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Worksheet Grade 11 Math

Marrow Gap Booster Pump Station: Flow, Head, and Precision

A grade 11 worksheet on a fictional booster pump station: pipe area, velocity, flow-rate conversions, well drawdown, and head-to-pressure with sensible precision.

Objectives

  1. Students will chain unit conversions from pipe area and velocity to a flow rate in gallons per minute and million gallons per day.
  2. Students will convert between feet of head and pressure in psi using the standard hydraulic constants.
  3. Students will report a computed flow to a precision appropriate to the measuring instrument.

Vocabulary

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.
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.

Also: velocity, head, significant figures

Teacher overview

This worksheet follows a single fictional system, the booster pump station in the town of Marrow Gap, which draws water from a well and pushes it uphill to a storage tank. Students start from the pipe's size and the water's speed, find the flow rate, and convert it step by step into the units a utility actually reports. The same station also lets students move between feet of head and pressure in psi, and decide how precisely a measured flow should be written. A common error to watch for is dropping or mixing units partway through the chain, which is exactly the discipline this worksheet builds.

Marrow Gap Booster Pump Station: Flow, Head, and Precision — Grade 11 Math
Name: ____________________________ Date: ______________

Student Worksheet

The town of Marrow Gap (a fictional community) runs one booster pump station. Its discharge pipe is 1.0 foot in diameter, and during normal operation the water inside it moves at 4.0 feet per second. Use these standard constants where needed: cross-sectional area = 0.785 × diameter², one cubic foot per second equals 448.8 gallons per minute, one million gallons per day equals 694 gallons per minute, one psi equals 2.31 feet of head, and one foot of head equals 0.433 psi.

  1. Find the cross-sectional area of the 1.0-foot-diameter discharge pipe, using area = 0.785 × diameter². Give your answer in square feet.
  2. Using flow rate Q = area × velocity, find the pipe's flow rate in cubic feet per second at a velocity of 4.0 feet per second.
  3. Convert the flow rate from problem 2 into gallons per minute.
  4. Convert the flow rate from problem 3 into million gallons per day. Round to the nearest hundredth.
  5. The station's well has a static water level of 40 feet below ground and a pumping water level of 70 feet below ground. Find the well drawdown, which is the pumping level minus the static level.
  6. A pressure gauge on the discharge line reads 52 psi. Convert this pressure to feet of head. Round to the nearest foot.
  7. The station's flow meter can only be read to the nearest 10 gallons per minute. Report your problem 3 flow rate to a precision that matches the meter, and explain in one sentence why writing more digits would be false precision.
  8. If the pump runs for 6 hours at the rounded flow of 1,409 gallons per minute, how many total gallons does it deliver?
  9. Marrow Gap's peak demand is 1.8 million gallons per day. Using your answer from problem 4, decide whether the pump can meet that demand, and state the margin in million gallons per day.
  10. Explain, in two or three sentences, why keeping the units attached at every step from problem 1 to problem 4 protects you from a wrong answer.

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. Area = 0.785 × diameter² = 0.785 × (1.0)² = 0.785 × 1.0 = 0.785 square feet.
  2. Answer: 3.14 cfs. Q = area × velocity = 0.785 sq ft × 4.0 ft/s = 3.14 cubic feet per second.
  3. Answer: 1,409 gpm. 3.14 cfs × 448.8 gpm per cfs = 1,409.2, which rounds to about 1,409 gallons per minute.
  4. Answer: 2.03 MGD. 1,409.2 gpm ÷ 694 gpm per MGD = 2.031, which rounds to 2.03 million gallons per day.
  5. Answer: 30 ft. Drawdown = pumping level − static level = 70 ft − 40 ft = 30 feet.
  6. Answer: 120 ft. 52 psi × 2.31 ft per psi = 120.12, which rounds to 120 feet of head.
  7. Answer: 1,410 gpm. The meter reads only to the nearest 10 gallons per minute, so 1,409.2 is reported as 1,410; writing 1,409.2 claims a tenth-of-a-gallon precision the meter cannot support.
  8. Answer: 507,240 gal. 1,409 gpm × 60 minutes per hour × 6 hours = 1,409 × 360 = 507,240 gallons.
  9. Answer: 0.23 MGD. The pump delivers 2.03 MGD and demand is 1.8 MGD, so it meets demand with a margin of 2.03 − 1.8 = 0.23 million gallons per day.
  10. Answer: Units carried through the chain act as a check. Each conversion factor cancels the old unit and leaves the new one, so a wrong setup shows up as a leftover or missing unit before the arithmetic hides the mistake.

Differentiation

Support

Give students a printed conversion strip listing the four constants (area factor 0.785, 448.8 gpm per cfs, 694 gpm per MGD, and the 2.31 and 0.433 head factors) with an arrow showing the direction each one converts. Work problems 1 and 2 together as a class before releasing the rest.

Extension

Ask students to find the velocity that would be required in the same 1.0-foot pipe to exactly meet the 1.8 million-gallons-per-day peak demand, working backward from that flow through the same conversion chain.

ELL supports

Provide a labeled cross-section diagram of the pump station showing the well, the static and pumping water levels, the pipe, and the discharge gauge, with each key term labeled. Pair the words "static," "pumping," "drawdown," and "head" with the specific place each is measured on the diagram.