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FLOWFLOW ININOPENOPEN CHANNELSCHANNELS

教授 楊文衡教授 楊文衡

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Contents 1. Introduction 1.1 Introduction channels 1.2 Types of channels 1.3 Classification of Flow 1.4 Velocity Distribution 1.5 One-Dimensional Method of Flow Analysis 1.6 Pressure Distribution 1.7 Pressure Distribution in Curvilinear Flows 1.8 Flows with Small Water-Surface Curvature

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1.9 Equation of continuity 1.10 Energy Equation

2. Energy Depth Relationship 2.1 Specific Energy 2.2 Critical Depth 2.3 Calculation of the Critical Depth 2.4 Section Factor Z 2.5 First Hydraulic Exponent M 2.6 Computations

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2.7 Transitions Reference Problems Objective Questions Appendix 2A 3. Uniform Flow 3.1 Introduction 3.2 Chezy Equation 3.3 Dracy-Weisbach Friction Factor 3.4 Mannings Formula 3.5 Other Resistance Formula 3.6 Velocity Distribution

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3.7 Shear Stress Distribution 3.8 Resistance Formula for Practical Use 3.9 Manning’s Roughness Coefficient n 3.10 Equivalent Roughness 3.11 Uniform Flow Computations 3.12 Standard Lined Canal Sections 3.13 Maximum Discharge of a Channel of the Second Kind 3.14 Hydraulically-Efficient Channel Section 3.15 The Section Hydraulic Exponent N 3.16 Compound Section 3.17 Generalised-Flow Realtion

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3.18 Design of Irrigation Canal 4. Gradually-varied Flow Theory 4.1 Introduction 4.2 Differential Equation of GVF 4.3 Classification of Flow Profile 4.4 Some Features of Flow Profiles 4.5 Control Sections 4.6 Analysis of Flow Profile 4.7 Transitional Depth

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5. Gradually-Varied Flow Computations 5.1 Introduction 5.2 Direct Integration of GVF Differential Equation 5.3 Estimation of N and M for Trapezoidal Channels 5.4 Bresse’s Solution 5.5 Channels with Considerable Variation in Hydraulic Exponents 5.6 Direct Integration for Circular Channels 5.7 Simple Numerical Solutions of GVF Problems

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5.8 Advanced Numerical Methods 5.9 Graphical Methods 5.10 Flow Profiles in Divided Channels 5.11 Role of End Conditions 6. Rapidly-varied Flow-1 - Hydraulic Jump 6.1 Introduction 6.2 The Momentum Equation for the Jump 6.3 Classification of Jumps 6.4 Characteristics of Jump in a Rectangular Channels

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6.5 Jumps in Non-Rectangular Channels 6.6 Jump on a Sloping Floor 6.7 Use of the Jump as an Energy Dissipator 6.8 Location of the Jump7. Rapidly-varied Flow-2-Flow Measurements 7.1 Introduction 7.2 Sharp-Crested Weir 7.3 Special Sharp-Crested weirs 7.4 Ogee Spillway 7.5 Broad-Crested Weir 7.6 Critical Depth Flumes

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7.7 End Depth in a Free Overfall 7.8 Sluice-Gate Flow 8. Spatially-varied Flow 8.1 Introduction 8.2 SVF with Increasing Discharge 8.3 SVF with Decreasing Discharge 8.4 Side Weir 8.5 Bottom Racks 9. Supercritical-flow Transitions 9.1 Introduction 9.2 Response to a Disturbance

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9.3 Gradual Change in the Boundary 9.4 Flow at a Corner 9.5 Wave Interactions and Reflections 9.6 Contractions 9.7 Supercritical Expansions 9.8 Stability of Supercritical Flows10. Unsteady Flows 10.1 Introduction 10.2 Gradually Varied Unsteady Flow (GVUF)

10.3 Uniformly Progressive Wave

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10.4 Numerical Methods 10.5 Rapidly-Varied Unsteady Flow11. Hydraulics of Mobile Bed Channels 11.1 Introduction 11.2 Initiation of Motion of Sediment 11.3 Bed Forms 11.4 Sediment Load 11.5 Design of Stable Channels Carrying Clear water 11.6 Regime Channels

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Chapter 1Chapter 1

IntroductionIntroduction

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1.11.1 INTRODUCTIONINTRODUCTION An open channel is a conduit in which a liquid flows wit

h a free surface. The free surface is actually an interface between the moving liquid and an overlying fluid medium and will have constant pressure. In civil engineering applications water is the most common liquid with air at atmospheric pressure as the overlying fluid. As such our attention will be chiefly focussed on the flow of water with a free surface. The prime motivating force for open channel flow is that due to gravity.

