cve 372 hydromechanics - 3. open channel flow 2users.metu.edu.tr/bertug/cve372/cve 372...

CVE 372 Hydromechanics 1/68

Dr. Bertuğ AkıntuğDepartment of Civil EngineeringMiddle East Technical University

Northern Cyprus Campus

CVE 372CVE 372HYDROMECHANICSHYDROMECHANICS

OPEN CHANNEL FLOWOPEN CHANNEL FLOWIIII

2. OPEN CHANNEL FLOW2. OPEN CHANNEL FLOW


Overview

3.4 Rapidly Varied Flow3.4.1 Specific Force and Conjugate Depth3.4.2 Hydraulic Jump

3.5 Gradually Varied Flow3.5.1 General Equation of Gradually Varied Flow3.5.2 Types of Slopes3.5.3 Longitudinal Flow Profiles

3.6 Design of Open Channels for Uniform Flow3.3.1 Hydraulic Efficiency of Cross-sections3.3.2 Design of Nonerodible Channels3.3.3 Design of Erodible Channels



Rapidly Varied Flow

3.4.1 Specific Force and Conjugate Depth

Specific Force

gAQAyF

2+=

Specific force for any channel section is given by



Rapidly Varied Flow

3.4.1 Specific Force and Conjugate Depth

Specific Energy Curve Channel Cross-Section Specific Force Curve

At Critical Depth:

,0=dydF ,T

dydA

= ,1=rF ,13

2=

gATQ

and

gAQAyF

2+=



Rapidly Varied Flow

3.4.1 Specific Force and Conjugate DepthSpecific force becomes minimum when the flow is critical.Lower limb AC corresponds to supercritical flow.Upper limb BC corresponds to subcritical flow.For a given F and Q, there are two possible flow regimes represented by the depths y1 and y2 which are called conjugate or sequent depths.For rectangular channels, the specific force can be written as specific force per unit width as gy

qybF 2

2

21

+=

gAQAyF

2+=



Rapidly Varied Flow

3.4.2 Hydraulic JumpHydraulic jump is a rapidly varied flow in which flow changes abruptly from supercritical flow to a subcritical flow accompanied by considerable turbulence and energy loss.

LhEE += 21

2

2

221

2

11

21

gAQAy

gAQAy

FF

+=+

=

( )18121 2

11

2 −+= rFyy

For rectangular channels:

( )21

312

4 yyyyhL

−=



Rapidly Varied Flow

3.4.2 Hydraulic JumpFor a simple jump, since F1=F2, y1 and y2 are conjugate (sequent) depths.



Rapidly Varied Flow

3.4.2 Hydraulic JumpPractical Applications of Hydraulic Jump

To dissipate energyTo recover head or rise the water levelTo increase weight on apronTo mix chemicals used for water purificationTo aerate water

Types of Hydraulic Jumps:



Rapidly Varied Flow

3.4.2 Hydraulic Jump



Rapidly Varied Flow




Rapidly Varied Flow

3.4.2 Hydraulic JumpExample 9: Determine the type of hydraulic jump for the information

given in the figure below.

(1) (2)

T

Q=? m3/s y1=0.8 m

y2=1.5 m y 1

1



Rapidly Varied Flow

3.4.2 Hydraulic JumpExample 10: A hydraulic jump take place in a rectangular channel of

10 m wide. The initial depth and the velocity of the jump are 0.6 m and 10.67 m/s, respectively. If the depth of flow on the step is 2.44 m, computea) the step height, ∆zb) the force on the step.

(1) (2) (3)

b = 10 m

V1=10.67 m/s y1=0.6 m ∆z

y3=2.44 m

Fb



Rapidly Varied Flow

3.4.2 Hydraulic JumpExample 11: Water is flowing in a rectangular channel of width of 1 m

shown in the figure below. a) compute y2, ∆z, and y4.b) express the alternative and conjugate depthsc) find the force exerted by the flow on the step by writing themomentum equations between section (1) and (3) and (1) and (2), and discuss the reason for different answer.

