Irrigation Flow Time
Calculate irrigation duration based on flow rate, area, and water requirements. Enter values for instant results with step-by-step formulas.
Formula
Time (min) = (Area × Depth × 27,154) / (Flow Rate × Efficiency)
Where Area is in acres, Depth is desired application in inches, 27,154 converts acre-inches to gallons, Flow Rate is in GPM, and Efficiency is expressed as a decimal. The result gives minutes of run time needed to apply the desired water depth.
Worked Examples
Example 1: Center Pivot Irrigation Scheduling
Problem: A 125-acre center pivot system has a flow rate of 800 GPM with 85% efficiency. Calculate the time needed to apply 1 inch of water and the application rate.
Solution: Step 1: Calculate gross water volume\nGross acre-inches = 125 × 1 = 125 acre-inches\n\nStep 2: Adjust for efficiency\nNet acre-inches = 125 / 0.85 = 147.1 acre-inches\n\nStep 3: Convert to gallons\nTotal gallons = 147.1 × 27,154 = 3,994,555 gallons\n\nStep 4: Calculate run time\nTime = 3,994,555 / 800 GPM = 4,993 minutes = 83.2 hours\n\nStep 5: Calculate application rate\nApplication rate = (800 × 60) / (125 × 27,154) = 0.0141 in/hr\n\nStep 6: Express as rotation time\nAt 100% timer: 83.2 hours per revolution for 1\" application
Result: Run Time: 83.2 hours (3.5 days) | Application Rate: 0.014 in/hr | 3.99 million gallons
Example 2: Drip Irrigation for Orchard
Problem: A 20-acre orchard uses drip irrigation at 95% efficiency. The well produces 150 GPM. How long to apply 2 inches of water per week?
Solution: Step 1: Calculate water volume needed\nGross = 20 acres × 2 inches = 40 acre-inches\nNet (adjusted) = 40 / 0.95 = 42.1 acre-inches\nGallons = 42.1 × 27,154 = 1,143,183 gallons\n\nStep 2: Calculate irrigation time\nTime = 1,143,183 / 150 GPM = 7,621 minutes\nTime in hours = 127 hours per week\n\nStep 3: Daily irrigation schedule\nHours per day = 127 / 7 = 18.1 hours/day\n\nStep 4: Evaluate feasibility\n18 hours/day is near continuous operation\nConsider: night irrigation, split applications\n\nStep 5: Cost estimate at $0.10/1000 gal\nWeekly cost = (1,143,183 / 1000) × $0.10 = $114.32\nCost per acre-inch = $114.32 / 40 = $2.86
Result: Weekly run time: 127 hours | Daily: 18 hrs/day | Cost: $114/week | Need larger well or longer cycle
Example 3: Flood Irrigation Set Time
Problem: A farmer flood irrigates a 5-acre basin from a canal delivering 2 CFS (cubic feet per second). With 55% efficiency, how long to apply 4 inches?
Solution: Step 1: Convert flow to GPM\n2 CFS × 448.83 = 897.66 GPM\n\nStep 2: Calculate water needed\nGross = 5 × 4 = 20 acre-inches\nNet = 20 / 0.55 = 36.4 acre-inches\nGallons = 36.4 × 27,154 = 988,406 gallons\n\nStep 3: Calculate set time\nTime = 988,406 / 897.66 = 1,101 minutes\nTime = 18.3 hours\n\nStep 4: Water losses at 55% efficiency\nWater applied = 988,406 gallons\nWater used by crop = 988,406 × 0.55 = 543,623 gal\nWater lost = 444,783 gallons (45%)\n\nStep 5: Comparison with improved method\nIf switched to 85% efficiency system:\nWould need only 639,224 gallons\nSavings = 349,182 gallons per irrigation
Result: Set time: 18.3 hours | 988,406 gallons applied | 45% loss to runoff/deep percolation
Frequently Asked Questions
How do I calculate irrigation run time?
Irrigation run time is calculated by: Time = (Area × Depth × 27,154) / (Flow Rate × Efficiency). First, determine the volume of water needed (area in acres × depth in inches = acre-inches). Convert to gallons (1 acre-inch = 27,154 gallons). Divide by your system's flow rate in GPM, then adjust for application efficiency. The result is time in minutes to apply the desired depth.
What is irrigation efficiency?
Irrigation efficiency is the percentage of applied water that's actually available for plant use. Losses occur through evaporation, wind drift, runoff, and deep percolation. Typical efficiencies: Flood/furrow irrigation: 40-60%, Sprinkler systems: 70-85%, Center pivot: 80-90%, Drip/micro irrigation: 85-95%. Higher efficiency means less water wasted and lower operating costs.
What flow rate do I need for my field?
Required flow rate depends on field size, water depth, and available irrigation time. Formula: GPM needed = (Acres × Inches × 27,154) / (Hours available × 60). For example, to apply 1 inch to 10 acres in 8 hours: (10 × 1 × 27,154) / (8 × 60) = 566 GPM. Higher flow rates allow shorter set times but require larger equipment and higher energy costs.
How do I convert between irrigation flow units?
Common conversions: 1 GPM = 0.00223 CFS (cubic feet/second), 1 CFS = 448.83 GPM, 1 GPM = 0.0631 liters/second, 1 acre-inch/hour = 452.57 GPM, 1 GPM continuous for 1 hour = 60 gallons = 0.0022 acre-inches. For well capacity, flow is often measured in GPM; for surface water rights, CFS or acre-feet are common.
How do I schedule irrigation efficiently?
Efficient scheduling matches water application to crop needs: 1) Monitor soil moisture (sensors, feel, appearance), 2) Track evapotranspiration (ET) from weather data, 3) Apply water before stress occurs (typically at 50% available water depletion), 4) Irrigate during low-wind, cool periods to reduce evaporation, 5) Consider deficit irrigation during less critical growth stages.
How does center pivot irrigation work?
Center pivots rotate around a central point, covering circular fields. The outer spans travel faster, so sprinkler packages are designed to apply more water at the ends. Key specs: Rotation time (typically 12-96 hours), system capacity (GPM), and percent timer setting. Application depth = (GPM × Hours × 60) / (Acres × 27,154). 100% timer means continuous rotation at design speed.