Math problem set Grade 12 Math

Halloway Standpipe: Storage, Fire Flow, and Pressure

A grade 12 problem set sizing a fictional town's standpipe, converting elevation to service pressure, and checking whether stored water meets a two-hour fire-flow demand.

Objectives

  1. Students will compute the storage volume of a cylindrical standpipe in gallons.
  2. Students will convert elevation head to service pressure and back using the standard constants.
  3. Students will test stored water against a fire-flow demand stated in gpm and duration.

Vocabulary

Standpipe
A standpipe is a tall, cylindrical water storage tank that sits directly on the ground instead of standing on legs like a water tower. It stores water and helps maintain steady pressure in the pipes nearby.
Total dynamic head (TDH)
Total dynamic head is the total amount of lift a pump has to provide to move water where it needs to go, adding up the height it must lift the water, the friction it must overcome in the pipes, and the pressure needed at the far end. Engineers calculate it to pick a pump that is powerful enough for the job.

Also: elevation head, service pressure, fire flow

Teacher overview

The fictional town of Halloway stores its water in a tall cylindrical standpipe and relies on elevation, not a running pump, to hold pressure at the tap. This set has students size the standpipe, translate its water-surface elevation into service pressure, and test whether the stored volume can sustain a two-hour fire flow. A common misconception is that a wider tank produces more pressure; students should see that pressure at a service connection depends only on the height of water above that point, so the tank's diameter changes the stored volume but not the pressure.

Answer Key

  1. Answer: 880,770 gallons.
    Volume = 0.785 x 50 squared x 60 = 0.785 x 2,500 x 60 = 117,750 cubic feet; 117,750 x 7.48 = 880,770 gallons.

  2. Answer: 56.29 psi.
    130 feet x 0.433 psi per foot = 56.29 psi at the Elm Street connection.

  3. Answer: 180,000 gallons.
    1,500 gpm x 2 hours x 60 minutes per hour = 1,500 x 120 = 180,000 gallons.

  4. Answer: 20.4%.
    180,000 / 880,770 x 100 = 20.44%, about 20.4 percent of the full tank.

  5. Answer: 587.2 minutes, or about 9.8 hours.
    880,770 gallons / 1,500 gpm = 587.18 minutes; 587.18 / 60 = 9.79 hours.

  6. Answer: 148 feet, or 64.08 psi.
    Total dynamic head = 130 + 18 = 148 feet; 148 x 0.433 = 64.08 psi.

  7. Answer: 51.09 psi, which stays above 20 psi.
    The water surface falls to 130 - 12 = 118 feet above Elm Street; 118 x 0.433 = 51.09 psi, well above the 20 psi minimum.

  8. Answer: Diameter sets volume, but only height sets pressure.
    The volume formula multiplies by the diameter squared, so a wider tank holds much more water. Pressure at a connection depends only on the vertical height of water above it, which the diameter does not change, so problem 2 is unaffected by the tank's width.

Differentiation

Support

Provide a printed constant card with the three formulas, and pre-compute the tank radius and squared diameter for problem 1 so students can focus on the volume and conversion steps.

Extension

Ask students to find the tank diameter that would double the storage volume in problem 1 while keeping the 60-foot depth, and confirm that this larger tank leaves the Elm Street pressure in problem 2 unchanged.

ELL supports

Provide a labeled diagram of the standpipe showing diameter, water depth, and the 130-foot height above the service connection as three separate measurements. Allow students to answer problem 8 with a labeled sketch and a short caption.

Extensions

Have students find the height and diameter of a real standpipe or elevated tank in a nearby town, estimate its storage and the pressure it delivers to a ground-level home, and compare those figures with the Halloway values.