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Worksheet Grade 8 Science

Willowbank Reclamation Plant: Removal Efficiency and Loads

A grade 8 science worksheet where students read a fictional treatment plant's BOD and TSS data, find removal efficiency, and write daily loads in scientific notation.

Objectives

  1. Students will calculate the removal efficiency of a wastewater treatment step from influent and effluent concentrations.
  2. Students will express daily pollutant loads in scientific notation and compare their magnitudes.

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

Teacher overview

The fictional Willowbank Reclamation Plant treats a town's wastewater before returning it to the Larch River. Secondary treatment rules require the finished water to reach a monthly-average BOD of 30 mg/L or less, about 85 percent removal, because raw sewage typically carries 200 to 300 mg/L of oxygen-demanding material. This worksheet has students read the plant's stage-by-stage data, calculate removal efficiency, and rewrite the daily pollutant loads in scientific notation. Watch for a common error: students sometimes divide by the effluent instead of the influent when finding removal efficiency.

Willowbank Reclamation Plant: Removal Efficiency and Loads — Grade 8 Science
Name: ____________________________ Date: ______________

Student Worksheet

The fictional Willowbank Reclamation Plant records the concentration of two pollutants at three points in its process. BOD (biochemical oxygen demand) measures oxygen-demanding organic material; TSS measures total suspended solids. Suppose the plant treats 6 million gallons of wastewater each day (6 MGD). Use this data table for the items below.

Stage BOD (mg/L) TSS (mg/L)
Raw influent 220 240
After primary settling 150 90
Final effluent 22 18

Use this formula where it applies: Removal efficiency (%) = (influent − effluent) ÷ influent × 100.

  1. Find the plant's overall BOD removal efficiency, using the raw influent BOD of 220 mg/L and the final effluent BOD of 22 mg/L.
  2. Find the plant's overall TSS removal efficiency, using the raw influent TSS of 240 mg/L and the final effluent TSS of 18 mg/L.
  3. Secondary rules require a monthly-average effluent BOD of 30 mg/L or less. Does the final effluent BOD of 22 mg/L meet that limit, and by how many mg/L is it under or over?
  4. The primary settling step alone lowers BOD from 220 mg/L to 150 mg/L. What percent of the BOD does primary settling remove on its own?
  5. The daily amount of a pollutant is Load (lbs/day) = concentration (mg/L) × flow (MGD) × 8.34. Find the BOD load entering the plant, using the raw influent BOD of 220 mg/L and the flow of 6 MGD.
  6. Find the BOD load leaving the plant in the final effluent, using 22 mg/L and 6 MGD.
  7. Write your answers to items 5 and 6 in scientific notation, each rounded to two significant figures.
  8. Using your scientific-notation answers, about how many times as much BOD enters the plant each day as leaves it?
  9. Fish need dissolved oxygen above about 5 milligrams per liter. Explain how removing BOD before the water reaches the Larch River helps protect fish there.
  10. The operators aim to remove at least 85 percent of the incoming BOD. Based on your answer to item 1, is the plant meeting that goal? Explain.

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key
  1. Answer: 90 %.
    (220 − 22) ÷ 220 × 100 = 198 ÷ 220 × 100 = 90 percent.
  2. Answer: 92.5 %.
    (240 − 18) ÷ 240 × 100 = 222 ÷ 240 × 100 = 92.5 percent.
  3. Answer: Yes, under by 8 mg/L.
    The effluent BOD of 22 mg/L is below the 30 mg/L limit; 30 − 22 = 8 mg/L of margin.
  4. Answer: 32 %.
    (220 − 150) ÷ 220 × 100 = 70 ÷ 220 × 100 = 31.8 percent, which rounds to about 32 percent.
  5. Answer: 11,009 lbs/day.
    220 × 6 × 8.34 = 1,320 × 8.34 = 11,008.8, which rounds to about 11,009 lbs/day.
  6. Answer: 1,101 lbs/day.
    22 × 6 × 8.34 = 132 × 8.34 = 1,100.88, which rounds to about 1,101 lbs/day.
  7. Answer: 1.1 × 10^4 lbs/day and 1.1 × 10^3 lbs/day.
    11,008.8 rounds to 1.1 × 10^4; 1,100.88 rounds to 1.1 × 10^3.
  8. Answer: 10 times.
    (1.1 × 10^4) ÷ (1.1 × 10^3) = 10, so about ten times as much BOD enters as leaves. This matches the concentrations, since 220 ÷ 22 = 10.
  9. Answer: It keeps river oxygen high enough for fish.
    Bacteria in the river feed on BOD and use up dissolved oxygen as they do. Releasing high-BOD water would pull oxygen below the roughly 5 milligrams per liter fish need, so removing most of the BOD first protects them.
  10. Answer: Yes.
    The overall BOD removal from item 1 is 90 percent, which is at or above the 85 percent goal, so the plant meets it.

Differentiation

Support

Provide the removal-efficiency formula worked through on a separate example (influent 100, effluent 20, so 80 percent) before students try items 1 and 2, and let them use a calculator for every step. For items 5 and 6, pre-fill the multiplication setup so students only carry out the arithmetic.

Extension

Ask students to find the TSS load leaving the plant each day in lbs/day, write it in scientific notation, and compare it to the BOD effluent load from item 6, explaining which pollutant the plant discharges more of by mass.

ELL supports

Pre-teach "influent" (water coming in) and "effluent" (water going out) with a simple left-to-right diagram of the plant, and provide a labeled example of a number written in scientific notation. Allow students to answer items 9 and 10 with a labeled sketch plus a short phrase rather than full sentences.