Water for Schools
Menu

Worksheet Grade 9 Math

Hollisford Well Field: Specific Capacity and Drawdown Worksheet

A grade 9 worksheet on well drawdown and specific capacity, where students build and solve a linear model for a fictional municipal well.

Objectives

  1. Students will calculate well drawdown and specific capacity from static and pumping water levels.
  2. Students will build and solve a linear equation relating drawdown to pumping rate.

Vocabulary

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: static water level, pumping water level, specific capacity

Teacher overview

The Hollisford Well Field (a fictional groundwater supply used for this worksheet) draws water from a deep well. When a pump runs, the water level in the well drops from its resting level; that drop is the drawdown. Dividing the pumping rate by the drawdown gives the specific capacity, a number operators track to judge a well's health. This worksheet has students find drawdown and specific capacity, then build a simple linear model that treats drawdown as proportional to the pumping rate. A common misconception is confusing the pumping water level (a depth below ground) with the drawdown (a change in level); the two are related but not the same.

Hollisford Well Field: Specific Capacity and Drawdown Worksheet — Grade 9 Math
Name: ____________________________ Date: ______________

Student Worksheet

Use these facts about the Hollisford well (a fictional municipal well) for every problem. Depths are measured below the ground surface.

  • Static (resting) water level: 40 feet below ground.
  • When the pump runs at 300 gpm, the pumping water level is 65 feet below ground.
  • Drawdown = pumping water level - static water level.
  • Specific capacity = pumping rate / drawdown (in gallons per minute per foot).
  • 1 million gallons per day (MGD) = 694 gallons per minute (gpm).
  1. Find the drawdown when the pump runs at 300 gpm.
  2. Find the specific capacity of the well, in gallons per minute per foot.
  3. Assume drawdown stays proportional to the pumping rate. Write an equation for drawdown d (in feet) as a function of pumping rate q (in gpm), using the specific capacity from item 2.
  4. Use your equation to predict the drawdown at 480 gpm, then state the pumping water level at that rate.
  5. The pump intake sits 95 feet below ground. Operators must keep the pumping water level at least 10 feet above the intake so the pump does not draw air. What is the deepest allowed pumping water level?
  6. Using your answer to item 5, what is the maximum drawdown the operators will allow?
  7. Use your equation from item 3 to find the largest pumping rate that keeps drawdown at or below the maximum from item 6.
  8. Convert the safe pumping rate from item 7 to million gallons per day (MGD). Round to two decimals.
  9. If the well runs 18 hours per day at the safe rate from item 7, how many gallons does it deliver in one day?
  10. After several years the specific capacity falls from your item 2 value to 9 gallons per minute per foot. Explain what that change tells operators about the well, and what happens to drawdown at a given pumping rate.

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: 25 ft. Drawdown = 65 - 40 = 25 feet.
  2. Answer: 12 gpm/ft. Specific capacity = 300 gpm / 25 ft = 12 gallons per minute per foot.
  3. Answer: d = q / 12. Because each foot of drawdown yields 12 gpm, the drawdown in feet equals the pumping rate divided by 12.
  4. Answer: 40 ft. d = 480 / 12 = 40 feet of drawdown. The pumping water level is 40 (static) + 40 (drawdown) = 80 feet below ground.
  5. Answer: 85 ft below ground. The intake is 95 feet down, and the level must stay 10 feet above it: 95 - 10 = 85 feet below ground.
  6. Answer: 45 ft. Maximum drawdown = deepest level - static level = 85 - 40 = 45 feet.
  7. Answer: 540 gpm. Solve d = q / 12 for q at d = 45: q = 12 x 45 = 540 gallons per minute.
  8. Answer: 0.78 MGD. 540 gpm / 694 gpm per MGD = 0.778, which rounds to 0.78 million gallons per day.
  9. Answer: 583,200 gallons. 540 gpm x 60 minutes x 18 hours = 540 x 1,080 = 583,200 gallons in one day.
  10. Answer: The lower specific capacity means the screen or aquifer near the well is clogging or the well is aging. At any given pumping rate the drawdown is now larger, because fewer gallons per minute come out for each foot the level drops.

Differentiation

Support

Provide a labeled side-view sketch of the well showing the static level, the pumping level, the drawdown, and the intake depth as separate labels. Work item 1 and item 2 together as a class before students continue, and allow a calculator throughout.

Extension

Ask students to graph drawdown versus pumping rate from the equation d = q / 12 across the full range up to the maximum safe rate found in item 7, mark that safe point, and explain what the slope of the line represents in the well's real behavior.

ELL supports

Pre-teach "static," "pumping," and "drawdown" with a two-picture card: one well at rest and one well while pumping, with the falling level marked by an arrow. Provide a sentence frame for item 10: "A smaller specific capacity means the well ___, so drawdown gets ___."