Math problem set Grade 10 Math

Redwing Filtration Plant: Filter Loading and Turbidity Removal

An 8-problem grade 10 set where students calculate filter loading rate and turbidity removal efficiency, checking results against real EPA compliance limits.

Objectives

  1. Students will calculate a filter's loading rate from flow and filter surface area.
  2. Students will calculate turbidity removal efficiency using influent and effluent readings.
  3. Students will compare calculated results against real EPA turbidity compliance limits.

Vocabulary

Turbidity
Turbidity is a measure of how cloudy or murky water looks because of tiny particles floating in it. Clear water has low turbidity, and muddy water has high turbidity.

Also: filter loading rate, removal efficiency, NTU

Teacher overview

Filtration is the step in drinking-water treatment that physically removes the fine particles responsible for turbidity, and both how fast water moves through a filter and how well the filter performs are measured with real formulas operators use every day. This problem set follows one fictional plant's filter bed through a loading-rate check and a turbidity-removal check, building from problem to problem toward a single question: how much cleaner does the filtered water need to be? A common misconception is assuming any percent removal above 90 is automatically safe; because the regulatory limit is measured in NTU, not percent, a high removal percentage can still leave a plant out of compliance if the influent was unusually dirty to begin with, which the Extension problem makes clear.

Answer Key

  1. Answer: 4.0 gpm/sq ft.
    4,800 gpm ÷ 1,200 sq ft = 4.0 gpm/sq ft.

  2. Answer: yes.
    4.0 gpm/sq ft falls within the typical 2-to-6 gpm/sq ft range for conventional filters.

  3. Answer: 5.0 gpm/sq ft.
    6,000 gpm ÷ 1,200 sq ft = 5.0 gpm/sq ft, which is still within the 2-to-6 range.

  4. Answer: 97.8 percent.
    (8.2 − 0.18) ÷ 8.2 × 100 = 97.8049, which rounds to 97.8%.

  5. Answer: 0.12 NTU.
    0.3 − 0.18 = 0.12 NTU under the limit.

  6. Answer: 0.05 NTU.
    1.05 − 1 = 0.05 NTU over the ceiling.

  7. Answer: 0.25 NTU.
    Let E = the maximum effluent turbidity. 0.97 = (8.2 − E) ÷ 8.2, so 8.2 − E = 7.954, and E = 8.2 − 7.954 = 0.246, which rounds to 0.25 NTU.

  8. Answer: Operating near the limit leaves no margin for a bad day.
    Influent turbidity varies with weather and season, so a filter that performs at exactly 0.3 NTU on an average day may exceed the limit during a storm; running well below the limit builds in a safety margin.

Differentiation

Support (below grade level)

Provide the filter loading rate formula and the removal efficiency formula as a printed reference card, and work problem 1 together as a class before students continue independently.

Extension (above grade level)

Ask students to find the minimum removal efficiency, as a percent, that Redwing's fictional filter would need to keep effluent turbidity at or below 0.3 NTU if a future storm raised influent turbidity to 15 NTU, and to check their answer using the removal-efficiency formula.

ELL supports

Pre-teach "influent," "effluent," and "turbidity" with a simple before-and-after diagram of cloudy water entering a filter and clear water leaving it, labeled with the English terms and, where possible, home-language equivalents.