Contents preface (VII) introduction 1—37
Table 8.4 Recommended values of
- Bu səhifədəki naviqasiya:
- Table 8.5 Values of C ′ and x in the Kennedy’s equation for different regions (7)
- Fig. 8.6
| Table 8.4 Recommended values of B/h for stable channels
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Q (m3/s) |
5.0 |
10.0 |
15.0 |
50.0 |
100.0 |
200.0 |
300.0 |
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B/h |
4.5 |
5.0 |
6.0 |
9.0 |
12.0 |
15.0 |
18.0 |
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Several investigations carried out on similar lines indicated that the constant C′ and the exponent x in the Kennedy’s equation, U = C′ mhx are different for different canal systems. Table 8.5 gives the value of C′ and x in the Kennedy’s equation for some regions.
Table 8.5 Values of C′ and x in the Kennedy’s equation for different regions (7)
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Region
C′
x
Egypt
0.25 to 0.31
0.64 to 0.73
Thailand
0.34
0.66
Rio Negro (Argentina)
0.66
0.44
Krishna River (India)
0.61
0.52
Chenab River (India)
0.62
0.57
Pennar River (India)
0.60
0.64
Shwebo (Burma)
0.60
0.57
Imperial Valley (USA)
0.64 to 1.20
0.61 to 0.64
The design procedure based on Kennedy’s theory involves trial. For known Q, n, m, and S, assume a trial value of h and obtain the critical velocity U from the Kennedy’s equation [Eq. (8.10)]. From the continuity equation [Eq. (8.11)] one can calculate the area of cross-section A and, thus, know the value of B for the assumed value of h. Using these values of B and h, compute the mean velocity from the Manning’s equation [Eq. (8.12)]. If this value of the mean velocity matches with the value of the critical velocity obtained earlier, the assumed value of h and the computed value of B provide channel dimensions. If the two velocities do not match, assume another value of h and repeat the calculations.
Ranga Raju and Misri (8) suggested a simplified procedure which does not involve trial. The method is based on the final side slope of 1H : 2V attained by an alluvial channel. During construction, the side slopes of a channel are kept flatter than the angle of repose of the soil. But, after some time of canal running, the side slopes become steeper due to the deposition of sediment. The final shape of the channel cross-section is approximately trapezoidal with side slopes 1H : 2V. For this final cross-section of channel, one can write,
where,
Now
and
A = Bh + 0.5h2 = h2(p + 0.5) P = B + 2.236h = h(p + 2.236)
p = B/h
R = h( p + 0.5) p + 2.236
Q
U = h2 ( p + 0.5)
(8.13)
(8.14)
(8.15)
(8.16)
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292 |
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IRRIGATION AND WATER RESOURCES ENGINEERING |
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Substituting the value of U and R in the Manning’s equation [Eq. (8.10)], one obtains |
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S = |
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Q 2 n 2 ( p + 2.236)4/3 |
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(8.17) |
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h 16/3 ( p + 0.5)10/3 |
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Similarly, substituting the value of U in the Kennedy’s equation [Eq. (8.10)], one gets, |
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1818.Q O0.378 |
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h = |
M |
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P |
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(8.18) |
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N |
( p + 0.5)mQ |
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On substituting the value of h from Eq. (8.18) in Eq. (8.17), one finally obtains |
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SQ0.02 |
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(B/ h + 2.236) |
1.333 |
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n 2 m2 |
= 0.299 |
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(8.19) |
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(B/ h + 0.5)1.313 |
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Figure 8.6 shows the graphical form of Eq. (8.19).
For given Q, n, m, and a suitably selected value of S, compute SQ0.02 /n2 m2 and read the value of B/h from the Fig. 8.6. From Table 8.4, check if this value of B/h is satisfactory for the given discharge. If the value of B/h needs modification, choose another slope. Having obtained B/h, calculate h from Eq. (8.18) and then calculate B.
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×100 |
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0.02 |
SQ |
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m |
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2 |
n |
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75
70
65
60
55
50
45
40
35
0 2 4 6 8 10 12 14 16 18 20 22 24
p = B/h
Fig. 8.6 Diagram for design of alluvial channels using the Kennedy’s equation (8)
Example 8.5 Design a channel carrying a discharge of 30 m3/s with critical velocity ratio and Manning’s n equal to 1.0 and 0.0225, respectively. Assume that the bed slope is equal to 1 in 5000.
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DESIGN OF STABLE CHANNELS |
293 |
Solution:
Kennedy’s method:
Assume h = 2.0 m. From Kennedy’s equation [Eq. (8.10)]
U = 0.55 mh0.64 = 0.55 × 1 × (2.0)0.64
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= 0.857 m/s
∴
A = Q/U =
30
= 35.01 m2
0.857
For a trapezoidal channel with side slope 1 H : 2V,
Bh +
h2
= B(2.0) +
2 × 2
= 2B + 2 = 35.01
2
2
∴
B = 16.51 m
∴
R =
35.01
= 16.7 m
16.51 + 2.0 5
Therefore, from the Manning’s equation [Eq. (8.12)],
1
2/3
1/2
1
2/3
F
1 I
1/ 2
U =
n
R
S
=
0.0225
(1.67)
G
J
= 0.885 m/s
H
5000K
Since the velocities obtained from the Kennedy’s equation and Manning’s equation are appreciably different, assume h = 2.25 m and repeat the above steps.
U = 0.55 × 1 × (2.25)0.64 = 0.924 m/s
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A =
30
= 32.47 m2
0.924
∴ B(2.25) + (0.5) (2.25)2 = 32.47
∴
B = 13.31 m
R =
32.47
= 1.77 m
13.31 × ( 5 × 2.25)
U =
1
(1.77)
2/3 F
1 I 1/2
G
J
0.0225
H
5000
K
= 0.92 m/s
Since the two values of the velocities are matching, the depth of flow can be taken as equal to 2.25 m and the width of trapezoidal channel = 13.31 m.
Ranga Raju and Misri’s method :
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SQ0.02 |
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1 |
(30)0.02 |
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5000 |
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n 2 m2 |
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(0.0225)2 × (1)2 |
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Hence, from Fig. 8.6, |
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B |
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p = |
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= 6.0 |
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h |
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1818. Q O0.378 |
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and |
h = M |
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P |
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N |
( p + 0.5)mQ |
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= 0.423
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1818. × 30 |
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0.378 |
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= G |
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= 2.235 m |
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(6.0 + 0.5) × 1K |
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294 |
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IRRIGATION AND WATER RESOURCES ENGINEERING |
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∴ |
B = 6.0 h = 13.41 m |
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Q |
30 |
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and |
U = |
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h 2 ( p + 0.5) |
(2.235) 2 ( 6.0 + 0.5) |
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= 0.924 m/s.
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