Math problem set Grade 10 Math
Talus River Well Field: Pumping and Tank Fill-Time Problems
A grade 10 math problem set where students use the continuity equation and cylinder volume formula to size pipe flow and tank fill time for a well field.
Objectives
- Students will apply the continuity equation to calculate water velocity in a pipe from a known flow rate and cross-sectional area.
- Students will calculate the volume of a cylindrical storage tank and convert it to gallons.
- Students will calculate and compare tank fill times under different pumping rates.
Vocabulary
Key words: continuity equation, cross-sectional area, well field, fill time
Teacher overview
The town of Windrow (a fictional town used for this problem set) gets its water from three wells in the Talus River Well Field, each pumping at the same steady rate. This problem set walks students through the well field's combined pumping rate, the water's velocity inside the pipe that carries it to town, and how long a new elevated storage tank takes to fill under normal and reduced pumping conditions. A common misconception in this math is using a pipe's diameter in place of its cross-sectional area; the continuity equation in problem 4 requires area, which must be calculated first using Area = 0.785 × Diameter², the same circle-area shortcut used throughout.
Problems
- Windrow (a fictional town used for this problem set) gets its water from three wells that each pump water at a steady 300 gallons per minute (gpm). What is the well field's combined pumping rate when all three wells run at once?
- Water leaves the well field through a single pipe with an inside diameter of 12 inches (1.0 foot). Using Area = 0.785 × Diameter², find the pipe's cross-sectional area in square feet.
- Convert the combined pumping rate from problem 1 into cubic feet per second, using the conversion 1 cubic foot per second = 448.8 gallons per minute. Round your reported answer to the nearest tenth, but keep the unrounded value on your calculator for problem 4.
- Using the continuity equation Q = A × V, your answer from problem 2, and the unrounded flow value from problem 3, calculate the water's velocity in the pipe in feet per second. Round to the nearest tenth.
- Windrow's new elevated tank is a cylinder 30 feet in diameter and 18 feet deep. Using Volume (cu ft) = 0.785 × Diameter² × depth, find the tank's volume in cubic feet.
- Convert your answer from problem 5 to gallons, using the conversion that one cubic foot holds 7.48 gallons.
- At the full pumping rate from problem 1, how long will it take to fill the empty tank? Give your answer in minutes, then convert to hours and minutes.
- One well goes offline for repairs, dropping the combined pumping rate to 600 gpm. How long will the same tank take to fill now, in minutes? By how many extra minutes does the repair delay the fill compared to problem 7?
- Challenge. The utility wants the tank to fill in 120 minutes or less even with one well offline. At 600 gpm, will they meet that goal? If not, what minimum pumping rate, in gpm, would fill the tank in exactly 120 minutes?
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: 900 gpm.
300 gpm × 3 wells = 900 gpm. - Answer: 0.785 sq ft.
Area = 0.785 × 1.0² = 0.785 sq ft. - Answer: 2.0 cfs.
900 gpm ÷ 448.8 gpm/cfs = 2.005, which rounds to 2.0 cfs. - Answer: 2.6 ft/s.
V = Q ÷ A = 2.005 ÷ 0.785 = 2.554, which rounds to 2.6 ft/s. - Answer: 12,717 cu ft.
Volume = 0.785 × 30² × 18 = 0.785 × 900 × 18 = 706.5 × 18 = 12,717 cu ft. - Answer: 95,123 gallons.
12,717 cu ft × 7.48 gal/cu ft = 95,123.16, which rounds to 95,123 gallons. - Answer: 106 minutes (about 1 hour 46 minutes).
95,123 gallons ÷ 900 gpm = 105.69, which rounds to 106 minutes; 106 ÷ 60 = 1.77 hours, or about 1 hour and 46 minutes. - Answer: 159 minutes; 53 extra minutes.
95,123 gallons ÷ 600 gpm = 158.5, which rounds to 159 minutes. 159 − 106 = 53 extra minutes. - Answer: No; about 793 gpm needed.
At 600 gpm the tank takes 159 minutes, missing the 120-minute goal. Minimum rate = 95,123 gallons ÷ 120 minutes = 792.7, which rounds to about 793 gpm.
Differentiation
Support (below grade level)
Provide the continuity equation and cylinder volume formula as labeled reference cards, and pre-calculate the pipe's cross-sectional area from problem 2 so students can focus on the velocity calculation in problem 4.
Extension (above grade level)
Ask students to find the pipe diameter that would keep the water velocity at or below 2.0 feet per second at the fictional well field's full 900 gpm pumping rate, working backward through the continuity equation.
ELL supports
Pre-teach "well field," "cross-sectional area," and "fill time" with a labeled diagram showing wells, a pipe, and a tank in sequence. Allow students to show work using the diagram and numbers rather than full sentences.