strujanje u otvorenim kanalima

8/13/2019 Strujanje u Otvorenim Kanalima

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11Flow in Open

Channels11.1 Basic Concepts and Definitions

11.2 Energy Equation for Open-Channel Flows

11.3 Localized Effect of Area Change (Frictionless Flow)

11.4 The Hydraulic Jump

11.5 Steady Uniform Flow

11.6 Flow with Gradually Varying Depth

11.7 Discharge Measurement Using Weirs

11.8 Summary and Useful Equations

Case Study in Energy and the Environment

Using a Reservoir as a Battery

We are all familiar with electric batteries;we have them in our cars, our laptops, our

cell phones, and our MP3 players, to mention just afew of their uses. Batteries are energy storage devicesthat allow us to generate energy at one time and placeand store it for use at a different time and in anotherplace. The figure shows a rather mundane-looking

dam (its the Ffestiniog Dam in north Wales), but its

actually part of a pretty exciting development, theFfestiniog Pumped Storage Scheme: its a battery!The idea of using reservoirs not only as a source of

power but as a way to store power is not new; effortswere made in the 19th century. But it is becoming veryimportant in optimizing power plant performance, aswell as in storing renewable energy generated by wind,

600


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In this chapter we introduce some of the basic concepts in the study of open-channelflows. The topic of open-channel flow is covered in much more detail in a number ofspecialized texts [18].

Many flows in engineering and in nature occur with a free surface. An example ofa human-made channel is shown in Fig. 11.1. This is a view of the 190-mile-longHayden-Rhodes Aqueduct, which is part of the Central Arizona Project (CAP). TheCAP is a 336-mi (541 km) diversion canal in Arizona used to redirect water from

wave, and ocean current farms, some of which we havereviewed in previous Case Studies in Energy and theEnvironment. Power companies have always hadthe problem that energy demand tends to have severepeaks and troughs; in the afternoon and evening thereis high demand; in the middle of the night, lowdemand. However, for best efficiency, plants shouldoperate at a steady energy output; in addition, thepower company needs to have on hand extra powergeneration capability just for those peaks. On the otherhand, renewable energy needs to be harvested when

its availablewhen the wind is blowing, when thereare waves or decent currents flowingand these timesdo not always correspond to the times when theenergy is needed. With schemes like the one at Ffes-tiniog, at times of low electrical demand, excess gen-eration capacity from the power company is used topump water into an upper reservoir; when there is highdemand, water is released back into a lower reservoirthrough a turbine, generating electricity. Reversibleturbine/generator assemblies act as pump and turbine(usually of a Francis turbine design; see Chapter 10).

The systems four water turbines can generate 360 MWof electricity within a minute of the need arising!

Some facilities worldwide are purely pumped-storage plants, which simply shift water between tworeservoirs, but combined pump-storage plants thatalso generate their own electricity like conventionalhydroelectric plants are becoming more common. Theprocess is reasonably efficient and is the only way thathuge amounts of energy can be stored (electric bat-teries are all relatively low capacity). Taking intoaccount losses in the turbine/generator system and

from evaporation loss at the exposed water surface, aswell as the possibility of losses due to hydraulic jumps(discussed in this chapter) occurring at outlets, about70 to 85 percent of the electrical energy used to pumpthe water into the elevated reservoir can be regained.In future years, increased effort will be placed onincreasing the efficiency of these systems, and theywill become much more common. The Ffestiniog sys-tem is for storing excess power plant energy, but inthe future we may expect to see pumped-storageplants adjacent to a number of wind farms.

The Ffestiniog dam.

Flow in Open Channels 6


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Fig. 11.1 Hayden-Rhodes Aqueduct, Central Arizona Project.[Courtesy of the U.S. Bureau of Reclamation (1985), photograph by Joe Madrigal Jr.]

602 Chapter11Flow in Open Channels


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the Colorado River into central and southern Arizona, and it is the largest and mostexpensive aqueduct system ever constructed in the United States.

Because free surface flows differ in several important respects from the flows inclosed conduits that we reviewed in Chapter 8, we treat them separately in thischapter. Familiar examples where the free surface is at atmospheric pressure includeflows in rivers, aqueducts, and irrigation canals, flows in rooftop or street gutters, anddrainage ditches. Human-made channels are given many different names, including

canal, flume, or culvert. A canalusually is excavated below ground level and may beunlined or lined. Canals generally are long and of very mild slope; they are used tocarry irrigation or storm water or for navigation. A flumeusually is built above groundlevel to carry water across a depression. A culvert, which usually is designed to flowonly part-full, is a short covered channel used to drain water under a highway orrailroad embankment.

In this chapter we shall develop, using control volume concepts from Chapter 4,some basic theory for describing the behavior and classification of flows in natural andhuman-made channels. We shall consider:

Flows for which the local effects of area change predominate and frictional forcesmay be neglected. An example is flow over a bump or depression, over the shortlength of which friction is negligible.

Flow with an abrupt change in depth. This occurs during a hydraulic jump (see

Fig. 11.12 for examples of hydraulic jumps).

Flow at what is called normal depth. For this, the flow cross section does not vary inthe flow direction; the liquid surface is parallel to the channel bed. This is analogousto fully developed flow in a pipe.

Gradually varied flow. An example is flow in a channel in which the bed slopevaries. The major objective in the analysis of gradually varied flow is to predict theshape of the free surface.

It is quite common to observe surface waves in flows with a free surface, the simplestexample being when an object such as a pebble is thrown into the water. The prop-agation speed of a surface wave is analogous in many respects to the propagation of asound wave in a compressible fluid medium (which we discuss in Chapter 12). Weshall determine the factors that affect the speed of such surface waves. We will see

that this is an important determinant in whether an open-channel flow is able togradually adjust to changing conditions downstream or a hydraulic jump occurs.

This chapter also includes a brief discussion of flow measurement techniques foruse in open channels.

11.1Basic Concepts and DenitionsBefore analyzing the different types of flows that may occur in an open channel, wewill discuss some common concepts and state some simplifying assumptions. We aredoing so explicitly, because there are some important differences between ourprevious studies of pipes and ducts in Chapter 8 and the study of open-channel

flows.One significant difference between flows in pipes and ducts is

The driving force for open-channel flows is gravity.(Note that some flows in pipes and ducts are also gravity driven (for example, flowdown a full drainpipe), but typically flow is driven by a pressure difference generatedby a device such as a pump.) The gravity force in open-channel flow is opposed byfriction force on the solid boundaries of the channel.

11.1 Basic Concepts and Definitions 6


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Simplifying Assumptions

The flow in an open channel, especially in a natural one such as a river, is often verycomplex, three-dimensional, and unsteady. However, in most cases, we can obtainuseful results by approximating such flows as being:

One-dimensional.

Steady.

A third simplifying assumption is:

The flow at each section in an open-channel flow is approximated as a uniformvelocity.

Typical contours of actual streamwise velocity for a number of open-channel sectionsare shown in Fig. 11.2. These would seem to indicate that the third assumption isinvalid, but in fact it is a reasonable approximation, as we shall now justify. Most flowsof interest are large in physical scale, so the Reynolds numbers generally are large.Consequently, open-channel flow seldom is laminar; in this chapter we will assumeturbulent flow. As we saw in earlier chapters, turbulence tends to smooth out thevelocity gradient (see Fig. 8.11 for turbulent pipe flow and Fig. 9.7a for turbulentboundary layers). Hence although the profiles, as shown in Fig. 11.2, are notuniform,

as a reasonable approximation we will assume uniform velocity at each section, withthe kinetic energy coefficient, , taken to be unity (the kinetic energy coefficient isdiscussed in Section 8.6). This is illustrated in Fig. 11.3a.

Figure 11.2 shows that the measured maximum velocity occurs below the freesurface, in spite of the fact that there is negligible shear stress due to air drag soone would expect the maximum velocity to occur at the free surface. Secondary flowsare also responsible for distorting the axial velocity profile; examples of secondaryflows are when a channel has a bend or curve or has an obstruction, such as a bridgepier. The high velocities that may be present in the vortices generated in such casescan seriously erode the bottom of a natural channel.

VIDEO

A Turbulent Channel (Animation).

VIDEO

The Glen Canyon Dam: A Source of

Turbulent Channel Flow.

Triangular channel

Circular channelNatural irregular channel

Trapezoidal channel

Shallow ditch Narrrow

rectangular

section

2.0

2.0

1.5

1.5

0.5 0.5

1.0

1.0

2.5

2.5

2.5

2.0

2.0

2.0

2.01.5

1.5

1.5

1.50.5

0.5

0.5

1.0

1.0

1.0

1.0

0.5

0.5

Fig. 11.2 Typical contours of equal velocity in open-channel sections. (From Chow [1], used bypermission.)



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The next simplifying assumption we make is:

The pressure distribution is approximated as hydrostatic.This is illustrated in Fig. 11.3band is a significant difference from the analysis of flowsin pipes and ducts of Chapter 8; for these we found that the pressure was uniform ateach axial location and varied in the streamwise direction. In open-channel flows, thefree surface will be at atmospheric pressure (zero gage), so the pressure at the surfacedoes not vary in the direction of flow. The major pressure variation occurs acrosseach

section; this will be exactly true if streamline curvature effects are negligible, which isoften the case.

As in thecase of turbulent flowin pipes,we must rely on empirical correlationsto relatefrictional effects to the average velocity of flow. The empirical correlation is includedthrough a head loss term in the energy equation (Section 11.2). Additional complicationsin many practical cases include the presence of sediment or other particulatematterin theflow, as well as the erosion of earthen channels or structures by water action.

Channel Geometry

Channels may be constructed in a variety of cross-sectional shapes; in many casesregular geometric shapes are used. A channel with a constant slope and cross section

is termed prismatic. Lined canals often are built with rectangular or trapezoidal sec-tions; smaller troughs or ditches sometimes are triangular. Culverts and tunnelsgenerally are circular or elliptical in section. Natural channels are highly irregular andnonprismatic, but often they are approximated using trapezoid or paraboloid sections.Geometric properties of common open-channel shapes are summarized in Table 11.1.

