River EngineeringJohn Fenton

Institute of Hydraulic and Water Resources EngineeringVienna University of Technology, Karlsplatz 13/222,

1040 Vienna, AustriaURL: http://johndfenton.com/

URL: mailto:JohnDFenton@gmail.com

1. Introduction1.1 The nature of what we will and will not do – illuminated by someaphorisms and some people

“There is nothing so practical as a good theory” – stated in 1951 by Kurt Lewin (D-USA,1890-1947): this is essentially the guiding principle behind these lectures. We want to solvepractical problems, both in professional practice and research, and to do this it is a big help to havea theoretical understanding and a framework.

“The purpose of computing is insight, not numbers” – the motto of a 1973 book on numericalmethods for practical use by the mathematician Richard Hamming (USA, 1915-1998). Thatstatement has excited the opinions of many people (search any three of the words in the Internet!).However, numbers are often important in engineering, whether for design, control, or other aspectsof the practical world. A characteristic of many engineers, however, is that they are often blindedby the numbers, and do not seek the physical understanding that can be a valuable addition to thenumbers. In this course we are not going to deal with many numbers. Instead we will deal withthe methods by which numbers could be obtained in practice, and will try to obtain insight intothose methods. Hence we might paraphrase simply: "The purpose of this course is insight into thebehaviour of rivers; with that insight, numbers can be often be obtained more simply and reliably".

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“It is EXACT, Jane” – a story told to the lecturer by a botanist colleague. The most importantriver in Australia is the Murray River, 2 375 km (Danube 2 850 km), maximum recorded ow3 950 m3s 1 (Danube at Iron Gate Dam: 15 400 m3s 1). It has many tributaries, ow measurementin the system is approximate and intermittent, there is huge biological and uvial diversity andirregularity. My colleague, non-numerical by training, had just seen the demonstration by anhydraulic engineer of a one-dimensional computational model of the river. She asked: “Just howaccurate is your model?”. The engineer replied intensely: "It is EXACT, Jane".

Nothing in these lectures will be exact. We are talking about the modelling of complex physicalsystems.

A further example of the sort of thinking that we would like to avoid: in the area of palaeo-hydraulics, some Australian researchers made a survey to obtain the heights of oods at individualtrees. This showed that the palaeo- ood reached a maximum height on the River Murray at acertain position of 18 01m (sic), Having measured the cross-section of the river, they applied theGauckler-Manning-Strickler Equation to determine the discharge of the prehistoric ood, stated tobe 7 686 m3s 1 ...

William of Ockham (England, c1288-c1348): Ockham’s razor is the principle that can bepopularly stated as “when you have two competing theories that make similar predictions, thesimpler one is the better”. The term razor refers to the act of shaving away unnecessary assumptionsto get to the simplest explanation, attributed to 14th-century English logician and Franciscan friar,William of Ockham. The explanation of any phenomenon should make as few assumptions aspossible, eliminating those that make no difference in the observable predictions of the explanatory

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hypothesis or theory. When competing hypotheses are equal in other respects, the principlerecommends selection of the hypothesis that introduces the fewest assumptions and postulates thefewest entities while still suf ciently answering the question. That is, we should not over-simplifyour approach.

In general, model complexity involves a trade-off between simplicity and accuracy of the model.Occam’s Razor is particularly relevant to modelling. While added complexity usually improves thet of a model, it can make the model dif cult to understand and work with.

The principle has inspired numerous expressions including “parsimony of postulates”, the“principle of simplicity”, the “KISS principle” (Keep It Simple, Stupid). Other commonrestatements are:

Leonardo da Vinci (I, 1452–1519, world’s most famous hydraulician, also an artist): hisvariant short-circuits the need for sophistication by equating it to simplicity “Simplicity is theultimate sophistication”.

