Worksheet Grade 10 Science

Marrow Creek Reclamation Plant: BOD Removal and Activated Sludge

A grade 10 science worksheet set at the fictional Marrow Creek plant, where students compute BOD and TSS removal efficiency and check secondary-treatment compliance.

Grade: Grade 10
Subject: Science
Material: Worksheet
Time: 40 min
Prep: No prep
Materials: None

Standards

HS-ESS3-4 NGSS

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

How this material aligns: Students evaluate how the plant's activated sludge process reduces the pollutant load discharged to a river, refining the design with removal-efficiency data in items 3 through 7.

All materials for HS-ESS3-4 →

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 milligram-per-liter and percent units through the multistep removal-efficiency calculations in items 3, 4, 6, and 7, interpreting each result against a stated standard.

All materials for HSN-Q.A.1 →

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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 explain the role of microorganisms in the activated sludge process.
Students will calculate BOD and TSS removal efficiency from influent and effluent concentrations.
Students will judge whether an effluent value meets a stated secondary-treatment standard.

Vocabulary

Activated sludge: Activated sludge is a living brown mixture of helpful bacteria and other microbes that wastewater plants grow on purpose. The microbes clean the water by eating the waste in it.
Biochemical oxygen demand (BOD): Biochemical oxygen demand, or BOD, measures how much oxygen tiny microbes will use up as they eat the waste in water. Dirtier water has a higher BOD.
Effluent: Effluent is the cleaned water that leaves a treatment plant at the end of the process. It is tested carefully before being released to a river, lake, or the ocean.
Aeration: Aeration means mixing air into water. At a wastewater plant, blowers bubble huge amounts of air through tanks so the helpful microbes have the oxygen they need to eat the waste.

Also: removal efficiency

Teacher overview

The Marrow Creek Water Reclamation Facility is a fictional plant that cleans sewage before returning it to a river. Its core step is the activated sludge process, in which a tank full of bacteria feeds on the dissolved waste. This worksheet has students measure how well that process works by computing removal efficiency for two pollutants, then checking the treated water against the secondary-treatment standard of 30 mg/L. Watch for the common error of subtracting in the wrong order in the removal-efficiency formula, which flips the percent into a negative or nonsense value.

Marrow Creek Reclamation Plant: BOD Removal and Activated Sludge — Grade 10 Science

Name: ____________________________ Date: ______________

Student Worksheet

The Marrow Creek Water Reclamation Facility (a fictional plant) reports these values for one day. Raw sewage entering the plant (the influent) has a BOD of 240 mg/L and a TSS of 220 mg/L. After the activated sludge process, the treated water leaving the plant (the effluent) has a BOD of 18 mg/L and a TSS of 15 mg/L. Use the formula: removal efficiency (%) = (influent − effluent) ÷ influent × 100.

In one or two sentences, explain what the bacteria in the activated sludge tank do to the sewage.
BOD stands for biochemical oxygen demand. Explain why water with a high BOD can lower the dissolved oxygen a river needs to keep fish alive, remembering that most fish need oxygen above 5 mg/L.
Calculate the BOD removal efficiency for the day, using influent 240 mg/L and effluent 18 mg/L. Round to the nearest tenth of a percent.
Calculate the TSS removal efficiency, using influent 220 mg/L and effluent 15 mg/L. Round to the nearest tenth of a percent.
The secondary-treatment standard is an effluent BOD of 30 mg/L or less. Does the plant's effluent BOD of 18 mg/L meet the standard, and by how many mg/L does it clear or miss it?
During a storm, the influent BOD rises to 300 mg/L while the removal efficiency stays at the value you found in item 3 (92.5 percent). Calculate the new effluent BOD, then state whether it still meets the 30 mg/L standard.
To always meet the 30 mg/L standard when the influent BOD is 300 mg/L, what is the minimum removal efficiency the plant must reach? Round to the nearest whole percent.
Explain why operators pump air into the aeration basin to keep dissolved oxygen high, given that the bacteria there are aerobic.
Some settled sludge is pumped back into the aeration basin instead of being thrown away. Explain what returning this sludge does for the process.
Suppose the effluent flowed into a small lake instead of the river. Explain how lowering the BOD before discharge helps protect the lake's dissolved oxygen and the fish that need it above 5 mg/L.

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: They break down the dissolved organic waste. The bacteria in the activated sludge tank use the sewage as food, consuming the organic material and settling out as a thicker sludge, which leaves the water cleaner.
Answer: High BOD means more organic waste for bacteria to consume, and those bacteria use up dissolved oxygen as they feed. If the oxygen falls below about 5 mg/L, fish can suffocate, so a high BOD load can indirectly kill fish.
Answer: 92.5 % BOD removal. (240 − 18) ÷ 240 × 100 = 222 ÷ 240 × 100 = 92.5 percent.
Answer: 93.2 % TSS removal. (220 − 15) ÷ 220 × 100 = 205 ÷ 220 × 100 = 93.18, which rounds to 93.2 percent.
Answer: Yes, it meets the standard by 12 mg/L. The effluent BOD of 18 mg/L is below the 30 mg/L limit, and 30 − 18 = 12 mg/L of margin.
Answer: 22.5 mg/L, which still meets the standard. Effluent = 300 × (1 − 0.925) = 300 × 0.075 = 22.5 mg/L, and 22.5 is below the 30 mg/L limit.
Answer: 90 %. The plant must cut 300 mg/L down to 30 mg/L, so (300 − 30) ÷ 300 × 100 = 270 ÷ 300 × 100 = 90 percent minimum removal.
Answer: The bacteria are aerobic, meaning they need oxygen to break down waste. Pumping air keeps enough dissolved oxygen in the basin for the bacteria to stay active and keep consuming the BOD.
Answer: Returning the sludge keeps a large, healthy population of the waste-eating bacteria in the aeration basin. Without that return, the bacteria would wash out of the tank and the process would slow down.
Answer: Lowering the BOD means the effluent carries far less organic waste into the lake, so lake bacteria use up less dissolved oxygen. That keeps the lake's oxygen above the 5 mg/L that fish need, protecting them.

Differentiation

Support

Provide the removal-efficiency formula already set up with labeled blanks for the influent and effluent values, and work item 3 together as a model before students attempt items 4, 6, and 7 on their own. Allow a calculator throughout.

Extension

Ask students to design a two-stage version of the plant in which a primary step removes 35 percent of the incoming BOD before the activated sludge stage. Have them calculate the combined removal efficiency needed for the effluent to reach 30 mg/L from an influent of 300 mg/L, and show every step.

ELL supports

Pre-teach "influent," "effluent," "removal efficiency," and "aeration" with a simple labeled flow diagram of water entering the plant, passing through the aeration basin, and leaving. Provide sentence frames for the explanation items, such as "The bacteria use oxygen because ___."