Management of Salt-Affected Soils - Practices of Irrigation and On-farm Water Management - page 316

Solution

Given, average infiltration rate, I f = 1.2 cm/h

= 28.4 cm/day

Infiltration time ( t i ) = 6h = 6/24 day = 0.25 day

ET during infiltration cycle = (80/120) × 10 = 6.67 cm

We know, leaching fraction (LF) = 1 − (ET / I f t i ) = 1 − 6.67 / (28.8 × 0.25)

= 0.074

LF D d / D i =EC iw /EC dw

Thus, EC iw =

=

LF × EC dw = 0.07 × 1.5 = 0.111 dS/m (Ans.)

Example 8.6

A wheat crop is planted in a silt loam soil and irrigated from a river water. The
EC of the river water at the time of irrigation is 1.5 dS/m. The crop ET during
the growing season is 76 cm. How much additional water must be applied for
leaching?

Solution

Given,

EC iw = 1.5 dS/m

For wheat, at 100% yield potential, EC e = 6.0

We know,

LR = EC iw /(5EC e − EC iw )

= 1.5 / (5 × 6.0 − 1.5)

= 0.0526

The required amount of water for both ET demand and leaching requirement:

(a) Using the relation, AW = ET/(1- LR)

= 76/(1 - 0.0526)

= 80.22 cm (Ans.)

(b) Using the relation, AW ET (1+LF)/ E f

and assuming E f = 90%, AW = 76 (1 + 0.056)/0.9 = 89.14mm (Ans.)

=