Math problem set Grade 10 Science

Quintara Heights Pressure Zones: Head, Pressure, and Storage

A grade 10 problem set where students convert elevation head to pressure, size a cylindrical tank in gallons, and find supply time for a fictional hilly utility.

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: Every problem carries feet, psi, MGD, gpm, or gallons through a multi-step conversion using 0.433, 2.31, 694, and 7.48, choosing consistent units as items 1, 6, 7, and 8 require.

All materials for HSN-Q.A.1 →

HSG-MG.A.1 Common Core Math

Use geometric shapes, their measures, and their properties to describe objects (e.g., modeling a tree trunk or a human torso as a cylinder).

How this material aligns: Problems 5 and 6 model the standpipe as a cylinder and apply the 0.785 x diameter-squared x depth volume formula to estimate its storage capacity in cubic feet and gallons.

All materials for HSG-MG.A.1 →

HS-ETS1-1 NGSS

Analyze a major global challenge to specify qualitative and quantitative criteria and constraints for solutions that account for societal needs and wants.

How this material aligns: Problem 3 tests the hillside home against the state minimum-pressure criterion, and problem 10 converts the utility's downtown pressure target into the quantitative water-surface elevation constraint any taller standpipe would have to meet.

All materials for HS-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 elevation head in feet to pressure in psi using 0.433 psi per foot.
Students will calculate a cylindrical tank's volume in cubic feet and gallons.
Students will find how long a full tank can supply a stated demand.

Vocabulary

Standpipe: A standpipe is a tall, cylindrical water storage tank that sits directly on the ground instead of standing on legs like a water tower. It stores water and helps maintain steady pressure in the pipes nearby.

Also: head, pressure, elevation, storage

Teacher overview

Pressure in a water system comes from the height of the water surface above the point being served, so a home high on a hill sees lower pressure than one down in the valley fed by the same tank. This set uses the fictional Quintara Municipal Water and its hilltop standpipe to practice converting elevation head to pressure, sizing the tank, and finding how long its stored water lasts. Operators use two constants throughout: every foot of head adds 0.433 psi, and every psi requires 2.31 feet of head. A common misconception this set corrects is treating the tank's height as the head; the head is the difference between the water-surface elevation and the elevation of the point being served, not the tank's own height.

Quintara Heights Pressure Zones: Head, Pressure, and Storage — Grade 10 Science

Name: ____________________________ Date: ______________

Problems

Suppose the fictional town of Quintara Heights is served by one standpipe. Use these scenario values for every problem. Each foot of head adds 0.433 psi; each psi needs 2.31 feet of head; a cylinder's volume in cubic feet is 0.785 x diameter squared x depth; one cubic foot holds 7.48 gallons; and one MGD equals 694 gpm.

Water-surface (overflow) elevation: 220 feet
Downtown service point elevation: 90 feet
Hillside home elevation: 150 feet
Standpipe tank: 50 feet in diameter, 40 feet of water depth
Average town demand: 0.5 MGD

Find the head at the downtown service point (the water surface minus the downtown elevation), then convert it to pressure in psi. Round to the nearest whole psi.
Find the head at the hillside home, then convert it to pressure in psi. Round to the nearest whole psi.
State rules require at least 20 psi at every connection. Does the hillside home meet that minimum, and by about how many psi does it clear or miss it?
A proposed taller standpipe would raise the water-surface elevation to 240 feet. What would the downtown pressure become? Round to the nearest whole psi.
Model the tank as a cylinder and find its volume in cubic feet using 0.785 x diameter squared x depth.
Convert the tank volume from problem 5 to gallons.
Convert the town's average demand of 0.5 MGD to gallons per minute.
If no water entered the tank, how many hours could the full tank from problem 6 supply the town at 0.5 MGD? Round to the nearest whole hour.
Explain why the hillside home has lower pressure than the downtown service point, even though both are fed by the same tank.
The utility wants at least 40 psi downtown. What is the minimum water-surface elevation needed, given the downtown elevation of 90 feet? Round up to the next whole foot.

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: 56 psi. Head = 220 - 90 = 130 feet. Pressure = 130 x 0.433 = 56.29 psi, which rounds to 56 psi.
Answer: 30 psi. Head = 220 - 150 = 70 feet. Pressure = 70 x 0.433 = 30.31 psi, which rounds to 30 psi.
Answer: Yes, by about 10 psi. The hillside pressure of about 30 psi is above the 20 psi minimum: 30.31 - 20 = 10.31, roughly a 10 psi margin.
Answer: 65 psi. New head = 240 - 90 = 150 feet. Pressure = 150 x 0.433 = 64.95 psi, which rounds to 65 psi.
Answer: 78,500 cubic feet. Volume = 0.785 x 50 squared x 40 = 0.785 x 2,500 x 40 = 78,500 cubic feet.
Answer: 587,180 gallons. 78,500 cubic feet x 7.48 gallons per cubic foot = 587,180 gallons.
Answer: 347 gpm. 0.5 MGD x 694 gpm per MGD = 347 gpm.
Answer: About 28 hours. At 0.5 MGD the town uses 500,000 gallons per day, so 587,180 / 500,000 = 1.174 days; 1.174 x 24 = 28.2, about 28 hours.
Answer: Pressure depends on the head, which is the drop from the water surface down to the service point. The hillside home sits higher, so its head (220 - 150 = 70 feet) is smaller than the downtown head (220 - 90 = 130 feet), giving it less pressure.
Answer: 183 feet. To get 40 psi the head must be 40 x 2.31 = 92.4 feet, so the water surface must be at least 90 + 92.4 = 182.4 feet. Rounding up to 183 feet keeps the pressure at or above 40 psi (93 x 0.433 = 40.3 psi), while 182 feet would give only 92 x 0.433 = 39.8 psi.

Differentiation

Support

Provide a labeled elevation sketch showing the water surface, the downtown point, and the hillside home, with the head marked as the vertical distance between the surface and each point. Give the two constants (0.433 and 2.31) on a reference strip.

Extension

Ask students to find the pressure at a valley connection 70 feet below the downtown point, then explain why the utility might install a pressure-reducing valve there, referring to the 40-to-80 psi comfort range.

ELL supports

Pre-teach "head," "elevation," and "pressure" with the labeled sketch in English and, where possible, students' home languages. Allow students to show work using the diagram plus numbers rather than full sentences on the conversion items.