Agriculture Reference
In-Depth Information
1.2.6 Sample Examples on Irrigation Channel Design
Example 1.1
Determine the velocity of flow and discharge capacity of an unlined canal branch
on a grade of 1 m in 800 m having depth of flow 1.5 m, bottom width 0.80 m, and
side slope 1:1.
Solution
Required:
=
(a) Velocity of flow,
V
?
(b) Discharge,
Q
=
?
Given,
Depth of flow,
d
=
1.5 m
Bed width,
b
0.8 m
Canal bottom slope,
S
=
1
800
=
=
0.00125
Side slope,
Z
:1
1:1
As the channel is unlined, assume roughness coefficient of Manning's formula,
N
=
=
0.023
Area,
A
(1.5)
2
3.45 m
2
=
bd
+
zd
2
=
(0.8
×
1.5)
+
1
×
=
2
d
√
1
Wetted perimeter,
P
=
b
+
+
z
2
=
5.042 m
A
We know, hydraulic radius,
R
=
P
=
0.684
1
1
0.023
n
R
2
/
3
S
1
/
22
(0.684)
2
/
3
We
know,
velocity,
V
=
=
×
×
√
0.00125 = 1.193 m/s (Ans.)
Discharge,
Q
4.116 m
3
/s (Ans.)
=
AV
=
3.45
×
1.193
=
Example 1.2
Design an earthen channel of trapezoidal section for the following conditions:
=
Discharge
2 cumec
Channel bottom slope: 1 in 1,200
Side slope: 1.3:1
Va l u e o f
N
in Manning's equation
=
0.02
Solution
Given,
2m
3
Discharge,
Q
s
Slope of the channel bed,
S
=
/
1
1200
=
Side slope,
z
:1
=
1.3:1
Manning's
N
=
0.02
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