How much flow does irrigation need?
A pump, well or water right has to move water fast enough to replace what the crop is using, spread across however many acres it irrigates at once. That rate is normally expressed in gallons per minute (gpm), and dividing by the acres gives gpm per acre, a figure that lets you compare systems and fields of different sizes on the same basis.
The flow needed depends on four things: how fast the crop is using water at its peak (peak ET, in inches or millimeters per day), how many acres that flow has to cover, how efficiently the irrigation system applies water, and how many hours a day the system actually runs. Kansas State Research and Extension publishes a standard conversion for this: about 452.57 gpm delivers one acre-inch of water in one hour — often rounded to 453 — which is the constant behind every flow calculation on this page.
Size for the peak, not the average. A system sized only for the season's average water use will fall behind during the hottest, driest stretch, when demand is highest and stored soil water runs out fastest.
If you already know your flow and want to check what it delivers instead of finding a target flow, switch the calculator above to Check my flow.
How this gpm per acre calculator works
The calculator has two modes that use the same underlying relationship between flow, area, time and depth of water.
1. Required flow (forward mode)
Flow needed per acre
GPM per acre = (Peak ET × 452.57) ÷ (Efficiency × Pumping hours per day)
Peak ET is in inches per day; efficiency is entered as a percent and used as a fraction (80% = 0.80). 452.57 is the gpm needed to deliver one acre-inch of water in one hour.
Total flow
Total GPM = GPM per acre × Acres
Gross depth pumped
Gross depth per day = Peak ET ÷ Efficiency
This is the depth the pump has to put out, including the share lost before it reaches the crop.
2. Checking an existing flow (reverse mode)
Gross depth delivered
Gross depth per day = (Flow ÷ Acres) × Pumping hours ÷ 452.57
Net depth reaching the crop
Net depth per day = Gross depth per day × Efficiency
Shown only when you enter an efficiency; without it, the calculator shows the gross depth only.
Acres this flow can fully cover
Acres covered = (Flow × Efficiency × Pumping hours) ÷ (452.57 × Peak ET)
Shown only when you enter both an efficiency and a peak ET.
Metric units
In metric, the calculator works the same way in liters per second per hectare (L/s/ha), using the fact that one millimeter of water over one hectare is 10 cubic meters (10,000 liters):
Metric flow per hectare
L/s per ha = (Peak ET, mm/day × 2.7778) ÷ (Efficiency × Pumping hours)
2.7778 is 10,000 ÷ 3,600, converting liters per hectare per day into liters per second.
Useful conversions
| Quantity | Value |
|---|---|
| 1 acre-inch | 27,154.29 gallons |
| 1 acre-inch delivered in 1 hour | about 452.57 gpm |
| 1 acre-foot | 325,851 gallons (12 acre-inches) |
| 1 mm over 1 hectare | 10,000 liters (10 m³) |
| 1 US gpm | 3.785 L/min (0.0631 L/s) |
| 1 acre | 0.4047 hectare |
For a one-off volume conversion instead of a flow rate, use the acre-inches calculator or the acre-feet calculator.
When this calculator isn't the right tool
It finds the flow rate a system needs, or what an existing flow delivers. It doesn't cover:
- Center pivot design specifics such as depth applied per revolution, travel speed or nozzle packages. See the center pivot calculator.
- Pump and pipe sizing, which also depend on pressure, elevation change and friction loss. See the mainline size calculator.
- Working out peak ET itself from weather data. See the reference ET calculator and crop evapotranspiration calculator.
- Well yield or aquifer capacity. See the irrigation well capacity calculator.
GPM per acre calculation example
A center pivot irrigates 130 acres of corn. The peak ET for the area is 0.30 inch per day, the pivot's application efficiency is 85%, and it runs 20 hours a day, leaving time for a daily maintenance check.
- Flow needed per acre: (0.30 × 452.57) ÷ (0.85 × 20) = 135.77 ÷ 17 = about 7.99 gpm/acre.
- Total flow: 7.99 × 130 ≈ about 1,038 gpm.
- Gross depth pumped: 0.30 ÷ 0.85 = 0.35 in/day.
- Volume per day: 0.35 × 130 × 27,154.29 ≈ about 1,245,900 gallons per day.