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1.2 TYPES OF CHANNELS1.2 TYPES OF CHANNELSPrismatic and Non-prismatic Channels A channel in which the cross-sectional shape and

size and also the bottom slop are constant is termed as a prismatic channel. Most of the man-made (artificial) channels are prismatic channels over long stretches. The rectangle, trapezoid, triangle and circle are some of the commonly-used shapes in man-made channels. All natural channels generally have varying cross-sections and consequently are non-prismatic.

Rigid and Mobile Boundary Channels On the basis of the nature of the boundary open

channels can be broadly classified into two types: (i) rigid channels and (ii) mobile boundary channels.

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1.3 CLASSIFICATION OF FLOWS1.3 CLASSIFICATION OF FLOWSSteady and Unsteady Flows A steady flows occurs when the flow properties,

such as the depth or discharge at a section do not change with time. As a corollary, if the depth or discharge changes with time the flow is termed unsteady.

Flood flows in rivers and rapidly-varying surges in canals are some example of unsteady flows. Unsteady flows are considerably more difficult to analysis than steady flows.

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Uniform and non-uniform Flows If the flow properties, say the depth of flow, in an

open channel remain constant along the length of channel, the flow is said to be uniform. As a corollary of this, a flow in which the flow properties vary along the channel is termed as non-uniform flow or varied flow.

A prismatic channel carrying a certain discharge with a constant velocity is an example of uniform flow (Fig. 1.1(a)).

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Flow in a non-prismatic channel and flow with varying

velocities in a prismatic channel are examples of varied flow. Varied flow can be either steady or unsteady.

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Gradually-varied and Rapidly –varied Flows If the change of depth in a varied flow is gradual

so that the curvature of streamlines is not excessive, such a flow is said to be a gradually –varied flow (GVF). The passage of a flood wave in a flood wave in a river is a case of unsteady GVF (Fig. 1.1(b)).

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A hydraulic jump occurring below a spillway or a sluice gate is an example of steady RVF. A surge , moving up a canal (Fig. 1.1(c)) and a bore traveling up a river are examples of unsteady RVF.

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Spatially-varied flow Varied flow classified as GVF and RVF assumes

that no flow is externally added to or taken out of the canal system. The volume of water in a known time interval is conserved in the channel system. In steady-varied flow the discharge is constant at all sections. However, if some flow is added to or abstracted from the system the resulting varied flow is known as a spatially varied flow (SVF).

SVF can be steady or unsteady. In the steady SVF the discharge while being steady-varies along the channel length. The flow over a bottom

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rack is an example of steady SVF (Fig 1.1(d)). The production of surface runoff due to rainfall, known as overland flows, is a typical example unsteady SVF.

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Classification Thus open channel flows are classified for purposes of identification and analysis.

Fig. 1.1(a) through (d) shows some typical examples of the above types of flows

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1.4 VELOCITY DISTRIBUTION1.4 VELOCITY DISTRIBUTION The presence of corners and boundaries in an open channel causes the velocity vectors of the flow to have components not only in the longitudinal direction but also in the lateral as well as normal direction to the flow. In a macro-analysis, one is concerned only with the major component, viz. the longitudinal component, . The other two component being small are ignored and is designated as . The distribution of in a channel is dependent on the geometry of channel. Figure 1.2(a) and (b) show isovels (contours of equal velocity) of for a natural and rectangular channel respectively.

xv

vxv v

v

25

Fig. 1.2Fig. 1.2

26

A typical velocity profile at a section in a plan normal to the direction of flow is presented in Fig. 1.2(c).

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Field observations in rivers and canals have show that the average velocity at any vertical , occurs at a level of 0.6 from the free surface, where =depth of flow. Further, it is found that

(1.1)

in which = velocity at depth of 0.2 form

the free surface, and = velocity at depth of 0.8 from the free surface. This property of the velocity distribution is commonly used in

0y0y

avv

28.02.0 vv

vav

2.0v8.0v

0y

0y

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Stream-gauging practice to determine the discharge using the area-velocity method. The surface velocity is related to the average velocity as

(1.2) where, = a coefficient with a value between 0.8

and 0.95. The proper value of depends on the channel section and has to be determined by field calibrations. Knowing , one can estimate the average velocity in an open channel by using floats and other surface velocity measuring devices.

svavv

sav kvv k

kk

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1.5 ONE-DIMENSIONAL METHOD OF 1.5 ONE-DIMENSIONAL METHOD OF FLOW ANALYSISFLOW ANALYSIS Flow properties, such as velocity and pressure gradient

open channel flow situation can be expected to have components in the longitudinal as well as in the normal directions. For purposes of obtaining engineering solutions, a majority of open channel flow problems are analysed by one-dimensional analysis where only the mean or representative properties of a cross-section are considered and their variations in the longitudinal direction analysed.

