CWR4202 Open Channel Flow Chap 5

Open channel flow

5.1 Flow with a free surface

Open channel flow is characterized by the existence of a free surface (thewater surface). In contrast to pipe flow, this constitutes a boundary atwhich the pressure is atmospheric, and across which the shear forces arenegligible. The longitudinal profile of the free surface defines the hydraulicgradient and determines the cross-sectional area of flow, as is shown inFigure 5.1. It also necessitates the introduction of an extra variable - thestage (see Fig. 5.1) - to define the position of the free surface at any pointin the channel.

In consequence, problems in open channel flow are more complex thanthose of pipe flow, and the solutions are more varied, making the study ofsuch problems both interesting and challenging. In this chapter the basicconcepts are introduced and a variety ofcommon engineering applicationsare discussed.

5.2 Flow classification

The fundamental types of flow have already been discussed in Chapter 2.However, it is appropriate here to expand those descriptions as applied toopen channels. Recalling that flow may be steady or unsteady and uniformor non-uniform, the major classifications applied to open channels are asfollows:

Steady uniform flow, inwhich the depth is constant, both with time anddistance. This constitutes the fundamental type of flow in an open channelin which the gravity forces are in equilibrium with the resistance forces. Itis considered in section 5.6.

Steady non-uniform flow, in which the depth varies with distance, but notwith time. The flow may be either (a) gradually varied or (b) rapidly varied.Type (a) requires the joint application of energy and frictional resistance

NATURAL AND ARTIFICIAL CHANNELS 123

(a) Pipe flow

hydraulic gradientenergy gradient

Cross section A-Aarea defined bypipe boundary

energy gradient_ A / watersurface

/ , and hydraulic--/^gradient stage (h)

Cross section A-Aarea defined by

water surface level(b) Channel flow and channel shape

Figure 5.1 Comparison between pipe flow and open channel flow.

equations, and is considered in section 5.10. Type (b) requires the application of energy and momentum principles, and is considered in sections 5.7and 5.8.

Unsteady flow, in which the depth varies with both time and distance(unsteady uniform flow is very rare). This is the most complex flow type,requiring the solution of energy, momentum and friction equations throughtime. It is considered in section 5.11.

The various flow types are all shown in Figure 5.2.

5.3 Natural and artificial channels and their properties

Artificial channels comprise all man-made channels, including irrigation andnavigation canals, spillway channels, sewers, culverts and drainage ditches.They are normally of regular cross-sectional shape and bed slope, and assuch are termed prismatic channels. Their construction materials are varied,but commonly used materials include concrete, steel and earth. The surfaceroughness characteristics of these materials are normally well defined withinengineering tolerances. In consequence, the application of hydraulic theoriesto flow in artificial channels will normally yield reasonably accurate results.

uJ

Mini

124 OPEN CHANNEL FLOW

constant depthvarying depth

Steady uniform flowGVF GVF

RVF GVF

Unsteady uniform flow

GVF RVF GVF

Steady varied flowRVF rapidly varied flowGVF gradually varied flow

flood wave (GVF) Unsteady flow bore (RVF)Figure 5.2 Types of flow.

In contrast, natural channels are normally very irregular in shape, andtheir materials are diverse. The surface roughness of natural channelschanges with time, distance and water surface elevation. Therefore, it ismore difficult to apply hydraulic theory to natural channels and obtain satisfactory results. Many applications involve man-made alterations to naturalchannels (e.g. rivercontrol structures and flood alleviation measures). Suchapplications require an understanding not only of hydraulic theory, butalso of the associated disciplines of sediment transport, hydrology and rivermorphology (refer to Chapters 9, 10 and 15).

Various geometric properties of natural and artificial channels need to bedetermined for hydraulic purposes. In the case of artificial channels, thesemay all be expressed algebraically in terms of the depth (y), as is shownin Table 5.1. This is not possible for natural channels, so graphs or tablesrelating them to stage (h) must be used.

NATURAL AND ARTIFICIAL CHANNELS 125

Table 5.1 Geometric properties of some common prismaticchannels.

area, A

wetted perimeter, P

top width, B

hydraulic radius, R

hydraulic mean depth, Dn

r B 1>=>- V

h t

Rectangle Trapezoid

by

b + 2y

b

byb + 2y

y

(b+ xy)y

>+ 2y^/l+x2

b + 2xy

(b+ xy)y;+ 2yvT+5?

(b+ xy)yb + 2xy

Circle

4)DD

(1/2*) A1 / d> sin \8\s',sin

The commonly used geometric properties are shown in Figure 5.3 anddefined as follows:

Depth (y) - the vertical distance of the lowest point of a channel sectionfrom the free surface;Stage (h) - the vertical distance of the free surface from an arbitrary datum;Area (A) - the cross-sectional area of flow normal to the direction of flow;Wetted perimeter (P) - the length of the wetted surface measured normalto the direction of flow;Surface width (B) - the width of the channel section at the free surface;Hydraulic radius (R) - the ratio of area to wetted perimeter (A/P);Hydraulic mean depth (Dm) - the ratio of area to surface width {A/B).

p ^M"-: '

Figure 5.3 Definition sketch of geometric channel properties.


5.4 Velocity distributions, energy and momentum coefficients

The point velocity in an open channel varies continuously across the cross-section because of friction along the boundary. However, the velocitydistribution is not axisymmetric (as in pipe flow) due to the presence ofthe free surface. One might expect to find the maximum velocity at thefree surface, where the shear stress is negligible, but the maximum velocitynormally occurs below the free surface. Typical velocity distributions areshown in Figure 5.4 for various channel shapes.

The depression of the point of maximum velocity below the free surfacemay be explained by the presence of secondary currents which circulate fromthe boundaries towards the channel centre. Detailed experiments on velocitydistributions have demonstrated the existence of such secondary currentsand recent theoretical studies concerning three-dimensional turbulence haveilluminated the mechanisms for their existence (refer to section 15.7 forfurther details).

The energy and momentum coefficients (a and 0) defined in Chapter 2can only be evaluated for a channel if the velocity distribution has beenmeasured. This contrasts with pipe flow, where theoretical velocity distributions for laminar and turbulent flow have been derived, which enabledirect integration of the defining equations to be made.

For turbulent flow in regular channels, a rarely exceeds 1.15 and (3 rarelyexceeds 1.05. In consequence, these coefficients are normally assumed tobe unity. However, in irregular channels where the flow may divide intodistinct regions, a may exceed 2 and should therefore be included in flowcomputations. Referring to Figure 5.5, which shows a natural channel withtwo flood banks, the flow may be divided into three regions. By making the

Figure 5.4 Velocity distributions in open channels. (Note: Contour numbersexpressed as a percentage ofthe maximum velocity.)

rLAMINAR AND TURBULENT FLOW 127

Figure 5.5 Division of a channel into a main channel and flood banks.

assumption that a = 1 for eachregion, the value of a for the whole channelmay be found as follows:

where

fu3 dA _ V3A1 + VjA2 + VjA3V3A V3(A1 +A2 +A3)

V:Q _V1A1 + V2A2 + V3A3A Al+A2 + A3

(5.1)

5.5 Laminar and turbulent flow

In section 5.2, the state of flow was not discussed. It may be either laminaror turbulent, as in pipe flow. The criterion for determining whether the flowis laminar or turbulent is the Reynolds Number (Re), which was introducedin Chapters 3 and 4:

for pipe flow

for laminar pipe flow

for turbulent pipe flow

Re = pDV/p.

