Open Channel Flow Fluid Mechanics Fundamentals and Applications-2th Edition

O P E N - C H A N N E L F L O W

Open-channel flow implies flow in a channel open to the atmosphere,but flow in a conduit is also open-channel flow if the liquid does notfill the conduit completely, and thus there is a free surface. The pipe

flow discussed in Chap. 8 involves closed conduits filled with a liquid or agas. An open-channel flow, however, involves liquids only (typically water orwastewater) exposed to a gas (usually air, which is at atmospheric pressure).

Flow in pipes is driven by a pressure difference, whereas flow in a chan-nel is driven naturally by gravity. Water flow in a river, for example, is dri-ven by the elevation difference between upstream and downstream. Theflow rate in an open channel is established by the dynamic balance betweengravity and friction. Inertia of the flowing liquid also becomes important inunsteady flow. The free surface coincides with the hydraulic grade line(HGL) and the pressure is constant along the free surface. But the height ofthe free surface from the channel bottom and thus all dimensions of theflow cross-section along the channel is not known a priori—it changesalong with average flow velocity. The pressure in a channel varies hydrosta-tically in the vertical direction when the flow is steady and fully developed.

In this chapter we present the basic principles of open-channel flows andthe associated correlations for steady one-dimensional flow in channels ofcommon cross sections. Detailed information can be obtained from severalbooks written on the topic, some of which are listed in the references.

679

CHAPTER

13OBJECTIVES

When you finish reading this chapter, youshould be able to

■ Understand how flow in openchannels differs from flow inpipes

■ Learn the different flow regimes in open channels and theircharacteristics

■ Predict if hydraulic jumps are tooccur during flow, and calculatethe fraction of energy dissipatedduring hydraulic jumps

■ Learn how flow rates in openchannels are measured usingsluice gates and weirs

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13–1 ■ CLASSIFICATION OF OPEN-CHANNEL FLOWS

Open-channel flow refers to the flow of liquids in channels open to theatmosphere or in partially filled conduits and is characterized by the pres-ence of a liquid–gas interface called the free surface (Fig. 13–1). Most nat-ural flows encountered in practice, such as the flow of water in creeks,rivers, and floods, as well as the draining of rainwater off highways, parkinglots, and roofs are open-channel flows. Human-made open-channel flowsystems include irrigation systems, sewer lines, drainage ditches, and gut-ters, and the design of such systems is an important application area ofengineering.

In an open channel, the flow velocity is zero at the side and bottom sur-faces because of the no-slip condition, and maximum at the midplane of thefree surface (when there are significant secondary flows, as in the bends ofnoncircular channels, the maximum velocity occurs below the free surfacesomewhere within the top 25 percent of depth, as shown in Fig. 13–2). Fur-thermore, flow velocity also varies in the flow direction in most cases. There-fore, the velocity distribution (and thus flow) in open channels is, in general,three-dimensional. In engineering practice, however, the equations are writ-ten in terms of the average velocity at a cross section of the channel. Sincethe average velocity varies only with streamwise distance x, V is a one-dimensional variable. The one-dimensionality makes it possible to solve sig-nificant real-world problems in a simple manner by hand calculations, andwe restrict our consideration in this chapter to flows with one-dimensionalaverage velocity. Despite its simplicity, the one-dimensional equations pro-vide remarkably accurate results and are commonly used in practice.

The no-slip condition on the channel walls gives rise to velocity gradients,and wall shear stress tw develops along the wetted surfaces. The wall shearstress varies along the wetted perimeter at a given cross section and offersresistance to flow. The magnitude of this resistance depends on the viscosityof the fluid as well as the velocity gradients at the wall surface.

Open-channel flows are also classified as being steady or unsteady. Aflow is said to be steady if there is no change with time at a given location.The representative quantity in open-channel flows is the flow depth (oralternately, the average velocity), which may vary along the channel. Theflow is said to be steady if the flow depth does not vary with time at anygiven location along the channel (although it may vary from one location toanother). Otherwise, the flow is unsteady. In this chapter we deal withsteady flow only.

Uniform and Varied FlowsFlow in open channels is also classified as being uniform or nonuniform(also called varied), depending on how the flow depth y (the distance of thefree surface from the bottom of the channel measured in the vertical direc-tion) varies along the channel. The flow in a channel is said to be uniformif the flow depth (and thus the average velocity) remains constant. Other-wise, the flow is said to be nonuniform or varied, indicating that the flowdepth varies with distance in the flow direction. Uniform flow conditions

680FLUID MECHANICS

FIGURE 13–1Natural and human-made open-channel flows are characterized by afree surface open to the atmosphere.© Vol. 16/PhotoDisc.

2.01.51.00.5

FIGURE 13–2Typical constant relative velocitycurves in an open channel oftrapezoidal cross section.

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are commonly encountered in practice in long straight sections of channelswith constant slope and constant cross section.

In open channels of constant slope and constant cross section, the liquidaccelerates until the head loss due to frictional effects equals the elevationdrop. The liquid at this point reaches its terminal velocity, and uniform flowis established. The flow remains uniform as long as the slope, cross section,and surface roughness of the channel remain unchanged. The flow depth inuniform flow is called the normal depth yn, which is an important charac-teristic parameter for open-channel flows (Fig. 13–3).

The presence of an obstruction in the channel, such as a gate or a changein slope or cross section, causes the flow depth to vary, and thus the flowto become varied or nonuniform. Such varied flows are common in bothnatural and human-made open channels such as rivers, irrigation systems,and sewer lines. The varied flow is called rapidly varied flow (RVF) if theflow depth changes markedly over a relatively short distance in the flowdirection (such as the flow of water past a partially open gate or over afalls), and gradually varied flow (GVF) if the flow depth changes gradu-ally over a long distance along the channel. A gradually varied flow regiontypically occurs between rapidly varied and uniform flow regions, as shownin Fig. 13–4.

In gradually varied flows, we can work with the one-dimensional averagevelocity just as we can with uniform flows. However, average velocity is notalways the most useful or most appropriate parameter for rapidly varyingflows. Therefore, the analysis of rapidly varied flows is rather complicated,especially when the flow is unsteady (such as the breaking of waves on theshore). For a known discharge rate, the flow height in a gradually variedflow region (i.e., the profile of the free surface) in a specified open channelcan be determined in a step-by-step manner by starting the analysis at across section where the flow conditions are known, and evaluating headloss, elevation drop, and then the average velocity for each step.

Laminar and Turbulent Flows in ChannelsLike pipe flow, open-channel flow can be laminar, transitional, or turbulent,depending on the value of the Reynolds number expressed as

(13–1)Re �rVRh

m�

VRh

n

681CHAPTER 13

V � constant

Slope: S0 � constant

y � yn � constant

Uniform flow

FIGURE 13–3For uniform flow in an open channel,the flow depth y and the average flow

velocity V remain constant.

GVFUF RVF GVF UF

FIGURE 13–4Uniform flow (UF), gradually variedflow (GVF), and rapidly varied flow

(RVF) in an open channel.

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Here V is the average liquid velocity, n is the kinematic viscosity, and Rh isthe hydraulic radius defined as the ratio of the cross-sectional flow area Acand the wetted perimeter p,

Hydraulic radius: (13–2)

Considering that open channels come with rather irregular cross sections,the hydraulic radius serves as the characteristic dimension and brings uni-formity to the treatment of open channels. Also, the Reynolds number isconstant for the entire uniform flow section of an open channel.

You might expect that the hydraulic radius would be defined as halfthe hydraulic diameter, but this is unfortunately not the case. Recall that thehydraulic diameter Dh for pipe flow is defined as Dh � 4Ac/p so thatthe hydraulic diameter reduces to the pipe diameter for circular pipes. Thenthe relation between hydraulic radius and hydraulic diameter becomes

Hydraulic diameter: (13–3)

So, we see that the hydraulic radius is in fact one-fourth, rather than one-half, of the hydraulic diameter (Fig. 13–5).

Therefore, a Reynolds number based on the hydraulic radius is one-fourthof the Reynolds number based on hydraulic diameter as the characteristicdimension. So it will come as no surprise that the flow is laminar for Re �2000 in pipe flow, but for Re � 500 in open-channel flow. Also, open-chan-nel flow is usually turbulent for Re � 2500 and transitional for 500 � Re �2500. Laminar flow is encountered when a thin layer of water (such as therainwater draining off a road or parking lot) flows at a low velocity.

The kinematic viscosity of water at 20°C is 1.00 � 10�6 m2/s, and theaverage flow velocity in open channels is usually above 0.5 m/s. Also, thehydraulic radius is usually greater than 0.1 m. Therefore, the Reynolds num-ber associated with water flow in open channels is typically above 50,000,and thus the flow is almost always turbulent.

Note that the wetted perimeter includes the sides and the bottom of thechannel in contact with the liquid—it does not include the free surface andthe parts of the sides exposed to air. For example, the wetted perimeter andthe cross-sectional flow area for a rectangular channel of height h and widthb containing water of depth y are p � b � 2y and Ac � yb, respectively.Then,

Rectangular channel: (13–4)

As another example, the hydraulic radius for the drainage of water of depthy off a parking lot of width b is (Fig. 13–6)

Liquid layer of thickness y: (13–5)

since b �� y. Therefore, the hydraulic radius for the flow of a liquid filmover a large surface is simply the thickness of the liquid layer.

Rh �Ac

p�

yb

b � 2y�

yb

b� y

Rh �Ac

p�

yb

b � 2y�

y

1 � 2y/b

Dh �4Ac

p� 4Rh

Rh �Ac

p (m)

682FLUID MECHANICS

I’ve known since grade school that radius is half of diameter. Now they tell me that hydraulic radius is one-fourth of hydraulic diameter!

???

FIGURE 13–5The relationship between the hydraulicradius and hydraulic diameter is notwhat you might expect.

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13–2 ■ FROUDE NUMBER AND WAVE SPEEDOpen-channel flow is also classified as tranquil, critical, or rapid, depend-ing on the value of the dimensionless Froude number discussed in Chap. 7and defined as

Froude number: (13–6)

where g is the gravitational acceleration, V is the average liquid velocity at across section, and Lc is the characteristic length, which is taken to be theflow depth y for wide rectangular channels. The Froude number is animportant parameter that governs the character of flow in open channels.The flow is classified as

(13–7)

Fr � 1 Supercritical or rapid flow

Fr � 1 Critical flow

Fr 1 Subcritical or tranquil flow

Fr �V

2gLc

�V

2gy

683CHAPTER 13

y

RR

u

Ac � R2(u � sin u cos u)

u � sin u cos u2u

u

Acp

p � 2Ru

Rh � R�

y(b � y/tan u)b � 2y/sin u

AcpRh � �

ybb � 2y

Acp

y1 � 2y/bRh � � �

y

b

ybb � 2y

y V b

(b) Trapezoidal channel

(d) Liquid film of thickness y

(a) Circular channel (u in rad)

(c) Rectangular channel

Acp

ybbRh � y� � �

b

y

b

FIGURE 13–6Hydraulic radius relations for various

open-channel geometries.

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This resembles the classification of compressible flow with respect to theMach number: subsonic for Ma 1, sonic for Ma � 1, and supersonic forMa � 1 (Fig. 13–7). Indeed, the denominator of the Froude number has thedimensions of velocity, and it represents the speed c0 at which a small dis-turbance travels in still liquid, as shown later in this section. Therefore, inanalogy to the Mach number, the Froude number is expressed as the ratio ofthe flow speed to the wave speed, Fr � V/c0, just as the Mach number isexpressed as the ratio of the flow speed to sound speed, Ma � V/c.

The Froude number can also be thought of as the square root of the ratioof inertia (or dynamic) force to gravity force (or weight). This can bedemonstrated by multiplying both the numerator and the denominator of thesquare of the Froude number V2/gLc by rA, where r is density and A is arepresentative area, which gives

(13–8)

Here LcA represents volume, rLcA is the mass of this fluid volume, and mgis the weight. The numerator is twice the inertial force rV2A, which can bethought of as the dynamic pressure rV2 times the cross-sectional area, A.

Therefore, the flow in an open channel is dominated by inertial forces inrapid flow and by gravity forces in tranquil flow.

It follows that at low flow velocities (Fr 1), a small disturbance travelsupstream (with a velocity c0 � V relative to a stationary observer) andaffects the upstream conditions. This is called tranquil or subcritical flow.But at high flow velocities (Fr � 1), a small disturbance cannot travelupstream (in fact, the wave is washed downstream at a velocity of V � c0relative to a stationary observer) and thus the upstream conditions cannot beinfluenced by the downstream conditions. This is called rapid or supercrit-ical flow, and the flow in this case is controlled by the upstream conditions.Therefore, a surface wave travels upstream when Fr 1, is swept down-stream when Fr � 1, and appears frozen on the surface when Fr � 1. Also,the surface wave speed increases with flow depth y, and thus a surface distur-bance propagates much faster in deep channels than it does in shallow ones.

Consider the flow of a liquid in an open rectangular channel of cross-sectional area Ac with a volume flow rate of V

.. When the flow is critical,

Fr � 1 and the average flow velocity is V � , where yc is the criti-cal depth. Noting that , the critical depth can beexpressed as

Critical depth (general): (13–9)

For a rectangular channel of width b we have Ac � byc, and the criticaldepth relation reduces to

Critical depth (rectangular): (13–10)

The liquid depth is y � yc for subcritical flow and y yc for supercriticalflow (Fig. 13–8).

As in compressible flow, a liquid can accelerate from subcritical to super-critical flow. Of course, it can also decelerate from supercritical to subcriti-

yc � aV#

��

2

gb2b1/3

yc �V#

2

gA2c

V#

� AcV � Ac1gyc

1gyc

12

12

Fr2 �V�

2

gLc

rA

rA�

2(12rV

2A)

mg

Inertia force

Gravity force

684FLUID MECHANICS

CompressibleCompressibleFlow

Open-ChannelOpen-ChannelFlow

Ma Ma � V/c Fr � V/c0

MaMa 1 Subsonic1 Subsonic Fr Fr 1 Subcritical 1 SubcriticalMaMa � 1 Sonic1 Sonic Fr Fr � 1 Critical 1 CriticalMaMa � 1 Supersonic1 Supersonic Fr Fr � 1 Supercritical 1 Supercritical

V � speed of flow speed of flowc � �kRTkRT � speed of sound (ideal gas) speed of sound (ideal gas)

c0 � �gygy � speed of wave (liquid) speed of wave (liquid)

FIGURE 13–7Analogy between the Mach number incompressible flow and the Froudenumber in open-channel flow.

Subcritical flow: y � yc

Supercritical flow: y yc

yc y

yc

y

FIGURE 13–8Definitions of subcritical flow andsupercritical flow in terms of criticaldepth.

cen72367_ch13.qxd 11/6/04 12:39 PM 84

cal flow, but it can do so by undergoing a shock. The shock in this case iscalled a hydraulic jump, which corresponds to a normal shock in com-pressible flow. Therefore, the analogy between open-channel flow and com-pressible flow is remarkable.

Speed of Surface WavesWe are all familiar with the waves forming on the free surfaces of oceans,lakes, rivers, and even swimming pools. The surface waves can be veryhigh, like the ones we see on the oceans, or barely noticeable. Some aresmooth; some break on the surface. A basic understanding of wave motionis necessary for the study of certain aspects of open-channel flow, and herewe present a brief description. A detailed treatment of wave motion can befound in numerous books written on the subject.

An important parameter in the study of open-channel flow is the wavespeed c0, which is the speed at which a surface disturbance travels througha liquid. Consider a long, wide channel that initially contains a still liquid ofheight y. One end of the channel is moved with speed dV, generating a sur-face wave of height dy propagating at a speed of c0 into the still liquid, asshown in Fig. 13–9a.

Now consider a control volume that encloses the wave front and moveswith it, as shown in Fig. 13–9b. To an observer traveling with the wavefront, the liquid to the right appears to be moving toward the wave frontwith speed c0 and the liquid to the left appears to be moving away from thewave front with speed c0 � dV. Of course the observer would think the con-trol volume that encloses the wave front (and herself or himself) is station-ary, and he or she would be witnessing a steady-flow process.

The steady-flow mass balance m.1 � m

.2 (or the continuity relation) for this

control volume of width b can be expressed as

(13–11)

We make the following assumptions: (1) the velocity is nearly constantacross the channel and thus the momentum flux correction factors (b1 andb2) are one, (2) the distance across the wave is short and thus friction at thebottom surface and air drag at the top are negligible, (3) the dynamic effectsare negligible and thus the pressure in the liquid varies hydrostatically; interms of gage pressure, P1, avg � rgh1, avg � rg(y/2) and P2, avg � rgh2, avg� rg(y � dy)/2, (4) the mass flow rate is constant with m

.1 � m

.2 � rc0yb,

and (5) there are no external forces or body forces and thus the only forcesacting on the control volume in the horizontal x-direction are the pressure

forces. Then, the momentum equation in the

x-direction becomes a balance between hydrostatic pressure forces andmomentum transfer,

(13–12)

Note that both the inlet and the outlet average velocities are negative sincethey are in the negative x-direction. Substituting,

(13–13)rg(y � dy)2b

2�rgy 2b

2� rc0yb(�c0 � dV) � rc0yb(�c0)

P2, avg A2 � P1, avg A1 � m#(�V2) � m

#(�V1)

a F→

� aoutbm#V→

� ainbm#V→

rc0� yb � r(c0 � dV)(y � dy)b → dV � c0

dy

y � dy

685CHAPTER 13

Movingposition

Movingwavefront

(a) Generation and propagation of a wave

Stillliquid

y

c0

dy

dV

y

Controlvolume

(b) Control volume relative to an observertraveling with the wave, with gage pressuredistributions shown

c0c0�dV

dy

rgyrg(y � dy) (1)(2)

FIGURE 13–9The generation and analysis of a wave

in an open channel.

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or,

(13–14)

Combining the momentum and continuity relations and rearranging give

(13–15)

Therefore, the wave speed c0 is proportional to the wave height dy. Forinfinitesimal surface waves, dy y and thus

Infinitesimal surface waves: (13–16)

Therefore, the speed of infinitesimal surface waves is proportional to thesquare root of liquid depth. Again note that this analysis is valid only forshallow water bodies, such as those encountered in open channels. Other-wise, the wave speed is independent of liquid depth for deep bodies ofwater, such as the oceans. The wave speed can also be determined by usingthe energy balance relation instead of the momentum equation together withthe continuity relation. Note that the waves eventually die out because of theviscous effects that are neglected in the analysis. Also, for flow in channelsof non-rectangular cross-section, the hydraulic depth defined as yh � Ac/Ltwhere Lt is the top width of the flow section should be used in the calcula-tion of Froude number in place of the flow depth y. For a half-full circularchannel, for example, the hydraulic depth is yh � (pR2/2)/2R � pR/4.

We know from experience that when a rock is thrown into a lake, the con-centric waves that form propagate evenly in all directions and vanish aftersome distance. But when the rock is thrown into a river, the upstream side ofthe wave moves upstream if the flow is tranquil or subcritical (V c0),moves downstream if the flow is rapid or supercritical (V � c0), and remainsstationary at the location where it is formed if the flow is critical (V � c0).

You may be wondering why we pay so much attention to flow being sub-critical or supercritical. The reason is that the character of the flow isstrongly influenced by this phenomenon. For example, a rock at the riverbedmay cause the water level at that location to rise or to drop, depending onwhether the flow is subcritical or supercritical. Also, the liquid level dropsgradually in the flow direction in subcritical flow, but a sudden rise in liquidlevel, called a hydraulic jump, may occur in supercritical flow (Fr � 1) asthe flow decelerates to subcritical (Fr 1) velocities.