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Regarding velocity, a mean velocity for the entire cross-section is defined on the basis of the longitudinal component of the velocity as

(1.3) This velocity is used as a representative

velocity at a cross-section. The discharge past a section can then be expressed as

(1.4)

V

v

A

vdAA

V1

V

A

VAvdAQ

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Kinetic Energy The flux of the kinetic energy flowing past a

section can also be expressed in terms of . But in this case, a correction factor will be needed as the kinetic energy per unit weight /2g will not be the same as /2g averaged over the cross-section area. An express for can be obtained as follows:

For an elemental area , the flux of kinetic energy through it is equal to

2V

V2v

dA

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or for discrete values of ,

The kinetic energy per unit weight of fluid can then be written as .

2

2vvdA

mass

KE

time

mass

AV

dAv3

3

vAV

Av3

3

g

V

2

2

AVdAvA

33

22

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Momentum Similarly, the flux of momentum at a section is

also expressed in terms of and a correction factor . Considering an elemental area , the flux of momentum in the longitudinal direction through this elemental area

vvdAvelocitytime

mass

AVdAv 22

AV

Av

AV

dAvA

2

2

2

2

V dA

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Values of and The coefficients and are both unity in the

case of a uniform velocity distribution. For any other variation > > 1.0. The higher the non-uniformity of velocity distribution, the greater will be the values of the coefficients.

Reliable data on the variation of and are not available. Generally, one can assume = =1.0 when the channels are straight, prismatic and uniform or GVF takes place. In local phenomenon, it is desirable to include estimated values of these coefficients. It is the practice to assume = =1.0 when no other specific information about the coefficients is available.

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EXAMPLE 1.1 The velocity distribution in a rectangular channel of width and depth of flow was approximated as in which = a constant. Calculate the average velocity for the cross-section and correction coefficients and .

Solution Area of cross-section Average velocity

B0y ykv 1

1k

0ByA

0

00

1 yBdyv

ByV

010 10 3

21 0

ykdyyky

y

36

Kinetic energy correction factor

Momentum correction factor

35.1

3

20

3

01

0

233

1

03

0

3 00

Byyk

Bdyyk

ByV

Bdyvyy

125.1

3

20

2

01

0

21

02

0

2 00

Byyk

yBdyk

BdyV

Bdyvyy

37

Pressure In some curvilinear flows, the piezometric pressure hav

e may head non-linear variations wit depth. The piezometric head at any depth from the free surface can be expressed as

in which = elevation of the bed, =pressure head at the bed if linear variation of pressure with depth existed and =deviation from the linear pressure head variation at any depth .

ph y

hyhyZhp 10

hhZhp 10

0Z 1h

hy

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For one-dimensional analysis, a representative piezometric head for the section called effective piezometric head, is defined as

Usually hydrostatic pressure variation is considered as the reference linear variation.

eph

1

0110

1 h

ep dyhh

hZh

hhZ 10

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1.6 PRESSURE DISTRIBUTION1.6 PRESSURE DISTRIBUTION The intensity of pressure for a liquid at its free

surface is equal to that of the surrounding atmosphere. Since the atmospheric pressure is commonly taken as a reference and of value equal to zero, the free surface of the liquid is thus a surface of zero. The distribution of pressure in an open channel flow is governed by the acceleration due to gravity and other accelerations and is

In any arbitrary direction :

g

s

sas

Zp

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and in the direction normal to direction, i.e. in the direction:

(1.13) in which =pressure, =acceleration

component in the direction, = acceleration in the direction and = elevation measured above a datum.

