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Math problem set Grade 12 Math

Yarrowdale Center Pivot: Irrigation Depth, Flow, and Run Hours

A grade 12 problem set sizing a fictional farm's center pivot, converting crop water need in inches to gallons, then to gpm, MGD, and cfs, and testing a well's capacity.

Objectives

  1. Students will convert an irrigation depth in inches over a known acreage into gallons.
  2. Students will express one irrigation flow rate in gallons per minute, million gallons per day, and cubic feet per second.
  3. Students will solve an equation for pumping hours and judge whether the result fits the available schedule.

Vocabulary

Evapotranspiration
Evapotranspiration is the total amount of water that leaves the land and goes into the air, both from wet surfaces and from plants. Scientists combine evaporation and transpiration into one word.

Also: center pivot, application efficiency, net depth, gross depth, acre-inch

Teacher overview

A center pivot waters a circle, and the farm manager sizes it by depth rather than by volume: crops need a certain number of inches per week, and the pump has to deliver that depth across every acre the machine reaches. This set works through one July rotation at the fictional Yarrowdale Farm, from the crop's daily water use to the flow the pivot needs and the flow the farm's well can actually supply. The common error at this grade is multiplying by the application efficiency instead of dividing by it, which makes the gross depth smaller than the net depth the crop requires. Watch also for students who treat the pivot's quarter-mile length as a diameter rather than a radius.

Yarrowdale Center Pivot: Irrigation Depth, Flow, and Run Hours — Grade 12 Math
Name: ____________________________ Date: ______________

Problems

All values below describe the fictional Yarrowdale Farm and its single center pivot.

The pivot arm is 1,320 feet long and waters a full circle. In July the corn uses 0.28 inches of water per day through evapotranspiration. The farm irrigates on a 7-day rotation and can run the pivot up to 22 hours per day. The pivot's application efficiency is 85 percent, meaning only 85 percent of the water pumped reaches the crop root zone. Use 3.1416 for pi, 43,560 square feet per acre, 325,851 gallons per acre-foot, 694 gallons per minute per MGD, and 448.8 gallons per minute per cfs.

  1. Find the area the pivot waters, in acres. Round to the nearest tenth of an acre.

  2. Find the net depth of water the corn needs over one 7-day rotation, in inches.

  3. Find the gross depth Yarrowdale must pump to deliver that net depth at 85 percent application efficiency. Round to the nearest hundredth of an inch.

  4. One acre-foot is 325,851 gallons. How many gallons are in one acre-inch? Round to the nearest gallon.

  5. Using your answers to problems 1, 3, and 4, find the total gallons the pivot must pump in one rotation. Round to the nearest thousand gallons.

  6. The pivot has 7 days and can run 22 hours per day. Find the flow rate in gallons per minute the pivot must deliver, to the nearest whole gpm.

  7. Express your problem 6 flow rate in million gallons per day. Round to the nearest hundredth.

  8. Express your problem 6 flow rate in cubic feet per second. Round to the nearest hundredth.

  9. Yarrowdale's well delivers 800 gallons per minute. Write an equation in one variable for the hours h the well must run to deliver the problem 5 volume, then solve it. State whether the rotation is possible within the 7-day window at 22 hours per day.

  10. Suppose the farm runs the 800 gpm well for the full 22 hours a day across all 7 days. Find the net depth in inches the crop would actually receive, 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: 125.7 acres.
    Area = pi x radius squared = 3.1416 x 1,320 x 1,320 = 3.1416 x 1,742,400 = 5,473,924 square feet. In acres: 5,473,924 / 43,560 = 125.66, which rounds to 125.7 acres. The 1,320-foot arm is the radius, so the watered circle is half a mile across.

  2. Answer: 1.96 inches.
    0.28 inches per day x 7 days = 1.96 inches of net water need for the rotation.

  3. Answer: 2.31 inches.
    Gross depth = net depth / efficiency = 1.96 / 0.85 = 2.3059, which rounds to 2.31 inches. The gross figure must be larger than the net figure, since some pumped water never reaches the root zone.

  4. Answer: 27,154 gal.
    One acre-foot is 12 acre-inches, so 325,851 / 12 = 27,154.25, which rounds to 27,154 gallons per acre-inch.

  5. Answer: 7,885,000 gal.
    Acre-inches applied = 125.7 acres x 2.31 inches = 290.367 acre-inches. In gallons: 290.367 x 27,154 = 7,884,626, which rounds to 7,885,000 gallons for the rotation.

  6. Answer: 853 gpm.
    Available minutes = 7 days x 22 hours x 60 minutes = 9,240 minutes. Flow = 7,885,000 / 9,240 = 853.4, which rounds to 853 gallons per minute.

  7. Answer: 1.23 MGD.
    853 gpm / 694 gpm per MGD = 1.229, which rounds to 1.23 million gallons per day.

  8. Answer: 1.90 cfs.
    853 gpm / 448.8 gpm per cfs = 1.9006, which rounds to 1.90 cubic feet per second. Reporting the same flow three ways is routine, since irrigation districts, drinking water plants, and stream gauges each favor a different unit.

  9. Answer: 164.3 hours.
    Equation: 800 gallons per minute x 60 minutes per hour x h = 7,885,000 gallons, so 48,000h = 7,885,000 and h = 164.27, or about 164.3 hours. The rotation window allows only 7 x 22 = 154 hours, so the well cannot deliver the full requirement. Spread over 7 days, 164.3 hours is 23.5 hours per day, which exceeds the 22-hour daily limit.

  10. Answer: 1.84 inches.
    Volume available = 800 gpm x 60 minutes x 154 hours = 7,392,000 gallons. Acre-inches = 7,392,000 / 27,154 = 272.2 acre-inches, spread over 125.7 acres gives 272.2 / 125.7 = 2.17 inches gross. Net depth = 2.17 x 0.85 = 1.84 inches, which is 0.26 inches per day against a crop need of 0.28. The corn falls behind by about 0.12 inches per rotation unless rain makes up the difference.

Differentiation

Support

Give students the fictional pivot's area of 125.7 acres already computed, along with a conversion card listing acre-inch, MGD, and cfs factors, so the work centers on the depth-to-volume chain in problems 3 through 6. Provide problem 3 as a fill-in equation with 1.96 and 0.85 already placed.

Extension

Ask students to find how many hours per day the fictional farm's 800 gallon-per-minute well would have to run to meet the crop's daily need, and whether any rotation length at 22 hours a day could close the gap. Have them explain why changing the rotation length changes the volume per rotation but not the required flow, since the daily need and the daily run hours stay the same.

ELL supports

Pre-teach "net," "gross," "efficiency," and "rotation" with a simple diagram showing water pumped, water lost to wind and evaporation, and water reaching the roots. Provide a units chart with three columns labeled inches, gallons, and gallons per minute so students can place each answer in the right column as they go.

Extensions

Have students compare the fictional Yarrowdale pivot with a hypothetical second field of the same acreage watered by furrows at 60 percent application efficiency. Students calculate the gross depth and total gallons the furrow field would need for the same 1.96-inch net depth, then express the difference as a percent of the pivot's demand. The comparison gives a concrete number behind the choice between irrigation methods, which farms weigh against equipment cost and field shape.