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Worksheet Grade 11 Science

Oakhaven Clarifier: Detention Time and Overflow Rate

A grade 11 worksheet sizing a fictional clarifier by volume, detention time, surface overflow rate, and weir loading, checking detention against the design range.

Objectives

  1. Students will calculate a rectangular basin's volume in cubic feet and gallons.
  2. Students will compute detention time, surface overflow rate, and weir loading for a clarifier.
  3. Students will judge whether a clarifier's detention time meets the standard design range at two different flows.

Vocabulary

Clarifier
A clarifier is a big round or rectangular tank where water sits calmly so solids can sink to the bottom and be scraped away. Treatment plants use clarifiers at more than one stage of cleaning.
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.

Also: surface overflow rate, weir loading

Teacher overview

A sedimentation basin, or clarifier, gives suspended particles time to settle before the water moves on to filtration. Operators check three design numbers to know whether a basin is doing its job: detention time (how long water stays in the basin), surface overflow rate (flow divided by the basin's surface area), and weir loading (flow divided by the length of the overflow weir). This worksheet has students compute all three for the fictional Oakhaven clarifier at two flows, then compare the detention times against the standard design range. A common student error is to divide flow by the basin's total volume when finding overflow rate; overflow rate uses only the surface area, because it describes how fast water rises toward the outlet, not how long it stays in the tank. Sedimentation basins are typically designed for 2 to 4 hours of detention.

Oakhaven Clarifier: Detention Time and Overflow Rate — Grade 11 Science
Name: ____________________________ Date: ______________

Student Worksheet

The Oakhaven treatment plant is a fictional facility. Its rectangular clarifier is 80 feet long, 20 feet wide, and holds water 12 feet deep, with a 60-foot overflow weir. Use volume = length x width x depth (cubic feet), 1 cubic foot = 7.48 gallons, detention time = volume divided by flow, surface overflow rate = flow divided by surface area, and weir loading = flow divided by weir length. Treat 1.5 MGD as 1,500,000 gallons per day.

  1. Calculate the volume of the clarifier in cubic feet.
  2. Convert that volume to gallons.
  3. At a flow of 1.5 MGD, calculate the detention time in hours. (Divide the gallon volume by the daily flow to get days, then multiply by 24.)
  4. Sedimentation basins are typically designed for 2 to 4 hours of detention. Does Oakhaven's detention time at 1.5 MGD fall in that range?
  5. Calculate the surface overflow rate at 1.5 MGD, in gallons per day per square foot. (Surface area is length times width.)
  6. Calculate the weir loading at 1.5 MGD, in gallons per day per foot of weir.
  7. During peak demand the flow rises to 2.0 MGD. Recalculate the detention time in hours at this flow.
  8. Compare your item 7 answer to the 2-to-4-hour design range, and explain what a shorter detention time does to particle removal in the basin.
  9. Calculate the surface overflow rate at the 2.0 MGD peak flow.
  10. Without any new calculation, explain why a basin with more surface area would settle particles better than a narrow one carrying the same flow.

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.
    Volume = 80 x 20 x 12 = 19,200 cubic feet.
  2. Answer: 143,616 gal.
    19,200 cu ft x 7.48 gal/cu ft = 143,616 gallons.
  3. Answer: 2.30 hr.
    143,616 gal / 1,500,000 gal per day = 0.09574 day; 0.09574 x 24 = 2.30 hours.
  4. Answer: Yes.
    The detention time of about 2.30 hours falls within the typical 2-to-4-hour design range.
  5. Answer: 937.5 gpd/sq ft.
    Surface area = 80 x 20 = 1,600 square feet; 1,500,000 gal per day / 1,600 sq ft = 937.5 gallons per day per square foot.
  6. Answer: 25,000 gpd/ft.
    1,500,000 gal per day / 60 ft of weir = 25,000 gallons per day per foot.
  7. Answer: 1.72 hr.
    143,616 gal / 2,000,000 gal per day = 0.07181 day; 0.07181 x 24 = 1.72 hours.
  8. Answer: It falls below the design range, reducing particle removal.
    At 1.72 hours the basin is now below the 2-to-4-hour range, so particles have less time to settle, more of them carry over to the filters, and the settled water leaves cloudier.
  9. Answer: 1,250 gpd/sq ft.
    2,000,000 gal per day / 1,600 sq ft = 1,250 gallons per day per square foot.
  10. Answer: More surface area lowers the overflow rate.
    Spreading the same flow over a larger surface means the water rises more slowly toward the outlet, giving particles more chance to settle before they reach it, which is why a wide, shallow basin often clarifies better than a narrow, deep one.

Differentiation

Support

Provide the surface-area and volume values already computed (1,600 square feet and 19,200 cubic feet) so students can focus on the rate and time calculations, and give the formulas on a labeled reference card that distinguishes volume from surface area.

Extension

Ask students to find the maximum flow, in MGD, at which the Oakhaven clarifier would still hold a 2.0-hour detention time, working backward from the gallon volume and the 2.0-hour target.

ELL supports

Pre-teach "detention time," "overflow rate," and "weir" with a labeled cross-section of a clarifier showing water entering, particles settling, and clarified water spilling over the weir. Allow items 8 and 10 to be answered with short phrases or a labeled sketch.