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Math problem set Grade 7 Careers & CTE

Kestleby's Pressure Zones: A Distribution Operator's Math

A grade 7 problem set on tank height and water pressure in a fictional town, using the constant of proportionality and percent change the way an operator does.

Objectives

  1. Students will identify the constant of proportionality linking feet of water to pressure.
  2. Students will convert between feet of water and pressure in both directions.
  3. Students will judge whether a computed pressure falls inside the range a utility maintains.

Vocabulary

Pressure zone
A pressure zone is a section of a water system that a utility manages separately because it sits at a different elevation than the rest of the town. Utilities create pressure zones so that water pressure stays safe and steady for everyone, whether they live at the bottom of a hill or near the top.

Also: head, psi, hydrant, elevation

Teacher overview

A distribution operator spends part of every shift checking pressure, because pressure is what keeps clean water moving out of a pipe instead of letting dirty water seep in. This set follows one operator through a morning of readings in the fictional town of Kestleby, where a hillside splits the town into two pressure zones. Height and pressure are related by a fixed constant, which makes the whole set a proportional relationship problem with a decimal constant. The misconception this set corrects is the belief that a taller tank gives every house the same pressure. What matters is the height of water above a given point, so a house partway up the hill sees less pressure than a hydrant at the bottom of the same hill.

Kestleby's Pressure Zones: A Distribution Operator's Math — Grade 7 Careers & CTE
Name: ____________________________ Date: ______________

Problems

The fictional town of Kestleby stores water in a tank on a ridge. One foot of water height equals 0.433 psi of pressure, and one psi equals 2.31 feet of water. Every elevation below was invented for this problem set.

  1. On Monday morning the tank's water surface sits 120 feet above a hydrant on Mill Street. What is the pressure at that hydrant, in psi?
  2. The operator records more readings: 40 feet gives 17.32 psi, 70 feet gives 30.31 psi, and 100 feet gives 43.3 psi. What is the constant of proportionality, in psi per foot?
  3. Write an equation for the pressure p, in psi, produced by a height h, in feet.
  4. A gauge on the other side of town reads 62 psi. How many feet of water does that pressure represent?
  5. Overnight the tank drops, and the water surface now sits 96 feet above the Mill Street hydrant instead of 120 feet. What is the percent decrease in that height?
  6. What is the pressure at the Mill Street hydrant when the water surface is 96 feet above it? Round to the nearest tenth of a psi.
  7. Distribution systems typically operate between 40 psi and 80 psi. Does your answer to problem 6 fall inside that range?
  8. Most states require at least 20 psi everywhere in the system. How many feet of water does 20 psi represent?
  9. A house on the hill sits 35 feet higher than the Mill Street hydrant. When the tank's water surface is 120 feet above that hydrant, what is the pressure at the house? Round to the nearest tenth of a psi.
  10. Using the overnight surface height of 96 feet above the hydrant, what is the pressure at that same house on the hill? Round to the nearest tenth of a psi.

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: 51.96 psi.
    Multiply the height by the constant: 120 feet times 0.433 psi per foot equals 51.96 psi. The units cancel to leave psi, which is a useful check.
  2. Answer: 0.433 psi per foot.
    Divide pressure by height for each reading. The first gives 17.32 divided by 40, which is 0.433. The second gives 30.31 divided by 70, which is 0.433. The third gives 43.3 divided by 100, which is 0.433. All three match, so the relationship is proportional with that constant.
  3. Answer: p = 0.433h.
    Pressure equals the constant times the height, with p measured in psi and h measured in feet.
  4. Answer: 143.22 feet.
    Multiply by the other constant: 62 psi times 2.31 feet per psi equals 143.22 feet. Checking the reverse direction, 143.22 times 0.433 is about 62 psi.
  5. Answer: 20 percent.
    The drop is 120 minus 96, which is 24 feet. Divide by the starting height: 24 divided by 120 equals 0.2, which is 20 percent. Dividing by 96 instead gives 25 percent, which answers a different question.
  6. Answer: 41.6 psi.
    Multiply the new height by the constant: 96 times 0.433 equals 41.568, which rounds to 41.6 psi.
  7. Answer: Yes.
    The value 41.6 psi sits between 40 psi and 80 psi, so the hydrant is still inside the range the fictional Kestleby system maintains. It is near the low end, which is why the operator keeps watching the tank level.
  8. Answer: 46.2 feet.
    Multiply the minimum pressure by the second constant: 20 psi times 2.31 feet per psi equals 46.2 feet of water.
  9. Answer: 36.8 psi.
    Subtract the elevations first: the water surface is 120 feet above the hydrant, and the house is 35 feet higher than the hydrant, so the water stands 85 feet above the house. Multiply: 85 times 0.433 equals 36.805, which rounds to 36.8 psi. Using 120 feet for the house is the common error, and it overstates the pressure by about 15 psi.
  10. Answer: 26.4 psi.
    The water surface is now 96 feet above the hydrant, so it stands 96 minus 35, or 61 feet, above the house. Multiply: 61 times 0.433 equals 26.413, which rounds to 26.4 psi. The house is still above the 20 psi minimum, yet it has lost about 10 psi overnight, which residents on the hill would notice in the shower.

Differentiation

Support

Give these students a drawing of the fictional Kestleby ridge with the tank, the hydrant, and the hillside house marked, and with the height of water above each point labeled. Work problem 1 aloud as a model, then let students use a filled-in conversion card showing both constants side by side. Allow a calculator for every item.

Extension

Ask these students to find the elevation at which a house on the fictional Kestleby hill would drop to exactly 20 psi when the water surface sits 96 feet above the hydrant, then to write one sentence recommending what the operator should do about homes above that point.

ELL supports

Pre-teach pressure, head, elevation, hydrant, and zone with a labeled side-view sketch of the fictional Kestleby hill. Provide the frame "___ feet of water gives ___ psi, because each foot gives 0.433 psi," and allow students to answer problem 7 with yes or no plus one reason.

Extensions

Ask students to find out whether their own town has more than one pressure zone and where the dividing line runs. Students can then sketch their own hill, mark a high house and a low house, and predict which one loses pressure first when a tank level drops, using the same constant the fictional Kestleby operator uses.