Thedepth of flow, y, is the perpendicular distance measured from the channel bedto the free surface. The flow area, A, is the cross section of the flow perpendicular tothe flow direction. The wetted perimeter, P, is the length of the solid channel cross-section surface in contact with the liquid. The hydraulic radius, Rh, is defined as

Rh 5A

P 11:1

For flow in noncircular closed conduits (Section 8.7), the hydraulic diameter was

defined as

Dh 54A

P 8:50

Thus, for a circular pipe, the hydraulic diameter, from Eq. 8.50, is equal to the pipediameter. From Eq. 11.1, the hydraulic radius for a circular pipe would then be halfthe actual pipe radius, which is a bit confusing! The hydraulic radius, as defined byEq. 11.1, is commonly used in the analysis of open-channel flows, so it will be used

V

(a) Approximate velocity profile (b) Approximate pressuredistribution (gage)

Fig. 11.3 Approximations for velocity profile and pressuredistribution.



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throughout this chapter. One reason for this usage is that the hydraulic radius of awide channel, as seen in Table 11.1, is equal to the actual depth.

For nonrectangular channels, the hydraulic depth is defined as

yh 5A

bs11:2

wherebsis the width at the surface. Hence the hydraulic depth represents the averagedepth of the channel at any cross section. It gives the depth of an equivalent rectan-gular channel.

Speed of Surface Waves and the Froude Number

We will learn later in this chapter that the behavior of an open-channel flow as itencounters downstream changes (forexample,a bumpof the bed surface, a narrowing ofthe channel, or a change in slope) is strongly dependent on whether the flow is slow or

Table 11.1Geometric Properties of Common Open-Channel Shapes

Shape

Trapezoidal

Triangular

Rectangular

Wide Flat

Circular

SectionFlow

Area,AWetted

Perimeter,PHydraulicRadius, Rh

y (bycot )

( sin)8

y2cot

D2

by b 2y

by b

y cos

y (bycot )b

2y

sin

2y

sin 2

by

b 2y

b2y

sin

y

2

D

41

D sin

bs

bs

bs

b

b

y

y

y

y

y

D

b>>y



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fast. A slow flow will have time to gradually adjust to changes downstream, whereasa fast flow will also sometimes gradually adjust but in some situations will do soviolently (i.e., there will be a hydraulic jump; see Fig. 11.12a for an example). Thequestion is what constitutes a slow or fast flow? These vague descriptions will bemade more precise now. It turns out that the speed at which surface waves travel alongthe surface is key to defining more precisely the notions of slow and fast.

To determine the speed (or celerity) of surface waves, consider an open channel

with movable end wall, containing a liquid initially at rest. If the end wall is given asudden motion, as in Fig. 11.4a, a wave forms and travels down the channel at somespeed, c(we assume a rectangular channel of width, b, for simplicity).

If we shift coordinates so that we are traveling with the wave speed, c, we obtain asteady control volume, as shown in Fig. 11.4b(where for now we assume c . V). Toobtain an expression for c, we will use the continuity and momentum equations forthis control volume. We also have the following assumptions:

1. Steady flow.

2. Incompressible flow.

3. Uniform velocity at each section.

4. Hydrostatic pressure distribution at each section.

5. Frictionless flow.

Assumption 1 is valid for the control volume in shifted coordinates. Assumption 2 isobviously valid for our liquid flow. Assumptions 3 and 4 are used for the entirechapter. Assumption 5 is valid in this case because we assume the area on which itacts,bx, is relatively small (the sketch is not to scale), so the total friction force isnegligible.

For an incompressible flow with uniform velocity at each section, we can use theappropriate form of continuity from Chapter 4,X

CS~V ~A 5 0 4:13b

Applying Eq. 4.13b to the control volume, we obtain

c 2 Vfy 1 ybg 2 cyb 5 0 11:3or

cy 2 Vy 1 cy 2 Vy 2 cy 5 0

Solving for V,

V 5 c y

y 1 y 11:4

V

(a) Absolute coordinates

Fluid

moving at

speed V

Fluid at

rest

Fluid moving

at speed

(cV)

Wave at rest

Wave moving

at speed c

(b) Coordinates at rest relative to wave

Fluid moving

at speed

c

Control

volume

x

y

y

y y+ y y

x

Fig. 11.4 Motion of a surface wave.



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For the momentum equation, again with the assumption of uniform velocity at eachsection, we can use the following form of the x component of momentum

Fx 5 FSx 1 FBx 5@

@t

ZCV

u dV--- 1X

CSu~V ~A 4:18d

The unsteady term@/@tdisappears as the flow issteady, and the body forceFBx is zerofor horizontal flow. So we obtain

FSx 5X

CSu~V ~A 11:5

The surface force consists of pressure forces on the two ends, and friction force on thebottom surface (the air at the free surface contributes negligible friction in open-channel flows). By assumption5we neglect friction. The gage pressure at the two endsis hydrostatic, as illustrated in Fig. 11.4b. We recall from our study of hydrostatics thatthe hydrostatic force FR on a submerged vertical surface of area A is given by thesimple result

FR 5 pcA 3:10bwherepc is the pressure at the centroid of the vertical surface. For the two verticalsurfaces of the control volume, then, we have

FSx 5 FRleft 2 FRright 5pcAleft 2 pcAright

5

(g

y 1 y

2

y 1 y b

)2

(g

y

2

yb

)

5gb

2 y 1 y2 2 gb

2 y2

Using this result in Eq. 11.5 and evaluating the terms on the right,

FSx 5gb

2 y 1 y2 2 gb

2 y2 5

XCS

u~V ~A

5 2 c 2 Vfc 2 Vy 1 ybg 2 cf 2 cybgThe two terms in braces are equal, from continuity as shown in Eq. 11.3, so themomentum equation simplifies to

gyy 1gy2

2 5 ycV

or

g 1 1y

2y

y 5 cV

Combining this with Eq. 11.4, we obtain

g 1 1y

2y

y 5 c 2

y

y 1 y

and solving for c,

c2 5 gy 1 1y

2y

1 1

y

y



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For waves of relatively small amplitude (yy), we can simplify this expression to

c 5ffiffiffiffiffi

gyp 11:6

Hence the speed of a surface disturbance depends on the local fluid depth. Forexample, it explains why waves crash as they approach the beach. Out to sea, thewater depths below wave crests and troughs are approximately the same, and hence so

are their speeds. As the water depth decreases on the approach to the beach, thedepth of crests start to become significantly larger than trough depths, causing creststo speed up and overtake the troughs.

Note that fluid properties do not enter into the speed: Viscosity is usually a minorfactor, and it turns out that the disturbance or wave we have described is due to theinteraction of gravitational and inertia forces, both of which are linear with density.Equation 11.6 was derived on the basis of one-dimensional motion (x direction); amore realistic model allowing two-dimensional fluid motion (xandydirections) showsthat Eq. 11.6 applies for the limiting case of large wavelength waves (Problem 11.6explores this). Also, there are other types of surface waves, such as capillary wavesdriven by surface tension, for which Eq. 11.6 does not apply (Problems 11.7 and 11.8explore surface tension effects).

Example11.1 SPEED OF FREE SURFACE WAVESYou are enjoying a summers afternoon relaxing in a rowboat on a pond. You decide to find out how deep the watis by splashing your oar and timing how long it takes the wave you produce to reach the edge of the pond. (The ponis artificial; so it has approximately the same depth even to the shore.) From floats installed in the pond, you knoyoure 20 ft from shore, and you measure the time for the wave to reach the edge to be 1.5 s. Estimate the podepth. Does it matter if its a freshwater pond or if its filled with seawater?

Given: Time for a wave to reach the edge of a pond.

Find: Depth of the pond.

Solution: Use the wave speed equation, Eq. 11.6.

Governing equation: c 5ffiffiffiffiffi

gyp

The time for a wave, speed c, to travel a distance L, is t 5L

c, so c 5

L

Wt. Using this and Eq. 11.6,

ffiffiffiffiffigy

p 5

L

Wt

wherey is the depth, or

y 5L2

gt2

Using the given data

y 5 202ft2 31

32:2

s2

ft 3

1

1:521

s2 5 5:52 ft

y

The pond depth is about 512

ft.

Theresultobtainedisindependenwhetherthewaterisfreshorsalinbecausethesp

eedofthesesurfacewavesisindependentoffluidproperties.



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The speed of surface disturbances given in Eq. 11.6 provides us with a more usefullitmus test for categorizing the speed of a flow than the terms slow and fast. Toillustrate this, consider a flow moving at speed V, which experiences a disturbanceat some point downstream. (The disturbance could be caused by a bump in the channelfloor or by a barrier, for example.) The disturbance will travel upstream at speed crelative to the fluid. If the fluid speed is slow, V, c, and the disturbance will travelupstream at absolute speed (cV). However, if the fluid speed is fast, V. c, and thedisturbance cannot travel upstreamand instead is washed downstream at absolute speed(V c). This leads to radicallydifferent responses of slow and fast flows to a downstreamdisturbance. Hence, recalling Eq. 11.6 for the speed c, open-channel flows may beclassified on the basis of Froude number first introduced in Chapter 7:

Fr 5Vffiffiffiffiffigy

p 11:7

Instead of the rather loose terms slow and fast, we now have the followingcriteria:

Fr, 1 Flow is subcritical, tranquil, or streaming. Disturbances can travel upstream;downstream conditions can affect the flow upstream. The flow can gradually

adjust to the disturbance.Fr= 1 Flow is critical.

Fr. 1 Flow is supercritical, rapid, or shooting. No disturbance can travel upstream;downstream conditions cannot be felt upstream. The flow may violentlyrespond to the disturbance because the flow has no chance to adjust to thedisturbance before encountering it.

