Chapter 9 Uniform Flow in Open Channel

7/29/2019 Chapter 9 Uniform Flow in Open Channel

1/33an Afnizan b Wan Mohamed

KAP/FKAAS

By:-

MR WAN AFNIZAN BIN WAN MOHAMED

DEPT. OF WATER & ENVIRONMENTAL ENGINEERING

FAC. OF CIVIL & ENVIRONMENTAL ENGINEERING

e-mail: [email protected]

HYDRAULICSHYDRAULICSHYDRAULICSHYDRAULICS

(DFC 2053)(DFC 2053)(DFC 2053)(DFC 2053)

CHAPTER 9CHAPTER 9CHAPTER 9CHAPTER 9

UNIFORM FLOW IN

OPEN CHANNEL



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CONTENT

CHANNEL GEOMETRY

TYPES OF FLOW

DISCHARGE FORMULA

APPLICATION OF NORMAL

DEPTH : CHART APPROACH

BEST EFFECTIVE SECTION

STATE OF FLOW

TYPES OF FLOW

Two criteria :-

A TIME AS THE CRITERION

(i) Steady Flow

Depth of flow does not change with time

(ii) Unsteady Flow

Depth of flow changes with time

B SPACE AS THE CRITERION

(i) Uniform Flow

Depth of flow is same at every section of channel

(ii) Non Uniform Flow

Depth of flow changes along of channel section



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.... Cont.... Cont.... Cont.... Cont

TYPES OF FLOW

B SPACE AS THE CRITERION

(iii) Rapidly Varied Flow

Depth of flow changes abruptly over a comparativelyshort distance (e.g : Hydraulic jump)

(iv) Gradually Varied Flow

Depth of flow changes gradually over a comparativelylong distance

C COMBINATION OF TWO CRITERION

(i) Steady Uniform

(ii) Steady Rapidly Varied

(iii) Steady Gradually Varied


TYPES OF FLOW

C COMBINATION OF TWO CRITERION

(iv) Unsteady Rapidly Varied

(v) Unsteady Gradually Varied

Open Channel Flow

Steady Flow Unsteady Flow

Uniform Flow Non Uniform Flow

Gradually Varied FlowRapidly Varied Flow

Figure 9.1 : Types of flow in open channel

Note : Unsteady Uniform

Flow Not exist



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STATE OF FLOW

Influenced by three factors :-

Inertial Force

Viscosity Force

Gravity Force

A EFFECT OF INERTIAL & VISCOSITY FORCES

)(RNumbereynoldsRForceViscosity

ForceInertiale====

or represent as :-

====

==== vRvRRe

. 9.1

STATE OF FLOW

State of flow ( based on Reynolds Number ) :-


where ;

v = Velocity (m/s)

= Dynamic viscosity (N.s/m2) = Water density (1000 kg/m3)= Kinematic viscosity (m2/s)R = Hydraulic radius (m)

P

AR ====

. 9.2

(i) Re 500 Laminar Flow

Flow : Slow & Shallow

Slippery surface



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STATE OF FLOW

Cont .. State of flow ( based on Reynolds Number ) :-


(ii) Re 12500 Turbulent Flow

Flow : Normally occurred in open channel

Transitional Flow(iii) 500 Re 12500

STATE OF FLOW

B EFFECT OF INERTIAL & GRAVITY FORCES

)(FNumberFroudeForceGravity

ForceInertialr====

or represent as :-

. 9.3


gD

vFr ====

where ;

v = Velocity (m/s)

g = Earths gravity (m/s2)

D = Hydraulic depth (m)

T

AD ==== . 9.4



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STATE OF FLOW

State of flow ( based on Froude Number ) :-


(i) Fr = 1 Critical Flow

(ii) Fr < 1 Sub Critical Flow

Super Critical Flow(iii) Fr >1

Flow in slow condition

Flow in turbulent condition

CHANNEL GEOMETRY

Three types of channel :-

A NATURAL CHANNEL

Figure 9.2 : Natural channel



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CHANNEL GEOMETRY

B ARTIFICIAL CHANNEL

Figure 9.3 : Artificial channel


CHANNEL GEOMETRY

C PRISMATIC CHANNEL

Uniform cross section & slope at whole channel

length.

Usually artificial channel.

CHANNEL GEOMETRY ELEMENT

y = Depth of water (m)T = Top width water surface (m)B = Base width water surface (m)P = Wetted perimeter (m)A = Wetted area (m2)R = Hydraulic radius R = A/P (m)D = Hydraulic depth D = A/T (m)



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CHANNEL GEOMETRY

CHANNEL GEOMETRIC ELEMENT

T

B

Py

Figure 9.4 : Channel geometric element


CHANNEL GEOMETRY


Figure 9.5 : Channels sides slope

I SIDES SLOPE, Z

1

z

Note : If slope, = 45 z = 1( z at left & right side is same )



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CHANNEL GEOMETRY


Figure 9.5 : Channels sides slope

II CHANNEL SLOPE, So ( unit less )


CHANNEL GEOMETRY


III RELATIONSHIP BETWEEN v & Q

AVQ ==== . 9.5where ;

Q = Discharge or flow rate (m3/s)

v = Velocity (m/s)



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CHANNEL GEOMETRY

CHANNEL GEOMETRIC ELEMENTDERIVATION OF CHANNEL FORMULA

B

y

1

z

1

z

31

2

L

Figure 9.6 : Derivation of channel formula


CHANNEL GEOMETRY


From Figure 9.6 :

T = Top width water surface

zy2BT ++++====

A = Wetted area

A = Area 1 + Area 2 + Area 3

(((( )))) (((( )))) (((( )))) (((( ))))[[[[ ]]]] (((( )))) (((( ))))

2zyByA

yzy2

1yByzy

2

1A

++++====

++++++++

====



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CHANNEL GEOMETRY


Cont . From Figure 9.6 :

P = Wetted perimeter

(((( ))))

2

22

222

22

z1yL

z1yL

yzyL

zyyL

++++====

++++====

++++====

++++====


CHANNEL GEOMETRY


Cont . From Figure 9.6 :

P = Wetted perimeter

2

z1y2BP

L2BP

++++++++====

++++====

therefore ;

Note :Use this trapezoidal formula (A, T & P) to find formulae forrectangular & triangular shape.

