Chap5 1 Kennedy

8/2/2019 Chap5 1 Kennedy

1/15

DESIGN OF HYDRAULIC STRUCTURES

Ashfaque Ahmed MemonDepartment of Civil Engineering


2/15

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.

Chap5 1 Kennedy

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