Consider the direction along the streamline and the direction across it. The direction of the normal towards the centre of curvature is considered as positive.

naZpn

sn

p sas na

n Z

sn

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We are interested in study the pressure distribution in the -direction. The normal acceleration of any streamline at a sections is given by

(1.14) Where = velocity of flow along the streamline

of radius of curvature .

n

r

van

2

vr

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Hydrostatic Pressure Distribution The normal acceleration will be zero (i) if = 0, i.e. when there is no motion, or (ii) if , i.e. when the streamlines are

straight lines. Consider the case of no motion, i.e. the still water

case, (Fig. 1.3(a))

v

r

na

43

From Eq. (1.13), since = 0, taking in the direction and integrating

=constant=C (1.15)

At the free surface [point 1 in Fig. 1.3(a)] =0 and = , giving = . At any point at a depth below the free surface,

i.e. (1.16)

Znna

Zr

p

1p

Z1Z C 1Z A

y

yP

yZZP

A

AA

1

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Channels with Small Slope Lets us consider a channel with a very small

value of the longitudinal slop . Let . For such channels the vertical section is practically the same as the normal section. If a flow takes place in this channel with the water surface parallel to the bed, i.e. uniform flow, the streamlines will be straight lines

1000/1~sin~

45

And as such in a vertical direction [section 0-1 in Fig.1.3(b)] the normal acceleration = 0.

Following the argument of the preview paragraph, the pressure distribution at the section 0-1 will be hydrostatic. At any point at a depth below the water surface,

and = Elevation of water surface

Thus the piezometric head at any point in the channel will be equal to the water-surface elevation. The hydraulic grade line will therefore lie essentially on the water surface.

na

A y

yp

1ZZ

p

46

Channels with Large Slope Figure 1.4 shows a uniform free-surface flow in a

channel with a large value of inclination . The flow is uniform, i.e. the water is parallel to the bed. An element of length and unit width is considered at the section 0-1.

L

47

At any point at a depth measured normal to the water surface, the weight of column

and acts vertically downwards. The pressure at supports the normal component of the column . Thus

(1.17) (1.18) (1.18a) The pressure varies linearly with the depth

but the constant of proportionality is . If = normal depth of flow, the pressure on the

bed at point 0, .

yA

1A LyA ''1'AA

1A ''1 A cosLyLpA

Ap cosy/Ap cosyAp cos

h cos0 hp

48

If = vertical depth to water surface measured at the point ,then and the pressure head at point ,on the bed is given by

(1.19) The piezometric height at any point . Thus for channels with large values of the slope, the co

nventionally-defined hydraulic gradient line does not lie on the water surface.

dO

Ocosdh

20 coscos dhp

coscos 0 hZyZA

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1.7 PRESSURE DISTRIBUTION IN 1.7 PRESSURE DISTRIBUTION IN CURVILINEAR FLOWSCURVILINEAR FLOWS Figure 1.5(a)shows a curvilinear flow in a vertical

plane on an upward convex surface. For simplicity consider a section in which

the direction and direction coincide. Replacing the direction in Eq.(113)by ( ) direction,

(1.20)

201AZ

rnr

g

aZ

p

rn

50

Let us assume a simple case in which =constant. Then, the integration Eq.(l.20)yields

na

51

(1.21) in which =constant. With the boundary condition th

at at point 2 which lies on the free surface, and and

,

(1.22) Let = depth below the free surface of any point

in the section . Then for point ,

Crg

aZ

p n

2rr 0p

2ZZ

rrg

aZZ

p n 22

ZZ 2

AA yA 201

C

52

and (1.23) Equation (1.23) shows that the pressure is less than th

e pressure obtained by the hydrostatic distribution [Fig.1.5(b)].

For any normal direction OBC in Fig.1.5(a), at point , , , and for any point at a radial distance tom the origin , by Eq.(1.22)

ZZyrr 22

yg

ay

p n

C 0p 2rrc O

rrg

aZZ

p nc 2

53

but

giving

(1.24) It may be noted that when =0, Eq.(124) is the same a

s Eq.(1.18a), for the flow down a steep slope. If the curvature is convex downwards, (i.e. direction is

opposite to direction) following the argument as above, for constant the pressure at any point at a depth below the free surface in a vertical section [Fig.1.6(a)]can be shown to be

cos2 rrZZc

rrg

arr

p n 22 cos

201A

na

rZ

naA y

54

The pressure distribution in a vertical section is as shown in Fig.1.6(b).

Thus it is seen that for a curvilinear flow in a vertical plane, an additional pressure will be imposed on the hydrostatic pressure distribution. The extra a pressure will be additive if the curvature is convex downwards and subtractive if it is convex upwards.

yg

ay

p n

55

Normal Acceleration In the previous discussion on curvilinear flows,

the normal acceleration on was assumed to be constant. However, it is known that at any point in a curvilinear flow, , where

= velocity and =radius of curvature of the streamline at that point.

rvan2 v

r

56

In general, one can write the pressure distribution can then be expressed by

(1.26) This expression can be evaluated if is known. For simple analysis, the following

functional forms are used in appropriate circumstances:

(i) =constant = =mean velocity of flow (ii) ,(free-vortex model) (iii) ,(forced-vortex model)

rfv

Constdrgr

vZ

p

2

v Vrcv

crv

rfv

57

(iv) =constant= , where =radius of curvature at mid-depth.

