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Worksheet Grade 10 Math

Hartshorn Heights Booster Pump: Pipe Flow and Elevation Pressure

A grade 10 worksheet where students model a pipe as a circle, apply the continuity equation, and find elevation-driven pressure loss for a fictional booster pump station.

Objectives

  1. Students will model a pipe's cross section as a circle to calculate its cross-sectional area.
  2. Students will apply the continuity equation to find water velocity from a flow rate and a pipe's cross-sectional area.
  3. Students will calculate pressure loss caused by elevation gain using the 0.433 psi-per-foot rule.

Vocabulary

Key words: continuity equation, cross-sectional area, head, booster pump, static pressure

Teacher overview

Hartshorn Heights (a fictional hillside subdivision used for this worksheet) sits well above the rest of its water system's service area, so the utility relies on a booster pump station to keep pressure adequate at every tap. This worksheet uses one section of the booster main to practice modeling a pipe as a circle, applying the continuity equation, and calculating how much pressure elevation gain removes from a water system.

Hartshorn Heights Booster Pump: Pipe Flow and Elevation Pressure — Grade 10 Math
Name: ____________________________ Date: ______________

Student Worksheet

The booster main serving Hartshorn Heights (a fictional pipeline used for this worksheet) has an inside diameter of 12 inches and carries a flow of 2 cubic feet per second (cfs). The pump station sits at the base of the hill, where static pressure from the main tank measures 58 psi. The top of Hartshorn Heights is 165 feet higher in elevation than the pump station. Use the fact that each foot of elevation gain removes 0.433 psi of pressure, and that 1 cfs equals 448.8 gpm, for all items.

  1. Convert the pipe's 12-inch diameter to feet, then calculate its cross-sectional area using Area (sq ft) = 0.785 × Diameter².
  2. Using the continuity equation Q = A × V, calculate the water's velocity in the pipe in feet per second. Round to the nearest hundredth.
  3. Convert the pipe's 2 cfs flow rate to gallons per minute (gpm).
  4. Convert the pipe's flow rate to million gallons per day (MGD), using the fact that 1 MGD equals 1.547 cfs. Round to the nearest hundredth.
  5. Calculate the pressure lost to elevation gain over the 165-foot climb to the top of Hartshorn Heights.
  6. If no booster pump were installed, what would the pressure be at the top of Hartshorn Heights? Explain what a negative result means for water service.
  7. The booster pump adds 60 psi to the line at the base of the hill, before the climb. Calculate the actual pressure at the top of Hartshorn Heights with the pump running.
  8. Most states require at least 20 psi at every connection. Does the boosted pressure from item 7 meet that minimum? By how many psi does it clear the requirement?
  9. A planned expansion adds homes 40 feet higher than the current top of Hartshorn Heights. Calculate the additional pressure loss this would cause, and find the new arrival pressure with the same 60 psi pump boost.
  10. In one or two sentences, explain why elevation gain reduces water pressure, using the 0.433 psi-per-foot relationship in your explanation.

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: 0.785 sq ft.
    12 inches = 1 foot. Area = 0.785 × 1² = 0.785 sq ft.

  2. Answer: 2.55 ft/s.
    V = Q ÷ A = 2 cfs ÷ 0.785 sq ft = 2.5478, which rounds to 2.55 ft/s.

  3. Answer: 898 gpm.
    2 cfs × 448.8 gpm/cfs = 897.6, which rounds to 898 gpm.

  4. Answer: 1.29 MGD.
    2 cfs ÷ 1.547 cfs/MGD = 1.2928, which rounds to 1.29 MGD.

  5. Answer: 71.4 psi.
    165 ft × 0.433 psi/ft = 71.445, which rounds to 71.4 psi.

  6. Answer: about −13.4 psi.
    58 psi − 71.445 psi = −13.445 psi. A negative result means the main cannot deliver water to the top of the hill without a pump; pressure would fall to zero about 134 feet up (58 ÷ 0.433), roughly 31 feet short of the top.

  7. Answer: 46.6 psi.
    58 psi + 60 psi − 71.445 psi = 46.555, which rounds to 46.6 psi.

  8. Answer: 26.6 psi.
    46.555 psi − 20 psi = 26.555, which rounds to a margin of about 26.6 psi above the state minimum.

  9. Answer: 29.2 psi.
    40 ft × 0.433 psi/ft = 17.32 psi of additional loss. New arrival pressure = 46.555 − 17.32 = 29.235, which rounds to 29.2 psi.

  10. Answer: Elevation gain reduces pressure because pressure depends on the height of water above a point.
    Each foot a delivery point sits above the source removes 0.433 psi, so the 165-foot climb to Hartshorn Heights alone removes more than 71 psi before any pump boost is added.

Differentiation

Support (below grade level)

Provide the diameter-to-area conversion (12 in = 1 ft, Area = 0.785 sq ft) already completed, and give students a labeled diagram showing the pump station, the 165-foot climb, and the top of the hill before they attempt items 5 through 9.

Extension (above grade level)

Ask students to find the minimum pump boost, in whole psi, that would guarantee at least 20 psi at the top of Hartshorn Heights under the item 9 expansion scenario (165 plus 40 feet of total climb), and to check their answer by substituting it back into the pressure equation.

ELL supports

Pre-teach "elevation," "boost," and "static pressure" using a simple side-view diagram of the hill with the pump station, the 165-foot rise, and an arrow showing pressure decreasing as height increases. Allow students to show work with labeled diagrams in place of full sentences for items 6 and 10.

Extensions

Ask students to sketch a graph of pressure (y-axis) against elevation above the pump station (x-axis) from 0 to 165 feet, plotting the pressure at 0, 50, 100, and 165 feet using the 0.433 psi-per-foot rate, and to mark the point where the unboosted line would cross zero pressure.