Worksheet Grade 10 Math
Aldermoss Carbon Vessels: PFAS Bed Volumes and Changeout
A grade 10 worksheet where students size a fictional carbon vessel as a cylinder, find contact time and bed volumes per day, and solve for the changeout date and cost.
Objectives
- Students will compute the volume of a cylindrical carbon bed in cubic feet and in gallons.
- Students will find empty bed contact time and bed volumes treated per day from a stated flow rate.
- Students will write and solve a one-variable equation for the number of days until a carbon changeout.
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: bed volume, empty bed contact time, breakthrough, changeout
Teacher overview
The Aldermoss Water District, its two carbon vessels, and every dimension, price, and reading on this worksheet are fictional. Granular activated carbon removes PFAS by adsorption, and the carbon is spent after a known number of bed volumes have passed through it, which is why operators plan changeouts by counting bed volumes rather than by counting days directly. Students model the bed as a cylinder, convert to gallons, and then turn the supplier's bed life into a date and a cost.
Student Worksheet
The Aldermoss Water District is fictional. Each of its two carbon vessels holds a bed of granular activated carbon 10 feet in diameter and 6 feet deep, and each vessel treats 700 gpm. Use 0.785 × diameter squared × depth for the volume of a cylinder in cubic feet, and use 7.48 gallons per cubic foot.
Find the volume of one carbon bed in cubic feet, using a diameter of 10 feet and a depth of 6 feet.
Convert the item 1 volume to gallons.
Empty bed contact time in minutes is the bed volume in gallons divided by the flow in gpm. Find it for one vessel at 700 gpm, rounded to the nearest tenth.
How many gallons pass through one vessel in a day at 700 gpm? Use 1,440 minutes in a day.
One bed volume is the gallon figure from item 2. How many bed volumes pass through one vessel in a day? Round to the nearest tenth.
The carbon supplier states that the bed treats 40,000 bed volumes before breakthrough. Let d be the number of days until changeout. Write an equation in d using your item 5 value.
Solve your equation for d. Round to the nearest tenth of a day.
Raw water at the fictional district measures 24 ppt of PFOA, and the treated water measures 2 ppt. Find the percent removed, rounded to the nearest tenth.
The carbon weighs 30 pounds per cubic foot. Find the mass of carbon in one vessel.
Carbon costs $2.10 per pound at this fictional district. Find the cost to refill one vessel.
Both vessels are refilled at the same changeout. Find the cost of that changeout.
Explain why doubling the bed depth from 6 feet to 12 feet would roughly double the empty bed contact time.
The federal maximum contaminant level for PFOA is 4.0 ppt. Is the treated water at 2 ppt below that level? Answer yes or no.
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: 471 cubic feet. The volume is 0.785 × 10 × 10 × 6, and 0.785 × 100 = 78.5, so 78.5 × 6 = 471 cubic feet.
Answer: 3,523.08 gallons. Multiplying gives 471 × 7.48 = 3,523.08 gallons.
Answer: 5.0 minutes. Dividing gives 3,523.08 ÷ 700 = 5.033, which rounds to 5.0 minutes of empty bed contact time.
Answer: 1,008,000 gallons. Multiplying gives 700 × 1,440 = 1,008,000 gallons in one day.
Answer: 286.1 bed volumes. Dividing gives 1,008,000 ÷ 3,523.08 = 286.11, which rounds to 286.1 bed volumes per day.
Answer: 286.1d = 40,000. The left side is bed volumes treated after d days at the fictional district's rate, and the right side is the supplier's stated bed life.
Answer: 139.8 days. Dividing gives 40,000 ÷ 286.1 = 139.81, which rounds to 139.8 days, or just over four and a half months.
Answer: 91.7 percent. The removal is (24 − 2) ÷ 24 = 22 ÷ 24 = 0.9167, which is 91.7 percent.
Answer: 14,130 pounds. Multiplying gives 471 × 30 = 14,130 pounds of carbon in one vessel.
Answer: $29,673. Multiplying gives 14,130 × 2.10 = 29,673 dollars for one vessel.
Answer: $59,346. Two vessels cost 2 × 29,673 = 59,346 dollars at one changeout.
Answer: Contact time is proportional to bed volume at a fixed flow. Doubling the depth doubles the volume of the cylinder, because the diameter and therefore the cross-sectional area stay the same. The flow of 700 gpm does not change, so the gallons in the bed divided by the flow doubles as well, giving about 10.1 minutes instead of about 5.0 minutes.
Answer: Yes. The treated value of 2 ppt is below the 4.0 ppt maximum contaminant level for PFOA, so the fictional district's finished water would meet that limit.
Differentiation
Support
Give students the item 1 and item 2 values already computed for the fictional vessel, along with a labeled cross-section showing the 10-foot diameter and the 6-foot bed depth, so the worksheet begins at item 3. Provide the item 6 equation with only the 40,000 missing.
Extension
Ask students to find the vessel diameter that would give a 10.0-minute empty bed contact time at 700 gpm with the bed depth held at 6 feet, and to explain why the answer is not simply twice the original diameter. A further challenge is to compute the cost per million gallons treated at this fictional district by combining the changeout cost from item 11 with the volume treated over the 139.8 days.
ELL supports
Provide a labeled diagram of the fictional vessel with the words diameter, depth, and bed volume placed on the drawing, beside a second picture showing water entering the top and leaving the bottom. Offer a sentence frame for item 12: "When the depth doubles, the volume ___, and the flow stays the same, so the contact time ___."
Extensions
Have students build a table of changeout intervals for the fictional Aldermoss vessels at flows of 500, 700, and 900 gpm, then describe in a sentence how the interval responds to flow. A short written recommendation on which flow the district should run, given the carbon cost from item 11, makes a useful bridge to a later lesson on operating cost per unit volume.