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Lesson plan Grade 9 Science

Cobble Flats Irrigation: Water, Energy, and the Cost of Efficiency

A 55-minute grade 9 lesson comparing furrow and center-pivot irrigation on a fictional farm, testing whether the water and energy savings pay for the new system.

Objectives

  1. Students will convert a crop water requirement into the gross depth of water an irrigation system must apply at a stated efficiency.
  2. Students will calculate the water and pumping energy saved by moving from furrow irrigation to a center pivot.
  3. Students will compute a benefit-cost ratio for an irrigation upgrade and interpret what a ratio near one means for a farm.

Vocabulary

Acre-foot
The amount of water it takes to cover one acre of land, about the size of a football field, one foot deep. Western states use acre-feet to measure large amounts of water.
Furrow irrigation
Watering crops by letting water run down small ditches plowed between the rows of plants. It is a simple, old method, but a lot of water can soak away or run off.

Also: application efficiency, center pivot, benefit-cost ratio

Teacher overview

A crop needs a certain depth of water at its roots during a season, and an irrigation system has to deliver more than that because some water evaporates, runs off the end of the field, or sinks below the root zone. The ratio of what reaches the roots to what leaves the pump is called application efficiency, and it drives both the water bill and the electricity bill. This lesson follows the fictional Cobble Flats farm in the Dunlin Valley as it decides whether to replace furrow irrigation with a center pivot. Students often assume that a more efficient system means the crop needs less water, when in fact the crop's requirement does not change at all; only the amount that must be pumped changes. A second misconception the lesson corrects is that any water saving is automatically worth the equipment, which the closing benefit-cost ratio tests directly.

Procedure

Hook (5 min): Ask students how much water a farmer has to pump to put 1 inch of water on a field. Take two or three answers, then explain that the answer depends on how much of the pumped water reaches the roots. Write the words "application efficiency" on the board and leave the question open.

Direct instruction with a worked example (15 min): Define gross applied depth as the crop requirement divided by the application efficiency. Work a drip example for the fictional Cobble Flats farm on the board. A crop needs 24 inches of water and a drip system runs at 90 percent efficiency, so the gross depth is 24 divided by 0.90, which equals 26.7 inches. On a 120-acre field that is 26.7 times 120, or 3,204 acre-inches, and dividing by 12 gives about 267 acre-feet for the season. Point out that the crop still received exactly 24 inches; the efficiency changed only the pumping. Introduce the two costs the farm pays on every acre-foot: the district water charge and the electricity to lift and pressurize the water.

Guided practice (15 min): Distribute the handout and work item 1 together, showing 24 divided by 0.60 on the board. Then have pairs complete items 2 through 5 while you circulate. Watch for the most common error, which is multiplying by the efficiency instead of dividing by it; a student who reports 14.4 inches for item 1 has made exactly that error.

Independent practice (15 min): Students complete items 6 through 10 on their own. These items move from volumes to energy and then to money, so remind students to carry their acre-foot answers forward rather than restarting.

Closure and formative check (5 min): On an index card, each student answers this question: a field needs 18 inches of water and the system applies it at 75 percent efficiency, so what gross depth must the system deliver? The answer is 18 divided by 0.75, which equals 24 inches. Collect the cards at the door. If more than about a quarter of the class multiplied instead of divided, reteach the efficiency relationship with a second worked example before assigning any related homework.

Cobble Flats Irrigation: Water, Energy, and the Cost of Efficiency — Grade 9 Science
Name: ____________________________ Date: ______________

Student Handout

Suppose the fictional Cobble Flats farm irrigates one 120-acre field. The crop needs 24 inches of water at the roots for the season. The farm's furrow system applies water at 60 percent efficiency. A center pivot would apply it at 80 percent efficiency. Pumping and pressurizing water on this farm takes 145 kWh per acre-foot. Electricity costs $0.11 per kWh, and irrigation district water costs $32 per acre-foot. The pivot system would cost $72,000 and last 15 years. There are 12 acre-inches in 1 acre-foot.

  1. Find the gross depth of water the furrow system must apply, in inches.
  2. Find the gross depth of water the center pivot must apply, in inches.
  3. Using your answer from item 1 and the 120-acre field, find the season's furrow water volume in acre-feet.
  4. Using your answer from item 2 and the 120-acre field, find the season's pivot water volume in acre-feet.
  5. How many acre-feet would the farm save in a season by switching to the pivot?
  6. Pumping takes 145 kWh per acre-foot. Using the water saved in item 5, find the electricity saved in a season, in kWh.
  7. District water costs $32 per acre-foot. Using the water saved in item 5, find the yearly water charge the farm avoids.
  8. Electricity costs $0.11 per kWh. Using the electricity saved in item 6, find the yearly electricity charge the farm avoids.
  9. The pivot costs $72,000 and lasts 15 years. Find the annualized cost of the system, meaning the purchase price spread evenly across its life.
  10. Add your answers from items 7 and 8 to get the yearly benefit, then divide that benefit by the annualized cost from item 9 to find the benefit-cost ratio. Round to the nearest hundredth.

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: 40 inches. 24 inches divided by 0.60 equals 40 inches of gross applied depth.
  2. Answer: 30 inches. 24 inches divided by 0.80 equals 30 inches.
  3. Answer: 400 acre-feet. 40 inches times 120 acres equals 4,800 acre-inches, and 4,800 divided by 12 equals 400 acre-feet.
  4. Answer: 300 acre-feet. 30 inches times 120 acres equals 3,600 acre-inches, and 3,600 divided by 12 equals 300 acre-feet.
  5. Answer: 100 acre-feet. 400 acre-feet minus 300 acre-feet equals 100 acre-feet saved.
  6. Answer: 14,500 kWh. 100 acre-feet times 145 kWh per acre-foot equals 14,500 kWh.
  7. Answer: $3,200. 100 acre-feet times $32 per acre-foot equals $3,200 in avoided water charges.
  8. Answer: $1,595. 14,500 kWh times $0.11 per kWh equals $1,595 in avoided electricity charges.
  9. Answer: $4,800 per year. $72,000 divided by 15 years equals $4,800 per year.
  10. Answer: 1.00. The yearly benefit is $3,200 plus $1,595, which equals $4,795, and $4,795 divided by $4,800 equals 0.999, which rounds to 1.00. A ratio at about 1 means the water and energy savings alone just cover the equipment, so the farm's decision turns on the other effects, such as labor, yield, and whether water becomes scarcer or more expensive later.

Differentiation

Support

Provide the acre-inch to acre-foot conversion worked out for one example and a labeled diagram contrasting water applied at the pump with water reaching the roots. Give students items 1 through 5 only, with the efficiency values already written as decimals, and pair them with a partner for items 3 and 4.

Extension

Have students find the electricity price at which the benefit-cost ratio would reach 1.25, holding every other value from the handout constant, and then explain what that result says about farms in places where power is expensive.

ELL supports

Pre-teach efficiency, gross, applied, and annualized with a diagram showing 100 units of water leaving a pump and only 60 arriving at the roots. Provide a sentence frame for item 10: "The ratio is ___, which means the savings are ___ than the yearly cost of the system."

Extensions

Have students repeat items 3 through 10 for the drip system from the board example, which applies water at 90 percent efficiency and would cost $138,000 over 15 years, and then write a short recommendation to the fictional farm comparing all three systems.