hydraulics topic 2 uniform flow in open channel

8/19/2019 Hydraulics Topic 2 Uniform Flow in Open Channel

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Hydraulics

Topic 2. Uniform Flow in Open Channel

Dr. Mohd Ariff bin Ahmad [email protected]

Assoc. Prof. Dr. Tan Lai [email protected]


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Learning Outcomes

At the end of this topic, students should be able to:

i. Understand the concept of uniform flow

ii. Calculate normal flow depth in variable channelsections using Chezy and Manning equations

iii. Determine the best hydraulic/effective section of

open channel

BFC21103 HydraulicsTan et al. ([email protected])


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Open Channel Flow

Classification

based on Time

Classification

based on Space

Steady Unsteady Uniform Non-Uniform

GVF RVF


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• Uniform flow is considered to be steady only, sinceunsteady uniform flow is practically does not exist.

• Steady uniform flow is rare in natural streams, only

happens in prismatic channels.

• We adopt / assume uniform flow for most flow

computations because uniform flow calculation is

simple, practical and provide satisfactory solution.


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The 132 km long All-American Canal links California's Imperial Valley to the Colorado River. This new

concrete-lined section saves about 3.8 million of water a year over its leaky earthen forerunner


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The concrete channel of Los Angeles River (NGM, 2010)


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The Klang River, Kuala Lumpur & Selangor


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• In uniform flow, the normal depth y o occurs when depth of water is

the same along the channel.

• Normal depth y o implies that the water depth, flow area, wetted

perimeter, velocity and discharge at every section of the channel

are constant within a prismatic channel.

• Thus, in uniform flow, the energy line, water surface and channel

bottom are parallel, i.e. the slopes are equal S f = Sw = So = S.

g

V

2

2

y o


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2.1 Velocity Distribution


Natural channel

0.84

0.820.800.76

0.70

0.620.48

V max

V max

0.52

0.50

0.450.400.35

y o

0.53

Rectangular channel

V average

0.6y o

V max0.2y

o

V

y

y o

Velocity distribution

Depends on the geometry of the channel and wetted boundary

roughness


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2.2 Chezy and Manning Equations

Two most common equations used in the uniform flow computations:

1. Chezy formula

2. Manning formula

2

1

2

1

oSCRV

2

1

3

21

oSRnV

x

o

x SRV constantThus, the general uniform flow equation:

C = Chezy roughness coefficient

n = Manning roughness coefficient


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6

11R

nC Difference between Chezy and Manning formulae

Factors determining the roughness are surface roughness, vegetation,

channel irregularity, channel alignment, silting and scouring,

obstruction, size and shape of channel, stage and discharge, seasonalchange, and suspended material and bed load.


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The Chezy two assumptions are:

1. The force resisting the flow per unit area of the channel bed is

proportional to the square of the velocity:

2. The effective component of the gravity force causing the flow

must be equal to the total force of resistance. This is also the

basic principle of uniform flow where uniform flow will be

developed if the resistance is balance by the gravity forces:

Derivation of Chezy equation

PLV k F f 2

sin ALF g


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g

V

2

2

y o

W

Datum

A

P


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sin2 ALPLV k

o

ALSPLV k 2

oSP

A

k V

2

2

1

2

12

1

oSRk V

2

1

2

1

oSRC V where C = Chezy coefficient

1221 sin MM pF W p f

Since for uniform flow, 2121 and MM p p

Total force of resistance is counter-balances with the

effective component of gravity, which is acts parallel

to the channel bed.

Fr = Force of resistance

W = Weight of the fluid = ALθ = Slope angle of the bed

= Specific weight of the fluid

A = Cross sectional area of the channel

L = Characteristic length of the channel

The resistance to flow is proportional to

the square of the velocity.

Fr = resistance to flow (N)

Aw = wetted area = PxL

P = wetted perimeter

L = length of the channel

K = constant of proportionality

V = mean velocity of flow


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A rectangular channel 2.0 m wide carries water at a depth of 0.5 m.

The channel is laid on a slope of 0.0004. The Chezy coefficient is 73.6.Compute the discharge of the channel.