EXAMPLE 1.2 At a section in a rectangular channel, the pressure distribution was recorded as shown in Fig.1.7. Determine the effective piezometric head for this section. Take the hydrostatic pressure distribution as the reference.

Solution = elevation of the bed of channel above the datum = depth of flow at the section OB. = piezometric head at point , depth below the fre

e surface.

na RV 2

0Z

1hA yph

R

58

yhkyZhp 12

0

hhZhp 10

ykyh 2

59

Effective piezometric head, by Eq.(1.11) is

1

01

10

1 h

ep dyhh

hZh

1

0

2

110

1 hdyyky

hhZ

22

211

0

khhZhep

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EXAMPLE 1.3 A spillway bucket has a radius of curvature R as shown in Fig. l.8 (a) Obtain an expression for the pressure distribution at a radial section of inclination to the vertical. Assume the velocity at any radial section to be uniform and the depth of now h to be constant. (b) what is the effective piezometric head for the above pressure distribution?

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Solution (a) Consider the section 012. Velocity = =

constant across 12. Depth of flow =h. From Eq.(1.26), since the curvature is convex downwards

(E.1) At point 1, , ,

V

Constdrgr

vZ

p

2

Crg

VZ

p ln

2

0p 1ZZ hRr

hRg

VZC ln

2

1

62

At any point , at radial distance from .

(E.2) But

(E.3) Equation (E. 3) represents the pressure distribution at a

ny point At point 2, . (b)Effective piezometric head, : From Eq(E2),the piezometric head at is

A r O

hR

r

g

VZZ

pln

2

1

cos1 hRrZZ

hR

r

g

VhRr

plncos

2

A

2, ppRr

pheph

63

Noting that and expressing in the form of Eq.(1.10)

where

The effective piezometric head From Eq.(1.11) is on integration

cos21 hZZ

ph

hR

r

g

VZ

AZ

php ln

2

1

hhZhp cos2

hR

r

g

Vh

ln

2

eph

R

hRep drhR

r

g

V

hhZh ln

cos

1cos

2

2

64

(E.4)

It may be noted that when and ,

hR

RRh

gh

VhZhep ln

coscos

2

2

cos

11

ln

cos

2

2

Rh

gR

RhRh

V

hZ

R 0Rh

cos2 hZhep

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1.8 FLOWS WITH SMALL WATER-1.8 FLOWS WITH SMALL WATER-SURFACE CURVATURESURFACE CURVATURE Consider a free-surface flow with a convex

upward water-surface over a horizontal bed (Fig. l .9). For this water surface, is negative. The radius of curvature of the free surface is given by

(1.27)

22 / dxhd

2

2

232

2

2

11

1

dx

hd

dxdh

dxhd

r

66

Assuming linear variation of the curvature with depth, at any point at a depth below the free surface, the radius of curvature is given by

(1.28)

A yr

h

yh

dx

hd

r 2

21

67

If the velocity at any depth is assumed to be constant and equal to the mean velocity in the section, the normal acceleration at point

is given by

(1.29) where . Taking the channel bed as the dat

um, the piezometric head at point is then by Eq. (1.26)

naV

A yhK

dx

hd

h

yhV

r

Van

2

222

222 dxhdhVK

Aph

Constdyyhg

kZ

php

68

i.e. (1.30) Using the boundary condition; at , and , leads to ,

(1.31) Equation (1.31) gives the variation of the piezometric h

ead with the depth below the free surface. Designing

0y0p hhp hC

Cy

hyg

khp

2

2

2

2yhy

g

khhp

yhhhp

2

2yhy

g

kh

69

The mean value of

The effective piezometric head at the section with the channel bed as the datum can now be expressed as

(1.32)

h

Ahdy

hh

0

1

g

Khdyyhy

gh

K A

32

2

0

2

eph

g

Khhhep 3

2

70

It may be noted that and hence is negative for convex upward curvature and positive for concave upward curvature. Substituting for , Eq. (1.32) reads as

(1.33) This equation, attributed to Boussinesq finds applicat

ion in solving problems with small departures from the hydrostatic pressure distribution due to the curvature of the water surface.

