Open Channel Hydraulics

(Review)

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Hydraulics

Dr. Mohsin Siddique

Outcome of Today’s Lecture

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� After completing this lecture…

� The students should be able to:

� Understand the concepts and basic equations used in open channel flow

� Determine the velocity and discharge using Chezy’s and Manning’s equation

� Understand the concept of most economical sections

Open Channel Flow

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� An open channel is the one in which stream is not complete enclosed by solid boundaries and therefore has a free surface subjected only to atmosphere pressure.

� The flow in such channels is not caused by some external head, but rather only by gravitational component along the slope of channel. Thus open channel flow is also referred to as free surface flow or gravity flow.

� Examples of open channel are

� Rivers, canals, streams, & sewerage system etc

Open Channel Flow

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Thal Canal Indus river

Comparison between open channel flow and

pipe flow

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Aspect Open Channel Pipe flow

Cause of flow Gravity force (provided by sloping bottom)

Pipes run full and flow takes place under hydraulic pressure.

Cross-sectional shape

Open channels may have any shape, e.g., triangular, rectangular, trapezoidal, parabolic or circular etc

Pipes are generally round in cross-section which is uniform along length

Surface roughness

Varies with depth of flow Varies with type of pipe material

Piezometrichead

(z+h), where h is depth of channel

(z+P/γ) where P is the pressure in pipe

Velocity distribution

Maximum velocity occurs at a little distance below the water surface. The shape of the velocity profile is dependent on the channel roughness.

The velocity distribution is symmetrical about the pipe axis. Maximum velocity occurs at the pipe center and velocity at pipe walls reduced to zero.

Types of Channels

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� Natural Channels: It is one with irregular sections of varying shapes, developing in natural way. .e.g., rivers, streams etc

� Artificial Channels: It is the one built artificially for carrying water for various purposes. e.g., canals,

� Open Channel: A channel without any cover at the top. e.g., canals, rivers streams etc

� Covered Channels: A channel having cover at the top. e.g., partially filled conduits carrying water

� Prismatic Channels: A channel with constant bed slope and cross-section along its length.

Types of flow in open channels

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� Steady and unsteady flow

� Uniform and non-uniform flow

� Laminar and Turbulent flow

� Subcritical, critical and supercritical flow

Same definition with pipe flows

� Laminar and Turbulent flow: For open channels, it is defined with Reynolds No. as;

νh

e

VRR =

ννh

e

VRVDR

4==

Remember in pipe flows

Therefore,For laminar flow: Re <= 500For transitional flow: 500 <Re< 1000For Turbulent flow: Re >= 1000

For laminar flow: Re <= 2000

Types of flow in open channels

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� Subcritical, Critical and Supercritical Flow. These are classified with Froude number.

� Froude No. (Fr). It is ratio of inertial force to gravitational force of flowing fluid. Mathematically, Froude no. is

If Fr. < 1, Flow is subcritical flowFr. = 1, Flow is critical flowFr. > 1, Flow is supercritical flow

gh

VFr =

Where, V is average velocity of flow, h is depth of flow and g is gravitational acceleration

Definitions

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� Depth of Flow: It is the vertical distance of the lowest point of a channel section(bed of the channel) from the free surface.

� Depth of Flow Section: It is depth of flow normal to bed of the channel.

� Top Width: It is the width of channel section at the free surface.

� Wetted Area: It is the cross-sectional area of the flow section of channel.

� Wetted Perimeter: It is the length of channel boundary in contact with the flowing water at any section.

� Hydraulic Radius: It is ratio of cross-sectional area of flow to wetted perimeter.

Open channel formulae for uniform flow

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� For uniform flow in open channels, following formulae are widely used

oRSCV =

2/13/21oSR

nV =

Here, V=Average flow velocityR=Hydraulic radiusSo=Channel bed slope

C= Chezy’s constant

n= Manning’s Roughness coefficient

1. Chezy’s Formula: Antoine de Chezy (1718-1798), a French bridge and hydraulic expert, proposed his formula in 1775.

2. Manning’s Formula: Rober Manning (An Irish engineer) proposed the following relation for Chezy’s coefficient C

( ) 6/1/1 RnC =

According to which Chezy’s equation can be written as

Derivation of Chezy’s formula

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� In uniform flow the cross-sectional through which flow occurs is constant along the channel and so also is the velocity. Thus y1=y2=yo and V1=V2 =V and the channel bed, water surface and energy line are parallel to one another.

