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Math problem set Grade 9 Math

Verwood Irrigation District: A Crop Water Budget Problem Set

A grade 9 problem set converting inches of applied water into acre-feet, gallons, and MGD to size a fictional irrigation district's seasonal water budget.

Objectives

  1. Students will convert an applied-water depth over an area into acre-feet and gallons.
  2. Students will convert a seasonal water volume into an average daily rate in MGD and gpm.

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.
Irrigation efficiency
The share of irrigation water that crops actually use compared with the amount taken from the river or well. Higher efficiency means less waste.
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: applied depth

Teacher overview

The Verwood Irrigation District (a fictional district used for this set) delivers water to farm fields by the acre-foot, the volume that covers one acre to a depth of one foot. Planners start from the depth of water a crop needs, spread over the acres being farmed, and then convert that volume into gallons and into an average daily flow. This set builds that chain one step at a time. A common misconception is forgetting that applied depth is often given in inches while area math uses feet, so students must convert inches to feet before multiplying by acreage.

Verwood Irrigation District: A Crop Water Budget Problem Set — Grade 9 Math
Name: ____________________________ Date: ______________

Problems

Use these facts about the Verwood district (a fictional irrigation district) for every problem. Round to the nearest whole unit unless told otherwise.

  • Area farmed: 640 acres.
  • Water the crop needs this season (applied depth): 24 inches.
  • Irrigation season length: 180 days.
  • 1 acre-foot = 325,851 gallons.
  • 1 MGD (million gallons per day) = 694 gpm (gallons per minute).
  1. Convert the 24-inch applied depth to feet, then find the total seasonal volume in acre-feet (area x depth).
  2. Convert the total seasonal volume from item 1 to gallons.
  3. Find the average delivery in gallons per day over the 180-day season.
  4. Convert the average daily delivery from item 3 to million gallons per day (MGD). Round to two decimals.
  5. Convert the MGD value from item 4 to gallons per minute (gpm).
  6. The district's canal can deliver at most 1,800 gpm. Does the required average rate from item 5 fit within capacity, and by how much?
  7. Write an equation for total seasonal volume V (in acre-feet) as a function of applied depth d (in feet) for the 640-acre district. Then find V if the depth is raised to 2.5 feet.
  8. Only 75 percent of delivered water reaches the crop roots; the rest is lost to evaporation and seepage. To get 1,280 acre-feet to the roots, how many acre-feet must the district actually supply?
  9. Convert the water lost in item 8 (the supplied acre-feet minus 1,280) to gallons.
  10. The district considers lining the canal to raise efficiency from 75 percent to 85 percent. Using the numbers, explain why an efficiency gain saves water even though the crop still needs the same 1,280 acre-feet at its roots.

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: 1,280 acre-feet. 24 inches / 12 = 2.0 feet; 640 acres x 2.0 feet = 1,280 acre-feet.
  2. Answer: 417,089,280 gallons. 1,280 acre-feet x 325,851 gallons per acre-foot = 417,089,280 gallons.
  3. Answer: 2,317,163 gallons per day. 417,089,280 gallons / 180 days = 2,317,162.7, which rounds to 2,317,163 gallons per day.
  4. Answer: 2.32 MGD. 2,317,163 gallons per day is 2.32 million gallons per day (2,317,163 / 1,000,000).
  5. Answer: 1,610 gpm. 2.32 MGD x 694 gpm per MGD = 1,610.08, about 1,610 gallons per minute.
  6. Answer: Yes, by about 190 gpm. The canal delivers up to 1,800 gpm, and the required average is about 1,610 gpm, leaving 1,800 - 1,610 = 190 gpm to spare.
  7. Answer: 1,600 acre-feet. The equation is V = 640d. At d = 2.5 feet, V = 640 x 2.5 = 1,600 acre-feet.
  8. Answer: 1,707 acre-feet. Supplied volume = 1,280 / 0.75 = 1,706.7, which rounds to 1,707 acre-feet.
  9. Answer: 139,138,377 gallons. Lost volume = 1,707 - 1,280 = 427 acre-feet; 427 x 325,851 = 139,138,377 gallons.
  10. Answer: Raising efficiency to 85 percent means the district only needs 1,280 / 0.85 = 1,505.9 acre-feet, about 1,506 acre-feet, instead of 1,707. That saves roughly 201 acre-feet of supplied water while the crop still receives the same 1,280 acre-feet at its roots, because less water is lost on the way.

Differentiation

Support

Provide the inches-to-feet conversion (24 inches = 2.0 feet) already done, and give a printed conversion strip listing "1 acre-foot = 325,851 gallons" and "1 MGD = 694 gpm." Let students keep a running units column beside each answer.

Extension

Ask students to find what applied depth, in inches, the district could supply if the canal ran at its full capacity from item 6 for the whole 180-day season, working backward from gpm to acre-feet to depth over 640 acres.

ELL supports

Provide a labeled picture of one acre-foot (a one-acre square filled one foot deep) and a small unit ladder showing inches, feet, acre-feet, and gallons. Pre-teach "applied depth," "acre-foot," and "efficiency" with short definitions and a matching sketch.