Re < 2000

Re > 4000

(3-2)

These results can be applied to open channel flow if a suitable form ofthe Reynolds Number can be found. This requires that the characteristiclength dimension, the diameter (for pipes), be replaced by an equivalentcharacteristic length for channels. The one adopted is termed the hydraulicradius (R) as defined in section 5.3.

Hence, the Reynolds Number for channels may be written as

Re(channel) = pRV/u. (5.2)

For a pipe flowing full, R = D/4, so

R-e(charmel) R-e(pipe)/4

1000

In practice, the upper limit of Re is not so well defined for channels as itis for pipes, and is normally taken to be 2000.

In Chapter 4, the Darcy-Weisbach formula for pipe friction was introduced, and the relationship between laminar, transitional and turbulentflow was depicted on the Moody diagram. Asimilar diagram for channelshas been developed. Starting from the Darcy-Weisbach formula,

hf = XLVz/2gD (4.8)and making the substitutions R= D/4 and hf/L = S0 (where S0 = bedslope), then for uniform flow in an open channel

' 2g4Ror

X= 8gRS0/V2 (5.3)

The \-Re relationship for pipes isgiven bythe Colebrook-White transitionlaw, and by substituting R= D/4 the equivalent formula for channels is

yr-218 k14.8J?Also combining (5.4a) with (5.3) yields

0.6275 \

ReVx /

0.6275vV= -2j8g~RS~0logt 14.8R R^RS~0/

(5.4a)

(5.4b)

A \-Re diagram for channels may be derived using (5.4a) and channelvelocities may be found directly from (5.4b). However, the applicationof these equations to a particular channel is more complex than the pipecase, due to the extra variables involved (i.e. for channels, Rchanges withdepth and channel shape). In addition, the validity of this approach isquestionable because the presence of the free surface has a considerableeffect on the velocity distributions, as previously discussed. Hence, the

UNIFORM FLOW 129

frictional resistance is non-uniformly distributed around the boundary, incontrastto pressurized pipeflow, wherethe frictional resistance is uniformlydistributed around the pipe wall.

In practice, the flow in openchannels is normally in the rough turbulentzone, and consequently it is possible to use simpler formulae to relatefrictional losses to velocity and channel shape, as is discussed in the nextsection.

5.6 Uniform flow

The development of friction formulaeHistorically, the development of uniform flow resistance equations precededthe detailed investigations of pipe flow resistance. These developments areoutlined below, and comparisons are made with pipe flow theory.

For uniform flow to occur, the gravity forces must exactly balance the frictional resistance forces which constitute the boundary shear force. Figure 5.6shows a small longitudinal section in which uniform flow exists.

The gravity force resolved in the direction of flow = pgAL sin 0 and theshear force resolved in the direction of flow = t0PL, where t0 is the meanboundary shear stress. Hence,

t0PL = pgALsinS

Considering channels of small slope only, then

sin 8 ~ tan 6 = Sn

hence,

T0 = pgAS0/P

or

t0 = pgRS0 (5.5)

Figure 5.6 Derivation of uniform flow equations.

1


The Chezy equation

To interpret (5.5), an estimate of the magnitude of t0 is required. Assuminga state of rough turbulent flow, then

t0ocV2 or t0 = KV2

Substituting into (5.5) for t0,

which may be written as

V = PgK

RSa

V = C^ (5.6)

This is known as the Chezyequation. It is named after the French engineerwho developed the formula when designing a canal for the Paris watersupply in 1768. The Chezy coefficient C is not, in fact, constant butdepends on the Reynolds Number and boundary roughness, as can bereadily appreciated from the previous discussions of the A-Re diagram. Adirect comparison between C and k can be found by substituting (5.6) into(5.3) to yield

C = j8g~/\In 1869, an elaborate formula for Chezy's C was published by two Swiss

engineers: GanguiUet and Kutter. It was based on actual discharge data fromthe River Mississippi and a wide range of natural and artificial channels inEurope. The formula (in metric units) is

C = 0.55241.6-r-l.811/w-f-0.00281/So

1+ [41.65+ (0.000281/S0)]/VR (5.7)

wheren is a coefficient known as Kutter's n, and is dependent solely on theboundary roughness.

The Manning equation

In 1889, the Irish engineer Robert Manning presented another formula (ata meeting of the Institution of Civil Engineers of Ireland) for the evaluationof the Chezy coefficient, which was later simplified to

C = R1/6/n (5.8)

UNIFORM FLOW 131

This formula was developed from seven different formulae, and was furtherverified by 170 observations. Other research workers in the field derivedsimilar formulae independently of Manning, including Hagen in 1876,Gauckler in 1868 and Strickler in 1923. In consequence, there is someconfusion as to whom the equation should be attributed to, but it is generally known as the Manning equation.

Substitution of (5.8) into (5.6) yields

V = (l/n)R2/3Sl/2and the equivalent formula for discharge is

y - P2/3 ^ (5.9)

where n is a constant known as Manning's n (it is numerically equivalentto Kutter's n).

Manning's formula has the twin attributes of simplicity and accuracy.It provides reasonably accurate results for a large range of natural andartificial channels, given that the flow is in the rough turbulent zone andthat an accurate assessment of Manning's n has been made. It has beenwidely adopted for use by engineers throughout the world.

Evaluation of Manning's n

The value of the roughness coefficient n determines the frictional resistanceof a given channel. It can be evaluated directly by discharge and stagemeasurements for a known cross-section and slope. However, for designpurposes, this information is rarely available, and it is necessary to rely ondocumented values obtained from similar channels.

For the case of artificially lined channels, n may be estimated with reasonable accuracy. For natural channels, the estimates are likely to be rather lessaccurate. In addition, the value of n may change with stage (particularlywith flood flows over flood banks) and with time (due to changes in bedmaterial as a result of sediment transport) or season (due to presence ofvegetation). In such cases, a suitably conservative design value is normallyadopted. Table 5.2 lists typical values of n for various materials and channelconditions. More detailed guidance is given by Chow (1959).

Uniform flow computations

Manning's formula may be used to determine steady uniform flow. Thereare two types of commonly occurring problems to solve. The first is to determine the discharge given the depth, and the second is to determine the depthgiven the discharge. The depth is referred to as the normal depth, which is

1


Table 5.2 Typical values of Manning's n.

Channel type

unlined canals

lined canals

models

Surface material and alignment

earth, straight 0.02-0.025earth, meandering 0.03-0.05gravel (75-150 mm), straight 0.03-0.04gravel (75-150 mm), winding or braided 0.04-0.08earth, straight 0.018-0.025rock, straight 0.025-0.045concrete 0.012-0.017

mortar 0.011-0.013

Perspex 0.009

Note: See Chow (1959) for a discussion regarding the units of Manning's n.

synonymous with steady uniform flow. As uniform flow can only occur ina channel of constant cross-section, natural channels should be excluded.However, in solving the equations of gradually varied flow applicable tonatural channels, it is still necessary to solve Manning's equation. Thereforeit is useful to consider the application of Manning's equation to irregularchannels in this section. The following examples illustrate the applicationof the relevant principles.

Example 5.1 Discharge from depth for a trapezoidal channel

The normal depth of flow in a trapezoidal concrete lined channel is 2 m. The channelbase width is 5 m and has side slopes of 1:2. Manning's n is 0.015 and the bedslope, S0, is 0.001. Determine the discharge (Q), mean velocity (V) and the ReynoldsNumber (Re).