(1) (2) (3) b = 1m

y1=1.5 m

∆z=? y3=0.3 m

(4)

y4=?y2=?



Overview






Gradually Varied Flow

3.5.1 General Equation of Gradually Varied Flow

0t

property)any(

0

=∂

∂

=∂∂xQ

Only y, z, and v changes along x




3.5.2 Types of Slopes




3.5.3 Longitudinal Flow Profile





Solution of GVF Equations1. Direct Integration Method2. Graphical Integration Method3. Numerical Step Methods

3.1. The Direct Step Method3.2. The Standard Step Method




3.5.3 Longitudinal Flow ProfileThe Direct Step Method

Consider a short channel reach of length ∆x. Equating the total heads at section 1 and section 2

xSEExS f ∆+=+∆ 210




3.5.3 Longitudinal Flow ProfileThe Direct Step Method



Rapidly Varied Flow

3.5.3 Longitudinal Flow ProfileExample 12: Determine the water surface profile for the following

rectangular channel with b = 3 mater, n=0.0125, and Q=10 m3/s

b = 3 mS0=0.0150

S0=0.0016




3.5.3 Longitudinal Flow ProfileExample 13:

A straight prismatic irrigation channel has a trapezoidal cross-section with bottom width b = 6 m and equal side slopes of 2 horizontal and 1 vertical. The bottom slope and the Manning’s roughness coefficient are S0 = 0.0002 and n = 0.014, respectively. At a certain point regulation works produce a depth of 1.80 m in the channel when the discharge is 9.81 m3/s. Using the direct step method find the distance between y = 1.80 m and y= 1.50 m in three steps.




3.5.3 Longitudinal Flow ProfileExample 14:

A reservoir is emptied through a sluice gate yielding a smallest depth of 0.40 m at vena-contracta. The channel bottom slope is S0=0.0100 with n = 0.02 while q = 1.793 m3/s/m. Determine the water depth at a point 60 m downstream (horizontally) of vena-contracta in two steps of equal distances by the standard step method. Assume a wide rectangular channel and neglect the eddy lossess.



Overview






Design of Open Channels for Uniform Flow

3.6.1 Hydraulic Efficiency of Cross-sections





Example 15: Between two open channel having rectangular and trapezoidal sections as shown, determine which section is hydraulically better. Assume both channels have the same roughness and slope.




3.6.1 Hydraulic Efficiency of Cross-sectionsRecommended Channel Cross-sections




3.6.2 Design of Nonerodible Channels




3.6.2 Design of Nonerodible ChannelsExample 16

A trapezoidal channel carrying 11.5 m3/s clear water is build with nonerodible (concrete) channel having a slope of 0.0016 and n=0.025. Proportion the section dimensions.

Example 17Design the channel given in Example 16 by using experience curve and take z=1.5.




3.6.3 Design of Erodible ChannelsChannels used as irrigation and drainage canals, are unlined for cost reasons and they must be so proportioned as to permit neither silting nor scouring in objectionable quantity.There are two methods of approach to the proper design of erodible channels:

Method of maximum permissible velocityMethod of permissible tractive force




3.6.3 Design of Erodible ChannelsMethod of Maximum Permissible Triactive Force

Average unit tractive forceFor a uniform flow the average unit tractive force acting on the wetted perimeter of the channel section:

τo = γRSo




3.6.3 Design of Erodible ChannelsMethod of Maximum Permissible Triactive ForceUnit tractive force

A typical distribution of unit tractive force in a trapezoidal channel is shown below.

Distribution of unit tractive force in a trapezoidal channel with b=4y




3.6.3 Design of Erodible ChannelsMethod of Maximum Permissible Tractive ForceMaximum unit tractive force

Following figures show the maximum unit tractive forces on the sides and bottom of various channel sections.