This phenomenon can occur downstream of a sluice gate as shown inFig. 13–10. The liquid approaches the gate with a subcritical velocity, butthe upstream liquid level is sufficiently high to accelerate the liquid to asupercritical level as it passes through the gate (just like a gas flowing in aconverging–diverging nozzle). But if the downstream section of the channelis not sufficiently sloped down, it cannot maintain this supercritical veloc-ity, and the liquid jumps up to a higher level with a larger cross-sectionalarea, and thus to a lower subcritical velocity. Finally, the flow in rivers,canals, and irrigation systems is typically subcritical. But the flow pastsluice gates and spillways is typically supercritical.

You can create a beautiful hydraulic jump the next time you wash dishes(Fig. 13–11). Let the water from the faucet hit the middle of a dinner plate. As

c0 � 2gy

c20 � gya1 �

dy

yb a1 �

dy

2yb

ga1 �dy

2yb dy � c0 dV

686FLUID MECHANICS

Subcriticalflow

Sluicegate

Hydraulicjump

Supercriticalflow

Subcriticalflow

FIGURE 13–10Supercritical flow through a sluicegate.

FIGURE 13–11A hydraulic jump can be observed ona dinner plate when (a) it is right-side-up, but not when (b) it is upside down.Photos by Abel Po-Ya Chuang. Used by permission.

(a)

(b)

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the water spreads out radially, its depth decreases and the flow is supercritical.Eventually, a hydraulic jump occurs, which you can see as a sudden increasein water depth. Try it!

13–3 ■ SPECIFIC ENERGYConsider the flow of a liquid in a channel at a cross section where the flowdepth is y, the average flow velocity is V, and the elevation of the bottom ofthe channel at that location relative to some reference datum is z. For sim-plicity, we ignore the variation of liquid speed over the cross section andassume the speed to be V everywhere. The total mechanical energy of thisliquid in the channel in terms of heads is expressed as (Fig. 13–12)

(13–17)

where z is the elevation head, P/rg � y is the gage pressure head, and V2/2gis the velocity or dynamic head. The total energy as expressed in Eq. 13–17is not a realistic representation of the true energy of a flowing fluid sincethe choice of the reference datum and thus the value of the elevation head zis rather arbitrary. The intrinsic energy of a fluid at a cross section can berepresented more realistically if the reference datum is taken to be the bot-tom of the channel so that z � 0 there. Then the total mechanical energy of afluid in terms of heads becomes the sum of the pressure and dynamic heads.The sum of the pressure and dynamic heads of a liquid in an open channelis called the specific energy Es and is expressed as (Bakhmeteff, 1932)

(13–18)

as shown in Fig. 13–12.Consider flow in an open channel of constant width b. Noting that the

volume flow rate is V.

� AcV � ybV, the average flow velocity can beexpressed as

(13–19)

Substituting into Eq. 13–18, the specific energy can be expressed as

(13–20)

This equation is very instructive as it shows the variation of the specificenergy with flow depth. During steady flow in an open channel the flow rateis constant, and a plot of Es versus y for constant V

.and b is given in

Fig. 13–13. We observe the following from this figure:

• The distance from a point on the vertical y-axis to the curve represents thespecific energy at that y-value. The part between the Es � y line and thecurve corresponds to dynamic head (or kinetic energy) of the liquid, andthe remaining part to pressure head (or flow energy).

• The specific energy tends to infinity as y → 0 (due to the velocityapproaching infinity), and it becomes equal to flow depth y for largevalues of y (due to the velocity and thus the kinetic energy becoming very

Es � y �V#

2

2gb2y 2

V �V#

yb

Es � y �V

2

2g

H � z �Prg

�V

2

2g� z � y �

V 2

2g

687CHAPTER 13

z

yEs

V2

2g

Energy line

Reference datum

FIGURE 13–12The specific energy Es of a liquid in an

open channel is the total mechanicalenergy (expressed as a head) relative

to the bottom of the channel.

y

EsEs, min

Es � y

Subcriticalflow, Fr 1

Fr � 1

Criticaldepth

Supercriticalflow, Fr � 1yc

y

V2

2g

.V � constant

FIGURE 13–13Variation of specific energy Es with

depth y for a specified flow rate.

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small). The specific energy reaches a minimum value Es, min at someintermediate point, called the critical point, characterized by the criticaldepth yc and critical velocity Vc. The minimum specific energy is alsocalled the critical energy.

• There is a minimum specific energy Es, min required to support thespecified flow rate V

.. Therefore, Es cannot be below Es, min for a given V

..

• A horizontal line intersects the specific energy curve at one point only,and thus a fixed value of flow depth corresponds to a fixed value ofspecific energy. This is expected since the velocity has a fixed value when V

., b, and y are specified. However, for Es � Es, min, a vertical line

intersects the curve at two points, indicating that a flow can have twodifferent depths (and thus two different velocities) corresponding to afixed value of specific energy. These two depths are called alternatedepths. For flow through a sluice gate with negligible frictional losses(and thus Es � constant), the upper depth corresponds to the upstreamflow, and the lower depth to the downstream flow.

• A small change in specific energy near the critical point causes a largedifference between alternate depths and may cause violent fluctuations inflow level. Therefore, operation near the critical point should be avoidedin the design of open channels.

The value of the minimum specific energy and the critical depth at whichit occurs can be determined by differentiating Es from Eq. 13–20 withrespect to y for constant b and V

., and setting the derivative equal to zero:

(13–21)

Solving for y, which is the critical flow depth yc, gives

(13–22)

The flow rate at the critical point can be expressed as V.

� ycbVc. Substitut-ing, the critical velocity is determined to be

(13–23)

which is the wave speed. The Froude number at this point is

(13–24)

indicating that the point of minimum specific energy is indeed the criticalpoint, and the flow becomes critical when the specific energy reaches itsminimum value.

It follows that the flow is subcritical at lower flow velocities and thushigher flow depths (the upper arm of the curve), supercritical at highervelocities and thus lower flow depths (the lower arm of the curve), and crit-ical at the critical point (the point of minimum specific energy).

Noting that , the minimum (or critical) specific energy can beexpressed in terms of the critical depth alone as

(13–25)Es, min � yc �V

2c

2g� yc �

gyc

2g�

3

2 yc

Vc � 1gyc

Fr �V

2gy�

Vc

2gyc

� 1

Vc � 2gyc

yc � a V#

2

gb 2b

1/3

dEs

dy�

d

dy ay �

V#

2

2gb2y 2b � 1 �V#

2

gb2y 3 � 0

688FLUID MECHANICS

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In uniform flow, the flow depth and the flow velocity, and thus the spe-cific energy, remain constant since . The head loss is madeup by the decline in elevation (the channel is sloped downward in the flowdirection). In nonuniform flow, however, the specific energy may increaseor decrease, depending on the slope of the channel and the frictional losses.If the decline in elevation across a flow section is more than the head loss inthat section, for example, the specific energy increases by an amount equalto the difference between elevation drop and head loss. The specific energyconcept becomes a particularly useful tool when studying varied flows.

EXAMPLE 13–1 Character of Flow and Alternate Depth

Water is flowing steadily in a 0.4-m-wide rectangular open channel at a rateof 0.2 m3/s (Fig. 13–14). If the flow depth is 0.15 m, determine the flowvelocity and if the flow is subcritical or supercritical. Also determine thealternate flow depth if the character of flow were to change.

SOLUTION Water flow in a rectangular open channel is considered. The char-acter of flow, the flow velocity, and the alternate depth are to be determined. Assumptions The specific energy is constant.Analysis The average flow velocity is determined from

The critical depth for this flow is

Therefore, the flow is supercritical since the actual flow depth is y � 0.15 m,and y yc. Another way to determine the character of flow is to calculatethe Froude number,

Again the flow is supercritical since Fr � 1. The specific energy for the givenconditions is

Then the alternate depth is determined from to be

Solving for y2 gives the alternate depth to be y2 � 0.69 m. Therefore, if thecharacter of flow were to change from supercritical to subcritical while holdingthe specific energy constant, the flow depth would rise from 0.15 to 0.69 m.Discussion Note that if water underwent a hydraulic jump at constant spe-cific energy (the frictional losses being equal to the drop in elevation), theflow depth would rise to 0.69 m, assuming of course that the side walls ofthe channel are high enough.

Es2 � y2 �V#

2

2gb2y 22

→ 0.7163 m � y2 �(0.2 m3/s)2

2(9.81 m/s2)(0.4 m)2y 22

Es1 � Es2

Es1 � y1 �V#

2

2gb2y 21

� (0.15 m) �(0.2 m3/s)2

2(9.81 m/s2)(0.4 m)2(0.15 m)2 � 0.7163 m

Fr �V

1gy�

3.33 m/s

2(9.81 m/s2)(0.15 m)� 2.75

yc � a V#

2

gb2b1/3

� a (0.2 m3/s)2

(9.81 m/s2)(0.4 m)2b1/3

� 0.294 m

V �V#

Ac

�V#

yb�

0.2 m3/s

(0.15 m)(0.4 m)� 3.33 m/s

Es � y � V 2/2g

689CHAPTER 13

0.2 m3/s

0.15 m

0.4 m

FIGURE 13–14Schematic for Example 13–1.

cen72367_ch13.qxd 11/6/04 12:39 PM Page 689

13–4 ■ CONTINUITY AND ENERGY EQUATIONSOpen-channel flows involve liquids whose densities are nearly constant, andthus the one-dimensional steady-flow conservation of mass or continuityequation can be expressed as

(13–26)

That is, the product of the flow cross section and the average flow velocityremains constant throughout the channel. The continuity equation betweentwo sections along the channel is expressed as

Continuity equation: (13–27)

which is identical to the steady-flow continuity equation for liquid flow in apipe. Note that both the flow cross section and the average flow velocitymay vary during flow, but, as stated, their product remains constant.

To determine the total energy of a liquid flowing in an open channel rela-tive to a reference datum, as shown in Fig. 13–15, consider a point A in theliquid at a distance a from the free surface (and thus a distance y � a fromthe channel bottom). Noting that the elevation, pressure (hydrostatic pressurerelative to the free surface), and velocity at point A are zA � z � (y � a),PA � rga, and VA � V, respectively, the total energy of the liquid in termsof heads is

(13–28)

which is independent of the location of the point A at a cross section. There-fore, the total mechanical energy of a liquid at any cross section of an openchannel can be expressed in terms of heads as

(13–29)

where y is the flow depth, z is the elevation of the channel bottom, and V isthe average flow velocity. Then the one-dimensional energy equation foropen-channel flow between an upstream section 1 and a downstream section2 can be written as

Energy equation: (13–30)

The head loss hL due to frictional effects is expressed as in pipe flow as

(13–31)

where f is the average friction factor and L is the length of channel betweensections 1 and 2. The relation Dh � 4Rh should be observed when using thehydraulic radius instead of the hydraulic diameter.

Flow in open channels is gravity driven, and thus a typical channel isslightly sloped down. The slope of the bottom of the channel is expressed as

(13–32)

where a is the angle the channel bottom makes with the horizontal. In gen-eral, the bottom slope S0 is very small, and thus the channel bottom is

S0 � tan a�z1 � z2

x2 � x1�

z1 � z2

L

hL � f L

Dh

V 2

2g� f

L

Rh

V

2

8g

z1 � y1 �V

21

2g� z2 � y2 �

V 22

2g� hL

H � z � y �V

2

2g

HA � zA �PA

rg�

V 2A

2g� z � (y � a) �

rga

rg�

V 2

2g� z � y �

V 2

2g

Ac1V1 � Ac2V2

V#

� AcV � constant

690FLUID MECHANICS

z

Ay

y � a

a

V2

V

2g

Energy lineH � z � y �

Reference datum

V2

2g

FIGURE 13–15The total energy of a liquid flowing inan open channel.

cen72367_ch13.qxd 11/6/04 12:39 PM Page 690

nearly horizontal. Therefore, L � x2 � x1, where x is the distance in the hor-izontal direction. Also, the flow depth y, which is measured in the verticaldirection, can be taken to be the depth normal to the channel bottom withnegligible error.

If the channel bottom is straight so that the bottom slope is constant, thevertical drop between sections 1 and 2 can be expressed as .Then the energy equation (Eq. 13–30) becomes


This equation has the advantage that it is independent of a reference datumfor elevation.

In the design of open-channel systems, the bottom slope is selected suchthat it provides adequate elevation drop to overcome the frictional head lossand thus to maintain flow at the desired rate. Therefore, there is a close con-nection between the head loss and the bottom slope, and it makes sense toexpress the head loss as a slope (or the tangent of an angle). This is done bydefining a friction slope as

Friction slope: (13–34)

Then the energy equation can be written as


Note that the friction slope is equal to the bottom slope when the head lossis equal to the elevation drop. That is, Sf � S0 when hL � z1 � z2.

Figure 13–16 also shows the energy line, which is a distance z � y �V 2/2g (total mechanical energy of the liquid expressed as a head) above thehorizontal reference datum. The energy line is typically sloped down likethe channel itself as a result of frictional losses, the vertical drop beingequal to the head loss hL. Note that if there were no head loss, the energyline would be horizontal even when the channel is not. The elevation andvelocity heads (z � y and V2/2g) would then be able to convert to eachother during flow in this case, but their sum would remain constant.

13–5 ■ UNIFORM FLOW IN CHANNELSWe mentioned in Sec. 13–1 that flow in a channel is called uniform flow ifthe flow depth (and thus the average flow velocity since V

.� AcV � con-

stant in steady flow) remains constant. Uniform flow conditions are com-monly encountered in practice in long straight runs of channels with constantslope, constant cross section, and constant surface lining. In the design ofopen channels, it is very desirable to have uniform flow in the majority ofthe system since this means having a channel of constant height, which iseasier to design and build.

The flow depth in uniform flow is called the normal depth yn, and theaverage flow velocity is called the uniform-flow velocity V0. The flowremains uniform as long as the slope, cross section, and surface roughnessof the channel remain unchanged (Fig. 13–17). When the bottom slope is

y1 �V

21

2g� y2 �

V 22

2g� (Sf � S0)L

Sf �hL

L

y1 �V

21

2g� S0L � y2 �

V 22

2g� hL

z1 � z2 � S0L

691CHAPTER 13

a

z

V2

V2

V1

L

2g

x1 x2

y1y2

z1 z2

Energy line

Horizontalreference datum

(1)(2)

Slope: S0 � constant

x

V2

hL

2

1

2g

FIGURE 13–16The total energy of a liquid at two

sections of an open channel.

a

z

x1 x2

y1y2

z1 z2

(1)(2)

Slope: S0 � tan a � constant

Head loss � elevation losshL � z1 � z2 � S0L

x2 � x1 �

V1 � V2 � V0y1 � y2 � yn

x

Lcosa� L

FIGURE 13–17In uniform flow, the flow depth y,

the average flow velocity V, and thebottom slope S0 remain constant, and

the head loss equals the elevation loss,hL � z1 � z2 � SfL � S0L.

cen72367_ch13.qxd 11/6/04 12:39 PM Page 691

increased, the flow velocity increases and the flow depth decreases. There-fore, a new uniform flow will be established with a new (lower) flow depth.The opposite occurs if the bottom slope is decreased.

During flow in open channels of constant slope S0, constant cross sectionAc, and constant surface friction factor f, the terminal velocity is reachedand thus uniform flow is established when the head loss equals the elevationdrop. Therefore,

(13–36)

since hL � S0L in uniform flow and Dh � 4Rh. Solving the second relationfor V0, the uniform-flow velocity and the flow rate are determined to be

(13–37)

where

(13–38)

is called the Chezy coefficient. The Eqs. 13–37 and the coefficient C arenamed in honor of the French engineer Antoine Chezy (1718–1798), whofirst proposed a similar relationship in about 1769. The Chezy coefficient isa dimensional quantity, and its value ranges from about 30 m1/2/s for smallchannels with rough surfaces to 90 m1/2/s for large channels with smoothsurfaces (or, 60 ft1/2/s to 160 ft1/2/s in English units).

The Chezy coefficient can be determined in a straightforward mannerfrom Eq. 13–38 by first determining the friction factor f as done for pipeflow in Chap. 8 from the Moody chart or the Colebrook equation for thefully rough turbulent limit (Re → �),

(13–39)

Here e is the mean surface roughness. Note that open-channel flow is typi-cally turbulent, and the flow is fully developed by the time uniform flow isestablished. Therefore, it is reasonable to use the friction factor relation forfully developed turbulent flow. Also, at large Reynolds numbers, the frictionfactor curves corresponding to specified relative roughness are nearly hori-zontal, and thus the friction factor is independent of the Reynolds number.The flow in that region is called fully rough turbulent flow.

Since the introduction of the Chezy equations, considerable effort hasbeen devoted by numerous investigators to the development of simplerempirical relations for the average velocity and flow rate. The most widelyused equation was developed independently by the Frenchman Philippe-Gaspard Gauckler (1826–1905) in 1868 and the Irishman Robert Manning(1816–1897) in 1889.

Both Gauckler and Manning made recommendations that the constant inthe Chezy equation be expressed as

(13–40)

where n is called the Manning coefficient, whose value depends on theroughness of the channel surfaces. Substituting into Eqs. 13–37 gives the fol-lowing empirical relations known as the Manning equations (also referredto as Gauckler–Manning equations since they were first proposed byPhilippe-Gaspard Gauckler) for the uniform-flow velocity and the flow rate,

C �an R1/6

h

f � [2.0 log(14.8Rh /e)]�2

C � 28g/f

V0 � C2S0Rh and V#

� CAc2S0Rh

hL � f L

Dh

V

2

2g or S0L � f

L

Rh

V

20

8g

692FLUID MECHANICS

cen72367_ch13.qxd 11/6/04 12:39 PM Page 692

Uniform flow: (13–41)

The factor a is a dimensional constant whose value in SI units is a � 1 m1/3/s.Noting that 1 m � 3.2808 ft, its value in English units is

(13–42)

Note that the bottom slope S0 and the Manning coefficient n are dimension-less quantities, and Eqs. 13–41 give the velocity in m/s and the flow rate inm3/s in SI units when Rh is expressed in m. (The corresponding units inEnglish units are ft/s and ft3/s when Rh is expressed in ft.)

Experimentally determined values of n are given in Table 13–1 for numer-ous natural and artificial channels. More extensive tables are available in theliterature. Note that the value of n varies from 0.010 for a glass channel to0.150 for a floodplain laden with trees (15 times that of a glass channel).There is considerable uncertainty in the value of n, especially in naturalchannels, as you would expect, since no two channels are exactly alike. Thescatter can be 20 percent or more. Also, the coefficient n depends on thesize and shape of the channel as well as the surface roughness.

Critical Uniform FlowFlow through an open channel becomes critical flow when the Froude num-ber Fr � 1 and thus the flow speed equals the wave speed ,where yc is the critical flow depth, defined previously (Eq. 13–9). When thevolume flow rate V

., the channel slope S0, and the Manning coefficient n are

known, the normal flow depth yn can be determined from the Manningequation (Eq. 13–41). However, since Ac and Rh are both functions of yn, theequation often ends up being implicit in yn and requires a numerical (or trialand error) approach to solve. If yn � yc, the flow is uniform critical flow,and bottom slope S0 equals the critical slope Sc in this case. When flowdepth yn is known instead of the flow rate V

., the flow rate can be deter-

mined from the Manning equation and the critical flow depth from Eq.13–9. Again the flow is critical only if yn � yc.

