Worksheet Grade 10 Engineering & Design

Pellworth Booster Pump Station: Run-Time and Energy Worksheet

A grade 10 engineering worksheet sizing a fictional booster pump's daily run-time, pipe velocity, and energy use, and writing a gallons-versus-time equation.

Grade: Grade 10
Subject: Engineering & Design
Material: Worksheet
Time: 40 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: Items 1, 5, 6, and 7 carry flow, volume, and energy units through pump run-time, unit-conversion, and pipe-velocity calculations, the unit-guided multi-step reasoning this standard names.

All materials for HSN-Q.A.1 →

HSA-CED.A.2 Common Core Math

Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales.

How this material aligns: Items 2 through 4 create and interpret a two-variable equation relating gallons pumped to elapsed time and read its slope on a graph, exactly this standard's modeling task.

All materials for HSA-CED.A.2 →

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This material is in the public domain (CC0) and may be copied, edited, and shared freely, without attribution. Updated 2026-07-17.

Objectives

Students will calculate a booster pump's daily run-time from a flow rate and a delivery target.
Students will write and interpret a linear equation relating gallons pumped to time.
Students will calculate pipe velocity and specific energy use for a pumping system.

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: booster pump, continuity, specific energy, run-time

Teacher overview

A booster pump lifts water into an elevated tank so gravity can serve the neighborhood the rest of the day. The fictional Pellworth Water District runs one booster pump to keep its tank filled, and operators need to know how long the pump must run, how fast water moves inside the discharge pipe, and how much energy the job takes. In this worksheet, students size the pump's daily work, write an equation for the gallons it delivers over time, and check its energy use against typical values. The pump moves water at a steady rate, so gallons delivered grow in a straight line with time, which lets students model the work with a simple linear equation.

Pellworth Booster Pump Station: Run-Time and Energy Worksheet — Grade 10 Engineering & Design

Name: ____________________________ Date: ______________

Student Worksheet

At Pellworth (a fictional district built for this worksheet), the booster pump delivers a steady 1,200 gpm. The tank needs 500,000 gallons delivered each day. The pump's discharge pipe has an inner diameter of 1.25 feet.

How many minutes must the pump run each day to deliver 500,000 gallons at 1,200 gpm? Convert your answer to hours.
Write an equation for the total gallons G the pump delivers after running t minutes at 1,200 gpm.
Use your equation to find how many gallons the pump delivers in 90 minutes of running.
If you graph gallons delivered on the vertical axis and minutes on the horizontal axis, what is the slope of the line, and what does that slope represent?
Convert the pump's flow of 1,200 gpm to million gallons per day (MGD), using the conversion that 1 MGD equals 694 gpm.
Find the discharge pipe's cross-sectional area in square feet, using Area = 0.785 × Diameter².
Find the water's velocity inside the pipe in feet per second. First convert 1,200 gpm to cubic feet per second using 1 cfs = 448.8 gpm, then use the continuity equation Velocity = Flow ÷ Area with your area from item 6.
During one full day's run, the pump draws 650 kWh to deliver the 500,000 gallons. Find the specific energy in kWh per 1,000 gallons.
Conventional treatment and delivery together typically use about 1 to 2 kWh per 1,000 gallons for the whole system. Compare the pump's specific energy from item 8 with that range. What does the comparison suggest about this one pump's share of the system's energy use?
Next summer, daily demand is expected to rise to 650,000 gallons. At the same 1,200 gpm, how many hours must the pump run to meet the new demand?
Explain why an operator might prefer to run this steady pump for more hours a day rather than install a much larger pump that fills the tank in a short burst.

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: about 6.94 hours.
500,000 gallons ÷ 1,200 gpm = 416.67 minutes; 416.67 ÷ 60 = 6.944, which rounds to about 6.94 hours.
Answer: G = 1,200t.
At 1,200 gpm, the gallons delivered equal 1,200 times the number of minutes t, so G = 1,200t.
Answer: 108,000 gallons.
G = 1,200 × 90 = 108,000 gallons in 90 minutes.
Answer: 1,200 gpm.
The slope is 1,200 gallons per minute, which is the pump's flow rate, because the line rises 1,200 gallons for every additional minute of running.
Answer: about 1.73 MGD.
1,200 gpm ÷ 694 gpm/MGD = 1.729, which rounds to about 1.73 MGD.
Answer: about 1.23 square feet.
Area = 0.785 × 1.25² = 0.785 × 1.5625 = 1.2266, which rounds to about 1.23 square feet.
Answer: about 2.18 feet per second.
1,200 gpm ÷ 448.8 gpm/cfs = 2.6738 cfs. Velocity = 2.6738 ÷ 1.2266 = 2.180, which rounds to about 2.18 feet per second.
Answer: 1.3 kWh per 1,000 gallons.
The 500,000 gallons is 500 units of 1,000 gallons: 650 kWh ÷ 500 = 1.3 kWh per 1,000 gallons.
Answer: It falls inside the whole-system range, so this one pump uses about as much as a typical entire system.
A specific energy of 1.3 kWh per 1,000 gallons is inside the 1-to-2 kWh per 1,000 gallons range, yet that range covers treatment and delivery for the whole system. This single booster therefore uses about as much energy per 1,000 gallons as a typical system uses in total, which is a large share. Accept any answer that makes this comparison and suggests a reasonable follow-up, such as checking how high the pump lifts the water or how efficiently it runs.
Answer: about 9.03 hours.
650,000 gallons ÷ 1,200 gpm = 541.67 minutes; 541.67 ÷ 60 = 9.028, which rounds to about 9.03 hours.
Answer: Accept any answer noting that a steady pump running longer keeps flow and pressure smooth, avoids the pressure surges and pipe stress of a large pump switching on and off, and often runs closer to its efficient design point, while a much larger pump costs more to buy and can waste energy in short bursts.

Differentiation

Support (below grade level)

Provide the run-time setup and the two conversions (694 gpm per MGD and 448.8 gpm per cfs) on a reference card, and work item 1 together so students see the gallons-divided-by-flow pattern before writing the equation in item 2.

Extension (above grade level)

Ask students to find what steady flow, in gpm, the pump would need to deliver the summer demand of 650,000 gallons in the same 6.94 hours it currently takes for 500,000 gallons, and to explain how that change would affect the pipe velocity from item 7.

ELL supports

Provide a labeled diagram of the pump, discharge pipe, and elevated tank, with flow rate, velocity, and energy marked. Provide a sentence frame for items 4, 9, and 11, and allow the calculations in items 1 through 8 and 10 to be shown as numbered steps.