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Flow in Open Channels

• Significance– essential for water resources development;– designs of irrigation, navigation, spillways, sewers,

culverts and drainage ditches.– Rivers and creeks management

• Aims– flow capacity, water depth– channel design

• Characteristics of open channel flows– Existence of free surface– Driven force: gravity– More complex and more challenging than pipe flow– Pressure distribution: hydrostatic distribution

(Please print the notes for yourself)

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Type and Geometry of Channels• Natural channels

– such as rivers and streams: normally irregular geometry

• Artificial channels– developed by human, such as navigation channels, power canals,

irrigation and drainage channels– regular geometric and hydraulic properties

• Energy line (EL)• Hydraulic gradient line (HGL)• Basic Eq: Bernoulli’s equation

Fundamentals

Horizontal

y1

(p/γ+z)1

Datum

HGL

ELV12/2g

V22/2g

hL1-2

(p/γ+z)2 y21 S0

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Main Contents of Open Channel Flows

Classification of Open Channel flows

Part I: Uniform Flow Part II: Non-uniform Flow(Weeks 6 & 7)

Gradually varied flow

Changing conditionstake place over a long distance

•Equations of motion

•Coefficient of friction

•Compound channels

•Section of maximum discharge

Rapidly varied flow

Changing conditions take place over a short distance and hydraulic jumps

(Week 5)

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Classification of Open Channel flows• Steady uniform flow-constant depth in time and distance• Steady non-uniform flow-depth varies with distance but not time• Unsteady flow-depth varies with both time and distance

Steady non-uniform

Constant depth

y

Steady uniform Flow

Varying depth

y

Unsteady Uniform Flow

Unsteady Flow

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In one flow, the flow type may change:

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Some commonly used terms• Depth (y): vertical distance from free surface to channel bottom• Stage (h): vertical distance of free surface from an arbitrary datum• Area (A): cross sectional area of flow normal to the flow direction• Wetted perimeter (P): length of wetted surface measured normal

to the direction of flow• Hydraulic radius (Rh): ratio of area to wetted perimeter A/P• Surface width (B): channel width at the free surface• Discharge (Q): volume of flow per unit time• Specific discharge (q): the discharge per unit width• Hydraulic mean depth: ratio of area to surface width D A Bh = /

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Geometric properties of some common prismatic channels

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Laminar and turbulent channel flowsIn channel flow, Reynolds number is defined as:

μρ /Re hchannel VR=

For laminar channel flow:

For turbulent channel flow:

500Re <channel

1000Re <channel

Note: Theoretically, a relation for friction loss in channel flow canbe developed similar to the Darcy-Weisdach formula. In practice, we always use much simpler formulae to relate looses to velocityand channel shape – as will be discussed later.

(5.1)

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Part I: Uniform Flow

• Flow depth is constant (termed as normal depth) both in time and space;

• Gravity force is balanced by friction forces

• Truly uniform flow rarely found in reality;

• Many flows can be treated as uniform flows;

• Uniform flow is considered as the base (reference) for all other types of flows;

• What do we want to know in open channel flow?

– Velocity and pressure distributions, discharge, water level …...

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1.1 Equations of motion

τ ρ θ0 PL gAL= sin

τ ρ θ ρ0 0= =gA P gR Shsin /

sin tanθ θ≈ = S0

Use the definition of the friction coefficient: fV

=τ

ρ02 8/

Vgf

R Sh=8

0

( / )f V gR Sh8 20 0ρ τ ρ= =

and let

V C R Sh= 0

Weisbach-Darcy

Chézy equation

Resistance = Driving Force

(Since θ is very small)

(5.2)

(5.3)Or

θ

L

τ 0

ρgALρgALsinθ

Channel slope S0

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Where C in Eq. (5.3) is the resistance coefficient of Chézy.

• In uniform regime, the flow depth h is defined as the normal flow depth, h = hn.

• Some common formulae for friction coefficient have been elaborated over the years:

1) coefficient of Weisbach - Darcy

2) coefficient of Chézy.

3) coefficient of Manning-Strickler

4) coefficient of friction for mobile bed

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1.2 Coefficient of friction• The Weisbach-Darcy equation (5.2)

– Useful for laminar and turbulent flow with circular cross section and standard roughness

• Chézy equation Eq. (5.3) – for truly turbulent flow (often the case);

• Accuracy of the two formulas is strongly dependent on the choice of the friction coefficient, f or C;

• Artificial and particularly natural channels have all types of form of the cross section. No parameter exists which would take care of the variability in forms;

• For open channel flows, Eq(5.3) is widely used.

