Chapter 3 channel design

CHAPTER 3: OPEN CHANNEL DESIGN

Dr. Mohsin Siddique

Assistant Professor

1

0401544-HYDRAULIC STRUCTURES

University of Sharjah

Dept. of Civil and Env. Engg.

Learning Outcome of Today’s Lecture

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� After completing this lecture…

� The students should be able to:

� Understand the concepts of open channel design

� Learn about various method applied in open channel design

� Design open channel section for both rigid boundary and erodible conditions

Open Channel Design

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� It is the process to obtain a shape, slope and geometry of channel/canal which should not have objectionable silting and scouring.

� For example for a trapezoidal channel, it consists of determining;

� (1) depth,

� (2) bed width,

� (3) side slope and

� (4) longitudinal slope of the channel so as to produce a non-silting and

non-scouring velocity for the given discharge and sediment load.

depth

Bed width

Side slope

Free board

Open Channel Design: Channel types

4

� Types of channels based on material

� Lined channels (Rigid boundary channels)

� Unlined channels (erodible or earthen channels)

� Types of channels based on shape

� Circular channel

� Triangular channel

� Rectangular channel

� Trapezoidal channel

Open Channel Design: Free Board

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� Free board is vertical distance between the design water surface and the top of the channel bank. It is provided to account for uncertainty in design, construction, and operation of the channel.

� The US Bureau of Reclamation recommends a minimum freeboard of 0.3m (1ft) for small channels and the following formula for estimating the freeboard (FB) for larger channels

� Where y is depth of channel in m (ft) and C is coefficient which varies from 0.7 (1.2) for small channels with capacity of 0.6 m3/s (20 ft3/s) to 0.9 (1.6) for larger channels with a capacity of 85m3/s (3000 ft3/s) or greater

� Sometime additional freeboard is required at out edge of curved section due to centrifugal force.

yCFB =

Channel Design: Rigid Boundary-Rectangular

Channels

6

� In rigid channels a layer of rigid material (e.g., Concrete, bricks and stone etc) is used at the periphery of channel to reduce seepage, to increase discharge capacity and prevent erosion.

� Dimension of rectangular channels are based on most efficient rectangular section. i.e (b=2y or y=b/2) where b and y are width and depths of channel respectively.

� For trapezoidal channels, side slopes varies from 1:1 for small channels to 1.5(H):1(v) for large channels.

� These channels can be used for both subcritical and supercritical flows.

� The design if primarily based on Manning’s Equation

� Where Q is discharge, n is roughness coefficient, A is area of flow, R is hydraulic radius, So is channel bed slope, and Co is coefficient (1 for SI unit and 1.49 for U.S. customary units)

( )2/13/22/13/2 / oooo SCnQARSAR

n

CQ =⇒=

7 Ref: Open channel hydraulics by Ven Te Chow

Channel Design: Rigid Boundary Channels

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� Design Procedure:

� 1. Select a value of roughness coefficient ,n, and bottom slope, So, for the flow surface

� 2. Compute section factor from AR2/3 = nQ/(CoSo0.5),

� 3. Determine the channel dimensions and the flow depth for which AR2/3 is equal to the value determined in step 2. For example, for a trapezoidal section, select a value for the side slope, z, and compute several different ratios of bottom width Bo and flow depth y for which AR2/3 is equal to that determined in step 2. Select a ratio Bo/y that gives a cross section near to the best hydraulic section

� 4. Check that the minimum velocity is not less than that required to carry the sediment to prevent silting.

� 5. Add a suitable amount of freeboard.

� 6. Make a sketch providing all the dimensions.

Geometric relation for most efficient

Section

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� Rectangular

� Trapezoidal 1Sy2Sy2b1Sy2

Sy2b 22+=+⇒+=

+

yb 2=

Depth=y

Bed width=b

Side slope=S

Free board

Depth=y

Bed width=b

Free board

Channel Design: Rigid Boundary

Channels

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� Example: Design a trapezoidal channel to carry a discharge of 10 m3/s. The channel will be excavated through rock by blasting. The topography in the area is such that a bottom slope of 1 in 4000 will be suitable.

