Worksheet Grade 7 Engineering & Design

Windrow Booster Pump: Pressure-Zone Design Constraints

Students size a fictional town's booster pump, turning water height into pressure and checking each neighborhood against a service pressure range.

Grade: Grade 7
Subject: Engineering & Design
Material: Worksheet
Time: 35 min
Prep: No prep
Materials: Calculator

Standards

7.RP.A.2 Common Core Math

Recognize and represent proportional relationships between quantities, including identifying the constant of proportionality and representing proportional relationships by equations.

How this material aligns: Items 1, 2, 7, and 8 use the constant 0.433 psi per foot to turn water height into pressure, the proportional relationship with a constant of proportionality this standard names.

All materials for 7.RP.A.2 →

MS-ETS1-1 NGSS

Define the criteria and constraints of a design problem with sufficient precision to ensure a successful solution, taking into account relevant scientific principles and potential impacts on people and the natural environment that may limit possible solutions.

How this material aligns: Items 3, 5, and 7 check each neighborhood's pressure against the 20 psi minimum and the 40 to 80 psi service range, defining the design's constraints with precision.

All materials for MS-ETS1-1 →

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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 convert water height in feet to pressure in psi using the rate 0.433 psi per foot.
Students will check each neighborhood's pressure against a service range and size a booster pump to fix a low zone.

Vocabulary

Key words: pressure, booster pump, psi, constraint

Teacher overview

Water pressure comes from height. The higher the water sits above a home, the more pressure that home gets. This worksheet uses the fictional town of Windrow, where a hilltop neighborhood keeps losing pressure. Students find the pressure at each spot, check it against the town's service range, and size a booster pump to fix the low zone. A common misconception is that a pump is needed everywhere; in fact, homes below the tower already have plenty of pressure, and only the high neighborhoods need a boost.

Windrow Booster Pump: Pressure-Zone Design Constraints — Grade 7 Engineering & Design

Name: ____________________________ Date: ______________

Student Worksheet

Use these values for the fictional Windrow system. Round each pressure to the nearest whole psi.

The water surface in the tower sits 100 feet above Main Street.
Pressure at any point = height of water above it (in feet) × 0.433 psi per foot.
The town's service range is 40 to 80 psi. The minimum allowed is 20 psi.

Find the water pressure at Main Street. The water surface is 100 feet above it.
The Hilltop neighborhood sits 70 feet above Main Street. So the water surface is only 30 feet above Hilltop. Find the pressure at Hilltop.
Does Hilltop meet the 20 psi minimum? By how many psi does it fall short?
Engineers add a booster pump that raises Hilltop's pressure by 30 psi. Find Hilltop's new pressure.
Is Hilltop's new pressure inside the 40 to 80 psi service range?
The pump raised Hilltop from 13 psi to 43 psi. Find the percent increase in pressure. Round to the nearest whole percent.
A steeper hill, Ridge Court, sits 90 feet above Main Street, so the water surface is only 10 feet above it. The pump can add up to 45 psi. Find Ridge Court's pressure after the pump, and state whether it lands in the 40 to 80 psi range.
A home in the valley sits 120 feet below the water surface. Find its pressure.
Explain why the high neighborhoods need a booster pump but the valley home does not. Use the idea that pressure depends on the height of water above a point.

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: 43 psi.
100 feet × 0.433 psi per foot = 43.3, which rounds to 43 psi.
Answer: 13 psi.
The water surface is 30 feet above Hilltop. 30 × 0.433 = 12.99, which rounds to 13 psi.
Answer: 7 psi short.
No, Hilltop does not meet the minimum. 20 psi − 13 psi = 7 psi short of the 20 psi minimum.
Answer: 43 psi.
13 psi + 30 psi from the pump = 43 psi.
Answer: Yes.
43 psi falls inside the 40 to 80 psi service range.
Answer: 231%.
The rise is 43 − 13 = 30 psi. 30 ÷ 13 = 2.31, which is about a 231 percent increase.
Answer: 49 psi.
The water surface is 10 feet above Ridge Court. 10 × 0.433 = 4.33, about 4 psi. With the pump: 4.33 + 45 = 49.33, about 49 psi, which lands inside the 40 to 80 psi range.
Answer: 52 psi.
120 feet × 0.433 = 51.96, which rounds to 52 psi.
Answer: High homes have little water height above them.
Accept any answer explaining that pressure depends on how much water height sits above a point. High neighborhoods have little water above them, so their pressure is low and a pump is needed. The valley home has a tall column of water above it, so it already has strong pressure.

Differentiation

Support

Give students the conversion as a labeled rule, "feet of water above a point × 0.433 = psi," on a printed strip. Work item 1 together as a model, then let students try the rest. Allow a calculator throughout.

Extension

Ask students to find the greatest height, in feet, a neighborhood could sit above Main Street and still reach the service range after the pump adds a 30 psi boost. Have them show the reasoning behind their answer.

ELL supports

Provide a labeled diagram of the tower and three homes at different heights, with the water surface marked. Pre-teach "pressure," "booster pump," and "psi." Allow item 9 to be answered with a sentence frame: "High homes have ___ water above them, so their pressure is ___."