Water for Schools
Menu

Lesson plan Grade 6 Science

Havenmoor Sedimentation Basin: Detention Time and Settling

A 40-minute grade 6 science lesson where students compute detention time for a fictional plant's settling basin and connect slower water to cleaner water.

Objectives

  1. Students will find detention time by dividing a settling basin's volume by its flow rate.
  2. Students will explain why water that moves more slowly through a basin comes out cleaner.

Vocabulary

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.
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: settling

Teacher overview

After a water plant mixes in chemicals that make dirt clump together, the water flows slowly through a large tank called a sedimentation basin, where the clumps sink to the bottom. How long the water spends in that basin is the detention time, found by dividing the basin's volume by the flow rate. Plants aim for a few hours so the heavy bits have time to settle. This lesson uses the fictional Havenmoor plant to make that idea concrete with grade-level division. A common misconception is that a faster flow gives the water more time in the basin; it is the reverse, because the same tank empties sooner when water rushes through.

Procedure

Hook (5 min): Shake a jar of muddy water, set it on a desk, and have students watch for fifteen seconds as the biggest bits begin to sink. Ask what the mud needs in order to settle, and steer the class to the idea of time.

Direct instruction (10 min): Explain that a sedimentation basin holds water still enough for clumps of dirt to sink. Introduce the rule: detention time = basin volume divided by flow rate. Tell students that Havenmoor aims for a detention time between 2 and 4 hours, long enough for good settling.

Model (8 min): On the board, work Havenmoor's basin: a volume of 120,000 gallons and a flow of 40,000 gallons per hour. Divide 120,000 by 40,000 to get 3 hours, and check that 3 hours lands inside the 2-to-4-hour target.

Guided and independent practice (12 min): Students complete the handout, finding detention time at three different flow rates and deciding each time whether the basin still meets the target. Circulate and check that students divide volume by flow, not flow by volume.

Closure and exit ticket (5 min): Each student answers one question on a card: "If the plant speeds up the flow, does the water get more time or less time to settle?" Collect and scan the cards; if more than a quarter answer "more time," reteach the relationship before the next class.

Havenmoor Sedimentation Basin: Detention Time and Settling — Grade 6 Science
Name: ____________________________ Date: ______________

Student Handout

Havenmoor is a fictional plant. Its sedimentation basin holds 120,000 gallons. Detention time = basin volume divided by flow rate. The target detention time is 2 to 4 hours.

  1. The flow into the basin is 40,000 gallons per hour. Find the detention time.
  2. Is that detention time within the 2-to-4-hour target?
  3. The flow rises to 60,000 gallons per hour. Find the new detention time.
  4. Is that detention time within the target?
  5. The flow rises again to 80,000 gallons per hour. Find the detention time.
  6. Is that detention time within the target?
  7. As the flow rate goes up, does the detention time go up or down?
  8. Explain why water that moves more slowly through the basin comes out cleaner.

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: 3 hours.
    120,000 gallons divided by 40,000 gallons per hour = 3 hours.
  2. Answer: Yes.
    3 hours falls between 2 and 4 hours, so it meets the target.
  3. Answer: 2 hours.
    120,000 gallons divided by 60,000 gallons per hour = 2 hours.
  4. Answer: Yes.
    2 hours is at the low edge of the 2-to-4-hour target, so it still meets it.
  5. Answer: 1.5 hours.
    120,000 gallons divided by 80,000 gallons per hour = 1.5 hours.
  6. Answer: No.
    1.5 hours is below the 2-hour minimum, so it is too short.
  7. Answer: Down.
    The basin holds the same volume, so a faster flow empties it sooner, which shortens the detention time.
  8. Answer: Slower water gives the heavy clumps of dirt more time to sink to the bottom before the water leaves the basin, so more dirt is removed and the water that flows out is clearer. Accept any answer that links more time to more settling.

Differentiation

Support

Give students the division set up as a labeled fraction, volume on top and flow on the bottom, with problem 1 already worked as a model. Allow a calculator, and provide a number line marked 2 to 4 hours so students can see at a glance whether each answer lands inside the target.

Extension

Ask students to find the highest flow rate, in gallons per hour, at which Havenmoor's basin still meets the 2-hour minimum, and to explain how they used the volume to work backward to that flow.

ELL supports

Pre-teach "sedimentation," "detention time," and "flow rate" with the jar demonstration and a simple drawing of water entering and leaving a tank. Provide the frame "When the flow is faster, the detention time is ___," and allow one-word answers of "yes" or "no" for items 2, 4, and 6 and "up" or "down" for item 7.