Solution

Using Table 5.1,

A= (5 + 2y)y P= 5+ 2yv/l~+22~Hence, applying (5.9) for y = 2m

0 * x K5-M12F x0M10.015 [5+ (2x2V5)]2/3

= 45m3/s

UNIFORM FLOW

To find the mean velocity, simply apply the continuity equation:

45V =

A (5 + 4)2The Reynolds Number is given by

= 2.5 m/s

Re = pPV/|xwhere R = A/P. In this case,

and

(5+ 4)2R = [5+ (2x24] = 1.29m

Re:103x 1.29x2.5

1.14xl0-3= 2.83 x10s

133

Note: Re is very high, and corresponds to the rough turbulent zone. Therefore,Manning's equation is applicable. The interested reader may care to check thevalidity of this statement by applying the Colebrook-White equation (5.4b). Firstlycalculate a ks value equivalent to n = 0.015 for y = 2m[s =2.225mm]. Then usingthesevaluesof ks and n compare the discharges as calculatedusingthe Manning andColebrook-White equations for a range of depths. Providedthe channel is operatingin the rough turbulent zone, the results are very similar.

Example 5.2 Depth from dischargefor a trapezoidal channelIf the discharge in the channel given in Example 5.1 were 30m3/s, find the normaldepth of flow.

Solution

From Example 5.1

Q- xj(5+2y0.015 [5+ 2V5y]2/3

Q = 2.108x5/3[(5+2y)y]

[5+ 2-v/Jy]2/3At first sight this may appear to be an intractable equation, and it will also bedifferent for different channel shapes. The simplest method of solution is to adopta trial-and-error procedure. Various values of y are tried, and the resultant Q iscompared with that required. Iteration ceaseswhen reasonable agreement is found.In this case y < 2 as Q < 45, so an initial value of 1.7 is tried.

yc

(b) For a given constant specific energy:

1. The depth-discharge curve shows that discharge is a maximum atthe critical depth.

2. For all other discharges there are two possible depths of flow (sub-and supercritical) for any particular value of Es.

The general equation of critical flow

Referring to Figure 5.11, the general equation for critical flow may bederived by determining Sc and Qmax, independently, from the specificenergy equation. It will be seen that the two methods both result in thesame solution.

For Q = constant

s=y + aV2/2g

or

s=y+2i^For a minimum value (i.e. c,

= 0dE^dy

aQ2 d / 1 \ dA^^dAU2")^ (b)

For s = constant

s= y+ aV2/2g

or

'2gQ 1/2-A(Es-y)

For Qmax with s = 0 = constant,

dQ= [2jfA(E0-y)-V2dy V a

^(0-y)^)=0 (b)


Since 8A = B by,then in the limit

6AAy

B

Since hA = B 8y,then in the limit

dAdy

substituting into (b)

A

E=2B+yand substituting into (a)

B

substituting, into (b)

aQ20 = 1-

2gor

aCj2cgAl

B, 2A;3

Comparing,

QL*b= ,gAl

-tmax

gA*

(5.22)

In other words, at the critical depth the discharge is a maximum and thespecific energy is a minimum.

Critical depth and critical velocity (for a rectangular channel)Determination of the critical depth (yc) in a channel is necessary for bothrapidly and gradually varied flow problems. The associated critical velocity(Vc) will be used in the explanation of the significance of the Froude Number.Both of these parameters may be derived directly from (5.22) for any shapeof channel. For illustrative purposes, the simplest artificial channel shape isused here, i.e. rectangular. For critical flow,

aQ2B~Ja^ = i

For a rectangular channel, Q = qb,B = b, and A = by. Substituting in (5.22)and taking a = 1, then

1/3yc = tflg) (5.23)

^c = 1

-jiIS-*'.~

THE USE OF ENERGY PRINCIPLES 147

then

Vc = Vgy~c (5.24)

Also, as

ESc = yc + V?/2g

then

sc = yc+yc/2

(5.25)

An application of the critical depth lineThe application of critical depth is illustrated in Figure 5.12, which showsthe effect of a local bed rise on the free surface elevation for various initialdepths of flow. It should be noted from (5.23) that the critical depth isdependent only on the discharge. Hence, for a given discharge the criticaldepth line may be drawn on the longitudinal channel section as shown inFigure 5.12. If the upstream flow is initially subcritical, then a depressionof the water surface will occur above the rise in bed level. Conversely, ifthe upstream water level is initially supercritical, then an increase in the

for (a) y, g> yc, Ay ~ Az(b) y, > yc, Ay > Az(c) yi Az(d) y, yC)Ay = 0

Figure 5.12 Effect of a local bed rise on the water surface elevation.$*mi


water surface level will occur. As the upstream water level approaches thecritical depth, thenthechange in elevation of the free surface becomes moremarked. Allof these resultsmay easily be verified by reference to the specificenergy diagram drawn alongside in Figure 5.12. A practical example ofthis phenomenon often occurs in rivers under flood conditions. The normaldepth of flow may be nearly critical, and any undulations in the bed resultin large standing wave formations.

A further result may be gleaned from this example. If the local bedrise is sufficiently large, then critical flow will occur. It has already beenshown that discharge is a maximum for critical flow and hence, underthesecircumstances, the local bed rise is acting as a 'choke' on the flow, so calledbecause it limitsthe discharge. It is left to the reader to decide what happensif Az is so large that s2 is less than any pointon the s curve. The questionwill be discussed again later in section 5.9 and in Example 5.10.

The Froude Number

The Froude Number is defined as

Fr = V/JgLwhere L is a characteristic dimension. It is attributed to William Froude(1810-97), who used such a relationship in model studies for ships. If isreplaced by Dm, the hydraulic mean depth, then the resulting dimensionlessparameter

FI=V/y/gD~n (5.26)

is applicable to open channel flow. This isextremely useful, as it defines theregime of flow, and as many of the energy and momentum equations maybe written in terms of the Froude Number.

The physical significance of Fr may be understood in two different ways.First, from dimensional analysis,

Frz =inertial force _ p2V2

gravitational force pg3V2

7lSecondly, by consideration of the speed of propagation, c, of a wave of lowamplitude and long wavelength. Such waves may be generated in a channelas oscillatory waves or surge waves. Oscillatory waves (e.g. ocean waves)are considered in Chapter 8, and surge waves in section 5.11. Both typesof wave lead to the result that c = ^/gy. For a rectangular channel Dm = y,and hence

Fr:V water velocitygy wave velocity

THE USE OF ENERGY PRINCIPLES 149

Also for a rectangular channel Vc = Jgy~c, and hence, for critical flow,

:-L=^=ifgylgy

This is a general result for all channels (as it can be shown that for non-rectangular channels Vc = /gtT and c= JgDJ.

For subcritical flow,

VVr and Fr > 1

The Froude Number therefore defines the regime of flow. There is a secondconsequence of major significance. Flow disturbances are propagated at avelocity of c= */gy. Hence, if the flow issupercritical, any flow disturbancecan only travel downstream as thewater velocity exceeds thewave velocity.The converse is true for subcritical flow. Such flow disturbances are introduced by all channel controls and local features. For example, if a pebble isdropped into a still lake, then small waves are generated in all directions atequal speeds, resulting in concentric wave fronts. Now imagine droppingthe pebble into a river. The wave fronts will move upstream slower thanthey do downstream, due to the current. If the water velocity exceeds thewave velocity, the wave will not move upstream at all.