Given B = 2.0 m, y = 0.5 m, So = 0.0004 and C = 73.6

A = By = 1 m2, P = B + 2y = 3 m, R = 1/3 m

B

y

oRS AC Q

0004.0

3

16.731 Q

/sm850.0 3Q

Activity 2.1


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Water flows in a triangular channel with side slope 1.5(H) : 1(V),

bottom slope 0.0002 and Chezy coefficient of 67.4. The depth offlow is 2.0 m. Find the flow rate and average velocity. Based on

Froude number, determine the state of flow.

z

y1

Given y = 2.0 m, z = 1.5, So = 0.0002 and C = 67.4

A = zy 2 = 6 m2, P = 2y = 7.211 m, R = A/P = 0.832 m, D = A/T = 6/2zy = 1 m

Activity 2.2

oRSC V

0002.0832.04.67 V

m/s869.0V


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AV Q

869.06Q

/sm217.5 3Q

gDV Fr

181.9

869.0Fr

flowlsubcritica277.0Fr


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Chezy resistance factor C

The following two equations can be used to determine Chezycoefficient:

1. Ganguillet-Kutter

2. Bazin

Rn

S

nSC

o

o

00155.0231

100155.023

R

mC

1

87

n = Kutter coefficient

m = Bazin coefficient


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Table 2.1a Values of Manning roughness coefficient n

Surface characteristics Range of n

(a) Lined channels with straight alignment

Concrete

i. formed, no finish 0.013 - 0.017

ii. trowel finish 0.011 - 0.015

iii. float finish 0.013 - 0.015

iv. gunite, good section 0.016 - 0.019

v. gunite, wavy section 0.018 - 0.022

Concrete bottom, float finish, sides as indicated

i. dressed stone in mortar 0.015 - 0.017

ii. random stone in mortar 0.017 - 0.020

iii. cement rubble masonry 0.020 - 0.025iv. cement rubble masonry, plastered 0.016 - 0.020

v. dry rubble (rip-rap) 0.020 - 0.030

Tile 0.016 - 0.018

Brick 0.014 - 0.017


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Table 2.1b Values of Manning roughness coefficient n


Sewers (concrete, asbestos-cement, vitrified-clay

pipes)

0.012 - 0.015

Asphalt

i. smooth 0.013

ii. rough 0.016

Concrete lined, excavated rock

i. good section 0.017 - 0.020

ii. irregular section 0.022 - 0.027

Laboratory flumes-smooth metal bed, glass or

perspex sides

0.009 - 0.010


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Manning roughness coefficient n

= 0.020 - 0.022


= 0.020 - 0.022


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Manning roughness coefficient n = 0.022 - 0.024


= 0.020


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(b) Unlined, non-erodible channels

Earth, straight and uniform

i. clean, recently completed 0.016 - 0.020

ii. clean, after weathering 0.018 - 0.025

iii. gravel, uniform section, clean 0.022 - 0.030

iv. with short grass, few weeds 0.022 - 0.033

Channels with weeds and brush, uncut

i. dense weeds, high as flow depth 0.050 - 0.120

ii. clean bottom, brush on sides 0.040 - 0.080

iii. dense weeds or aquatic plants in deep

channels

0.030 - 0.035

iv. grass, some weeds 0.025 - 0.033

Rock 0.025 - 0.045

Table 2.1c Values of Manning roughness coefficient n


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(c) Natural channels

Smooth natural earth channels, free from growth,

little curvature

0.020

Earth channels, considerably covered with small

growth

0.035

Mountain streams in clean loose cobbles, rivers

with variable section with some vegetation on thebanks

0.040 - 0.050

Rivers with fairly straight alignment, obstructed

by small trees, very little under brush

0.060 - 0.075

Rivers with irregular alignment and cross-section,

covered with growth of virgin timber and

occasional patches of bushes and small trees

0.125

Table 2.1d Values of Manning roughness coefficient n


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Manning roughness coefficient n = 0.11

Manning roughness coefficient n = 0.20


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Grassed swale

Table 2.2 Values of Manning

roughness coefficient for

grassed swale

Surface

cover

Manning n

Short grass 0.030 - 0.035

Tall grass 0.035 - 0.050


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Table 2.3 Proposed values of Bazin coefficient m

Description of channel Bazin coefficient m

Very smooth cement of planed wood 0.11

Unplaned wood, concrete, or brick 0.21

Ashlar, rubble masonry, or poor brickwork 0.83

Earth channels in perfect condition 1.54

Earth channels in ordinary condition 2.36

Earth channels in rough condition 3.17


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Calculate the velocity and discharge in a trapezoidal channel having a

bottom width of 20 m, side slopes 1(H) : 2(V), and a depth of water 6

m. Given Kutter's n = 0.015 and So = 0.005.

