Agriculture Reference
In-Depth Information
3.5.3.4 Sample Workout Problems
Example 3.8
In a sprinkler irrigation system, the lateral spacing along the mainline is 20 m
and sprinkler spacing along laterals is 15 m. The application rate for fulfilling the
peak demand of the proposed crop should be 8 mm/d. Find the discharge rate per
sprinkler.
Solution
3600
×
q s (l/s)
We know application rate, I (mm/hr)
=
S m ( m )
×
S l ( m )
I
×
S m ×
S l
Or, q s =
3600
Given,
I
8 mm/d
Assuming that the sprinkler will operate 12 h a day
Then, I
=
=
8 mm/12 h
=
0.667 mm/h
S m =
20 m
S l =
15 m
Putting the values, q s =
0.055 l/s
Or, 200 l/h (Ans.)
Another Mode of Operation
If we consider that the daily demand should be provided within a certain practical
irrigation period, say in 4 h, to avoid excessive evaporation loss, then the application
rate would be I
=
8 mm/4 h
=
2 mm/h
Thus, q s =
0.1667 l/s or 600 l/h (Ans.)
Example 3.9
In a sprinkler irrigation system, the required total capacity of the system is 0.5 m 3 /s.
Determine the pump capacity. Assume that head loss in pipe and bends and velocity
head required
=
3mofwater.
Solution
Pump capacity, P
=
( Q
×
9.81
×
H )[KW]
0.5 m 3 /s
Here Q
=
Total head
3m
Putting the values, P
=
=
0.5
×
9.81
×
3
=
14.7 KW (Ans.)
 
Search WWH ::




Custom Search