Wolfang A. Mozart (A, 1756–1791): “Gewaltig viel Noten, lieber Mozart”, soll Kaiser Josef II.über die erste der großen Wiener Opern, die “Entführung”, gesagt haben, und Mozart antwortete:“Gerade so viel, Eure Majestät, als nötig ist.” (Emperor Joseph II said about the rst of the greatVienna operas, “Die Entführung aus dem Serail”, “Far too many notes, dear Mozart”, to whichMozart replied “Your Majesty, there are just as many notes as are necessary”). The truthfulness ofthe story is questioned – Josef was more sophisticated than that ...

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Albert Einstein (D-USA,1879-1955): “Make everything as simple as possible, but not simpler.”This is a better and shorter statement than Ockham!

Karl Popper (A-UK, 1902-1994) argued that we prefer simpler theories to more complex ones“because their empirical content is greater; and because they are better testable”. In other words, asimple theory applies to more cases than a more complex one, and is thus more easily falsi able.Popper coined the term critical rationalism to describe his philosophy. The term indicates hisrejection of classical empiricism, and of the classical observationalist-inductivist account ofscience that had grown out of it. Logically, no number of positive outcomes at the level ofexperimental testing can con rm a scienti c theory (Hume’s “Problem of Induction”), but a singlecounterexample is logically decisive: it shows the theory, from which the implication is derived, tobe false. For example, consider the inference that “all swans we have seen are white, and thereforeall swans are white”, before the discovery of black swans in Australia. Popper’s account of thelogical asymmetry between veri cation and falsi ability lies at the heart of his philosophy ofscience. It also inspired him to take falsi ability as his criterion of demarcation between what isand is not genuinely scienti c: a theory should be considered scienti c if and only if it is falsi able.This led him to attack the claims of both psychoanalysis and contemporary Marxism to scienti cstatus, on the basis that the theories enshrined by them are not falsi able.

Thomas Kuhn (USA, 1922-1996): In The Structure of Scienti c Revolutions argued that scientistswork in a series of paradigms, and found little evidence of scientists actually following afalsi cationist methodology. Kuhn argued that as science progresses, explanations tend to becomemore complex before a sudden paradigm shift offers radical simpli cation. For example Newton’s

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classical mechanics is an approximated model of the real world. Still, it is quite suf cient formost ordinary-life situations. Popper’s student Imre Lakatos (H-UK, 1922-1974) attempted toreconcile Kuhn’s work with falsi cationism by arguing that science progresses by the falsi cationof research programs rather than the more speci c universal statements of naive falsi cationism.

Another of Popper’s students Paul Feyerabend (A-USA, 1924-1994) ultimately rejected anyprescriptive methodology, and argued that the only universal method characterising scienti cprogress was "anything goes!"

1.2 Summary• We will use theory, but we will try to keep things simple, rather simpler than is often the case inthis eld, especially in numerical methods.

• Often our knowledge of physical quantities is limited, and approximation is justi ed.• We will recognise that we are modelling.• An approximate model can often reveal to us more about the problem.• It might be thought that the lectures show a certain amount of inconsistency – in occasionalplaces the lecturer will develop a more generalised and “accurate” model, paradoxically toemphasise that we are just modelling.

• We will attempt to obtain insight and understanding – and a sense of criticality.

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1.3 Types of channel ow to be studiedAn important part of this course will be the study of different types of channel ow.

(b) Steady gradually-varied ow

(a) Steady uniform ow

Normal depth

(d) Unsteady ow

(c) Steady rapidly-varied ow

Figure 1.1: Different types of ow in an open channel

Case (a) – Steady uniform ow:

Steady ow is where there is no changewith time, 0. Distant from controlstructures, gravity and resistance are inbalance, and if the cross-section is constant,the ow is uniform, 0. This is thesimplest model, and often is used as thebasis and a rst approximation for others.