Note that for nonrectangular channels we use the hydraulic depth yh,

Fr 5Vffiffiffiffiffiffiffigyh

p 11:8

These regimes of flow behavior are qualitatively analogous to the subsonic, sonic, andsupersonic regimes of gas flow that we will discuss in Chapter 12. (In that case we are

also comparing a flow speed, V, to the speed of a wave, c, except that the wave is asound wave rather than a surface wave.)

We will discuss the ramifications of these various Froude number regimes later inthis chapter.

11.2 Energy Equation for Open-Channel FlowsIn analyzing open-channel flows, we will use the continuity, momentum, and energyequations. Here we derive the appropriate form of the energy equation (we will usecontinuity and momentum when needed). As in the case of pipe flow, friction in open-channel flows results in a loss of mechanical energy; this can be characterized by ahead loss. The temptation is to just use one of the forms of the energy equation forpipe flow we derived in Section 8.6, such as

p1

g 1 1

V2

2

2g 1 z1

!2

p2

g 1 2

V2

2

2g 1 z2

! 5

hlTg

5 HlT 8:30

The problem with this is that it was derived on the assumption of uniform pressure ateach section, which is not the case in open-channel flow (we have a hydrostatic



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pressure variation at each location); we do not have a uniform p1 at section 1 anduniformp2at section 2!

Instead we need to derive an energy equation for open-channel flows from firstprinciples. We will closely follow the steps outlined in Section 8.6 for pipe flows butuse different assumptions. You are urged to review Section 8.6 in order to be aware ofthe similarities and differences between pipe flows and open-channel flows.

We will use the generic control volume shown in Fig. 11.5, with the following

assumptions:1. Steady flow

2. Incompressible flow

3. Uniform velocity at a section

4. Gradually varying depth so that pressure distribution is hydrostatic

5. Small bed slope

6. _Ws 5 _Wshear 5 _Wother 5 0

We make a fewcomments here. We have seen assumptions14 already; theywill alwaysapply in this chapter. Assumption 5 simplifies the analysis so that depth,y, istakento bevertical and speed,V, is taken to be horizontal, rather than normal and parallel to thebed, respectively. Assumption6states that there is no shaft work, no work due to fluid

shearing at the boundaries, andno other work. There is no shear work at the boundariesbecause oneach part of thecontrol surface thetangentialvelocityis zero (onthe channelwalls) or the shear stress is zero (the open surface), so no work can be done. Note thatthere can still be mechanical energy dissipation within the fluid due to friction.

We have chosen a generic control volume so that we can derive a generic energyequation for open-channel flows, that is, an equation that can be applied to a varietyof flows such as ones with a variation in elevation, or a hydraulic jump, or a sluice gate,and so on, between sections 1 and 2. Coordinate z indicates distances measured inthe vertical direction; distances measured vertically from the channel bed are denotedbyy. Note thaty1andy2are the flow depths at sections 1 and 2, respectively, andz1andz2 are the corresponding channel elevations.

The energy equation for a control volume is

QWsWshearWother

0(6) 0(6) 0(6) 0(1)

t CVe dV

CS

(e p) V

V2

2

dA . . . .

e u

(4.56)

gz

Control

volume

z1

y1

y

z2

y2

z

Fig. 11.5 Control volume and coordinates forenergy analysis of open-channel flow.

11.2 Energy Equation for Open-Channel Flows 6


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Recall that u is the thermal specific energy and v = 1/ is the specific volume. Afterusing assumptions1 and 6, and rearranging, with _m 5

R~Vd ~A, anddA = bdywhere

b(y) is the channel width, we obtain

_Q 5 2

Z1

p

1

V2

2 1gz

0

@

1

AVbdy 2

Z1

uVbdy 1

Z2

p

1

V2

2 1gz

0

@

1

AVbdy 1

Z2

uVbdy

5

Z1

p

1V2

2 1gz

0@ 1AVbdy 1Z2

p

1V2

2 1gz

0@ 1AVbdy 1 _mu2 2 u1or

Z1

p

1

V2

2 1gz

Vbdy 2

Z2

p

1

V2

2 1gz

Vbdy 5 _mu2 2 u1 2 _Q 5 _mhlT

11:9This states that the loss in mechanical energies (pressure, kinetic and potential)through the control volume leads to a gain in the thermal energy and/or a loss of heatfrom the control volume. As in Section 8.6, these thermal effects are collected into thehead loss term hl

T

.The surface integrals in Eq. 11.9 can be simplified. The speed,V, is constant at each

section by assumption3. The pressure,p, does vary across sections 1 and 2, as doesthe potential, z. However, by assumption 4, the pressure variation is hydrostatic.Hence, for section 1, using the notation of Fig. 11.5

p 5 gy1 2y

[so p = gy1 at the bed and p = 0 (gage) at the free surface] and

z 5z1 1yConveniently, we see that the pressure decreases linearly with y while z increaseslinearly with y, so the two terms together are constant,

p

1gz

1

5 gy1 2y 1gz1 1y 5 gy1 1 z1

Using these results in the first integral in Eq. 11.9,

Z1

p

1

V2

2 1gz

Vbdy 5

Z1

V2

2 1gy1 1 z1

Vbdy 5

V212

1gy1 1gz1

_m

We find a similar result for section 2, so Eq. 11.9 becomes

V222

1gy2 1gz2

2

V212

1gy1 1gz1

5 h lT

Finally, dividing byg (withHl 5 h lT=g) leads to an energy equation for open-channelflow

V212g

1y1 1 z1 5V222g

1y2 1 z2 1 Hl 11:10

This can be compared to the corresponding equation for pipe flow, Eq. 8.30, presentedat the beginning of this section. (Note that we Hluse rather thanHlT; in pipe flow wecan have major and minor losses, justifyingTfor total, but in open-channel flow we do



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not make this distinction.) Equation 11.10 will prove useful to us for the remainder ofthe chapter and indicates that energy computations can be done simply from geometry(y and z) and velocity, V.

Thetotal head or energy head, H, at any location in an open-channel flow can bedefined from Eq. 11.10 as

H 5

V2

2g 1

y1

z 11:11

whereyandz are the localflow depthandchannel bed elevation, respectively (they nolonger represent the coordinates shown in Fig. 11.5). This is a measure of themechanical energy (kinetic and pressure/potential) of the flow. Using this in theenergy equation, we obtain an alternative form

H1 2 H2 5 Hl 11:12From this we see that the loss of total head depends on head loss due to friction.

Specific Energy

We can also define the specific energy (or specific head), denoted by the symbol E,

E 5V2

2g 1y 11:13

This is a measure of the mechanical energy (kinetic and pressure/potential) of the flowabove and beyond that due to channel bed elevation; it essentially indicates the energydue to the flows speed and depth . Using Eq. 11.13 in Eq. 11.10, we obtain another formof the energy equation,

E1 2 E2 1 z1 2 z2 5 Hl 11:14From this we see that the change in specific energy depends on friction and on channelelevation change. While the total head must decrease in the direction of flow (Eq.11.12), the specific head may decrease, increase, or remain constant, depending on thebed elevation, z.

From continuity, V= Q/A, so the specific energy can be written

E 5Q2

2gA2 1y 11:15

For all channelsA is a monotonically increasing function of flow depth (as Table 11.1indicates); increasing the depth must lead to a larger flow area. Hence, Eq. 11.15indicates that the specific energy is a combination of a hyperbolic-type decrease withdepth and a linear increase with depth. This is illustrated in Fig. 11.6. We see that fora given flow rate, Q, there is a range of possible flow depths and energies, but one

depth at which the specific energy is at a minimum. Instead ofEversusy we typicallyplot y versus Eso that the plot corresponds to the example flow section, as shown inFig. 11.7.

Recalling that the specific energy, E, indicates actual energy (kinetic plus potential/pressure per unit mass flow rate) being carried by the flow, we see that a given flow, Q,can have a range of energies, E, and corresponding flow depths, y. Figure 11.7 alsoreveals some interesting flow phenomena. For a given flow, Q, and specific energy, E,there are two possible flow depths, y; these are called alternate depths. For example,



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we can have a flow at depthy1or depthy2. The first flow has large depth and is movingslowly, and the second flow is shallow but fast moving. The plot graphically indicates

this: For the first flow,E1is made up of a largey1and smallV21=2g; for the second flow,E2is made up of a small y2and largeV

22=2g. We will see later that we can switch from

one flow to another. We can also see (as we will demonstrate in Example 11.2 for arectangular channel) that for a given Q, there is always one flow for which the specificenergy is minimum, E= Emin; we will investigate this further after Example 11.2 andshow that Emin = Ecrit, where Ecritis the specific energy at critical conditions.

E

y

2gA2

Q2

y

Fig. 11.6 Dependence of specificenergy on flow depth for a given flow rate.

Critical flow

Constant Q

E1=E2 Q2

2gA2E= y+

y

yc

y2

y2

y1

y1

2V22g

2V1

2g

Fig. 11.7 Specific energy curve for a given flow rate.

Example11.2 SPECIFIC ENERGY CURVES FOR A RECTANGULAR CHANNELFor a rectangular channel of widthb = 10 m, construct a family of specific energy curves for Q = 0, 2, 5, and 10 m3/s.What are the minimum specific energies for these curves?

Given: Rectangular channel and range of flow rates.

Find: Curves of specific energy. For each flow rate, find the minimum specific energy.

Solution: Use the flow rate form of the specific energy equation (Eq. 11.15) for generating the curves.

Governing equation:E 5

Q2

2gA2 1y 11:15



16/57

For the specific energy curves, express Eas a function of depth, y .

E 5Q2

2gA2 1y 5

Q2

2gby2 1y 5

Q2

2gb2

1

y2 1y

The table and corresponding graph were generated from this equation using Excel.