During uniform critical flow, S0 � Sc and yn � yc. Replacing V.

and S0 inthe Manning equation by and Sc, respectively, and solving forSc gives the following general relation for the critical slope,

Critical slope (general): (13–43)

For film flow or flow in a wide rectangular channel with b �� yc, Eq.13–43 simplifies to

Critical slope (b �� yc): (13–44)

This equation gives the slope necessary to maintain a critical flow of depthyc in a wide rectangular channel having a Manning coefficient of n.

Superposition Method for Nonuniform PerimetersThe surface roughness and thus the Manning coefficient for most naturaland some human-made channels vary along the wetted perimeter and evenalong the channel. A river, for example, may have a stony bottom for its

Sc �gn2

a2y 1/3c

Sc �gn2yc

a2R 4/3h

V#

� Ac1gyc

Vc � 1gyc

a � 1 m1/3/s � (3.2808 ft)1/3/s � 1.486 ft1/3/s

V0 �an R2/3

h S 1/20 and V

#�

an AcR

2/3h S 1/2

0

693CHAPTER 13

TABLE 13–1

Mean values of the Manningcoefficient n for water flow in open channels*

From Chow (1959).

Wall Material n

A. Artificially lined channelsGlass 0.010Brass 0.011Steel, smooth 0.012Steel, painted 0.014Steel, riveted 0.015Cast iron 0.013Concrete, finished 0.012Concrete, unfinished 0.014Wood, planed 0.012Wood, unplaned 0.013Clay tile 0.014Brickwork 0.015Asphalt 0.016Corrugated metal 0.022Rubble masonry 0.025

B. Excavated earth channelsClean 0.022Gravelly 0.025Weedy 0.030Stony, cobbles 0.035

C. Natural channelsClean and straight 0.030Sluggish with deep pools 0.040Major rivers 0.035Mountain streams 0.050

D. FloodplainsPasture, farmland 0.035Light brush 0.050Heavy brush 0.075Trees 0.150

* The uncertainty in n can be � 20 percent ormore.

cen72367_ch13.qxd 11/6/04 12:39 PM Page 693

regular bed but a surface covered with bushes for its extended floodplain.There are several methods for solving such problems, either by finding aneffective Manning coefficient n for the entire channel cross section, or byconsidering the channel in subsections and applying the superposition prin-ciple. For example, a channel cross section can be divided into N subsec-tions, each with its own uniform Manning coefficient and flow rate. Whendetermining the perimeter of a section, only the wetted portion of theboundary for that section is considered, and the imaginary boundaries areignored. The flow rate through the channel is the sum of the flow ratesthrough all the sections.

EXAMPLE 13–2 Flow Rate in an Open Channel in Uniform Flow

Water is flowing in a weedy excavated earth channel of trapezoidal cross sec-tion with a bottom width of 0.8 m, trapezoid angle of 60°, and a bottomslope angle of 0.3°, as shown in Fig. 13–18. If the flow depth is measuredto be 0.52 m, determine the flow rate of water through the channel. Whatwould your answer be if the bottom angle were 1°?

SOLUTION Water is flowing in a weedy trapezoidal channel of given dimen-sions. The flow rate corresponding to a measured value of flow depth is to bedetermined. Assumptions 1 The flow is steady and uniform. 2 The bottom slope is con-stant. 3 The roughness of the wetted surface of the channel and thus thefriction coefficient are constant.Properties The Manning coefficient for an open channel with weedy sur-faces is n � 0.030.Analysis The cross-sectional area, perimeter, and hydraulic radius of thechannel are

The bottom slope of the channel is

Then the flow rate through the channel is determined from the Manningequation to be

The flow rate for a bottom angle of 1° can be determined by using S0� tan a � tan 1° � 0.01746 in the last relation. It gives V

.� 1.1 m3/s.

Discussion Note that the flow rate is a strong function of the bottom angle.Also, there is considerable uncertainty in the value of the Manning coeffi-cient, and thus in the flow rate calculated. A 10 percent uncertainty in nresults in a 10 percent uncertainty in the flow rate. Final answers are there-fore given to only two significant digits.

V#

�an

AcR2 3h S 1 2

0 �1 m1/3 s

0.030(0.5721 m2)(0.2859 m)2/3(0.005236)1 2 � 0.60 m3/s

S0 � tan a� tan 0.3� � 0.005236

Rh �Ac

p�

0.5721 m2

2.991 m� 0.2859 m

p � b �2y

sin u� 0.8 m �

2 � 0.52 m

sin 60�� 2.001 m

Ac � yab �y

tan ub � (0.52 m)a0.8 m �

0.52 m

tan 60�b � 0.5721 m2

694FLUID MECHANICS

y � 0.52 m

u � 60°

b � 0.8 m


cen72367_ch13.qxd 11/6/04 12:39 PM Page 694

EXAMPLE 13–3 The Height of a Rectangular Channel

Water is to be transported in an unfinished-concrete rectangular channelwith a bottom width of 4 ft at a rate of 51 ft3/s. The terrain is such that thechannel bottom drops 2 ft per 1000 ft length. Determine the minimumheight of the channel under uniform-flow conditions (Fig. 13–19). Whatwould your answer be if the bottom drop is just 1 ft per 1000 ft length?

SOLUTION Water is flowing in an unfinished-concrete rectangular channelwith a specified bottom width. The minimum channel height correspondingto a specified flow rate is to be determined.Assumptions 1 The flow is steady and uniform. 2 The bottom slope is con-stant. 3 The roughness of the wetted surface of the channel and thus thefriction coefficient are constant.Properties The Manning coefficient for an open channel with unfinished-concrete surfaces is n � 0.014.Analysis The cross-sectional area, perimeter, and hydraulic radius of thechannel are

The bottom slope of the channel is S0 � 2/1000 � 0.002. Using the Man-ning equation, the flow rate through the channel can be expressed as

which is a nonlinear equation in y. Using an equation solver such as EES oran itirative approach, the flow depth is determined to be

If the bottom drop were just 1 ft per 1000 ft length, the bottom slope wouldbe S0 � 0.001, and the flow depth would be y � 3.3 ft.Discussion Note that y is the flow depth, and thus this is the minimumvalue for the channel height. Also, there is considerable uncertainty in thevalue of the Manning coefficient n, and this should be considered whendeciding the height of the channel to be built.

EXAMPLE 13–4 Channels with Nonuniform Roughness

Water flows in a channel whose bottom slope is 0.003 and whose cross sec-tion is shown in Fig. 13–20. The dimensions and the Manning coefficientsfor the surfaces of different subsections are also given on the figure. Deter-mine the flow rate through the channel and the effective Manning coefficientfor the channel.

SOLUTION Water is flowing through a channel with nonuniform surfaceproperties. The flow rate and the effective Manning coefficient are to bedetermined.

y � 2.5 ft

51 ft3/s �1.486 ft1/3 s

0.014 (4y ft2)a 4y

4 � 2y ftb 2 3

(0.002)1 2

V#

�an

AcR2 3h S 1 2

0

Ac � by � (4 ft)y p � b � 2y � (4 ft) � 2y Rh �Ac

p�

4y

4 � 2y

695CHAPTER 13

y

b � 4 ft

V � 51 ft3/s.


cen72367_ch13.qxd 11/6/04 12:39 PM Page 695

Assumptions 1 The flow is steady and uniform. 2 The bottom slope is con-stant. 3 The Manning coefficients do not vary along the channel.Analysis The channel involves two parts with different roughnesses, andthus it is appropriate to divide the channel into two subsections as indi-cated in Fig. 13–20. The flow rate for each subsection can be determinedfrom the Manning equation, and the total flow rate can be determined byadding them up.

The side length of the triangular channel is .Then the flow area, perimeter, and hydraulic radius for each subsection andthe entire channel become

Subsection 1:

Subsection 2:

Entire channel:

Using the Manning equation for each subsection, the total flow rate throughthe channel is determined to be

Knowing the total flow rate, the effective Manning coefficient for the entirechannel can be determined from the Manning equation to be

neff �aAcR

2 3h S 1 2

0

V# �

(1 m1/3 s)(37 m2)(1.806 m)2/3(0.003)1 2

84.8 m3 s� 0.035

� 84.8 m3/s � 85 m3/s

� (1 m1/3 s) c(21 m2)(2 m)2/3

0.030�

(16 m2)(1.60 m)2/3

0.050d(0.003)1 2

V#

� V#

1 � V#

2 �an1

Ac1R2 3h1 S 1 2

0 �an2

Ac2R2 3h2 S 1 2

0

Ac � 37 m2 p � 20.486 m Rh �Ac

p�

37 m2

20.486 m� 1.806 m

Ac2 � 16 m2 p2 � 10 m Rh2 �Ac2

p2�

16 m2

10 m� 1.60 m

Ac1 � 21 m2 p1 � 10.486 m Rh1 �Ac1

p1�

21 m2

10.486 m� 2.00 m

s � 132 � 32 � 4.243 m

696FLUID MECHANICS

6 m

3 m

2 m

8 m

Light brushn2 � 0.050

Clean naturalchannel

n1 � 0.030 s

1 2


cen72367_ch13.qxd 11/6/04 12:39 PM Page 696

Discussion The effective Manning coefficient neff of the channel turns out tolie between the two n values, as expected. The weighted average of the Man-ning coefficient of the channel is navg � (n1p1 � n2p2)/p � 0.040, which isquite different than neff. Therefore, using a weighted average Manning coeffi-cient for the entire channel may be tempting, but it would not be so accurate.

13–6 ■ BEST HYDRAULIC CROSS SECTIONSOpen-channel systems are usually designed to transport a liquid to a loca-tion at a lower elevation at a specified rate under the influence of gravity atthe lowest possible cost. Noting that no energy input is required, the cost ofan open-channel system consists primarily of the initial construction cost,which is proportional to the physical size of the system. Therefore, for agiven channel length, the perimeter of the channel is representative of thesystem cost, and it should be kept to a minimum in order to minimize thesize and thus the cost of the system.

From another perspective, resistance to flow is due to wall shear stress twand the wall area, which is equivalent to the wetted perimeter per unit chan-nel length. Therefore, for a given flow cross-sectional area Ac, the smallerthe wetted perimeter p, the smaller the resistance force, and thus the largerthe average velocity and the flow rate.

From yet another perspective, for a specified channel geometry with a spec-ified bottom slope S0 and surface lining (and thus the roughness coefficient n),the flow velocity is given by the Manning formula as . There-fore, the flow velocity is proportional to the hydraulic radius, and thehydraulic radius must be maximized (and thus the perimeter must be mini-mized since Rh � Ac/p) in order to maximize the average flow velocity orthe flow rate per unit cross-sectional area. Thus we conclude the following:

The best hydraulic cross section for an open channel is the one with themaximum hydraulic radius or, equivalently, the one with the minimum wettedperimeter for a specified cross section.

The shape with the minimal perimeter per unit area is a circle. Therefore,on the basis of minimum flow resistance, the best cross section for an openchannel is a semicircular one (Fig. 13–21). However, it is usually cheaper toconstruct an open channel with straight sides (such as channels with trape-zoidal or rectangular cross sections) instead of semicircular ones, and thegeneral shape of the channel may be specified a priori. Thus it makes senseto analyze each geometric shape separately for the best cross section.

As a motivational example, consider a rectangular channel of finishedconcrete (n � 0.012) of width b and flow depth y with a bottom slope of 1°(Fig. 13–22). To determine the effects of the aspect ratio y/b on thehydraulic radius Rh and the flow rate V

.per unit cross-sectional area (Ac

� 1 m2), Rh and V.

are evaluated from the Manning formula. The results aretabulated in Table 13–2 and plotted in Fig. 13–23 for aspect ratios from 0.1to 5. We observe from this table and the plot that the flow rate V

.increases

as the flow aspect ratio y/b is increased, reaches a maximum at y/b � 0.5,

V � aR2 3h S 1 2

0 n

697CHAPTER 13

Ry

FIGURE 13–21The best hydraulic cross section for an

open channel is a semicircular onesince it has the minimum wetted

perimeter for a specified cross section,and thus the minimum flow resistance.

y

b

FIGURE 13–22A rectangular open channel of width

b and flow depth y. For a given cross-sectional area, the highest flow rate occurs when y � b/2.

cen72367_ch13.qxd 11/6/04 12:39 PM Page 697

and then starts to decrease (the numerical values for V.

can also be inter-preted as the flow velocities in m/s since Ac � 1 m2). We see the same trendfor the hydraulic radius, but the opposite trend for the wetted perimeter p.These results confirm that the best cross section for a given shape is the onewith the maximum hydraulic radius, or equivalently, the one with the mini-mum perimeter.

698FLUID MECHANICS

TABLE 13–2

Variation of the hydraulic radius Rh and the flow rate V.with aspect ratio y/b for

a rectangular channel with Ac � 1 m2, S0 � tan 1°, and n � 0.012

Aspect Channel Flow Perimeter Hydraulic Flow RateRatio Width Depth p, m Radius V

.,

y/b b, m y, m Rh, m m3/s

0.1 3.162 0.316 3.795 0.264 4.530.2 2.236 0.447 3.130 0.319 5.140.3 1.826 0.548 2.921 0.342 5.390.4 1.581 0.632 2.846 0.351 5.480.5 1.414 0.707 2.828 0.354 5.500.6 1.291 0.775 2.840 0.352 5.490.7 1.195 0.837 2.869 0.349 5.450.8 1.118 0.894 2.907 0.344 5.410.9 1.054 0.949 2.951 0.339 5.351.0 1.000 1.000 3.000 0.333 5.291.5 0.816 1.225 3.266 0.306 5.002.0 0.707 1.414 3.536 0.283 4.743.0 0.577 1.732 4.041 0.247 4.344.0 0.500 2.000 4.500 0.222 4.045.0 0.447 2.236 4.919 0.203 3.81

03.75

4.15

4.55

4.95

5.35

5.75

1 2 3

Aspect ratio r � y/b

Flow

rat

e V

, m3 /s

.

4 5

FIGURE 13–23Variation of the flow rate in arectangular channel with aspect ratio r � y/b for Ac � 1 m2 and S0 � tan 1�.

cen72367_ch13.qxd 11/6/04 12:39 PM Page 698

Rectangular ChannelsConsider liquid flow in an open channel of rectangular cross section ofwidth b and flow depth y. The cross-sectional area and the wetted perimeterat a flow section are

(13–45)

Solving the first relation of Eq. 13–45 for b and substituting it into the sec-ond relation give

(13–46)

Now we apply the criterion that the best hydraulic cross section for an openchannel is the one with the minimum wetted perimeter for a given cross sec-tion. Taking the derivative of p with respect to y while holding Ac constantgives

(13–47)

Setting dp/dy � 0 and solving for y, the criterion for the best hydrauliccross section is determined to be

Best hydraulic cross section (rectangular channel): (13–48)

Therefore, a rectangular open channel should be designed such that the liq-uid height is half the channel width to minimize flow resistance or to maxi-mize the flow rate for a given cross-sectional area. This also minimizes theperimeter and thus the construction costs. This result confirms the findingfrom Table 13–2 that y � b/2 gives the best cross section.

Trapezoidal ChannelsNow consider liquid flow in an open channel of trapezoidal cross section ofbottom width b, flow depth y, and trapezoid angle u measured from the hor-izontal, as shown in Fig. 13–24. The cross-sectional area and the wettedperimeter at a flow section are

(13–49)

Solving the first relation of Eq. 13–49 for b and substituting it into the sec-ond relation give

(13–50)

Taking the derivative of p with respect to y while holding Ac and u constantgives

(13–51)

Setting dp/dy � 0 and solving for y, the criterion for the best hydrauliccross section for any specified trapezoid angle u is determined to be

Best hydraulic cross section (trapezoidal channel): (13–52)y �b sin u

2(1 � cos u)

dp

dy� �

Ac

y 2 �1

tan u�

2

sin u� �

b � y tan u

y�

1

tan u�

2

sin u

p �Ac

y�

y

tan u�

2y

sin u

Ac � ab �y

tan uby and p � b �

2y

sin u

y �b

2

dp

dy� �

Ac

y 2 � 2 � �by

y 2 � 2 � �by

� 2

p �Ac

y� 2y

Ac � yb and p � b � 2y

699CHAPTER 13

y

u

b

Rh � �Acp

y(b � y/tan u)b � 2y/sin u

s

FIGURE 13–24Parameters for a trapezoidal channel.

cen72367_ch13.qxd 11/6/04 12:39 PM Page 699

For the special case of u � 90° (a rectangular channel), this relation reducesto y � b/2, as expected.

The hydraulic radius Rh for a trapezoidal channel can be expressed as

(13–53)

Rearranging Eq. 13–52 as b sin u � 2y(1 � cos u), substituting into Eq.13–53 and simplifying, the hydraulic radius for a trapezoidal channel withthe best cross section becomes

Hydraulic radius for the best cross section: (13–54)

Therefore, the hydraulic radius is half the flow depth for trapezoidal chan-nels with the best cross section regardless of the trapezoid angle u.

Similarly, the trapezoid angle for the best hydraulic cross section is deter-mined by taking the derivative of p (Eq. 13–50) with respect to u whileholding Ac and y constant, setting dp/du � 0, and solving the resulting equa-tion for u. This gives

Best trapezoid angle: (13–55)

Substituting the best trapezoid angle u � 60° into the best hydraulic crosssection relation y � b sin u/(2 � 2 cos u) gives

Best flow depth for u � 60°: (13–56)

Then the length of the side edge of the flow section and the flow area become

(13–57)

(13–58)

(13–59)

since . Therefore, the best cross section for trapezoidal chan-nels is half of a hexagon (Fig. 13–25). This is not surprising since a hexagonclosely approximates a circle, and a half-hexagon has the least perimeter perunit cross-sectional area of all trapezoidal channels.

Best hydraulic cross sections for other channel shapes can be determinedin a similar manner. For example, the best hydraulic cross section for a cir-cular channel of diameter D can be shown to be y � D/2.

EXAMPLE 13–5 Best Cross Section of an Open Channel

Water is to be transported at a rate of 2 m3/s in uniform flow in an openchannel whose surfaces are asphalt lined. The bottom slope is 0.001. Deter-mine the dimensions of the best cross section if the shape of the channel is(a) rectangular and (b) trapezoidal (Fig. 13–26).

tan 60� � 23

Ac � ab �y

tan uby � ab �

b23 2

tan 60�b(b23 2) �

323

4 b2

p � 3b

s �y

sin 60��

b23 2

23 2� b

y �23

2 b

u � 60�

Rh �y

2

Rh �Ac

p�

y(b � y tan u)

b � 2y sin u�

y(b sin u � y cos u)

b sin u � 2y

700FLUID MECHANICS

y

b

b

Rh � b�y2

34

60°

b�32

Ac � b234

3

FIGURE 13–25The best cross section for trapezoidalchannels is half of a hexagon.

cen72367_ch13.qxd 11/6/04 12:39 PM Page 700

SOLUTION Water is to be transported in an open channel at a specifiedrate. The best channel dimensions are to be determined for rectangular andtrapezoidal shapes.Assumptions 1 The flow is steady and uniform. 2 The bottom slope is con-stant. 3 The roughness of the wetted surface of the channel and thus thefriction coefficient are constant.Properties The Manning coefficient for an open channel with asphalt liningis n � 0.016.Analysis (a) The best cross section for a rectangular channel occurs whenthe flow height is half the channel width, y � b/2. Then the cross-sectionalarea, perimeter, and hydraulic radius of the channel are

Substituting into the Manning equation,

which gives b � 1.84 m. Therefore, Ac � 1.70 m2, p � 3.68 m, and thedimensions of the best rectangular channel are

(b) The best cross section for a trapezoidal channel occurs when the trape-zoid angle is 60° and flow height is . Then,

Substituting into the Manning equation,

which yields b � 1.12 m. Therefore, Ac � 1.64 m2, p � 3.37 m, and thedimensions of the best trapezoidal channel are

Discussion Note that the trapezoidal cross section is better since it has asmaller perimeter (3.37 m versus 3.68 m) and thus lower construction cost.