Case (b) – Steady gradually-varied ow:

Where all inputs are steady but wherechannel properties may vary and/or acontrol may be introduced which imposesa water level at a certain point. The heightof the surface varies along the channel.For this case we will study the governingdifferential equation that describes howconditions vary along the waterway, and we

will obtain an approximate mathematical solution to solve general problems approximately.7

(b) Steady gradually-varied ow

(a) Steady uniform ow

Normal depth

(d) Unsteady ow

(c) Steady rapidly-varied ow

Figure 1.2: Different types of ow in an open channel

Case (c) – Steady rapidly-varied ow:

Figure (c) shows three separate gradually-varied ow regions separated by tworapidly-varied regions: (1) ow under asluice gate and (2) a hydraulic jump. Thebasic hydraulic approximation that variationis gradual breaks down in those regions.We can analyse them by considering energyor momentum conservation locally. In thiscourse we will not be considering these –earlier courses at TUW have.

Case (d) – Unsteady ow:

Here conditions vary with time and positionas a ood wave traverses the waterway. Wewill consider ood wave motion at somelength.

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1.4 Some possibly-surprising results

Effects of turbulence on dynamicsWhere the uid ow uctuates in time, apparently randomly, about some mean condition, e.g.the ow of wind, water in pipes, water in a river. In practice we tend to work with mean owproperties, however in this course we will adopt empirical means of incorporating some of theeffects of turbulence. Consider the component of velocity at a point written as a sum of themean (¯) and uctuating ( 0) components:

= ¯ + 0

By de nition, the mean of the uctuations, which we write as 0, is

0 =1Z0

0 = 0 (1.1)

where is some time period much longer than the uctuations.

Now let us compute the mean value of the square of the velocity, such as we might nd in

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computing the mean pressure on an object in the ow:2 = (¯ + 0)2 = ¯2 + 2¯ 0 + 02 expanding,= ¯2 + 2¯ 0 + 02, considering each term in turn,= ¯2 + 2¯ 0 + 02, but, as 0 = 0 from(1.1),= ¯2 + 02 (1.2)

hence we see that the mean of the square of the uctuating velocity is not equal to the square of themean of the uctuating velocity, but that there is also a component 02, the mean of the uctuatingcomponents. We will need to incorporate this.

Pressure in open channel ow – effects of resistance on ows over steep slopes

Isobars

sin

Resistance

cos

Figure 1.3: Channel ow showing isobars and forces perunit mass on a uid particle

An almost-universal assumption in river en-gineering is that the pressure distribution ishydrostatic, equivalent to that of water whichis not moving, such that pressure at a point isgiven by the height of water above, = ,where is uid density ( 1000 kgm 3 forfresh water), 9 8 ms 2 is gravitationalacceleration, and is the vertical height of thesurface above the point. This is not necessarily

the case in owing water, and needs to be known for cases such as spillways or block ramps, whichare steep.

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Consider gure 1.3 showing an open channel ow with forces per unit mass acting on a particle.The gure is drawn, showing that in general, the depth is not constant, and the bed is not parallelto the free surface. The surface is an isobar, a line of constant pressure, = 0. In the ow, otherisobars will approximately be parallel to this, while the channel bed is not necessarily an isobar.We consider the vector Euler equation for the motion of a uid particle

Acceleration =1 × Pressure gradient + Body forces per unit mass

In a direction parallel to the free surface, the pressure is constant and there is no pressure gradient.The acceleration of the particle will be given by the difference between the component of gravitysin and the resistance force per unit mass. We usually do not know the details of that, so there islittle that we can say. Now considering a direction perpendicular to that, given by the co-ordinateon the gure, there is very little acceleration, so we assume it to be zero, and so we obtain the

result

0 =1

cos .

Now integrating this with respect to between a general point, such as = at the particleshown, and = 0 on the surface where = 0 we obtain

= cos × .

