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Worksheet Grade 11 Math

Corriston Valley Almond Orchard: An Irrigation Water-Budget Worksheet

A grade 11 worksheet using dimensional analysis to size an irrigation water budget for a fictional almond orchard, converting acre-feet to gallons, gpm, and MGD.

Objectives

  1. Students will convert an applied water depth and field area into a seasonal volume in acre-feet and gallons.
  2. Students will express a seasonal water volume as an average flow rate in gallons per minute and million gallons per day.

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.
Flow rate
Flow rate measures how much water moves past a point in a certain amount of time, like gallons per minute through a garden hose. Bigger pipes and stronger pushes give bigger flow rates.

Also: application efficiency, dimensional analysis

Teacher overview

This worksheet follows the water an orchard uses across one growing season, from a rainfall-style depth to a steady flow rate a pump would have to deliver. The setting is the fictional Corriston Valley Irrigation District, so every figure is invented for practice. The work is pure dimensional analysis: students keep track of units at each step and convert among acre-feet, gallons, gallons per minute, and million gallons per day. A common error is dropping or mixing units partway through, which is why each item asks for a specific unit in its answer.

Corriston Valley Almond Orchard: An Irrigation Water-Budget Worksheet — Grade 11 Math
Name: ____________________________ Date: ______________

Student Worksheet

The Corriston Valley Irrigation District (a fictional district created for this worksheet) manages a 120-acre almond orchard. Over one irrigation season the trees need a net applied water depth of 42 inches spread across the whole field. Use these conversions throughout:

  • 1 acre-foot = 325,851 gallons
  • 1 day = 1,440 minutes
  • 1 million gallons = 1,000,000 gallons
  1. Convert the 42-inch net application depth to feet.
  2. Find the net seasonal volume in acre-feet using volume = area x depth (area in acres, depth in feet).
  3. Convert the net seasonal volume from item 2 to gallons.
  4. The irrigation season lasts 180 days. If the water were applied at a constant rate, what average flow in gallons per day would deliver the net volume?
  5. Convert the daily flow from item 4 to gallons per minute (gpm), rounding to the nearest whole gpm.
  6. Express the daily flow from item 4 in million gallons per day (MGD), rounded to the nearest hundredth.
  7. The drip system is only 85 percent efficient, so the district must divert more water than the trees actually use. Find the gross seasonal volume in acre-feet the district must divert (net volume divided by 0.85).
  8. The district pays a wholesale water charge of 95 dollars per acre-foot for the gross diverted water. Find the season's total water cost, using your gross volume from item 7.
  9. Explain, in two or three sentences, why the district must state every quantity in consistent units before it can compare a field's demand to a pump's delivery rate.
  10. A second orchard has the same 42-inch net depth but only 40 acres. Without repeating every step, find its net seasonal volume in acre-feet and explain how it compares to the 120-acre orchard.

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: 3.5 ft. 42 inches divided by 12 inches per foot equals 3.5 feet.
  2. Answer: 420 acre-feet. 120 acres x 3.5 feet = 420 acre-feet.
  3. Answer: 136,857,420 gallons. 420 acre-feet x 325,851 gallons per acre-foot = 136,857,420 gallons (about 136.9 million gallons).
  4. Answer: about 760,319 gallons per day. 136,857,420 gallons divided by 180 days = 760,319 gallons per day.
  5. Answer: about 528 gpm. 760,319 gallons per day divided by 1,440 minutes per day = 528 gpm.
  6. Answer: 0.76 MGD. 760,319 gallons per day is 0.760 million gallons per day, which rounds to 0.76 MGD.
  7. Answer: about 494 acre-feet. 420 acre-feet divided by 0.85 = 494.1 acre-feet of gross diverted water.
  8. Answer: about 46,941 dollars. 420 divided by 0.85 = 494.118 acre-feet (unrounded) x 95 dollars per acre-foot = 46,941 dollars for the season. Accept 46,930 to 46,940 dollars if the student uses the rounded 494 or 494.1 acre-feet from item 7.
  9. Answer: Consistent units let quantities be compared and combined without error. A field demand stated in acre-feet cannot be matched against a pump rated in gallons per minute until both are converted to a shared basis, so unit tracking prevents mistakes that would missize the pump.
  10. Answer: 140 acre-feet. 40 acres x 3.5 feet = 140 acre-feet, which is one-third of the 120-acre orchard's 420 acre-feet because the depth is the same and 40 is one-third of 120.

Differentiation

Support

Provide a printed conversion strip listing each factor as a labeled arrow rule (inches to feet, acre-feet to gallons, per day to per minute) so students can see which direction each conversion runs. Work item 1 and item 2 together as a model before students continue independently, and allow unrounded calculator values.

Extension

Ask students to find the flow in gpm the district would need if the gross diverted volume were delivered only during a 12-hour daily irrigation window rather than continuously across 24 hours, and to explain why a shorter window raises the required flow rate.

ELL supports

Pre-teach "acre-foot," "depth," "divert," and "efficiency" with a simple labeled sketch of a field holding water to a depth. Provide a sentence frame for item 9: "The district must use the same units because ..." so students can focus on the reasoning rather than the sentence structure.