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Math problem set Grade 9 Engineering & Design

Fennbrook Carbon Adsorber: Sizing a PFAS Removal Vessel

A grade 9 set that sizes a fictional granular activated carbon vessel for PFAS removal, using cylinder volume, empty bed contact time, and bed volumes.

Objectives

  1. Students will calculate the media volume of a cylindrical carbon vessel in cubic feet and gallons.
  2. Students will calculate empty bed contact time and bed volumes treated before breakthrough.

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.

Also: empty bed contact time, bed volume, breakthrough

Teacher overview

Fennbrook is a fictional water system that found PFOA, one of the PFAS "forever chemicals," in a well above the federal limit of 4.0 parts per trillion. The utility plans to add a granular activated carbon vessel, which passes water slowly through a bed of carbon that traps the contaminant. This set has students size that vessel and check how long the water stays in contact with the carbon.

A common misconception is that a faster flow through the same vessel cleans more water without any cost. In fact, faster flow shortens the empty bed contact time, and too little contact time lets the contaminant slip through. Students who ignore contact time will design a vessel that runs fast but fails to remove enough PFAS.

Fennbrook Carbon Adsorber: Sizing a PFAS Removal Vessel — Grade 9 Engineering & Design
Name: ____________________________ Date: ______________

Problems

Use these facts about the fictional Fennbrook carbon vessel for every problem.

  • The carbon bed is a cylinder 10 feet across and 8 feet deep.
  • Cylinder volume (cu ft) = 0.785 × diameter² × depth. One cubic foot holds 7.48 gallons.
  • Empty bed contact time (minutes) = bed volume in gallons ÷ flow in gpm.
  • The design flow is 350 gpm.

  1. Find the volume of the carbon bed in cubic feet. Round to the nearest whole cubic foot.

  2. Convert the bed volume to gallons. Round to the nearest whole gallon.

  3. Find the empty bed contact time at the design flow of 350 gpm. Round to the nearest tenth of a minute.

  4. The design guideline calls for a contact time of at least 10 minutes. What is the highest flow, in gpm, that still meets the 10-minute rule? Round down to a whole gpm so the rule is still met.

  5. Testing shows the carbon can treat about 30,000,000 gallons before the contaminant breaks through. How many bed volumes is that? (Bed volumes = total gallons treated ÷ bed volume in gallons.) Round to the nearest whole bed volume.

  6. At the design flow of 350 gpm, how many days can the vessel run before it reaches 30,000,000 gallons treated? (One day at 350 gpm is 350 × 1,440 gallons.) Round to the nearest tenth of a day.

  7. The well measures 25 ppt of PFOA and must reach the 4.0 ppt limit or lower. What removal efficiency, as a percent, does the vessel need to reach exactly 4.0 ppt? (Removal % = (influent − effluent) ÷ influent × 100.) Round to the nearest whole percent.

  8. Water systems must meet the 4.0 ppt limits by April 2029. In May 2026 EPA proposed letting eligible systems apply for a two-year exemption, to April 2031, but the proposal was not final as of September 2026. In one or two sentences, explain why a utility might still install carbon now rather than seek the later date.

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: 628 cubic feet.
    Volume = 0.785 × 10² × 8 = 0.785 × 100 × 8 = 628 cu ft.

  2. Answer: 4,697 gallons.
    628 cu ft × 7.48 gal/cu ft = 4,697.44, which rounds to 4,697 gallons.

  3. Answer: 13.4 minutes.
    Contact time = 4,697.44 ÷ 350 = 13.421, which rounds to 13.4 minutes.

  4. Answer: 469 gpm.
    The shortest allowed contact time is 10 minutes, so the highest flow is 4,697.44 ÷ 10 = 469.7 gpm. Round down to 469 gpm, because 470 gpm would give only 9.99 minutes.

  5. Answer: 6,387 bed volumes.
    30,000,000 ÷ 4,697 = 6,387.1, which rounds to 6,387 bed volumes. Students who divide by the unrounded 4,697.44 get 6,386; accept either.

  6. Answer: 59.5 days.
    One day at 350 gpm is 350 × 1,440 = 504,000 gallons. 30,000,000 ÷ 504,000 = 59.52, which rounds to 59.5 days.

  7. Answer: 84 percent.
    Removal % = (25 − 4) ÷ 25 × 100 = 21 ÷ 25 × 100 = 84%.

  8. Answer: Utilities often act early because the contamination is already above the limit and continues reaching customers. Installing carbon now protects public health, spreads out construction work, and avoids relying on an exemption the utility may not receive.

Differentiation

Support

Provide the cylinder volume for problem 1 of the fictional Fennbrook vessel already worked out, so students can focus on the contact-time and bed-volume steps. Give a printed reference card with the two formulas and the 7.48 conversion so nothing must be recalled from memory.

Extension

Ask students to design a second vessel that treats the same 350 gpm but holds a contact time of at least 15 minutes. They should choose a new diameter or depth, show the volume math, and confirm the contact time meets the target.

ELL supports

Pre-teach "contact time," "bed volume," and "breakthrough" with a simple diagram of water moving down through a carbon bed. Provide a sentence frame for problem 8: "A utility might install carbon now because ___."