Worksheet Grade 11 Math

Quillan Bend Filtration Plant: Loading Rate and Detention Time

A grade 11 worksheet on a fictional conventional treatment plant: filter loading rate as a density, basin volume, detention time, and a turbidity compliance check.

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

Standards

HSG-MG.A.2 Common Core Math

Apply concepts of density based on area and volume in modeling situations (e.g., persons per square mile, BTUs per cubic foot).

How this material aligns: Items 2 and 4 apply a density-style rate of gallons per minute per square foot of filter area to judge whether the filter is loaded within its safe operating range.

All materials for HSG-MG.A.2 →

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: Items 5 through 10 chain rectangular-basin volume in cubic feet to gallons, then divide by flow to find detention time in consistent units of minutes and hours.

All materials for HSN-Q.A.1 →

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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 compute a filter loading rate as a density of flow per unit area and compare it to an operating range.
Students will calculate a rectangular basin's volume in cubic feet and gallons.
Students will find detention time by dividing volume by flow in consistent units.

Vocabulary

Detention time: Detention time is how long water stays inside a tank or basin before flowing out. Water must stay long enough for each treatment step, like settling or disinfecting, to finish its job.
Sedimentation: Sedimentation is the step where water sits almost still in a big basin so the heavy clumps of dirt can sink to the bottom. The cleaner water on top then moves on to the filters.
Turbidity: Turbidity is a measure of how cloudy or murky water looks because of tiny particles floating in it. Clear water has low turbidity, and muddy water has high turbidity.

Also: loading rate

Teacher overview

This worksheet uses one fictional plant, the Quillan Bend water treatment plant, to connect two ideas that are usually taught separately: a rate expressed as an amount per unit area, and a volume divided by a flow. Students find how heavily a filter is loaded, then size a sedimentation basin and find how long water rests in it. Real operating ranges anchor the work, including the 2-to-6 gallons per minute per square foot band for conventional filters and the 0.3 NTU turbidity standard for finished water. A common misconception this worksheet corrects is dividing a volume in gallons by a flow in a different unit without first making the units agree.

Quillan Bend Filtration Plant: Loading Rate and Detention Time — Grade 11 Math

Name: ____________________________ Date: ______________

Student Worksheet

The Quillan Bend plant (a fictional facility) treats river water through sedimentation and rapid sand filtration. One of its filters is a rectangle 20 feet by 15 feet. The sedimentation basin ahead of it is 60 feet long, 20 feet wide, and 12 feet deep. Use these facts: filter loading rate = flow in gallons per minute ÷ filter surface area in square feet; conventional filters run at 2 to 6 gallons per minute per square foot; one cubic foot holds 7.48 gallons; detention time = basin volume ÷ flow; and finished water must stay at or below 0.3 NTU in 95% of monthly samples.

Find the surface area of the filter, in square feet.
The flow to the filter is 900 gallons per minute. Find the filter loading rate in gallons per minute per square foot.
Is the loading rate from problem 2 within the conventional range of 2 to 6 gallons per minute per square foot? Answer yes or no and state the value.
During a high-demand period the flow rises to 1,500 gallons per minute. Find the new loading rate and state whether it is still within the conventional range.
Find the volume of the sedimentation basin in cubic feet.
Convert the basin volume from problem 5 into gallons.
At a flow of 900 gallons per minute, find the detention time of the sedimentation basin in minutes, then convert it to hours.
Sedimentation basins are usually designed for 2 to 4 hours of detention. Rounded to one decimal place, does the detention time from problem 7 fall in that range?
During low demand the flow drops to 600 gallons per minute. Find the new detention time in hours.
Explain, in two or three sentences, why the flow must be in gallons per minute and the volume in gallons before you divide to get a detention time in minutes.
In one month, 38 of the plant's 40 turbidity samples were at or below 0.3 NTU. Find the percent of samples that met the limit, and state whether the plant met the 95% requirement.

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: 300 sq ft. Surface area = 20 ft × 15 ft = 300 square feet.
Answer: 3.0 gpm/sq ft. Loading rate = 900 gpm ÷ 300 sq ft = 3.0 gallons per minute per square foot.
Answer: yes, 3.0 gpm/sq ft. The value 3.0 falls between 2 and 6 gallons per minute per square foot, so the filter is within the conventional range.
Answer: 5.0 gpm/sq ft. Loading rate = 1,500 gpm ÷ 300 sq ft = 5.0 gallons per minute per square foot, which is still within the 2-to-6 range.
Answer: 14,400 cu ft. Volume = length × width × depth = 60 ft × 20 ft × 12 ft = 14,400 cubic feet.
Answer: 107,712 gal. 14,400 cu ft × 7.48 gallons per cubic foot = 107,712 gallons.
Answer: 2.0 hr. Detention time = 107,712 gal ÷ 900 gpm = 119.68 minutes, and 119.68 ÷ 60 = 1.99 hours, which is about 2.0 hours.
Answer: yes. Rounded to one decimal place, 1.99 hours is 2.0 hours, which falls at the very low end of the usual 2-to-4-hour design range.
Answer: 3.0 hr. Detention time = 107,712 gal ÷ 600 gpm = 179.52 minutes, and 179.52 ÷ 60 = 2.99 hours, which is about 3.0 hours.
Answer: Detention time is a volume divided by a flow, so the volume and the flow must use the same volume unit. Dividing gallons by gallons per minute cancels the gallons and leaves minutes. If the volume were in cubic feet or the flow in MGD, the units would not cancel and the result would not be a detention time in minutes.
Answer: 95%. 38 ÷ 40 = 0.95, or 95 percent, which exactly meets the requirement that 95% of monthly samples stay at or below 0.3 NTU.

Differentiation

Support

Provide the basin already outlined on grid paper so students can see the length, width, and depth before computing volume, and give the loading-rate and detention-time formulas printed with labeled blanks for flow, area, and volume.

Extension

Ask students to find the maximum flow, in gallons per minute, at which the sedimentation basin would still provide at least 2 hours of detention, working backward from the basin volume.

ELL supports

Pair "loading rate" with a picture of water spread over a filter's surface and "detention time" with a picture of water resting in a basin, so the two rates are visually distinct. Provide the units for each answer printed next to the blank.