Math problem set Grade 9 Engineering & Design

Ash Creek Treatment Plant: Detention, Filters, Chlorine

A grade 9 engineering set on a surface water plant: students size a sedimentation basin's detention time, a filter loading rate, and a chlorine feed in pounds per day.

Objectives

  1. Students will calculate a sedimentation basin's detention time and a filter's loading rate from a design flow.
  2. Students will use the pounds formula to find a chlorine feed rate and compare a residual to the regulatory limit.

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.

Also: filter loading rate, chlorine dose, residual

Teacher overview

A conventional surface water plant slows water down in a sedimentation basin so heavy particles settle, filters the water, then adds chlorine to disinfect it. Engineers size each step from the plant's design flow. This set uses one fictional plant to find a basin's detention time, a filter's loading rate, and a chlorine feed rate in pounds per day, then checks each result against a normal design range.

A common error in this math is leaving detention time in days when the design range is stated in hours, which makes a basin look far too small. Detention time comes out in days when volume in gallons is divided by flow in gallons per day, so the result must be multiplied by 24 to compare it with the usual 2 to 4 hour range for sedimentation.

Answer Key

  1. Answer: 24,000 cu ft.
    Volume = 80 ft x 25 ft x 12 ft = 24,000 cubic feet.

  2. Answer: 179,520 gallons.
    24,000 cubic feet x 7.48 gallons per cubic foot = 179,520 gallons.

  3. Answer: 2.2 hours.
    Detention in days = 179,520 gallons / 2,000,000 gallons per day = 0.08976 day. In hours: 0.08976 x 24 = 2.15, about 2.2 hours. This falls inside the 2 to 4 hour range.

  4. Answer: 4.6 gpm/sq ft.
    Flow = 2.0 MGD x 694 = 1,388 gpm. Total filter area = 2 x (10 ft x 15 ft) = 300 square feet. Loading = 1,388 / 300 = 4.63, about 4.6 gpm per square foot, inside the 2 to 6 range.

  5. Answer: 36.7 lb/day.
    Pounds per day = 2.2 mg/L x 2.0 MGD x 8.34 = 36.7 pounds per day.

  6. Answer: 1.2 mg/L.
    Chlorine demand = dose - residual = 2.2 mg/L - 1.0 mg/L = 1.2 mg/L. The 1.0 mg/L residual is below the 4.0 mg/L maximum residual disinfectant level.

  7. Answer: 2.15 MGD.
    Set detention to 2.0 hours = 2.0 / 24 day, about 0.0833 day. Flow = volume / time = 179,520 gallons / (2.0 / 24 day) = 179,520 x 12 = 2,154,240 gallons per day, about 2.15 MGD. Above this flow the detention time drops below 2.0 hours.

  8. Answer: 3.1 gpm/sq ft.
    Three filters give 3 x 150 = 450 square feet. Loading = 1,388 / 450 = 3.08, about 3.1 gpm per square foot. Adding filter area lowers the loading rate, so water passes through more slowly and the filters remove particles more effectively.

Differentiation

Support

Provide the basin volume in gallons (179,520) already computed so students can begin at problem 3, and give a worked reminder that a detention time in days is multiplied by 24 to reach hours. Allow a calculator for every step.

Extension

Ask students to find the detention time and filter loading rate if the plant's design flow rose to 3.0 MGD, and to explain which result moves outside its recommended range first as flow increases, and what upgrade would bring it back in range.

ELL supports

Pre-teach "detention time," "sedimentation," "loading rate," "dose," and "residual" with a labeled side-view diagram of a basin, a filter, and a chlorine feed point. Allow students to record answers as numbers with unit labels rather than full sentences.