Irrigation System Designing - Practices of Irrigation and On-farm Water Management - page 73

3.2.6 Sample Workout Problems

Example 3.1

Design a border strip with the following characteristics:

Field length, L = 180 m

Field slope, S 0 = 0.003

Infiltration family, IF = 0.5

0.033 m/h a

k =

a 0.63

Roughness coefficient, n

=

0.15
Required depth of infiltration, D req =

=

0.08 m

Solution
We get the unit flow rate to be applied:

L 1.0562 n 0.1094 k 1.225 a 3.832 × × ×

q apl =

CU q

S 0.09 × D 0.823

req

Here, given

L = 180 m

n = 0.15

0.033 m/h a k =

a = 0.62

S 0.003 m/m

D req =

=

0.08 m

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

(i) Putting the values, we obtain,

(180) 1.0562 (0.15) 0.1094 (0.033) 1.225 (0.63) 3.832
(0.003) 0.09 (0.08) 0.823

q apl = 0.642 ×

4.157 m 3 = / h-m (Ans.)

(ii) We get application time for designed flow rate:

L 1.1 n 0.0093 S 0.0203

0

k 0.387 D 0.952

req

× × × ×

T apl = CU T ×

q 1.0885
appl

a 0.75

×

Given,

4.157 m 3 /h-m
CU T =

q apl =

2.5 (coefficient to convert into SI unit)