Specific Energy, E(m)

y(m) Q 5 0 Q 5 2 Q 5 5 Q 5 10

0.100 0.10 0.92 5.20 20.490.125 0.13 0.65 3.39 13.170.150 0.15 0.51 2.42 9.210.175 0.18 0.44 1.84 6.830.200 0.20 0.40 1.47 5.300.225 0.23 0.39 1.23 4.250.250 0.25 0.38 1.07 3.510.275 0.28 0.38 0.95 2.970.30 0.30 0.39 0.87 2.570.35 0.35 0.42 0.77 2.01

0.40 0.40 0.45 0.72 1.670.45 0.45 0.49 0.70 1.460.50 0.50 0.53 0.70 1.320.55 0.55 0.58 0.72 1.220.60 0.60 0.62 0.74 1.170.70 0.70 0.72 0.80 1.120.80 0.80 0.81 0.88 1.120.90 0.90 0.91 0.96 1.151.00 1.00 1.01 1.05 1.201.25 1.25 1.26 1.28 1.381.50 1.50 1.50 1.52 1.592.00 2.00 2.00 2.01 2.052.50 2.50 2.50 2.51 2.53

To find the minimum energy for a given Q, we differentiate Eq. 1,

dE

dy 5

Q2

2gb2

2

2

y3

1 1 5 0

Hence, the depth yEmin for minimum specific energy is

yEmin 5Q2

gb2

13

Using this in Eq. 11.15:

Emin 5Q2

2gA2 1yEmin 5

Q2

2gb2y2Emin

1"Q2

gb2#13

51

2"Q2

gb2#"gb2

Q2#23

1"Q2

gb2#13

53

2"Q2

gb2#13

Emin 53

2

"Q2

gb2

#13

53

2yEmin

0

1

2

3

0 1 2

E(m)

y(m)

Q= 0

Q= 2 m3/s

Q= 5 m3/s

Q= 10 m3/s

Emin



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with

Vc 5ffiffiffiffiffiffiffi

gycp

5gQ

b

1=311:24

For the rectangular channel, a particularly simple result for the minimum energy isobtained when Eq. 11.24 is used in Eq. 11.15,

E 5 Emin 5V2c2g

1yc 5gyc

2g 1yc

or

Emin 53

2yc 11:25

This is the same result we found in Example 11.2. The critical state is an importantbenchmark. It will be used in the next section to help determine what happens when aflow encounters an obstacle such as a bump. Also, near the minimum E, as Fig. 11.7shows, the rate of change ofy withEis nearly infinite. This means that for critical flow

conditions, even small changes in E, due to channel irregularities or disturbances, cancause pronounced changes in fluid depth. Thus, surface waves, usually in an unstablemanner, form when a flow is near critical conditions. Long runs of near-critical flowconsequently are avoided in practice.

Example11.3 CRITICAL DEPTH FOR TRIANGULAR SECTIONA steep-sided triangular section channel (= 60) has a flow rate of 300 m3/s. Find thecritical depth for this flow rate. Verify that the Froude number is unity.

Given: Flow in a triangular section channel.

Find: Critical depth; verify thatFr= 1.

Solution:Use the critical flow equation, Eq. 11.21.

Governing equations: Q2 5gA3c

bscFr 5

Vffiffiffiffiffiffiffigyh

p

The given data is: Q 5 300m3=s 5 60

From Table 11.1 we have the following:

A 5 y2 cot

and from basic geometry

tan 5 ybs=2

so bs 5 2y cot

Using these in Eq. 11.21

Q2 5gA3c

bsc5

gy2c cot 32yc cot

51

2gy5c cot

2

bs

y



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11.3Localized Effect of Area Change (Frictionless Flow)We will next consider a simple flow case in which the channel bed is horizontal and forwhich the effects of channel cross section (area change) predominate: flow over abump. Since this phenomenon is localized (it takes place over a short distance), theeffects of friction (on either momentum or energy) may be reasonably neglected.

The energy equation, Eq. 11.10, with the assumption of no losses due to frictionthen becomes

V212g

1y1 1 z1 5V222g

1y2 1 z2 5V2

2g 1y 1 z 5 const 11:26

(Note that Eq. 11.26 could also have been obtained from by applying the Bernoulliequation between two points 1 and 2 on the surface, because all of the requirementsof the Bernoulli equation are satisfied here.) Alternatively, using the definition ofspecific energy

Hence

yc 52Q2tan2

g

1=5

Using the given data

yc 5 2 3 3002 m3

s 2

3 tan2 603

180

3 s

2

9:81m" #1=5

55:51 3 104m51=5

Finally

yc 5 8:88 m yc

To verify that Fr=1, we need Vand yh.

From continuity

Vc 5Q

Ac5

Q

y2c cot 5 300

m3

s 3

1

8:882m2 3

1

cot 60 3

180

5 6:60 m=s

and from the definition of hydraulic depth

yhc 5Ac

bsc5

y2c cot

2yc cot 5

yc

2 5 4:44 m

Hence

Frc 5Vcffiffiffiffiffiffiffiffiffigyhc

p 56:60

m

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi9:81

m

s2 3 4:44 m

s 5 1 Frc 5 1

We have verified that at critical depth the Froude number is unity.

Aswiththerectangularchannel,triangularsectionchannelanalysileadstoanexplicitequationforycEq.11.21.Othermorecomplicatedchannelcrosssectionsoftenleadtoimplicitequationthatneedstobesolvednumerically.

11.3 Localized Effect of Area Change (Frictionless Flow) 6


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E1 1 z1 5 E2 1 z2 5 E1 z 5 const

We see that the specific energy of a frictionless flow will change only if there is achange in the elevation of the channel bed.

Flow over a Bump

Consider frictionless flow in a horizontal rectangular channel of constant width, b,with a bump in the channel bed, as illustrated in Fig. 11.9. (We choose a rectangularchannel for simplicity, but the results we obtain will apply generally.) The bumpheight above the horizontal bed of the channel is z = h(x); the water depth, y(x), ismeasured from the local channel bottom surface.

Note that we have indicated two possibilities for the free surface behavior: Perhapsthe flow gradually rises over the bump; perhaps it gradually dips over the bump.(There are other possibilities too!) One thing we can be sure of, however, is that if itrises, it will not have the same contour as the bump. (Can you explain why?) Applyingthe energy equation (Eq. 11.26) for frictionless flow between an upstream point 1and any point along the region of the bump,

V212g 1y1 5 E1 5

V2

2g 1y 1 h 5 E1 hx 5 const 11:27

Equation 11.27 indicates that the specific energy must decrease through the bump,then increase back to its original value (ofE1 = E2),

Ex 5 E1 2 hx 11:28From continuity

Q 5 bV1y1 5 bVy

Using this in Eq 11.27

Q2

2gb2y211

y1 5

Q2

2gb2y2 1

y1

h 5

const 11:29We can obtain an expression for the variation of the free surface depth by differ-entiating Eq. 11.29:

2Q2

gb2y3dy

dx 1

dy

dx 1

dh

dx 5 0

Free

surface

z0

y1 y(x)

x

y2y1

zh (x)

Fig. 11.9 Flow over a bump in a horizontal channel.



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some type of flow restriction, the flow downstream of the bump will return to itsoriginal subcritical state. Note that as we mentioned earlier, when a flow is at thecritical state the surface behavior tends to display dramatic variations in behavior.Finally, Fig. 11.11 indicates that a supercritical flow (point d) that encounters abump would increase in depth over the bump (to pointcat the bump peak), and thenreturn to its supercritical flow at pointd. We also see that if the bump is high enougha supercritical flow could slow down to critical (point e) and then either return tosupercritical (pointd) or become subcritical (point a). Which of these possibilitiesactually occurs obviously depends on the bump shape, but also on upstream and

downstream conditions (the last possibility is somewhat unlikely to occur in prac-tice). The alert reader may ask, What happens if the bump is so big that the specificenergy wants to decrease below the minimum shown at pointe? The answer is thatthe flow will no longer conform to Eq. 11.26; the flow will no longer be frictionless,because a hydraulic jump will occur, consuming a significant amount of mechanicalenergy (see Section 11.4).

y

E

a

b

cd

e

(E=Emin)

Fig. 11.11 Specific energy curve for flow overa bump.

Flow regime> 0

FlowSubcritical

Fr< 1

Supercritical

Fr> 1

dh

dx

Flow

< 0dy

dx

> 0dh

dx> 0

dy

y

h

dx

< 0dh

dx> 0

dy

dx

< 0dh

dx< 0

dy

dx

FlowFlow

Fig. 11.10 Effects of bed elevation changes.



24/57

Example11.4 FLOW IN A RECTANGULAR CHANNEL WITH A BUMP OR A NARROWINGA rectangular channel 2 m wide has a flow of 2.4 m 3/s at a depth of 1.0 m. Determine whether critical depth occurs(a) a section where a bump of height h = 0.20 m is installed across the channel bed, (b) a side wall constriction (wno bumps) reducing the channel width to 1.7 m, and (c) both the bump and side wall constrictions combined. Neglehead losses of the bump and constriction caused by friction, expansion, and contraction.

Given: Rectangular channel with a bump, a side wall constriction, or both.

Find: Whether critical flow occurs.

Solution: Compare the specific energy to the minimum specific energy for the given flow rate in each case toestablish whether critical depth occurs.

Governing equations: E 5Q2

2gA2 1y 11:15 yc 5 Q

gb2

1=311:23

Emin 53

2yc 11:25 E 5 E1 2 h 11:28

(a) Bump of height h = 0.20 m:

The initial specific energy, E1, is

E1 5 y1 1Q2

2gA2 5 y1 1

Q2

2gb2y21

5 1:0 m 1 2:42

m3

s

23

1

2 3

s2

9:81 m 3

1

22 m2 3

1

12 m2

E1 5 1:073 m

Then the specific energy at the peak of the bump, Ebump, is obtained from Eq. 11.28

Ebump 5 E1 2 h 5 1:073 m 2 0:20 mEbump 5 0:873 m

We must compare this to the minimum specific energy for the flow rate Q. First, the critical depth is

yc 5

Q2

gb2

1=3

5

"2:42

m3

s

2

3s2

9:81 m 3

1

22 m2

#1=3

yc 5 0:528 m

(Note that we have y1 .yc, so we have a subcritical flow.)Then the minimum specific energy is

Emin 53

2yc 5 0:791 m

Comparing Eqs. 1 and 2 we see that with the bump wedonotattain critical conditions.