Anexample of how channel controls transmit disturbances upstream anddownstream is shown in Figure 5.13. To summarize:

for Fr > 1 - supercritical flow- water velocity > wave velocity- disturbances travel downstream- upstream water levels are unaffected by downstream control

for Fr < 1 - subcritical flow- water velocity < wave velocity- disturbances travel upstream and downstream- upstream water levels are affected by downstream control

When the Froude Number is close to one in a channel reach, flow conditions tend to become unstable, resulting in wave formations. If the channelis a compound channel, for example flood flows in a main channel and itsflood plains, then some very interesting and little investigated phenomena


(a) Caused by the weir is transmitted upstream as Fr < 1 and yn > yc

(b) Caused by a change of slope is not transmitted upstream as Fr > 1 and y< yc

Figure 5.13 Flow disrurbance.

can occur. For example, it is possible to imagine that flow in the mainchannel is very close to critical flow and that flow on the flood plains is stillsubcritical (due to much lower velocities). In fact, what happens is that atthe interface between flood plain and main channel, flow on the flood plainside of the interface becomes critical first because the velocity is close tothat of the main channel but the depth is much less. Studies by Knight andYuen (1990) have shown that both sub- and supercritical flow can existsimultaneously in a compound channel.

5.8 Rapidly varied flow: the use of momentum principles

The hydraulic jumpThe hydraulic jump phenomenon is an important type of rapidly variedflow. It is also an example of a stationary surge wave. A hydraulic jumpoccurs when a supercritical flow meets a subcritical flow. The resulting flowtransition is rapid, and involves a large energy loss due to turbulence. Underthese circumstances, a solution to the hydraulic jump problem cannot befound using a specific energy diagram. Instead, the momentum equation isused. Figures 5.14(a) and (b) depict a hydraulic jump and the associatedspecific energy and force-momentum diagrams. Initially, As is unknown,

I THE USE OF MOMENTUM PRINCIPLES 151

=yi

V, yc

i

As

F-i

(a) With associated specific energy diagram

(b) With associated force-momentum diagram

Figure 5.14 The hydraulic jump.

F+ M

as only the discharge and upstream depth are given. Byusing the momentumprinciple, the sequent depth y2 may be found in terms of the initial depthy, and the upstream Froude Number (FrJ.

The momentum equationThe momentum equation derived in Chapter 2 may be written as

, = AM


In this case, the forces acting are the hydrostatic pressure forces upstream(FA and downstream (2) of the jump. Their directions of action are asshown in Figure 5.14(b). This is a point which often causes difficulty. It maybe understood most easily by considering the jump to be inside a controlvolume, and to consider the external forces acting on the control volume(as described in Chapter 2).

Ignoring boundary friction, and for small channel slopes:

net force in x-direction = F1F2

momentum change = M2 M1

F1-F2 = M2- Mj

or

F1+M1=F2+M2 = constant (for constant discharge)

If depth (y) is plotted against force + momentum ( + M) for a constantdischarge, as in Figure 5.14(b), then for a stable hydraulic jump

F + M = constant (5.27)

Therefore, for anygiven initialdepth, the sequent depth is the correspondingdepth on the force-momentum diagram.

Solution of the momentum equation for a rectangular channelFor a rectangular channel, (5.27) may be evaluated as follows:

Fi = Pg(yi/2)Vib F2 = pg(y2/2)y2bM1=pQV1 M2 = PQV2

Q= PQyxb

Substituting in (5.27) and rearranging,

pQ y2b

M^)=eOi(i_A2 b \y2 yx

THE USE OF MOMENTUM PRINCIPLES

Substituting q = Q/b and simplifying,

(yl-yl) = i2 (L-!yi Vi

2(yi+y2)(yi"y2) =7"^TxWi+Vi) = -2 g yi^i

Substituting g = V^ and dividing byy\,

Jaf1+2s)_a..R2 yi V yi / g?i

This is a quadratic equation in y2/yi> whose solution is

y2 = (y1/2)(v/l *Fr2-Alternatively it may be shown that

yi = (fc/2)(Vl+ 8F*2-l)

153

(5.28a)

(5.28b)

Energy dissipation in a hydraulic jumpUsing (5.28a), y2 may be evaluated in terms of yx and hence the energy lossthrough the jump determined:

A = Ex - E2 = yt2g

y2 +

Substituting q = Vy,

A = y]-y2 +2g \yj y\

Equation (5.28a) may be rewritten as

2yi \yi

V2

2g

(5.29)

(5.30)>

**-


The upstream Froude Number is related to q as follows:

Fri = WgyT = (4/yi)/Vgy7

Fr? =

if y2 = y2,if y2 > y2,

if y2 < y2,

CRITICAL DEPTH METERS

. "--

p

yi

***

Figure 5.15 Stability of the hydraulic jump.

155

F + M

then a stable jump forms;then the downstream force + momentum is greater thanthe upstream force + momentum, and the jump movesupstream;then the downstream force + momentum is less than theupstream force + momentum and the jump movesdownstream.

Even if a stable jump forms, it may occur anywhere between the two sluicegates.

The occurrence and uses of a hydraulic jumpThe hydraulic jump phenomenon has important applications in hydraulicengineering. Jumps may form downstream of hydraulic structures suchas spillways, sluice gates and venturi flumes. They also sometimes occurdownstream of bridge piers (normally under flood flow conditions).

Jumps may be deliberately induced to act as an energy dissipation device(e.g. in stilling bays) and to localize bed scour. These uses are discussed inChapter 13.

5.9 Critical depth meters

The effect of a local bed rise on the flow regime has previously beendiscussed. It was concluded that critical flow over the bed rise would occurif the changein specific energy was such that it was reduced to the minimum

*


H t Az2

Az,

yc

/

Figure 5.16 The broad-crested weir.

specific energy. This situation is shown in Figure 5.16. This condition ofcritical depth over the bed rise, at first sight, may be thought only to occurfor one particular step height (Az-A and upstream depth (yA.

Consider what would happen if Azt was further increased to Az2. Thechange in energy from (1) to (2) would be such that no position on thespecific energy curve corresponding to (2) could be found. This apparentimpossibility may be explained by reconsidering the specific energy curve.It was drawn for a particular fixed value of discharge (Qx). In Figure 5.16,a second (thinner) curve is drawn for a smaller discharge (Q2) for whichthe specific energy at position (2) is a minimum. Hence, if the step heightwas Az2, then the discharge would initially be limited to Q2. However, ifthe discharge in the channel was Qu then the difference in the discharges(Qt - Q2) would have to be stored in the channel upstream, resulting inan increase in the depth yv After a short time, the new upstream depth(y'A would be sufficient to pass the required discharge Qr over the bed rise.In other words, the upstream energy will adjust itself such that the givendischarge (QA can pass over the bed risewith the minimum specific energyobtained at the critical depth. Of course, this will only apply provided thatAz is sufficiently large. Under these conditions, the local bed rise is actingas a control on the discharge by providing a choke. The upstream waterdepth is controlled by the bed rise, not by the channel. Such a bed rise isknown as a broad-crested weir. Its usefulness lies in its ability to controlthe discharge and hence it may be used as a discharge measuring device.

Derivation of the discharge equation for broad-crested weirsReferring to Figure 5.16, and assuming that the depth is critical at position(2), then from (5.24)

V, = V, =.

CRITICAL DEPTH METERS

and

Q = VA = JWtby,

where b is the width of the channel at position (2). Also, from (5.25)

yc =^Assuming no energy losses between (1) and (2), then

Substituting for yc

or

*|h/2\3/2Q-s1,2b(j) h3/

157

bH1'1 (5.33)

In practice, there are energy losses and it is the upstream depth rather thanenergy that is measured. Equation (5.33) is modified by the inclusion of thecoefficient of discharge (Cd) to account for energy losses and the coefficientof velocity (C) to account for the upstream velocity head, giving

Broad-crested weirs are discussed in more detail in Chapter 13.

