Agriculture Reference
In-Depth Information
Example 1.3
Design a concrete lined canal from the following data:
Discharge
=
400 cumec
Slope
1in4,000
Side slope
=
1.5:1 (
H
:
V
)
Limiting velocity
=
2.5 m/s
Va l u e o f
N
in Manning's equation
=
=
0.013
Solution
Applying Manning's equation, the velocity,
V
is given by
1
N
R
2
/
3
S
1
/
2
V
=
(A)
Here, limiting velocity,
V
=
2.5 m/s
=
N
0.013
1
4000
S
=
Substituting the above values in Eq. (A), we get
1
4000
1
/
2
1
R
2
/
3
2.5
=
0.013
×
×
√
4000 = 2.554
or
R
2
/
3
=
×
×
2.5
0.013
(2.489)
3
/
2
=4.08m
R
=
Cross-sectional area,
A
=
V
400/2.5 = 160 m
2
We know, for most economical section in trapezoidal canal,
R
=
d
/
2
Then,
d
=
2
R
=
2
×
4.08
=
8.16 m
zd
2
We get,
A
=
bd
+
8.16 + 1.5 (8.16)
2
That is, 160 =
b
×
Or,
b
7.36 m
Considering freeboard as 15% of flow depth, depth of canal,
dc
=8.16
=
×
1.15 = 9.384 m
Top width,
T
=
b
+2
zd
c
=7.36
+
2
×
1.5
×
9.384 = 35.51 m
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