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Lesson plan Grade 8 Science

A Shift at Ironwood Water Plant: Sizing Flow and Pumps

A 50-minute grade 8 science lesson where students run a fictional water plant's numbers: converting MGD to gpm, sizing a pump for peak demand, and setting criteria.

Objectives

  1. Students will convert a plant's demand between million gallons per day and gallons per minute.
  2. Students will model pumped volume as a proportional function of time and compare two pump rates.
  3. Students will state the criteria and constraints an operator uses to size a pump for peak demand.

Vocabulary

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: MGD, high-service pump, criteria, constraint

Teacher overview

A water plant operator has to keep enough water moving to meet the town's demand at every hour, including the busy morning peak, while holding pressure in a safe range. This lesson puts students in that role at the fictional Ironwood Water Plant, where they convert the plant's demand between million gallons per day (MGD) and gallons per minute (gpm) and then size a pump for the peak. Students use the conversion that 1 MGD equals 694 gpm, and the guideline that distribution systems typically run between 40 and 80 psi, and most states require at least 20 psi. A common misconception is that a pump sized for average demand is enough; the plant must instead meet the higher peak demand, or pressure falls when everyone uses water at once.

Procedure

Hook (5 min): Ask students what would happen if a town's water plant pumped only enough water for an average hour but everyone showered at 7 a.m. at the same time. Collect ideas about pressure dropping, then introduce the fictional Ironwood Water Plant and its operator's job.

Direct instruction (13 min): Define flow rate, MGD, and gpm. Model the conversion for Ironwood's average demand of 2 MGD: 2 MGD × 694 gpm per MGD = 1,388 gpm. Explain that pumped volume over time is proportional, so gallons = rate × minutes, and show that 1,388 gpm × 60 minutes = 83,280 gallons in an hour. Introduce the typical range of 40 to 80 psi and the 20 psi minimum most states require.

Guided practice (10 min): As a class, find Ironwood's peak demand, which is 1.5 times the average, in MGD and then in gpm. Confirm 3 MGD converts to 3 × 694 = 2,082 gpm, and discuss why the pump must meet the peak rather than the average.

Independent practice (15 min): Students complete the handout, converting units, writing the proportional pump equation, and comparing two candidate pumps against the peak-demand target.

Closure and exit ticket (7 min): Students answer handout question 6 on an index card, naming which pump meets peak demand and why. Confirm at least 80 percent chose the pump whose rate exceeds 2,082 gpm before dismissal.

A Shift at Ironwood Water Plant: Sizing Flow and Pumps — Grade 8 Science
Name: ____________________________ Date: ______________

Student Handout

Use the following information about the fictional Ironwood Water Plant. Use the conversion 1 MGD = 694 gpm.

  1. Ironwood's average demand is 2 MGD. Convert 2 MGD to gallons per minute.
  2. The plant's peak demand is 1.5 times its average demand. What is the peak demand in MGD?
  3. Convert the peak demand from question 2 to gallons per minute.
  4. Pumped volume is proportional to time. Write an equation for the gallons g a pump delivers in m minutes when it runs at 1,388 gpm.
  5. Using your equation from question 4, how many gallons does the pump deliver in 60 minutes?
  6. Ironwood must choose one high-service pump. Pump X delivers 1,500 gpm and Pump Y delivers 2,200 gpm. Which pump can meet the peak demand from question 3, and why?
  7. State two criteria and one constraint Ironwood's engineers should set before choosing a pump.

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: 1,388 gpm.
    2 MGD × 694 gpm per MGD = 1,388 gallons per minute.
  2. Answer: 3 MGD.
    Peak demand = 1.5 × 2 MGD = 3 MGD.
  3. Answer: 2,082 gpm.
    3 MGD × 694 gpm per MGD = 2,082 gallons per minute.
  4. Answer: g = 1,388m.
    At a steady 1,388 gallons per minute, gallons delivered equals the rate times the minutes, so g = 1,388m.
  5. Answer: 83,280 gallons. g = 1,388 × 60 = 83,280 gallons delivered in one hour.
  6. Answer: Pump Y.
    Pump Y's 2,200 gpm is greater than the 2,082 gpm peak demand, so it can keep up. Pump X's 1,500 gpm is below 2,082 gpm, so it would fall short and pressure would drop during the peak.
  7. Answer: Criteria: meet peak flow and hold pressure; constraint: budget.
    Two criteria are delivering at least 2,082 gpm at peak and keeping system pressure between 40 and 80 psi. One constraint is a fixed budget or energy limit that rules out an oversized, costly pump.

Differentiation

Support (below grade level)

Provide the conversion set up for question 1 as 2 × 694, and give a filled example of the proportional equation for a different rate (such as g = 1,000m) before students write their own in question 4.

Extension (above grade level)

Ask students to find how many minutes Pump Y running at 2,200 gpm would need to deliver one million gallons, and to explain how that time compares with running at the fictional plant's average rate of 1,388 gpm.

ELL supports

Pre-teach "flow rate," "peak," "pump," and "pressure" with a picture of a faucet at low and high flow. Provide a conversion card showing "1 MGD = 694 gpm" and sentence starters for question 6, such as "Pump ___ can meet peak demand because ___."

Extensions

Have students research a real nearby water utility's reported daily demand, convert it to gallons per minute, and estimate the peak flow its pumps must meet, then compare that scale to Ironwood's.