BFC21103 Hydraulics Chapter 2. Uniform Flow in Open Channel

BFC21103 Hydraulics

Chapter 2. Uniform Flow in Open Channel

Tan Lai Wai, Wan Afnizan & Zarina Md [email protected]

Updated: September 2014

Learning Outcomes

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

i. Understand the concept of uniform flow

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

iii. Determine the best hydraulic/effective section of open channel

BFC21103 Hydraulics Tan et al. ([email protected])


Open Channel Flow

Classification based on Time

Classification based on Space

Steady Unsteady Uniform Non-Uniform

GVF RVF


• Uniform flow is considered to be steady only, since unsteady 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.


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


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


The Klang River, Kuala Lumpur & Selangor


• In uniform flow, the normal depth yo occurs when depth of water is the same along the channel.

• Normal depth yo 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 Sf = Sw = So = S.

g

V

2

2

yo

2.1 Velocity Distribution


Natural channel

0.840.820.800.760.700.620.48

Vmax

Vmax

0.52

0.500.450.400.35

yo

0.53

Rectangular channel

Vaverage

0.6yo

Vmax

0.2yo

V

y

yo

Velocity distribution

Depends on the geometry of the channel and wetted boundary roughness


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

oSRn

V

xo

xSRV constantThus, the general uniform flow equation:

C = Chezy roughness coefficient

n = Manning roughness coefficient


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, seasonal change, and suspended material and bed load.


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

PLVkFf2

sinALFg


g

V

2

2

yo

W

Datum

A

P


sin2 ALPLVk

oALSPLVk 2

oSP

A

kV

2

2

1

2

12

1

oSRk

V

2

1

2

1

oSRCV where C = Chezy coefficient

1221 sin MMpFWp f

Since for uniform flow, 2121 and MMpp

Total force of resistance is counter-balances with the

effective component of gravity, which is acts parallel

to the channel bed.

Fr = Force of resistanceW = Weight of the fluid = ALθ = Slope angle of the bed = Specific weight of the fluidA = Cross sectional area of the channelL = 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 = PxLP = wetted perimeterL = length of the channel K = constant of proportionalityV = mean velocity of flow


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

oRSACQ

0004.03

16.731 Q

/sm 850.0 3Q

Activity 2.1


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 onFroude number, determine the state of flow.

z

y1

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

A = zy2 = 6 m2, P = 2y = 7.211 m, R = A/P = 0.832 m, D = A/T = 6/2zy = 1 m

Activity 2.2

oRSCV

0002.0832.04.67 V

m/s 869.0V


AVQ

869.06Q

/sm 217.5 3Q

gD

VFr

181.9

869.0Fr

flow lsubcritica277.0Fr


Chezy resistance factor C

The following two equations can be used to determine Chezy coefficient:

1. Ganguillet-Kutter

2. Bazin

R

n

S

nSC

o

o

00155.0

231

100155.023

R

mC

1

87

n = Kutter coefficient

m = Bazin coefficient


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 - 0.025

iv. 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


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


Manning roughness coefficient n= 0.020 - 0.022




Manning roughness coefficient n= 0.020



(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



(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 the banks

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





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


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


Calculate the velocity and discharge in a trapezoidal channel having abottom width of 20 m, side slopes 1(H) : 2(V), and a depth of water 6m. 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 + zy2 = 138 m2,

P = B + 2y = 33.42 m,

R = A/P = 4.13 m


R

n

S

nSC

o

o

00155.0

231

100155.023

Ganguillet-Kutter

13.4

015.0

005.0

00155.0231

015.0

1

005.0

00155.023

C

769.76C


oRSCV Chezy velocity

005.013.4769.76 V

m/s 03.11V

AVQDischarge

03.11138Q

/sm 14.1522 3Q


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

87Bazin

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

13.4

21.01

87

C

Kutter)-Ganguillet (from 76.769 Bazin) (from 852.78 C

Activity 2.4


A trapezoidal channel is 10.0 m wide and has a side slope of1.5(H) : 1(V). The bed slope is 0.0003. The channel is lined withsmooth concrete n = 0.012. Compute the mean velocity anddischarge 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 + zy2 = 43.5 m2,

P = B + 2y = 20.817 m,

R = A/P = 2.090 m

21 z

Activity 2.5


2

1

3

21

oSRn

V Manning velocity

2

1

3

2

0003.0090.2012.0

1V

m/s 359.2V

AVQDischarge

359.25.43

/sm 625.102 3


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

oSARn

Q Manning discharge

2

1

3

2

09.25.43012.0

150 oS

0000712.0oS

51012.7 oS


A triangular channel with an apex angle of 75 carries a flow of1.2 m3/s at a depth of 0.80 m. If the bed slope is 0.009, find theroughness 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 = zy2 = 0.491 m2, P = 2y = 2.017 m,

R = A/P = 0.2435 m

21 z

z

y1 75

2tan

z

2

75tan

0.767


2

1

3

21

oSARn

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


A trapezoidal channel of bottom width 25 m and side slope2.5(H):1(V) carries a discharge of 450 m3/s with a normal depth of3.5 m. The elevations at the beginning and end of the channel are685 m and 650 m, respectively. Determine the length of thechannel if n = 0.02.