The same calculation in metric
A 50-hectare field has a peak ET of 8 mm per day, sprinkler efficiency of 80%, and pumps 22 hours a day:
- Flow needed per hectare: (8 × 2.7778) ÷ (0.80 × 22) = 22.22 ÷ 17.6 = about 1.3 L/s/ha.
- Total flow: 1.3 × 50 ≈ about 63 L/s.
- Gross depth pumped: 8 ÷ 0.80 = 10.0 mm/day.
- Volume per day: 10 mm × 50 ha × 10,000 = 5,000,000 liters (5,000 m³) per day.
Checking an existing flow
The same 130-acre pivot actually has a well rated at 1,000 gpm. Switching to Check my flow with 130 acres, 20 hours a day and 85% efficiency: gross depth pumped is (1,000 ÷ 130) × 20 ÷ 452.57 ≈ 0.34 in/day, and net depth reaching the crop is 0.34 × 0.85 ≈ 0.29 in/day — short of the 0.30 in/day peak ET in the first example, so this well can't quite keep up on the very hottest days.
Using the result
If the required flow is more than your source can deliver
A shortfall between required and available flow generally means one of four adjustments: irrigate fewer acres in each set or pumping cycle, add pumping hours where the water right and power rates allow it, improve application efficiency so less water is wasted, or accept a planned deficit and rely on stored soil water and rainfall to cover part of the gap. The available water capacity calculator shows how many days of stored soil water a field has in reserve.
Measuring your actual flow and efficiency
Flow can be read from a flow meter, taken from the pump curve at your system's actual operating pressure, or measured directly with a bucket test on a single outlet. Application efficiency is harder to measure directly; a catch-can or distribution uniformity test gives a good working estimate for sprinkler systems.
See the catch-can test calculator and distribution uniformity calculator for that measurement, and the irrigation efficiency calculator to turn it into an efficiency figure for this calculator.
Typical application efficiency by system
These are potential efficiencies for well-designed, well-maintained systems, published by University of Nebraska-Lincoln and University of Wyoming Extension. An older or poorly maintained system of the same type will run below its potential.
| System | Potential efficiency |
|---|---|
| Subsurface drip | above 95% |
| Surface drip / micro-spray | 85–95% |
| Low energy precision application (LEPA) | 80–90% |
| Center pivot / linear move | 75–85% |
| Solid set / hand move / side roll sprinkler | 65–85% |
| Traveling gun | 65–75% |
| Furrow with tailwater reuse | 60–80% |
| Conventional furrow or basin | 45–65% |
What changes the required flow
The four inputs above set the number, and these are the main reasons each one moves.
| Factor | Effect on required flow | What to adjust |
|---|---|---|
| Peak ET | Higher crop water use on hot, dry, windy days raises the flow needed directly. | Use a peak-week or peak-day figure, not a seasonal average. |
| Application efficiency | A less efficient system has to pump proportionally more to deliver the same amount to the crop. | Measure it rather than assume the top of a typical range; maintain equipment to keep it there. |
| Pumping hours per day | Fewer hours available means each hour has to deliver more, raising the flow rate needed. | Pump the maximum hours your water right, power rates and labor allow during peak demand. |
| Acres irrigated at once | More acres on one flow need a proportionally higher total flow at the same rate per acre. | Split large areas into smaller sets if the source can't cover them all at once. |
| Crop growth stage | Peak ET usually occurs during rapid vegetative growth or reproductive stages, not at planting or maturity. | Check whether your peak ET figure reflects the crop's most water-demanding stage. |
| Conveyance losses | Water lost in unlined canals or leaking pipe between the source and the field isn't counted in application efficiency. | Add a separate allowance if water travels far before reaching the field. |
If several fields draw from the same source at different times, the source only has to meet the flow for whichever field is irrigating at that moment, not the sum of every field's peak flow.
Frequently asked questions
How many gpm per acre do I need for irrigation?
It depends on peak crop water use, your system's application efficiency and how many hours a day you can pump. As one calculated example, a crop using 0.30 inch per day at 85% efficiency needs about 7.99 gpm per acre when pumped 20 hours a day, but about 9.98 gpm per acre at the same efficiency pumped only 16 hours a day. Lower efficiency, fewer pumping hours or higher peak ET all raise the number, so there is no single correct figure without your own values.