z

y1

B

Activity 2.3

Given B = 20 m, y = 6.0 m, z = 0.5, So = 0.005 and n = 0.015

A = By + zy 2 = 138 m2,

P = B + 2y = 33.42 m,

R = A/P = 4.13 m


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R

n

S

nSC

o

o

00155.0

231

100155.023

Ganguillet-Kutter

13.4

015.0

005.0

00155.0231

015.01

005.000155.023

C

769.76C


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oRSC V Chezy velocity

005.013.4769.76 V

m/s03.11V

AV Q Discharge

03.11138Q

/sm14.1522 3Q


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Find the equivalent Bazin coefficient m for the question in Activity 2.3

and compare the Chezy coefficients obtained from Kutter n & Bazin m.

Assume that for concrete with Kutter n = 0.015, Bazin m = 0.21

R

mC 1

87

Bazin

Known A = 138 m2, P = 33.42 m, R = 4.13 m

13.4

21.01

87

C

Kutter)-Ganguillet(from76.769Bazin)(from852.78 C

Activity 2.4


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A trapezoidal channel is 10.0 m wide and has a side slope of

1.5(H) : 1(V). The bed slope is 0.0003. The channel is lined with

smooth concrete n = 0.012. Compute the mean velocity and

discharge for a depth of flow of 3.0 m.

z

y1

B

Given B = 10 m, y = 3.0 m, z = 1.5, So = 0.0003 and n = 0.012

A = By + zy 2 = 43.5 m2,

P = B + 2y = 20.817 m,

R = A/P = 2.090 m

21 z

Activity 2.5


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2

1

3

2

1oSR

nV Manning velocity

2

1

3

2

0003.0090.2012.0

1V

m/s359.2V

AV Q Discharge

359.25.43

/sm625.102 3


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In the channel of Example 2.5, find the bottom slope necessary to

carry only 50 m3/s of the discharge at a depth of 3.0 m.

Activity 2.6

Given B = 10 m, y = 3.0 m, z = 1.5 and n = 0.012

and A = 43.5 m2, P = 20.817 m, R = 2.090 m

2

1

3

21

oS ARnQ Manning discharge

2

1

3

2

09.25.43012.0

150 oS

0000712.0oS

51012.7 oS


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A triangular channel with an apex angle of 75 carries a flow of

1.2 m3/s at a depth of 0.80 m. If the bed slope is 0.009, find the

roughness coefficient C and n of the channel.

Activity 2.7

Given y = 0.80 m, So = 0.009, = 75, and Q = 1.2 m3/s

and A = zy 2 = 0.491 m2, P = 2y = 2.017 m,

R = A/P = 0.2435 m

21 z

z

y1 75

2

tan

z

2

75tan

0.767


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2

1

3

21

oS AR

n

Q Using Manning equation

2

1

3

2

009.02435.0491.01

2.1 n

0151.0n

2

1

2

1

oSCARQ Using Chezy equation

2

1

2

1

009.02435.0491.02.1 C

197.52C


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A trapezoidal channel of bottom width 25 m and side slope

2.5(H):1(V) carries a discharge of 450 m3/s with a normal depth of

3.5 m. The elevations at the beginning and end of the channel are

685 m and 650 m, respectively. Determine the length of the

channel if n = 0.02.

Given B = 25 m, z = 2.5, y o = 3.5, n = 0.02, and Q = 450 m3/s

z

y1

B

A = By + zy 2 = 118.125 m2

P = B + 2y = 43.848 m 21 z

R = A/P = 2.694 m

Activity 2.8


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2

1

3

21

oS AR

n

Q Manning equation,

2

1

3

2

694.2125.11802.0

1450 oS

00155.0oS

H

oL

zS

HL65068500155.0

m13.22601HL

Manning equation,


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2.3 Conveyance

Conveyance K of a channel section is a measure of the carrying

capacity of the channel section per unit longitudinal slope. It is

directly proportional to discharge Q.

1. Chezy formula

2. Manning formula

2

1

2

1

oSCARQ

2

1

3

21

oS ARnQ

2

1

CARK

3

21 ARnK


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Section factor Z in the Manning formula is AR2/3, which is a function of

the depth of flow.

