Math problem set Grade 8 Engineering & Design
Sablerock's Pressure Filters: A Cylinder Volume and Loading-Rate Set
A grade 8 engineering set that sizes cylindrical pressure filters, converts volume to gallons, and checks loading rates against a safe design range.
Objectives
- Students will use the cylinder formula to find a filter's surface area and volume, then convert cubic feet to gallons.
- Students will calculate filter loading rate and check it against a safe design range to judge a treatment design.
Vocabulary
- Backwash
- Backwash is when a treatment plant reverses the flow of water through a filter to flush out the dirt and particles it has trapped. It is like rinsing out a dirty sponge so the filter can keep working well.
Also: pressure filter, surface area, loading rate, cubic feet
Teacher overview
The Sablerock Water Plant is a fictional treatment plant adding new pressure filters, each shaped like a tall cylinder. This set walks students through the design math an engineer uses to size one: first the plan area of the top circle, then the volume, then the gallons it holds, and finally the loading rate that decides how much water each filter can clean. Each design is checked against a safe loading window of 2 to 6 gallons per minute for each square foot of filter surface. A common misconception in this math is using the full diameter in the area formula instead of squaring it correctly, so watch for students who write 0.785 times 8 rather than 0.785 times 8 times 8 in problem 1.
Problems
Use these design facts for every problem. Suppose the Sablerock plant, a fictional plant, builds filters that are vertical cylinders 8 feet across (diameter) with a straight-side water depth of 9 feet. The plant wants each filter to run at a loading rate of 5 gpm for every square foot of surface. A safe filter runs between 2 and 6 gpm per square foot. One cubic foot of water holds 7.48 gallons.
Find the plan (top) surface area of one filter. Use area = 0.785 x diameter². Round to the nearest hundredth.
Find the volume of one filter in cubic feet. Use volume = 0.785 x diameter² x depth.
Convert the filter volume from problem 2 into gallons.
At a loading rate of 5 gpm per square foot, find the design flow through one filter, in gpm. (Flow = loading rate x surface area.)
Suppose Sablerock must deliver 1,000 gpm of clean water. Using the flow from problem 4, what is the smallest whole number of filters that can meet the demand?
With 4 filters all running at the design flow from problem 4, what is the total flow in gpm, and does it reach the 1,000 gpm demand?
During a backwash, one filter is shut off and only 3 remain. If those 3 filters must still deliver 1,000 gpm, what loading rate (gpm per square foot) would each need? Is that inside the safe 2-to-6 window?
Challenge. To keep each of the 3 filters at or below 6 gpm per square foot while still delivering 1,000 gpm, what minimum surface area does each filter need, and about what diameter does that require? Round the diameter to the nearest tenth of a foot.
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: 50.24 sq ft.
Area = 0.785 x 8² = 0.785 x 64 = 50.24 square feet. The number 8 is the diameter, and it must be multiplied by itself before multiplying by 0.785.Answer: 452.16 cu ft.
Volume = 0.785 x 8² x 9 = 0.785 x 64 x 9 = 50.24 x 9 = 452.16 cubic feet.Answer: 3,382 gal.
452.16 cubic feet x 7.48 gallons per cubic foot = 3,382.16 gallons, which rounds to about 3,382 gallons.Answer: 251 gpm.
Flow = 5 gpm per square foot x 50.24 square feet = 251.2 gpm, or about 251 gpm through one filter.Answer: 4 filters.
1,000 gpm divided by 251.2 gpm per filter = 3.98, and a plant cannot build part of a filter, so it rounds up to 4 filters.Answer: 1,005 gpm.
4 filters x 251.2 gpm = 1,004.8 gpm, about 1,005 gpm, which does reach the 1,000 gpm demand with roughly 5 gpm to spare.Answer: 6.6 gpm per square foot.
Each of the 3 filters would carry 1,000 divided by 3 = 333.33 gpm. Loading = 333.33 divided by 50.24 = 6.63, about 6.6 gpm per square foot, which is above the safe ceiling of 6, so 3 filters cannot safely cover the full demand.Answer: About 55.56 sq ft per filter and a diameter of about 8.4 ft.
Each filter needs at most 333.33 gpm at 6 gpm per square foot, so area = 333.33 divided by 6 = 55.56 square feet. Then diameter² = 55.56 divided by 0.785 = 70.78, and the diameter is the square root of 70.78, about 8.4 feet, slightly wider than the 8-foot design. Because 8.4 ft gives about 6.02 gpm per square foot, a designer would round up to 8.5 ft to stay at or below 6 even during backwash.
Differentiation
Support
Give students a printed reference strip with the three formulas already labeled: area = 0.785 x diameter², volume = area x depth, and flow = loading rate x area. Work problem 1 together so the squaring step is modeled before students continue. Allow calculator use for every step.
Extension
Ask students to redesign the plant with 5 smaller filters instead of 4, choosing a diameter that keeps every filter inside the safe 2-to-6 loading window even when one filter is offline. Students should show the loading rate for both the normal case and the backwash case.
ELL supports
Provide a labeled cylinder diagram showing diameter, surface area (the top circle), and depth as separate measurements before problem 1. Pre-teach the words surface, loading, and backwash with a simple sketch, and let students record their work as labeled numbers rather than full sentences.
Extensions
Invite a local water operator or public works staff member to describe how many filters the town runs and how they schedule backwashing so service never drops below demand. Students can compare the real filter count and schedule with the design they built here, which turns the problem set into a short piece of authentic local research.