Irrigation System Designing - Practices of Irrigation and On-farm Water Management - page 74

Putting the values,

(180) 1.1 (0.1) 0.0093 (0.003) 0.0203 (0.033) 0.387 (0.08) 0.952
(4.157) 1.0885 (0.62) 0.75

T apl =

×

2.5

= 4.76 h (Ans.)

Example 3.2

In an irrigation command, it is decided to implement border irrigation system. From
the field observation, the following field characteristics are gathered:

• Field length, L = 250 m

• Field slope, S 0 = 0.005

• Infiltration family, IF = 0.5

0.04 m/h a

k =

a = 0.6

• Roughness coefficient, n = 0.12

• Required depth of infiltration, D req = 0.11 m

Design a border strip with the above information.

Solution

We get unit flow rate to be applied:

CU q L 1.0562

n 0.1094 k 1.225 a 3.832 × × ×

q apl =

S 0.09 D 0.823

req

×

Here, given

L = 250 m

n = 0.12

0.04 m/h a

k =

a = 0.6

S 0.005 m/m

D req =

=

0.09 m

CU q = conversion factor. To convert into SI unit, CU q = 0.642

(a) Putting the values, we obtain

(250) 1.0562 (0.12) 0.1094 (0.04) 1.225 (0.60) 3.832
(0.005) 0.09 (0.09) 0.823

q apl = 0.642 ×

5.5546 m 3 /h - m (Ans.)

=

= 1.543 l/s - m