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

z

y1

B

A = By + zy2 = 118.125 m2

P = B + 2y = 43.848 m21 z

R = A/P = 2.694 m

Activity 2.8


2

1

3

21

oSARn

Q Manning equation,

2

1

3

2

694.2125.11802.0

1450 oS

00155.0oS

H

oL

zS

HL

65068500155.0

m 13.22601HL

Manning equation,


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

oSARn

Q

2

1

CARK

3

21AR

nK

(unit K = m3/s)


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

In Manning formula 2

1

3

21

oSARn

Q

Therefore,2

13

2

oS

QnAR

Section factor AR2/3 is normally used to compute the normal depth yowhen the discharge Q, bottom slope So and Manning roughness coefficient n are provided.

Computation of yo could be through either direct trial-and-error computation, based on graph, or through provided design chart.

2.4 Section Factor


Normal depth, (yo) determination

approaches :

Trial & Error

Graph

Chart

NORMAL DEPTH CALCULATION


TRIAL & ERROR METHOD

Formula used (from conveyence factor formulae):-

o

2

1

SC

QAR If chezy coefficient

is given

If Manning

coefficient is given

…. 2.1

…. 2.2


1st

APPROACH.... Cont ‘

o

3

2

S

nQ=AR

RHS value is normally

given

BFC21103 Hydraulics Tan et al. ([email protected]

BFC21103 Hydraulics Tan et al. ([email protected]


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

z

y1

BArranging Manning equation as a function of section factor,

2

13

2

oS

QnAR

2

1

3

21

oSARn

Q

From manning equation,

Remember trapezoidal

formulae :

A = Byo + zyo2

P = B +2yo(1+z2)1/2


2

1

3

22

2

00035.0

015.020

25.325

5.155.15

o

oooo

y

yyyy

036.16

25.325

5.15

3

2

3

52

o

oo

y

yy

Therefore, yo = 1.820 m

yo (m)

32

3

52

25.325

5.15

o

oo

y

yy

By trial-and-error:

1

2

1.8

1.820

5.391

19.159

15.706

16.035

Expand LHS equation (AR) according to its shape,

[ ]o

3

2

2o

2

oo2

oo S

Qn=

z+1y2+B

zy+By zy+By

≈ 16.04

(Right Hand Side equation) OK!


GRAPHICAL METHOD

Using previous equation ( Eq. 2.1 & 2.2) yo can be solved by plotting graph (normal depth, yo

against section factor AR2/3)


2nd

APPROACH

.... Cont ‘


Graphically,

036.16

25.325

5.15

3

2

3

52

o

oo

y

yy

yo (m)

32

3

52

25.325

5.15

o

oo

y

yy

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

AR2/3

yo (

m)

yo = 1.82 m

16.036



CHART METHOD

Section factor, Zo is divided with B8/3 ( B = Channel width ) or d8/3 ( D = Channel diameter )

Limited if channel is triangular

Formula used :-

o3

8

3

8

3

2

SB

nQ

B

AR

o3

8

o3

8

o

3

2

Sd

nQ

d

AR

Rectangular &

Trapezoidal shape

Circular shape

…. 2.3

…. 2.4

3rd

APPROACH.... Cont ‘


Design Chart is available,

Circular

Rectangular (z = 0)

B

3

8

3

2

3

8

3

2

and

od

AR

B

AR

od

y

B

y and

0.2194

0.37

o38 SB

nQ


036.163

2

AR

3

8

3

8

3

2

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


Design chart for lined open drain from Urban Stormwater Management Manual for Malaysia (Department of Irrigation and Drainage, 2000)


Activity 2.10

A concrete-lined trapezoidal channel with n = 0.015 is to have aside 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 ofdischarge at a normal depth of 2.50 m.

z

y1

B

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

A = By + zy2 = 2.5B + 6.25

P = B + 2y = B + 7.07121 z

071.7

25.65.2

B

B

P

AR


Manning equation as a function of section factor,

2

13

2

oS

QnAR

2

1

3

2

0004.0

015.0100

071.7

25.65.225.65.2

B

BB

75

071.7

25.65.2

3

2

3

5

B

B

By trial-and-error, B = 16.33 m


Activity 2.11

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

yR

3

3


2

13

2

oS

QnAR

2

1

3

2

0001.0

013.010

3

36

o

oo

y

yy

167.23

3 3

2

o

oo

y

yy

By trial-and-error, yo = 1.942 m


A sewer pipe of 2.0 m diameter is laid on a slope of 0.0004 withn = 0.014. Find the depth of flow when the discharge is 2 m3/s.

2r D

yo

sin228

2

D

Area A =

Perimeter P = D

Activity 2.12


2

13

2

oS

QnAR Manning equation:

2

13

2

0004.0

014.02AR

3

8

3

8

3

2

2

4.1

D

AR

2205.03

8

3

2

D

AR

For design chart:

6.0D

yo

26.0 oy = 1.20 m

Intersecting at circular section gives


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


Simplification for Wide Rectangular Channel

Wide channel: 02.0B

yo

For wide channel, is small, thereforeB

yooyR

Or simply, oyR

Discharge per unit width

Normally used in rectangular channels.