13–7 ■ GRADUALLY VARIED FLOWTo this point we considered uniform flow during which the flow depth y andthe flow velocity V remain constant. In this section we consider graduallyvaried flow (GVF), which is a form of steady nonuniform flow character-ized by gradual variations in flow depth and velocity (small slopes and noabrupt changes) and a free surface that always remains smooth (no disconti-nuities or zigzags). Flows that involve rapid changes in flow depth andvelocity, called rapidly varied flows (RVF), are considered in Section 13–8.A change in the bottom slope or cross section of a channel or an obstruction

b � 1.12 m y � 0.973 m and u � 60�

V#

�an AcR

2 3h S 1 2

0 → b � a (0.016)(2 m3/s)

0.7523a23/4b2/3(1 m1/3 s)20.001b 3/8

p � 3b Rh �y

2�23

4 b

Ac � y(b � b cos u) � 0.523b2(1 � cos 60�) � 0.7523b2

y � b13 2

b � 1.84 m and y � 0.92 m

V#

�an AcR

2 3h S 1 2

0 → b � a2nV#42/3

a2S0

b 3/8

� a2(0.016)(2 m3/s)42/3

(1 m1/3 s)20.001b 3/8

Ac � by �b2

2 p � b � 2y � 2b Rh �

Ac

p�

b

4

701CHAPTER 13

y

b

b60°

b�32

�b2y �

b

b2


cen72367_ch13.qxd 11/6/04 12:39 PM Page 701

in the path of flow may cause the uniform flow in a channel to becomegradually or rapidly varied flow.

Rapidly varied flows occur over a short section of the channel with rela-tively small surface area, and thus frictional losses associated with wallshear are negligible. Head losses in RVF are highly localized and are due tointense agitation and turbulence. Losses in GVF, on the other hand, are pri-marily due to frictional effects along the channel and can be determinedfrom the Manning formula.

In gradually varied flow, the flow depth and velocity vary slowly, and thefree surface is stable. This makes it possible to formulate the variation offlow depth along the channel on the basis of the conservation of mass andenergy principles and to obtain relations for the profile of the free surface.

In uniform flow, the slope of the energy line is equal to the slope of thebottom surface. Therefore, the friction slope equals the bottom slope, Sf� S0. In gradually varied flow, however, these slopes are different.

Consider steady flow in a rectangular open channel of width b, andassume any variation in the bottom slope and water depth to be rather grad-ual. We again write the equations in terms of average velocity V and assumethe pressure distribution to be hydrostatic. From Eq. 13–17, the total head ofthe liquid at any cross section is H � zb � y � V 2/2g, where zb is the verti-cal distance of the bottom surface from the reference datum. DifferentiatingH with respect to x gives

(13–60)

But H is the total energy of the liquid and thus dH/dx is the slope of theenergy line (negative quantity), which is equal to the negative of the frictionslope, as shown in Fig. 13–27. Also, dzb /dx is the negative of the bottomslope. Therefore,

(13–61)

Substituting Eq. 13–61 into Eq. 13–60 gives

(13–62)

The continuity equation for steady flow in a rectangular channel is V.

� ybV� constant. Differentiating with respect to x gives

(13–63)

Substituting Eq. 13–63 into Eq. 13–62 and noting that is the Froudenumber,

(13–64)

Solving for dy/dx gives the desired relation for the rate of change of flowdepth (or the surface profile) in gradually varied flow in an open channel,

(13–65)dy

dx�

S0 � Sf

1 � Fr2

S0 � Sf �dy

dx�

V 2

gy dy

dx�

dy

dx� Fr2

dy

dx

V 1gy

0 � bV dy

dx� yb

dV

dx →

dV

dx� �

Vy

dy

dx

S0 � Sf �dy

dx�

Vg

dV

dx

dH

dx� �

dhL

dx� �Sf and dzb

dx� �S0

dH

dx�

d

dx azb � y �

V 2

2gb �

dzb

dx�

dy

dx�

Vg

dV

dx

702FLUID MECHANICS

z , H

V2

V � dV

y � dy

V

2g

x x � dx

y

zb

dx

zb � dzb

Energy line, H

Friction slope SfHorizontal

Horizontalreference datum

Bottom slope S0

x

dhL

(V � dV)2

2g

FIGURE 13–27Variation of properties over adifferential flow section in an openchannel under conditions of graduallyvaried flow (GVF).

cen72367_ch13.qxd 11/6/04 12:39 PM Page 702

which is analogous to the variation of flow area as a function of the Machnumber in compressible flow. This relation is derived for a rectangularchannel, but it is also valid for channels of other constant cross sections pro-vided that the Froude number is expressed accordingly. An analytical ornumerical solution of this differential equation gives the flow depth y as afunction of x for a given set of parameters, and the function y(x) is the sur-face profile.

The general trend of flow depth—whether it increases, decreases, orremains constant along the channel—depends on the sign of dy/dx, whichdepends on the signs of the numerator and the denominator of Eq. 13–65.The Froude number is always positive and so is the friction slope Sf(except for the idealized case of flow with negligible frictional effects forwhich both hL and Sf are zero). The bottom slope S0 is positive for down-ward-sloping sections (typically the case), zero for horizontal sections, andnegative for upward-sloping sections of a channel (adverse flow). The flowdepth increases when dy/dx � 0, decreases when dy/dx 0, and remainsconstant (and thus the free surface is parallel to the channel bottom, as inuniform flow) when dy/dx � 0 and thus S0 � Sf. For specified values of S0and Sf , the term dy/dx may be positive or negative, depending on whetherthe Froude number is less than or greater than 1. Therefore, the flow behav-ior is opposite in subcritical and supercritical flows. For S0 � Sf � 0, forexample, the flow depth increases in the flow direction in subcritical flow,but it decreases in supercritical flow.

The determination of the sign of the denominator 1 � Fr2 is easy: it ispositive for subcritical flow (Fr 1), and negative for supercritical flow (Fr� 1). But the sign of the numerator depends on the relative magnitudes ofS0 and Sf. Note that the friction slope Sf is always positive, and its value isequal to the channel slope S0 in uniform flow, y � yn. Noting that head lossincreases with increasing velocity, and that the velocity is inversely propor-tional to flow depth for a given flow rate, Sf � S0 and thus S0 � Sf 0when y yn, and Sf S0 and thus S0 � Sf � 0 when y � yn. The numeratorS0 � Sf is always negative for horizontal (S0 � 0) and upward-sloping (S0 0) channels, and thus the flow depth decreases in the flow direction duringsubcritical flows in such channels.

Liquid Surface Profiles in Open Channels, y(x)Open-channel systems are designed and built on the basis of the projectedflow depths along the channel. Therefore, it is important to be able to pre-dict the flow depth for a specified flow rate and specified channel geometry.A plot of flow depths gives the surface profile of the flow. The generalcharacteristics of surface profiles for gradually varied flow depend on thebottom slope and flow depth relative to the critical and normal depths.

A typical open channel involves various sections of different bottomslopes S0 and different flow regimes, and thus various sections of differentsurface profiles. For example, the general shape of the surface profile in adownward-sloping section of a channel is different than that in an upward-sloping section. Likewise, the profile in subcritical flow is different than theprofile in supercritical flow. Unlike uniform flow that does not involve inertia

703CHAPTER 13

cen72367_ch13.qxd 11/6/04 12:39 PM Page 703

forces, gradually varied flow involves acceleration and deceleration of liq-uid, and the surface profile reflects the dynamic balance between liquidweight, shear force, and inertial effects.

Each surface profile is identified by a letter that indicates the slope of thechannel and by a number that indicates flow depth relative to the criticaldepth yc and normal depth yn. The slope of the channel can be mild (M),critical (C), steep (S), horizontal (H), or adverse (A). The channel slope issaid to be mild if yn � yc, steep if yn yc, critical if yn � yc, horizontal ifS0 � 0 (zero bottom slope), and adverse if S0 0 (negative slope). Notethat a liquid flows uphill in an open channel that has an adverse slope(Fig. 13–28).

The classification of a channel section depends on the flow rate and thechannel cross section as well as the slope of the channel bottom. A channelsection that is classified to have a mild slope for one flow can have a steepslope for another flow, and even a critical slope for a third flow. Therefore,we need to calculate the critical depth yc and the normal depth yn before wecan assess the slope.

The number designation indicates the initial position of the liquid surfacefor a given channel slope relative to the surface levels in critical and uni-form flows, as shown in Fig. 13–29. A surface profile is designated by 1 ifthe flow depth is above both critical and normal depths (y � yc and y � yn),by 2 if the flow depth is between the two (yn � y � yc or yn y yc), andby 3 if the flow depth is below both the critical and normal depths (y ycand y yn). Therefore, three different profiles are possible for a specifiedtype of channel slope. But for channels with zero or adverse slopes, type 1flow cannot exist since the flow can never be uniform in horizontal andupward channels, and thus normal depth is not defined. Also, type 2 flowdoes not exist for channels with critical slope since normal and criticaldepths are identical in this case.

The five classes of slopes and the three types of initial positions discussedgive a total of 12 distinct configurations for surface profiles in GVF, all tab-ulated and sketched in Table 13–3. The Froude number is also given foreach case, with Fr � 1 for y yc, as well as the sign of the slope dy/dx ofthe surface profile determined from Eq. 13–65, dy/dx � (S0 � Sf)/(1 � Fr2).Note that dy/dx � 0, and thus the flow depth increases in the flow directionwhen both S0 � Sf and 1 � Fr2 are positive or negative. Otherwise dy/dx 0 and the flow depth decreases. In type 1 flows, the flow depth increases inthe flow direction and the surface profile approaches the horizontal planeasymptotically. In type 2 flows, the flow depth decreases and the surfaceprofile approaches the lower of yc or yn. In type 3 flows, the flow depthincreases and the surface profile approaches the lower of yc or yn. Thesetrends in surface profiles continue as long as there is no change in bottomslope or roughness.

Consider the first case in Table 13–3 designated M1 (mild channel slopeand y � yn � yc). The flow is subcritical since y � yc and thus Fr 1 and1 � Fr2 � 0. Also, Sf S0 and thus S0 � Sf � 0 since y � yn, and thus theflow velocity is less than the velocity in normal flow. Therefore, the slopeof the surface profile dy/dx � (S0 � Sf)/(1 � Fr2) � 0, and the flow depthy increases in the flow direction. But as y increases, the flow velocitydecreases, and thus Sf and Fr approach zero. Consequently, dy/dx

704FLUID MECHANICS

Horizontal

A

S

H

M

C

Mild

Steep

Critical

Adverse

FIGURE 13–28Designation of the letters S, C, M, H,and A for liquid surface profiles fordifferent types of slopes.

3

Channel bottom

Free surface incritical flow

Free surface inuniform flow

yn

yc

y

2

1

FIGURE 13–29Designation of the numbers 1, 2, and 3for liquid surface profiles based on thevalue of the flow depth relative to thenormal and critical depths.

cen72367_ch13.qxd 11/6/04 12:39 PM Page 704

705CHAPTER 13

TABLE 13–3

Classification of surface profiles in gradually varied flow

Channel Profile Froude Profile SurfaceSlope Notation Flow Depth Number Slope Profile

Mild (M) yc yn M1 y � yn Fr 1S0 Sc

M2 yc y yn Fr 1

M3 y yc Fr � 1

Steep (S) yc � yn S1 y � yc Fr 1S0 Sc

S2 yn y yc Fr � 1

S3 y yn Fr � 1

Critical (C) yc � yn C1 y � yc Fr 1S0 Sc

C3 y yc Fr � 1

Horizontal (H) yn → � H2 y � yc Fr 1S0 � 0

H3 y yc Fr � 1

Adverse (A) S0 0 A2 y � yc Fr 1

A3 y yc Fr � 1dydx

� 0

dydx

0

dydx

� 0

dydx

0

dydx

� 0

dydx

� 0

dydx

� 0

dydx

0

dydx

� 0

dydx

� 0

dydx

0

dydx

� 0

yc � yn

Channel bottom, S0 � Sc

C1

C3

yc

Channel bottom, S0 0

A2

A3

yc

Channel bottom, S0 � 0

H2

H3

yn

yc

Channel bottom, S0 � Sc

S1

S2

S3

Normaldepth

Surface profile y(x)

Startingpoint

Horizontal

yc

yn

Criticaldepth

Channel bottom, S0 Sc

M1

M2

M3

Horizontal

Horizontal

yn: doesnot exist

M3M2

M1

S2

S1

S1

S3

C1C3

H2 H2H3

A2

A3

A2

cen72367_ch13.qxd 11/6/04 2:51 PM Page 705

706FLUID MECHANICS

approaches S0 and the rate of increase in flow depth becomes equal to thechannel slope. This requires the surface profile to become horizontal atlarge y. Then we conclude that the M1 surface profile first rises in theflow direction and then tends to a horizontal asymptote.

As y → yc in subcritical flow (such as M2, H2, and A2), we have Fr → 1and 1 � Fr2 → 0, and thus the slope dy/dx tending to negative infinity. Butas y → yc in supercritical flow (such as M3, H3, and A3), we have Fr → 1and 1 � Fr2 → 0, and thus the slope dy/dx, which is a positive quantity,tending to infinity. That is, the free surface rises almost vertically and theflow depth increases very rapidly. This cannot be sustained physically, andthe free surface breaks down. The result is a hydraulic jump. The one-dimensional assumption is no longer applicable when this happens.

Some Representative Surface ProfilesA typical open-channel system involves several sections of different slopes,with connections called transitions, and thus the overall surface profile offlow is a continuous profile made up of the individual profiles describedearlier. Some representative surface profiles commonly encountered in openchannels, including some composite profiles, are given in Fig. 13–30. Foreach case, the change in surface profile is caused by a change in channelgeometry such as an abrupt change in slope or an obstruction in the flowsuch as a sluice gate. More composite profiles can be found in specializedbooks listed in the references. A point on a surface profile represents theflow height at that point that satisfies the mass, momentum, and energy con-servation relations. Note that dy/dx 1 and S0 1 in gradually variedflow, and the slopes of both the channels and the surface profiles in thesesketches are highly exaggerated for better visualization. Many channels andsurface profiles would appear nearly horizontal if drawn to scale.

Figure 13–30a shows the surface profile for gradually varied flow in achannel with mild slope and a sluice gate. The subcritical upstream flow(note that the flow is subcritical since the slope is mild) slows down as itapproaches the gate (such as a river approaching a dam) and the liquid levelrises. The flow past the gate is supercritical (since the height of the openingis less than the critical depth). Therefore, the surface profile is M1 beforethe gate and M3 after the gate prior to the hydraulic jump.

A section of an open channel may have a negative slope and involveuphill flow, as shown in Fig. 13–30b. Flow with an adverse slope cannot bemaintained unless the inertia forces overcome the gravity and viscous forcesthat oppose the fluid motion. Therefore, an uphill channel section must befollowed by a downhill section or a free outfall. For subcritical flow with anadverse slope approaching a sluice gate, the flow depth decreases as thegate is approached, yielding an A2 profile. Flow past the gate is typicallysupercritical, yielding an A3 profile prior to the hydraulic jump.

The open-channel section in Fig. 13–30c involves a slope change fromsteep to less steep. The flow velocity in the less steep part is lower (asmaller elevation drop to drive the flow), and thus the flow depth is higherwhen uniform flow is established again. Noting that uniform flow withsteep slope must be supercritical (y yc), the flow depth increases from theinitial to the new uniform level smoothly through an S3 profile.

cen72367_ch13.qxd 11/6/04 12:39 PM Page 706

Figure 13–30d shows a composite surface profile for an open channel thatinvolves various flow sections. Initially the slope is mild, and the flow isuniform and subcritical. Then the slope changes to steep, and the flowbecomes supercritical when uniform flow is established. The critical depthoccurs at the break in grade. The change of slope is accompanied by asmooth decrease in flow depth through an M2 profile at the end of the mild

707CHAPTER 13

yn1

yc

yc

yn2

(a) Flow through a sluice gate in an open channel with mild slope

(b) Flow through a sluice gate in an open channel with adverse slope and free outfall

(c) Uniform supercritical flow changing from steep to less steep slope

(d) Uniform subcritical flow changing from mildto steep to horizontal slope with free outfall

Mild

Adverse

Uniform flow

Uniform flow

Less steep

Horizontal

Freeoutfall

Hydraulicjump

Uniformflow

Uniformflow

H3H2

M2

S2

Steep

Mild

Steep

Uniformflow

Uniformflow

Hydraulicjump

Hydraulicjump

M1

A2

A3

A2

M3

yn2

yn2

yn2

yc

yc

yn1

yc

yn1

y yn2

S3

FIGURE 13–30Some common surface profiles

encountered in open-channel flow. All flows from left to right.

cen72367_ch13.qxd 11/6/04 12:39 PM Page 707

section, and through an S2 profile at the beginning of the steep section. Inthe horizontal section, the flow depth increases first smoothly through anH3 profile, and then rapidly during a hydraulic jump. The flow depth thendecreases through an H2 profile as the liquid accelerates toward the end ofthe channel to a free outfall. The flow becomes critical before reaching theend of the channel, and the outfall controls the upstream flow past thehydraulic jump. The outfalling flow stream is supercritical. Note that uni-form flow cannot be established in a horizontal channel since the gravityforce has no component in the flow direction, and the flow is inertia-driven.

Numerical Solution of Surface ProfileThe prediction of the surface profile y(x) is an important part of the designof open-channel systems. A good starting point for the determination of thesurface profile is the identification of the points along the channel, calledthe control points, at which the flow depth can be calculated from a knowl-edge of flow rate. For example, the flow depth at a section of a rectangularchannel where critical flow occurs, called the critical point, can be deter-mined from yc � (V

.2/gb2)1/3. The normal depth yn, which is the flow depth

reached when uniform flow is established, also serves as a control point.Once flow depths at control points are available, the surface profileupstream or downstream can be determined usually by numerical integra-tion of the nonlinear differential equation

(13–66)

The friction slope Sf is determined from the uniform-flow conditions, andthe Froude number from a relation appropriate for the channel cross section.

EXAMPLE 13–6 Classification of Channel Slope

Water is flowing uniformly in a rectangular open channel with unfinished-concrete surfaces. The channel width is 6 m, the flow depth is 2 m, and thebottom slope is 0.004. Determine if the channel should be classified asmild, critical, or steep for this flow (Fig. 13–31).