According to force balance along the direction of flow; we can write,

PLALFF oτθγ =+− sin21

F1= Pressure force at section 1

F2= Pressure force at section 2

W= Weight of fluid between section 1 and 2=So= slope of channel

θ= Inclination of channel with horizontal line

τo= shearing stress

P= Wetted perimeter

L= length between sections

V= Avg. Flow velocityyo= depth of flow

ALγ

Derivation of Chezy’s formula

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θγθγθγ

τ sinsinsin

RP

A

PL

ALo ===

x

zzSo

∆

−= 21

( ) ( )x

yzyzSw

∆

+−+= 2211

( ) ( )

x

hS

x

gvyzgvyzS

L

∆=

∆

++−++=

2/2/ 222111

θsin=≈= SSS wo

For channels with So<0.1, we can safely assume that

oo RSγτ =

Therefore;

Derivation of Chezy’s formula

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τo (shearing stress) can also be expressed as (already discussed)

2

2V

C fo ρτ =

Comparing both equations of τo we get;

o

foo

f

of

RSCV

fCRSf

gRS

C

gV

RSV

C

=

===

=

4/82

2

2

Q

γρ

Where C is Chezy’s Constant whose value depend upon the type of channel surface

f

gC

8=Q

Relation b/w f and C

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� As f and C are related, the same consideration that are present for determination of friction factor, f, for pipe flows also applies here.

4/82

fCf

g

C

gC f

f

=== Q

Empirical Relations for Chezy’s Constant, C

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� Although Chezy’s equation is quite simple, the selection of a correct value of C is rather difficult. Some of the important formulae developed for Chezy’s Constant C are;

� 1. Bazin Formula: A French hydraulic engineer H. Bazin (1897) proposed the following empirical formula for C

RKC

/181

6.157

+=

R= Hydraulic RadiusK=Bazin Constant

The value of K depends upon the type of channel surface

Empirical Relations for Chezy’s Constant, C

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� 2. Kutter’s Formula: Two Swiss engineers Ganguillet and Kutterproposed following formula for determination of C

R= Hydraulic Radiusn=Manning’s roughness coefficient

� 3. Manning’s Formula:Rober Manning (An Irish engineer) proposed the following relation for Chezy’s coefficient C

2/13/21oSR

nV =

n= Manning’s Roughness coefficient

( ) 6/1/1 RnC =

The values of n depends upon nature of channel surface

BG units SI units

Empirical Relations for Chezy’s Constant, C

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Relation b/w f and n

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� Since

� Also

� It mean n and f can also be related with each other.

� Hence

4/82

fCf

g

C

gC f

f

=== Q

( ) 6/1/1 RnC =

g

fRn

8486.1 6/1=

g

fRn

8

6/1= SI

BG

Chezy’s and Manning’s Equations in SI and BG System

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� Chezy’s Equation � Manning’s Equation

2/13/21oSR

nV =

( ) 2/13/23 1/ oSAR

nsmQ =

( ) 2/13/2486.1oSAR

ncfsQ =

SI

BG

oRSCV =

oRSCAQ =

Value of C is determine from respective BG or SI Kutter’sformula.

C= Chezy’s Constant

A= Cross-sectional area of flow A= Cross-sectional area of flow

Problem-1

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� Water is flowing in a 2-m-wide rectangular, brick channel (n=0.016) at a depth of 120 cm. The bed slope is 0.0012. Estimate the flow rate using the Manning’s equation.

� Solution: First, calculate the hydraulic radius

� Manning’s equation (for SI units) provides

Problem-2

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� Compute the flow rate for a depth of 2, 4, 6 and 8ft.