(5.34)

Venturi flumesThis is a second example of a critical depth meter. In this case, the channelwidth is contracted to choke the flow, as shown in Figure 5.17. To demonstrate that critical flow is produced, it is necessary to draw two specificenergy curves. Although the discharge (Q) isconstant, the discharge per unitwidth (q) is greater in the throat of the flume than it is upstream. To forcecritical depth to occur in the throat, the specific energy in the throat mustbe a minimum. Neglecting energy losses, then the upstream specific energy(ES1) will have the same value. Providing that the upstream energy (En) for

S0 when y < ynSf < S0 when y > yn

?

GRADUALLY VARIED FLOW 165

horizontal asymptote

region 1

region 2

region 3

Figure 5.19 Profile types for a mild slope.

and

Fr > 1 when y < yc

Fr < 1 when y > yc

These inequalities may now be used to find the sign of dy/dx in (5.38) forany condition.

Figure 5.19 shows a channel of mild slope with the critical and normaldepths of flow marked. For gradually varied flow, the surface profile mayoccupy the three regions shown, and the sign of dy/dx can be found foreach region:

Region 1y > yn> yc, Sf < S0 and Fr < 1, hence dy/dx is positive.Region 2yn > y > yc, Sf > S0 and Fr < 1, hence dy/dx is negative.

Region 3ya > yc > y, Sf > S0 and Fr > 1, hence dy/dx is positive.The boundary conditions for each region may be determined similarly.

Region 1. As y - oo, Sf and Fr -> 0 and dy/dx S0, hence the watersurface is asymptotic to a horizontal line (as y is referred to the channelbed).

Fory > yn, Sf> S0 and dy/dx > 0, hence the water surface isasymptoticto the line y = yn.

Thiswater surface profile is termed an Ml profile. It is the type of profilewhich would form upstream of a weir or reservoir, and is known as abackwater curve.

Regions 2 and 3. The profiles may be derived in a similar manner. Theyare shown in Figure 5.19. However, there are two anomalous results for

>i


Regions 2 and 3. First, for the M2 profile, dy/dx oo as y -> yc. This isphysically impossible, and may be explained by the fact that as y > yc thefluid enters a region of rapidly varied flow, and hence (5.37) and (5.38) areno longer valid. The M2 profile is known as a drawdown curve, and wouldoccur at a free overfall.

Secondly, for the M3 profile, dy/dx > oo as y -* yc. Again, this is impossible, and in practice a hydraulic jump will form before y = yc.

So far, the discussion of surface profiles has been restricted to channelsof mild slope. For completeness, channels of critical, steep, horizontal andadverse slopes must be considered. The resulting profiles can all be derivedby similar reasoning, and are shown in Figure 5.20.

Outlining surface profiles and determining control points

An understanding of these flow profiles and how to apply them is an essential prerequisite for numerical solution of the equations. Before particulartypes of flow profiles can be determined for any given situation, two thingsmust be ascertained:

(a) Whether the channel slope is mild, critical or steep. To determine itscategory, the critical and normal depth of flow must be found for theparticular design discharge.

(b) The position of the control point or points must be established. Acontrol point is defined as any point where there is a known relationshipbetween head and discharge. Typical examples are weirs, flumes andgates or, alternatively, any point in a channel where critical depth occurs(e.g. at the brink of a free overfall), or the normal depth of flow at asuitably remote distance from the point of interest.

Having established the slope category and the position of any controlpoints, the flow profile(s) may then be sketched. For subcritical flow, theprofiles are controlled from a point downstream. For supercritical flow, theprofiles are controlled from upstream.

Figure 5.21 shows three typical flow profiles. Figure 5.21 (a) shows theinfluence of a broad-crested weir on upstream water levels. Numerical solution of this problem proceeds upstream from the weir (refer to Example 5.7).

Figure 5.21(b) shows the influence of bridge piers under flood conditions.Many old masonry bridges (with several bridge piers) act in a similar mannerto venturi flumes in choking the flow (particularly at high discharges). Flowthrough the bridge is rapidly varied and, on exit from the bridge piers, theflow is supercritical. However, supercritical flow cannot exist for long, asthe downstream slope is mild and the downstream flow uniform (assumingthat it is unaffected by downstream control). A hydraulic jump must formto return the flow to the subcritical condition. The position and height of

"his isycthe8) arewould

mpos-

innelsi\ anderived

essen-

Icularthings

ine its

or the

ed. A>nship:s andDCCUtS

n at a

ontrol

iv, thew, the

vs the1 solu-e5.7).itions.tanner

. Flowrs, theng, astiming: fotmght of

2

D.O u^ to

o on


Region 1 Region 2 Region 3

Ml

None

n

None

Figure 5.20 Classification of gradually varied flow profiles.

the jump are determined as shown in Figure 5.21(b). An example of thistype of problem is given in Example 5.8.

Figure 5.21(c) shows the flow profile for a side channel spillway andstilling basin. In this example, it is assumed that the side channel is not

*+

0*Z

* m >

1 6 8

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

M l

y c

' 1 S < . 9 . .

( a ) I n f l u e n c e o f b r o a d c r e s t e d w e i r o n r i v e r f l o w

b r i d g e d e c k

c o n j u g a t e d e p t h c u r v e

" g ^ p g g j ^ ^ P ^ a ^ p ^ ^ ^ i ^ s ^ j w ^ g ^ p ^ s B ^ ^ b

S < 5 C

( b ) I n f l u e n c e o f b r i d g e p i e r s o n r i v e r f l o w u n d e r f l o o d c o n d i t i o n s

c o n j u g a t e d e p t h c u r v e

5 = 0 y 2 I I y c

( c ) S i d e c h a n n e l s p i l l w a y a n d s t i l l i n g b a s i n

s < s r

F i g u r e 5 . 2 1 E x a m p l e s o f t y p i c a l s u r f a c e p r o f i l e s . ( N o t e : d e n o t e s a c o n t r o l

p o i n t . )

s o s t e e p a s t o i n v a l i d a t e t h e e q u a t i o n s o f g r a d u a l l y v a r i e d f l o w , a n d t h a t

a e r a t i o n d o e s n o t o c c u t ( r e f e r t o C h a p t e r 1 3 f o r m o r e d e t a i l s ) .

M e t h o d s o f s o l u t i o n o f t h e g r a d u a l l y v a r i e d f l o w e q u a t i o n

T h e t h r e e f o r m s o f t h e g r a d u a l l y v a r i e d f l o w e q u a t i o n a r e :

d H

~ d x ~ - - S f

^ - S . - S ,

d x

* 0 ~ * f

( 5 . 3 6 )

( 5 . 3 7 )

GRADUALLY VARIED FLOW

dydx 1-Fr2

There are three types of solution to the above equations:

169

(5.38)

1. direct integration;2. graphical integration;3. numerical integration.

These three methods of solution are discussed, and examples are givenwhere appropriate.

Direct integration. Equation(5.38) may besolved directly onlyfor regularchannels. Integration methods have been developed by various workers,starting with Dupuit in 1848, who found a solution for wide rectangularchannels. He used the Chezy resistance equation and ignored changes ofkinetic energy. In 1860, Bresse found a solution includingchanges of kineticenergy (again for wide rectangular channels using the Chezy equation).Bakhmeteff, starting in 1912, extended the work to trapezoidal channels.Using Manning's resistance equation, various other workers have extendedBakhmeteff's method.