SOLUTION Water is flowing in an open channel uniformly. It is to be deter-mined whether the channel slope is mild, critical, or steep for this flow.Assumptions 1 The flow is steady and uniform. 2 The bottom slope is con-stant. 3 The roughness of the wetted surface of the channel and thus thefriction coefficient are constant.Properties The Manning coefficient for an open channel with unfinished-concrete surfaces is n � 0.014.Analysis The cross-sectional area, perimeter, and hydraulic radius are

Rh �Ac

p�

12 m2

10 m� 1.2 m

p � b � 2y � 6 m � 2(2 m) � 10 m

Ac � yb � (2 m)(6 m) � 12 m2

dy

dx�

S0 � Sf

1 � Fr2

708FLUID MECHANICS

�b2

b � 6 m

S0 � 0.004

y � 2 m


cen72367_ch13.qxd 11/6/04 12:39 PM Page 708

The flow rate is determined from the Manning equation to be

Noting that the flow is uniform, the specified flow rate is the normal depthand thus y � yn � 2 m. The critical depth for this flow is

This channel at these flow conditions is classified as steep since yn yc ,and the flow is supercritical.Discussion If the flow depth were greater than 2.2 m, the channel slopewould be said to be mild. Therefore, the bottom slope alone is not sufficientto classify a downhill channel as being mild, critical, or steep.

13–8 ■ RAPIDLY VARIED FLOW AND HYDRAULIC JUMP

Recall that flow in open channels is called rapidly varied flow (RVF) if theflow depth changes markedly over a relatively short distance in the flowdirection (Fig. 13–32). Such flows occur in sluice gates, broad- or sharp-crested weirs, waterfalls, and the transition sections of channels for expan-sion and contraction. A change in the cross section of the channel is animportant reason for the occurrence of rapidly varied flow. But some rapidlyvaried flows, such as flow through a sluice gate, occur even in regionswhere the channel cross section is constant.

Rapidly varied flows are typically complicated by the fact that they mayinvolve significant multidimensional and transient effects, backflows, andflow separation. Therefore, rapidly varied flows are usually studied experi-mentally or numerically. But despite these complexities, it is still possible toanalyze some rapidly varied flows using the one-dimensional flow approxi-mation with reasonable accuracy.

The flow in steep channels can be supercritical, and the flow can changeto subcritical if the channel can no longer sustain supercritical flow due to areduced slope of the channel or increased frictional effects. Any suchchange from supercritical to subcritical flow occurs through a hydraulicjump. A hydraulic jump involves considerable mixing and agitation, andthus a significant amount of mechanical energy dissipation.

Consider steady flow through a control volume that encloses the hydraulicjump, as shown in Fig. 13–33. To make a simple analysis possible, we makethe following assumptions:

1. The velocity is nearly constant across the channel at sections 1 and 2,and therefore the momentum-flux correction factors are b1 � b2 � 1.

2. The pressure in the liquid varies hydrostatically, and we consider gagepressure only since atmospheric pressure acts on all surfaces and itseffect cancels out.

yc �V#

2

gA2c

�(61.2 m3 s)2

(9.81 m/s2)(12 m2)� 2.2 m

V#

�an AcR

2 3h S 1 2

0 �1 m1/3 s

0.014 (12 m2)(1.2 m)2/3(0.004)1 2 � 61.2 m3/s

709CHAPTER 13

FIGURE 13–32Rapidly varied flow occurs when thereis a sudden change in flow, such as an

abrupt change in cross section.

x

EsEs1

Es2 � y2 �V 2

2g

y

rgy1

hL

y1

y2

V2

V1

Energy lineControlvolume

(1) (2)

1

2

Subcritical

Supercritical

rgy2

2

FIGURE 13–33Schematic and flow depth-specific

energy diagram for a hydraulic jump(specific energy decreases).

cen72367_ch13.qxd 11/6/04 12:39 PM Page 709

3. The wall shear stress and the associated losses are negligible relative tothe losses that occur during the hydraulic jump due to the intenseagitation.

4. The channel is wide and horizontal.5. There are no external or body forces other than gravity.

For a channel of width b, the conservation of mass or continuity relationm.2 � m

.1 can be expressed as ry1bV1 � ry2bV2 or

(13–67)

Noting that the only forces acting on the control volume in the horizontal x-

direction are the pressure forces, the momentum equation

in the x-direction becomes a balance between hydrostatic pressure

forces and momentum transfer,

(13–68)

where P1, avg � rgy1/2 and P2, avg � rgy2/2. For a channel width of b, we haveA1 � y1b, A2 � y2b, and m

.� m

.2 � m

.1 � rA1V1 � ry1bV1. Substituting and

simplifying, the momentum equation reduces to

(13–69)

Eliminating V2 by using V2 � (y1/y2)V1 from the continuity equation gives

(13–70)

Canceling the common factor y1 � y2 from both sides and rearranging give

(13–71)

where . This is a quadratic equation for y2/y1, and it has tworoots—one negative and one positive. Noting that y2/y1 cannot be negativesince both y2 and y1 are positive quantities, the depth ratio y2/y1 is deter-mined to be

Depth ratio: (13–72)

The energy equation for this horizontal flow section can be expressed as

(13–73)

Noting that V2 � (y1/y2)V1 and , the head loss associated withhydraulic jump is expressed as

(13–74)

The energy line for a hydraulic jump is shown in Fig. 13–33. The drop inthe energy line across the jump represents the head loss hL associated withthe jump.

hL � y1 � y2 �V 2

1 � V 22

2g� y1 � y2 �

y1Fr21

2 a1 �

y 21

y 22

b

Fr1 � V1 1gy1

y1 �V

21

2g� y2 �

V 22

2g� hL

y2

y1� 0.5a�1 � 21 � 8Fr2

1 b

Fr1 � V1 1gy1

ay2

y1b 2

�y2

y1� 2Fr2

1 � 0

y 21 � y 2

2 �2y1V 1

2

gy2 (y1 � y2)

y 21 � y 2

2 �2y1V1

g (V2 � V1)

P1, avg A1 � P2, avg A2 � m#V2 � m

#V

1

ainbm#

V→

a F→

� aoutbm#V→

�

y1V1 � y2V2

710FLUID MECHANICS

cen72367_ch13.qxd 11/6/04 12:39 PM Page 710

For a given Fr1 and y1, the downstream flow depth y2 and the head loss hLcan be calculated from Eqs. 13–72 and 13–74, respectively. Plotting hLagainst Fr1 would reveal that hL becomes negative for Fr1 1, which isimpossible (it would correspond to negative entropy generation, whichwould be a violation of the second law of thermodynamics). Thus we con-clude that the upstream flow must be supercritical (Fr1 � 1) for a hydraulicjump to occur. In other words, it is impossible for subcritical flow toundergo a hydraulic jump. This is analogous to gas flow having to be super-sonic (Mach number greater than 1) to undergo a shock wave.

Head loss is a measure of the mechanical energy dissipated via internalfluid friction, and head loss is usually undesirable as it represents themechanical energy wasted. But sometimes hydraulic jumps are designed inconjunction with stilling basins and spillways of dams, and it is desirable towaste as much of the mechanical energy as possible to minimize themechanical energy of the water and thus its potential to cause damage. Thisis done by first producing supercritical flow by converting high pressure tohigh linear velocity, and then allowing the flow to agitate and dissipate partof its kinetic energy as it breaks down and decelerates to a subcritical veloc-ity. Therefore, a measure of performance of a hydraulic jump is its fractionof energy dissipation.

The specific energy of the liquid before the hydraulic jump is Es1 � y1� V1

2/2g. Then the energy dissipation ratio (Fig. 13–34) can beexpressed as

(13–75)

The fraction of energy dissipation ranges from just a few percent for weakhydraulic jumps (Fr1 2) to 85 percent for strong jumps (Fr1 � 9).

Unlike a normal shock in gas flow, which occurs at a cross section andthus has negligible thickness, the hydraulic jump occurs over a considerablechannel length. In the Froude number range of practical interest, the lengthof the hydraulic jump is observed to be 4 to 7 times the downstream flowdepth y2.

Experimental studies indicate that hydraulic jumps can be considered infive categories as shown in Table 13–4, depending primarily on the value ofthe upstream Froude number Fr1. For Fr1 somewhat higher than 1, the liquidrises slightly during hydraulic jump, producing standing waves. At largerFr1, highly damaging oscillating waves occur. The desirable range of Froudenumbers is 4.5 Fr1 9, which produces stable and well-balanced steadywaves with high levels of energy dissipation within the jump. Hydraulicjumps with Fr1 � 9 produce very rough waves. The depth ratio y2/y1 rangesfrom slightly over 1 for undular jumps that are mild and involve small risesin surface level to over 12 for strong jumps that are rough and involve highrises in surface level.

In this section we limit our consideration to wide horizontal rectangularchannels so that edge and gravity effects are negligible. Hydraulic jumps innonrectangular and sloped channels behave similarly, but the flow character-istics and thus the relations for depth ratio, head loss, jump length, and dis-sipation ratio are different.

Dissipation ratio �hL

Es1�

hL

y1 � V 21 2g

�hL

y1(1 � Fr21 2)

711CHAPTER 13

hL

y1

y2V2

V1

V 2

2g

Energy line

Dissipation ratio � �hL

Es1

hL

y1 � V 2/2g

(1) (2)

V 2

2g2

1

1

FIGURE 13–34The energy dissipation ratio represents

the fraction of mechanical energydissipated during a hydraulic jump.

cen72367_ch13.qxd 11/6/04 12:39 PM Page 711

EXAMPLE 13–7 Hydraulic Jump

Water discharging into a 10-m-wide rectangular horizontal channel from asluice gate is observed to have undergone a hydraulic jump. The flow depthand velocity before the jump are 0.8 m and 7 m/s, respectively. Determine(a) the flow depth and the Froude number after the jump, (b) the head lossand the dissipation ratio, and (c) the wasted power production potential dueto the hydraulic jump (Fig. 13–35).

SOLUTION Water at a specified depth and velocity undergoes a hydraulicjump in a horizontal channel. The depth and Froude number after the jump,

712FLUID MECHANICS

TABLE 13–4

Classification of hydraulic jumps

Source: U.S. Bureau of Reclamation (1955).

Depth Fraction ofUpstream Ratio Energy Surface

Fr1 y2/y1 Dissipation Description Profile

1 1 0 Impossible jump. Would violate the second law of thermodynamics.

1–1.7 1–2 5% Undular jump (or standing wave). Small rise in surface level. Low energy dissipation. Surface rollers develop near Fr � 1.7.

1.7–2.5 2–3.1 5–15% Weak jump. Surface rising smoothly, with small rollers. Low energy dissipation.

2.5–4.5 3.1–5.9 15–45% Oscillating jump. Pulsations caused by entering jets at the bottom generate large waves that can travel for miles and damage earth banks. Should be avoided in the design of stilling basins.

4.5–9 5.9–12 45–70% Steady jump. Stable, well-balanced,and insensitive to downstream conditions. Intense eddy motion and high level of energy dissipation within the jump. Recommended range for design.

�9 �12 70–85% Strong jump. Rough and intermittent. Very effective energy dissipation, but may be uneconomical compared to other designs.

hL

V1 � 7 m/s V2

y1 � 0.8 my2

Energy line

(1) (2)


y1V1 V2 y2

cen72367_ch13.qxd 11/6/04 12:39 PM Page 712

the head loss and the dissipation ratio, and the wasted power potential areto be determined.Assumptions 1 The flow is steady or quasi-steady. 2 The channel is suffi-ciently wide so that the end effects are negligible.Properties The density of water is 1000 kg/m3.Analysis (a) The Froude number before the hydraulic jump is

which is greater than 1. Therefore, the flow is indeed supercritical before thejump. The flow depth, velocity, and Froude number after the jump are

Note that the flow depth triples and the Froude number reduces to aboutone-fifth after the jump.

(b) The head loss is determined from the energy equation to be

The specific energy of water before the jump and the dissipation ratio is

Therefore, 17.3 percent of the available head (or mechanical energy) of theliquid is wasted (converted to thermal energy) as a result of frictional effectsduring this hydraulic jump.

(c) The mass flow rate of water is

Then the power dissipation corresponding to a head loss of 0.572 m becomes

Discussion The results show that the hydraulic jump is a highly dissipativeprocess, wasting 314 kW of power production potential in this case. That is,if the water is routed to a hydraulic turbine instead of being released from

� 314,000 N � m/s � 314 kW

E#dissipated � m# ghL � (56,000 kg/s)(9.81 m/s2)(0.572 m)a 1 N

1 kg � m/s2b

m# � rV#

� rby1V1 � (1000 kg/m3)(0.8 m)(10 m)(7 m/s) � 56,000 kg/s


Es1�

0.572 m

3.30 m� 0.173

Es1 � y1 �V

21

2g� (0.8 m) �

(7 m/s)2

2(9.81 m/s2)� 3.30 m

� 0.572 m

hL � y1 � y2 �V

21 � V

22

2g� (0.8 m) � (2.46 m) �

(7 m/s)2 � (2.28 m/s)2

2(9.81 m/s2)

Fr2 �V2

2gy2

�2.28 m/s

2(9.81 m/s2)(2.46 m)� 0.464

V2 �y1

y2V1 �

0.8 m

2.46 m (7 m/s) � 2.28 m/s

y2 � 0.5y1a�1 � 21 � 8Fr21b � 0.5(0.8 m)a�1 � 21 � 8 � 2.502b � 2.46 m

Fr1 �V1

2gy1

�7 m/s

2(9.81 m/s2)(0.8 m)� 2.50

713CHAPTER 13

cen72367_ch13.qxd 11/6/04 12:39 PM Page 713

the sluice gate, up to 314 kW of power could be generated. But this poten-tial is converted to useless thermal energy instead of useful power, causing atemperature rise of

for water. Note that a 314-kW resistance heater would cause the same tem-perature rise for water flowing at a rate of 56,000 kg/s.

13–9 ■ FLOW CONTROL AND MEASUREMENTThe flow rate in pipes and ducts is controlled by various kinds of valves.Liquid flow in open channels, however, is not confined, and thus the flowrate is controlled by partially blocking the channel. This is done by eitherallowing the liquid to flow over the obstruction or under it. An obstructionthat allows the liquid to flow over it is called a weir, and an obstructionwith an adjustable opening at the bottom that allows the liquid to flowunderneath it is called an underflow gate. Such devices can be used to con-trol the flow rate through the channel as well as to measure it.

Underflow GatesThere are numerous types of underflow gates to control the flow rate, eachwith certain advantages and disadvantages. Underflow gates are located atthe bottom of a wall, dam, or an open channel. Two common types of suchgates, the sluice gate and the drum gate, are shown in Fig. 13–36. A sluicegate is typically vertical and has a plane surface, whereas a drum gate has acircular cross section with a streamlined surface.

When the gate is opened, the upstream liquid accelerates as it approachesthe gate, reaches the critical speed at the gate, and accelerates further tosupercritical speeds past the gate. Therefore, an underflow gate is analogousto a converging–diverging nozzle in gas dynamics. The discharge from anunderflow gate is called a free outflow if the liquid jet streaming out of thegate is open to the atmosphere, and it is called a drowned (or submerged)outflow if the discharged liquid flashes back and submerges the jet, as shownin Fig. 13–36. In drowned flow, the liquid jet undergoes a hydraulic jump,and thus the downstream flow is subcritical. Also, drowned outflow involvesa high level of turbulence and backflow, and thus a large head loss hL.

The flow depth-specific energy diagram for flow through underflow gateswith free and drowned outflow is given in Fig. 13–37. Note that the specificenergy remains constant for idealized gates with negligible frictional effects(from point 1 to point 2a), but decreases for actual gates. The downstream issupercritical for a gate with free outflow (point 2b), but subcritical for onewith drowned outflow (point 2c) since a drowned outflow also involves ahydraulic jump to subcritical flow, which involves considerable mixing andenergy dissipation.

Assuming the frictional effects to be negligible and the upstream (orreservoir) velocity to be low, it can be shown by using the Bernoulli equa-tion that the discharge velocity of a free jet is (see Chap. 5 for details)

(13–76)V � 22gy1

�T �E#dissipated

m# cp

�314 kJ/s

(56,000 kg/s)(4.18 kJ/kg � �C)� 0.0013�C

714FLUID MECHANICS

cen72367_ch13.qxd 11/6/04 12:39 PM Page 714

The frictional effects can be accounted for by modifying this relation with adischarge coefficient Cd. Then the discharge velocity at the gate and theflow rate become

(13–77)

where b and a are the width and the height of the gate opening, respectively.The discharge coefficient Cd � 1 for idealized flow, but Cd 1 for actual

flow through the gates. Experimentally determined values of Cd for under-flow gates are plotted in Fig. 13–38 as functions of the contraction coeffi-cient y2/a and the depth ratio y1/a. Note that most values of Cd for free out-flow from a vertical sluice gate range between 0.5 and 0.6. The Cd valuesdrop sharply for drowned outflow, as expected, and the flow rate decreasesfor the same upstream conditions. For a given value of y1/a, the value of Cddecreases with increasing y2/a.

V � Cd22gy1 and V#

� Cdba22gy1

715CHAPTER 13

y1

a

V1

(a) Sluice gate with free outflow

Sluice gate

Vena contracta

V2y2

FIGURE 13–36Common types of underflow gates to control flow rate.

y1

a

V1

(b) Sluice gate with drowned outflow

Sluice gate

V2y2

y1V1

(c) Drum gate

V2y2

Drum

Es

Es1 � y1 �

Es1 � Es2a

V 2

2g

y

2a2b

1

2c

Subcriticalflow

Drownedoutflow

Frictionlessgate

Supercriticalflow

1


energy diagram for flow throughunderflow gates.

00

0.1

0.2

0.3

0.4

0.5

0.6

2 4 6 8

y1/a

Cd

10 12 14 16

Free outflow

Drowned outflow

y2/a � 2 3 4 5 6 7 8

FIGURE 13–38Discharge coefficients for drownedand free discharge from underflow

gates.From Henderson, Open Channel Flow, 1st Edition,

© 1966. Reprinted by permission of PearsonEducation, Inc., Upper Saddle River, NJ.

cen72367_ch13.qxd 11/6/04 12:39 PM Page 715

EXAMPLE 13–8 Sluice Gate with Drowned Outflow

Water is released from a 3-m-deep reservoir into a 6-m-wide open channelthrough a sluice gate with a 0.25-m-high opening at the channel bottom.The flow depth after all turbulence subsides is measured to be 1.5 m. Deter-mine the rate of discharge (Fig. 13–39).

SOLUTION Water is released from a reservoir through a sluice gate into anopen channel. For specified flow depths, the rate of discharge is to bedetermined.Assumptions 1 The flow is steady or quasi-steady. 2 The channel is suffi-ciently wide so that the end effects are negligible.Analysis The depth ratio y1/a and the contraction coefficient y2/a are

The corresponding discharge coefficient is determined from Fig. 13–38 to beCd � 0.47. Then the discharge rate becomes

Discussion In the case of free flow, the discharge coefficient would be Cd� 0.59, with a corresponding flow rate of 6.78 m3/s. Therefore, the flowrate decreases considerably when the outflow is drowned.

Overflow GatesRecall that the total mechanical energy of a liquid at any cross section of anopen channel can be expressed in terms of heads as H � zb � y � V2/2g,where y is the flow depth, zb is the elevation of the channel bottom, and V isthe average flow velocity. During flow with negligible frictional effects(head loss hL � 0), the total mechanical energy remains constant, and theone-dimensional energy equation for open-channel flow between upstreamsection 1 and downstream section 2 can be written as

(13–78)

where Es � y � V2/2g is the specific energy and �zb � zb2 � zb1 is the ele-vation of the bottom point of flow at section 2 relative to that at section 1.Therefore, the specific energy of a liquid stream increases by |�zb| duringdownhill flow (note that �zb is negative for channels inclined down),decreases by �zb during uphill flow, and remains constant during horizontalflow. (The specific energy also decreases by hL for all cases if the frictionaleffects are not negligible.)

