Worksheet Grade 9 Math

Brindlemoor Pressure Zones: Elevation and PSI Worksheet

A grade 9 worksheet on the fictional Brindlemoor Water District, converting ground elevation to water pressure in psi across three streets with the 0.433 rule.

Objectives

  1. Students will convert feet of water height above a point into pressure in psi.
  2. Students will write and solve a one-variable equation for pressure at a given elevation.
  3. Students will compare pressure-zone results against minimum and comfort service standards.

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: head, elevation, psi, booster pump

Teacher overview

The fictional Brindlemoor Water District pumps water into one tall tank, then serves homes at three different ground elevations. All three streets are in one pressure zone because they share one tank's water surface; the booster pump in item 8 creates a separate, higher zone for Summit Lane. Pressure at any home depends only on the height of water standing above that home. This worksheet has students turn elevation into pressure with the operator rule that each foot of water height adds 0.433 psi, then work the rule backward to find the elevation that meets a pressure target.

Answer Key

  1. Answer: 180 ft. Head = 660 minus 480 = 180 feet of water above River Road.
  2. Answer: 78 psi. 180 ft times 0.433 = 77.94, which rounds to 78 psi.
  3. Answer: 52 psi. Head = 660 minus 540 = 120 ft; 120 times 0.433 = 51.96, about 52 psi.
  4. Answer: 26 psi. Head = 660 minus 600 = 60 ft; 60 times 0.433 = 25.98, about 26 psi.
  5. Answer: Yes, by about 6 psi. Summit Lane is near 26 psi; 25.98 minus 20 = 5.98, roughly a 6 psi margin.
  6. Answer: 65 psi. The equation is P = 0.433 times (660 minus E). At E = 510: P = 0.433 times 150 = 64.95, about 65 psi.
  7. Answer: 568 ft. Solve 40 = 0.433 times (660 minus E). Then 660 minus E = 40 divided by 0.433 = 92.38 ft, so E = 660 minus 92.38 = 567.62, about 568 feet.
  8. Answer: 43 psi. New head = 700 minus 600 = 100 ft; 100 times 0.433 = 43.3, about 43 psi.
  9. Answer: 141 ft. 61 psi times 2.31 = 140.91, about 141 feet of water above the gauge.
  10. Answer: Lower homes have more feet of water standing above them, so their head is larger. Because pressure equals head times 0.433, a larger head gives a higher pressure at the same tank.

Differentiation

Support

Give students a printed strip with two labeled arrow rules, "feet to psi: multiply by 0.433" and "psi to feet: multiply by 2.31," and one worked example that uses numbers not on this worksheet. Provide the head values for items 1 through 4 already subtracted so students can focus on the multiplication.

Extension

Ask students to write a single equation for the pressure at any elevation E and graph P against E for elevations from 480 to 660 feet. Then have them read the graph to find the elevation where pressure drops to exactly 30 psi and confirm it algebraically.

ELL supports

Pre-teach "elevation," "head," "pressure," and "booster pump" with a labeled side-view sketch of a tank on a hill feeding three houses at different heights. Provide a sentence frame for item 10: "Lower homes have more ___ above them, so they have ___ pressure."