Figure 3: Maximum unit tractive forces in terms of γyS0




3.6.3 Design of Erodible ChannelsMethod of Maximum Permissible Triactive ForcePermissible tractive force

The permissible tractive force is the maximum tractiveforce that will not cause serious erosion of the material forming the channel bed on a level surface. This unit tractive force is determined by laboratory experiments and the value thus obtained is known as the critical tractive force.




3.6.3 Design of Erodible ChannelsMethod of Maximum Permissible Tractive ForcePermissible tractive force

For noncohesive materials the permissible tractive force is a function of the average particle diameter.

Figure 4: Unit tractive forces for canals in noncohesive materials




3.6.3 Design of Erodible ChannelsMethod of Maximum Permissible Triactive ForcePermissible tractive force

For cohesive materials triactiveforce is function of the void ratio

Figure 5: Unit tractive forces for canals in cohesive materials




3.6.3 Design of Erodible ChannelsMethod of Maximum Permissible Triactive ForceMotion of the particle on the sideaτs = the tractive force on

the side slopeWs.sinφ = the gravity force

componentwhere a: the effective area of the particle, τ : the unit tractive force on the side of the channel, Ws : the submerged weight of the particle.

: the resultant force

Ws cosφ tanθ = the resistance to motion of the particle tanθ = the coefficient of friction (θ is the angle of repose)

( ) ( )[ ] 2/122sin ss aW τφ +




3.6.3 Design of Erodible ChannelsMethod of Maximum Permissible Triactive ForceMotion of the particle on the side (continued)

2222sss asinWtancosW τ+φ=θφ

θφ

−θφ=τ 2

21

tantantancos

aWs

s




3.6.3 Design of Erodible ChannelsMethod of Maximum Permissible Triactive ForceMotion of the particle on the bed

= unit tractive force on the bottom.

The ratio of τs to τb is called the tractive force ratio.

b ,tan ττθ bs aW =

θτ tanaWs

b =

θφφ

ττ

2

2

tantan1cos −==

b

sKθφ

2

2

sinsin1−=K




3.6.3 Design of Erodible Channels

Design Procedure:Given: Design discharge Q, type of soil and

channel bed slope So;1. Side slope z, roughness coefficient n,

and angle of repose θ are selected from the figure.

Figure 6: Angle of repose of noncohesive materials




3.6.3 Design of Erodible Channels

Design Procedure:2. The permissible tractive force,τp, is

determined according to cohessiveness(Noncohesive or Cohesive)




3.6.3 Design of Erodible ChannelsDesign Procedure:3. Any b/y ratio is assumed and tractive force of water;

On the channel bed, C1γySo, and On the sides of the channel, C2γySo are determined.




3.6.3 Design of Erodible ChannelsDesign Procedure:4. The value of y is determined from the following inequalities and its

smaller value is accepted.

( )

−==≤

θφτττγ 2

2

epermissibl 2 sinsin1. ppbo KySC

( ) epermissiblp 1 boySC ττγ =≤




3.6.3 Design of Erodible ChannelsDesign Procedure:5. Using the assumed b/y ratio and the computed value of y the

channel capacity (discharge) is determined by the Manning equation

6. If the computed channel capacity is different from the design discharge, a new value for b/y ratio is assumed and the procedure is repeated until the computed discharge is equal to the given design discharge.

7. Critical flow conditions are checked8. A freeboard is added on the water depth.

2/13/2oSR

nAQ =




3.6.3 Design of Erodible ChannelsNote:

For cohesive soils, only the tractive force of bottom is critical

For noncohesive soils, one must compute the value of K from its equation. On the basis of stability criteria, if:

K < 0.78 side shear controls the required depthK > 0.78 bed shear controls the required depth




3.6.3 Design of Erodible ChannelsExample 18:

Design an unlined trapezoidal canal using the tractive force approach to carry Q = 4.6 m3/s. Given are z = 2.0, n = 0.02 andSo = 0.0016. The canal material is moderately rounded noncohesive soil that 25% material by weight is larger than 25 mm.

cve 372 hydromechanics - 3. open channel flow 2users.metu.edu.tr/bertug/cve372/cve 372...

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