Full details of these methods may be found in Henderson (1966) orChow (1959). They are not discussed here as they have been supersededby numerical integration methods which may be used for both regular andirregular channels.

Graphical integration. Equation (5.38) may be rewritten as

dx 1-Fr2

Hence,

or

where

dy S0 - S

x ryi 1-Fr2/ dx=fJO Jyy\ ^o ~ Sf

X= [nf(y)dyjyi

f(y)1-Fr2S0 Sf

dy

0a:


If a graph of y against f(y) is plotted, then the area under the curve isequivalent to X. The value of the function f(y) may be found by substitutionof A, P, S0 and Sf for various y for a given Q. Hence, the distance Xbetween given depths (yx and y2) may be found graphically.

This method was quite popular until the widespread use of computersfacilitated the use of the more versatile numerical methods.

Numerical integration. Using simple numerical techniques, all types ofgraduallyvaried flowproblems may be quickly and easilysolvedusingonlya microcomputer. A single program may be written which will solve mostproblems.

However, as an aid to understanding, the numerical methods arediscussed under three headings:

(a) the direct step method (distance from depth for regular channels);(b) the standard step method, regular channels (depth from distance for

regular channels); and(c) the standard step method, natural channels (depth from distance for

natural channels).

Thedirectstep method. Equation (5.38) may be rewritten in finite differenceform as

Ax = Ay1-FrzS0 Sf (5.39)

where 'mean' refers to the mean value for the interval (Ax). This formof the equation may be used to determine, directly, the distance betweengiven differences of depth for any trapezoidal channel. The method is bestillustrated by an example.

Example 5.7 Determining a backwater profile by the direct step methodUsing Figure 5.21(a), showing a backwater curve, determine the profile for thefollowing (flood) conditions:

Q = 600 m3/s S0 = 2 m/km n = 0.04channel: rectangular, width 50 m

weir: Q = 0.88]tlt sill height: Ps = 2.5m

*


Solution

First, establish a control point as follows.(a) Find normal depth (ya) from Manning's equation:

lA5/3

In this case,

O--^ 51/2

600 =~ x .}:0y^'L, x0.0021/20.04 (50 + 2yn)2/3

Solution by trial and error yields

yn -4.443 m

(b) Find the critical depth (yc):(a2\113yc = 1 (for a rectangular channel)\gj

In this case,

/(600/50)2\1/3yc = (- -1 =2.448 m(c) Find the depth over the weir (yw):

Q = 1.705CdBh3/2In this case,

600 = 1.705 x 0.88x50 x b3'2

h = 4m

171

yw = ^ + Ps

yw = 6.5m

Hence yw > yn > yc (i.e. Region 1), and S < SQ (i.e. a mild slope).This confirms (a) that the control is at the weir and (b) that there is an Ml type

profile.The profile isfound bytakingyw = 6.5m as the initialdepth and y= 4.5 m (slightly

greater than yn) as the final depth, and proceeding upstream at small intervals of

0*c

40MI

172

8

6

4exit

a

2

0

2S00

OPEN CHANNEL FLOW

2000 J500 1000

Distance (m)

water surface profile

500

Figure 5.22 Backwater profile for Example 5.7.

depth Ay. Fr, SQ and Sf are evaluated at each intermediate depth, and a solutionis found for Ax using the finite difference equation. A tabular solution is shown inTable 5.3, and the resulting profile is shown in Figure 5.22.

Table 5.3 is self-explanatory, but the following points should be kept inmind:

(a) Signs.

AyAy

is positive in the direction of flowis positive (except for adverse slopes)is positive by definitionis positive if final depth > initial depthis negative if initial depth > final depth

In Example 5.7, Ay is negative and S0 is positive, which makes Axnegative.

(b) Accuracy. Normal depth is always approached asymptotically, so fordepths approaching normal depthS0 Sf isvery small andmustbecalculatedaccurately. Of course, it is not possible to calculate the distance of normaldepth from the control point as this is theoretically infinite. In Example 5.7,a depth slightly larger than normal depth was chosen for this reason.

In regions of large curvature (i.e. approaching critical depth) equation (5.38) is no longer valid as the pressure distribution departs fromhydrostatic pressure. Thus the accuracy of the solution is impaired ascritical depth is approached and the solution should be terminated beforecritical depth is reached.

\Table 5.3 Computer solution of Example 5.7.

Discharge (cumecs)?600Channel width (m)?50Mannings's ?0.04Number of intervals (max 99)?10Slope (in decimals)?0.002SLOPE+ OR-P1Normal depth 4.4333Critical depth 2.4483

Initial depth?6.5Final depth?4.5

y(m) A(m2) P(m) Fr (1-Fr2)mean

Sf (So -Sf)mean

x (m)

6.5000 325.0000 63.0000 0.2312

0.9439

0.0006

0.0014

0.0000

6.3000 315.0000 62.6000 0.2423

0.9383

0.0007

0.0013

-139.1034

6.1000 305.0000 62.2000 0.2543

0.9319

0.0007

0.0012

-284.4225

5.9000 295.0000 61.8000 0.2673

0.9246

0.0008

0.0011

-437.6747

5.7000 285.0000 61.4000 0.2815

0.9163

0.0009

0.0010

-601.3459

5.5000 275.0000 61.0000 0.2970

0.9066

0.0010

0.0009

-779.2213

5.3000 265.0000 60.6000 0.3140

0.8954

0.0011

0.0008

-977.4688

5.1000 255.0000 60.2000 0.3327

0.8823

0.0013

0.0006

-1207.1478

4.9000 245.0000 59.8000 0.3532

0.8669

0.0015

0.0004

-1491.1912

4.7000 235.0000 59.4000 0.3760

0.8488

0.0017

0.0002

-1890.7515

4.5000 225.0000 59.0000 0.4014 0.0019 -2695.5699

,4WIM


(c) Choice of step interval. Numerical solutions always involve approximations, and here the choice of step interval affects the solution. The smallerthe step interval, the greater the accuracy. If the calculations are carriedout by computer, then successively smaller steps may be chosen until thesolution converges to the desired degree of accuracy. In Example 5.7, suchan exercise yielded the following results:

No. ofsteps Ay Length of reach Percentage change

5

10

20

30

60

125

250

500

1000

2000

-0.4 2587.0

-0.2 2696.0

-0.1 2757.0

-0.067 2775.0

-0.033 2788.4

-0.016 2792.2

-0.008 2793.1

-0.004 2793.3

-0.002 2793.4

-0.001 2793.4

4.0

2.2

0.6

0.5

0.14

0.03

0.007

0.004

0.000

The results ofthis exercise suggest that1000 steps are necessary toachieveconvergence. However the 10 steps originally selected give a solution towithin 3.5% of the converged solution.(d) Validity of solution. It has been suggested that the identification of acontrol point is of paramount importance in determining gradually variedflow profiles. This is reasonable if abroad understanding of the flow patternis the aim. However, the equations may be solved between any two depthsprovided they are within the same region of flow. Thus, in the case ofExample 5.7, the solution may proceed upstream or downstream, providedthat both the initial and final depths are greater than the normal depth. Ifthiscondition is not met, then the sign of Ax will change at some intermediate


point, demonstrating that the solution has passed through an asymptoteand the solution is no longer valid.(e) Composite profiles. Channels may have more than one control point,as already shown in Figure 5.21(b) and (c). In the case of Figure 5.21(b), ahydraulic jump effects the transition from supercritical to subcritical flow,but its height and position are determined by the upstream and downstreamcontrol points. An example of how to solve such problems follows.

