Math problem set Grade 9 Math

Copperton Booster Pump Station: Horsepower and Energy Cost

A grade 9 problem set that sizes a fictional booster pump in water horsepower, then works through kilowatts and daily and yearly pumping energy cost.

Grade: Grade 9
Subject: Math
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: Problems 1 through 5 carry units through the water-horsepower, brake-horsepower, and kilowatt chain, the multi-step unit tracking this standard requires.

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 sets a daily cost cap and asks students to write and solve an inequality for run hours, matching this standard's create-and-solve requirement.

All materials for HSA-CED.A.1 →

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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 for a pump from flow, head, and efficiency.
Students will convert pump power to kilowatts and calculate daily and yearly energy cost.

Vocabulary

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

Teacher overview

Copperton is a fictional town whose booster pump station lifts water from a low pressure zone up to a hilltop neighborhood. Every pump costs money to run, and the electric bill depends on how much power the motor draws. This set walks students through the same power math a design engineer uses, moving from flow and head to horsepower, then to kilowatts, and finally to dollars.

A common misconception is that the pump's water horsepower equals the power the motor actually pulls from the grid. It does not. Water horsepower is the useful power delivered to the water, while the motor must draw more than that because the pump and motor are never fully efficient. Students who skip the efficiency step will always underestimate the energy cost.

Copperton Booster Pump Station: Horsepower and Energy Cost — Grade 9 Math

Name: ____________________________ Date: ______________

Problems

Use these facts about the fictional Copperton booster station for every problem.

The pump moves 1,200 gpm against a total head of 210 feet.
Water horsepower (WHP) = (flow in gpm × head in feet) ÷ 3,960.
Brake horsepower (BHP) = WHP ÷ pump efficiency. The pump efficiency is 0.75.
Electrical power in kilowatts = BHP × 0.746 ÷ motor efficiency. The motor efficiency is 0.90.
The station runs 18 hours per day. Electricity costs $0.12 per kWh.

Find the pump's water horsepower. Round to the nearest tenth.
Find the brake horsepower the motor must supply to the pump. Round to the nearest tenth.
Find the brake power in kilowatts, using kW = BHP × 0.746. Round to the nearest tenth. (This is the power at the pump shaft, before the motor loss.)
Find the electrical power the motor draws from the grid, dividing your problem 3 answer by the motor efficiency of 0.90. Round to the nearest tenth.
How many kilowatt-hours does the station use in one day? Multiply your problem 4 answer by the 18 run hours. Round to the nearest whole kWh.
What is the station's electricity cost for one day? Round to the nearest cent.
What is the station's electricity cost for a full year of 365 days? Round to the nearest dollar.
The utility wants to spend no more than $120 per day on this pump. Using the motor power from problem 4, write and solve an inequality for the number of run hours that keeps the daily cost at or under $120. Round down to the nearest tenth of an hour.

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: 63.6 hp.
WHP = (1,200 × 210) ÷ 3,960 = 252,000 ÷ 3,960 = 63.636, which rounds to 63.6 hp.
Answer: 84.8 hp.
BHP = 63.636 ÷ 0.75 = 84.848, which rounds to 84.8 hp.
Answer: 63.3 kW. kW = 84.848 × 0.746 = 63.297, which rounds to 63.3 kW.
Answer: 70.3 kW.
Motor draw = 63.297 ÷ 0.90 = 70.33, which rounds to 70.3 kW.
Answer: 1,265 kWh.
Daily energy = 70.3 × 18 = 1,265.4, which rounds to 1,265 kWh per day. A student who carries the unrounded 70.33 kW gets 1,266 kWh; accept either.
Answer: $151.80.
Daily cost = 1,265 kWh × $0.12 = $151.80. Carrying unrounded values through the chain gives up to $151.91, so accept answers from $151.80 to $151.91.
Answer: $55,407.
Yearly cost = $151.80 × 365 = $55,407. Carrying unrounded values through the chain gives up to $55,448, so accept answers from $55,407 to $55,448.
Answer: 14.2 hours.
The hourly cost is 70.3 kW × $0.12 = $8.436 per hour. The inequality is 8.436 × h ≤ 120, so h ≤ 120 ÷ 8.436 = 14.22. Rounding down, the pump can run about 14.2 hours per day and stay within the $120 budget.

Differentiation

Support

Give students a printed flow chart of the four power steps, in order, with the divide or multiply operation labeled on each arrow. Let them fill in one number per box so the chain from horsepower to dollars stays visible. Allow calculator use throughout.

Extension

Ask students to find the yearly savings if the utility replaces the pump with a more efficient one, raising pump efficiency from 0.75 to 0.85 while all other values stay the same. They should rework problems 2 through 7 and report the difference in yearly cost.

ELL supports

Pre-teach the words "efficiency," "horsepower," and "kilowatt" with a simple labeled diagram of a motor turning a pump. Provide a sentence frame for problem 8: "The pump can run ___ hours because the daily cost must stay at or under ___ dollars."