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Math problem set Grade 12 Engineering & Design

Saxby Terrace Pressure Zone: Elevation and Tank Head Problems

A grade 12 problem set where students convert tank and street elevations into static pressures, find pressure-zone limits, and size a tank's usable storage.

Objectives

  1. Students will convert between elevation head and pressure using 0.433 psi per foot and 2.31 feet per psi.
  2. Students will solve for the highest street elevation a tank can serve at a stated minimum pressure.
  3. Students will calculate a tank's total and usable storage and express usable storage as hours of supply.

Vocabulary

Pressure zone
A pressure zone is a section of a water system that a utility manages separately because it sits at a different elevation than the rest of the town. Utilities create pressure zones so that water pressure stays safe and steady for everyone, whether they live at the bottom of a hill or near the top.

Also: static pressure, head, overflow elevation, usable storage

Teacher overview

The fictional town of Saxby is extending service up a hillside neighborhood called Saxby Terrace, and every problem in this set comes from the questions its engineers must answer: what pressure each home will see, how high the zone can reach, and how much of the new tank's water is actually usable. The mathematics is the elevation-pressure relationship that operator certification exams test, using 0.433 psi per foot of head and its inverse, 2.31 feet of head per psi. A common error in this set is using the wrong conversion factor, multiplying feet of head by 2.31 when the situation calls for 0.433; a quick sense check helps, because 100 feet of head is about 43.3 psi, not 231 psi.

Saxby Terrace Pressure Zone: Elevation and Tank Head Problems — Grade 12 Engineering & Design
Name: ____________________________ Date: ______________

Problems

Use these facts about the fictional Saxby system for every problem. The new Saxby Terrace tank holds its full water surface at an elevation of 1,240 feet. One foot of head equals 0.433 psi, and one psi equals 2.31 feet of head. One cubic foot holds 7.48 gallons.

  1. A home sits at a street elevation of 1,020 feet. With the tank full, how much static pressure does that home receive, to the nearest tenth of a psi?
  2. A second home sits higher on the hill at 1,150 feet. Find its static pressure with the tank full, to the nearest tenth of a psi.
  3. Most states require a minimum of 20 psi everywhere in a distribution system. With the tank full, what is the highest street elevation, in feet, that still receives exactly 20 psi?
  4. Saxby's design goal for new construction is 40 psi. What is the highest street elevation that meets the 40 psi goal with the tank full, in feet?
  5. The tank is a cylinder 60 feet in diameter with a full water depth of 40 feet. Using Volume (cu ft) = 0.785 × Diameter² × depth, find its full volume in cubic feet.
  6. Convert the tank's full volume to gallons.
  7. Operators never drain the tank completely; they allow the level to drop only 20 feet below full before pumps refill it. How many gallons of usable storage does that 20-foot operating band hold?
  8. Saxby Terrace is expected to demand 1.2 MGD (1,200,000 gallons per day). If refill pumps failed with the tank full, about how many hours could the usable band alone meet demand, to the nearest half hour?
  9. When the water surface drops the full 20 feet of the operating band, by how many psi does the pressure fall at the home from problem 2, to the nearest tenth?

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

Answer Key
  1. Answer: 95.3 psi. Head = 1,240 − 1,020 = 220 feet. Pressure = 220 × 0.433 = 95.26, which rounds to 95.3 psi.
  2. Answer: 39.0 psi. Head = 1,240 − 1,150 = 90 feet. Pressure = 90 × 0.433 = 38.97, which rounds to 39.0 psi.
  3. Answer: 1,193.8 feet. A pressure of 20 psi requires 20 × 2.31 = 46.2 feet of head. The street can therefore sit no higher than 1,240 − 46.2 = 1,193.8 feet.
  4. Answer: 1,147.6 feet. The 40 psi goal requires 40 × 2.31 = 92.4 feet of head, so the highest qualifying elevation is 1,240 − 92.4 = 1,147.6 feet.
  5. Answer: 113,040 cubic feet. Volume = 0.785 × 60² × 40 = 0.785 × 3,600 × 40 = 2,826 × 40 = 113,040 cubic feet.
  6. Answer: 845,539 gallons. 113,040 cubic feet × 7.48 gallons per cubic foot = 845,539.2, or about 845,539 gallons.
  7. Answer: 422,770 gallons. The band is the same cylinder with a 20-foot depth: 0.785 × 3,600 × 20 = 56,520 cubic feet, and 56,520 × 7.48 = 422,769.6, about 422,770 gallons.
  8. Answer: 8.5 hours. 422,769.6 gallons ÷ 1,200,000 gallons per day = 0.352 day. Multiplying by 24 hours per day gives 8.46, which rounds to about 8.5 hours.
  9. Answer: 8.7 psi. A 20-foot drop in water surface removes 20 × 0.433 = 8.66 psi of head, so the home's pressure falls from 39.0 psi to about 30.3 psi, a drop of about 8.7 psi.

Differentiation

Support

Provide a conversion card showing both factors side by side, with the sense check written out: multiply feet by 0.433 to get psi, and multiply psi by 2.31 to get feet. Set up problems 1 and 3 as fill-in equations so students supply only the subtraction and the multiplication.

Extension

Ask students to work backward for the fictional Saxby system: if the utility wanted the 1,150-foot home to receive 60 psi with the tank full, at what elevation would the tank's water surface need to sit? (60 × 2.31 = 138.6 feet of head, so 1,150 + 138.6 = 1,288.6 feet.) Have them state one practical drawback of raising the tank.

ELL supports

Provide a labeled side-view sketch of the hill showing the tank, the two homes, and vertical arrows for head, so "elevation," "head," and "pressure" each attach to a visible distance. Offer a sentence frame for problem 9: "When the level drops ___ feet, the pressure falls by ___ psi."