Example 11.1 Design a suitable cross-drainage structure for the following data:
Discharge in the canal,
= 357.0 m3/s
Bed width of the canal, Bc
= 23.0 m
Side slope of the canal, m0
= 2.0
Bed level of the canal,
= 267.0 m
Depth of flow in the canal, hc
= 6.7 m
Bed width of the flume, Bf
= 15.0 m
High flood discharge of the stream
= 500 m3/s
High flood level of the stream
= 268.0 m
Bed level of the stream
= 265.0 m
Solution: Using Eq. (11.21), the length of expansion transition,
L = 2.35 (Bc– Bf) + 1.65 m0hc = 2.35 (23 – 15) + 1.65 × 2 × 6.7 = 40.91 ≅ 41 m
Let the transition be divided into 9 sub-reaches of 4.0 m each and the remaining 5 m length be considered as the tenth sub-reach at the end.
Bed width, b is calculated using Eqs. (11.19) and (11.20).
∴
or
or
n = 0.8 – 0.26 m01/2 = 0.8 – 0.26 (2)1/2 = 0.432
B − Bf
x
L
F
xIn O
=
M1
− G 1
−
J
P
Bc− Bf
L M
H
LK
P
N
Q
B −15
x
L
F
x I
0.432 O
=
M1
− G 1
−
J
P
23 − 15
41 M
H
41K
P
N
Q
8x
L
F
x I
0.432 O
B = 15 +
41
M1
− G 1
−
J
P
M
H
41K
P
N
Q
For example, at section 1-1,
8
×4L
F
4 I
0.432 O
B
= 15 +
41
M1
− G 1
−
J
P
1 – 1
M
H
41K
P
= 15.034 m
N
Q
Values of B for different values of x have been similarly calculated and are as shown in col. 3 of Table 11.5.
400
IRRIGATION AND WATER RESOURCES ENGINEERING
Side slopes at various sections are calculated using Eq. (11.22). Thus
m
F
xI 1/ 2
= 1 − G
1 −
J
m0
H
LK
L
F
x
I
1/ 2 O
or
m =2M1
− G 1
−
J
P
M
H
41K
P
For example, at section 1-1,
N
Q
L
1/ 2 O
F
4
I
m
= 2 M1
− G 1
−
J
P = 0.100
1 – 1
M
H
41K
P
N
Q
Values of m for different values of x have been similarly calculated and are tabulated in col. 4 of Table 11.5.
For constant specific energy condition:
v 2
E = E = h +
c
2g
f
c
c
and
vc= 357/[6.7 (23 + 2 × 6.7)] = 1.464 m/s
∴
E = E =6.7+
(1464.)
2
= 6.809 m
f
c
2 × 9.81
hf+
Q2
= 6.809
2gB
2 h
2
f
f
or
hf+
(357)2
= 6.809
2g(15)
2
2
hf
Solving for hf , one obtains hf = 6.006 m
Similarly,
E1 – 1 =h1 – 1 +
Q2
= Ec = 6.809 m
2g(b
− 1
+ m
h
− 1
)2h2
1
1 −
1 1
1 − 1
∴
h1 – 1 +
(357)2
= 6.809
2 × 9.81 (15.034 + 0.1 h
1
)
2 h2
1 −
1 − 1
On solving this equation,
h1 – 1= 6.088 m
Further,
vf
=
357
= 3.963 m/s
15
× 6.006
and
v1 – 1
=
357
= 3.749 m/s
(15.034
+
0.1 × 6.088) 6.088
Using Eq. (11.10),
hL
= 0.3 ×
1
(3.9632
− 3.7492) = 0.025 m
f , 1−1
2
×
9.81
Using Eq. (11.23),
∆zf,1 – 1 = 0.025 m
The positive value of ∆z indicates fall in the bed elevation with respect to the flume bed. Values of ∆z for other sections can be similarly obtained. The values of ∆z and h have been tabulated in cols. 5 and 6 of Table 11.5.