In Manning formula 21

3

21

oS ARnQ

Therefore,2

132

oS

Qn AR

Section factor AR2/3 is normally used to compute the normal depth y o

when the discharge Q, bottom slope So and Manning roughnesscoefficient n are provided.

Computation of y o could be through either direct trial-and-error

computation, based on graph, or through provided design chart.

2.4 Section Factor


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Activity 2.9

A trapezoidal channel 5.0 m wide and having a side slope of 1.5(H) :

1(V) is laid on a slope of 0.00035. The roughness coefficient n = 0.015.Find the normal depth for a discharge of 20 m3/s through this channel.

Given B = 5.0 m, z = 1.5, So = 0.00035, n = 0.015, and Q = 20 m3/s

z

y1

B

A = By + zy 2 = 5y + 1.5y 2

P = B + 2y = 5 + 2 y 21 z 25.3

y

y y

P

AR

25.325

5.15 2


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Arranging Manning equation as a function of section factor,

2

13

2

oS

Qn AR

2

1

3

21

oS ARnQ Manning equation,

2

1

3

2

22

00035.0

015.020

25.325

5.155.15

o

oooo

y

y y y y

036.16

25.325

5.15

32

3

52

o

oo

y

y y

Therefore, y o = 1.820 m

y o (m)

32

3

52

25.325

5.15

o

oo

y

y y

By trial-and-error:

1

2

1.8

1.820

5.391

19.159

15.706

16.035


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Graphically,

036.16

25.325

5.15

3

2

3

52

o

oo

y

y y

y o (m)

32

3

52

25.325

5.15

o

oo

y

y y

1

2

1.5

1.7

5.391

19.159

11.198

14.115

1.8

1.9

15.706

17.387 0

0.5

1

1.5

2

2.5

0 5 10 15 20 25

AR 2/3

y o

( m )

y o = 1.82 m

16.036



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Design Chart is available,

Circular

Rectangular ( z = 0)

B

3

8

3

2

3

8

3

2

and

od

AR

B

AR

o

d

y

B

y and

0.2194

0.37


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036.1632

AR

3

8

3

8

32

5

036.16

B

AR2194.0

37.0B

y

537.0 y


At the x -axis,

Intersecting at z = 1.5 of trapezoidal channel gives


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Design chart for lined open drain fromUrban Stormwater Management Manual

for Malaysia (Department of Irrigation

and Drainage, 2000)


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Activity 2.10

A concrete-lined trapezoidal channel with n = 0.015 is to have a

side slope of 1(H) : 1(V). The bottom slope is to be 0.0004. Find thebottom width of the channel necessary to carry 100 m3/s of

discharge at a normal depth of 2.50 m.

z

y1

B

Given y o = 2.5 m, z = 1, So = 0.0004, n = 0.015, and Q = 100 m3/s

A = By + zy 2 = 2.5B + 6.25

P = B + 2y = B + 7.071 21 z

071.7

25.65.2

B

B

P

A

R


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Manning equation as a function of section factor,

2

13

2

oS

Qn AR

2

1

3

2

0004.0

015.0100071.725.65.225.65.2

BBB

75

071.7

25.65.2

3

2

3

5

B

B

By trial-and-error, B = 16.33 m


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Activity 2.11

Water flows uniformly at 10 m3/s in a rectangular channel with a base

width of 6.0 m, channel slope of 0.0001 and Manning's coefficient n =0.013. Using trial-and-error method, find the normal depth.

B

y

Given Q = 10 m3/s, B = 6.0 m, So = 0.0001 and n = 0.013

A = By = 6y

P = B + 2y = 6 + 2y

y

y R

3

3


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2

1

3

2

oS

Qn

AR

2

1

3

2

0001.0

013.010

3

36

o

ooy

y y

167.23

3 32

o

ooy

y y

By trial-and-error, y o = 1.942 m


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A sewer pipe of 2.0 m diameter is laid on a slope of 0.0004 with

n = 0.014. Find the depth of flow when the discharge is 2 m3

/s.

2 r

D

y o

sin228

2

D

Area A =

Perimeter P = D

Activity 2.12


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2

13

2

oS

Qn AR Manning equation:

2

13

2

0004.0

014.02 AR

3

8

3

8

3

2

2

4.1

D

AR

2205.0

3

8

3

2

D

AR

For design chart:

6.0D

y o

26.0 oy = 1.20 m

Intersecting at circular section gives


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Design Chart:

Circular

Rectangular (z = 0)

B

3

8

3

2

3

8

3

2

and

od

AR

B

AR

od

y

B

y and

0.2205

0.6


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Simplification for Wide Rectangular Channel

Wide channel: 02.0B

y o

For wide channel, is small, thereforeB

y ooy R

Or simply, oy R

Discharge per unit width

Normally used in rectangular channels.