22 dxhd

K

K

2

22

3

1

dx

d

g

hVhhep

4

71

1.9 EQUATION OF CONTINUITY1.9 EQUATION OF CONTINUITY The continuity equation is a statement of the law of co

nservation of matter. In open-channel flows, since we deal with incompressible fluids, this equation is relatively simple and much more so for cases of steady flow.

Steady Flow In a steady flow the volumetric rate of flow (discharge i

n ) past various section must be the same. Thus in a varied flow, if = discharge,

= mean velocity and =area of cross-section with suffixes representing the sections to which they refer

(1.34)

sm3

QV A

...2211 AVAVVAQ

72

If the velocity distribution is given, the discharge is obtained by integration as in Eq.(1.4).

It should be kept in mind that the area element and the velocity through this area element must be perpendicular to each other.

In a steady spatially-varied flow, the discharge at various sections will not be the same. budgeting of inflows and outflows of a reach is necessary. Consider, for example, an SVF with increasing discharge as in Fig.1.10.The rate of addition of discharge = .

The discharge at any section at a distance from section 1

*qdxdQ x

73

IF =constant, and .

xdxqQQ

0 *1

xq xqQQ *1 LqQQ *12

74

Unsteady Flow In the unsteady flow of incompressible fluids, if

we consider a reach of the channel, the continuity equation states that the net discharge going out of all the boundary surfaces of the reach is equal to the rate of depletion of the storage within it.

In Fig. 1.11, if , more flow goes out than what is coming into section 1. The excess volume of outflow in a time is made good by the depletion of storage within the reach bounded by sections l and 2. As a result of this the water surface will start falling. If =distance between sections land 2,

12 QQ

t

x

75

The excess volume rate of flow in a time . If the top width of the canal at any depth

is , . The storage volume at depth . The rate of decrease of storage = .

The decrease in storage in the time

. By continuity . Or (1.36)

xx

QQQ

12

txxQQ y T TyA

xAy .

t

yxT

t

y

y

Ax

tt

yxTt

txt

yTtx

x

Q

0

t

yT

x

Q

76

Equation (1.36) is the basic equation of continuity for unsteady, open-channel flow.

77

EXAMPLE 1.4 The velocity distribution in the plane of a vertical sluice gate discharging free is shown in Fig.1.12. Calculate the discharge per unit width of the gate.

Solution The component of velocity normal to the axis

is calculated as . This is considered as average velocity over an elemental height .

ycosVVn

y

78

The discharge per unit width . The velocity is zero at the boundaries, i.e. at points 0 t

o 7, and should be noted in calculating the average velocities for the end sections.

per meter width

yVq n

15cos6.205.010cos5.205.05cos3.205.0q

30cos10.205.025cos5.205.020cos6.205.0

0909.01133.01222.01256.01231.01146.0

smq 36897.0

79

EXAMPLE 1.5 while measuring the discharge in a small stream it was found that the depth of flow increases at the rate of . If the discharge at that section was and the surface width of the stream was , estimate the discharge at a section 1 km upstream.

hm10.0sm325

m20

80

Solution This is a case of unsteady flow and the continuity equa

tion (Eq. 1.36) will be used.

By Eq. (1.36),

discharge at the upstream section

000556.06060

10.020

t

yT

t

yT

x

Q

x

QQ

12

1Q

000556.010000.252

xt

yTQ

sm3556.25

81

1.10 ENERGY EQUATION1.10 ENERGY EQUATION In the one-dimensional analysis of steady open-chann

el flow, the energy equation in the form of the Bernoulli equation is used. According to this equation, the total energy at a downstream section differs from the total energy at the upstream section by an amount equal to the loss of energy between the sections.

Figure 1.13 shows a steady varied flow in a channel. If the effect of the curvature on the pressure distribution is neglected, the total energy head (in N.m / newton of fluid) at any point at a depth below the water surface is

(1.37)A d

g

VdZH A 2cos

2

82

This total energy will be constant for all values of from zero to at a normal section through point (i.e. section ), where =depth of flow measured normal to the bed. Thus the total energy at any section whose bed is at an elevation above the datum is

(1.38)

Ad y

OA y

ZgVyZH 2cos 2

83

In Fig. l.l3 the total energy at a point on the bed is plotted along the vertical through that point. Thus the elevation of energy line on the line 1-1 represents the total energy at any point on the normal section through point 1. The total energies at normal sections through l and 2 are therefore

g

VyZH

2cos

21

1111

g

VyZH

2cos

22

2222

84

respectively. The term represents the elevation of the hydraulic grade line above the datum.

