Math problem set Grade 11 Math

Kestrel Bend PFAS: GAC Contactor Sizing and Contact-Time Problems

A grade 11 math problem set on removing PFAS with granular activated carbon: percent reduction, flow conversions, bed volume, and empty bed contact time.

Grade: Grade 11
Subject: Math
Material: Math problem set
Time: 45 min
Prep: No prep
Materials: Calculator

Standards

HSN-Q.A.1 Common Core Math

Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays.

How this material aligns: Problems 3 through 7 carry units through a multi-step chain from million gallons per day to cubic feet per minute to compute empty bed contact time, the dimensional discipline this standard targets.

All materials for HSN-Q.A.1 →

HSN-Q.A.3 Common Core Math

Choose a level of accuracy appropriate to limitations on measurement when reporting quantities.

How this material aligns: Problem 8 asks students to report a treated PFOA reading only to the accuracy the test method supports, matching the reported precision to the instrument as this standard requires.

All materials for HSN-Q.A.3 →

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This material is in the public domain (CC0) and may be copied, edited, and shared freely, without attribution. Updated 2026-07-23.

Objectives

Students will calculate the minimum percent reduction of PFOA needed to meet the maximum contaminant level.
Students will convert plant flow through consistent units to find the empty bed contact time of a carbon vessel.

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.
Parts per trillion: A unit for measuring extremely small amounts of a chemical in water. PFAS limits are set in parts per trillion because these chemicals matter even in tiny doses.

Also: empty bed contact time, removal efficiency

Teacher overview

The fictional Kestrel Bend Water District found PFOA and PFOS in one of its wells above the federal maximum contaminant level of 4.0 parts per trillion each, so it installed two granular activated carbon (GAC) contactors in series to adsorb the contaminants. This set walks students through the numbers an engineer checks before and after that installation, which are the percent reduction the carbon must deliver and the empty bed contact time each vessel provides. A common misconception in this math is computing contact time with mismatched units, dividing a bed volume in cubic feet by a flow left in gallons per minute; the flow must first be converted to cubic feet per minute so the units cancel to time.

Kestrel Bend PFAS: GAC Contactor Sizing and Contact-Time Problems — Grade 11 Math

Name: ____________________________ Date: ______________

Problems

Use these values for the fictional Kestrel Bend well and its GAC system:

Raw well water: PFOA at 12 ppt and PFOS at 9 ppt.
The maximum contaminant level (MCL) for PFOA is 4.0 ppt.
Plant flow through the carbon: 1.5 MGD.
Each GAC vessel is a cylinder 10 feet in diameter with a carbon bed 8 feet deep.
Conversions: 1 MGD = 694 gpm; 1 cubic foot = 7.48 gallons.
Cylinder bed volume (cubic feet) = 0.785 x diameter squared x depth.

The raw PFOA is 12 ppt and the MCL is 4.0 ppt. How many ppt above the MCL is the raw PFOA?
To just meet the MCL, the carbon must lower PFOA from 12 ppt to 4.0 ppt. Using removal efficiency = (in minus out) divided by in, times 100, find the minimum percent reduction the carbon must provide. Round to the nearest whole percent.
Convert the plant flow of 1.5 MGD to gallons per minute, using 1 MGD = 694 gpm.
Convert that flow to cubic feet per minute, using 1 cubic foot = 7.48 gallons. Round to the nearest tenth.
Find the carbon bed volume of one vessel in cubic feet, using volume = 0.785 x diameter squared x depth with a 10-foot diameter and an 8-foot depth.
Empty bed contact time equals bed volume divided by flow. Using your answers from problems 4 and 5, find the contact time of one vessel in minutes. Round to the nearest tenth.
The two vessels sit in series, so water passes through both carbon beds. What is the total empty bed contact time for the pair?
A lab reports the treated PFOA as "0.8" ppt using a method reliable to the nearest 0.1 ppt. Should the operator record it as 0.8 ppt or as 0.80000 ppt, and why?
After a year the treated PFOA rises to 3.6 ppt. Is the water still below the 4.0 ppt MCL, and by how many ppt?
Suppose the treated PFOA keeps rising at a steady 0.3 ppt per month from 3.6 ppt. How many months until it reaches the 4.0 ppt MCL? Round to the nearest tenth.

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: 8.0 ppt. The raw PFOA is 12 ppt and the MCL is 4.0 ppt, so 12 minus 4.0 equals 8.0 ppt above the limit.
Answer: 67 percent. Removal efficiency is (12 minus 4.0) divided by 12, times 100, which is 8 divided by 12 times 100, or 66.7 percent, rounding to 67 percent.
Answer: 1,041 gpm. 1.5 MGD times 694 gpm per MGD equals 1,041 gpm.
Answer: 139.2 cubic feet per minute. 1,041 gallons per minute divided by 7.48 gallons per cubic foot equals 139.17, which rounds to 139.2 cubic feet per minute.
Answer: 628 cubic feet. Volume equals 0.785 times 10 squared times 8, which is 0.785 times 100 times 8, equal to 628 cubic feet.
Answer: 4.5 minutes. Contact time equals 628 cubic feet divided by 139.2 cubic feet per minute, which is 4.51, rounding to 4.5 minutes.
Answer: 9.0 minutes. Two vessels in series give 2 times 4.51 minutes, which is 9.02, or about 9.0 minutes of total contact time.
Answer: 0.8 ppt. The method is reliable only to the nearest 0.1 ppt, so writing 0.80000 ppt claims a precision the test cannot support; 0.8 ppt matches the instrument.
Answer: yes, by 0.4 ppt. The treated PFOA of 3.6 ppt is below the 4.0 ppt MCL, and 4.0 minus 3.6 equals a 0.4 ppt margin.
Answer: 1.3 months. The gap to the MCL is 4.0 minus 3.6, or 0.4 ppt, and 0.4 divided by 0.3 ppt per month equals 1.33, or about 1.3 months.

Differentiation

Support

Provide a units map that lays the conversion chain out in order, MGD to gallons per minute to cubic feet per minute, with the matching conversion factor printed beside each arrow. Let students use it while they work problems 3 through 6 so the sequence is visible rather than recalled.

Extension

Ask students to find the flow, in gallons per minute, at which one vessel's empty bed contact time would drop to 3.0 minutes, working backward from the fixed 628-cubic-foot bed volume. Have them explain what a shorter contact time means for how well the carbon adsorbs PFOA.

ELL supports

Pre-teach "adsorb," "bed volume," and "contact time" with a labeled diagram of a carbon vessel showing water entering the top and leaving the bottom. Provide a sentence frame for problem 8: "The operator should record ___ ppt because the test is only accurate to ___."