Water for Schools
Menu

Math problem set Grade 12 Math

Grayling Ford Filter Backwash: Rates, Volumes, and Tank Sizing

A grade 12 problem set on filter operations at a fictional plant: loading rates with a filter offline, backwash flow and volume, rise rate, and sizing a wash-water tank.

Objectives

  1. Students will calculate filter loading rates under normal operation and with one filter out of service.
  2. Students will compute backwash flow, wash volume, and rise rate from a stated backwash rate.
  3. Students will size a cylindrical wash-water tank from a required storage volume and depth.

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: filter loading rate, rise rate, media, wash-water tank

Teacher overview

A granular media filter runs in two directions. For most of the day it strains particles out of settled water, and once a day or so the flow reverses while clean water scours the trapped material back out. This set follows one shift at the fictional Grayling Ford water treatment plant through both directions of that flow, from routine loading-rate checks to sizing the elevated tank that supplies wash water. A common misconception in this math is treating the service loading rate and the backwash rate as the same quantity; both are expressed in gpm per square foot, yet one describes water filtering downward in service and the other a much stronger upward wash flow, and mixing them up produces answers that are wrong by a factor of ten.

Grayling Ford Filter Backwash: Rates, Volumes, and Tank Sizing — Grade 12 Math
Name: ____________________________ Date: ______________

Problems

Use these facts about the fictional Grayling Ford plant throughout. The plant filters 4 MGD through four identical filters. Each filter box measures 20 feet by 25 feet, giving 500 square feet of media surface. One MGD equals 694 gpm, and one cubic foot holds 7.48 gallons.

  1. Convert the plant's 4 MGD production to gallons per minute.
  2. With all four filters in service and flow split evenly, how many gpm does each filter carry?
  3. Calculate each filter's loading rate in gpm per square foot, to the nearest hundredth. Conventional plants typically run 2 to 6 gpm per square foot; is Grayling Ford inside that range?
  4. One filter is taken offline for backwashing while the plant keeps producing 4 MGD through the remaining three. Find the loading rate on each remaining filter, to the nearest hundredth of a gpm per square foot.
  5. The backwash system washes one 500-square-foot filter at a rate of 15 gpm per square foot. What total backwash flow, in gpm, does that require?
  6. A full wash lasts 10 minutes at that flow. How many gallons of wash water does one backwash use?
  7. Convert the 15 gpm-per-square-foot wash rate into a rise rate in feet per minute, using the fact that one cubic foot holds 7.48 gallons. Round to the nearest tenth.
  8. All four filters are washed once each day. What percent of the plant's 4,000,000-gallon daily production goes to backwashing? Round to the nearest tenth of a percent.
  9. Engineers want the elevated wash-water tank to hold enough for two complete washes, 150,000 gallons, in a cylindrical tank with 30 feet of usable depth. Using Volume (cu ft) = 0.785 × Diameter² × depth, find the required 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
  1. Answer: 2,776 gpm. 4 MGD × 694 gpm per MGD = 2,776 gpm.
  2. Answer: 694 gpm. 2,776 gpm ÷ 4 filters = 694 gpm per filter.
  3. Answer: 1.39 gpm per square foot. 694 gpm ÷ 500 square feet = 1.388, which rounds to 1.39 gpm per square foot. That is below the typical 2 to 6 range, so Grayling Ford's filters are lightly loaded.
  4. Answer: 1.85 gpm per square foot. 2,776 gpm ÷ 3 filters = 925.3 gpm per filter, and 925.3 ÷ 500 = 1.85 gpm per square foot, still below the typical range.
  5. Answer: 7,500 gpm. 15 gpm per square foot × 500 square feet = 7,500 gpm.
  6. Answer: 75,000 gallons. 7,500 gpm × 10 minutes = 75,000 gallons.
  7. Answer: 2.0 feet per minute. Each square foot receives 15 gallons per minute, and 15 ÷ 7.48 gallons per cubic foot = 2.005 cubic feet per minute over each square foot, which is a rise of about 2.0 feet per minute.
  8. Answer: 7.5 percent. Four washes × 75,000 gallons = 300,000 gallons per day. 300,000 ÷ 4,000,000 = 0.075, or 7.5 percent of production.
  9. Answer: 29.2 feet. Required volume = 150,000 ÷ 7.48 = 20,053.5 cubic feet. Dividing by the 30-foot depth gives a base area of 668.4 square feet. Then Diameter² = 668.4 ÷ 0.785 = 851.5, and the square root of 851.5 is 29.18, so the tank needs a diameter of about 29.2 feet.

Differentiation

Support

Give students a formula card listing the loading-rate, backwash-volume, and cylinder-volume formulas in the order the problems use them. Before assigning the set, work one full example on a different fictional filter, such as a 400-square-foot box carrying 800 gpm, so the division pattern is modeled once.

Extension

Suppose Grayling Ford's demand grows to 5 MGD. Have students recompute problems 3 and 4 at the higher flow and state whether the plant can still wash a filter while staying inside the typical 2 to 6 gpm-per-square-foot range, showing the calculation that supports their answer.

ELL supports

Provide a labeled cross-section of one filter showing water moving down through the media in service and up during backwash, with the two rates written beside opposite arrows. Allow answers to problems 3 and 4 to be recorded in a two-column table labeled "four filters" and "three filters" rather than in sentences.