Math problem set Grade 12 Science
Dellacre Farm Irrigation Demand: Net, Gross, and Efficiency
A grade 12 problem set on irrigation water math: students convert crop needs to acre-feet, apply system efficiency to find gross demand, and size the delivery flow.
Objectives
- Students will convert a crop's seasonal water depth over a field area into a net volume in acre-feet and gallons.
- Students will use application efficiency to convert net crop need into gross water that must be diverted.
- Students will size an average delivery flow in MGD and gpm and compare irrigation methods.
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.
- Drip irrigation
- Watering plants through tubes that release water slowly, drop by drop, right at the roots. Very little water is wasted along the way.
Also: net irrigation requirement, application efficiency, gross demand
Teacher overview
The fictional Dellacre Farm irrigates a 150-acre field from a canal, and this problem set works through how much water the farm actually has to divert to meet its crop. Students turn a seasonal crop need into a volume, then account for the water a sprinkler system loses so they can find the larger gross amount that must be delivered. A common error is multiplying the net crop need by the efficiency instead of dividing by it. Because a system loses water, the gross applied amount is always larger than the net need, so the efficiency goes in the denominator.
Problems
Use these values and conversions for the fictional Dellacre Farm:
- Field area: 150 acres
- Net crop water requirement for the season: 24 inches
- Sprinkler system application efficiency: 75 percent
- Irrigation season length: 100 days
- Conversions: 1 acre-foot = 325,851 gallons; depth in feet = inches / 12; 1 MGD = 694 gpm.
Convert the 24-inch seasonal crop requirement to a depth in feet.
Calculate the net water volume the crop needs over the 150-acre field, in acre-feet.
Convert the net requirement from problem 2 to gallons.
Calculate the gross water that must be applied, accounting for the 75 percent application efficiency (gross = net / efficiency), in acre-feet.
Calculate the extra water lost to inefficiency, in acre-feet, as the difference between the gross and net volumes.
Convert the gross applied water from problem 4 to gallons.
The gross water is delivered evenly over the 100-day season. Calculate the average delivery flow in MGD. Round to the nearest hundredth.
Express the average delivery flow from problem 7, before rounding, in gallons per minute, using 1 MGD = 694 gpm. Round to the nearest whole gpm.
The farm switches to drip irrigation at 90 percent application efficiency for the same 300 acre-feet of net need. Calculate the new gross demand in acre-feet and the water saved compared with the sprinkler system in problem 4.
Explain why raising application efficiency lowers the water a farm must divert from the canal, and name one method that raises efficiency.
From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.
Answer Key
Answer: 2.0 feet.
24 inches / 12 inches per foot = 2.0 feet.Answer: 300 acre-feet.
150 acres x 2.0 feet = 300 acre-feet of net crop need.Answer: 97,755,300 gallons.
300 acre-feet x 325,851 gallons per acre-foot = 97,755,300 gallons, about 97.8 million gallons.Answer: 400 acre-feet.
Gross = net / efficiency = 300 acre-feet / 0.75 = 400 acre-feet.Answer: 100 acre-feet.
400 acre-feet gross minus 300 acre-feet net = 100 acre-feet lost to inefficiency.Answer: 130,340,400 gallons.
400 acre-feet x 325,851 gallons per acre-foot = 130,340,400 gallons, about 130.3 million gallons.Answer: 1.30 MGD.
130,340,400 gallons / 100 days = 1,303,404 gallons per day, which is 1.30 MGD.Answer: 905 gpm.
1.3034 MGD x 694 gpm per MGD = 904.6, which rounds to 905 gpm.Answer: 333.33 acre-feet gross, saving about 66.67 acre-feet.
Gross for drip = 300 acre-feet / 0.90 = 333.33 acre-feet. Compared with the 400 acre-feet the sprinkler system needed, the drip system saves 400 - 333.33 = 66.67 acre-feet, about 67 acre-feet.Answer: Higher efficiency means less water is lost, so less must be diverted.
When more of the applied water reaches the crop's root zone, a smaller gross amount meets the same net need, so the farm diverts less from the canal. Drip irrigation, better scheduling with soil-moisture sensors, or lining canals to cut seepage all raise efficiency.
Differentiation
Support
Provide problem 4 with the division set up (300 / 0.75 = ___) and a printed reminder that gross is larger than net. Work problems 1 and 2 together as a class so the depth-times-area step is clear before students continue.
Extension
Ask students to find the annual water savings in gallons if the drip system in problem 9 is used every season for five years, and to weigh that savings against a reason a farm might still hesitate to install drip irrigation.
ELL supports
Pre-teach "net," "gross," and "efficiency" with a picture of water leaving a sprinkler where some reaches the crop and some is lost. Provide a sentence frame for problem 10 and allow a labeled diagram in place of a paragraph.
Extensions
Have students look up the main crop grown in their region and its typical seasonal water requirement, then repeat problems 1 through 4 with that figure to estimate the gross water a local field of the same size would need.