� Solution: For the blasted rock surface, n = 0.030 and let us select a value for the side slope, Z, as 1 horizontal to 4 vertical. The substitution of these values into the Manning equation yields

yB

yyB

ZyB

Z

o

o

o

56.1

)25.0(2125.02

21Zy2

1Zy2y2B

2

2

2

o

=

−+=

−+=

+=+

( )

( )

yy

y

P

AR

yyyyBP

yyyyyyBA

o

o

495.066.3

81.1

66.306.256.114/12

81.125.056.1)4

1(

2

2

2

===

=+=−+=

=+=+=

my

y

yy

AR

72.2

97.1831.1

97.18)495.0(81.1

97.18

67.2

3/22

3/2

=

=

=

=


Channels

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� Solving this equation for y, we get

y = 2.72 m.

Then, Bo = 1.56×2.72 = 4.24m.

� Based on Eq. of Freeboard,

FB =0.8 × (2.72)0.5 = 1.32m.

Therefore, total depth = 2.72 + 1.32 = 4.04 =4.05 m.

� The flow area for a flow depth of 2.72 m is 13.31 m2.

Therefore, the flow velocity = 10/13.31 = 0.75m/s

Froude No. = V/(gy)0.5=0.75/(9.81*2.72)0.5=0.14


Channels

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Make a clear sketch of channel showing all dimensions

Depth=

Bed width=

Side slope=

Free board=

Channel Design: Erodible Channels

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� If the channel bottom or sides are erodible, then the design requires that the channel size and bottom slope are selected so that channel is not eroded.

� Erodible channels are designed for subcritical flow conditions with value of Froude number less than 1.

� A trapezoidal channel section is usually used for erodible channels. To design these channels, first an appropriate value for the side slope is elected so that the sides are stable under all conditions.

Basic Definitions

� Froude’s Number (FN)

� It is the ratio of inertial forces to gravitational forces.

� For a rectangular channel it may be written as

� FN= 1 Critical Flow

> 1 Super-Critical Flow

< 1 Sub-Critical Flow

gy

VNF =

William Froude (1810-79)

Born in England and engaged

in shipbuilding. In his sixties

started the study of ship

resistance, building a boat

testing pool (approximately 75

m long) near his home. After his

death, this study was continued

by his son, Robert Edmund

Froude (1846-1924). For

similarity under conditions of

inertial and gravitational forces,

the non-dimensional number

used carries his name.

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Channel Design: Erodible Channels

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� There are many method available but the following two methods are discussed here.

� Permissible velocity method and

� Tractive force method.

Channel Design: Erodible Channels-

Permissible velocity Method

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� PermissibleVelocity Method

� In permissible velocity method, channel size is selected such that mean flowvelocity for design discharge under uniform flow conditions is less thanpermissible velocity.

� Permissible Velocity is defined as the mean velocity at or below whichbottom and sides of channels are not eroded.

� Permissible velocity depends upon:

� Type of soil

� Size of particles

� Depth of flow

� Curvature of channel



17



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� Design Procedure:

� 1. For the specified material, select value of n, side slope, z, the permissible velocity, V, (from Tables).

� 2. Determine the required hydraulic radius, R, from Manning formula, and the required flow area, A, from the continuity equation, A = Q/V.

� 3. Compute the wetted perimeter, P = A/R.

� 4. Determine the channel bottom width, Bo, and the flow depth, y, for which the flow area A is equal to that computed in step 2 and the wetted perimeter, P, is equal to that computed in step 3.

� 5. Add a suitable value for the freeboard.

� 6. Make a sketch providing all the dimensions.

( )212 zyBP

yzyBA

o

o

++=

+=



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� Design a channel to carry a flow of 6.91 m3/s. The channel will be excavated through stiff clay at a channel bottom slope of 0.00318.

� Solution:

� For stiff clay, n = 0.025, suggested side slope, z = 1 : 1 (from Table ), and the permissible flow velocity (from Table ) is 1.8 m/s.

� Hence,

A = Q/V=6.91/1.8 = 3.83 m2

� Substituting values for V, n, and So into Manning equation,

� and solving for R, we get R = 0.713 m. Hence,

P = A/R=3.83/0.713= 5.37m

2/13/2

oo SAR

n

CQ =



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� Substitution into expressions for P and A and equating them to the values computed above, we obtain

� Eliminating Bo from these equations yields

� Solution of above quadratic equation yields y=1.22m and hence Bo=1.9m

� Freeboard (FB)=0.8x(1.22)0.5=0.99

� Total depth of channel section=1.22+0.99=2.21m

( ) ( )

yByB

yyByyB

oo

oo

83.211237.5

.183.3

2 +=++=

+=+=

083.337.583.1 2 =+− yy



21


Depth=

Bed width=

Side slope=

Free board=



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� Problem: A grass-lined drainage channel is to carry a discharge of 2000 cfsat a maximum velocity of 4.0 fps. The side slopes of the channel will be 4:1 and the longitudinal slope of the channel will be 0.001. Design the channel for Manning’s n values of 0.03 and 0.035.