B

Qq Discharge per unit width

Unit is m3/s/m.

yV q or


55/74

BFC21103 Hydraulics

Tan et al. ([email protected])

Water flows through a very wide channel at a rate of 2.5 m3/s/m.

The channel has a base width of 60 m, channel slope of 0.005 and

Manning's coefficient of 0.013. What is the normal depth?

Given: q = 2.5 m3/s/m, B = 60 m, So = 0.005, n = 0.013

For a wide rectangular channel, R = y

V y q oManning equation:

Activity 2.13

2

1

3

21

oo SRny q

2

1

3

51

oo Sy nq

2

1

3

5

005.0013.0

15.2 oy

m6272.0oy


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BFC21103 Hydraulics


2.5 Best Hydraulic Section (Most Effective Section)

3

21

ARnK

A non-erodible channel should be designed for the best hydraulic

efficiency.

Best hydraulic section gives minimum area for a given discharge.

Referring to the channel conveyance,

for a constant flow area A, the conveyance increases with increase

in hydraulic radius R or decrease in the wetted perimeter P.

Simply, Qmax, Rmax and P min gives best hydraulic section.

Pmin - reduces construction cost (less lining material), and

- reduces friction force.


57/74

BFC21103 Hydraulics


Cross

section

Side

slope zArea A

Wetted

perimeter P

Hydraulic

radius R

Top width

T

Hydraulic

depth D

Section

factor Z

Trapezoid

Rectangle - 2y

2

4y 2y y

Triangle 1 y 2 2y

Semicircle - 2y

Parabola -

Table. Best hydraulic sections

23y

2

2y

2

3

24y

y 32

y 22

y

y 3

28

2

y

2

y

4

2y

2

y

2

y

y 3

34

y 22

y 4

3

2

y

y 4

y 3

2

5.2

2

3y

5.22y

5.2

2

2y

5.2

4y

5.2

9

38y

3

1


58/74

BFC21103 Hydraulics


What is the best hydraulic section for a rectangular channel?

B

y

By AFor a rectangular channel,

y BP 2

Let's first assume A to be constant:

y y AP 2

22 y

A

dy

dP

For best hydraulic section 0d

d

y

P

Activity 2.14


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BFC21103 Hydraulics


For best hydraulic section 022

ey

A

22 ey A

2

2 ee y By ey B 2

ey BP 2ee y y P 22

ey P 4

P

A

R

e

e

y

y R

4

2 2

2

ey

R


60/74

BFC21103 Hydraulics


Show that the best hydraulic trapezoidal section is one-half

of a hexagon.

60 1

3

1z

For a trapezoid,

2zy By A

212 zy BP

Activity 2.15

Let's first assume A and z to be constant:

For best hydraulic section 0d

d

y

P

zy y

AB


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BFC21103 Hydraulics


22 12 zzy A

dy dP

212 zy zy

y

AP Substituting B

For best hydraulic section 012 2

2 zz

y

A

e 22

12 ey zz A

212 zy zy

y

AP And,

zzy P e 2122

P

AR Therefore,

zzy zzy R

e

e

2

22

12212

2

ey R


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BFC21103 Hydraulics


If z is allowed to vary,2212 ey zz A

zz

A

y e 212

zzy P e 2122Substitute into P,

zzzz A

P

2

2 12122

zz AP 2122

0d

d

z

P

3

1ez

When


63/74

BFC21103 Hydraulics


zzy P e 2122

2

12 zy BP

212 zy PB

ey B3

2

ey P 32

2212 ey zz A

23 ey A

3

1ezWhen ,


64/74

BFC21103 Hydraulics


A slightly rough brick-lined trapezoidal channel carrying a

discharge of 25.0 m3

/s is to have a longitudinal slope of 0.0004.Analyse the proportions of

(a) an efficient trapezoidal channel section having a side of

1.5(H) : 1(V),

(b) the most efficient-channel section of trapezoidal shape.

