Design Of Hydraulic Structure - Kennedy's Theory

Kennedy's Theory
Civil Engineering
Irrigation Engineering

1. DESIGN OF HYDRAULIC
STRUCTURES

2. ALLUVIAL CHANNEL
DESIGN

3. KENNEDY’S THEORY
R. G. Kennedy – 1895
Non-silting non-scouring reaches for 30 years in Upper Bari Doab
Canal (UBDC) system.
Vertical eddies generated from the bed are responsible for keeping
silt in suspension.
Critical velocity
Mean velocity which keeps the channel free from silting and
scouring.
V
o
= 0.55 D
0.64

which can be written in general form as,
V
o
= C D
n

where,V
o
= critical velocity,D = depth of water
C = constant and n = index number

4. Later on realizing the channel material (sandy silt in UBDC),
he modified the equation as
V
o = 0.55 m D
0.64

or V
o
= C m D
n

where,
m = C.V.R = Critical Velocity Ratio = V/V
o ; V = actual velocity
m = 1.1 – 1.2 coarser sand
m = 0.7 – 0.9 finer sand
m = 0.85 Sindh canals
Values of C (the constant in Kennedy’s Eq.) and m (the
Critical Velocity Ratio, CVR) for various grades of silt
Type of silt grade C m
Coarser silt 0.7 1.3
Sandy loam silt 0.65 1.2
Coarse light sandy silt 0.59 1.1
Light sandy silt 0.53 1.0

5. Rugosity coefficient
Kennedy used Kutters equation for determining the mean velocity of flow in
the channel
Where N depends upon the boundary material
Channel condition N
Very good 0.0225
Good 0.025
Indifferent 0.0275
Poor 0.03
Discharge (cumec) N (in ordinary soil)
14 – 140 0.025
140 – 280 0.0225
> 280 0.02
RS
R
N
S
SN
V
÷
ø
ö
ç
è
æ
++
++
=
00155.0
231
00155.01
23

6. Water Surface Slope
No relationship by Kennedy.
Governed by available ground slope.
Different sections for different slopes.
Wood’s normal design table for B/D ratio.
Silt Carrying Capacity of Channel
Q
t
= K B V
o
0.25
where
Q
t
= total quantity of silt transported
B = bed width
V
o
= critical velocity
K = constant, whose value was not determined by Kennedy

9. Modification for Sindh Canals
In 1940, while designing Guddu Barrage project canals, K. K. Framji
proposed B/D ratio for Sindh canals as:
5.15.3
61
-=Q
D
B
Discharge
m
3
/sec
B/D ratio for
standard section
B/D ratio for
limiting section
100 6.0 4.0
1000 8.4 5.0
5000 13.0 8.0

10. Design Procedure
Case I : Given Q, N, m and S (from L-section)
1.Assume D
2.Calculate velocity from Kennedy’s equation, V
K
= 0.55 m D
0.64

3.Calculate area, A = Q / V
K
4.Calculate B from A = B D + z D
2
; assume side slope 1(V) : ½(H),
if not given.
5.Calculate wetted perimeter and hydraulic mean depth from;
6.Determine mean velocity from Chezy’s equation,V
c
=C √(RS)
if V
c = V
k then O.K.
otherwise repeat the above procedure with another value of D
until V
c
= V
k.

Note: increse D if V
k < V
c
decrease D if V
k
> V
c

DBP 5+=
DB
DBD
P
A
R
5
5.0
2
+
+
==

11. Problem:
Design an irrigation channel for the following data using Kennedy’s
theory:
Full Supply Discharge (F.S.Q) = 14.16 cumec
Slope, S = 1/5000
Kutter’s rugosity coefficient, N = 0.0225
Critical velocity ratio, m =1
Side slope, z = ½
Solution:
1. Assume D = 1.72 m
2. V
k
= 0.55 m D
0.64
=0.55(1)(1.72)
0.64
= 0.778 m
3. A = Q/V
k
= 14.16/0.778 = 18.2 m
2

