Determining How Long to Run Drip Irrigation Systems for Vegetables
Posted: June 26, 2013
As a general rule, vegetable crops require 1 – 1.5 acre-inches of water per week. When plants are small aim for the lower end and when they are large, the upper end. Additionally, when it is excessively hot and/or windy, plants may need more than 1.5 inches of water in a week. When using a plastic mulch the entire 1 – 1.5 inches will need to be applied weekly regardless of rain events. With bare ground or when using water permeable mulches (for example, straw or paper mulch), the amount of water applied by irrigation should be reduced by the amount of rainfall the field receives. For example, if it rains ¼ inch, reduce the amount of irrigation water applied by ¼ inch for that week.
The time to run a drip system to apply 1 inch of water to the soil depends on the drip tube flow rate and width the roots extend which is generally the same as the width of most beds: about 30 inches. The table below shows how long, in hours, to run a drip system to apply 1 inch of water to the 30 inch width based on the drip tube flow rate.
|Emmitter Spacing||inches||Drip tube flow rate @ 8 psi
|(gpm/100 ft)||Time to apply 1 inch of water
If you have a drip tape with a flow rate of 0.45 gpm per 100 ft it would take 5.8 hours to apply 1 inch of water to the 30 inch beds. Typically this is applied through several weekly applications.
Want to know how the values in the table were determined?
Visualize an acre with beds on 6 ft centers.
The values needed are the:
- width of the bed, which is generally 30 inches or 2.5 feet;
- number of gallons in an acre inch of water: 27,154;
- number of square feet in an acre: 43,560 and;
- drip tube flow rate: this varies, for the example we’ll use 0.45 gpm/100 ft.
First, determine how much drip tape is needed by dividing the area in an acre by the row spacing. 43,560 ft2 ÷ 6 ft = 7,260 ft of drip tape needed.
Next, determine the area of the acre to which the water will be applied. 7,260 ft drip tape × 2.5 ft wide beds = 18,150 ft2. (0.42 acre.)
Determine the number of gallons of water needed to apply. 27,154 gal/acre-in × 0.42 acre = 11,405 gal.
Determine the number of 100 feet units of drip tape used. 7,260 ft ÷ 100 ft = 72.6 units.
Determine the number of gallons per minute needed using the drip tube flow rate. 72.6 units × 0.45 gpm = 32.67 gpm. If your well does not have this capacity, you will need to water in zones.
Lastly, determine the amount of time to run the system. 11,405 gal needed/32.67 gpm = 349 minutes or 5.8 hours.
Soil texture and available water holding capacity determine the rate at which water moves through the soil and therefore how long to run the drip system per application. To determine how long to run the drip system at one time, first find the available water holding capacity of the soil using a table such as the one below.
|Soil Texture||Available water holding capacity (inch of water/inch depth of soil)|
|Coarse sand / compacted sands||0.02-0.06|
|Fine sandy loam / compacted loam||0.14-0.18|
|Loam and silt loam||0.17-0.23|
|Clay loam and silty clay loam||0.14-0.21|
|Silty clay and clay||0.13-0.18|
For example, if you have a sandy loam, available water holding capacity is 0.11 to 0.15 inch of water/inch depth of soil. Pick a number within the range, say 0.12.
Then, using the table below and the drip tube flow rate, find the maximum time in minutes to run the drip system at one time.
For our example the available water holding capacity is 0.12 inch of water/inch depth of soil and the drip tube flow rate is 0.45 gpm per 100 ft. Using the table below the drip system would be run 110 minutes for each irrigation event, typically in a 24 hour period, to avoid leaching and runoff. Repeat events until the system is run for 5.8 hours in a week to apply 1 inch of water.
Maximum Minutes per Application
|Tubing flow rate (gpm per 100 ft)||0.2||0.3||0.4||0.45||0.5||0.6|
|Available water holding capacity (inch of water/inch depth of soil)|