(b) A side wall constriction (with no bump) reducing the channel width to 1.7 m:

In this case the specific energy remains constant throughout (h = 0), even at the constriction; so

Econstriction 5 E1 2 h 5 E1 5 1:073m However, at the constriction, we have a new value for b, (bconstriction= 1.7 m), and so a new critical depth

11.3 Localized Effect of Area Change (Frictionless Flow) 6


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ycconstriction 5

" Q2

gb2constriction

#1=35 2:42

m3

s

!23

s2

9:81 m 3

1

1:72 m2

24

351=3

ycconstriction 5 0:588m

Then the minimum specific energy at the constriction is

Eminconstriction 5 32ycconstriction 5 0:882m 4

Comparing Eqs. 3 and 4 we see that with the constrictionwe do notattain critical conditions.

We might enquire as to what constriction would cause critical flow. To find this, solve

E 5 1:073m 5 Emin 53

2yc 5

3

2

Q2

gb2c

1=3

for the critical channel width bc.Hence

Q2

gb2c5

2

3Emin24 35

3

bc 5Qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

8

27gE3min

s

5

27

8

!1=23 2:4

m3

s

!3

s

9:811=2 m1=23

1

1:0733=2 m3=2

bc 5 1:27 m

To make the given flow attain critical conditions, the constriction should be 1.27 m; anything wider, and criticalconditions are not reached.

(c) For a bump ofh =0.20 m andthe constriction to b =1.7 m:

We have already seen in case (a) that the bump (h = 0.20 m) was insufficient by itself to create critical conditions. Fromcase (b) we saw that at the constriction the minimum specific energy is Emin = 0.882 m rather than Emin = 0.791 m in themain flow. When we have both factors present, we can compare the specific energy at the bump and constriction,

Ebump 1 constriction 5 Ebump 5 E1 2 h 5 0:873m 5and the minimum specific energy for the flow at the bump and constriction,

Eminconstriction 53

2ycconstriction 5 0:882m 6

From Eqs. 5 and 6 we see that with both factors the specific energy isactuallylessthan the minimum. The fact that we must have a specific energythat is less than the minimum allowable means something has to give! Whathappens is that the flow assumptions become invalid; the flow may nolonger be uniform or one-dimensional, or there may be a significant energyloss, for example due to a hydraulic jump occurring. (We will discusshydraulic jumps in the next section.)

Hence the bump and constriction together are sufficient to make the flowreach critical state.

ThisExampleillustrateshowtodeter-

minewhethera channelbumporconstriction,orboth,leadtocriticalflowconditions.



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11.4The Hydraulic JumpWe have shown that open-channel flow may be subcritical (Fr, 1) or supercritical(Fr. 1). For subcritical flow, disturbances caused by a change in bed slope or flowcross section may move upstream and downstream; the result is a smooth adjustmentof the flow, as we have seen in the previous section. When flow at a section issupercritical, and downstream conditions will require a change to subcritical flow, the

need for this change cannot be communicated upstream; the flow speed exceedsthe speed of surface waves, which are the mechanism for transmitting changes. Thus agradual change with a smooth transition through the critical point is not possible. Thetransition from supercritical to subcritical flow occurs abruptly through a hydraulicjump. Hydraulic jumps can occur in canals downstream of regulating sluices, at thefoot of spillways (see Fig. 11.12a), where a steep channel slope suddenly becomesflatand even in the home kitchen (see Fig. 11.12b)! The specific energy curve andgeneral shape of a jump are shown in Fig. 11.13. We will see in this section that thejump always goes from a supercritical depth (y1 ,yc) to a subcritical depth (y2 .yc)and that there will be a drop Ein the specific energy. Unlike the changes due tophenomena such as a bump, the abrupt change in depth involves a significant loss ofmechanical energy through turbulent mixing.

The control volume for a hydraulic jump is sketched in Fig. 11.14.We shall analyze the jump phenomenon by applying the basic equations to the

control volume shown in the sketch. Experiments show that the jump occurs over arelatively short distanceat most, approximately six times the larger depth (y2) [9]. Inview of this short length, it is reasonable to assume that friction force Ffacting on thecontrol volume is negligible compared to pressure forces. Note that we are thereforeignoring viscous effects for momentum considerations, but notfor energy considera-tions (as we just mentioned, there is considerable turbulence in the jump). Althoughhydraulic jumps can occur on inclined surfaces, for simplicity we assume a horizontalbed, and rectangular channel of width b; the results we obtain will apply generally tohydraulic jumps.

Hence we have the following assumptions:

1. Steady flow


3. Uniform velocity at each section

(a) The Burdekin dam in Australia (b) The Kitchen Sink

(James Kilfiger)

Fig. 11.12 Examples of a hydraulic jump.

VIDEO

A Laminar Hydraulic Jump.

11.4The Hydraulic Jump 6


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4. Hydrostatic pressure distribution at each section

5. Frictionless flow (for the momentum equation)

These assumptions are familiar from previous discussions in this chapter. For anincompressible flow withuniform velocityat each section, we can use the appropriateform of continuity from Chapter 4,X

CS~V ~A 5 0 4:13b

Applying Eq. 4.13b to the control volume we obtain

2 V1by1 1 V2by2 5 0

or

V1y1 5 V2y2 11:31This is the continuity equation for the hydraulic jump. For the momentum equation,again with the assumption of uniform velocity at each section, we can use the fol-lowing form for the x component of momentum

Fx 5 FSx 1 FBx 5@

@t

ZCV

u dV--- 1X

CSu~V ~A 4:18d

The unsteady term@/@tdisappears as the flow is steady, and the body force FBx is zerofor horizontal flow. So we obtain

FS~x 5X

CSu~V ~A 11:32

E

Specific energy curve Hydraulic jump

E

yc

y1

y

yc

y2

Fig. 11.13 Specific energy curve for flow through a hydraulic jump.

Control

volume

Flow

y

x

Ff

y1

y2

Fig. 11.14 Schematic of hydraulic jump, showingcontrol volume used for analysis.



28/57

The surface force consists of pressure forces on the two ends and friction force on thewetted surface. By assumption5we neglect friction. The gage pressure at the two endsis hydrostatic, as illustrated in Fig. 11.3b. We recall from our study of hydrostatics thatthe hydrostatic force, FR, on a submerged vertical surface of area, A , is given by thesimple result

FR 5 pcA 3:10b

where pc is the pressure at the centroid of the vertical surface. For the two verticalsurfaces of the control volume, then, we have

FSx 5 FR1 2 FR2 5pcA1 2 pcA2 5fgy1y1bg 2 fgy2y2bg

5gb

2 y21 2y22

Using this result in Eq. 11.32, and evaluating the terms on the right,

FSx 5gb

2 y21 2y22 5

XCS

u~V ~A 5 V1f2V1y1bg 1 V2fV2y2bg

Rearranging and simplifying

V21y1

g 1

y212

5V22y2

g 1

y222

11:33

This is the momentum equation for the hydraulic jump. We have already derived theenergy equation for open-channel flows,

V212g

1y1 1 z1 5V222g

1y2 1 z2 1 Hl 11:10

For our horizontal hydraulic jump, z1 = z2, so

E1 5V212g

1y1 5V222g

1y2 1 Hl 5 E2 1 Hl 11:34

This is the energy equation for the hydraulic jump; the loss of mechanical energy is

E 5 E1 2 E2 5 Hl

The continuity, momentum, and energy equations (Eqs. 11.31, 11.33, and 11.34,respectively) constitute a complete set for analyzing a hydraulic jump.

Depth Increase Across a Hydraulic Jump

To find the downstream or, as it is called, the sequentdepth in terms of conditionsupstream from the hydraulic jump, we begin by eliminating V2 from the momentumequation. From continuity, V2 = V1y1/y2(Eq. 11.31), so Eq. 11.33 can be written

V21y1

g 1

y212

5V21y1

g

y1

y2

1

y222

Rearranging

y22 2y21 5

2V21y1g

1 2y1

y2

5

2V21y1g

y2 2y1

y2

11.4The Hydraulic Jump 6


29/57

Dividing both sides by the common factor (y2y1), we obtain

y2

1y1

52V21y1

gy2Next, multiplying by y2 and dividing by y

21 gives

y2

y1

21

y2

y1

5

2V21gy1

5 2Fr21 11:35

Solving for y2/y1 using the quadratic formula (ignoring the physically meaninglessnegative root), we obtain

y2y1

51

2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 1 8Fr21

q 2 1

11:36

Hence, the ratio of downstream to upstream depths across a hydraulic jump is only a

function of the upstream Froude number. Equation 11.36 has been experimentally wellverified, as can be seen in Fig. 11.15a. Depthsy1 andy2 are referred to as conjugate depths.From Eq. 11.35, we see that an increase in depth (y2 .y1) requires an upstream Froudenumbergreater than one(Fr1 . 1).We have notyet established that we musthave Fr1 . 1,just that it must be for an increase in depth (theoretically we could haveFr1 , 1 andy2 ,y1); we will now consider the head loss to demonstrate that we musthaveFr1 . 1.

Head Loss Across a Hydraulic Jump

Hydraulic jumps often are used to dissipate energy below spillways as a means ofpreventing erosion of artificial or natural channel bottom or sides. It is therefore ofinterest to be able to determine the head loss due to a hydraulic jump.

From the energy equation for the jump, Eq. 11.34, we can solve for the head loss

Hl 5 E1 2 E2 5V212g

1y1 2V222g

1y2

From continuity, V2 = V1y1/y2, so

Hl 5V212g

1 2y1y2

2" #1 y1 2y2

00

2

4

6

8

10

12

14

1 2 3 4

Upstream Froude number, Fr1

5 6 7 8 9

(a) Depth ratio (b) Head loss

Eq. 11.39Eq. 11.36

10

Depthratio,

y2

y1

00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

2 4

Upstream Froude number, Fr1

6 8 10

Dimension

lessheadloss,

Hl

E1

Fig. 11.15 Depthratio andhead loss fora hydraulic jump. (Data from Peterka [9].)