It is much more convenient to measure all elevations vertically, and so we use , such that= cos , and we obtain the general expression for pressure

= cos2 (1.3)11

Hydrostatic approximation

That result is really only important on structures such as spillways and block ramps. In almostall open channels the slope is small enough such that cos2 1, and we can use the hydrostaticapproximation, obtained from a static uid, where the surface is horizontal,

= (1.4)

Substituting = , where is the free surface elevation and is the elevation of an arbitrarypoint in the uid,

= ( ) (1.5)From this we have at a speci c vertical cross-section,

+ = (1.6)

so that anywhere on a vertical section + is constant, given by the free surface elevation.The general result of equation (1.3) for ow on a nite slope seems to have beenforgotten by many. In general, pressure in owing water is not “hydrostatic”.However in this course, bed slopes are small enough that we will use it.

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2. Resistance in river and other open channel owsThe resistance to the ow of a stream is probably the most important problem in river mechanics.

Page 14: We consider a simple theory based on force balance and some classical uid mechanicsexperiments to obtain a ow formula for a wide channel.

Page 16: To obtain the equivalent formula for channels of any section we consider velocitydistributions in real streams and develop an approximation giving a general ow formula.

Page 20: We consider an approximation to that formula and nd that we have obtained theGauckler-Manning-Strickler formula, including a theoretical prediction of Strickler’s formula forthe effect of boundary grain size.

Page 25: Comparison with a series of experiments validates the approach, giving an explicit owformula for a variety of channel boundaries.

Page 31: For more general river problems, considering the nature of the bed particles and bedforms, vegetation, meandering, and possibly obstacles, it is better to use a formulation in whichforces and the mechanics are clearer: the Chézy-Weisbach ow formula.

Page 33: A large number of stream-gauging results are considered and the values of the Weisbachresistance coef cient, its dependence on grain size, and on the state of the bed are obtained.Empirical formulae are considered.

Page 38: The common problem of calculating the water depth for a given ow rate is considered.A computational method is developed and applied.

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2.1 The channel ow formulaHere, the fundamental ow formula for steady uniform ow in channels is developed from theoryand experimental results. We show that the traditional ow formulae of Gauckler-Manning andChézy-Weisbach are simply different approximations to that.

Resistance

sin

Figure 2.1: Uniform ow in a channel, showing resistanceand gravity forces on a nite length, plus cross-sectionquantities

Consider a horizontal length of uniformchannel ow, inclined at a small angle tothe horizontal, with cross-sectional area .The volume of the element is , the verticalgravitational force on the water is ,where is uid density and is gravitationalacceleration. The component of this along theslope is sin . The resistance force along

the slope, of length cos is cos , where is the mean resistance shear stress, assumeduniformly distibuted around the wetted perimeter around which it acts. Equating gravitationaland resistance components gives cos = sin . To high accuracy for small ,cos 1 and sin tan = , the slope, giving

= (2.1)

Our problem is now to express shear stress in terms of ow quantities.

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Figure 2.2: Idealisedlogarithmic velocitypro le in turbulent owover rough bed

One of the most famous series of experiments in uid mechanics wasperformed by Johann Nikuradse at Göttingen in the 1930s, who studied theow of uid over uniformly-rough sand grains. The uid was actually air,and the sand grains were actually in circular pipes, but the results are stillvalid enough.

With those results, for a wide channel of depth with sand grains of sizes, the velocity distribution for fully rough ow (no effects of viscosity), theuniversal velocity distribution can be written:

= ln30

s(2.2)

in terms of the shear velocity =p

, the von Kármán constant 0 4,the vertical co-ordinate , and where the factor of 30 is for closely-packed uniform sand grains.It varies with other types of boundary roughness. The mean velocity is obtained by integratingbetween 0 and , such that

=1Z0

= ln30 e

s(2.3)

where e = exp(1) = 2 718 is Euler’s number. The result has been obtained in terms of relativeroughness s . We replace =

pusing equation (2.1) to give

=1r μ

ln30 e

s

¶(2.4)

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We have obtained something very useful – a formula for the mean ow velocity in a wide channelof constant depth , slope , and relative roughness s . We have used simple mechanics plus anempirical laboratory result. Surprisingly, the formula is explicit in terms of physical quantities – wehave not had to assume a value like the Manning coef cient or Strickler coef cient St = 1 !

