Worksheet Grade 10 Engineering & Design

Cinderhill Wellfield Capacity: A Grade 10 Drawdown Worksheet

A grade 10 engineering worksheet on groundwater supply: students find well drawdown and specific capacity, combine well yields, and model yield as a function of drawdown.

Objectives

  1. Students will calculate well drawdown and specific capacity from static and pumping water levels.
  2. Students will combine well yields and write an equation relating a well's yield to its drawdown.

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, specific capacity, yield, wellfield

Teacher overview

When a pump pulls water from a well, the water level in the well drops, and how far it drops for a given yield tells an engineer how productive the well is. This worksheet has students analyze the fictional Cinderhill wellfield, finding each well's drawdown and specific capacity, combining the wells' yields, and writing an equation that links a well's yield to its drawdown. A common misconception in this math is treating a deeper pumping level as a sign of a better well; a large drawdown for a small yield actually signals a less productive well, which is why engineers compare specific capacity, the yield per foot of drawdown, rather than the pumping level alone.

Answer Key

  1. Answer: 33 feet.
    Drawdown = 78 − 45 = 33 feet.

  2. Answer: 10 gpm per foot.
    Specific capacity = 330 gpm ÷ 33 feet = 10 gpm per foot of drawdown.

  3. Answer: 45 feet.
    Drawdown = 95 − 50 = 45 feet.

  4. Answer: 8 gpm per foot.
    Specific capacity = 360 gpm ÷ 45 feet = 8 gpm per foot of drawdown.

  5. Answer: Well 1.
    Well 1's specific capacity of 10 gpm per foot is higher than Well 2's 8 gpm per foot, so Well 1 delivers more water for each foot the level drops, making it the more productive well per foot of drawdown.

  6. Answer: 690 gpm.
    Combined yield = 330 + 360 = 690 gpm.

  7. Answer: about 0.99 MGD.
    690 gpm ÷ 694 gpm/MGD = 0.994, which rounds to about 0.99 MGD.

  8. Answer: 993,600 gallons per day.
    690 gpm × 1,440 minutes/day = 993,600 gallons per day.

  9. Answer: 25 feet.
    The equation is Y = 10d. Solving 250 = 10d gives d = 250 ÷ 10 = 25 feet of drawdown.

  10. Answer: 40 feet of drawdown.
    Drawdown = 400 gpm ÷ 10 gpm per foot = 40 feet, so the new pumping water level is 45 + 40 = 85 feet.

  11. Answer: Lower static level, less water.
    Accept any answer explaining that pumping starts from a lower level, so the pump reaches its usable limit sooner and the same drawdown yields less water. Over-pumping or drought that lowers the static level shrinks the amount each well can safely deliver.

Differentiation

Support (below grade level)

Provide the two formulas (drawdown and specific capacity) on a reference card and complete Well 1's drawdown and specific capacity together, so students repeat the same two steps for Well 2 on their own.

Extension (above grade level)

Ask students to find the specific capacity a third fictional well would need to deliver 500 gpm at a drawdown of no more than 40 feet, and to explain what a low specific capacity would mean for the pump size and energy needed to reach that yield.

ELL supports

Provide a labeled well diagram showing the static water level, the pumping water level, and the drawdown between them, with yield and specific capacity defined beside it. Allow students to show problems 1 through 10 as numbered calculations and to answer problem 11 with a labeled sketch plus one sentence.