( ) 2/13/2486.1oSAR

ncfsQ =

For BG units

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� Let’s consider a trapezoidal channel having bottom width, b, depth of flow, d, and side slope, S.

Trapezoidal section

h

b

s

Sh Sh

1

( ) 2222

2

22

sec

hShShA/hhShbPimeterWetted Per

ShbhAa of flowtional areCross-

++−=++==

+==

Shh

Ab

ShbhA

−=

+= 2

b+2Sh

1Sh 2 +θ

Problem-3

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� Solution

( ) 2/13/23 1/ oSAR

nsmQ =

For SI units 4

y

Problem-4

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2/13/2486.1oSR

nV =

For BG units

6

3

Problem-5

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( ) 2/13/23 1/ oSAR

nsmQ =

For SI units

321 QQQQ ++=

Problem-7

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( ) 2/13/23 1/ oSAR

nsmQ =

For SI units

Solution: (a)

Problem-7

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� (b) ( ) 2/13/23 1/ oSAR

nsmQ =For SI units

Most Economical Section

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� From Manning’s formula, we can write that

� For a given channel of slope, So, area of cross-section, A, and roughness, n, we can simplify above equation as

� It emphasis that discharge will be maximum, when Rh is maximum and for a given cross-section, Rh will be maximum if perimeter is minimum.

� Therefore, the most economical section (also called best section or most efficient section) is the one which gives maximum discharge for a given area of cross-section (say excavation for channel shape).

ohSARn

Q1

∝

PQ

P

AQRQ h

1∝⇒∝⇒∝

Most economical rectangular section

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� Let’s consider a rectangular channel as shown in figure in which width of channel is b and depth of flow is h.

b

h

2h/A2hbPPerimeter Wetted

bhAflow of area sectional-Cross

+=+==

==

h

� For most economical section, perimeter should be minimum. i.e.,

( ) ( ) 02h/A2hbP/dh

0P/dh

=+=+=

=

hdh

d

dh

dd

d

2/2

2

202

2

2

2

bhorhb

hbh

hAh

A

==

=

=⇒=+−

� Hence for most economical rectangular section, width is twice the depth of channel

Problem

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� Let’s consider a trapezoidal channel having bottom width, b, depth of flow is d, and side slope, S, as shown in figure

Most economical trapezoidal section

h

b

s

Sh Sh

1

( ) 2222

2

hSh2/AhSh2bPPerimeter Wetted

ShbhAflow of area sectional-Cross

++−=++==

+==

Shh

� For most economical section, perimeter should be minimum. i.e.,

( ) 0hSh2/A0 22 =++−⇒= Shhdh

d

dh

dP

Shh

b −=

+=

A

ShbhA 2

b+2Sh

1Sh 2 +θ

Most economical trapezoidal section

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� Hence for most economical trapezoidal section, top width is twice the length of one sloping side or half of

top width is equal to length of one sloping side

( ) 01S201S2h/A 2

2

2 =++−−⇒=++− Sh

AShh

dh

d

1Sh2Sh2b1Sh2

Sh2b

1S2Sh2b

1S2Shb

1S2Shb

1S2Shbh

1S2

22

2

22

2

2

22

2

+=+⇒+=+

+=+

+=++

⇒+=++

+=++

⇒+=+

h

h

Sh

hS

h

Sh

Sh

A

Most economical trapezoidal section

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� For given width, b, and depth, h, perimeter becomes only the function of side slope, S,. So if we estimate value of S that provide minimum P then we have;

( ) ( ) 01S2h/A0hhS2/A0 2222 =++−⇒=++−⇒= ShhdS

dShh

dS

d

dS

dP

( ) ( )

( ) ( ) SSh

ShSh

21Sh21S

01Sh2021S2

12h

22

2/1212/12

=+−⇒=+−

=++−⇒=

×++−

−−

Squaring both sides of equation, we get

3

1

3

141S 222 =⇒=⇒=+− SSS

If sloping sides make an angle θ with the horizontal than S=tanθ

oS 60

3

1tan =⇒== θθ

Thank you

� Questions….

� Feel free to contact:

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