Worksheet Grade 12 Science

Ferngate High-Service Pump: Tank Fill, Pressure, and Energy

A grade 12 science worksheet using operator math to find a fictional booster station's tank fill time, discharge pressure, and daily energy cost.

Objectives

  1. Students will calculate the daily volume a booster pump delivers and the time it needs to fill an elevated tank.
  2. Students will convert tank height to discharge pressure using the standard 0.433 psi per foot rule.
  3. Students will calculate the pump's daily energy use and the energy cost per million gallons delivered.

Vocabulary

Key words: discharge pressure, elevated storage, kilowatt-hour, pump run time, million gallons per day

Teacher overview

The fictional town of Ferngate runs a high-service booster station that pumps treated water up into an elevated storage tank, which then supplies homes by gravity. This worksheet has students size the pump's daily output, find how long it takes to fill the tank, convert the tank's height into discharge pressure, and price out the electricity the pump uses. A common error to watch for is treating the pump's full-speed flow rate as the whole day's production. The pump only runs part of each day, so the daily volume comes from the running hours, not from a full 24 hours.

Answer Key

  1. Answer: 1.73 MGD.
    1,200 gpm / 694 gpm per MGD = 1.729, which rounds to 1.73 MGD.

  2. Answer: 720,000 gallons.
    1,200 gpm x 60 minutes per hour x 10 hours = 720,000 gallons in one day.

  3. Answer: 0.72 MGD.
    720,000 gallons is 0.72 million gallons delivered in one day, so 0.72 MGD.

  4. Answer: 67,824 cu ft.
    Volume = 0.785 x 60 x 60 x 24 = 0.785 x 3,600 x 24 = 67,824 cubic feet.

  5. Answer: 507,324 gallons.
    67,824 cu ft x 7.48 gallons per cu ft = 507,323.52, which rounds to 507,324 gallons.

  6. Answer: 423 minutes.
    507,324 gallons / 1,200 gpm = 422.8, which rounds to 423 minutes, or about 7.0 hours.

  7. Answer: 56 psi.
    0.433 psi per foot x 130 feet = 56.29, which rounds to 56 psi.

  8. Answer: 750 kWh.
    75 kW x 10 hours = 750 kilowatt-hours per day.

  9. Answer: 22,500 kWh.
    750 kWh per day x 30 days = 22,500 kilowatt-hours in the month.

  10. Answer: $2,700.
    22,500 kWh x $0.12 per kWh = $2,700 for the month.

  11. Answer: $125 per million gallons.
    Daily energy cost = 750 kWh x $0.12 = $90. Daily volume = 0.72 million gallons. $90 / 0.72 = $125 per million gallons.

  12. Answer: The pump only runs 10 of the 24 hours in a day.
    Because the pump is off for 14 hours, it moves water for less than half the day. Running all 24 hours at 1,200 gpm would deliver 1,728,000 gallons (1.73 MGD), more than twice the 720,000 gallons it actually delivers in 10 hours.

Differentiation

Support

Provide the four formulas and conversions on a printed reference card, and work items 1 and 2 together as a class before students continue. Accept unrounded calculator answers alongside the rounded values, and note beside item 4 that the 0.785 formula uses the full diameter, not the radius.

Extension

Ask students to find the electricity cost of delivering water for a full year at the same daily schedule, then estimate how much the town would save each year by shifting the pump's run time to overnight hours billed at a lower rate they choose and state.

ELL supports

Pre-teach "discharge pressure," "kilowatt-hour," and "run time" with a labeled diagram of a pump lifting water into a tall tank. Allow students to answer item 12 with a labeled sketch and numbers rather than full sentences.

Extensions

Have students find out whether their local utility uses elevated tanks or ground-level tanks with pumps that run on demand, and write one question for the public works department asking how the system keeps pressure steady overnight when most pumps are off.