Silt Theories (Kennedy's Theory-1)

1. 10. Alluvial Channel Design-Silt Theories
[Kennedy’s Theory-II]
B.Sc. Civil Engineering 8th Semester
Muhammad Ajmal (PhD)
Lecturer
Agri. Engg. Deptt.
CE-402 Irrigation Engineering
1

2. Case II : Given Q, N, m and B/D
1. Determine A in terms of D
let B/D = y
therefore, B = y D
2. Substitute eq. (1) and kennedy’s equation into continuity equation and solve for D, i.e.
Q = A V
3. Knowing D, calculate B and R
B = y D
4. Determine V from Kennedy’s equation
V = 0.55 m D0.64
5. Determine slope from Kutter’s equation by trial and error
2
2
2
5
.
0
5
.
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yD
D
BD
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5
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y
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BD
R
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Kennedy’s Theory

3. Problem
Using Kennedy’s theory design an irrigation channel to carry a discharge of 56.63 cumec. Assume N =
0.0225, m = 1.03 and B/D = 11.3.
Solution
1. B/D = 11.3, therefore B = 11.3 D
A = B D + 0.5 D2 =11.3 D2 + 0.5 D2 = 11.8 D2
2. V = 0.55 m D0.64 = 0.55 (1.03) D0.64 = 0.5665 D0.64
3. Q = A V
56.63 = (11.8 D2 ) (0.5665 D0.64 )
D = 2.25 m
4. B = 11.3 (2.25) = 25.43 m
5. R = A / P
A = B D + 0.5 D2 = (25.43)(2.25) + 0.5 (2.25)2 = 59.75 m2
P = B + √5 D = 25.43 + √5 (2.25) = 30.46 m
R = 59.75 / 30.46 = 1.96 m

4. 6. V = 0.55 m D0.64 = 0.55 (1.03) (2.25)0.64 = 0.95 m/sec
7.
Simplifying, we get;
67.44 S3/2 – 0.93 S + 1.55x10-3 S1/2 = 1.68×105
Solving by trial and error, we get
S = 1 in 5720
Results
B = 25.43 m
D = 2.25 m
S = 1 / 5720
RS
R
N
S
S
N
V

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00155
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0
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0225
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0
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

5. Case III : Given S, N, m and B/D
1. From the B/D ratio, determine B in terms of D.
2. Determine A, P and R in terms of D.
3. From Kennedy’s equation, determine velocity (Vk) in terms of D.
4. Putting values of N, S and R in the Chezy’s equation and Kutter’s formula, determine velocity (Vc).
Simplify the expression, and solve it by trail and error for D.
5. Knowing D, calculate B, A and Vk.
6. Using continuity equation, determine the discharge (Q).

6. Problem:
Design a section by Kennedy’s theory, given B/D = 5.7, S = 1/5000 and N = 0.0225. Also determine the
discharge carried by the channel.
Solution:
B/D = 5.7, B = 5.7 D
Assuming z = ½
Since V = 0.55 m D0.64
Assuming m =1
V = 0.55 D0.64 ---------- (1)
Also
D
D
D
D
D
D
D
D
B
D
BD
R 78
.
0
94
.
7
2
.
6
5
7
.
5
5
.
0
7
.
5
5
5
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0 2
2
2
2

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
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RS
R
N
S
S
N
V
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
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
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00155
.
0
23
1
00155
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0
1
23
   
2
783
.
0
939
.
0
5000
1
78
.
0
78
.
0
0225
.
0
5000
1
00155
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0
23
1
5000
1
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0
0225
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0
1
23
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



D
D
D
D
V

7. Equating equation (1) and (2)
0.55 D1.14 – 0.939 D + 0.43 D0.64 = 0
By trial and error
D = 2.1 m
B = 5.7 x 2.1 = 11.97 m
A = B D + z D2 = (11.97 x 2.1) + 0.5 (2.1)2 = 14.175 m2
V = 0.55 (2.1)0.64 = 0.884 m/sec
Q = A V = (14,175)(0.884) = 12.53 m3/sec.
Results
B = 11.97 m
D = 2.1 m
Q = 12.53 cumec
D
D
D


783
.
0
939
.
0
55
.
0 64
.
0

8. Shortcoming/ Drawbacks in Kennedy’s Theory
Following are the drawbacks in Kennedy’s theory
1. Kennedy did not realise the importance of B/D ratio.
2. He did not define the silt grade and silt charge
3. He did not give any relation between velocity of flow and water surface slope
4. He did not give any relation for the bed slope of the canal. The bed slope could be adopted as per ground
condition.
5. He did not account for silt concentration and bed load separately and accounted all factors in one single
factor “m”.
6. For the design of a canal he aimed to achieve the average regime condition
7. For the determination of mean velocity of flow, he solely depend upon Kutter’s equation. Thus the
limitation of Kutter’s equation were incorporated in his theory also.

9. 9