would keep the peak flow within the 700 gallons per day per square foot limit.

ELL supports

Provide a card matching each rate to the thing it is divided by: surface overflow rate to square feet of water surface, weir overflow rate to feet of weir, and detention time to the flow, since detention time is the tank's volume divided by the flow. Include a labeled sketch of the fictional tank with the words diameter, depth, and weir written directly on the drawing.

Extensions

Have students find a photograph or aerial image of a circular clarifier at a treatment plant, estimate its diameter from a scale bar, and repeat problems 1 and 2 using a plant flow their teacher supplies. Comparing that surface overflow rate with the 623.9 figure from the fictional Scarrow Bend tank shows how consistent these design rates are across real facilities.