Water for Schools
Menu

Math problem set Grade 9 Careers & CTE

Gullwing Ridge Pump Station: Flow, Head, and Drawdown Math

A grade 9 careers math set where students do a distribution operator's calculations for one pump station: drawdown, pressure and head, flow conversions, and fill time.

Objectives

  1. Students will calculate well drawdown, pressure, and head using stated conversions.
  2. Students will convert a pump's flow between gpm, MGD, and cfs and use it to find a tank's fill time.

Vocabulary

Drawdown
Drawdown is how far the water level in a well drops when a pump pulls water out. The harder and longer you pump, the bigger the drawdown gets.
Flow rate
Flow rate measures how much water moves past a point in a certain amount of time, like gallons per minute through a garden hose. Bigger pipes and stronger pushes give bigger flow rates.

Also: head, continuity

Teacher overview

A distribution operator keeps water moving at the right pressure, and that work rests on a handful of formulas an operator uses every week. This set walks students through the calculations for one pump station and its storage tank, building from a well's drawdown up to a tank's fill time. A common mistake in the tank-volume step is to square the radius instead of the diameter; the formula V = 0.785 x D squared x depth already uses the full diameter, so students who first halve the diameter get a volume four times too small. Keep units written at every step.

Gullwing Ridge Pump Station: Flow, Head, and Drawdown Math — Grade 9 Careers & CTE
Name: ____________________________ Date: ______________

Problems

All problems describe the fictional Gullwing Ridge pump station. Use these facts:

  • Well drawdown = pumping water level - static water level
  • Pressure: one foot of head equals 0.433 psi; one psi equals 2.31 feet of head
  • Continuity: flow Q = area x velocity
  • One MGD equals 694 gpm; one cubic foot per second equals 448.8 gpm
  • Tank volume in cubic feet = 0.785 x diameter squared x depth; one cubic foot holds 7.48 gallons
  • Normal service pressure runs 40 to 80 psi
  1. At the Gullwing Ridge well, the static water level sits 40 feet below the surface. While the pump runs, the level drops to 95 feet below the surface. Find the drawdown.
  2. The water surface in the elevated storage tank stands 120 feet above a pressure gauge at the base of the tower. Using 0.433 psi per foot, find the pressure at the gauge to the nearest psi.
  3. A hydrant on the system needs 60 psi to work well. Using 2.31 feet per psi, how many feet of head does 60 psi represent? Round to the nearest foot.
  4. The station pump is rated at 1,388 gpm. Express that flow in MGD.
  5. Water enters a pipe with a cross-sectional area of 0.8 square feet at a velocity of 3 feet per second. Find the flow in cubic feet per second, then convert it to gpm.
  6. The storage tank is a cylinder 30 feet across and 24 feet deep. Find its volume in cubic feet, then convert to gallons.
  7. Using the pump rate from problem 4 in gpm, how long, to the nearest minute, would it take to fill the empty tank from problem 6? Then state that time in hours and minutes.
  8. Explain why the volume formula uses the full diameter, not the radius, and what happens to the answer if a student halves the diameter first by mistake.
  9. The tank's water surface sits 110 feet above a home. Find the pressure at the home to the nearest psi, and state whether it falls within the normal 40 to 80 psi service band.

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: 55 ft. Drawdown = pumping level - static level = 95 ft - 40 ft = 55 ft.
  2. Answer: 52 psi. 120 ft x 0.433 psi/ft = 51.96 psi, which rounds to 52 psi.
  3. Answer: 139 ft. 60 psi x 2.31 ft/psi = 138.6 ft, which rounds to 139 ft of head.
  4. Answer: 2.0 MGD. 1,388 gpm / 694 gpm per MGD = 2.0 MGD.
  5. Answer: 1,077 gpm. First, Q = area x velocity = 0.8 sq ft x 3 ft/s = 2.4 cfs. Then 2.4 cfs x 448.8 gpm/cfs = 1,077.12 gpm, or about 1,077 gpm.
  6. Answer: 126,831 gal. Volume = 0.785 x 30 squared x 24 = 0.785 x 900 x 24 = 16,956 cu ft. Then 16,956 cu ft x 7.48 gal/cu ft = 126,830.88 gal, or about 126,831 gallons.
  7. Answer: 91 min. The pump runs at 1,388 gpm, so 126,831 gal / 1,388 gpm = 91.4 minutes, about 91 minutes, which is roughly 1 hour and 31 minutes.
  8. Answer: The factor 0.785 already accounts for the circle, so the formula is built for the full diameter. If a student halves the diameter to a radius first and then squares it, the volume comes out about one fourth of the correct value, because squaring a halved number cuts it to one quarter.
  9. Answer: 48 psi. 110 ft x 0.433 psi/ft = 47.63 psi, about 48 psi, which is inside the normal 40 to 80 psi service band.

Differentiation

Support

Provide the two pressure rules as labeled arrow cards ("feet to psi: multiply by 0.433" and "psi to feet: multiply by 2.31") and give students the tank volume from problem 6 already computed so they can focus on the fill-time step in problem 7.

Extension

Ask students to find what pump rate in gpm would fill the tank from problem 6 in exactly 60 minutes, working backward from the tank's gallon volume, and to express that rate in MGD.

ELL supports

Provide a labeled diagram of the well and tank showing static level, pumping level, drawdown, and head as separate measurements. Allow students to show their work as labeled numbers and arrows rather than full sentences.