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
- Students will calculate a sedimentation basin's detention time and a filter's loading rate from a design flow.
- 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.
Problems
The Ash Creek Water Treatment Plant (a fictional plant used for this set) treats river water. Use these values for all problems:
- Design flow: 2.0 MGD.
- Sedimentation basin (rectangular): length 80 feet, width 25 feet, water depth 12 feet.
- Filters: two filters, each 10 feet by 15 feet.
- Conversions: 1 cubic foot = 7.48 gallons; 1 MGD = 694 gpm; 1 gallon of water weighs 8.34 pounds.
- Design ranges: sedimentation detention 2 to 4 hours; filter loading 2 to 6 gpm per square foot.
Find the sedimentation basin volume in cubic feet. Volume = length x width x depth.
Convert the basin volume to gallons.
Find the detention time in hours at the design flow. Detention time in days = volume in gallons / flow in gallons per day; multiply by 24 for hours. State whether it falls in the 2 to 4 hour range.
Find the filter loading rate in gpm per square foot. First convert 2.0 MGD to gpm, then divide by the total filter area (two filters). State whether it falls in the 2 to 6 range.
The operator feeds chlorine at a dose of 2.2 mg/L. Use the pounds formula, pounds per day = dose (mg/L) x flow (MGD) x 8.34, to find the chlorine feed in pounds per day.
The plant targets a chlorine residual of 1.0 mg/L leaving the plant. Chlorine demand = dose - residual. Find the demand, and confirm the 1.0 mg/L residual stays below the 4.0 mg/L maximum residual disinfectant level.
The state asks the plant to keep sedimentation detention at 2.0 hours or more. Using detention time = volume / flow, find the flow in MGD at which the detention time drops to exactly 2.0 hours.
The plant plans to add a third filter of the same size (10 feet by 15 feet). Recompute the filter loading rate at 2.0 MGD with three filters, and explain how adding filter area changes the loading rate.
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: 24,000 cu ft.
Volume = 80 ft x 25 ft x 12 ft = 24,000 cubic feet.Answer: 179,520 gallons.
24,000 cubic feet x 7.48 gallons per cubic foot = 179,520 gallons.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.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.Answer: 36.7 lb/day.
Pounds per day = 2.2 mg/L x 2.0 MGD x 8.34 = 36.7 pounds per day.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.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.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.