Math problem set Grade 12 Science

Primary Clarifier Detention and Overflow at Netherford

A grade 12 science problem set sizing a fictional primary clarifier: volume, detention time, surface overflow rate, BOD removal, and the pounds of BOD removed per day.

Grade: Grade 12
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 units through the volume, detention-time, removal-efficiency, and pounds-formula steps in problems 1 through 9, the multi-step dimensional analysis this standard requires.

All materials for HSN-Q.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: Problem 5 treats surface overflow rate as a density, gallons per day spread over square feet of clarifier surface, the density-modeling skill this standard names directly.

All materials for HSG-MG.A.2 →

HS-ESS3-4 NGSS

Evaluate or refine a technological solution that reduces impacts of human activities on natural systems.

How this material aligns: Problems 6 through 10 evaluate the clarifier and the biological stage that follows it as a pollution-control solution, computing BOD removal efficiency and the pounds of BOD kept out of the receiving water each day against the 30 mg/L secondary-treatment limit.

All materials for HS-ESS3-4 →

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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 volume and detention time of a rectangular primary clarifier.
Students will calculate surface overflow rate and BOD removal efficiency from plant data.
Students will use the pounds formula to find the pounds of BOD removed per day.

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.
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, removal efficiency, BOD

Teacher overview

A primary clarifier is a large settling tank where the heaviest solids drop out of wastewater before biological treatment. Operators size and check it with a few standard numbers: how long water stays in the tank (detention time), how fast water rises through it (surface overflow rate), and how much organic load it removes. This set uses the fictional Netherford Water Reclamation Facility to walk through those calculations and finish with the pounds formula, which turns a concentration and a flow into a daily mass.

A common misconception is that detention time depends only on the tank. It depends on flow as well: the same tank gives water less contact time as flow rises. Another is forgetting that removal efficiency is measured against the influent, not the final limit. Problems 6 and 7 keep those two ideas distinct.

Primary Clarifier Detention and Overflow at Netherford — Grade 12 Science

Name: ____________________________ Date: ______________

Problems

The Netherford Water Reclamation Facility (a fictional plant used for this problem set) runs a rectangular primary clarifier 80 feet long, 20 feet wide, and 12 feet deep, treating a flow of 1.5 MGD. Raw sewage enters with a BOD of 220 mg/L. Use these formulas and facts for every problem.

Rectangular volume (cu ft) = length x width x depth
Detention time (days) = volume (gallons) / flow (gallons per day); multiply by 24 for hours
Surface overflow rate = flow (gallons per day) / surface area (sq ft)
Removal efficiency (%) = (influent - effluent) / influent x 100
Pounds formula: pounds per day = concentration (mg/L) x flow (MGD) x 8.34
Conversions: 1 cubic foot = 7.48 gallons; 1 MGD = 1,000,000 gallons per day

Calculate the clarifier volume in cubic feet, using the 80 by 20 by 12 foot dimensions.
Convert the volume from problem 1 to gallons, using 1 cubic foot = 7.48 gallons.
Calculate the detention time in hours at the flow of 1.5 MGD. Round to the nearest tenth.
Netherford's design standard calls for 1.5 to 2.5 hours of detention in its primary clarifier. Is the clarifier within that range? Answer yes or no.
Calculate the surface overflow rate in gallons per day per square foot, using the tank's 80 by 20 foot surface. Round to the nearest tenth.
The clarifier lowers BOD from the raw 220 mg/L to 150 mg/L. Calculate the clarifier's BOD removal efficiency as a percentage. Round to the nearest tenth.
After the downstream biological process, the final effluent BOD is 18 mg/L. Calculate the overall removal efficiency from the raw 220 mg/L to the final 18 mg/L. Round to the nearest tenth.
Using the pounds formula, calculate the pounds per day of BOD entering the plant at 220 mg/L and 1.5 MGD. Round to the nearest tenth.
Calculate the pounds per day of BOD leaving in the final effluent at 18 mg/L and 1.5 MGD, then the pounds of BOD removed per day. Round to the nearest tenth.
The secondary-treatment standard is a monthly-average effluent BOD of 30 mg/L or less. Does the 18 mg/L final effluent meet it, and by what margin? Explain in one sentence.

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: 19,200 cu ft.
Volume = length x width x depth = 80 x 20 x 12 = 19,200 cubic feet.
Answer: 143,616 gallons.
19,200 cubic feet x 7.48 gallons per cubic foot = 143,616 gallons.
Answer: 2.3 hours.
Detention time = volume / flow = 143,616 gallons / 1,500,000 gallons per day = 0.09574 days; x 24 = 2.298, which rounds to 2.3 hours.
Answer: Yes.
2.3 hours falls within Netherford's 1.5 to 2.5 hour design range.
Answer: 937.5 gal/day per sq ft.
Surface area = 80 x 20 = 1,600 sq ft. Surface overflow rate = 1,500,000 gallons per day / 1,600 sq ft = 937.5 gallons per day per square foot.
Answer: 31.8 %.
Removal = (220 - 150) / 220 x 100 = 70 / 220 x 100 = 31.82, which rounds to 31.8 percent.
Answer: 91.8 %.
Removal = (220 - 18) / 220 x 100 = 202 / 220 x 100 = 91.82, which rounds to 91.8 percent.
Answer: 2,752.2 lbs/day.
Pounds per day = 220 mg/L x 1.5 MGD x 8.34 = 2,752.2 pounds per day entering the plant.
Answer: 2,527.0 lbs/day.
Effluent load = 18 mg/L x 1.5 MGD x 8.34 = 225.2 pounds per day. Removed = 2,752.2 - 225.2 = 2,527.0 pounds per day.
Answer: Yes, by 12 mg/L.
The 18 mg/L final effluent is below the 30 mg/L standard, clearing it by 30 - 18 = 12 mg/L of margin.

Differentiation

Support

Provide the formulas on a reference card and complete problem 1 and problem 2 together. Give a labeled place for each intermediate value (volume in cubic feet, then gallons) so the detention-time step has its input ready.

Extension

Ask students to find the flow, in MGD, at which the clarifier's detention time would drop to exactly 2.0 hours, holding the volume at 143,616 gallons. They should set up and solve the detention-time equation for the unknown flow.

ELL supports

Pre-teach "clarifier," "detention time," "overflow rate," and "removal efficiency" with a diagram of a settling tank showing solids dropping to the bottom. Allow students to answer problem 10 with a labeled calculation rather than a full sentence.

Extensions

Have students compare the clarifier's 31.8 percent BOD removal with the 91.8 percent overall removal and write a short note explaining why the primary clarifier alone cannot meet the 30 mg/L standard, so a biological stage must follow it.