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8/2/2019 Chap5 1 Kennedy
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DESIGN OF HYDRAULIC STRUCTURES
Ashfaque Ahmed MemonDepartment of Civil Engineering
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ALLUVIAL CHANNEL DESIGN
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KENNEDYS THEORY
R. G. Kennedy 1895
Non-silting non-scouring reaches for 30 years in Upper BariDoab Canal (UBDC) system.
Vertical eddies generated from the bed are responsible forkeeping silt in suspension.
Critical velocity
Mean velocity which keeps the channel free from silting and
scouring.Vo = 0.55 D
0.64
or Vo = C Dn
where, Vo = critical velocity, D = depth of water
C = constant and n = index number
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Later on realizing the channel material (sandy silt inUBDC) he modified the equation as
Vo = 0.55 m D0.64
where,V = actual velocity; andm = C.V.R = critical velocity ratio = V/Vo
m = 1.1 1.2 coarser sand
m = 0.7 0.9 finer sandm = 0.75 Sindh canals
Values of C and m for various grades of silt
Type of silt grade C mCoarser silt 0.7 1.3Sandy loam silt 0.65 1.2Coarse light sandy silt 0.59 1.1Light sandy silt 0.53 1.0
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Rugosity coefficient
Kennedy used Kutters equation for determining the mean velocity offlow 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
SNV
++
++= 00155.0
231
00155.0123
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Water Surface Slope
No relationship by Kennedy.
Governed by available ground slope.
Different sections for different slopes.Woods normal design table for B/D ratio.
Silt Carrying Capacity of Channel
Qt = K B Vo0.25
where
Qt = total quantity of silt transported
B = bed width
Vo = critical velocityK = constant, whose value was not determined by
Kennedy
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Design ProcedureCase I : Given Q, N, m and S (from L-section)
1. Assume D
2. Calculate velocity from Kennedys equation, VK= 0.55 mD0.64
3. Calculate area, A = Q / VK
4. Calculate B from A = B D + z D2 ; assume side slope 1(V) :(H), if not given.
5. Calculate wetted perimeter and hydraulic mean depthfrom;
6. Determine mean velocity from Chezys equation, Vc =C(RS)
if Vc = Vk then O.K.
otherwise repeat the above procedure with another valueof D until Vc = Vk.
DBP 5+=
DB
DBD
P
AR
5
5.0 2
++
==
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Case II : Given Q, N, m and B/D
1. DetermineA in terms ofD
let B/D = y
therefore, B = y D
2. Substitute eq. (1) and kennedys equation into continuity equation andsolve for D, i.e.
Q = A V
3. Knowing D, calculate B and R
B = y D
4. Determine V from Kennedys equation
V = 0.55 m D0.64
5. Determine slope from Kutters 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
+
=ym
QD
DB
DBDR
5
5.0 2
++=
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Problem:
Design an irrigation channel for the following data usingKennedys theory:
Full Supply Discharge (F.S.Q) = 14.16 cumecSlope, S = 1/5000
Kutters rugosity coefficient, N = 0.0225
Critical velocity ratio, m =1
Side slope, z =
Solution:
1. Assume D = 1.72 m
2. Vk= 0.55 m D0.64
=0.55(1)(1.72)0.64
= 0.778 m3. A = Q/Vk = 14.16/0.778 = 18.2 m
2
4. A = B D + 0.5 D2 for z =1/2 or 0.5
18.2 = 1.72 B + 0.5(1.72)2
B = 9.72 m
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5.R = A / P = 18.2 / 13.566 = 1.342 m
6.
Vc = 0.771 m
0.778 m
Result:B = 9.72 mD = 1.72 m
m566.13)72.1(572.95 =+=+= DBP
RS
R
N
S
SNVc
++
++
= 00155.0231
00155.0123
( )50001342.1
342.10225.0
5000100155.0231
50001
00155.0
0225.0
123
++
++=
cV
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Problem:Using Kennedys theory design an irrigation channel to carrya discharge of 56.63 cumec. Assume N = 0.0225, m = 1.03and 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 / PA = 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
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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 + 1.55x10-3S1/2 = 1.68x105
Solving by trial and error, we getS = 1 in 5720
Results:B = 25.43 mD = 2.25 m
S = 1 / 5720
RS
RN
S
SNV
++
++=
00155.0231
00155.0123
S
S
S )96.1(
96.1
0225.000155.0231
00155.0
0225.0
123
95.0
++
++=
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Problem:
Design a section by Kennedys theory, given B/D = 5.7, S = 1/5000and N = 0.0225. Also determine the discharge carried by the
channel.Solution:
B/D = 5.7, B = 5.7 D
Assuming z =
Since V = 0.55 m D0.64
Assuming m =1
V = 0.55 D0.64 ---------- (1)
Also
DD
D
DD
DD
DB
DBDR 78.0
94.7
2.6
57.5
5.07.5
5
5.02222
==++
=++
=
RS
R
N
S
SNV
++
++=00155.0
231
00155.01
23
( ) ( )2
783.0
939.05000178.0
78.00225.0
5000100155.0231
50001
00155.0
0225.0
123
+
=
++
++=
D
DD
D
V
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Equating equation (1) and (2)
0.55 D1.14
0.939 D + 0.43 D0.64
= 0By 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
DD
+=
783.0
939.055.0 64.0
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Shortcomings of Kennedys theory
1. The method involves trial and error.
2. Shape of section i.e. B/D is not known in
advance.
3. Kutters equation is used instead ofMannings equation. Therefore limitations ofKutters formula are also incorporated inKennedys theory. Moreover it involves morecomputations.
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