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Bradenton AREC Research Report GC1979-12 Augu 1 19799199
CALCULATION OF IRRIGATION WATER FOR SEEP IRRIGATED LAND FROM RISER FLOW RATES
Alexander A. Csizinszky .F.A.S. Univ. of Florida
Alexander A. Csizinszky
The determination of theoretical irrigation requirements for Florida
vegetable crops grown with the seep irrigated system was calculated by G. A.
Marlowe, Jr. and J. S. Rogers (1), and by G. A. Marlowe, Jr. and A. J. Overman
(2). Growers, however, may like to know how much water they are applying
for the crop at any specific time and at what rate, by measuring the rate of
flow through the risers at the heads of the lateral irrigation ditches. In
this paper an example is worked out for a tomato field to illustrate the method
of calculation. It must be pointed out that water flow rates, number of
irrigation days, number of plants, and the yield of tomatoes per acre may vary
from farm to farm and the examples in the calculations should not be used as
recommended water flow ratesor.plant numbers per acre. It is also assumed
that there is no fluctuation in water pressure, and that rainfall for the season
is zero. We use a 300 ft. long and 370 ft. wide land with a lateral irrigation
furrow spacing at 37 ft., i.e. 10 irrigation furrows to the land. The rate of
water flow is 1.75 gallons per minute (gpm) per riser except for the first and
last risers which have a flow rate of 2.6 gpm to compensate for the lateral
seepage loss to adjoining areas. Irrigation may be started 35 days before
setting plants in the field in order to provide adequate moisture in a zero
rainfall year for land preparation, formation of plant beds, fumigation and
laying of plastic mulch over the plant beds. The length of season from
transplanting to last picking is 110 days, which gives a total of 145 irrigation
days for the season. How many acre inches of irrigatin-water was applied to
the land during the season?
The following units are used in the calculations: 1 acre (A) = 43,560
sq ft.; 1 A inch (1 A in.) = 27,154 gals water.
1) Acre inches of irrigation water for the season 2
= flow rate of irrigation water x number of risers x total flow time x 43,560 ft /A
27,154 gal x sq ft of area irrigated
(2.6 gpm x 2 risers x 60 min 2
=[(1.75 gpm x 8 risers x 60 min x 24 hrs x 145 days) + x 24 hrs x 145 days)l x 43.560 ft
27,154 gal/A in. x 111,000 sq ft
=57.94 (58.00) A. in. of irrigation water applied for the season and this is
equal to 58.00 A. in.
145 days = 0.4 A in. per day.
Expressed in gallons: 48.00 A in. x 27, 154 gal/A in. = 1,573,300 gallons per season
or 10,850 gal per day (gpd).
To calculate the total amount of water required to raise a crop the amount
of water needed to grow the transplants and the water used for the application of
pesticides also have to be considered.
2) If the rate of irrigation is required for any particular time when the
flow rate through the risers is known, then the same method of calculation is used,
changing only the flow time in the above formula. For example, we want to know
how many acre inches of irrigation water the land received at the flow rate of
2 gpm per riser for a 1 day period.
Irrigation water in A in. for 1 day
= 2 gpm x 10 risers x 60 min x 24 hrs x 43,560 sq ft/A
27,154 gal/A in. x 111,000 sq ft
= 0.416 A in. per day
in gallons: 0.416 A in. x 27,154 gal/A in = 11,196 gal per day (gpd).
In metric units the calculations will be as follows:
ft = 0.305 meter (m)
gal = 3.785 liters (L)
acre = 4,047 m2
1 m = 10.764 sq ft 2
1 hectare (ha) = 10,000 m
1 cubic meter (m3) = 1,000 L
To transform the parameters from the U.S. system to metric
example #1 above: area of 111,000 sq ft = 10,312 m2, flow rate
6.624 liters per minute (Lpm) and 2.60 gpm = 9.841 Lpm.
of 1.75 gpm =
Liters of irrigation water per m per season :
(6.624 Lpm x 8 risers x 60 min x 24 hrs x 145 days) +
(9.841 Lpm x
x 24 hrs
2 risers x 60 min
x 145 days)
= 1,472 L of irrigation water per m2 per season.
This is equal to 1,472 L/m2 10.15 L/m2/day.
145 days15 L/m/day
For a 1 ha area the amount of irrigation water applied in m3/season:
1,472 L/m2/season x 10,000 m2
= 14,720 m3 of irrigation water per ha per
14,75 days = 101.52 m3/ha/day.
aln Europe, where the metric system is extensively used, the irrigation
water and rainfall is expressed in millimeters (mm) only. The reason is that
1 mm deep water on a 1 m? area is equal to 1 L/m2. Reporting the water in mm
per season one can calculate the volume of water applied for a particular area.
The calculation from mm to L/m2 is as follows: 1 m = 1,000 mm; 1 m2 = 1,000 mm
x 1,000 mm = 1,000,000 mm2; 1 mm deep water on 1 m2 = 1,000,000 mm2 x 1 mm =
1,000,000 mm3 = 1,000 cm3 = 1 L/m2.
x 24 hrs x 145 davs)
If we want to convert acre inches of water to liters per m2 then the
following formula can be used:
A in. applied x gal per A in. x liters per gallon
m2 per acre
In example #1 we applied 58.00 A in. of irrigation water per season.
many liters of water did we apply per m2?
58.00 x 27,154 x 3.785 L = 1,472 L/m2/season
In short form we can use the factor of 25.38 to multiply the number of
A in. to get the result in L/m2. From the above example:
58.00 A in. x 25.38 = 1,472 L/m2/season.
1. Marlowe, G. A., Jr., and J. S. Rogers.
vegetable crops. Vegetable Crop Extension
University of Florida.
Water use by Florida
No. 16-1976. IFAS,
2. Marlowe, G. A., Jr., and A. J. Overman. 1977. A rationale for the deter-
mination of irrigation needs for vegetable crops grown with seep irrigation.
Bradenton AREC Research Report GC1977-78.