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Math problem set Grade 10 Engineering & Design

Kestrel Junction PFAS Adsorber: A Grade 10 GAC Sizing Problem Set

Grade 10 engineering math sizing a granular activated carbon adsorber for PFAS removal: flow splits, loading rate, bed volume, contact time, and the 4.0 ppt limit.

Objectives

  1. Students will size a granular activated carbon adsorber by finding its loading rate, bed volume, and empty-bed contact time.
  2. Students will compare a treated PFAS concentration and a required removal efficiency against the federal maximum contaminant level.

Vocabulary

Granular activated carbon
Small black grains of specially treated carbon that grab contaminants as water flows through them. Utilities use big tanks of it to remove PFAS, bad tastes, and odors.
PFAS
A large family of human-made chemicals used in things like nonstick pans, waterproof fabric, and firefighting foam. They are called forever chemicals because they take a very long time to break down.
Maximum contaminant level (MCL)
A maximum contaminant level, or MCL, is the highest amount of a certain substance that federal rules allow in drinking water. Water systems test regularly to prove their water stays under each limit.

Also: empty-bed contact time, loading rate

Teacher overview

The fictional Kestrel Junction Water Treatment Plant is adding a granular activated carbon (GAC) adsorber to pull PFAS out of its drinking water. This set walks students through the numbers an engineer checks before ordering the vessels: how the plant flow splits between two carbon beds, how fast water loads onto each bed, how much carbon each vessel holds, how long the water stays in contact with it, and whether the treated water clears the federal PFAS limit. A common misconception in this math is treating the empty-bed contact time as a property of the carbon alone; it actually depends on both the bed volume and the flow through that bed, so a faster flow shortens the contact time even when the vessel does not change.

Kestrel Junction PFAS Adsorber: A Grade 10 GAC Sizing Problem Set — Grade 10 Engineering & Design
Name: ____________________________ Date: ______________

Problems

Kestrel Junction (a fictional plant built for this set) treats a steady 2.0 MGD. The GAC stage uses two identical vessels running in parallel, so each vessel receives half of the plant flow. Each vessel is a cylinder 14 feet in diameter holding a carbon bed 10 feet deep. Testing shows the raw water carries 12 ppt of PFOA, and the federal maximum contaminant level (MCL) for PFOA is 4.0 ppt.

  1. Convert the plant flow of 2.0 MGD to gallons per minute, using the conversion that 1 MGD equals 694 gpm.

  2. The two vessels split the flow equally. What is the flow to each vessel in gpm?

  3. Find the bed surface area of one vessel, meaning the circular area the water passes down through, using Area = 0.785 × Diameter². The vessel is 14 feet in diameter.

  4. Find the hydraulic loading rate on one vessel in gpm per square foot, dividing the per-vessel flow from problem 2 by the surface area from problem 3.

  5. Conventional media beds are usually run between 2 and 6 gpm per square foot. Is the loading rate from problem 4 inside that range?

  6. Find the carbon bed volume of one vessel in cubic feet, using Volume = 0.785 × Diameter² × depth with a depth of 10 feet.

  7. Convert the bed volume from problem 6 to gallons, using the conversion that one cubic foot holds 7.48 gallons.

  8. Find the empty-bed contact time for one vessel in minutes by dividing the bed volume in gallons (problem 7) by the per-vessel flow in gpm (problem 2).

  9. The carbon removes 95 percent of the PFOA. Find the treated PFOA concentration in ppt, and state whether it meets the 4.0 ppt MCL.

  10. Find the minimum removal efficiency, as a percentage, the carbon would need so that the 12 ppt raw water just reaches the 4.0 ppt MCL. Use removal efficiency = (influent − effluent) ÷ influent × 100.

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: 1,388 gpm.
    2.0 MGD × 694 gpm/MGD = 1,388 gpm.

  2. Answer: 694 gpm.
    The two vessels share the flow equally: 1,388 gpm ÷ 2 = 694 gpm to each vessel.

  3. Answer: 153.86 square feet.
    Area = 0.785 × 14² = 0.785 × 196 = 153.86 square feet.

  4. Answer: about 4.51 gpm per square foot.
    694 gpm ÷ 153.86 square feet = 4.510, which rounds to about 4.51 gpm per square foot.

  5. Answer: Yes.
    A loading rate of about 4.51 gpm per square foot falls between 2 and 6 gpm per square foot, so it is inside the usual range.

  6. Answer: 1,538.6 cubic feet.
    Volume = 0.785 × 14² × 10 = 0.785 × 196 × 10 = 1,538.6 cubic feet.

  7. Answer: about 11,509 gallons.
    1,538.6 cubic feet × 7.48 gallons/cubic foot = 11,508.7, which rounds to about 11,509 gallons.

  8. Answer: about 16.6 minutes.
    Empty-bed contact time = 11,508.7 gallons ÷ 694 gpm = 16.58, which rounds to about 16.6 minutes.

  9. Answer: 0.6 ppt.
    12 ppt × (1 − 0.95) = 12 × 0.05 = 0.6 ppt. Because 0.6 ppt is below the 4.0 ppt MCL, the treated water meets the limit.

  10. Answer: about 66.7 percent.
    Removal efficiency = (12 − 4.0) ÷ 12 × 100 = 8 ÷ 12 × 100 = 66.67, which rounds to about 66.7 percent. The carbon must remove at least this share for the raw water to reach the 4.0 ppt MCL.

Differentiation

Support (below grade level)

Provide the four formulas (MGD-to-gpm, cylinder area, cylinder volume, and cubic-feet-to-gallons) on a printed reference card with each constant already written in, and work problem 3's area calculation together as a class before students carry the value forward into problems 4 and 6.

Extension (above grade level)

Ask students to find what steady plant flow, in MGD, would shorten the per-vessel empty-bed contact time to exactly 10 minutes, working backward from the bed volume in problem 7, and to explain what that shorter contact time would mean for how well the carbon captures PFAS.

ELL supports

Provide a labeled cross-section of one GAC vessel showing the carbon bed, the diameter, and the bed depth, with the terms loading rate, bed volume, and contact time marked beside a short definition. Allow students to show problems 1 through 8 as numbered calculations and to answer problems 9 and 10 with a labeled sketch plus one sentence.