4. A = B D + 0.5 D
2
for z =1/2 or 0.5
18.2 = 1.72 B + 0.5(1.72)
2

B = 9.72 m

12. 5.
R = A / P = 18.2 / 13.566 = 1.342 m
6.
V
c
= 0.771 m
≈ 0.778 m
Result:
B = 9.72 m
D = 1.72 m
m 566.13)72.1(572.95 =+=+= DBP
RS
R
N
S
SN
V
c
÷
ø
ö
ç
è
æ
++
++
=
00155.0
231
00155.01
23
( )50001342.1
342.1
0225.0
50001
00155.0
231
50001
00155.0
0225.0
1
23
÷
÷
ø
ö
ç
ç
è
æ
++
++
=\
cV

13. Example Problem
Q = 80 m
3
/sec
S = 1:5500 = 0.00018 m/m
m = 1
Assume D = 2.5 m
V = 0.55 D
0.64
= 0.989 m/sec
A = 80.918 m
2
Side Slope = 1V:1.5H
n = 0.0225
A = B D+ 1.5D
2
B = 28.617 m
P = 32.223 m
R = A/ P = 2.511 m
Using Kutter’s Formula in S.I. Units
C = 52.479
V = C√RS = 1.121 m/sec
Keep on trailing till Vc = V
D
B
1
1.5
1.803

14. 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 D
0.64

5.Determine slope from Kutter’s equation by trial and error
222
5.05.0 DyDDBDA +=+=
)1()5.0(
2
------+=yDA
)55.0).(5.0(
64.02
mDyDQ +=
( )
64.21
5.055.0
ú
û
ù
ê
ë
é
+
=\
ym
Q
D
DB
DBD
R
5
5.0
2
+
+
=

15. 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 D
2
=11.3 D
2
+ 0.5 D
2
= 11.8 D
2

2.V = 0.55 m D
0.64
= 0.55 (1.03) D
0.64
= 0.5665 D
0.64

3.Q = A V
56.63 = (11.8 D
2
) (0.5665 D
0.64
)
D = 2.25 m
4.B = 11.3 (2.25) = 25.43 m
5.R = A / P
A = B D + 0.5 D
2
= (25.43)(2.25) + 0.5 (2.25)
2
= 59.75 m
2

P = B + √5 D = 25.43 + √5 (2.25) = 30.46 m
R = 59.75 / 30.46 = 1.96 m

16. 6.V = 0.55 m D
0.64
= 0.55 (1.03) (2.25)
0.64
= 0.95 m/sec

7.
Simplifying, we get;
67.44 S
3/2
– 0.93 S + 1.55x10
-3
S
1/2
= 1.68x10
5
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
SN
V
÷
ø
ö
ç
è
æ
++
++
=
00155.0
231
00155.01
23
S
S
S
)96.1(
96.1
0225.000155.0
231
00155.0
0225.0
1
23
95.0
÷
ø
ö
ç
è
æ
++
++
=\

17. 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 (V
k
) in terms of D.
4.Putting values of N, S and R in the Chezy’s equation and Kutter’s
formula, determine velocity (V
c
). Simplify the expression, and solve it
by trail and error for D.
5.Knowing D, calculate B, A and V
k
.
6.Using continuity equation, determine the discharge (Q).

18. 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 D
0.64

Assuming m =1
V = 0.55 D
0.64
---------- (1)
Also
D
D
D
DD
DD
DB
DBD
R 78.0
94.7
2.6
57.5
5.07.5
5
5.0
2222
==
+
+
=
+
+
=
RS
R
N
S
SN
V
÷
ø
ö
ç
è
æ
++
++
=
00155.0
231
00155.01
23
( ) ()2
783.0
939.0
5000178.0
78.0
0225.0
50001
00155.0
231
50001
00155.0
0225.0
1
23
-----
+
=
÷
÷
ø
ö
ç
ç
è
æ
++
++
=
D
D
D
D
V

19. Equating equation (1) and (2)
0.55 D
1.14
– 0.939 D + 0.43 D
0.64
= 0
By trial and error
D = 2.1 m
B = 5.7 x 2.1 = 11.97 m
A = B D + z D
2
= (11.97 x 2.1) + 0.5 (2.1)
2
= 14.175 m
2

V = 0.55 (2.1)
0.64
= 0.884 m/sec
Q = A V = (14,175)(0.884) = 12.53 m
3
/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

20. Shortcomings of Kennedy’s theory
1.The method involves trial and error.
2.Shape of section i.e. B/D is not known in
advance.
3.Kutter’s equation is used instead of Manning’s
equation. Therefore limitations of Kutter’s
formula are also incorporated in Kennedy’s
theory. Moreover it involves more
computations.