Example 5.8 Composite profilesFor the situation shown inFigure 5.21(b), determine the distance ofthe jump fromthe bridge for the following conditions:

Q = 600m3/s S0 = 3m/km = 0.04channel: rectangular, width 50 m

depth at exit from the bridge = 1.2 m

Solution

First, normal and critical depth must be found using the same methods as inExample 5.7. These are:

yn = 3.897m yc= 2.448m

Inthis case, thesequent depth of thejump equals thenormal depth offlow. Utilizingthe hydraulic jump equation (5.28b)

y^(y1/2)(yll +m22-l)gives yt = 1.41m for y2 = ya = 3.897m.

Hence, to find the distance of the jump from the bridge an M3 profile must becalculated starting from the bridge (at y- 1.2) and ending at the initial depth ofthe jump (y= 1.41). This is most easily achieved by using the direct step method.Table 5.4 shows the calculations using a step length of 0.05 m. The position ofy = 1.41 m is founded by interpolation:

y = 1.4 x = 11.368

y=1.45 x = 14.151

(1.41-1.4) (1.45-1.4)(X-11.368) (14.151-11.368)

X= 11.925m

\*

*


Table 5.4 Computer solution of Example 5.8.

Discharge (cumecs)?600Channel width (m)?50Mannings's ?0.04Number of intervals (max 99)?5Slope (in decimals)?0.003SLOPE + OR- ?1

Normal depth 3.8967Critical depth 2.4483

Initial depth? 1.2Final depth? 1.45

y (m) A (m2) P (m) Fr (1-Fr2) Sf (SQ-Sf) x (m)mean mean

1.2000 60.0000 52.4000 2.9146 0.1336 0.0000

-7.0052 -0.1222

1.2500 62.5000 52.5000 2.7415 0.1169 2.8657

-6.0985 -0.1068

1.3000 65.0000 52.6000 2.5848 0.1028 5.7196

-5.3237 -0.0939

1.3500 67.5000 52.7000 2.4426 0.0909 8.5558

-4.6578 -0.0828

1.4000 70.0000 52.8000 2.3129 0.0807 11.3682

-4.0822 -0.0734

1.4500 72.5000 52.9000 2.1943 0.0720 14.1507

An alternative approach, necessary in cases where the jump exists between twoprofiles, is to plot the conjugate depth curve for the jump on the profile to find thepoints of intersection, as shown in Figure 5.23.

The standard step method (for regular channels). Equation (5.37) may berewritten in finite difference form as

As = Ax(S0-S/)meanwhere 'mean' refers to the mean values for the interval Ax.

(5.40)

'6

5

1"8-3o

2

conjugate depth curve


hydraulic jump

20 25

Distance (m)30

177

>y

35 45 50

Figure 5.23 Composite profiles for Example 5.8.

This form of the equation may be used to determine the depth at givendistance intervals. The solution method isan iterative procedure as follows:

1. assume a value for depth (y);2. calculate the corresponding specific energy (s^3. calculate the corresponding friction slope (Sf);4. calculate As over the interval Ax using (5.40);5. calculate Si+Ax = Sx + As;6. compare si+4:t and EsxGi7. if c ^ Ec , then return to 1.

Hence, to determine the depth at a given distance may require severaliterations. However, this method has the advantage over the direct stepmethod that depth is calculated from distance, which is the more usualproblem. Example 5.9 illustrates the method by solving again the problemof Example 5.7.

Example 5.9 Determining a backwater profile by the standard stepmethod

Solve Example 5.7 again by using a standard step length of 150 m.The establishment of a control point and the normal and critical depths are as inExample 5.7. A tabular solution is shown in Table 5.5. Only the correct values ofdepth (y) are shown in the table (to save space). The solution is most easily carriedout by computer.

*

Mi

>m,


Solution

Table 5.5 Computer solution of Example 5.9.

INPUT Q,N,SO,B,Yl,DX,L 600,.04,.002,50,6.5,-150,1950

Y (m) A (m2) P (m) EG (m) Sf (S0 - Sf) &.E (m) EC (m)mean

6.5000 325.0000 63.0000 6.6737 0.0006

6.2848 314.2386 62.5695 6.4706 0.0007

6.0795 303.9756 62.1590 6.2781 0.0008

5.8852 294.2583 61.7703 6.0971 0.0008

5.7027 285.1330 61.4053 5.9283 0.0009

5.5328 276.6424 61.0657 5.7726 0.0010

5.3765 268.8232 60.7529 5.6304 0.0011

5.2351 261.7527 60.4701 5.5029 0.0012

5.1078 255.3910 60.2156 5.3891 0.0013

4.9958 249.7889 59.9916 5.2899 0.0014

4.8975 244.8755 59.7950 5.2035 0.0015

4.8134 240.6717 59.6269 5.1302 0.0015

4.7416 237.0775 59.4831 5.0680 0.0016

0.0014

0.0013

0.0012

0.0011

0.0010

0.0009

0.0009

0.0008

0.0007

0.0006

0.0005

0.0004

6.6737

-0.2032

6.4705

-0.1928

6.2778

-0.1814

6.0967

-0.1692

5.9279

-0.1561

5.7722

-0.1424

5.6302

-0.1283

5.5020

-0.1141

5.3887

-0.1002

5.2890

-0.0867

5.2032

-0.0740

5.1295

-0.0623

5.0679

0.0003 -0.0518

X(m)

0

-150

-300

-450

-600

-750

-900

-1050

-1200

-1350

-1500

-1650

-1800

4.6818 234.0911 59.3636 5.0167 0.0017 5.0162 -1950

The standard step method (for natural channels). A common applicationof flow profiles is in determining the effects of channel controls in naturalchannels. For several reasons this application is more complex than thepreceding cases. The discharge is generally more variable and difficult to


quantify, and the assessment of Manning's n is less accurate. These aspectsare dealt with elsewhere (see section 5.6 and Chapter 10).

Another main difficulty lies inrelating areas and perimeters todepth. Thiscan only be accomplished by a detailed cross-sectional survey at knownlocations. In this way, tables of area and perimeter for a given stage maybe prepared. Notice that depth must be replaced by stage, as depth is not ameaningful quantity for natural sections. Alternatively, the cross-sectionaldata may be stored on a computer, values of area and perimeter beingcomputed as required for a given stage.

The solution technique is to use (5.36) in finite difference form as

AH = Ax(-S^)n (5.41)

where 'mean' refers to the mean value over the interval Ax.This may be solved iteratively to find the stage at a given distance in a

similar manner to the standard step method for regular channels, if depthis replaced by stage and specific energy by total energy.

Two further complications arise in applying equation (5.41) to naturalchannels. The first one is in determining the mean friction slope (Sfmean).At any particular cross-section Sf may be determined either directly fromManning's equation (see page 155) or by using the conveyance function(equation (5.16) with Sf replacing S0).

However, in a natural channel each cross-section is likely to be differentand hence it is necessary to find a representative friction slope for eachreach from the cross-sectional data at each end of the reach. It is possibleto conceive the representative friction slope as being some kind of weightedaverage value derived either directly from the friction slopes at eachcross-section or directly from a weighted average value of the channelconveyances.

Cunge et al. (1980, pages 129-30) compare four methods of finding arepresentative friction slope. Differences between results from the methodscan be very large if the upstream and downstream cross-sections are verydifferent. He concludes that if this is the case then a smaller distance stepshould be introduced. In equation (5.41) the mean value of Sf has beenchosen as it is the simplest representation conceptually.