How do I convert gpm per acre to inches of water per day?
Multiply gpm per acre by the hours pumped per day, then divide by 452.57, the gpm it takes to deliver one acre-inch per hour. That gives the gross depth pumped; multiply by your application efficiency to get the depth that actually reaches the crop. The calculator's second mode does this conversion automatically from your flow, acres, hours and efficiency.
How many gallons are in an acre-inch?
One acre-inch is the volume that covers one acre to a depth of 1 inch: 43,560 square feet times 144 square inches per square foot times 1 inch, divided by 231 cubic inches per US gallon, which comes to 27,154.29 gallons. Delivering one acre-inch in one hour takes about 452.57 gpm, a constant used throughout this calculator.
For acre-inch and acre-foot conversions on their own, see the acre-inches calculator and the acre-feet calculator.
How much does application efficiency change the flow I need?
Required flow is inversely proportional to efficiency, so the effect is larger than it sounds. Dropping from 85% to 65% efficiency raises the required flow by about 31%, and dropping from 85% to 50% raises it by about 70%. Because of this, potential application efficiency ranges from University of Nebraska-Lincoln and University of Wyoming Extension are worth checking against your actual system rather than assuming the top of the range.
To estimate your own system's efficiency from field measurements, see the irrigation efficiency calculator.
Should I size pumping capacity for peak ET or average ET?
This calculator works from peak ET, the highest-demand stretch of the season, because a system sized only for the seasonal average falls behind during the hottest, driest days. Stored soil water can cover short gaps between irrigations, but it can't make up a sustained shortfall when demand stays above supply. If your source can't reach the peak figure, the usual options are irrigating fewer acres at once, adding pumping hours, or improving application efficiency.
Soil water storage between irrigations is covered in the available water capacity calculator.
How do I find my peak crop water use (ET)?
Peak ET is usually reported by a state or regional agricultural weather network, or can be worked out from reference ET and a crop coefficient for the crop's growth stage. It is normally highest during the hottest, sunniest, driest weeks of the growing season and lower earlier and later. Use your local network's peak-week or peak-day figure rather than a seasonal average, since this calculator is built around the peak.
To calculate ET from weather data and a crop coefficient, see the reference ET calculator and the crop evapotranspiration calculator.
My well or pump can't deliver the flow the calculator shows. What can I do?
Switch to “Check my flow” and enter your actual flow, acres, pumping hours and efficiency. It shows the net depth per day that flow delivers to the crop and, if you also enter a peak ET, how many acres that flow can fully cover at that ET. From there, the usual choices are irrigating fewer acres per pumping cycle, adding pumping hours where allowed, or improving efficiency. It's also worth confirming the well's flow under sustained pumping, since output can fall over the season.
See the irrigation well capacity calculator for well-side sizing.
Does this calculator cover center pivot design or pump sizing?
It covers the flow a center pivot or any other system needs to keep up with peak ET across its acreage, but not pivot-specific details such as depth per revolution or nozzle packages, and not pump sizing, which also depends on pressure, lift and pipe friction losses. Use the required flow from this calculator as an input to those calculations.
For pivot-specific figures, see the center pivot calculator. For pipe sizing at this flow, see the mainline size calculator.
Who built and checked this calculator
Licensed Professional Engineer and Irrigation Association Certified Irrigation Designer with more than 20 years in irrigation design, system auditing and landscape water management. Full profile
How this page was checked
- Formula, unit conversions and both worked examples verified by hand against the calculator's code.
- Efficiency ranges checked against the extension source below.
- Constants cross-checked against Kansas State Research and Extension's published gpm-per-acre delivery rate table.
Sources
- Danny H. Rogers and Mahub Alam, Kansas State Research and Extension. Irrigation Formulas and Conversions (MF2404).
- Daran R. Rudnick and Suat Irmak, University of Nebraska-Lincoln, edited by Milt Geiger, University of Wyoming Extension. System Performance and Efficiency (B-1264, E3A-IE.4).
- USDA Natural Resources Conservation Service. National Engineering Handbook, Part 623, Irrigation [VERIFY URL].
- Irrigation Association. Irrigation.org.