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31/57

Given: Rectangular channel with hydraulic jump in which flow depth changes from 0.6 m to 1.6 m.

Find: Flow rate, critical depth, and head loss in the jump.

Solution: Use the equation that relates depths y1and y2in terms of the Froude number (Eq. 11.36); then use theFroude number (Eq. 11.7) to obtain the flow rate; use Eq. 11.23 to obtain the critical depth; and finally computethe head loss from Eq. 11.38b.

Governing equations:y2y1

5 12

2 1 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1 1 8Fr21q 11:36

Fr 5Vffiffiffiffiffigy

p 11:7

yc 5Q2

gb2

1=311:23

Hl 5y2 2y13

4y1y211:38b

(a) From Eq. 11.36

Fr1 5

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 1 2y2y1

22 1

8

vuuuuut

5

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 1 2 3

1:6 m

0:6 m

22 1

8

vuuuuutFr1 5 2:21

As expected, Fr1 . 1 (supercritical flow). We can now use the definition of Froude number for open-channelflow to find V1

Fr1 5 V1ffiffiffiffiffiffiffigy1

pHence

V1 5 Fr1ffiffiffiffiffiffiffi

gy1p

5 2:21 3

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi9:81 m

s2 3 0:6 m

r 5 5:36 m=s

From this we can obtain the flow rate, Q.

Q 5 by1V1 5 3:0 m 3 0:6 m 35:36 m

s

Q 5 9:65 m3=s Q

(b) The critical depth can be obtained from Eq. 11.23.

yc 5Q2

gb2

24

351=3

5

9:652

m6

s2 3

s2

9:81 m 3

1

3:02 m2

1=3

yc 5 1:02 m yc



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11.5Steady Uniform FlowAfter studying local effects such as bumps and hydraulic jumps, and defining somefundamental quantities such as the specific energy and critical velocity, we are readyto analyze flows in long stretches. Steady uniform flow is something that is to beexpected to occur for channels of constant slope and cross section; Figs. 11.1 and 11.2

show examples of this kind of flow. Such flows are very common and important, andhave been extensively studied.

The simplest such flow is fully developed flow; it is analogous to fully developedflow in pipes. A fully developed flow is one for which the channel is prismatic, that is, achannel with constant slope and cross section that flows at constant depth. This depth,yn, is termed the normal depth and the flow is termed a uniform flow. Hence theexpression uniform flow in this chapter has a different meaning than in earlierchapters. In earlier chapters it meant that the velocity was uniformat a section of theflow; in this chapter we use it to mean that, but in addition specifically that the flow isthe same at all sections. Hence for the flow shown in Fig. 11.16, we have A1 =A2 =A(cross-section areas),Q1 = Q2 = Q(flow rates),V1 = V2 = V(average velocity,V= Q/A),andy1 =y2 = yn (flow depth).

As before (Section 11.2), we use the following assumptions:

1. Steady flow


3. Uniform velocity at a section

4. Gradually varying depth so that pressure distribution is hydrostatic

5. Bed slope is small

6. _Ws 5 _Wshear 5 _Wother 5 0

Note that as illustrated in Fig. 11.13, y1 ,yc ,y2.

(c) The head loss can be found from Eq. 11.38b.

Hl 5y2 2y13

4y1y2

51

4

1:6 m 2 0:6 m3

1:6 m3

0:6 m

5 0:260m Hl

As a verification of this result, we use the energy equation directly,

Hl 5 E1 2 E2 5 y1 1V212g

2 y2 1

V222g

withV2 5 Q/(by2) 5 2.01 m/s,

Hl 5

0:6 m 1 5:362

m2

s2 3

1

2 3

s2

9:81 m

2

1:6 m 1 2:012

m2

s2 3

1

2 3

s2

9:81 m

Hl 5 0:258m

ThisExampleillustratescomputatofflowrate,criticaldepth,andheloss,fora hydraulicjump.

11.5 Steady Uniform Flow 6


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Note that assumption5 means that we can approximate the flow depthy to be verticaland flow speed horizontal. (Strictly speaking they should be normal and parallel to thechannel bottom, respectively.)

The continuity equation is obvious for this case.

Q 5

V1A1 5

V2A2 5

VA

For the momentum equation, again with the assumption of uniform velocity at eachsection, we can use the following form for the x component of momentum

Fx 5 FSx 1 FBx 5@

@t

ZCV

u dV--- 1X

CSu ~V ~A 4:18d

The unsteady term @/@t disappears as the flow is steady, and the control surfacesummation is zero because V1 = V2; hence the right hand side is zero as there is nochange of momentum for the control volume. The body force FBx 5 Wsin whereWis the weight of fluid in the control volume;is the bed slope, as shown in Fig. 11.16.The surface force consists of the hydrostatic force on the two end surfaces at 1 and 2and the friction force Ff on the wetted surface of the control volume; however,

because we have the same pressure distributions at 1 and 2, the netx component ofpressure force is zero. Using all these results in Eq. 4.18d we obtain

2 Ff 1 Wsin 5 0

or

Ff 5 Wsin 11:40We see that for flow at normal depth, the component of the gravity force driving theflow is just balanced by the friction force acting on the channel walls. This is incontrast to flow in a pipe or duct, for which (with the exception of pure gravity drivenflow) we usually have a balance between an applied pressure gradient and the friction.The friction force may be expressed as the product of an average wall shear stress, w,and the channel wetted surface area,PL (whereLis the channel length), on which thestress acts

Ff 5 wPL 11:41The component of gravity force can be written as

Wsin 5 gALsin gALgALSb 11:42

Control

volume

z1

y1=yn

z2

Sb= tan

y2=yn

Fig. 11.16 Control volume for uniform channelflow.



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which is valid for SI units. Mannings equation in SI units can also be expressed as

Q 51

nAR

2=3h S

1=2b 11:48a

ForVin ft/s andRhin feet (English Engineering units), Eq. 11.47a can be rewritten as

V 51:49

n R

2=3h S

1=2b 11:47b

and Eq. 11.48a can be written as

Q 51:49

n AR

2=3h S

1=2b 11:48b

whereA is in square feet. Note that a number of these equations, as well as many thatfollow, are engineering equations; that is, the user needs to be aware of the requiredunits of each term in the equation. In Table 11.1 we have previously listed data on AandR

hfor various channel geometries.

The relationship among variables in Eqs. 11.48 can be viewed in a number of ways.For example, it shows that the volume flow rate through a prismatic channel of givenslope and roughness is a function of both channel size and channel shape. This isillustrated in Examples 11.6 and 11.7.

Table 11.2A Selection of Mannings Roughness Coefficients

Manningsn

Depth Ranges

Lining Category Lining Type 00.5 ft (015 cm) 0.52.0 ft (1560 cm) .2.0 ft (.60 cm)Rigid Concrete 0.015 0.013 0.013

Grouted riprap 0.040 0.030 0.028Stone masonry 0.042 0.032 0.030Soil cement 0.025 0.022 0.020Asphalt 0.018 0.016 0.016

Unlined Bare soil 0.023 0.020 0.020Rock cut 0.045 0.035 0.025

Temporary Woven paper net 0.016 0.015 0.015Jute net 0.028 0.022 0.019Fiberglass roving 0.028 0.021 0.019Straw with net 0.065 0.033 0.025Curled wood mat 0.066 0.035 0.028

Synthetic mat 0.036 0.025 0.021Gravel riprap 1 in (2.5 cm) D50 0.044 0.033 0.0302 in (5 cm) D50 0.066 0.041 0.034

Rock riprap 6 in (15 cm) D50 0.104 0.069 0.03512 in (30 cm) D50 0.078 0.040

Source:Linsley et al. [11] and Chen and Cotton [12].



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Example11.6 FLOW RATE IN A RECTANGULAR CHANNELAn 8-ft-wide rectangular channel with a bed slope of 0.0004 ft/ft has a depth of flow of 2 ft. Assuming steady uniforflow, determine the discharge in the channel. The Manning roughness coefficient is n = 0.015.

Given: Geometry of rectangular channel and flow depth.

Find: Flow rateQ.Solution: Use the appropriate form of Mannings equation. For a problem in English Engineering units, this iEq. 11.48b.

Governing equations:

Q 51:49

n AR

2=3h S

1=2b Rh 5

by

b 1 2y Table 11:1

Using this equation with the given data

Q 51:49

n AR

2=3h S

1=2b

5

1:49

0:015 3 8 ft 3 2 ft 38 ft 3 2 ft

8 ft 1 2 3 2 ft0@ 1A2=3

3 0:004ft

ft0@ 1A1=2

Q 5 38:5 ft3=s Q

ThisExampledemonstratesuseManningsequationtosolveforrate, Q.Notethatbecausethisisengineeringequ

ation,theunitsnotcancel.

Example11.7 FLOW VERSUS AREA THROUGH TWO CHANNEL SHAPESOpen channels, of square and semicircular shapes, are being considered for carrying flow on a slope ofSb = 0.001; t

channel walls are to be poured concrete with n = 0.015. Evaluate the flow rate delivered by the channels for maimum dimensions between 0.5 and 2.0 m. Compare the channels on the basis of volume flow rate for given crosectional area.

Given: Square and semicircular channels;Sb = 0.001 andn = 0.015. Sizes between0.5 and 2.0 m across.

Find: Flow rate as a function of size. Compare channels on the basis of volumeflow rate, Q, versus cross-sectional area, A.

Solution: Use the appropriate form of Mannings equation. For a problem in SIunits, this is Eq. 11.48a.

Governing equation:

Q 51

nAR2=3h S

1=2b 11:48a

Assumption: Flow at normal depth.