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1.2 Coefficient of ChèzyFor rough turbulent flow, the formula of Chézy can be used:

V C R Sh= 0 (5.3)

The coefficient C [m1/2/s] is a dimensional expression and can be calculated using various empirical formulas. In practice, the so-called Manning (Manning-Strickler) formula is often used:

Cn

Rh=1 1 6/ (5.4)

Where n is the coefficient of Manning.

The average velocity and discharge are:

Vn

R Sh=1 2 3

01 2/ / (5.5)

Q An

R Sh=1 2 3

01 2/ / (5.6)

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Manning coefficient n

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Example 1A channel is to be built of medium-quality concrete to convey a discharge of 80 m3/s. The channel should have a trapezoidal cross section with a bottom width of 5 m and side slopes of 3. The channel slope is S0 =0.1%. Suppose the flow is uniform, calculate the flow depth using Manning coefficient.

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Example 2.The normal depth of flow in a trapezoidal concrete lined channel is 2 m. The channel has a base width of 5 m and side slopes of 1:2. The bed slope S0 is 0.001. Determine the discharge Q and the mean velocity V.

h1

m=2 b

BSolution:

B=b+2mh=5+8=13Area: A=(b+mh)h=18m2

mmhbP 9.1312 2 =++=

mPARh 29.1/ ==

smASRn

Q h /521 32/10

3/2 ==

For concrete, the Manning coefficient n=0.013

smAQV /9.218/52/ ===

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Example 3The discharge in a trapezoidal concrete lined channel is 30 m3/s. The channel has a base width of 5 m and side slopes of 1:2 while the bed slope S0 is 0.001. Determine the depth y of the flow.

y=?1

m=2 b

B

B=b+2my

Area: A=(5+2y)y

ymybP 52512 2 +=++=

PARh /=

( ) 30)001.0(525)25(25

013.011 2/1

3/22/1

03/2 =⎟

⎠

⎞⎜⎝

⎛+++==

yyyyyASR

nQ h

Using trial and error to solve for y=1.509m.

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1.3 Compound channels

•In reality, channel sections normally have different roughness. The use of (5.5) and (5.6) in such cases will lead to large errors. One procedure for this is to separate the cross section into several parts. It is assumed there is no resistance along the dashed vertical lines. The total discharge can be calculated using the following formula:

•If the entire channel has an uniform roughness, eqs. (5.5) and (5.6) can be applied as they are.

Q An

R S An

R S An

R Sh h h= + +11

12 3

01 2

22

22 3

01 2

33

32 3

01 21 1 1/ / / / / /

Where R A Ph1 1 1= / R A Ph2 2 2= / R A Ph3 3 3= /

(5.6)

A1

A2

A3

p3

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Example 4: The bed slope of a compound channel is 0.001. Other conditions are as in the figure. Find the discharge of the channel.

n=0.02

n=0.015

n=0.03

1 m

3m 2m 3m

1.5m

∑==

3

1iiQQ

2/10

3/2 SRnAQ

ini

ii =

?0.0314.54.53?0.0151.25452?0.0214.54.51QiniRniPiAii

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1.4 Section of Maximum Discharge

• The construction of a channel with a given discharge, a given slope and a given roughness will be less expensive if the cross section is the smallest.

• Take the formula of discharge:

For a constant A, if Rh is maximal or P is minimal ⇒ Q maximal.

• Amongst all geometrical forms possible, the cross section of a semi-circular form will give a Pmin for a given constant A.

•For a channel in an alluvium, one should take into account the angle of repose as well as various constraints due to construction.

•Consequently a trapezoidal form may be the most reasonable one.

3/2

3/52/1

02/1

03/2 11

PAS

nSR

nAQ h ==

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What is the best trapezoidal form?We can prove that for a trapezoidal cross-section channel, we have the following optimised dimensions:

,3 2nyA =

32 nyb =

ynyn yn

b b=2yn

L

and L = b

For a rectangular channel, we have

nyb 2=

The optimum cross-sections for some channels are given below:

(5.7)

(5.8)

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Properties of optimum open-channel sections (From Fox and McDonald)

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Example 5Find the dimensions of the most-efficient cross section for a rectangular channel that is to convey a uniform flow of 10 m3/s if the channel is lined with gunite concrete and is laid on a slope of 0.001.

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Example 6Find the dimensions of the most-efficient cross section for a trapezoidal channel that is to convey a uniform flow of 10 m3/s if the channel is lined with gunite concrete and is laid on a slope of 0.001. The side slopes of the channel cross section are 2.