Worksheet Grade 9 Engineering & Design
Reed Valley Wastewater Plant: Removal Efficiency Worksheet
A grade 9 engineering worksheet on a wastewater plant: students compute BOD and TSS removal efficiency and pollutant loads and check them against the secondary standard.
Objectives
- Students will calculate BOD and TSS removal efficiency from influent and effluent concentrations.
- Students will calculate a pollutant load in pounds per day and check results against the secondary treatment standard.
Vocabulary
- 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: removal efficiency, BOD, TSS, load
Teacher overview
A wastewater plant is judged on how much pollution it removes before returning water to a river. Two common measures are BOD, a gauge of oxygen-demanding waste, and TSS, the total suspended solids. Removal efficiency compares the incoming (influent) concentration with the outgoing (effluent) concentration. This worksheet uses one fictional plant's numbers to compute those efficiencies and the daily pollutant loads, then checks them against the secondary treatment standard.
Student Worksheet
The Reed Valley Water Resource Recovery Facility (a fictional plant used for this worksheet) treats the town's sewage. Use these values:
- Influent BOD: 240 mg/L; effluent BOD: 18 mg/L.
- Influent TSS: 260 mg/L; effluent TSS: 22 mg/L.
- Plant flow: 1.2 MGD.
- Removal efficiency = (influent - effluent) / influent x 100.
- Secondary standard: monthly-average BOD and TSS of 30 mg/L or less, and at least 85 percent removal.
- One gallon of water weighs 8.34 pounds.
Calculate the BOD removal efficiency as a percent. Round to the nearest tenth.
Calculate the TSS removal efficiency as a percent. Round to the nearest tenth.
Does the effluent BOD of 18 mg/L meet the 30 mg/L secondary standard? Answer yes or no and give the comparison.
Does the effluent TSS of 22 mg/L meet the 30 mg/L secondary standard? Answer yes or no and give the comparison.
The secondary standard also requires at least 85 percent removal. Do both the BOD and TSS removals exceed 85 percent?
Find the BOD load removed in pounds per day. Use pounds per day = removed concentration (mg/L) x flow (MGD) x 8.34, where the removed concentration is influent minus effluent BOD.
Find the BOD load still discharged to the river in pounds per day, using the effluent BOD of 18 mg/L.
A storm doubles the flow to 2.4 MGD, but rain dilutes the sewage so the influent BOD drops to 120 mg/L. Find the BOD load entering the plant during the storm, compare it with the normal influent load (240 mg/L at 1.2 MGD), and explain what you notice.
In your own words, define removal efficiency using the plant's BOD numbers as your example.
Explain why the facility reports both a concentration in mg/L and a load in pounds per day, rather than just one of them.
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: 92.5%.
(240 - 18) / 240 x 100 = 222 / 240 x 100 = 92.5 percent.Answer: 91.5%.
(260 - 22) / 260 x 100 = 238 / 260 x 100 = 91.5 percent.Answer: Yes.
The effluent BOD of 18 mg/L is below the 30 mg/L standard, so it meets the limit by 12 mg/L.Answer: Yes.
The effluent TSS of 22 mg/L is below the 30 mg/L standard, so it meets the limit by 8 mg/L.Answer: Yes.
BOD removal (92.5 percent) and TSS removal (91.5 percent) both exceed the 85 percent minimum.Answer: 2,222 lb/day.
Removed BOD = 240 - 18 = 222 mg/L. Load = 222 x 1.2 x 8.34 = 2,221.8, about 2,222 pounds per day.Answer: 180 lb/day.
Effluent load = 18 mg/L x 1.2 MGD x 8.34 = 180.1, about 180 pounds per day.Answer: 2,402 lb/day.
Storm load = 120 x 2.4 x 8.34 = 2,401.9, about 2,402 pounds per day. Normal influent load = 240 x 1.2 x 8.34 = 2,401.9, about 2,402 pounds per day. The two loads are the same, because the concentration was cut in half while the flow doubled, so the total pounds entering the plant did not change.Answer: Removal efficiency is the share of a pollutant taken out.
It compares influent and effluent. For BOD, the plant lowered 240 mg/L to 18 mg/L, removing 222 of every 240 units, which is 92.5 percent of the incoming BOD.Answer: Concentration and load answer different questions.
A permit often limits the concentration (mg/L) the plant may discharge, but the river responds to the total pounds it receives each day. Load equals concentration times flow, so a low concentration at a high flow can still send many pounds to the river.
Differentiation
Support
Provide the removal-efficiency formula worked through with a third example (influent 100 mg/L, effluent 10 mg/L gives 90 percent) before students try items 1 and 2, and give a filled-in unit reminder for the pounds formula. Allow a calculator throughout.
Extension
Ask students to find what effluent BOD concentration would drop the plant to exactly 85 percent removal at an influent of 240 milligrams per liter, and to explain why a plant aims to stay well above the minimum required removal rather than right at it.
ELL supports
Provide a visual glossary card for "influent," "effluent," "removal efficiency," "load," "BOD," and "TSS" with a simple in-and-out arrow diagram of the plant. Allow items 9 and 10 to be answered with a sentence frame such as "Removal efficiency is ___ because ___."