For a channel of constant width b, V.

� AcV � byV � constant in steadyflow and V � V

./Ac. Then the specific energy can be expressed as

(13–79)

The variation of the specific energy Es with flow depth y for steady flow ina channel of constant width b is replotted in Fig. 13–40. This diagram isextremely valuable as it shows the allowable states during flow. Once theupstream conditions at a flow section 1 are specified, the state of the liquid

Es � y �V#

2

2gb2y 2

zb1 � y1 �V

21

2g� zb2 � y2 �

V 22

2g or Es1 � �zb � Es2

V#

� Cdba22gy1 � 0.47(6 m)(0.25 m)22(9.81 m/s2)(3 m) � 5.41 m3/s

y1

a�

3 m

0.25 m� 12 and y2

a�

1.5 m

0.25 m� 6

716FLUID MECHANICS

y1 � 3 m

a � 0.25 m

y2 � 1.5 m

Sluice gate


Es

yc

Emin

Es � y

V 2

2g

y

ySupercriticalflow, Fr � 1

Fr � 1Criticaldepth

Subcriticalflow, Fr 1

V � constant.

FIGURE 13–40Variation of specific energy Es withdepth y for a specified flow rate.

cen72367_ch13.qxd 11/6/04 12:39 PM Page 716

at any section 2 on an Es-y diagram must fall on a point on the specificenergy curve that passes through point 1.

Flow over a Bump with Negligible FrictionNow consider steady flow with negligible friction over a bump of height �zbin a horizontal channel of constant width b, as shown in Fig. 13–41. Theenergy equation in this case is, from Eq. 13–78,

(13–80)

Therefore, the specific energy of the liquid decreases by �zb as it flows overthe bump, and the state of the liquid on the Es-y diagram shifts to the left by�zb, as shown in Fig. 13–41. The continuity equation for a channel of largewidth is y2V2 � y1V1 and thus V2 � (y1/y2)V1. Then the specific energy ofthe liquid over the bump can be expressed as

(13–81)

Rearranging,

(13–82)

which is a third-degree polynomial equation in y2 and thus has three solu-tions. Disregarding the negative solution, it appears that the flow depth overthe bump can have two values.

Now the curious question is, does the liquid level rise or drop over thebump? Our intuition says the entire liquid body will follow the bump andthus the liquid surface will rise over the bump, but this is not necessarily so.Noting that specific energy is the sum of the flow depth and dynamic head,either scenario is possible, depending on how the velocity changes. The Es-ydiagram in Fig. 13–41 gives us the definite answer: If the flow before thebump is subcritical (state 1a), the flow depth y2 decreases (state 2a). If thedecrease in flow depth is greater than the bump height (i.e., y1 � y2 � �zb),the free surface is suppressed. But if the flow is supercritical as it approachesthe bump (state 1b), the flow depth rises over the bump (state 2b), creating abigger bump over the free surface.

y 32 � (Es1 � �zb)y

22 �

V 21

2g y 2

1 � 0

Es2 � y2 �V

22

2g → Es1 � �zb � y2 �

V 21

2g y 2

1

y 22

Es2 � Es1 � �zb

717CHAPTER 13

EsEmin � Ec

y

�zb Supercriticalflow

Subcriticalflow

2b

2a

1a

1bV2V1

y1 y2

�zb

Subcriticalupstream flow

Supercriticalupstream flow

Bump


energy diagram for flow over a bumpfor subcritical and supercritical

upstream flows.

cen72367_ch13.qxd 11/6/04 12:39 PM Page 717

The situation is reversed if the channel has a depression of depth �zbinstead of a bump: The specific energy in this case increases (so that state 2is to the right of state 1 on the Es-y diagram) since �zb is negative. There-fore, the flow depth increases if the approach flow is subcritical anddecreases if it is supercritical.

Now let’s reconsider flow over a bump with negligible friction, as dis-cussed earlier. As the height of the bump �zb is increased, point 2 (either 2aor 2b for sub- or supercritical flow) continues shifting to the left on the Es-ydiagram, until finally reaching the critical point. That is, the flow over thebump is critical when the bump height is �zc � Es1 � Esc � Es1 � Emin,and the specific energy of the liquid reaches its minimum level.

The question that comes to mind is, what happens if the bump height isincreased further? Does the specific energy of the liquid continue decreas-ing? The answer to this question is a resounding no since the liquid isalready at its minimum energy level, and its energy cannot decrease any fur-ther. In other words, the liquid is already at the furthest left point on the Es-y diagram, and no point further left can satisfy conservation of mass,momentum, and energy. Therefore, the flow must remain critical. The flowat this state is said to be choked. In gas dynamics, this is analogous to theflow in a converging nozzle accelerating as the back pressure is lowered,and reaching the speed of sound at the nozzle exit when the back pressurereaches the critical pressure. But the nozzle exit velocity remains at thesonic level no matter how much the back pressure is lowered. Here again,the flow is choked.

Broad-Crested WeirThe discussions on flow over a high bump can be summarized as follows:The flow over a sufficiently high obstruction in an open channel is alwayscritical. Such obstructions placed intentionally in an open channel to mea-sure the flow rate are called weirs. Therefore, the flow velocity over a suffi-ciently broad weir is the critical velocity, which is expressed as ,where yc is the critical depth. Then the flow rate over a weir of width b canbe expressed as

(13–83)

A broad-crested weir is a rectangular block of height Pw and length Lwthat has a horizontal crest over which critical flow occurs (Fig. 13–42). Theupstream head above the top surface of the weir is called the weir head andis denoted by H. To obtain a relation for the critical depth yc in terms ofweir head H, we write the energy equation between a section upstream anda section over the weir for flow with negligible friction as

(13–84)

Cancelling Pw from both sides and substituting give

(13–85)

Substituting into Eq. 13–83, the flow rate for this idealized flow case withnegligible friction is determined to be

yc �2

3 aH �

V 21

2gb

Vc � 1gyc

H � Pw �V

21

2g� yc � Pw �

V 2c

2g

V#

� AcV � ycb2gyc � bg1 2y 3 2c

V � 1gyc

718FLUID MECHANICS

V1

Pw

HDischarge

Vc

Lw

yc

Broad-crestedweir

FIGURE 13–42Flow over a broad-crested weir.

cen72367_ch13.qxd 11/6/04 12:39 PM Page 718

(13–86)

This relation shows the functional dependence of the flow rate on the flowparameters, but it overpredicts the flow rate by several percent because itdoes not consider the frictional effects. These effects are properly accountedfor by modifying the theoretical relation (Eq. 13–86) by an experimentallydetermined weir discharge coefficient Cwd as

Broad-crested weir: (13–87)

where reasonably accurate values of discharge coefficients for broad-crestedweirs can be obtained from (Chow, 1959)

(13–88)

More accurate but complicated relations for Cwd, broad are also available inthe literature (e.g., Ackers, 1978). Also, the upstream velocity V1 is usuallyvery low, and it can be disregarded. This is especially the case for highweirs. Then the flow rate can be approximated as

Broad-crested weir with low V1: (13–89)

It should always be kept in mind that the basic requirement for the use ofEqs. 13–87 to 13–89 is the establishment of critical flow above the weir,and this puts some limitations on the weir length Lw. If the weir is too long(Lw � 12H), wall shear effects dominate and cause the flow over the weir tobe subcritical. If the weir is too short (Lw 2H), the liquid may not be ableto accelerate to critical velocity. Based on observations, the proper length ofthe broad-crested weir is 2H Lw 12H. Note that a weir that is too longfor one flow may be too short for another flow, depending on the value ofthe weir head H. Therefore, the range of flow rates should be known beforea weir can be selected.

Sharp-Crested WeirsA sharp-crested weir is a vertical plate placed in a channel that forces theliquid to flow through an opening to measure the flow rate. The type of theweir is characterized by the shape of the opening. A vertical thin plate witha straight top edge is referred to as rectangular weir since the cross sectionof the flow over it is rectangular; a weir with a triangular opening is referredto as a triangular weir; etc.

Upstream flow is subcritical and becomes critical as it approaches the weir.The liquid continues to accelerate and discharges as a supercritical flowstream that resembles a free jet. The reason for acceleration is the steadydecline in the elevation of the free surface, and the conversion of this eleva-tion head into velocity head. The flow-rate correlations given below arebased on the free overfall of liquid discharge past the weir, called nappes,being clear from the weir. It may be necessary to ventilate the space underthe nappe to assure atmospheric pressure underneath. Empirical relations fordrowned weirs are also available.

V#

� Cwd, broadb2ga23b 3 2

H 3 2

Cwd, broad �0.65

21 � H Pw

V#

� Cwd, broadb2ga23b 3 2aH �

V 21

2gb 3 2

V#

ideal � b2ga23b 3 2aH �

V 21

2gb 3 2

719CHAPTER 13

cen72367_ch13.qxd 11/6/04 12:39 PM Page 719

Consider the flow of a liquid over a sharp-crested weir placed in a hori-zontal channel, as shown in Fig. 13–43. For simplicity, the velocityupstream of the weir is approximated as being nearly constant through verti-cal cross section 1. The total energy of the upstream liquid expressed as ahead relative to the channel bottom is the specific energy, which is the sumof the flow depth and the velocity head. That is, y1 � V1

2/2g, where y1 � H� Pw. The flow over the weir is not one-dimensional since the liquid under-goes large changes in velocity and direction over the weir. But the pressurewithin the nappe is atmospheric.

A simple relation for the variation of liquid velocity over the weir can beobtained by assuming negligible friction and writing the Bernoulli equationbetween a point in upstream flow (point 1) and a point over the weir at adistance h from the upstream liquid level as

(13–90)

Cancelling the common terms and solving for u2, the idealized velocity dis-tribution over the weir is determined to be

(13–91)

In reality, water surface level drops somewhat over the weir as the waterstarts its free overfall (the drawdown effect at the top) and the flow separa-tion at the top edge of the weir further narrows the nappe (the contractioneffect at the bottom). As a result, the flow height over the weir is consider-ably smaller than H. When the drawdown and contraction effects are disre-garded for simplicity, the flow rate is obtained by integrating the product ofthe flow velocity and the differential flow area over the entire flow area,

(13–92)

where w is the width of the flow area at distance h from the upstream freesurface.

In general, w is a function of h. But for a rectangular weir, w � b, whichis constant. Then the integration can be performed easily, and the flow ratefor a rectangular weir for idealized flow with negligible friction and negligi-ble drawdown and contraction effects is determined to be

(13–93)

When the weir height is large relative to the weir head (Pw �� H), theupstream velocity V1 is low and the upstream velocity head can beneglected. That is, V1

2/2g H. Then,

(13–94)

Therefore, the flow rate can be determined from a knowledge of two geo-metric quantities: the crest width b and the weir head H, which is the verti-cal distance between the weir crest and the upstream free surface.

This simplified analysis gives the general form of the flow-rate relation,but it needs to be modified to account for the frictional and surface tensioneffects, which play a secondary role as well as the drawdown and contraction

V#

ideal, rec �2

3 b22gH 3 2

V#

ideal �2

3 b22g caH �

V 21

2gb 3 2

� aV 21

2gb 3 2d

V#

� �Ac

u2 dAc2 � �H

h�0

22gh � V 21 w dh

u2 � 22gh � V 21

H � Pw �V 2

1

2g� (H � Pw � h) �

u22

2g

720FLUID MECHANICS

y

x

V1

u2(h)h

Nappe

Weir(2)

(1)

H

Pw

21

FIGURE 13–43Flow over a sharp-crested weir.

cen72367_ch13.qxd 11/6/04 12:39 PM Page 720

effects. Again this is done by multiplying the ideal flow-rate relations by anexperimentally determined weir discharge coefficient Cwd. Then the flow ratefor a sharp-crested rectangular weir is expressed as

Sharp-crested rectangular weir: (13–95)

where, from Ref. 1 (Ackers, 1978),

(13–96)

This formula is applicable over a wide range of upstream Reynolds numbersdefined as Re � V1H/n. More precise but also more complex correlationsare also available in the literature. Note that Eq. 13–95 is valid for full-widthrectangular weirs. If the width of the weir is less than the channel width sothat the flow is forced to contract, an additional coefficient for contractioncorrection should be incorporated to properly account for this effect.

Another type of sharp-crested weir commonly used for flow measurementis the triangular weir (also called the V-notch weir) shown in Fig. 13–44.The triangular weir has the advantage that it maintains a high weir head Heven for small flow rates because of the decreasing flow area with decreasingH, and thus it can be used to measure a wide range of flow rates accurately.

From geometric consideration, the notch width can be expressed as w �2(H � h) tan(u/2), where u is the V-notch angle. Substituting into Eq. 13–92and performing the integration give the ideal flow rate for a triangular weirto be

(13–97)

where we again neglected the upstream velocity head. The frictional andother dissipative effects are accounted for conveniently by multiplying theideal flow rate by a weir discharge coefficient. Then the flow rate for asharp-crested triangular weir becomes

Sharp-crested triangular weir: (13–98)

where the values of Cwd typically range between 0.58 and 0.62. Therefore,the fluid friction, the constriction of flow area, and other dissipative effectscause the flow rate through the V-notch to decrease by about 40 percentcompared to the ideal case. For most practical cases (H � 0.2 m and 45° u 120°), the value of the weir discharge coefficient is about Cwd� 0.58. More precise values are available in the literature.

EXAMPLE 13–9 Subcritical Flow over a Bump

Water flowing in a wide horizontal open channel encounters a 15-cm-highbump at the bottom of the channel. If the flow depth is 0.80 m and thevelocity is 1.2 m/s before the bump, determine if the water surface isdepressed over the bump (Fig. 13–45).

SOLUTION Water flowing in a horizontal open channel encounters a bump.It will be determined if the water surface is depressed over the bump.

V#

� Cwd, tri8

15 tanau

2b22gH 5 2

V#

ideal, tri �8

15 tanau

2b22gH 5 2

Cwd, rec � 0.598 � 0.0897H

Pw

for H

Pw

� 2

V#

rec � Cwd, rec 2

3 b22gH 3 2

721CHAPTER 13

Upstream freesurface

Weirplate

uH

w h

Pw

FIGURE 13–44A triangular (or V-notch) sharp-crestedweir plate geometry. The view is from

downstream looking upstream.

EsEs1Es2

y2

y1

y

�zb

Subcriticalflow2

1

V1 � 1.2 m/s

y1 � 0.80 m y2�zb � 0.15 m

Depressionover the bump

Bump

FIGURE 13–45Schematic and flow depth-specificenergy diagram for Example 13–9.

cen72367_ch13.qxd 11/6/04 12:39 PM Page 721

Assumptions 1 The flow is steady. 2 Frictional effects are negligible so thatthere is no dissipation of mechanical energy. 3 The channel is sufficientlywide so that the end effects are negligible.Analysis The upstream Froude number and the critical depth are

The flow is subcritical since Fr 1 and therefore the flow depth decreasesover the bump. The upstream specific energy is

The flow depth over the bump can be determined from

Substituting,

or

Using an equation solver, the three roots of this equation are determined tobe 0.59 m, 0.36 m, and �0.22 m. We discard the negative solution asphysically impossible. We also eliminate the solution 0.36 m since it is lessthan the critical depth, and it can occur only in supercritical flow. Thus theonly meaningful solution for flow depth over the bump is y2 � 0.59 m. Thenthe distance of the water surface over the bump from the channel bottom is�zb � y2 � 0.15 � 0.59 � 0.74 m, which is less than y1 � 0.80 m. There-fore, the water surface is depressed over the bump in the amount of

Discussion Note that having y2 y1 does not necessarily indicate that thewater surface is depressed (it may still rise over the bump). The surface isdepressed over the bump only when the difference y1 � y2 is larger than thebump height �zb. Also, the actual value of depression may be differentthan 0.06 m because of the frictional effects that are neglected in theanalysis.

EXAMPLE 13–10 Measuring Flow Rate by a Weir

The flow rate of water in a 5-m-wide horizontal open channel is being mea-sured with a 0.60-m-high sharp-crested rectangular weir of equal width. If thewater depth upstream is 1.5 m, determine the flow rate of water (Fig. 13–46).

SOLUTION The water depth upstream of a horizontal open channelequipped with a sharp-crested rectangular weir is measured. The flow rate isto be determined.

Depression � y1 � (y2 � �zb) � 0.80 � (0.59 � 0.15) � 0.06 m

y 32 � 0.723y 2

2 � 0.0470 � 0

y 32 � (0.873 � 0.15 m)y 2

2 �(1.2 m/s)2

2(9.81 m/s2) (0.80 m)2 � 0

y 32 � (Es1 � �zb)y

22 �

V 21

2g y 2

1 � 0

Es1 � y1 �V 2

1

2g� (0.80 m) �

(1.2 m/s)2

2(9.81 m/s2)� 0.873 m

yc � a V#

2

gb2b1 3

� a(by1V1)2

gb2 b1 3

� ay 21V

21

gb 1 3

� a(0.8 m)2(1.2 m s)2

9.81 m/s2 b 1 3

� 0.455 m

Fr1 �V1

2gy1

�1.2 m/s

2(9.81 m2/s)(0.80 m)� 0.428

722FLUID MECHANICS

V1

Sharp-crestedrectangular weir

y1 � 1.5 m

Pw � 0.60 m

b � 5 m


cen72367_ch13.qxd 11/6/04 12:39 PM Page 722

Assumptions 1 The flow is steady. 2 The upstream velocity head is negligi-ble. 3 The channel is sufficiently wide so that the end effects are negligible.Analysis The weir head is

The discharge coefficient of the weir is

The condition H/Pw 2 is satisfied since 0.9/0.6 � 1.5. Then the waterflow rate through the channel becomes

Discussion The upstream velocity and the upstream velocity head are

This is 8.6 percent of the weir head, which is significant. When theupstream velocity head is considered, the flow rate becomes 10.2 m3/s,which is about 10 percent higher than the value determined. Therefore, it isgood practice to consider the upstream velocity head unless the weir heightPw is very large relative to the weir head H.

