Math problem set Grade 10 Science

Crestview Booster Pump: Horsepower, Energy, and Operating Cost

A grade 10 problem set where students chain the water-horsepower formula through efficiency to find a fictional pump station's daily energy use and cost.

Grade: Grade 10
Subject: Science
Material: Math problem set
Time: 45 min
Prep: No prep
Materials: Calculator

Standards

HSN-Q.A.1 Common Core Math

Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays.

How this material aligns: Students carry gpm, feet of head, horsepower, kilowatts, and kilowatt-hours consistently through the multi-step energy calculation in problems 1 through 7.

All materials for HSN-Q.A.1 →

HSA-CED.A.1 Common Core Math

Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions.

How this material aligns: Problem 8 has students create a one-variable cost equation from the pump's hourly energy cost and solve it against a daily electricity budget.

All materials for HSA-CED.A.1 →

HS-ETS1-3 NGSS

Evaluate a solution to a complex real-world problem based on prioritized criteria and trade-offs that account for a range of constraints, including cost, safety, reliability, and aesthetics as well as possible social, cultural, and environmental impacts.

How this material aligns: Problems 6 and 7 evaluate the booster pump on operating cost, and problem 8 solves for the run hours that satisfy Crestview's daily electricity budget, making the cost constraint the deciding criterion.

All materials for HS-ETS1-3 →

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This material is in the public domain (CC0) and may be copied, edited, and shared freely, without attribution. Updated 2026-07-23.

Objectives

Students will calculate water horsepower and brake horsepower using pump flow, head, and efficiency.
Students will convert a pump's power to daily energy use and operating cost.

Vocabulary

Total dynamic head (TDH): Total dynamic head is the total amount of lift a pump has to provide to move water where it needs to go, adding up the height it must lift the water, the friction it must overcome in the pipes, and the pressure needed at the far end. Engineers calculate it to pick a pump that is powerful enough for the job.

Also: water horsepower, brake horsepower, efficiency, kilowatt-hour

Teacher overview

The fictional town of Crestview runs a booster pump station to push water uphill to a high neighborhood. This set walks students from the pump's flow and lift through its efficiency to a daily energy bill, using the same water-horsepower formula operators use on certification exams. A common misconception in this math is confusing water horsepower, the useful work done on the water, with brake horsepower, the larger power the motor must actually supply once efficiency losses are included. Students who forget to divide by efficiency will underestimate the energy cost.

Crestview Booster Pump: Horsepower, Energy, and Operating Cost — Grade 10 Science

Name: ____________________________ Date: ______________

Problems

The Crestview booster station is a fictional pump, and all values below are given for it. Use these numbers and formulas:

Flow: 1,200 gpm.
Total dynamic head (the lift plus friction losses): 198 ft.
Pump efficiency: 75 percent (0.75).
Motor efficiency: 90 percent (0.90).
The pump runs 20 hours per day.
Electricity costs $0.12 per kWh.
Water horsepower = (flow in gpm times head in ft) divided by 3,960.
Brake horsepower = water horsepower divided by pump efficiency.
Motor input horsepower = brake horsepower divided by motor efficiency.
1 horsepower = 0.746 kW.

Calculate the water horsepower the pump delivers to the water.
Calculate the brake horsepower the pump requires, using the 0.75 pump efficiency.
Calculate the motor input horsepower, using the 0.90 motor efficiency.
Convert the motor input horsepower to kilowatts.
The pump runs 20 hours per day. Calculate the kilowatt-hours of electricity it uses each day.
At $0.12 per kWh, what is the pump's electricity cost per day? Round to the nearest dollar.
Using your daily cost, estimate the electricity cost of running this pump for a full year of 365 days.
Crestview wants to hold this pump's electricity to a budget of $120 per day. Using your kilowatt figure from problem 4, write an equation for the daily cost as a function of hours run, then solve for the number of hours that would cost exactly $120.

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key

Answer: 60 whp. Water horsepower = (1,200 gpm times 198 ft) divided by 3,960 = 237,600 divided by 3,960 = 60 water horsepower.
Answer: 80 bhp. Brake horsepower = 60 divided by 0.75 = 80 brake horsepower.
Answer: about 88.9 hp. Motor input = 80 divided by 0.90 = 88.89 horsepower.
Answer: about 66.3 kW. 88.89 horsepower times 0.746 kW per horsepower = 66.31 kW.
Answer: about 1,326 kWh. 66.31 kW times 20 hours = 1,326.2 kWh used per day.
Answer: about $159 per day. 1,326.2 kWh times $0.12 per kWh = $159.15, which rounds to about $159 per day.
Answer: about $58,090 per year. $159.15 per day times 365 days = $58,088, or about $58,090 for the year. Using the rounded $159 per day gives $58,035; accept either.
Answer: about 15 hours. Each hour costs 66.31 kW times $0.12 = about $7.96, so the equation is C = 7.96h; setting 7.96h = 120 gives h = 15.1, or about 15 hours.

Differentiation

Support

Provide the four formulas on a printed strip with each variable labeled, and pre-compute the water horsepower in problem 1 so students can practice the efficiency and energy steps without stalling on the first calculation.

Extension

Ask students to find how much Crestview would save per year if a rebuilt pump reached 85 percent efficiency instead of 75 percent, holding every other value the same, and to express the saving as a percentage of the original yearly cost.

ELL supports

Pair "horsepower," "efficiency," and "kilowatt-hour" with a simple diagram of energy flowing from the wall outlet, through the motor and pump, and into lifted water, with a little lost at each step. Provide the chain of formulas as labeled arrows so students can follow the order of operations.