B

Qq Discharge per unit width

Unit is m3/s/m.

yVq or

B

y

R

Note:

The bigger B, R will approach to y.

HENCE FOR A VERY WIDE

CHANNEL R = yo.


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 andManning'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

Manning equation:

Activity 2.13

o

32

S

Qn=AR

[ ][ ]o

3

2

oo S

)B.q(n=y By

o

3

5

o S

qn=y

5

3

oo

S

qny

5

3

o005.0

(2.5) (0.013)y

yo = 0.63 m

Remember for a

very wide channel ;

R = yo


2.5 Best Hydraulic Section (Most Effective Section)

3

21AR

nK

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 Pmin gives best hydraulic section.

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

- reduces friction force.


Table 2.1 Best hydraulic sections

SHAPE

AREA

(A)

WETTED

PERIMETER

(P)

HYDRAULIC

RADIUS

(R)

TOP WIDTH

(T)

Rectangular

Triangular

Trapezoidal

To sum up …. (this formula need to remember !!)

For another z or sides slope angle use this formula

to get new A value

2y2 y4

2

yy2

2y 2y2

4

2y

3y2

y2

3y22

y

3

3y4

3/1z

zy2z1y2B2

@ = 60

3

32y @

2

TB


What is the best hydraulic section for a rectangular channel?

B

y

ByAFor a rectangular channel,

yBP 2

Let's first assume A to be constant:

yy

AP 2 2

2

y

A

dy

dP

For best hydraulic section, 0d

d

y

P

Activity 2.14

(i)

From (i),y

A = B (ii)

(iii)

Substitude (ii) into (iii),

y2+y

A = P or P = Ay-1 + 2y


For best hydraulic section 022

ey

A

22 eyA

22 ee yBy

eyB 2

eyBP 2

ee yyP 22

eyP 4

P

AR

e

e

y

yR

4

2 2

2eyR


Show that the best hydraulic trapezoidal section is one-halfof a hexagon.

120

1

3

1z

For a trapezoid,

2zyByA

212 zyBP

Activity 2.15

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

For best hydraulic section 0d

d

y

P

zyy

AB

(i)

(ii)

Devide (i) with y ,

(iii)


2

212 zz

y

A

dy

dP

212 zyzyy

AP Substituting (iii) into (i),

For best hydraulic section 012 2

2 zz

y

A

e

2212 eyzzA

212 zyzyy

AP And,

zzyP e 2122

P

AR Therefore,

zzy

zzyR

e

e

2

22

122

12

2eyR


If z is allowed to vary, 2212 eyzzA

zz

Aye

212

zzyP e 2122Substitute ye into P,

zzzz

AP

2

212

122

zzAP 2122

0d

d

z

P

3

1ez

When


zzyP e 2122

212 zyBP

212 zyPB

eyB3

2

eyP 32

2212 eyzzA

23 eyA

3

1ezWhen ,


A slightly rough brick-lined trapezoidal channel carrying adischarge 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 of1.5(H) : 1(V),

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

Rough brick-lined gives Manning roughness n = 0.017

Activity 2.16


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

For best hydraulic section 2212 eyzzA 2eyR

21056.2 eyA

From Manning equation, 2

1

3

21

oSARn

Q

2

13

2

2 0004.02

1056.2017.0

125

e

e

yy

m 8298.2ey2zyByA

e

e

e yy

yB 5.1

1056.2 2

m 7137.1B 1.52.830 m1

1.714 m

and


(b) If the side slope is not fixed, the side slope and other channelcharacteristics for most-efficient trapezoidal section are

3

1ez

eyB3

2

23 eyA2eyR

5774.0ez

m 045.3ey

m 516.3eB

From Manning equation, 2

1

3

21

oSARn

Q

2

13

2

2 0004.02

1056.2017.0

125

e

e

yy

0.57743.045 m1

3.516 m


2.6 Channels of Compound Sections

Compound sections channel - channels that are composed of several distinct subsections with each subsection different in roughness from others.

Manning equation is applied separately to each subsection to determine the mean velocity.

n

i

iiAVQ1

A

SK

Vo

n

i

i2

1

1

Or


[Final exam question, Semester I, Session 2013/2014]A composite channel as shown is designed to convey 19.8 m3/s ofwater. The channel on a longitudinal slope So = 1:2000 is to belined with concrete (n = 0.017). Determine the normal depth offlow based on graphical method.

4 m

3

23 m

Activity 2.17


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 Cand velocity of flow if flowrate is 10 m3/s.

1.5 m

0.2 m

0.5 mFigure Q1(b)


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.


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

yyyyZ

yo

yo

2yo

45

6010 m

Figure Q2(a)

2oy


Q2. (b) Determine the depth of flow yo 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

yo

y1

y2

Figure Q2(b)


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 is needed 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 -

THANK YOU


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BFC21103 Hydraulics Chapter 2. Uniform Flow in Open Channel

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