Figure 2.3: Cross-section of ow showing isovels and, fora number of points on the bed, where the fastest-likelyuid comes from and how far it travels, the effectivelength scale for resistance calculations.

That was for a wide channel with an idealisedlogarithmic velocity distribution. In nature, forchannels of any general cross-section there isthe problem that the velocity has a maximumsomewhere below the surface, and in generalthe isovels are something like Figure 2.3.Even in straight channels there are longitudinalvortices such that in the centre of the channelthe maximum in velocity, which would beexpected to be at the surface, is actually at alower position.

To obtain a ow formula for channels of any cross-section, we hypothesise that the effectivedepth for resistance calculations is the typical distance from points with the highest velocityto the nearest point on the bed, as suggested by the red arrows on the gure. The uid owon the boundary, where resistance occurs, would be similar to that in a channel, not of theactual mean depth, but the mean length of the red arrows.Typical length scales as shown by thearrows are somewhat smaller than the overall mean depth of ow. Our problem is then, how to

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approximate that distance? We examine the approach suggested by the lecturer (Fenton 2011).

max

Figure 2.4: Experimental determination ofvelocity maxima in rectangular channel

We consider the experimental data for the vertical position ofthe locus of velocity maxima in rectangular channels fromYang, Tan & Lim (2004). They presented a formula for theheight above the bed of the velocity maximum as a functionof position across the channel. If the mean value of this iscalculated by integration, a formula for the mean elevationof the velocity maximum max is obtained as a function ofaspect ratio (channel width divided by depth ).

0 0

0 2

0 4

0 6

0 8

1 0

1 2 10 100

Experimentalrange

max

and( )

Aspect ratio

max — experiment

( ) , eqn (2.5)

Figure 2.5: Rectangular channels: dimension-less mean elevation of max and the effectivedepth ( )

There is another length scale in equation (2.4) which isthe ratio of area to perimeter , which, as ,should be smaller than the mean depth , so it mightbe a candidate for the depth scale as experienced by thebed. We calculate the ratio:

=( + 2 )

=+ 2

(2.5)

Both this and the experimental formula for max areplotted in Figure 2.5. Remarkably, the two coincideclosely over a wide range of aspect ratios, so that we havefound the quantity mimics the behaviour of max,which we have suggested is, instead of , the apparent

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depth that the ow on the bed experiences. This suggests that in equation (2.4), instead of in theterm from the velocity distribution, we can use . We cannot claim that this is a justi cation asstrong as it looks, but we have seen that already appears in the equation, appearing naturallyin the simple mechanical equilibrium calculation.

For we will not use the conventional and misleading term “hydraulic radius” (German afterStrickler – “Pro l- oder hydraulischer Radius”).

For channels that are not rectangular we have presented no results. Our suggestion is thatwill still be a plausible approximation, and it already appears in equation (2.4). The use ofwas justi ed by Keulegan (1938), however while that work mathematically correctly integratedlogarithmic velocity distributions perpendicular to parts of various shapes of cross-section, it didnot give any attention to the real velocity distributions, especially ignoring the phenomenon of thevelocity maximum being below the surface.

We have shown that is approximately equal to the mean distance of the maximum velocityfrom the bed, so that the ow on the bed is similar to that of a channel of depth . In channelsthat are wide, which is most, and is about the same as the geometric mean depth

. Our suggested channel ow formula, replacing by in equation (2.4) is

= =1r μ

ln30 e

s ( )

¶(2.6)

If we knew an accurate value of s, this is probably the formula that we should use, as it isin terms of phyical quantities that we know or we can approximate, including the equivalent

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grain size s. For example, if we were studying the Danube, we might simply use the typicalgrain size s = = 0 02 m. Traditional practice, however is often to use Chézy-Weisbachand Gauckler-Manning-Strickler formulae, which introduce resistance coef cients which, whilemore general, and allow for other forms of resistance such as vegetation and bed forms, are moreempirical and their physical signi cance less clear.

