Water for Schools
Menu

Lesson plan Grade 10 Math

Millhollow Sedimentation Basin: Volume and Detention Time Lesson

A grade 10 lesson where students find a sedimentation basin's volume, convert it to gallons, and use the detention-time formula to size a treatment plant's flow rate.

Objectives

  1. Students will calculate the volume of a rectangular sedimentation basin in cubic feet and convert it to gallons.
  2. Students will use the detention-time formula to find the flow rate a basin can support at a target detention time.
  3. Students will explain the relationship between flow rate and detention time.

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.
Flow rate
Flow rate measures how much water moves past a point in a certain amount of time, like gallons per minute through a garden hose. Bigger pipes and stronger pushes give bigger flow rates.

Also: sedimentation basin, rectangular prism

Teacher overview

Before water can be filtered, most treatment plants let it sit in a sedimentation basin long enough for heavier particles to settle out, and the size of that basin determines how much flow the plant can treat. This lesson uses a fictional plant's rectangular basin to connect a geometry formula students already know, volume equals length times width times depth, to the detention-time formula operators use to size real treatment flow rates. A common misconception is that a bigger basin can handle any flow rate; in fact, a basin's volume and the target detention time together set a firm limit on flow, which this lesson makes concrete with numbers.

Procedure

Hook (5 min): Ask students how long they think a treatment plant should let water sit still before filtering it, and take a few guesses. Reveal that sedimentation basins are typically designed for 2 to 4 hours of detention time, and ask why a plant would not want that time to be much shorter or much longer.

Direct instruction (15 min): Introduce Millhollow Water Treatment Plant (a fictional plant used for this lesson), whose sedimentation basin measures 60 feet long, 20 feet wide, and 12 feet deep. Model finding the volume: 60 × 20 × 12 = 14,400 cubic feet. Convert to gallons using 1 cubic foot = 7.48 gallons: 14,400 × 7.48 = 107,712 gallons. Introduce the detention-time formula, t = Volume ÷ Flow, and rearrange it to solve for flow: Flow = Volume ÷ time. Using a target detention time of 3 hours, model the full chain: 107,712 gallons ÷ 3 hours = 35,904 gallons per hour, then × 24 hours = 861,696 gallons per day, or about 0.86 MGD.

Guided practice (15 min): In pairs, students work with a second basin, 50 feet long, 15 feet wide, and 10 feet deep: 1. Find that basin's volume in cubic feet, then convert it to gallons. 2. Using that gallon capacity, calculate the flow rate (in MGD) the basin could support at a 2.5-hour detention time. Circulate and check that pairs convert cubic feet to gallons before dividing by time, not after.

Independent practice (10 min): Each student solves two problems on the handout: 3. Working backward from a target flow of 0.75 MGD and a required 3-hour detention time, find the minimum basin volume in gallons. 4. Find the detention time of an 80,000-gallon basin at that same 0.75 MGD flow, and check whether it meets the 2-to-4-hour detention-time range.

Closure / exit ticket (5 min): Students answer one question (5) on an index card: "If a plant's flow rate increases but the basin's volume stays the same, does detention time increase or decrease? Use the formula t = Volume ÷ Flow to justify your answer." Collect and scan for the correct reasoning (an unchanged numerator divided by a larger denominator produces a smaller result) before dismissal.

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: 56,100 gal.
    Guided practice basin volume: 50 × 15 × 10 = 7,500 cu ft. 7,500 × 7.48 = 56,100 gallons.

  2. Answer: 0.54 MGD.
    56,100 gallons ÷ 2.5 hours = 22,440 gallons per hour. 22,440 × 24 = 538,560 gallons per day, or about 0.54 MGD.

  3. Answer: 93,750 gal.
    0.75 MGD = 750,000 gallons per day, or 31,250 gallons per hour. Minimum volume = 31,250 gallons/hour × 3 hours = 93,750 gallons.

  4. Answer: 2.56 hours.
    80,000 gallons ÷ 31,250 gallons/hour = 2.56 hours, which falls within the 2-to-4-hour range.

  5. Exit ticket answer: Detention time decreases.
    Because t = Volume ÷ Flow, keeping the volume (numerator) fixed while the flow (denominator) grows produces a smaller quotient, so detention time drops as flow increases.

Differentiation

Support (below grade level)

Provide the guided-practice basin's volume already calculated (7,500 cu ft), so students only need to complete the gallon conversion and the flow-rate calculation, and give students a labeled diagram showing length, width, and depth as three distinct measurements.

Extension (above grade level)

Ask students to find how many additional feet of basin depth the fictional Millhollow plant's original basin (60 ft by 20 ft) would need to add, keeping the same footprint, to support a flow of 1.2 MGD at a 3-hour detention time, working backward through the full formula chain.

ELL supports

Pre-teach "length," "width," "depth," and "detention time" using a simple 3D box diagram with each dimension labeled, and provide the volume and detention-time formulas as a single reference card with each variable defined in plain language.