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Math problem set Grade 11 Careers & CTE

Saltmarch Ridge Irrigation: Acre-Feet and Run-Time Math

A grade 11 careers problem set in which an irrigation technician converts a net application depth to acre-feet and gallons, then solves for pivot run time and water cost.

Objectives

  1. Students will convert a net irrigation depth over a stated acreage into acre-feet and gallons.
  2. Students will write and solve an equation for the pump run time a stated gross volume requires.
  3. Students will explain why an irrigation technician orders water by gross volume rather than net volume.

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.

Also: distribution uniformity, net application depth, gross application, center pivot

Teacher overview

A grower asks the irrigation technician a simple-sounding question, which is how long to run the pivot, and answering it takes a chain of conversions that has to hold together from end to end. This set follows a technician for the fictional Saltmarch Ridge Irrigation District through one pivot order, from the agronomist's requested depth in inches to a run time in hours and a bill in dollars. Distribution uniformity, the ratio of the average depth received by the least-watered quarter of the field to the average depth across the whole field, enters partway through and changes every number after it. The misconception this set corrects is treating the requested depth as the amount to pump; because no system applies water perfectly evenly, gross volume is always larger than net volume, and ordering the net amount leaves part of the field short.

Saltmarch Ridge Irrigation: Acre-Feet and Run-Time Math — Grade 11 Careers & CTE
Name: ____________________________ Date: ______________

Problems

The fictional Saltmarch Ridge Irrigation District serves growers from a canal turnout. Use these conversions throughout: one acre-foot equals 325,851 gallons, one acre-foot equals 12 acre-inches, and one million gallons per day equals 694 gallons per minute. A grower's center pivot covers 132 acres, the pivot's supply pump delivers 1,150 gallons per minute, and the district charges $46.50 per acre-foot of water delivered.

  1. The agronomist asks for a net application depth of 1.2 inches across the full 132 acres. Find the net volume in acre-feet.
  2. Convert your answer from problem 1 to gallons. Round to the nearest gallon.
  3. The pivot's distribution uniformity is 0.82, so gross volume equals net volume divided by 0.82. Find the gross volume in gallons, rounded to the nearest gallon.
  4. Let t stand for the pump run time in minutes. Write an equation setting 1,150 times t equal to your gross volume from problem 3, then solve for t. Round to the nearest minute.
  5. Convert your run time from problem 4 to hours, rounded to the nearest tenth of an hour.
  6. Convert the pump rate of 1,150 gallons per minute to million gallons per day, rounded to the nearest hundredth.
  7. Convert the gross volume from problem 3 to acre-feet, rounded to the nearest hundredth.
  8. Using your rounded answer from problem 7 and the district rate of $46.50 per acre-foot, find the grower's water charge for this irrigation. Round to the nearest cent.
  9. The grower rebuilds the pivot's sprinkler package and raises distribution uniformity to 0.90. Write and solve an equation for the new gross volume in gallons needed to deliver the same net volume from problem 2. Round to the nearest gallon.
  10. Explain, in two or three sentences, why the technician orders water by gross volume rather than net volume, and state what happens to part of the field if the grower orders the net amount instead.

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: 13.2 acre-feet.
    A depth of 1.2 inches over 132 acres equals 132 times 1.2, or 158.4 acre-inches. Dividing by 12 acre-inches per acre-foot gives 13.2 acre-feet.
  2. Answer: 4,301,233 gallons.
    13.2 acre-feet times 325,851 gallons per acre-foot equals 4,301,233.2 gallons, which rounds to 4,301,233 gallons.
  3. Answer: 5,245,406 gallons.
    Gross equals net divided by distribution uniformity: 4,301,233.2 divided by 0.82 equals 5,245,406.34 gallons, which rounds to 5,245,406 gallons.
  4. Answer: 4,561 minutes.
    The equation is 1,150t equals 5,245,406.34. Dividing both sides by 1,150 gives t equals 4,561.22 minutes, which rounds to 4,561 minutes.
  5. Answer: 76.0 hours.
    4,561.22 minutes divided by 60 minutes per hour equals 76.02 hours, which rounds to 76.0 hours.
  6. Answer: 1.66 MGD.
    1,150 gallons per minute divided by 694 gallons per minute per million gallons per day equals 1.657, which rounds to 1.66 million gallons per day.
  7. Answer: 16.10 acre-feet.
    5,245,406.34 gallons divided by 325,851 gallons per acre-foot equals 16.0976 acre-feet, which rounds to 16.10 acre-feet.
  8. Answer: $748.65.
    16.10 acre-feet times $46.50 per acre-foot equals $748.65.
  9. Answer: 4,779,148 gallons.
    The equation is 0.90G equals 4,301,233.2, where G is the new gross volume. Dividing both sides by 0.90 gives G equals 4,779,147.99, which rounds to 4,779,148 gallons. That is about 466,258 gallons less than the gross volume at a uniformity of 0.82.
  10. Answer: Because no pivot applies water evenly, the technician must pump more than the net depth so the driest part of the field still receives it.
    Distribution uniformity compares the depth reaching the least-watered part of the field with the field average, and at 0.82 roughly one gallon in six of the gross volume goes to wetter areas that did not need it. Ordering the net volume means the wettest parts of the circle get their 1.2 inches while the driest parts fall short, which shows up later as uneven yield in the same field. Ordering the gross volume brings the low spots up to the requested depth, at the cost of over-applying elsewhere until the uniformity is improved.

Differentiation

Support

Give students a printed conversion card with the three conversions used at the fictional district, and provide the gross volume from problem 3 already computed so a student who stalls there can still write and solve the run-time equation in problem 4.

Extension

Ask students to find what pump rate, in gallons per minute, the fictional grower would need in order to finish the same gross volume within a single 48-hour window, then judge whether the district's canal turnout could reasonably supply it alongside neighboring growers.

ELL supports

Pre-teach "acre-foot," "net," "gross," and "distribution uniformity" with a labeled overhead sketch of the fictional pivot circle showing wetter and drier zones. Provide the sentence frame "Gross volume is larger than net volume because ___" for problem 10.

Extensions

Have students compute the district charge for the improved uniformity in problem 9, compare it to the charge in problem 8, and estimate how many irrigations at that saving would be needed to cover a sprinkler package cost that the teacher supplies.