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DESIGN OF HYDRAULIC STRUCTURES
ALLUVIAL CHANNEL DESIGN
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 velocityMean velocity which keeps the channel free from silting and scouring.
Vo = 0.55 D0.64 which can be written in general form as,
Vo = C Dn where, Vo = critical velocity, D = depth of water
C = constant and n = index number
Later on realizing the channel material (sandy silt in UBDC), he modified the equation as
Vo = 0.55 m D0.64 or Vo = C m Dn
where, m = C.V.R = Critical Velocity Ratio = V/Vo ; V = actual velocity
m = 1.1 – 1.2 coarser sandm = 0.7 – 0.9 finer sandm = 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 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
Rugosity coefficientKennedy used Kutters equation for determining the mean velocity of flow in the channel
Where N depends upon the boundary material
Channel condition NVery good 0.0225Good 0.025Indifferent 0.0275Poor 0.03
Discharge (cumec) N (in ordinary soil)14 – 140 0.025140 – 280 0.0225> 280 0.02
RS
RN
S
SNV
00155.0231
00155.0123
Water Surface SlopeNo 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 ChannelQt = K B Vo
0.25
whereQt = total quantity of silt transportedB = bed widthVo = critical velocityK = constant, whose value was not determined by
Kennedy
Modification for Sindh CanalsIn 1940, while designing Guddu Barrage project canals, K. K. Framji proposed B/D ratio for Sindh canals as:
5.15.3 61 QDB
Discharge
m3/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
Design Procedure Case I : Given Q, N, m and S (from L-section)1. Assume D2. Calculate velocity from Kennedy’s equation, VK = 0.55 m D0.64 3. Calculate area, A = Q / VK 4. Calculate B from A = B D + z D2 ; 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, Vc =C √(RS)if Vc = Vk then O.K.otherwise repeat the above procedure with another value of D until Vc = Vk. Note: increse D if Vk < Vc decrease D if Vk > Vc
DBP 5
DBDBD
PAR
55.0 2
Problem:Design an irrigation channel for the following data using Kennedy’s theory:Full Supply Discharge (F.S.Q) = 14.16 cumecSlope, S = 1/5000Kutter’s rugosity coefficient, N = 0.0225Critical velocity ratio, m =1Side slope, z = ½
Solution:1. Assume D = 1.72 m2. 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 m2 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
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
m 566.13)72.1(572.95 DBP
RS
RN
S
SNVc
00155.0231
00155.0123
50001342.1
342.10225.0
5000100155.0231
5000100155.0
0225.0123
cV
Example ProblemQ = 80 m3/secS = 1:5500 = 0.00018 m/mm = 1
Assume D = 2.5 mV = 0.55 D0.64 = 0.989 m/secA = 80.918 m2
Side Slope = 1V:1.5Hn = 0.0225
A = B D+ 1.5D2
B = 28.617 mP = 32.223 mR = A/ P = 2.511 mUsing Kutter’s Formula in S.I. Units
C = 52.479 V = C√RS = 1.121 m/secKeep on trailing till Vc = V
D
B
1 1.5
1.803
Case II : Given Q, N, m and B/D 1. Determine A in terms of D
let B/D = ytherefore, 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 equationV = 0.55 m D0.64
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
ymQD
DBDBDR
55.0 2
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 m5. 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 mR = 59.75 / 30.46 = 1.96 m
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 = 1.68x105
Solving by trial and error, we getS = 1 in 5720
Results:B = 25.43 mD = 2.25 mS = 1 / 5720
RS
RN
S
SNV
00155.0231
00155.0123
S
S
S )96.1(
96.10225.000155.0231
00155.00225.0123
95.0
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).
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 DAssuming z = ½
Since V = 0.55 m D0.64 Assuming m =1V = 0.55 D0.64 ---------- (1)
Also
DDD
DDDD
DBDBDR 78.0
94.72.6
57.55.07.5
55.0 2222
RS
RN
S
SNV
00155.0231
00155.0123
2783.0939.05000178.0
78.00225.0
5000100155.0231
5000100155.0
0225.0123
DDD
D
V
Equating equation (1) and (2)
0.55 D1.14 – 0.939 D + 0.43 D0.64 = 0By trial and errorD = 2.1 mB = 5.7 x 2.1 = 11.97 mA = 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/secQ = A V = (14,175)(0.884) = 12.53 m3/sec.
Results:B = 11.97 mD = 2.1 mQ = 12.53 cumec
DDD
783.0939.055.0 64.0
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.
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