� Answer: for n=0.03, y=4.9ft, Bo=83ft and FB=3.1ft

� for n=0.035, y=7.7ft, Bo=34.2ft and FB=3.9ft


Tractive Force Method

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� Tractive Force Method

� Scour and erosion process can be viewed in rational way by consideringforces acting on particles lying on channel bottom or sides.The channel is eroded if resultant of forces tending to move particles isgreater than resultant of forces resisting motion. This concept is referred astractive force approach.

� Tractive Force

� The force exerted by flowing water on bottom and sides of channel iscalled tractive force. In uniform flow, this force is equal to component ofweight acting in direction of flow and is given by

ooo ySRS γγτ ==



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� CriticalTractive Force

� The force at which channel material begins to move from stationerycondition is called critical tractive force.

� Distribution ofTractive Force

� Distribution of tractive force or shear stress over channel perimeter is notuniform. For trapezoidal channels, unit tractive force at channel bottommay be assumed equal to (γ y So) and at channel sides equal to 0.76 γ y So

� Reduction Factor for Channel Sides

� Reduction factor (tractive force ratio) for critical tractive force on channelsides is:

� i.e K=Permissible Tractive force on side slope/Critical Tractiveforce



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� Effect of angle of repose should be considered only for coarse non cohesive materials and can be neglected for fine cohesive materials.

� Critical shear stress for cohesive and non cohesive materials is given in the figures on next slides. These values are for straight channels and should be reduced for sinuous channels as below:

� Slightly sinuous channels = 10%

� Moderately sinuous channels = 25%

� Highly sinuous channels = 40%

Basic Definitions

Channel classification

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Straight Channel

Braided Channel

Meandering Channel

Sinuosity is ratio of actual path length to shortest path length



27 Angles of repose for non-cohesive material (After U.S. Bureau of Reclamation)



28

Permissible shear stress for noncohesivematerials (After U.S. Bureau of Reclamation)



29Permissible shear stress for cohesive materials (After U.S. Bureau of Reclamation)



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� Example: Design a straight trapezoidal channel for a design discharge of 10 m3/s. The bottom slope is 0.00025 and the channel is excavated through fine gravel having particle size of 8 mm. Assume the particles are moderately rounded and the water carries fine sediment at low concentrations.

� Solution:

� Q = 10 m3/s;

� So = 0.00025;

� Material: Fine gravel, moderately rounded; and

� Particle size = 8 mm.

� Determine: Bo = ? And Flow depth, y = ?

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� For fine gravel,

n = 0.024 and

z = 1V:3H.

� Therefore,

� θ = tan−1 (1/3) = 18.4o

� Practical size=8/25.4=0.315 in

� From Fig., φ = 24o

� Hence,

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� Critical shear stress from Fig = 0.15 lbs/ft2 = 7.18 N/m2.

� Since the channel is straight, we do not have to make a correction for the alignment.

� The permissible shear stress for the channel side is

� =7.18× 0.63 = 4.52 N/m2.

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� Now, the unit tractive force on the side = 0.76γySo

= 0.76×999×9.81y×0.00025

= 1.862y

� By equating the unit tractive force to the permissible stress, we obtain

1.862y = 4.52

Or y = 2.43m

� The channel bottom width, Bo, needed to carry 10 m3/s may be determined from the Mannings’ equation

( )( )

QSyzB

yzyByzyB

n

QSARn

o

oo =

++

++

=

0

3/2

2

0

3/2

12

1

1

34

� By substituting n = 0.024, z = 3, y = 2.43, So = 0.00025, and Q = 10 m3/s, and solving for Bo, we obtain Bo = 8.24m

� Free Board, FB, = 0.8 × (2.43)0.5 = 1.40 m.

� For a selected freeboard of 1.4 m, the depth of section = 2.43 + 1.4 = 3.82 m.

� For ease of construction, select a bottom width, Bo = 8.25 m.

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Depth=

Bed width=

Side slope=

Free board=

Thank you

� Questions….

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Engineering

Chapter 3 channel design