09. Silt Theories [Kennedy's Theory].pdf

Silt Theories (Kennedy's Theory)

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

2. Silt Theories
Canalssufferingeitherfromscouringorsilting.
Siltingreducesthedischargecapacityofcanalbyreducingitssection,whilescouringwiden
thesectioncausingthefullsupplylevel(FSL)lower.
Toavoidthisproblem,stablecanalswhichneitherscournorsilt,shouldbedesigned.
Totacklethisproblem,therearetwopopulartheoriesknownas:
(1)Kennedy’sSilttheory,and
(2)Lacy’sSilttheory.

3. Silt Theories
➢Kennedy’sSiltTheory
Studiednon-siltingnon-scouringreachesfor30yearsinUpperBariDoabCanal(UBDC)
system.
Accordingtothistheory,verticaleddiesgeneratedfromthebedareresponsiblefor
keepingsiltinsuspension.
Theseeddiesrisesduetothefrictionofflowingwateragainstthebedandworkupagainst
thedepthofcanal.
Thesiltsupportingpoweristhereforeproportionaltothebedwidthofthestreamandnotto
itswettedperimeter.

4. Kennedy’s Theory
Kennedy’s further defined critical velocity as non silting nor scouring velocity, and the
following relation was derived.
Critical velocity
Mean velocity which keeps the channel free from silting and scouring.
V
o= 0.55 D
0.64
In general form it is as,
V
o= C D
n
where,V
o= critical velocity,D = depth of water
C = constant and
n = index number

5. 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
Kennedy’s Theory

6. Rugositycoefficient
❖Kennedy used Kuttersequation for determining the mean velocity of flow in the channel
Where N depends upon the boundary material, R is the hydraulic radius (A/P)
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 0.0210.00155
23
0.00155
123
NS
VCRS RS
N
S R






Kennedy’s Theory
�=
23+
1
??????
+0.00155/�
1+23+
0.00155
�
??????/√�

7. 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
Kennedy’s Theory

8. Modification for Sindh Canals
❖In 1940, while designing GudduBarrage project canals, K. K. Framjiproposed 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
Kennedy’s Theory

9. Design Procedure
Case I : Given Q, N, m and S (from L-section)
1.Assume a trial value of depth (D) in meters;
2.Calculate the velocity using the equation V
o= 0.55 m D
0.64
3.Calculate area by equation A = Q/V
o
4.Assume side slope of channel as ½ : 1 (Z:1) [1(V) : ½(H)] then A = BD +ZD
2
= BD + ½ D
2
5.Calculate wetted perimeter and Hydraulic mean depth by applying formulas
P = B + 2D(??????
2
+1)=�+2�
1
2
2
+1=�+�√5
6.Determine mean velocity from Chezy’sequation,V
c= C √(RS)
7.if V
c= V
othen O.K. otherwise repeat the above procedure with another value of D until V
c= V
o.
Note: Increase D if V
o< V
c
Decrease D if V
o> V
cDBP 5 DB
DBD
P
A
R
5
5.0
2



Kennedy’s Theory

10. Problem 1
DesignanirrigationchannelforthefollowingdatausingKennedy’stheory:FullSupply
Discharge(F.S.Q)=14.16cumec,Slope,S=1/5000,Kutter’srugositycoefficient,N=0.0225,
Criticalvelocityratio,m=1,Sideslope,z=½
Solution:
1. Assume D = 1.72 m
2. V
o= 0.55 m D
0.64
=0.55(1)(1.72)
0.64
= 0.778 m
3. A = Q/V
o= 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
Kennedy’s Theory

11. 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 mm 566.13)72.1(572.95  DBP RS
R
N
S
SN
V
c









00155.0
231
00155.01
23  50001342.1
342.1
0225.0
50001
00155.0
231
50001
00155.0
0225.0
1
23











cV
Kennedy’s Theory

12. Kennedy’s Silt Theory Example
Problem 2: Design an irrigation channel with the following data by Kennedy’s method. Given that Q = 50
cumecs; S = 1/5000; N = 0.0225 and CVR = m =1.0
Solution:
Let the assumed depth = D = 2.3 m
The Kennedy’s critical velocity V
0= 0.55 (D)
0.64
= 0.55 (2.3)
0.64
=0.9372 m/sec
Area (A) = Q/V
0= 50/0.9372 = 52.15 m
2
;
Assuming Z : 1= ½ :1 then 52.15 = BD + ZD
2
= 52.15 = 2.3 B+ 0.5(2.3)
2
=>
B = 52.15-0.5 (2.3)
2
/2.3 = 21.524 m
Wetted Perimeter (P) = &#3627408437;+2&#3627408439;(√Z
2
+1) = 21.525 +2*2.3(1+1/4)
2
= 26.667 m
Hydraulic mean depth = A/P = 52.15/ 26.667 = 1.955 m
According to Kutter’s equation ??????=
23+
1
??????
+0.00155/&#3627408454;
1+23+
0.00155
??????
??????/√&#3627408453;
√&#3627408453;&#3627408454;
Plugging the value of N, R and S in the above equation V =0.9945 m/s

13. Kennedy’s Silt Theory Example
CVR = m = V/V
0= 0.9945/0.9472 = 1.06. As the CVR is greater than the given value
so use the 2
nd
trial.
Difference in the value of CVR = (1.06-1.0)100 = 6% ; Let increase the depth 12% then new depth D = 2.3
+.12*2.3 = 2.576 m then V
0= 0.55 (2.576)
0.64
=1.0077 m/s = 1.01m/s
•A =
&#3627408452;
??????0
=
50
1.01
= 49.5 m
2
Also A = BD + ZD
2
=> B =49.5-0.5*2.576
2
/2.576 = 17.93 m
•P = B + 2D (??????
2
+1)= 17.93 + 2*2.58 (0.5
2
+1)
0.5
= 23.69 m
•R =
49.50
23.69
= 2.08 m
•Use kutter’sequation for &#3627408438;=
23+
1
0.0225
+
0.00155
2∗10^−4
1+23+
0.00155
2∗10
−4
0.0225/√2.08
=50.6
•V = C √&#3627408453;&#3627408454;= 50.6 (2.08*1/5000)
0.5
=1.03
•CVR = 1.03/1.01 = 1.01 OK
•Thus B = 17.93 and D = 2.58 Answer
Note: Other cases to be continued in Next class

14. 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’sFormula 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

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