Irrigation System Designing - Practices of Irrigation and On-farm Water Management - page 75

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

S 0.0203
0

L 1.1 n 0.0093 k 0.387 D 0.952

req

× × × ×

T apl = CU T ×

q 1.0885
appl

× a 0.75

Given,

5.5546 m 3 /h-m
CU T =

q apl =

2.5 (coefficient to convert into SI unit)
Putting the values,

(250) 1.1 (0.12) 0.0093 (0.005) 0.0203 (0.04) 0.387 (0.09) 0.952
(5.5546) 1.0885 (0.6) 0.75

T apl = 2.5 ×

= 6.22h (Ans.)

Example 3.3

In a wheat field, a farmer has made border strip of 150 m length. The roughness of
the field is estimated as n 0.10, and the average field slope along the border is

0.08%. Determine the required flow rate per unit width of the border.

=

Solution

Minimum flow rate based on length of run, field slope, and roughness is (SCS):

S 0.5
0

L ×

10 − 6 ) q min = (5.95 × ×

n

Given,

=

L 150 m

S 0 =

0.08% = 0.008

n = 0.10

Putting the values in the above equation, we get

0.000798m 3 /s-m(Ans.)

q min =

Example 3.4

A soybean field is to be designed for irrigation through border method. The field
configuration allows 300 m long border strip. The grade of the field toward the field
drainage channel is 0.3%. Calculate the unit flow rate based on minimum criteria.