If the slope of the channel θ is small, ,the normal section is practically the same as the

vertical section and the total energy at any section can be written as

(1.39) Since most of the channels in practice happen

to have small values of , the term usually neglected.

hyZ cos

0.1cos

g

VyZH

2

2

10cos

85

Thus the energy equation is written as Eq. (1.40) in subsequent sections of this book, with the realisation that the slope term will be included if

is appreciably different from unity. Due to energy losses between sections 1 and 2,the en

ergy head will be larger than and head loss. Normally, the head loss ( ) can

be considered to be made up of frictional losses ( ) and eddy or form loss ( ) such that .

For prismatic channels, .

cos

H2H

LhHH 21

Lhfh

eh efL hhh 0eh

86

One can observe that for channels of small dope the piezometric head line essentially coincides with the free surface. The energy line which is a plot of vs is a droopi1g line in the longitudinal ( ) direction. The difference of the ordinates between the energy line and free surface represents the velocity head at that section. In general, the bottom profile, water-surface and energy line will have distinct slopes at a given section. The bed slope is a geometric parameter of the channel.

In designating the total energy by Eq. (l .37) or (l.38), hydrostatic pressure distribution was assumed.

xHx

87

However, if the curvature effects in a vertical plane arc appreciable, the pressure distribution at a section may have a non-linear variation with the depth . In such cases the effective piezometric head as defined Eq. (1.11) will be used to represent the total energy at a section as

(1.40)

deph

g

VhH ep 2

2

88

EXAMPLE 1.6 The width of a horizontal rectangular channel is reduced from 3.5m to 2.5m and the floor is raised by 0.25m in elevation at a given section. At the upstream section, the depth of flow is 2.0 m and the kinetic energy correction factor α is 1.15. If the drop in the water surface elevation at the contraction is 0.20 m, calculate the discharge if (a) the energy loss is neglected, and (b) the energy loss is one-tenth of the upstream velocity head. [The kinetic energy correction factor at the contracted section may be assumed to be unity].

89

Solution

90

Referring to Fig. 1.14

By continuity

(a) When there is no energy loss: By energy equation applied to sections 1 to 2,

my 0.21 my 55.120.025.00.22

222111 VyBVyB

221 5536.00.25.3

55.15.2VVV

g

VyZZ

g

VyZ

22

22

221

21

111

0.115.1 21 and

91

Discharge

Zyy

g

VV

21

21

22

2

15.1

25.055.100.25536.015.112

222 g

V

2.081.92

6476.022

V

smV 462.22 smQ 354.9462.255.15.2

92

(b) when there is then an energy loss:

By energy equation:

Substituting

g

V

g

VH L 2

115.02

1.021

21

1

LHg

VyZZ

g

VyZ

22

22

221

21

111

ZyyHg

V

g

VL

21

21

1

22

2 22

g

VHand L 2

115.015.1,0.121

12

25.055.100.22

115.02

15.12

21

21

22

g

V

g

V

g

V

93

Since

Discharge

21 5536.0 VV

2.05536.015.19.012

222 g

V

2.081.92

6826.0 22

V

sm397.2

smQ 3289.9397.255.15.2

94

1.11 MOMENTUM EQATION1.11 MOMENTUM EQATIONSteady Flow Momentum is a vector quantity. The momentum

equation commonly used in most of the open channel flow problems is the linear-momentum equation. This equation states that the algebraic sum of all external forces acting in a given direction on a fluid mass equals the time rate of change of linear-momentum of the fluid mass in that direction. In a steady flow the rate of change of momentum in a given direction will be equal to the net flux of momentum in that direction.

95

Figure 1.15 shows a control volume (a volume fixed in s

pace) bounded by sections 1 and 2, the boundary and a surface lying above the free surface.

96

The various forces acting on the control volume in the longitudinal direction are:

(i) Pressure forces acting on the control surfaces, and . (ii) Tangential force on the bed, , (iii) Body force, i.e. the component of the weight of the

fluid in the longitudinal direction, . By the linear-momentum equation in the longitudinal

direction for a steady-flow discharge of ,

1F 2F

3F

4F

Q

124321 MMFFFFF

97

in which =momentum flux entering the control volume =momentum flux leaving the control volume.

In practical applications of the momentum equation, the proper identification of the geometry of the control volume and the various forces acting on it are very important. The momentum equation is a particularly useful tool in analysing rapidly varied flow (RVF) situations where energy losses are complex and cannot be easily estimated. It is also very helpful in estimating forces on a fluid mass.

