Agriculture Reference
In-Depth Information
Solution
Given,
L
=
120 m
W
=
0.70 m
S
0.3%
q
1
=
=
0.005 m
3
/s
t
1
=
40 min
=
40
×
60 s
=
2,400 s
q
×
t
We get,
d
=
W
×
L
Thus, putting the values, depth of infiltration from initial stream,
d
1
=
0.1428 m
0.0025 m
3
/s
Now,
q
2
=
q
1
/2
=
0.005/2
=
t
2
=
30 min
=
30
×
60 s
=
1,800 s
Putting the values,
d
2
=
0.0535 m
Thus, total depth,
d
=
d
1
+
d
2
=
0.1428 m + 0.0535 m
=
0.1964 m (Ans.)
Example 3.7
Water is applied in a furrow using non-erosive maximum stream size. The length,
width, and slope of the furrow are 150, 0.75 m, and 0.4%, respectively. The stream
is continued for 2 h. Estimate the average depth of irrigation.
Solution
We get non-erosive maximum stream size,
q
max
=
0.60/S l/s
0.0015 m
3
/s
Here,
S
=
0.4%, thus
q
max
=
0.60/0.4
=
1.5 l/s
=
q
×
t
We get,
d
=
W
×
L
Given,
L
=
150 m
=
W
0.75 m
=
=
×
×
=
T
2h
2
(60
60) s
7,200 s
0.0015 m
3
/s
Putting the values,
d
and
q
=
=
0.096 m (Ans.)
3.5 Design of Sprinkler System
3.5.1 Design Aspects
Design aspects of sprinkler irrigation system are as follows:
•
System layout
•
Operating pressure, nozzle diameter, sprinklers discharge, and wetted diameter
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