V1 �V#

by1�

9.24 m3/s

(5 m)(1.5 m)� 1.23 m/s and V 2

1

2g�

(1.23 m/s)2

2(9.81 m/s2)� 0.077 m

� 9.24 m3/s

� (0.733) 2

3 (5 m)22(9.81 m/s2)(0.90 m)3 2

V#

rec � Cwd, rec 2

3 b22gH 3 2

Cwd, rec � 0.598 � 0.0897 H

Pw

� 0.598 � 0.0897 0.90

0.60� 0.733

H � y1 � Pw � 1.5 � 0.60 � 0.90 m

723CHAPTER 13

SUMMARY

Open-channel flow refers to the flow of liquids in channelsopen to the atmosphere or in partially filled conduits. Theflow in a channel is said to be uniform if the flow depth (andthus the average velocity) remains constant. Otherwise, theflow is said to be nonuniform or varied. The hydraulic radiusis defined as Rh � Ac/p. The dimensionless Froude number isdefined as

The flow is classified as subcritical for Fr 1, critical for Fr� 1, and supercritical for Fr � 1. Flow depth in critical flowis called the critical depth and is expressed as

where b is the channel width for wide channels.The speed at which a surface disturbance travels through a

liquid of depth y is the wave speed c0, which is expressed as

yc �V#

2

gA2c

or yc � a V#

2

gb2b1 3

Fr �V

2gLc

�V

2gy

. The total mechanical energy of a liquid in a chan-nel is expressed in terms of heads as

where zb is the elevation head, P/rg � y is the pressure head,and V2/2g is the velocity head. The sum of the pressure anddynamic heads is called the specific energy Es,

The continuity equation is Ac1V1 � Ac2V2. The energy equa-tion is expressed as

Here hL is the head loss and S0 � tan u is the bottom slope ofa channel. The friction slope is defined as Sf � hL/L.

y1 �V

21

2g� S0L � y2 �

V 22

2g� hL

Es � y �V

2

2g

H � zb � y �V

2

2g

c0 � 1gy

cen72367_ch13.qxd 11/6/04 12:39 PM Page 723

The flow depth in uniform flow is called the normal depthyn, and the average flow velocity is called the uniform-flowvelocity V0. The velocity and flow rate in uniform flow aregiven by

where n is the Manning coefficient whose value depends onthe roughness of the channel surfaces, and a � 1 m1/3/s� (3.2808 ft)1/3/s � 1.486 ft1/3/s. If yn � yc, the flow is uni-form critical flow, and the bottom slope S0 equals the criticalslope Sc expressed as

for film flow or flow in a wide rectangular channel with b �� yc.

The best hydraulic cross section for an open channel is theone with the maximum hydraulic radius, or equivalently, theone with the minimum wetted perimeter for a specified crosssection. The criteria for best hydraulic cross section for a rec-tangular channel is y � b/2. The best cross section for trape-zoidal channels is half of a hexagon.

In rapidly varied flow (RVF), the flow depth changesmarkedly over a relatively short distance in the flow direc-tion. Any change from supercritical to subcritical flow occursthrough a hydraulic jump, which is a highly dissipativeprocess. The depth ratio y2/y1, head loss, and energy dissipa-tion ratio during hydraulic jump are expressed as

�hL

y1(1 � Fr21 2)


Es1�

hL

y1 � V 21 2g

� y1 � y2 �y1Fr2

1

2a1 �

y 21

y 22

b

hL � y1 � y2 �V

21 � V

22

2g

y2

y1� 0.5a�1 � 21 � 8Fr2

1b

Sc �gn2yc

a2R4 3h

which simplifies to Sc �gn2

a2y 1 3c

V0 �an

R2 3h S 1 2

0 and V#

�anAcR

2 3h S 1 2

0

724FLUID MECHANICS

An obstruction that allows the liquid to flow over it iscalled a weir, and an obstruction with an adjustable openingat the bottom that allows the liquid to flow underneath it iscalled an underflow gate. The flow rate through a sluice gateis given by

where b and a are the width and the height of the gate open-ing, respectively, and Cd is the discharge coefficient, whichaccounts for the frictional effects.

A broad-crested weir is a rectangular block that has a hori-zontal crest over which critical flow occurs. The upstreamhead above the top surface of the weir is called the weirhead, H. The flow rate is expressed as

where the discharge coefficient is

The flow rate for a sharp-crested rectangular weir isexpressed as

where

For a sharp-crested triangular weir, the flow rate is given as

where the values of Cwd, tri typically range between 0.58 and0.62.

Open-channel analysis is commonly used in the design ofsewer systems, irrigation systems, floodways, and dams.Some open-channel flows are analyzed in Chap. 15 usingcomputational fluid dynamics (CFD).

V#

� Cwd, tri8

15 tanau

2b22gH 5 2

Cwd, rec � 0.598 � 0.0897H

Pw

for H

Pw

� 2

V#

rec � Cwd, rec2

3b22gH 3 2

Cwd, broad �0.65

21 � H Pw

V#

� Cwd, broadb2ga23b 3 2aH �

V 21

2gb 3 2

V#

� Cdba22gy1

REFERENCES AND SUGGESTED READING

1. P. Ackers et al. Weirs and Flumes for Flow Measurement.New York: Wiley, 1978.

2. B. A. Bakhmeteff. Hydraulics of Open Channels. NewYork: McGraw-Hill, 1932.

3. M. H. Chaudhry. Open Channel Flow. Upper SaddleRiver, NJ: Prentice Hall, 1993.

4. V. T. Chow. Open Channel Hydraulics. New York:McGraw-Hill, 1959.

5. C. T. Crowe, J. A. Roberson, and D. F. Elger. EngineeringFluid Mechanics, 7th ed. New York: Wiley, 2001.

6. R. H. French. Open Channel Hydraulics. New York:McGraw-Hill, 1985.

cen72367_ch13.qxd 11/6/04 12:39 PM Page 724

7. F. M. Henderson. Open Channel Flow. New York:Macmillan, 1966.

8. C. C. Mei. The Applied Dynamics of Ocean SurfaceWaves. New York: Wiley, 1983.

9. B. R. Munson, D. F. Young, and T. H. Okiishi.Fundamentals of Fluid Mechanics, 4th ed. New York:Wiley, 2002.

10. M. C. Potter and D. C. Wiggert. Mechanics of Fluids, 2nded. Upper Saddle River, NJ: Prentice Hall, 1997.

11. F. M. White. Fluid Mechanics, 5th ed. New York:McGraw-Hill, 2003.

12. U. S. Bureau of Reclamation. “Research Studies onStilling Basins, Energy Dissipaters, and AssociatedAppurtenances,” Hydraulic Lab Report Hyd.-399,June 1, 1955.

725CHAPTER 13

* Problems designated by a “C” are concept questions, andstudents are encouraged to answer them all. Problems designatedby an “E” are in English units, and the SI users can ignore them.Problems with the icon are solved using EES, and completesolutions together with parametric studies are included on theenclosed DVD. Problems with the icon are comprehensive innature and are intended to be solved with a computer, preferablyusing the EES software that accompanies this text.

PROBLEMS*

Classification, Froude Number, and Wave Speed

13–1C How does open-channel flow differ from internalflow?

13–2C What is the driving force for flow in an open chan-nel? How is the flow rate in an open channel established?

13–3C How does the pressure change along the free surfacein an open-channel flow?

13–4C Consider steady fully developed flow in an openchannel of rectangular cross section with a constant slope of5° for the bottom surface. Will the slope of the free surfacealso be 5°? Explain.

13–5C How does uniform flow differ from nonuniformflow in open channels? In what kind of channels is uniformflow observed?

13–6C What is normal depth? Explain how it is establishedin open channels.

13–7C What causes the flow in an open channel to be var-ied (or nonuniform)? How does rapidly varied flow differfrom gradually varied flow?

13–8C In open channels, how is hydraulic radius defined?Knowing the hydraulic radius, how can the hydraulic diame-ter of the channel be determined?

13–9C Given the average flow velocity and the flow depth,explain how you would determine if the flow in open chan-nels is tranquil, critical, or rapid.

13–10C What is the Froude number? How is it defined?What is its physical significance?

13–11C What is critical depth in open-channel flow? For agiven average flow velocity, how is it determined?

13–12C The flow in an open channel is observed to haveundergone a hydraulic jump. Is the flow upstream from thejump necessarily supercritical? Is the flow downstream fromthe jump necessarily subcritical?

13–13 Consider the flow of water in a wide channel. Deter-mine the speed of a small disturbance in the flow if the flowdepth is (a) 10 cm and (b) 80 cm. What would your answerbe if the fluid were oil?

13–14 Water at 20°C is flowing uniformly in a wide rectan-gular channel at an average velocity of 2 m/s. If the waterdepth is 0.2 m, determine (a) whether the flow is laminaror turbulent and (b) whether the flow is subcritical orsupercritical.

13–15 Water at 20°C flows in a partially full 2-m-diametercircular channel at an average velocity of 2 m/s. If the maxi-mum water depth is 0.5 m, determine the hydraulic radius,the Reynolds number, and the flow regime.

0.5 m

R � 1 m

FIGURE P13–15

13–16 Water at 15°C is flowing uniformly in a 2-m-widerectangular channel at an average velocity of 4 m/s. If thewater depth is 8 cm, determine whether the flow is subcriticalor supercritical. Answer: supercritical

13–17 After heavy rain, water flows on a concrete surfaceat an average velocity of 1.3 m/s. If the water depth is 2 cm,determine whether the flow is subcritical or supercritical.

cen72367_ch13.qxd 11/6/04 12:39 PM Page 725

13–18E Water at 70°F is flowing uniformly in a wide rec-tangular channel at an average velocity of 6 ft/s. If the waterdepth is 0.5 ft, determine (a) whether the flow is laminar orturbulent and (b) whether the flow is subcritical or super-critical.

13–19 Water at 10°C flows in a 3-m-diameter circularchannel half-full at an average velocity of 2.5 m/s. Determinethe hydraulic radius, the Reynolds number, and the flowregime (laminar or turbulent).

13–20 A single wave is initiated in a sea by a strong joltduring an earthquake. Taking the average water depth to be2 km and the density of seawater to be 1.030 kg/m3, deter-mine the speed of propagation of this wave.

Specific Energy and the Energy Equation

13–21C How is the specific energy of a fluid flowing in anopen channel defined in terms of heads?

13–22C Consider steady flow of water through two identi-cal open rectangular channels at identical flow rates. If theflow in one channel is subcritical and in the other supercriti-cal, can the specific energies of water in these two channelsbe identical? Explain.

13–23C For a given flow rate through an open channel, thevariation of specific energy with flow depth is studied. Oneperson claims that the specific energy of the fluid will beminimum when the flow is critical, but another person claimsthat the specific energy will be minimum when the flow issubcritical. What is your opinion?

13–24C Consider steady supercritical flow of water throughan open rectangular channel at a constant flow rate. Someoneclaims that the larger is the flow depth, the larger the specificenergy of water. Do you agree? Explain.

13–25C During steady and uniform flow through an openchannel of rectangular cross section, a person claims that thespecific energy of the fluid remains constant. A second per-son claims that the specific energy decreases along the flowbecause of the frictional effects and thus head loss. Withwhich person do you agree? Explain.

13–26C How is the friction slope defined? Under what con-ditions is it equal to the bottom slope of an open channel?

13–27C Consider steady flow of a liquid through a widerectangular channel. It is claimed that the energy line of flowis parallel to the channel bottom when the frictional lossesare negligible. Do you agree?

13–28C Consider steady one-dimensional flow through awide rectangular channel. Someone claims that the totalmechanical energy of the fluid at the free surface of a crosssection is equal to that of the fluid at the channel bottom ofthe same cross section. Do you agree? Explain.

13–29C How is the total mechanical energy of a fluid dur-ing steady one-dimensional flow through a wide rectangular

726FLUID MECHANICS

channel expressed in terms of heads? How is it related to thespecific energy of the fluid?

13–30C Express the one-dimensional energy equation foropen-channel flow between an upstream section 1 and down-stream section 2, and explain how the head loss can be deter-mined.

13–31 Water flows steadily in a 0.8-m-wide rectangularchannel at a rate of 0.7 m3/s. If the flow depth is 0.25 m,determine the flow velocity and if the flow is subcritical orsupercritical. Also determine the alternate flow depth if thecharacter of flow were to change.

13–32 Water at 15°C flows at a depth of 0.4 m with anaverage velocity of 6 m/s in a rectangular channel. Determinethe specific energy of water and whether the flow is subcriti-cal or supercritical.

13–33 Water at 15°C flows at a depth of 0.4 m withan average velocity of 6 m/s in a rectangular

channel. Determine (a) the critical depth, (b) the alternatedepth, and (c) the minimum specific energy.

13–34 Water at 10°C flows in a 6-m-wide rectangular chan-nel at a depth of 0.55 m and a flow rate of 12 m3/s. Deter-mine (a) the critical depth, (b) whether the flow is subcriticalor supercritical, and (c) the alternate depth. Answers:(a) 0.742 m, (b) supercritical, (c) 1.03 m

13–35E Water at 65°F flows at a depth of 0.8 ft with anaverage velocity of 14 ft/s in a wide rectangular channel.Determine (a) the Froude number, (b) the critical depth, and(c) whether the flow is subcritical or supercritical. Whatwould your response be if the flow depth were 0.2 ft?

13–36E Repeat Prob. 13–35E for an average velocity of10 ft/s.

13–37 Water flows through a 4-m-wide rectangular channelwith an average velocity of 5 m/s. If the flow is critical,determine the flow rate of water. Answer: 51.0 m3/s

13–38 Water flows half-full through a 50-cm-diameter steelchannel at an average velocity of 2.8 m/s. Determine the vol-ume flow rate and whether the flow is subcritical or super-critical.

13–39 Water flows half-full through a hexagon channel ofbottom width 2 m at a rate of 45 m3/s. Determine (a) theaverage velocity and (b) whether the flow is subcritical andsupercritical.

13–40 Repeat Prob. 13–39 for a flow rate of 30 m3/s.

Uniform Flow and Best Hydraulic Cross Sections13–41C When is the flow in an open channel said to beuniform? Under what conditions will the flow in an openchannel remain uniform?

13–42C Consider uniform flow through a wide rectangularchannel. If the bottom slope is increased, the flow depth will(a) increase, (b) decrease, or (c) remain constant.

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13–43C During uniform flow in an open channel, someoneclaims that the head loss can be determined by simply multi-plying the bottom slope by the channel length. Can it be thissimple? Explain.

13–44C During uniform flow in open channels, the flowvelocity and the flow rate can be determined from theManning equations expressed as V0 � (a/n)Rh

2/3S01/2 and V

.

� (a/n)AcRh2/3S0

1/2. What is the value and dimension of the con-stant a in these equations in SI units? Also, explain how theManning coefficient n can be determined when the frictionfactor f is known.

13–45C Show that for uniform critical flow, the general

critical slope relation reduces to for

film flow with b �� yc.

13–46C Which is a better hydraulic cross section for anopen channel: one with a small or a large hydraulic radius?

13–47C Which is the best hydraulic cross section for anopen channel: (a) circular, (b) rectangular, (c) trapezoidal, or(d) triangular?

13–48C The best hydraulic cross section for a rectangularopen channel is one whose fluid height is (a) half, (b) twice,(c) equal to, or (d) one-third the channel width.

13–49C The best hydraulic cross section for a trapezoidalchannel of base width b is one for which the length of the sideedge of the flow section is (a) b, (b) b/2, (c) 2b, or (d) .

13–50C Consider uniform flow through an open channellined with bricks with a Manning coefficient of n � 0.015. Ifthe Manning coefficient doubles (n � 0.030) as a result ofsome algae growth on surfaces while the flow cross sectionremains constant, the flow rate will (a) double, (b) decreaseby a factor of , (c) remain unchanged, (d) decrease byhalf, or (e) decrease by a factor of 21/3.

13–51 Water is flowing uniformly in a finished-concretechannel of trapezoidal cross section with a bottom width of0.6 m, trapezoid angle of 50°, and a bottom angle of 0.4°. Ifthe flow depth is measured to be 0.45 m, determine the flowrate of water through the channel.

12

13b

Sc �gn2

a2y 1 3c

Sc �gn2yc

a2R4 3h

13–53E A 6-ft-diameter semicircular channel made ofunfinished concrete is to transport water to a distance of 1 miuniformly. If the flow rate is to reach 150 ft3/s when thechannel is full, determine the minimum elevation differenceacross the channel.

13–54 A trapezoidal channel with a bottom width of 5 m,free surface width of 10 m, and flow depth of 2.2 m dis-charges water at a rate of 120 m3/s. If the surfaces of thechannel are lined with asphalt (n � 0.016), determine the ele-vation drop of the channel per km. Answer: 8.52 m

727CHAPTER 13

y � 0.45 m

u � 50°

b � 0.6 m

FIGURE P13–51

2.2 m

10 m

5 m

FIGURE P13–54

13–55 Reconsider Prob. 13–54. If the maximum flowheight the channel can accommodate is 2.4 m, determine themaximum flow rate through the channel.

13–56 Consider water flow through two identical channelswith square flow sections of 3 m � 3 m. Now the two chan-nels are combined, forming a 6-m-wide channel. The flowrate is adjusted so that the flow depth remains constant at 3m. Determine the percent increase in flow rate as a result ofcombining the channels.

3 m

3 m

3 m

3 m

FIGURE P13–56

13–57 A trapezoidal channel made of unfinished concretehas a bottom slope of 1°, base width of 5 m, and a side sur-face slope of 1:1, as shown in Fig. P13–57. For a flow rate of25 m3/s, determine the normal depth h.

13–52 Water flows uniformly half-full in a 2-m-diametercircular channel that is laid on a grade of 1.5 m/km. If thechannel is made of finished concrete, determine the flow rateof the water.

45° 45°

5 m

h

FIGURE P13–57

cen72367_ch13.qxd 11/6/04 12:39 PM Page 727

13–58 Repeat Prob. 13–57 for a weedy excavated earthchannel with n � 0.030.

13–59 A cast iron V-shaped water channel shown in Fig.P13–59 has a bottom slope of 0.5°. For a flow depth of 1 mat the center, determine the discharge rate in uniform flow.Answer: 3.59 m3/s

728FLUID MECHANICS

For a flow depth of 0.25 m at the center, determine the flowrate of water through the channel.

45° 45°1 m

FIGURE P13–59

13–60E Water is to be transported in a cast iron rectangularchannel with a bottom width of 6 ft at a rate of 70 ft3/s.The terrain is such that the channel bottom drops 1.5 ft per1000 ft length. Determine the minimum height of the channelunder uniform-flow conditions.

y

b � 6 ft

V � 70 ft3/s.

FIGURE P13–60E

13–61 Water flows in a channel whose bottom slope is0.002 and whose cross section is as shown in Fig. P13–61.The dimensions and the Manning coefficients for the surfacesof different subsections are also given on the figure. Deter-mine the flow rate through the channel and the effectiveManning coefficient for the channel.

6 m

1.5 m

2 m

2 m

10 m

Light brushn2 � 0.050

Concretechannel

n1 � 0.014

1 2

FIGURE P13–61

13–62 Consider a 1-m-internal-diameter water channel madeof finished concrete (n � 0.012). The channel slope is 0.002.

y � 0.25 m

R � 0.5 m

FIGURE P13–62

13–63 Reconsider Prob. 13–62. By varying the flowdepth-to-radius ratio y/R from 0.1 to 1.9 while

holding the flow area constant and evaluating the flow rate,show that the best cross section for flow through a circularchannel occurs when the channel is half-full. Tabulate andplot your results.

13–64 A clean-earth trapezoidal channel with a bottomwidth of 1.5 m and a side surface slope of 1:1 is to drainwater uniformly at a rate of 8 m3/s to a distance of 1 km. Ifthe flow depth is not to exceed 1 m, determine the requiredelevation drop.

13–65 A water draining system with a constant slope of0.0015 is to be built of three circular channels made of fin-ished concrete. Two of the channels have a diameter of 1.2 mand drain into the third channel. If all channels are to runhalf-full and the losses at the junction are negligible, deter-mine the diameter of the third channel. Answer: 1.56 m

13–66 Water is to be transported in an open channel whosesurfaces are asphalt lined at a rate of 4 m3/s in uniform flow.The bottom slope is 0.0015. Determine the dimensions of thebest cross section if the shape of the channel is (a) circularof diameter D, (b) rectangular of bottom width b, and (c)trapezoidal of bottom width b.

