Math problem set Grade 9 Math

Oakmere Pump Station: Wet Well Flow and Run Time

A grade 9 problem set at the fictional Oakmere pump station, finding wet-well volume, using Q equals A times V, converting cfs and MGD, and modeling the emptying time.

Grade: Grade 9
Subject: Math
Material: Math problem set
Time: 45 min
Prep: No prep
Materials: Calculator

Standards

HSN-Q.A.1 Common Core Math

Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays.

How this material aligns: Problems 4, 5, and 8 carry cfs, gpm, and MGD through the continuity and conversion steps so each flow lands in the unit the problem asks for.

All materials for HSN-Q.A.1 →

HSA-CED.A.2 Common Core Math

Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales.

How this material aligns: Problem 9 has students write the two-variable equation for gallons remaining against pumping time and evaluate it at a chosen minute.

All materials for HSA-CED.A.2 →

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This material is in the public domain (CC0) and may be copied, edited, and shared freely, without attribution. Updated 2026-07-23.

Objectives

Students will find the volume of a rectangular wet well in cubic feet and gallons.
Students will apply Q equals area times velocity and convert between cfs, gpm, and MGD.
Students will model the net emptying rate and remaining volume of the wet well over time.

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.

Also: wet well, continuity, net rate, cfs

Teacher overview

The fictional Oakmere pump station holds incoming water in a rectangular tank called a wet well, then lifts it out with a pump. This set walks through the well's volume, the flow through the discharge pipe using Q equals area times velocity, and the time the pump needs to empty the well. A common misconception in this math is treating the pump's discharge rate as the emptying rate; water keeps flowing in while the pump runs, so the net rate is what empties the well.

Oakmere Pump Station: Wet Well Flow and Run Time — Grade 9 Math

Name: ____________________________ Date: ______________

Problems

Oakmere (a fictional station) has a rectangular wet well 10 feet long, 8 feet wide, and 12 feet deep. Use these facts: one cubic foot holds 7.48 gallons; one cubic foot per second equals 448.8 gpm; one MGD equals 694 gpm.

Find the volume of the wet well in cubic feet.
Convert the volume from problem 1 to gallons.
Inflow to the empty well is 300 gpm and the pump is off. How many minutes does it take to fill the well? Round to the nearest whole minute.
The pump's discharge pipe has a cross-section area of 0.6 square feet, and water moves through it at 4 feet per second. Using Q equals area times velocity, find Q in cfs.
Convert the discharge from problem 4 to gpm.
While the pump runs at the problem-5 rate, 300 gpm still flows into the well. Find the net rate, in gpm, at which the well empties.
Starting from a full well (use your gallons from problem 2), how many minutes does the pump take to empty it at the net rate from problem 6? Round to the nearest tenth.
Express the pump's discharge from problem 5 in MGD. Round to the nearest hundredth.
Write an equation for the gallons G left in the well after t minutes of pumping, starting full and dropping at the net rate from problem 6. Then find G at t equals 5 minutes.
The pump switches on when the well is full and off when it is nearly empty. Explain why the inflow rate, not just the pump rate, sets how often the pump must start.

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

Answer Key

Answer: 960 cu ft. Volume = 10 times 8 times 12 = 960 cubic feet.
Answer: 7,180.8 gal. 960 cubic feet times 7.48 gallons per cubic foot = 7,180.8 gallons.
Answer: 24 min. 7,180.8 gallons divided by 300 gpm = 23.94, about 24 minutes.
Answer: 2.4 cfs. Q = area times velocity = 0.6 square feet times 4 feet per second = 2.4 cfs.
Answer: 1,077 gpm. 2.4 cfs times 448.8 gpm per cfs = 1,077.12, about 1,077 gpm.
Answer: 777 gpm. The pump removes 1,077 gpm while 300 gpm flows in; 1,077 minus 300 = 777 gpm net.
Answer: 9.2 min. 7,180.8 gallons divided by 777 gpm = 9.24, about 9.2 minutes.
Answer: 1.55 MGD. 1,077 gpm divided by 694 gpm per MGD = 1.552, about 1.55 MGD.
Answer: 3,295.8 gal. The equation is G = 7,180.8 minus 777t. At t = 5: G = 7,180.8 minus 3,885 = 3,295.8 gallons.
Answer: The pump rests while the inflow refills the well (volume divided by inflow) and runs while the net rate empties it (volume divided by pump rate minus inflow). Both times depend on the inflow, so the inflow sets how long each on-off cycle lasts and therefore how often the pump must start.

Differentiation

Support

Give students a labeled diagram of the wet well with length, width, and depth marked, and a reference card listing the three conversions. Provide the gallons from problem 2 already computed so students can attempt problems 7 and 9 even if problem 2 gave them trouble.

Extension

Ask students to find the inflow rate that would make the pump run exactly as long as it rests, if the fictional pump moves 1,077 gpm and the well fills and empties between the same two marks. Have them justify the answer with the net-rate idea.

ELL supports

Pre-teach "wet well," "discharge," "net rate," and "continuity" with a diagram showing water flowing in one pipe and out another at the same time. Provide a sentence frame for problem 10: "The inflow sets how often the pump starts because ___."