111 QVM 222 QVM

98

EXAMPLE 1.7 Estimate the force. on a sluice gate shown in Fig. 1.16

99

Solution Consider a unit width of the channel. The force exerted

on the fluid by the gate is , as shown in the figure. This is equal and opposite to the force exerted by the fluid on the gate, .

Consider the control volume as shown by dotted lines in the figure. Section 1 is sufficiently far away from the efflux section and hydrostatic pressure distribution can be assumed. The frictional force on the bed between sections 1 and 2 is neglected. Also assumed are .

Section 2 is at the vena contracta of the jet where the streamlines are parallel to the bed.

'F

F

0.121

100

The forces acting on the control volume in the longitudinal direction are:

= pressure force on the control surface at section = pressure force on the control surface at section acting in a direction opposing = reaction force of the gate on the section . By the momentum equation, Eq.(1.41),

(1.42)

1F

2F

3F

212

1'11 y

222

1'22 y

1F

'33

1222

21 2

1

2

1VVqFyy

101

in when = discharge per unit width . Simplifying Eq. (1.42),

(1.43) If the loss of energy between sections 1 and 2 is assumed

to be negligible, by the energy equation with

(1.44) Substituting

q 2211 yVyV

g

qyyyy

yy

yyF

2

212121

21 2

2

1

0.121

g

Vy

g

Vy

22

22

2

21

1

22

11 y

qVand

y

qV

102

and by Eq. (1.43)

(1.45) The force on the gate would be equal and opposite t

o .

21

22

21

2 2

yy

yy

g

q

21

321

2

1

yy

yyF

F

'F

103

EXAMPLE 1.8 Figure 1.17 shows a hydraulic jump in a horizontal apron aided by a two dimensional block on the apron. Obtain an expression for the drag force per unit length of the block.

Solution Consider a control volume surrounding the block

as shown in Fig.1.17. A unit width of apron is considered. The drag force on the block would have a reaction force = on the control surface, acting in the upstream direction as shown in Fig. 1.17. Assume, a frictionless, horizontal channel and hydrostatic pressure distribution at sections 1 and 2.

DF

104

By momentum equation, Eq. (1.41), in the direction of the flow

where =discharge per unit width of apron

1221 MMPFP D 1122

22

21 2

1

2

1VVqyFy D

q

2211 VyVy

105

Assuming 0.121

12

222

21

11

2

1

2

1

yyqyyFD

21

122

22

21

2

2 yy

yy

g

qyy

106

Unsteady Flow In unsteady flow the linear-momentum equation

will have an additional term over and above that of the steady flow equation to include the rate of change of momentum in the control volume. The momentum equation would then state that in an unsteady flow the algebraic sum of all external forces in a given direction on a fluid mass equals the net change of the linear-momentum flux of the fluid mass in that direction plus the time rate of increase of momentum in that direction within the control volume.

107

Specific Force The steady-state momentum equation (Eq. (1.41)) take

s a simple form if the tangential force and body force are both zero. In that case

or Denoting

The term is known as the specific force and represents the sum of the pressure force and momentum flux per unit weight of the fluid at a section.

4F3F

1221 MMFF 2211 MFMF

sPMF 1

21 ss PP

sP

108

Equation (1.46) states that the specific force is constant in a horizontal, frictionless channel. This fact can be advantageously used to solve some flow situations. An application of the specific force relationship to obtain an expression for the depth at the end of a hydraulic jump is given in Sec. 6.5. In a majority of applications the force

is taken as due to hydrostatic pressure distribution and hence is given by,

where is the depth of the centre of gravity of the flow area.

F

yAF y

109

習題習題1.19 Figure 1.26 shows a sluice gate in a rectangular

channels. Fill the missing data in the following table:

1.26 In problem 1.19 (a, b, c and d), determine the force per m width of the sluice gate.

110

1.25 A high-velocity flow from a hydraulic structure has a velocity of 6.0 and a depth of 0.40 . It is deflected upwards at the end of a horizontal apron through an angle of into the atmosphere as a jet by an end sill. Calculate the force on the sill per unit width.

1.28 Figure 1.28 shows a submerged flow over a sharp-crested weir in a rectangular channels. If the discharge per unit width is 1.8 , estimate the energy loss due to the weir. What is the force on the weir plate?

sm m

45

msm3

111

1.29 A hydraulic jump assisted by a two dimensional block is formed on a horizontal apron as shown in Fig. 1.29. Estimate the force in kN/m width on the block when a discharge of 6.64 per m width enters the apron at a depth of 0.5 m and leave it at a depth of 3.6 m.

DFsm3