13–67E A rectangular channel with a bottom slope of0.0005 is to be built to transport water at a rate of 800 ft3/s.Determine the best dimensions of the channel if it is to bemade of (a) unfinished concrete and (b) finished concrete.

13–68 Consider uniform flow in an asphalt-lined rec-tangular channel with a flow area of 2 m2 and a

bottom slope of 0.0003. By varying the depth-to-width ratioy/b from 0.1 to 2.0, calculate and plot the flow rate, and con-firm that the best flow cross section occurs when the flowdepth-to-width ratio is 0.5.

Gradually and Rapidly Varied Flows and Hydraulic Jump

13–69C How does nonuniform or varied flow differ fromuniform flow?

13–70C How does gradually varied flow (GVF) differ fromrapidly varied flow (RVF)?

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13–71C Someone claims that frictional losses associatedwith wall shear on surfaces can be neglected in the analysisof rapidly varied flow, but should be considered in the analy-sis of gradually varied flow. Do you agree with this claim?Justify your answer.

13–72C Consider steady flow of water in a horizontal chan-nel of rectangular cross section. If the flow is subcritical, theflow depth will (a) increase, (b) remain constant, or (c)decrease in the flow direction.

13–73C Consider steady flow of water in a downward-sloped channel of rectangular cross section. If the flow issubcritical and the flow depth is greater than the normaldepth (y � yn), the flow depth will (a) increase, (b) remainconstant, or (c) decrease in the flow direction.

13–74C Consider steady flow of water in a horizontal chan-nel of rectangular cross section. If the flow is supercritical,the flow depth will (a) increase, (b) remain constant, or (c)decrease in the flow direction.

13–75C Consider steady flow of water in a downward-sloped channel of rectangular cross section. If the flow issubcritical and the flow depth is less than the normal depth(y yn), the flow depth will (a) increase, (b) remain con-stant, or (c) decrease in the flow direction.

13–76C Consider steady flow of water in an upward-slopedchannel of rectangular cross section. If the flow is supercriti-cal, the flow depth will (a) increase, (b) remain constant, or(c) decrease in the flow direction.

13–77C Is it possible for subcritical flow to undergo ahydraulic jump? Explain.

13–78C Why is the hydraulic jump sometimes used to dis-sipate mechanical energy? How is the energy dissipation ratiofor a hydraulic jump defined?

13–79 Water flows uniformly in a rectangular channel withfinished-concrete surfaces. The channel width is 3 m, theflow depth is 1.2 m, and the bottom slope is 0.002. Deter-mine if the channel should be classified as mild, critical, orsteep for this flow.

13–80 Consider uniform water flow in a wide brick channelof slope 0.4°. Determine the range of flow depth for whichthe channel is classified as being steep.

13–81E Consider the flow of water through a 12-ft-wideunfinished-concrete rectangular channel with a bottom slopeof 0.5°. If the flow rate is 300 ft3/s, determine if the slope ofthis channel is mild, critical, or steep. Also, for a flow depthof 3 ft, classify the surface profile while the flow develops.

13–82 Water is flowing in a 90° V-shaped cast iron channelwith a bottom slope of 0.002 at a rate of 3 m3/s. Determine ifthe slope of this channel should be classified as mild, critical,or steep for this flow. Answer: mild

13–83 Water discharging into an 8-m-wide rectangu-lar horizontal channel from a sluice gate is

observed to have undergone a hydraulic jump. The flowdepth and velocity before the jump are 1.2 m and 9 m/s,respectively. Determine (a) the flow depth and the Froudenumber after the jump, (b) the head loss and the dissipationratio, and (c) the mechanical energy dissipated by thehydraulic jump.

729CHAPTER 13

y � 1.2 m

b � 3 m

FIGURE P13–79

V1 � 9 m/s V2

y1 � 1.2 my2

(1) (2)

FIGURE P13–83

13–84 Water flowing in a wide horizontal channel at a flowdepth of 35 cm and an average velocity of 12 m/s undergoesa hydraulic jump. Determine the head loss associated withhydraulic jump.

13–85 During a hydraulic jump in a wide channel, the flowdepth increases from 0.6 to 3 m. Determine the velocities andFroude numbers before and after the jump, and the energydissipation ratio.

13–86 Consider the flow of water in a 10-m-wide channelat a rate of 70 m3/s and a flow depth of 0.50 m. The waternow undergoes a hydraulic jump, and the flow depth after thejump is measured to be 4 m. Determine the mechanicalpower wasted during this jump. Answer: 4.35 MW

13–87 The flow depth and velocity of water after undergo-ing a hydraulic jump are measured to be 2 m and 3 m/s,respectively. Determine the flow depth and velocity beforethe jump, and the fraction of mechanical energy dissipated.

13–88E Water flowing in a wide channel at a depth of 2 ftand a velocity of 40 ft/s undergoes a hydraulic jump. Deter-mine the flow depth, velocity, and Froude number after thejump, and the head loss associated with the jump.

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Flow Control and Measurement in Channels

13–89C Draw a flow depth-specific energy diagram forflow through underwater gates, and indicate the flow throughthe gate for cases of (a) frictionless gate, (b) sluice gate withfree outflow, and (c) sluice gate with drowned outflow(including the hydraulic jump back to subcritical flow).

13–90C For sluice gates, how is the discharge coefficientCd defined? What are typical values of Cd for sluice gateswith free outflow? What is the value of Cd for the idealizedfrictionless flow through the gate?

13–91C What is the basic principle of operation of a broad-crested weir used to measure flow rate through an open chan-nel?

13–92C Consider steady frictionless flow over a bump ofheight �z in a horizontal channel of constant width b. Willthe flow depth y increase, decrease, or remain constant as thefluid flows over the bump? Assume the flow to be subcritical.

13–93C Consider the flow of a liquid over a bump duringsubcritical flow in an open channel. The specific energy andthe flow depth decrease over the bump as the bump height isincreased. What will the character of flow be when the spe-cific energy reaches its minimum value? Will the flow becomesupercritical if the bump height is increased even further?

13–94C What is a sharp-crested weir? On what basis arethe sharp-crested weirs classified?

13–95 Water is released from a 14-m-deep reservoir into a5-m-wide open channel through a sluice gate with a 1-m-highopening at the channel bottom. If the flow depth downstreamfrom the gate is measured to be 3 m, determine the rate ofdischarge through the gate.

730FLUID MECHANICS

13–97 Consider the uniform flow of water in a wide chan-nel with a velocity of 8 m/s and flow depth of 0.8 m. Nowwater flows over a 30-cm-high bump. Determine the change(increase or decrease) in the water surface level over thebump. Also determine if the flow over the bump is sub- orsupercritical.

13–98 The flow rate of water in a 4-m-wide horizontalchannel is being measured using a 0.75-m-high sharp-crestedrectangular weir that spans across the channel. If the waterdepth upstream is 2.2 m, determine the flow rate of water.Answer: 15.9 m3/s

a � 1 m

Sluice gate

y2 � 3 m

y1 � 14 m

FIGURE P13–95

13–96 Water flowing in a wide channel encounters a 22-cm-high bump at the bottom of the channel. If the flow depthis 1.2 m and the velocity is 2.5 m/s before the bump, deter-mine if the flow is chocked over the bump, and discuss.

V1 � 2.5 m/s

y1 � 1.2 m y2�zb � 0.22 m

Depressionover the bump

Bump

FIGURE P13–96

V1

Sharp-crestedrectangular weir

y1 � 2.2 m

Pw � 0.75 m

FIGURE P13–98

13–99 Repeat Prob. 13–98 for the case of a weir heightof 1 m.

13–100 Water flows over a 2-m-high sharp-crested rectan-gular weir. The flow depth upstream of the weir is 3 m, andwater is discharged from the weir into an unfinished-concretechannel of equal width where uniform-flow conditions areestablished. If no hydraulic jump is to occur in the down-stream flow, determine the maximum slope of the down-stream channel.

13–101E A full-width sharp-crested weir is to be used tomeasure the flow rate of water in a 10-ft-wide rectangularchannel. The maximum flow rate through the channel is150 ft3/s, and the flow depth upstream from the weir is not toexceed 5 ft. Determine the appropriate height of the weir.

13–102 Consider uniform water flow in a wide rectangularchannel with a depth of 2 m made of unfinished concrete laidon a slope of 0.0022. Determine the flow rate of water per m

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width of channel. Now water flows over a 15-cm-high bump.If the water surface over the bump remains flat (no rise ordrop), determine the change in discharge rate of water permeter width of the channel. Hint: Investigate if a flat surfaceover the bump is physically possible.

13–103 Consider uniform water flow in a wide channelmade of unfinished concrete laid on a slope of 0.0022. Nowwater flows over a 15-cm-high bump. If the flow over thebump is exactly critical (Fr � 1), determine the flow rate andthe flow depth over the bump per m width. Answers:20.3 m3/s, 3.48 m

13–108E Consider water flow through a wide channel at aflow depth of 8 ft. Now water flows through a sluice gatewith a 1-ft-high opening, and the freely discharged outflowsubsequently undergoes a hydraulic jump. Disregarding anylosses associated with the sluice gate itself, determine theflow depth and velocities before and after the jump, and thefraction of mechanical energy dissipated during the jump.

13–109 The flow rate of water flowing in a 3-m-wide chan-nel is to be measured with a sharp-crested triangular weir0.5 m above the channel bottom with a notch angle of 60°. Ifthe flow depth upstream from the weir is 1.5 m, determinethe flow rate of water through the channel. Take the weir dis-charge coefficient to be 0.60. Answer: 0.818 m3/s

731CHAPTER 13

y1 y2

�zb � 15 cm

Bump

Slope � 0.0022

FIGURE P13–103

13–104 The flow rate of water through a 5-m-wide (into thepaper) channel is controlled by a sluice gate. If the flowdepths are measured to be 1.1 and 0.45 m upstream anddownstream from the gates, respectively, determine the flowrate and the Froude number downstream from the gate.

Sluice gate

y2 � 0.45 m

y1 � 1.1 m

FIGURE P13–104

13–105E Water flows through a sluice gate with a 1.1-ft-high opening and is discharged with free outflow. If theupstream flow depth is 5 ft, determine the flow rate per unitwidth and the Froude number downstream the gate.

13–106E Repeat Prob. 13–105E for the case of a drownedgate with a downstream flow depth of 3.3 ft.

13–107 Water is to be discharged from a 6-m-deep lakeinto a channel through a sluice gate with a 5-m wide and 0.6-m-high opening at the bottom. If the flow depth downstreamfrom the gate is measured to be 3 m, determine the rate ofdischarge through the gate.

0.5 m

3 m

Weirplate

1 m 60°

Upstreamfree surface

FIGURE P13–109

13–110 Repeat Prob. 13–109 for an upstream flow depth of1.2 m.

13–111 A sharp-crested triangular weir with a notch angleof 100° is used to measure the discharge rate of water from alarge lake into a spillway. If a weir with half the notch angle(u � 50°) is used instead, determine the percent reduction inthe flow rate. Assume the water depth in the lake and theweir discharge coefficient remain unchanged.

13–112 A 1-m-high broad-crested weir is used to measurethe flow rate of water in a 5-m-wide rectangular channel. Theflow depth well upstream from the weir is 1.6 m. Determinethe flow rate through the channel and the minimum flowdepth above the weir.

Discharge

1.6 m

1 m Broad-crested weir

FIGURE P13–112

13–113 Repeat Prob. 13–112 for an upstream flow depth of2.2 m.

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13–114 Consider water flow over a 0.80-m-high suffi-ciently long broad-crested weir. If the minimum flow depthabove the weir is measured to be 0.50 m, determine the flowrate per meter width of channel and the flow depth upstreamof the weir.

Review Problems

13–115 A trapezoidal channel with a bottom width of 4 mand a side slope of 45° discharges water at a rate of 18 m3/s.If the flow depth is 0.6 m, determine if the flow is subcriticalor supercritical.

13–116 A rectangular channel with a bottom width of 2 mdischarges water at a rate of 8 m3/s. Determine the flow depthbelow which the flow is supercritical.

13–117 Water flows in a canal at an average velocity of4 m/s. Determine if the flow is subcritical or supercritical forflow depths of (a) 0.2 m, (b) 2 m, and (c) 1.63 m.

13–118 Water flows through a 1.5-m-wide rectangularchannel with a Manning coefficient of n � 0.012. If thewater is 0.9 m deep and the bottom slope of the channel is0.6°, determine the rate of discharge of the channel in uni-form flow.

13–119 A 5-m-wide rectangular channel lined withfinished concrete is to be designed to trans-

port water to a distance of 1 km at a rate of 12 m3/s. UsingEES (or other) software, investigate the effect of bottomslope on flow depth (and thus on the required channelheight). Let the bottom angle vary from 0.5 to 10° in incre-ments of 0.5°. Tabulate and plot the flow depth against thebottom angle, and discuss the results.

13–120 Repeat Prob. 13–119 for a trapezoidal channelthat has a base width of 5 m and a side sur-

face angle of 45°.

13–121 A trapezoidal channel with brick lining has a bot-tom slope of 0.001 and a base width of 4 m, and the side sur-faces are angled 30° from the horizontal, as shown in Fig.P13–121. If the normal depth is measured to be 2 m, estimatethe flow rate of water through the channel. Answer: 36.4 m3/s

732FLUID MECHANICS

13–123 Consider water flow through a V-shaped channel.Determine the angle u the channel makes from the horizontalfor which the flow is most efficient.

30° 30°

4 m

2 m

FIGURE P13–121

13–122 A 2-m-internal-diameter circular steel storm drain(n � 0.012) is to discharge water uniformly at a rate of12 m3/s to a distance of 1 km. If the maximum depth is to be1.5 m, determine the required elevation drop.

1.5 mR � 1 m

FIGURE P13–122

u uy

FIGURE P13–123

13–124E A rectangular channel with unfinished concretesurfaces is to be built to discharge water uniformly at a rateof 200 ft3/s. For the case of best cross section, determine thebottom width of the channel if the available vertical drop is(a) 8 and (b) 10 ft per mile. Answers: (a) 7.86 and (b) 7.54 ftper mile

13–125E Repeat Prob. 13–124E for the case of a trape-zoidal channel of best cross section.

13–126 Water flows in a channel whose bottom slope is0.5° and whose cross section is as shown in Fig. P13–126.The dimensions and the Manning coefficients for the surfacesof different subsections are also given on the figure. Deter-mine the flow rate through the channel and the effectiveManning coefficient for the channel.

6 m

1 m

3 m

1 m

10 m

Heavy brushn2 � 0.075

Clean earthchannel

n1 � 0.022

FIGURE P13–126

13–127 Consider two identical channels, one rectangular ofbottom width b and one circular of diameter D, with identicalflow rates, bottom slopes, and surface linings. If the flowheight in the rectangular channel is also b and the circularchannel is flowing half-full, determine the relation between band D.

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13–128 Consider the flow of water through a parabolicnotch shown in Fig. P13–128. Develop a relation for the flowrate, and calculate its numerical value for the ideal case inwhich the flow velocity is given by Toricelli’s equation

. Answer: 0.246 m3/sV � 12g(H � y)

13–133 Consider water flow through a wide rectangularchannel undergoing a hydraulic jump. Show that the ratio ofthe Froude numbers before and after the jump can beexpressed in terms of flow depths y1 and y2 before and afterthe jump, respectively, as .

13–134 A sluice gate with free outflow is used to controlthe discharge rate of water through a channel. Determine theflow rate per unit width when the gate is raised to yield a gapof 30 cm and the upstream flow depth is measured to be 1.8m. Also determine the flow depth and the velocity down-stream.

13–135 Water flowing in a wide channel at a flow depth of45 cm and an average velocity of 8 m/s undergoes a hydraulicjump. Determine the fraction of the mechanical energy of thefluid dissipated during this jump. Answer: 36.8 percent

13–136 Water flowing through a sluice gate undergoes ahydraulic jump, as shown in Fig. P13–136. The velocity ofthe water is 1.25 m/s before reaching the gate and 4 m/s afterthe jump. Determine the flow rate of water through the gateper meter of width, the flow depths y1 and y2, and the energydissipation ratio of the jump.

Fr1/Fr2 � 1(y2 y1)3

733CHAPTER 13

y

x

H � 0.5 m

b � 0.4 m

y � cx2

FIGURE P13–128

13–129 In practice, the V-notch is commonly used tomeasure flow rate in open channels. Using the

idealized Toricelli’s equation for velocity,develop a relation for the flow rate through the V-notch interms of the angle u. Also, show the variation of the flow ratewith u by evaluating the flow rate for u � 25, 40, 60, and75°, and plotting the results.

V � 12g(H � y)

u

H � 25 cm

y

FIGURE P13–129

13–130 Water flows uniformly half-full in a 1.2-m-diametercircular channel laid with a slope of 0.004. If the flow rate ofwater is measured to be 1.25 m3/s, determine the Manningcoefficient of the channel and the Froude number.

13–131 Water flowing in a wide horizontal channelapproaches a 20-cm-high bump with a velocity of 1.25 m/sand a flow depth of 1.8 m. Determine the velocity, flowdepth, and Froude number over the bump.

V2

y2

V1 � 1.25 m/s 20 cm

y1 � 1.8 m

FIGURE P13–131

13–132 Reconsider Prob. 13–131. Determine the bumpheight for which the flow over the bump is critical (Fr � 1).

y3 � 3 m

y1

y2

V1 � 1.25 m/s

V3 � 4 m/s

Sluice gate

FIGURE P13–136

13–137 Repeat Prob. 13–136 for a velocity of 2 m/s afterthe hydraulic jump.

13–138 Water is discharged from a 5-m-deep lake into afinished concrete channel with a bottom slope of 0.004through a sluice gate with a 0.5-m-high opening at the bot-tom. Shortly after supercritical uniform-flow conditions areestablished, the water undergoes a hydraulic jump. Determinethe flow depth, velocity, and Froude number after the jump.Disregard the bottom slope when analyzing the hydraulicjump.

13–139 Water is discharged from a dam into a wide spill-way to avoid overflow and to reduce the risk of flooding. Alarge fraction of the destructive power of water is dissipatedby a hydraulic jump during which the water depth rises from

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0.50 to 4 m. Determine the velocities of water before andafter the jump, and the mechanical power dissipated permeter width of the spillway.

13–140 The flow rate of water in a 6-m-wide rectangularchannel is to be measured using a 1.1-m-high sharp-crestedrectangular weir that spans across the channel. If the headabove the weir crest is 0.60 m upstream from the weir, deter-mine the flow rate of water.

13–141E Consider two identical 12-ft-wide rectangularchannels each equipped with a 2-ft-high full-width weir,except that the weir is sharp-crested in one channel andbroad-crested in the other. For a flow depth of 5 ft in bothchannels, determine the flow rate through each channel.Answers: 244 ft3/s, 79.2 ft3/s

734FLUID MECHANICS

Design and Essay Problems

13–142 Using catalogs or websites, obtain informationfrom three different weir manufacturers. Compare the differ-ent weir designs, and discuss the advantages and disadvan-tages of each design. Indicate the applications for which eachdesign is best suited.

13–143 Consider water flow in the range of 10 to 15 m3/sthrough a horizontal section of a 5-m-wide rectangular chan-nel. A rectangular or triangular thin-plate weir is to beinstalled to measure the flow rate. If the water depth is toremain under 2 m at all times, specify the type and dimen-sions of an appropriate weir. What would your response be ifthe flow range were 0 to 15 m3/s?

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Open Channel Flow Fluid Mechanics Fundamentals and Applications-2th Edition

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