Rough brick-lined gives Manning roughness n = 0.017

Activity 2.16


65/74

BFC21103 Hydraulics


(a) Fixed side slopes of 1.5(H) : 1(V),

For best hydraulic section2212 ey zz A

2

ey R

21056.2 ey A

From Manning equation, 21

3

21

oS AR

n

Q

2

13

2

20004.0

21056.2

017.0

125

eey

y

m8298.2ey 2

zy By A

e

e

e y y

y B 5.1

1056.2 2

m7137.1B 1.52.830 m1

1.714 m

and


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BFC21103 Hydraulics


(b) If the side slope is not fixed, the side slope and other channel

characteristics for most-efficient trapezoidal section are

31ez

ey B3

2

23 ey A 2

ey R

5774.0ez

m045.3ey

m516.3eB

From Manning equation, 21

321oS AR

nQ

2

13

2

20004.0

21056.2

017.0

125

eey

y

0.5774

3.045 m1

3.516 m


67/74

BFC21103 Hydraulics


2.6 Channels of Compound Sections

Compound sections channel - channels that are composed of several

distinct subsections with each subsection different in roughness fromothers.

Manning equation is applied separately to each subsection to

determine the mean velocity.

n

i

i i AV Q1

A

SK

V

o

n

i

i 2

1

1

Or

A i i 2 17


68/74

BFC21103 Hydraulics


[Final exam question, Semester I, Session 2013/2014]

A composite channel as shown is designed to convey 19.8 m3/s of

water. The channel on a longitudinal slope So = 1:2000 is to belined with concrete (n = 0.017). Determine the normal depth of

flow based on graphical method.

4 m

3

23 m

Activity 2.17

A i #2


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BFC21103 Hydraulics


Assignment #2

Q1. [Final Exam Sem II, Session 2008/2009]

(a) What is conveyance factor K ?

(b) Figure Q1(b) shows a compound channel and its dimensions.

The channel has bottom slope of 0.0036 and side slope of 1.5(H)

: 0.75(V). Determine the value of Chezy resistance coefficient C

and velocity of flow if flowrate is 10 m3

/s.

1.5 m

0.2 m

0.5 m

Figure Q1(b)


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Q1. (c) A very wide rectangular channel has a slope of 0.0004 and

Manning n = 0.02. If 2.54 m3/s/m flow is to be conveyed in this

channel, estimate the normal depth.

(d) A trapezoidal channel is to carry 18 m3/s of flowrate on a

bottom slope of 0.0009. Given that Manning's n is 0.026 and the

sides of channel are inclined 63.44° to the vertical, determine

the bottom width, depth and velocity for the best hydraulic

section.


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Q2. [Final Exam Sem I, Session 2010/2011]

(a) Utilizing the concept of section factor, prove that the section in

Figure Q2(a) gives

when the discharge of the uniform flow is 33.6 m3/s, bed slope

So = 0.001 and Manning coefficient n = 0.015.

94.1512.8

1058.41058.4

3

22

2

y

y y y y Z

y o

y o

2y o

45

60 10 m

Figure Q2(a)

2

oy


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Q2. (b) Determine the depth of flow y o of the channel if the best

hydraulic section is needed for a composite section as in Figure

Q2(b) to convey 6.5 m3/s of flow. Manning coefficient n and bed

slope are 0.015 and 0.0015, respectively.

4.5 m

y o

y 1

y 2

Figure Q2(b)

Q3 [Final Exam Sem I Session 2007/2008]


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BFC21103 Hydraulics


Q3. [Final Exam, Sem I, Session 2007/2008]

(a) Water flows at a depth of 2.5 m in a rectangular concrete

channel (n = 0.013) of width 12 m and bed slope 0.0028. Find

the velocity and rate of flow.

(b) A housing area needs a channel to convey 9.8 m3/s of runoff. A

trapezoidal channel is proposed with 3 m width and side slope

3(horizontal) : 4(vertical). If the channel is concrete-lined (n =

0.013) and bottom slope So is 1 : 2000, determine the normal

depth using graphical method.

Q4. [Final Exam, Sem I, Session 2007/2008]

(a) Prove that the most efficient cross section for triangular channel

is half of a square.

(b) A concrete-lined irrigation channel with Manning's n = 0.020 isneeded to convey 12.5 m3/s of flow. The channel has a

trapezoidal section with bottom slope So = 0.0015. Determine

the most effective size of the channel if the side slope is

restricted to 3(horizontal) : 1(vertical).

- End of Question -


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THANK YOU

hydraulics topic 2 uniform flow in open channel

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