For the square channel,

P 5 3b and A 5 b2 so Rh 5b

3

b

y



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Using this in Eq. 11.48a

Q 51

nAR

2=3h S

1=2b 5

1

nb2

b

3

2=3S

1=2b 5

1

32=3nS

1=2b b

8=3

Forb = 1 m,

Q 51

32=30:015 0:001

1=2

1

8=35 1:01 m3=s

Q

Tabulating for a range of sizes yields

b(m) 0.5 1.0 1.5 2.0

A(m2) 0.25 1.00 2.25 4.00

Q (m3/s) 0.160 1.01 2.99 6.44

For the semicircular channel,

P 5D

2 and A 5

D2

8

so Rh 5 D2

82D

5 D4

Using this in Eq. 11.48a

Q 51

nAR

2=3h S

1=2b 5

1

n

D2

8

D

4

0@

1A2=3S1=2b

5

45=32n S1=2b D

8=3

ForD = 1 m,

Q 5

45=320:015 0:001

1=2

1

8=35 0:329m3=s

Q

Tabulating for a range of sizes yields

D(m) 0.5 1.0 1.5 2.0

A(m2) 0.0982 0.393 0.884 1.57

Q (m3/s) 0.0517 0.329 0.969 2.09

For both channels, volume flow rate varies as

QBL8=3 or QBA4=3

sinceABL2. The plot of flow rate versus cross-sectional area shows thatthe semicircular channel is more efficient.

Performance of the two channels may be compared at any specifiedarea. At A 5 1 m2, Q/A 5 1.01 m/s for the square channel. For thesemicircularchannel with A 5 1 m2, then D 5 1.60 m, and Q 5 1.15 m3/s;so Q/A 5 1.15 m/s. Thus the semicircular channel carries approximately14 percent more flow per unit area than the square channel.

0.1

1.0

10.0

0.1 1.0 10.0

Cross-sectional area, A(m2)

Volumeflowr

ate,

Q(m

3/s) Semicircular

Square

D

Thecomparisononcross-sectionalareaisimportantindeterminingtheamountofexcavationrequiredtobuildthechannel.Thechannelshapesalsocouldbecomparedonthebasisofperimeter,whichwouldindicatetheamountofconcreteneededtofinishthechannel.

TheExcelworkbookforthisproblemcanbeusedforcom-

putingdataandplottingcurvesforothersquareandsemicircularchannels.



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We have demonstrated that Eqs. 11.48 mean that, for normal flow, the flow ratedepends on the channel size and shape. For a specified flow rate through a prismaticchannel of given slope and roughness, Eqs. 11.48 also show that the depth of uniformflow is a function of both channel size and shape, as well as the slope. There is onlyone depth for uniform flow at a given flow rate; it may be greater than, less than, orequal to the critical depth. This is illustrated in Examples 11.8 and 11.9.

Example11.8 NORMAL DEPTH IN A RECTANGULAR CHANNELDetermine the normal depth (for uniform flow) if the channel described in Example 11.6 has a flow rate of 100 c

Given: Geometric data on rectangular channel of Example 11.6.

Find: Normal depth for a flow rateQ =100 ft3/s.

Solution: Use the appropriate form of Mannings equation. For a problem in English Engineering units,this is Eq. 11.48b.

Governing equations:

Q 51:49

n

AR2=3h S

1=2b Rh 5

byn

b1

2yn Table 11:1

Combining these equations

Q 51:49

n AR

2=3h S

1=2b 5

1:49

n byn byn

b 1 2yn

2=3S

1=2b

Hence, after rearranging

Q n

1:49b5=3S1=2b

!3b 1 2yn2 5 y5n

SubstitutingQ =100 ft3/s, n = 0.015,b = 8 ft, and Sb = 0.0004 and simplifying(remembering this is an engineering equation, in which we insert values

without units),

3:898 1 2yn2 5 y5nThis nonlinear equation can be solved for yn using a numerical methodsuch as the Newton-Raphson method (or better yet using your calculatorssolving feature or Excels Goal Seek or Solver!). We find

yn 5 3:97ft yn

Note that there are five roots, but four of them are complexmathematically correctbut physically meaningless.

ThisExampledemonstratesthofManningsequationforfindthenormaldepth.Thisrelativelysimplephysicalproblemstillinvolvedsolving

a

nonlinearalgebraicequation.TheExcelworkbookforthisproblemcanbeusedforsolv

similarproblems.

Example11.9 DETERMINATION OF FLUME SIZEAn above-ground flume, built from timber, is to convey water from a mountain lake to a small hydroelectric plaThe flume is to deliver water at Q = 2 m3/s; the slope isSb = 0.002 andn = 0.013. Evaluate the required flume size (a) a rectangular section with y/b = 0.5 and (b) an equilateral triangular section.

Given: Flume to be built from timber, with Sb = 0.002,n = 0.013, andQ =2.00 m3/s.



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Find: Required flume size for:(a) Rectangular section with y/b =0.5.(b) Equilateral triangular section.

Solution: Assume flume is long, so flow is uniform; it is at normal depth. Then Eq. 11.48a applies.

Governing equation:

Q 5 1n

AR2=3h S1=2b 11:48a

The choice of channel shape fixes the relationship between RhandA; so Eq. 11.48a may be solved for normal depth,yn, thus determining the channel size required.

(a) Rectangular section

P 5 2yn 1 b; yn=b 5 0:5 sob 5 2yn

P 5 2yn 1 2yn 5 4yn A 5 ynb 5 yn2yn 5 2y2nsoRh 5

A

P 5

2y2n4yn

5 0:5yn

Using this in Eq. 11.48a,

Q 51

nAR

2=3h S

1=2b 5

1

n2y2n0:5yn2=3S1=2b 5

20:52=3n

y8=3n S1=2b

Solving for yn

yn 5nQ

20:52=3S1=2b

" #3=85

0:0132:0020:52=30:0021=2" #3=8

5 0:748m

The required dimensions for the rectangular channel are

yn 5 0:748m A 5 1:12 m2

b 5 1:50 m p 5 3:00 m Flume size

(b) Equilateral triangle section

P 5 2s 52yn

cos 30o A 5

yns

2 5

y2n2cos30o

so Rh 5A

P 5

yn

4

Using this in Eq. 11.48a,

Q 51

nAR

2=3h S

1=2b 5

1

n

y2n2cos30o

yn

4

2=3S

1=2b 5

1

2cos30o42=3ny8=3n S

1=2b

Solving for yn

yn 5 2cos30o

42=3

nQS

1=2b

" #3=85 2cos30

o

42=3

0:0132:000:0021=2" #3=8

5 1:42 m

The required dimensions for the triangular channel are

yn 5 1:42 m A 5 1:16 m2

bs 5 1:64 m p 5 3:28 m Flumesize

yns

b

yn



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EGL 5V2

2g 1 z 1y 11:50

and

HGL 5 z 1y 11:51

Hence, using Eqs. 11.50 and 11.51 in Eqs. 11.10, between points 1 and 2 we obtain

EGL1 2 EGL2 5 Hl 5 z1 2 z2

and (because V1 = V2)

HGL1 2 HGL2 5 Hl 5 z1 2 z2

For normal flow, the energy grade line, the hydraulic grade line, and the channel bed are

all parallel. The trends for the energy grade line, hydraulic grade line, and specificenergy, are shown in Fig. 11.17.

Optimum Channel Cross Section

For given slope and roughness, the optimum channel cross section is that for which weneed the smallest channel for a given flow rate; this is when Q/Ais maximized. FromEq. 11.48a (using the SI version, although the results we obtain will apply generally)

Q

A 5

1

nR

2=3h S

1=2b 11:52

Thus the optimum cross section has maximum hydraulic radius,Rh. SinceRh =A/P,Rh

is maximum when the wetted perimeter is minimum. Solving Eq. 11.52 for A (withRh =A/P) then yields

A 5nQ

S1=2b

" #3=5P2=5 11:53

From Eq. 11.53, the flow area will be a minimum when the wetted perimeter is aminimum.

EGL1

V

2g

21

HGL1

E1

V

2g

22

E2

EGL2

HGL2

EGL line

HGL line

Fig. 11.17 Energy grade line, hydraulic grade line, andspecific energy for uniform flow.



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Wetted perimeter, P, is a function of channel shape. For any given prismaticchannel shape (rectangular, trapezoidal, triangular, circular, etc.), the channel crosssection can be optimized. Optimum cross sections for common channel shapes aregiven without proof in Table 11.3.

Once the optimum cross section for a given channel shape has been determined,expressions for normal depth, yn, and area, A, as functions of flow rate can beobtained from Eq. 11.48. These expressions are included in Table 11.3.

11.6Flow with Gradually Varying DepthMost human-made channels are designed to have uniform flow (for example, see

Fig. 11.1). However, this is not true in some situations. A channel can have nonuniformflow, that is,a flowfor whichthe depth andhence speed, andso on vary along thechannelfor a number of reasons. Examples include when an open-channel flow encounters achange in bed slope, geometry, or roughness, or is adjusting itself back to normal depthafter experiencing an upstream change (such as a sluice gate). We have already studiedrapid, localized changes, such as that occurring in a hydraulic jump, but here we assumeflow depth changes gradually. Flow with gradually varying depth is analyzed by applyingthe energy equation to a differential control volume; the result is a differential equation

Table 11.3Properties of Optimum Open-Channel Sections (SI Units)

Shape Section

Optimum

Geometry

yn

Qn

Sb

60

b

Normal

Depth,yn

Cross-Sectional

Area,A

20.968

1/2

3/8

3

Qn

Sb

1.6221/2

3/4

Qn

Sbb 2yn

0.9171/2

3/8

Qn

Sb

1.6821/2

3/4

Qn

SbD 2yn

1.001/2

3/8

Qn

Sb

1.5831/2

3/4

Qn

Sb

yn

yn

yn

b

b

D

b>>y

yn

yn

45 1.297 1/2

3/8

Qn

Sb

1.6821/2

3/4

(Q/b)nNone 1.00

Sb1/2

3/8

Trapezoidal

Triangular

Rectangular

Wide Flat

Circular

11.6 Flow with Gradually Varying Depth 6


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that relates changes in depth to distance along the flow. The resulting equation may besolved analytically or, more typically numerically, ifwe approximate the head loss ateach section as being the same as that for flow at normal depth, using the velocity and

hydraulic radius of the section. Water depth and channel bed height are assumed tochange slowly. As in the case of flow at normal depth, velocity is assumed uniform, andthe pressure distribution is assumed hydrostatic at each section.

