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.)
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