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

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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

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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

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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

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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.