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Worksheet Grade 11 Math

Halstead Springs Sedimentation Basin: Detention and Overflow

A grade 11 worksheet computing basin volume, detention time, surface overflow rate, and weir overflow rate for a fictional plant's rectangular settling basin.

Objectives

  1. Students will calculate a rectangular basin's volume in cubic feet and gallons.
  2. Students will find detention time and compare it against the design range.
  3. Students will compute surface overflow and weir overflow rates from plant flow.

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.
Weir
A weir is a low wall that water flows over, used to control water levels and to measure flow. The deeper the water pouring over the weir, the more water is flowing.

Also: surface overflow rate, sedimentation basin

Teacher overview

A sedimentation basin gives suspended particles time to settle before water reaches the filters, and operators judge a basin by three numbers: detention time, surface overflow rate, and weir overflow rate. This worksheet has grade 11 students compute all three for the Halstead Springs Water Treatment Plant, a fictional plant, then test what happens when demand pushes more flow through the same basin. A common student error is to compare volume in cubic feet directly against a flow measured in gallons; the detention-time calculation only works once both quantities use the same unit.

Halstead Springs Sedimentation Basin: Detention and Overflow — Grade 11 Math
Name: ____________________________ Date: ______________

Student Worksheet

The Halstead Springs Water Treatment Plant is a fictional plant. Its rectangular sedimentation basin and flow are:

  • Basin length: 80 ft. Basin width: 20 ft. Water depth: 12 ft.
  • Plant flow: 1.5 MGD.
  • Total effluent weir length: 120 ft.
  • Design guideline: sedimentation basins usually hold water for 2 to 4 hours.
  • Conversions: 1 cubic foot = 7.48 gallons; 1 MGD = 694 gpm.
  1. Find the basin's volume in cubic feet (volume = length x width x depth).
  2. Convert that volume to gallons.
  3. Convert the plant flow of 1.5 MGD to gallons per day.
  4. Convert 1.5 MGD to gallons per minute.
  5. Find the detention time in hours (detention time = basin volume in gallons divided by flow in gallons per day, then multiplied by 24).
  6. Sedimentation basins are usually designed for 2 to 4 hours of detention. Does this basin fall inside that range?
  7. Find the basin's surface area (length x width).
  8. Find the surface overflow rate in gallons per day per square foot (plant flow divided by surface area).
  9. The effluent weirs total 120 feet. Find the weir overflow rate in gallons per day per foot (plant flow divided by weir length).
  10. Suppose demand rises and the plant pushes 2.0 MGD through the same basin. Recalculate the detention time in hours.
  11. Using your item 10 answer, explain what problem a detention time below the 2-hour design minimum could cause for the filters downstream.

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: 19,200 cu ft.
    80 ft x 20 ft x 12 ft = 19,200 cubic feet.
  2. Answer: 143,616 gal.
    19,200 cubic feet x 7.48 gallons per cubic foot = 143,616 gallons.
  3. Answer: 1,500,000 gpd.
    1.5 MGD means 1.5 million gallons per day, or 1,500,000 gallons per day.
  4. Answer: 1,041 gpm.
    1.5 MGD x 694 gpm per MGD = 1,041 gallons per minute.
  5. Answer: 2.3 hr.
    143,616 gallons divided by 1,500,000 gallons per day = 0.0957 days; 0.0957 x 24 = 2.30 hours.
  6. Answer: yes.
    A detention time of about 2.3 hours falls inside the 2-to-4-hour design range.
  7. Answer: 1,600 sq ft.
    80 ft x 20 ft = 1,600 square feet of surface area.
  8. Answer: 937.5 gpd per sq ft.
    1,500,000 gallons per day divided by 1,600 square feet = 937.5 gallons per day per square foot.
  9. Answer: 12,500 gpd per ft.
    1,500,000 gallons per day divided by 120 feet of weir = 12,500 gallons per day per foot.
  10. Answer: 1.7 hr.
    At 2.0 MGD the flow is 2,000,000 gallons per day; 143,616 divided by 2,000,000 = 0.0718 days; 0.0718 x 24 = 1.72 hours.
  11. Answer: Particles carry over to the filters.
    A detention time below 2 hours gives suspended solids less time to settle, so more turbidity reaches the filters, shortens filter runs, and can push finished-water turbidity higher.

Differentiation

Support

Provide a labeled sketch of the basin showing length, width, and water depth, and give students a conversion card listing "1 cubic foot = 7.48 gallons" and "1 MGD = 694 gpm" so they do not recall them from memory. Work item 1 and item 2 together as a class.

Extension

Ask students to find the maximum flow, in MGD, at which the basin still holds exactly 2 hours of detention, then explain why that number sets a practical ceiling on how much the plant can expand without building a second basin.

ELL supports

Pre-teach "detention," "settle," "overflow rate," and "weir" with a short demonstration of muddy water clearing in a jar over time. Provide a three-column table so students can record each item's number, formula, and result without writing full sentences.

Extensions

Have students find the design flow and basin dimensions of a real nearby treatment plant from its published consumer confidence report or a public works fact sheet, then compute the same three numbers and compare them to the fictional Halstead Springs basin.