Irrigation Management of Strawberry. Dr. John R. Duval Asst. Professor of Horticulture University of Florida http://strawberry.ifas.ufl.edu. Importance of proper Irrigation. Optimization of resources Minimization of inputs Social issues. Irrigation (The Basics).
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
Dr. John R. Duval
Asst. Professor of Horticulture
University of Florida
Works well in saturated range
Easy to install and maintain
Can operate for extended periods
Can be used in an automated system
Difficult to translate data into water volume content
Requires regular maintenance
Subject to breakage
Automated system costly
Soil Type Tension (cbars)
60 cbars limit of readily available water.
-Measures soil water content, measures at any depth, high precision
-Costly, stability questionable
-Measures soil water content non- destructive, can make real time measurements
-Costly, dangerous, dependent on soil characteristics
DWU = R + I – (CF X ETo) – (D + RO)
CF=0.15 + 0.018DAT + 0.0001DAT2
Assume a sandy loam with a soil water holding capacity of 1.5 inches of water
Water use estimate
Day 1 0.28 in.
Day 2 0.37 in.
Day 3 0.15 in.
Day 4 0.30 in.
Total 1.00 in. (2/3 Soil water)
So irrigate to replace 1 inch of water on the 5th day
4 ft bed spacing (10,890 linear bed feet)
2 ft bed top
Emitter rate of 0.38 GPH
Emitter spacing of 18 in (1.5ft)
So determine total number of emitters and multiply by discharge rate
(10,890 ft / 1.5 ft) X 0.38GPH =
1 acre inch of water = 27,154 Gallons
27,154 Gal / 2759GPH = 9hrs 50 min
We only want to water the bed. Which is only 2 ft wide.
2 ft bed / 4ft bed spacing = 0.5
And therefore we water for:
9hrs 50 min X 0.5 =4hrs 55 min
Flow rate Acre-In. water GPM/Acre applied in 1hr.
Assume flow rate of 300 GPM/A
1 inch of water needed divided by the Acre inches of water put out in a hour by the system 0.66 and multiply by 1 + the % loss in the system (25%)
(1 A/in / 0.66 A/in/hr) X 1.25 = 1hr 54 min