The second complication in solving equation (5.41) in natural channelsarises from the presence of flood banks. When flood discharges are beingconsidered, the flow will overtop the main channel and flow on the floodplain will occur. Under these circumstances the energy coefficient (a) mustbe computed. Where natural channels have well defined flood plains (asshown in Figure 5.5), then a is conveniently estimated by equation (5.14)which uses the channel conveyance function.

,/HM


An iterative solution method for two-stage natural sections can, therefore,be set up as follows:1. assume a value for stage (hc) at x + Ax;2 calculate the corresponding values of A P, and K, at x+ Ax from the

tabulated cross-sectional data and hence find a (using (5.14)) at x+ Axfor the assumed stage (hG); _3. calculate the mean cross-sectional velocity (V = Q^A) at x+Ax andhence find the total energy at x+ Ax (HG = hG + aV /2g);

4. calculate the friction slope SfG (x+Ax) at x+Ax using equation (5.16) andhence find the mean friction slope S^mean = (SfG (x+Ax) + ^/(x))/25

5. calculate AH over the interval Ax using (5.41);6. calculate Hx+Ax = Hx + AH;7. compare Hx+Ax and HG;8- if Hx+Ax # HG then repeat from 1until suitable convergence is obtained.

Several commercially available mathematical models have been developedfor application to natural channels. One such model (FLUCOMP1 &2),developed by the Hydraulics Research Station (England), allows for calibration of Manning's n values from recorded discharge and water evelsincorporates the effects of weirs and bridge piers, and includes lateralinflows. This model also computes one-dimensional gradually variedunsteady flow (see section 5.11 for further details). Complete details of themodel are given in Samuels and Gray (1982).

5.11 Unsteady flow

Types of unsteady flowUnsteady flow is the normal state of affairs in nature, but for manyengineering applications the flow may be considered to be steady. Howeverunder some circumstances, it is necessary to consider unsteady flow. Suchcircumstances may include the following:Translatory waves. The movement of flood waves down rivers.Surges and bores. Produced by sudden changes in depth and/or discharge(e.g. tidal effects or control gates).Oscillatory waves. Waves produced by vertical movement rather thanhorizontal movement (e.g. ocean waves).

Translatory waves are considered in a simplified way in Chapter 10(hydrology, flood routing) and they are considered in more detail later in thissection as an example ofgradually varied unsteady flow. Surge waves are anexample of rapidly varied unsteady flow and asimple treatment is presentedin the following section. Oscillatory waves are considered in Chapter 8.

UNSTEADY FLOW 181

Surge wavesAtypical surge wave is shown inFigure 5.24(a), inwhich a sudden increasen the downstream depth has produced a steep-fronted wave movingostream at velocity V. The wave is, in fact, a moving hydraulic jump, and

the solution for the speed ofthe wave ismost easily obtained by transposingthe problem to that of the hydraulic jump. This may be achieved by theuse of the technique of the travelling observer, as shown in Figure 5.24(b).To an observer travelling on the wave at velocity V, the wave is stationarybut the upstream and downstream velocities are increased to Vx + V andV + V respectively (and the river bed is moving at velocity V). The travelling observer sees a (stationary) hydraulic jump. Hence, the hydraulic jumpequation may be applied as follows:

y_i l + 8Fr2-l) (5.28a)

or

c 2 lyiFrj = - 2)/j

V2 1 1(see derivation of (5.28a))

i.e.

(Vi + V)2gyi

ly,2y1

y_i (5.42)

This equation maybesolved for knownupstream conditions and the downstream depth. If y2 is unknown, then the continuity equation may be used:

(V1 + V)y1 = (V2 + V)y2

In this case, Q2 must be known for solution of V.Considering (5.42) in more detail, three useful results may be obtained.

First, as y1/y1 tends to unity, then (Vt + V) tends to ^/gy. As (Vt + V) = c (the

--v, n 1

'^*ii^-5>-::

LVi + V

+ y\

j*yi V2 + V

Figure 5.24 Surge waves.

1

401 )M


wave speed relative to the water) this confirms the result given in section 5.7that small disturbances are propagated at a speed c= Vgy (for rectangularchannels). , ,. .Secondly, for y2/yx > 1, then (Vx + V) > Jgfx- This implies that surgewaves can travel upstream even for supercritical flow. This is confirmedby Example 5.10. However, after the passage of such a surge wave, theresulting flow is always subcritical.

Thirdly, for (V1 + V)> Jgy[, then the surge wave will overtake anyupstream'disturbances, and conversely any downstream disturbances willovertake the surge. Hence, this type of surge wave remains stable and steepfronted. ,

There are in fact, four possible types of surge waves; upstream anddownstream', each of which may be positive or negative. Positive surgesresult in an increase in depth and are stable and steep fronted (as discussedabove). Negative surges result in a decrease in depth. These are unstableand tend to die out as disturbances travel faster than the surge wave.

Example 5.10 Speed ofpropagation ofan upstream positive surge waveUniform flow in a steep rectangular channel is interrupted by the presence of ahump in the channel bed which produces critical flow for the initial dischargeof 4m3/s If the flow is suddenly reduced to 3m3/s some distance upstream,determine the new depth and speed of propagation of the surge wave. (Channeldata: b= 3m, = 0.015, 50 = 0.01, Az = 0.083 m).

Solution

Referring to Figure 5.12, this problem is an example of what happens whenAz >Es^Esz and yt

. = 0.701m;

Sn = 0.769 m;

UNSTEADY FLOW 183

Es-Esc= 0.068

puifi'


Strictly speaking, the equations of motion apply only to truly one-dimensional flow in which the water surface is horizontal at any cross-section and the velocity is uniform over the cross-section. This is only anapproximation to natural river flow. Supplementary coefficients (e.g. themomentum coefficient (3) are often introduced into the equations to simulate quasi two-dimensional flow. For flood flows in natural channels withflood plains further approximations have to be made.

The derivation and solution of the gradually varied unsteady flow equations is a complicated matter, even for the simplest case of a rectangularchannel. General solutions are made practicable only by the use of acomputer. An introductory discussion of some of the techniques used maybe found in Chapter 14, and the Reference list there includes some of themore advanced texts, which give details of the range of numerical techniques in current use. Practical aspects of computational modelling of riverflows are discussed in Chapter 15.

References and further reading

Ackers, P. (1992) HydraulicDesign ofTwo-stage Channels, Proc. Instn. Civil Engrs.,Water, Maritime and Energy, 96, Dec, 247-57.

Chow Ven te (1959) Open Channel Hydraulics, McGraw-Hill, Tokyo.Cunge, J. A., Holly, F. H. (Jr) and Verwey, A. (1980) Practical Aspects of Compu

tational River Hydraulics, Pitman, London.French, R. H. (1986) Open Channel Hydraulics, McGraw-Hill, Singapore.Henderson, F. M. (1966) Open Channel Flow, Macmillan, New York.Jansen, P.Ph. et al. (eds) (1979) Principles of River Engineering, Pitman, London.Knight, D. (1989) Hydraulics of flood channels, Chapter 6 in Floods: Hydrolog-

ical, Sedimentological and Geomorphological Implications (eds K. Beven andP. Carling), Wiley, Chichester.

Knight, D. W. and Yuen, K. W. H. (1990) Critical flow in a two stage channel, inInternational Conference on River Flood Hydraulics (ed. W. R. White), Wiley,Chichester.

Ramsbottom, D. M. (1989) Flood Discharge Assessment - Interim Report, ReportSRI 95, Hydraulics Research, Wallingford.

Townson, J. M. (1991) Free Surface Hydraulics, Unwin Hyman, London.

CWR4202 Open Channel Flow Chap 5

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