Math problem set Grade 6 Science

Copper Basin Treatment Plant: Filter Loading Rate Problems

A grade 6 math problem set where students calculate filter loading rates at a fictional treatment plant and compare them to the typical operating range.

Objectives

  1. Students will calculate filter loading rate as flow divided by filter surface area.
  2. Students will compare calculated loading rates against a typical operating range and convert a flow rate using a single-step conversion.

Vocabulary

Flow rate
Flow rate measures how much water moves past a point in a certain amount of time, like gallons per minute through a garden hose. Bigger pipes and stronger pushes give bigger flow rates.

Also: filter loading rate, surface area

Teacher overview

Copper Basin Water Treatment Plant, a fictional facility used for this problem set, filters its water through sand and gravel beds before sending it out to homes. Operators track each filter's loading rate, the flow of water divided by the filter's surface area, because conventional plants typically run between 2 and 6 gallons per minute for every square foot of filter surface. This set has students calculate loading rates for two of Copper Basin's filters and decide whether each falls inside that typical range. A common misconception in this math is dividing area by flow instead of flow by area; watch for that reversed division, since it produces an answer smaller than 1 that does not match the typical range at all.

Answer Key

  1. Answer: 5 gpm/sq ft.
    3,000 gpm ÷ 600 sq ft = 5 gpm/sq ft.
  2. Answer: Yes.
    5 gpm/sq ft falls between 2 and 6, so it is within the typical range.
  3. Answer: 3 gpm/sq ft.
    1,500 gpm ÷ 500 sq ft = 3 gpm/sq ft.
  4. Answer: Yes.
    3 gpm/sq ft falls between 2 and 6, so it is within the typical range.
  5. Answer: 4,500 gpm.
    3,000 gpm + 1,500 gpm = 4,500 gpm total flow.
  6. Answer: 67 percent.
    3,000 ÷ 4,500 = 0.6667, or about 67%.
  7. Answer: 4,320,000 gallons per day.
    3,000 gpm × 1,440 min/day = 4,320,000 gallons per day.
  8. Answer: 1.5 gpm/sq ft.
    750 gpm ÷ 500 sq ft = 1.5 gpm/sq ft.
  9. Answer: No.
    1.5 gpm/sq ft is below the typical range of 2 to 6, meaning the filter would be under-loaded compared with typical operating conditions, even though under-loading is generally not a safety concern the way over-loading can be.
  10. Answer: Keeping every filter's loading rate close to the typical range helps the plant treat water evenly across all its filters, so no single filter is pushed to work much harder than the others, which could wear it out faster or let water pass through before it is fully filtered.

Differentiation

Support

Provide the loading-rate formula as a labeled fraction (flow on top, area on bottom) with problem 1 worked through step by step as a model before students attempt problems 3 and 8 on their own.

Extension

Ask students to suppose a fourth filter receives 2,400 gpm, and to find the surface area it would need in order to run exactly at the middle of the typical range, 4 gpm/sq ft, and to show the division that produces that area.

ELL supports

Provide a labeled diagram of a filter bed showing "flow in" at the top and "surface area" as the footprint of the filter bed, with the loading-rate formula written directly on the diagram. Allow students to show their work using the diagram and numbers rather than full sentences for problems 1 through 5.