Worksheet Grade 9 Science
Ashvale Reclamation Plant: BOD Loading and Removal Efficiency
A grade 9 science worksheet using the pounds formula and removal-efficiency formula on a fictional plant's flow to judge how well it cleans wastewater.
Objectives
- Students will calculate pollutant loading in pounds per day from a concentration and a flow using the pounds formula.
- Students will calculate BOD and TSS removal efficiency and compare effluent values to the secondary treatment standard.
Vocabulary
- 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.
- 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.
- Influent
- Influent is the incoming dirty water that arrives at a treatment plant, ready to be cleaned. Operators test it to see how much work the plant has to do.
- 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.
Also: loading, removal efficiency
Teacher overview
The Ashvale Reclamation Plant is a fictional wastewater facility that treats the town's sewage and returns the cleaned water to the Millrace River. Operators judge the plant two ways: by concentration, measured in milligrams per liter, and by loading, the actual weight of pollutant handled each day. This worksheet has students move between those two views with the pounds formula and the removal-efficiency formula, then check the plant's effluent against the 30 mg/L secondary treatment standard. Watch for students who compare only concentrations and forget that a larger flow moves more pollutant at the same concentration.
Student Worksheet
Use these facts about the Ashvale Reclamation Plant, a fictional facility, for every item. Show your work and keep units attached to each number.
- Average flow treated: 2.5 MGD (million gallons per day).
- Water weighs 8.34 pounds per gallon.
- Pounds per day of a substance = concentration (mg/L) x flow (MGD) x 8.34.
- Removal efficiency (%) = (influent - effluent) / influent x 100.
- Secondary treatment must bring monthly-average BOD and TSS to 30 mg/L or less.
- BOD readings: 220 mg/L entering (influent), 15 mg/L leaving (effluent).
- TSS readings: 240 mg/L entering, 16 mg/L leaving.
- Convert the plant's flow of 2.5 MGD to gallons per minute, using the fact that 1 MGD equals 694 gpm. Round to the nearest whole gpm.
- Find the BOD loading entering the plant each day, in pounds per day. Use the pounds formula with the influent BOD of 220 mg/L.
- Find the BOD loading leaving the plant each day, in pounds per day, using the effluent BOD of 15 mg/L. Round to the nearest whole pound.
- Subtract your answer to item 3 from your answer to item 2 to find how many pounds of BOD the plant removes each day.
- Calculate the plant's BOD removal efficiency as a percent. Round to the nearest whole percent.
- Does the effluent BOD of 15 mg/L meet the 30 mg/L secondary standard? State by how many mg/L it clears or misses the limit.
- Calculate the plant's TSS removal efficiency as a percent, using 240 mg/L entering and 16 mg/L leaving. Round to the nearest whole percent.
- Find how many pounds of TSS the plant removes each day. Round to the nearest whole pound.
- Explain why loading in pounds per day is a more useful number than concentration in mg/L alone when a state inspector compares two treatment plants of very different sizes.
- Suppose a wet-weather day pushes the flow up to 3.5 MGD while the influent BOD stays at 220 mg/L. Find the new BOD loading entering the plant, then explain why the loading rises even though the concentration did not change.
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: 1,735 gpm. 2.5 MGD x 694 gpm/MGD = 1,735 gpm.
- Answer: 4,587 lb/day. 220 mg/L x 2.5 MGD x 8.34 = 550 x 8.34 = 4,587 pounds per day.
- Answer: 313 lb/day. 15 mg/L x 2.5 MGD x 8.34 = 37.5 x 8.34 = 312.75, which rounds to 313 pounds per day.
- Answer: 4,274 lb/day. 4,587 - 313 = 4,274 pounds per day. A check using the difference in concentration gives the same result: (220 - 15) x 2.5 x 8.34 = 205 x 2.5 x 8.34 = 4,274.
- Answer: 93 %. (220 - 15) / 220 x 100 = 205 / 220 x 100 = 93.18, which rounds to 93 percent.
- Answer: Yes, by 15 mg/L. The effluent BOD of 15 mg/L is below the 30 mg/L limit, clearing it by 30 - 15 = 15 mg/L.
- Answer: 93 %. (240 - 16) / 240 x 100 = 224 / 240 x 100 = 93.33, which rounds to 93 percent.
- Answer: 4,670 lb/day. (240 - 16) x 2.5 x 8.34 = 224 x 2.5 x 8.34 = 560 x 8.34 = 4,670.4, which rounds to 4,670 pounds per day.
- Answer: Loading counts the actual weight of pollutant handled, not just how strong the water is. A small plant and a large plant could both report 220 mg/L of BOD, but the large plant moves far more pounds each day because it treats more flow, so loading is what shows the true size of the job and the true impact on the river.
- Answer: 6,422 lb/day. 220 mg/L x 3.5 MGD x 8.34 = 770 x 8.34 = 6,421.8, which rounds to 6,422 pounds per day. The loading rises because loading depends on both concentration and flow; the same strength of water carried in a larger daily flow delivers more total pounds of BOD to the plant.
Differentiation
Support
Provide a printed formula card with the pounds formula and the removal-efficiency formula already written out, plus one fully worked example using different numbers (for instance, 100 mg/L at 1 MGD). Let students use a calculator and check items 2 and 3 against each other before attempting item 4.
Extension
Ask students to find the flow at which this plant's BOD loading would double from its item 2 value while the influent concentration stays unchanged, and to explain what that says about the relationship between flow and loading.
ELL supports
Pre-teach "influent" (water coming in) and "effluent" (water going out) with a simple labeled arrow diagram of a plant, and provide the sentence frames "Loading is more useful than concentration because ___" for item 9 and "The loading rises because ___" for item 10.
Extensions
Invite a local wastewater operator or a public-works staff member to share, by letter or a short visit, the average flow and typical BOD readings for the class's own community plant. Students can rerun items 2 through 5 with those real numbers and compare the local removal efficiency with the fictional plant in this worksheet.