Irrigation System Designing - Practices of Irrigation and On-farm Water Management - page 105

Example 3.10

A farm of 25 ha is planned to be brought under sprinkler irrigation. The textural
class of the soil is loam-to-silt loam, having moisture content at field capacity (FC)
and permanent wilting point (WP) of about 42% (by volume) and 26% (by volume),
respectively. An infiltration test data showed that constant (basic) infiltration rate is
2 mm/h. A hardpan (relatively impervious layer) exists at a depth of 2.0 m below the
soil surface. Long-term average reference evapotranspiration (ET 0 ) rate in that area
is 4.5 mm/d. Vegetable crops are planned to grow in the farm, and the crop coef-
ficient ( K c ) at maximum vegetative period is 1.1. The climate is moderately windy
in a part of the season. Design the sprinkler irrigation system (various components)
for the farm. Assume standard value of any missing data.

Solution

Given,

250,000 m 2 Area, A = 25 ha =

FC = 42% (by vol.)

WP = 26%

I c = 2 mm/h

ET 0 = 4.5 mm/d

K c = 1.1

D imp =

2 m below soil surface
Wind status: moderately windy
Now, the solution steps:

(1) ET max = 4.95 mm/d

(Assuming depletion of soil-moisture up to readily available level, so that ET
occurs at its maximum rate, i.e., soil moisture stress factor, K s =

ET 0 × K c = 4.5 × 1.1 =

1)
(2) Daily water requirement for the area, A (i.e., for whole farm here) is

A × ET max

E a

V A =

Assuming application efficiency, E a = 80%, i.e., 0.8

× /

250, 000 (4.95 1000)

1,546.875 m 3

Then, V A = =

0.8

V A
t

(3) Discharge rate, Q t =

Here, t = irrigation period = 4 h (assuming for the prevailing windy condition)

0.1074 m 3 /s
(4) Discharge rate of individual sprinkler, q

Thus, Q t = 1,546.875/(4 × 3,600) =

= S m × S l × I

S m = D mm (1 - F /2)

S l = D ml (1 - F /2)