For Rectangular z = 0 for Triangular B = 0



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.... Cont.... Cont.... Cont.... Cont CHANNEL GEOMETRY

B + 2zyBy + zy2

2zyzy2

B + 2yBBy

PTASHAPE

B

y

T

zz11 y

T

y1z

T

B

1z

d

T

y

2z1y2 ++++

2z1y2B ++++++++

)sin(8

d2

2

sind

2

d

( in radian ) ( in radian )( in angle )

To sum up .. Table 9.1 : Channels geometric elements

LetLetLetLets take as take as take as take a

break!!!break!!!break!!!break!!!



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Based on the figure given find :-

i) Top width water surface (T), wettedarea (A), wetted perimeter (P) &hydraulic radius (R).

ii) If Q = 2.4 m3/s, determine the flowstate.

iii) If inclined length (L) = 50 m, find thecost to construct this channel (Given

excavation cost = RM 3/m3

and liningcost = RM 5/m2)

EXAMPLE 9.1EXAMPLE 9.1EXAMPLE 9.1EXAMPLE 9.1

3 m

2 m

1 m

60


SOLUTION:SOLUTION:SOLUTION:SOLUTION:



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SOLUTIONSOLUTIONSOLUTIONSOLUTION

SOLUTION .... ContSOLUTION .... ContSOLUTION .... ContSOLUTION .... Cont



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Water flows 0.8 m depth inside a 1.2 mdiameter culvert. Calculate top widthwater surface (T), wetted area (A),wetted perimeter (P) and hydraulic radius(R).






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DISCHARGE FORMULA

Several discharge formula includes :

oRSACQ ==== Chezy formula

oSARn

1Q 3

2

==== Manning formula

Where ;

A = Channel area (m2)

C = Chezys coefficient (m0.5/s)

n = Mannings coefficient

R = Hydraulic radius (m)

So = Channel slope

Most popular !!

. 9.6

. 9.7



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.... Cont.... Cont.... Cont.... Cont DISCHARGE FORMULA

0.013xii) Unreap Wood

0.012xi) Reap Wood

0.012x) Steel

0.016ix) Aspalt (rough)

0.013viii) Aspalt (smooth)

0.015vii) Masonry (inside cement mortar)

0.013vi) Stone (covered)

0.017v) Concrete

0.013iv) Cement (mortar)

0.011iii) Cement (plane / smooth)

0.027 0.035ii) Earth ground (vegetation)

0.022i) Earth ground (clean)

Artificial Channel:

0.040 0.050iii) Mountain River

0.100ii) Vegetation

0.030i) Clean and Straight

Natural Channel:

MANNING VALUE,nTYPES OF CHANNEL SURFACE

Water flows inside trapezoidal channelhaving sides slope 1.5, base width 2.5 m,channel slope 0.0016 and Manningscoefficient 0.012. If normal depth insidethe channel is 3 m, what is the discharge?





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APPLICATION OF NORMAL DEPTH

Normal depth determination approaches :

Trial & Error

Graph

ChartOnly this part will be

focused in this course !!



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APPLICATION OF NORMAL DEPTH


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 :-

3

8

o3

8

3

2

BS

nQ

B

AR====

3

8

oo3

8

o

32

dS

nQ

d

AR====

Rectangular &

Trapezoidal shape

Circular shape

. 9.8

. 9.9

APPLICATION OF NORMAL DEPTH.... Cont.... Cont.... Cont.... Cont

Figure 9.7 : Determination of normal depth using chart method

38oBS

nQ====



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Water flows through rectangular channelat 10 m3/s having base width 6 m, channelslope 0.0001 and Mannings coefficient0.013. Find normal depth using chartmethod ?






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Shape that will be studied :

Rectangular

Trapezoidal

Triangular



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What

is

Effective Section



BEST EFFECTIVE SECTION Can convey Dischargemaximum (Qmax) or Hydraulic Radius maximum (Rmax) withWetted Perimeter minimum (Pmin)

How it is derived ??? P is differentiated from Manningformula.


oSARn

1Q 3

2

====

o

3

2S

P

A

n

1Q

35

====

Note :A, n & So value - fix

Attempt to get Qmax Pmin 0y

PminP ====



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Advantages of P minimum :

Reduced flow friction

Reduced construction cost

Get P equation in

y value

How to determine best effective section ???

y

P

0

y

P====

Trapezoidal

Triangular

Rectangular

TOP WIDTH

(T)

HYDRAULIC

RADIUS

(R)

WETTED

PERIMETER

(P)

AREA

(A)

SHAPE

.... Cont.... Cont.... Cont.... Cont BEST EFFECTIVE SECTION

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

2y2 y42

yy2

2

y2y2

4

2y

3y2

y2

3y22

y

3

3y4

3/1z ==== For another z or sides s lope angle use this formulato get new A valuezy2z1y2B

2++++====@ = 60



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Water flows 14.5 m3/s inside rectangularchannel having channel slope 0.002 andMannings coefficient 0.017. Find the besteffective section for this channel.






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Water flows 3.45 m3/s inside trapezoidalchannel having sides slopes angle 45,channel slope 1:1200 and Manningscoefficient 0.014. Find the best effectivesection for this channel.






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Chapter 9 Uniform Flow in Open Channel

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