The energy equation (Eq. 11.10) for open-channel flow was applied to a finite

control volume in Section 11.2,

V212g

1y1 1 z1 5V222g

1y2 1 z2 1 Hl 11:10

We apply this equation to the differential control volume, of length dx, shown inFig. 11.18. Note that the energy grade line, hydraulic grade line, and channel bottom allhave different slopes, unlike for the uniform flow of the previous section!

The energy equation becomes

V2

2g 1y 1 z 5

V2

2g 1 d

V2

2g

1y 1 dy 1 z 1 dz 1 dHl

or after simplifying and rearranging

2 d V2

2g

2 dy 2 dz 5 dHl 11:54

This is not surprising. The differential loss of mechanical energy equals the differentialhead loss. From channel geometry

dz 52 Sbdx 11:55We also have the approximation that the head loss in this differential nonuniform flowcan be approximated by the head loss that uniform flow would have at the sameflow rate, Q, at the section. Hence the differential head loss is approximated by

dHl 5 Sdx 11:56

whereS is the slope of the EGL (see Fig. 11.18). Using Eqs. 11.55 and 11.56 in Eq.11.54, dividing by dx , and rearranging, we obtain

V2

2g

y

EGL line

HGL line

+

V2d

V2

2g2g

z

y+dy

z+ dz

dx

Slope Sb

Slope S

Fig. 11.18 Control volume for energy analysis of gradually varying flow.



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d

dx

V2

2g

1

dy

dx 5 Sb 2 S 11:57

To eliminate the velocity derivative, we differentiate the continuity equation, Q =VA = const, to obtain

dQ

dx 5 0 5 A

dV

dx 1 V

dA

dx

or

dV

dx 5 2

V

A

dA

dx 5 2

Vbs

A

dy

dx 11:58

where we have used dA = bsdy (Eq. 11.17), where bs is the channel width at the freesurface. Using Eq. 11.58 in Eq. 11.57, after rearranging

d

dx

V2

2g

1

dy

dx 5

V

g

dV

dx 1

dy

dx 5 2

V2bs

gA

dy

dx 1

dy

dx 5 Sb 2 S 11:59

Next, we recognize that

V2bs

gA 5

V2

gA

bs

5

V2

gyh 5 Fr2

whereyh is the hydraulic depth (Eq. 11.2). Using this in Eq. 11.59, we finally obtainour desired form of the energy equation for gradually varying flow

dy

dx 5

Sb 2 S

1 2 Fr2 11:60

This equation indicates how the depth y of the flow varies. Whether the flow becomesdeeper (dy/dx . 0) or shallower (dy/dx , 0) depends on the sign of the right-hand side.For example, consider a channel that has a horizontal section (Sb =0):

dydx

5 2 S1 2 Fr2

Because of friction the EGL always decreases, so S . 0. If the incoming flow is sub-critical (Fr, 1), the flow depth will gradually decrease (dy/dx , 0); if the incomingflow is supercritical (Fr. 1), the flow depth will gradually increase (dy/dx . 0). Notealso that for critical flow (Fr= 1), the equation leads to a singularity, and graduallyflow is no longer sustainablesomething dramatic will happen (guess what).

Calculation of Surface Profiles

Equation 11.60 can be used to solve for the free surface shapey(x); the equation lookssimple enough, but it is usually difficult to solve analytically and so is solved

numerically. It is difficult to solve because the bed slope, Sb, the local Froude number,Fr, andS, the EGL slope equivalent to uniform flow at rate Q, will in general all varywith location, x. For S, we use the results obtained in Section 11.5, specifically

Q 51

nAR

2=3h S

1=2 11:48a

or for English Engineering units



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Q 51:49

n AR

2=3h S

1=2 11:48b

Note that we have used S rather thanSbin Eq. 11.48 as we are using the equation toobtain an equivalentvalue ofS for a uniform flow at rate Q! Solving for S ,

S 5n2Q2

A2R4=3h

11:61a


S 5n2Q2

1:492A2R4=3h

11:61b

We can also express the Froude number as a function ofQ,

Fr 5Vffiffiffiffiffiffiffigyh

p 5 QA ffiffiffiffiffiffiffi

gyhp 11:62

Using Eqs. 11.61a (or 11.61b) and 11.62 in Eq. 11.60

dy

dx 5Sb 2 S

1 2 Fr2 5

Sb 2n2Q2

A2R4=3

h

1 2Q2

A2gyh

11:63a


dy

dx 5

Sb 2n2Q2

1:492A2R4=3h

1 2Q2

A2gyh

11:63b

For a given channel (slope, Sb, and roughness coefficient, n, both of which may varywithx) and flow rateQ, the areaA, hydraulic radiusRh, and hydraulic depthyhare allfunctions of depthy (see Section 11.1). Hence Eqs. 11.63 are usually best solved using

a suitable numerical integration scheme. Example 11.10 shows such a calculation forthe simplest case, that of a rectangular channel.

Example11.10 CALCULATION OF FREE SURFACE PROFILEWater flows in a 5-m-wide rectangular channel made from unfinished concrete withn 5 0.015. The channel contains along reach on whichSbis constant at Sb 5 0.020. At one section, flow is at depth y1 5 1.5 m, with speed V1 5 4.0 m/s.Calculate and plot the free surface profile for the first 100 m of the channel, and find the final depth.

Given: Water flow in a rectangular channel.

Find: Plot of free surface profile; depth at 100 m.

Solution: Use the appropriate form of the equation for flow depth, Eq. 11.63a.

Governing equation:

dy

dx 5

Sb 2 S

1 2 Fr2 5

Sb 2n2Q2

A2R4=3h

1 2Q2

A2gyh

11:63a



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We use Eulers method (see Section 5.5) to convert the differential equation to a difference equation.In this approach, the differential is converted to a difference,

dy

dxy

x 1

where xand yare small but finite changesin x and y, respectively. Combining Eqs.

11.63a and 1, and rearranging,

y 5 x

Sb 2n2Q2

A2R4=3h

1 2Q2

A2gyh

0BBB@

1CCCA

Finally, we let y 5yi11 yi, where yi andyi11are the depths at point i and a point (i+1)distance x further downstream,

yi 1 1 5 yi 1 x

Sbi 2n2i Q

2

A2i R4=3hi

1 2 Q2A2igyhi

0BBBB@

1CCCCA 2

Equation 2 computes the depth, yi11, given data at point i. In the current application, Sband n are constant, but Rh, and yh will, of course, vary with x because they are functions of y. For a rectangular channel we have tfollowing:

Ai 5 byi

Rhi 5byi

b 1 2yi

yhi 5Ai

bs5

Ai

b 5

byi

bs5 yi

The calculations are conveniently performed and results plotted using Excel. Note that partial results are showin the table, and that for the first meter, over which there is a rapid change in depth, the step size is x 5 0.05.

i x(m) y(m) A(m2) Rh (m) yh(m)

1 0.00 1.500 7.500 0.938 1.500

2 0.05 1.491 7.454 0.934 1.491

3 0.10 1.483 7.417 0.931 1.483

4 0.15 1.477 7.385 0.928 1.477

5 0.20 1.471 7.356 0.926 1.471... ..

. ... ..

. ... ..

.

118 98 0.916 4.580 0.670 0.916

119 99 0.915 4.576 0.670 0.915120 100 0.914 4.571 0.669 0.914

The depth at location x 5 100 m is seen to be 0.914 m.

y100m 5 0:914m y100m

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

0 20 40 60 80 1

Distance along channel, x(m)

Waterdepth,y(m)



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11.7Discharge Measurement Using Weirs

Aweiris a device (or overflow structure) that is placed normal to the direction of flow.The weir essentially backs up water so that, in flowing over the weir, the water goesthrough critical depth. Weirs have been used for the measurement of water flow inopen channels for many years. Weirs can generally be classified as sharp-crested weirsand broad-crested weirs. Weirs are discussed in detail in Bos [14], Brater [15], andReplogle [16].

A sharp-crested weir is basically a thin plate mounted perpendicular to the flowwith the top of the plate having a beveled, sharp edge, which makes the nappe springclear from the plate (see Fig. 11.19).

The rate of flow is determined by measuring the head, typically in a stilling well(see Fig. 11.20) at a distance upstream from the crest. The head His measured using agage.

Suppressed Rectangular Weir

These sharp-crested weirs are as wide as the channel and the width of the nappe is thesame length as the crest. Referring to Fig. 11.20, consider an elemental area dA = bdhand assume the velocity is V 5

ffiffiffiffiffiffiffiffi2gh

p ; then the elemental flow is

dQ 5 bdhffiffiffiffiffiffiffiffi

2ghp

5 bffiffiffiffiffi

2gp

h1=2dh

Note (following the solution procedure of Example 11.8) that the normaldepth for this flow is yn 5 0.858 m; the flow depth is asymptoticallyapproaching this value. In general, this is one of several possibilities,depending on the values of the initial depth and the channel properties (slopeand roughness). A flow may approach normal depth, become deeper anddeeper, or eventually become shallower and experience a hydraulic jump.

Theaccuracyoftheresultsobtainedobviouslydependsonthenumericalmodelused;forexample,a moreaccuratemodelistheRK4method.Also,forthefirstmeterorso,therearerapidchangesindepth,bringing

into

questionthevalidityofseveralassumptions,forexample,uniform

flowandhydrostaticpressure.TheExcelworkbookforthisproblemcanbemodifiedforuseinsolvingsimilarproblems.

Crest V

Nappe V2g

DrawdownV0

2

V0

2g

H

P

Fig. 11.19 Flow over sharp-crested weir.



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The discharge is expressed by integrati

strujanje u otvorenim kanalima

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