Math problem set Grade 9 Science

Pump Station Horsepower at Haldane: A Grade 9 Energy Problem Set

A grade 9 problem set where students find water horsepower, brake horsepower, motor power, and the daily energy cost of a fictional high-service pump station.

Objectives

  1. Students will calculate water horsepower from a pump's flow and total head.
  2. Students will use pump and motor efficiency to find brake horsepower and motor input power.
  3. Students will convert horsepower to kilowatts and calculate a daily energy cost.

Vocabulary

Key words: water horsepower, brake horsepower, total head, efficiency, kilowatt-hour

Teacher overview

A high-service pump station lifts treated water up to a reservoir or into the distribution system, and the electricity to run it is often one of a utility's largest operating costs. Operators track the water horsepower a pump adds to the water, the brake horsepower the pump shaft actually needs, and the electrical power the motor draws, because efficiency losses grow the bill at every step. This set follows the fictional Haldane pump station through that chain to a daily energy cost. A common misconception is that a pump's electrical draw equals its water horsepower; because no pump or motor is perfectly efficient, the electrical power drawn is always larger than the water horsepower delivered.

Answer Key

  1. Answer: 63.6 hp.
    Water horsepower = (1,400 gpm x 180 ft) divided by 3,960 = 252,000 divided by 3,960 = 63.6 water horsepower.

  2. Answer: 84.8 hp.
    Brake horsepower = 63.6 divided by 0.75 = 84.8 brake horsepower.

  3. Answer: 94.2 hp.
    Motor input = 84.8 divided by 0.90 = 94.2 horsepower.

  4. Answer: 70.3 kW.
    94.2 hp x 0.746 kW per hp = 70.3 kilowatts.

  5. Answer: 1,406 kWh.
    70.3 kW x 20 hours = 1,406 kilowatt-hours in one day.

  6. Answer: $168.72.
    1,406 kWh x $0.12 per kWh = $168.72 per day.

  7. Answer: 67.5 percent.
    Wire-to-water efficiency = 0.75 x 0.90 = 0.675, or 67.5 percent. It means only about 68 percent of the electrical energy drawn ends up as useful water horsepower; the rest is lost as heat and friction in the pump and motor.

  8. Answer: 74.8 hp.
    New brake horsepower = 63.6 divided by 0.85 = 74.8 brake horsepower. Because the more efficient pump needs less brake horsepower for the same water horsepower, the motor draws less electricity, which lowers the daily energy bill.

Differentiation

Support

Provide the four formulas as a labeled ladder, with a blank beside each rung, so students can see that each answer feeds the next step. Work items 1 and 2 together, then let students continue the ladder on their own.

Extension

Ask students to find how many dollars per day the more efficient pump from item 8 would save, by carrying its brake horsepower through the motor, kilowatt, and cost steps and comparing the result with item 6.

ELL supports

Pre-teach "horsepower," "efficiency," and "kilowatt-hour" with a simple diagram: energy goes in at the motor, some is lost, and the rest lifts the water. Provide a sentence frame for item 7: "Wire-to-water efficiency means that out of all the electricity used, ___."