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
- Students will calculate a filter's loading rate from flow and filter surface area.
- Students will calculate turbidity removal efficiency using influent and effluent readings.
- 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.
Problems
Redwing Filtration Plant (a fictional plant used for this problem set) filters water at a flow of 4,800 gallons per minute (gpm) through filter beds with a combined surface area of 1,200 square feet. On a separate sampling day, the plant measures an influent turbidity of 8.2 NTU and an effluent (filtered) turbidity of 0.18 NTU. Use the formula Filter loading rate = flow ÷ filter surface area, and Removal efficiency (%) = (influent − effluent) ÷ influent × 100, for all problems below.
- Calculate the plant's filter loading rate in gallons per minute per square foot (gpm/sq ft).
- Conventional filters typically operate between 2 and 6 gpm/sq ft. Is Redwing's loading rate (problem 1) within that typical range?
- If plant flow increases to 6,000 gpm with the same 1,200 square feet of filter area, what is the new loading rate? Is it still within the typical range?
- Using the influent (8.2 NTU) and effluent (0.18 NTU) readings, calculate the filter's removal efficiency as a percent. Round to the nearest tenth of a percent.
- The Surface Water Treatment Rule requires finished-water turbidity at or below 0.3 NTU in at least 95 percent of each month's samples. Is Redwing's effluent turbidity (0.18 NTU) at or below 0.3 NTU? By what margin, in NTU?
- A different sample on a stormy day shows effluent turbidity of 1.05 NTU. The rule also states finished water must never exceed 1 NTU. By how many NTU does this sample exceed that ceiling?
- Suppose Redwing wants to guarantee at least 97 percent removal efficiency with the same 8.2 NTU influent turbidity. Write and solve a one-variable equation to find the maximum effluent turbidity this would allow.
- In one or two sentences, explain why a plant would aim to operate well below the 0.3 NTU limit rather than treating 0.3 NTU as a safe operating target.
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: 4.0 gpm/sq ft.
4,800 gpm ÷ 1,200 sq ft = 4.0 gpm/sq ft.Answer: yes.
4.0 gpm/sq ft falls within the typical 2-to-6 gpm/sq ft range for conventional filters.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.Answer: 97.8 percent.
(8.2 − 0.18) ÷ 8.2 × 100 = 97.8049, which rounds to 97.8%.Answer: 0.12 NTU.
0.3 − 0.18 = 0.12 NTU under the limit.Answer: 0.05 NTU.
1.05 − 1 = 0.05 NTU over the ceiling.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.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.