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Lesson plan Grade 11 Engineering & Design

Sundale Farm Irrigation: Comparing Sprinkler and Drip Efficiency

A grade 11 engineering lesson where students compare sprinkler and drip irrigation efficiency for a drought-hit farm and compute the pumped flow each system needs.

Objectives

  1. Students will calculate the pumped flow an irrigation system needs from its delivered flow and efficiency.
  2. Students will recommend an irrigation method by weighing water savings against cost and reliability trade-offs.

Vocabulary

Drip irrigation
Watering plants through tubes that release water slowly, drop by drop, right at the roots. Very little water is wasted along the way.
Sprinkler irrigation
Watering crops by spraying water through the air, the way a lawn sprinkler works but on a much larger scale.

Also: application efficiency, pumped flow

Teacher overview

Sundale Farm, in the fictional Redwillow Valley, wants to cut its irrigation water use during a drought by switching part of a field from sprinklers to drip lines. This lesson has students treat application efficiency as an engineering quantity: the fraction of pumped water that actually reaches the crop rather than being lost to wind, evaporation, or runoff. A common misconception in this math is multiplying the delivered water by the efficiency to find the pumped amount; because some water is lost, the pump must supply more than the crop receives, so the delivered flow is divided by the efficiency, not multiplied.

Procedure

Hook (5 min): Ask students to estimate what share of sprinkler water reaches the plant roots on a hot, windy day. Then give them the scale of the question: the Food and Agriculture Organization puts agriculture at roughly 70 percent of global freshwater withdrawals, so a small gain in irrigation efficiency adds up to a large volume of water saved.

Direct instruction (15 min): Define application efficiency as delivered water divided by pumped water. Explain why sprinklers lose water to wind and evaporation while buried drip lines deliver water straight to the root zone. Introduce the fictional Sundale Farm scenario and the idea that the pump must supply more than the crop needs.

Guided practice (15 min): Model a worked example that is not on the handout: to deliver 600 gallons per minute to a crop with a system that is 75 percent efficient, the pump must supply 600 divided by 0.75, which is 800 gallons per minute. Show how dividing by a number below 1 gives a larger pumped flow.

Independent practice (15 min): Students complete the Student Handout using the fictional Sundale Farm numbers. Circulate and check that students divide, not multiply, by the efficiency on items 1 and 2.

Closure / exit ticket (5 min): Each student answers on a card: "For a farm facing a severe drought, would you recommend drip or sprinkler irrigation, and why?" Collect and scan for a reason grounded in the efficiency numbers before dismissal.

Sundale Farm Irrigation: Comparing Sprinkler and Drip Efficiency — Grade 11 Engineering & Design
Name: ____________________________ Date: ______________

Student Handout

For the fictional Sundale Farm, the crop root zone needs 900 gallons per minute of water delivered during irrigation. The sprinkler system is 70 percent efficient, and the drip system is 90 percent efficient. The pump runs 6 hours per day.

  1. The sprinkler system is 70 percent efficient. To deliver 900 gpm to the crop, how many gpm must the pump supply? Round to the nearest whole number.
  2. The drip system is 90 percent efficient. To deliver the same 900 gpm to the crop, how many gpm must the pump supply?
  3. How many gpm does the drip system save at the pump compared with the sprinkler system?
  4. What percent reduction in pumped water does switching to drip represent? Round to the nearest whole number.
  5. The pump runs 6 hours a day. Using your answer to item 3, how many more gallons per day does the sprinkler system pump than the drip system?
  6. Drip lines cost more to install and can clog, while sprinklers are cheaper but lose more water. Recommend one system for Sundale Farm during a severe drought and justify your choice with one trade-off.

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: 1,286 gpm. Delivered flow divided by efficiency is 900 divided by 0.70, which equals 1,285.7, or about 1,286 gallons per minute.
  2. Answer: 1,000 gpm. 900 divided by 0.90 equals 1,000 gallons per minute.
  3. Answer: 286 gpm. The sprinkler pump supplies 1,286 gpm and the drip pump supplies 1,000 gpm, so the drip system saves 1,286 minus 1,000, which is 286 gallons per minute.
  4. Answer: 22 percent. Dividing the 286 gpm saved by the sprinkler's 1,286 gpm gives 0.222, or about 22 percent.
  5. Answer: 102,960 gal/day. The 286 gpm difference times 60 minutes times 6 hours is 286 times 360, which equals 102,960 gallons per day.
  6. Answer: Accept any recommendation with a real trade-off, such as: choose drip because it cuts pumped water by about 22 percent during a drought, accepting the higher install cost and the need to prevent clogging.

Differentiation

Support

Provide the efficiency equation with the operation labeled, "delivered divided by efficiency equals pumped," and the worked guided-practice example on paper. Let students use a calculator for every step.

Extension

Ask students to find the pumped flow a system that is only 60 percent efficient would need to deliver the same 900 gallons per minute to the fictional Sundale Farm crop, and to explain why a lower efficiency raises the pumped flow.

ELL supports

Pre-teach "efficiency," "delivered," and "pumped" with a labeled diagram showing water lost to wind above a sprinkler versus water reaching roots below a drip line. Provide a sentence frame for item 6: "I recommend ___ because it ___, even though ___."