Our ow formula (2.6) can be written

= =

rwhere (2.7)

=1ln

30 e

s ( )=1ln30 e 1

ln12

(2.8)

where we have introduced the symbol for the relative roughness

=s=Equivalent grain sizeHydraulic mean depth

Grain sizeDepth

(2.9)

and, although 30 e 11 0, it is usually written as 12, quite acceptably because our knowledgeof s is usually not good. We now have a ow formula for steady uniform ow in a channelbased on simple theory, experimental observations, and a bold approximation. We will soon seehow the most common ow formulae in practice are an approximation to this, but this has theadvantage that it is an explicit formula for the resistance coef cient in terms of the equivalentrelative roughness of boundary grains.

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Relative unimportance of grain sizeIn fact, , although all-important for us, is relatively slowly varying with grain size. Consider asmall change in the relative roughness (1 + ). The relative change in the factor is

=ln (12 ( (1 + )))

ln (12 )1

ln (12 )

having expanded the logarithm as a power series ln (1 + ) = + . Now for a value of= 0 001 (a 1mm grain in 1m of water), a relative change of = 50% gives a relative change inthe factor in the equation of only 5%. Even for a much rougher case of = 0 1, the same relativechange of 50% in grain size changes the left side by just 10%. It seems that it does not matter somuch if we cannot specify the bed conditions all that accurately.

2.2 The Gauckler-Manning-Strickler formula

The Gauckler-Manning formulaThe most widely-used resistance formula in river engineering is the Gauckler-Manning formula.

= =1μ ¶2 3

= St

μ ¶2 3(2.10)

where is the Manning coef cient and St = 1 is the Strickler coef cient, used in German-speaking countries. This was originally suggested by Gauckler and then by Manning in thenineteenth century, having observed that variation with seemed to be of this simple power

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form, compared with our equation (2.6) obtained from theory and experiments of the earlytwentieth century, which we wrote as equations (2.7) and (2.8) as

= =

rwhere

1ln12

s=1ln12

(2.11)

An approximation to our formulaWe now show that the Gauckler-Manning formula is an approximation to the theoretical andexperimental expression (2.11).

0

5

10

15

20

25

0.001 0.01 0.1

( )

Relative roughness = ( )

Logarithmic function, eqn (2.8)

Power function, eqn (2.12), = 1 7

Power function, eqn (2.13), = 1 6

Figure 2.6:

On Figure 2.6 is shown how the dimension-less factor varies as a function of relativeroughness , given by equation (2.8) fromexperimental uid mechanics. It is actuallypossible to approximate that curve closelyusing a monomial function . The bestvalues of and can be found by performinga least-squares t. Using 11 points equally-spaced in log between = 0 001 and 0 1,the result obtained is = 1 7 00 (which isa surprising coincidence), and 8 9. Thisgives close agreement with the expression from

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the logarithmic velocity distribution shown in the gure, giving

= 8 9

μs

¶1 7(2.12)

Using this gives us another ow formula

= =8 91 7s

μ ¶9 14where appears simply raised to a power, similar to the Gauckler-Manning formula (2.10).However unlike that, this gives an explicit formula for the coef cient in terms of bed grain size.We might consider this to be an advance, however if we modify our approach slightly we obtain asimilar approximation, which actually gives the well-known and widely-used Gauckler-Manningformula. The logarithmic function is now approximated again, this time by a function 1 6,where is a constant. It can also be determined by performing a least-squares t, giving a value of

7 8 such that

= 7 8

μs

¶1 6(2.13)

with results shown in Figure 2.6, showing that this is also quite a good approximation to thelogarithmic function. Substituting into the ow formula, equation (2.11) and re-writing, we obtain

= =7 81 6s

μ ¶2 3(2.14)

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which is simply the Gauckler-Manning equation but with an explicit expression for the Strick-ler/Manning coef cient:

St =1=7 81 6s

(2.15)

A similar result was obtained by Strickler (nearly a century ago, without optimising software!),based entirely on experiment on boundary roughnesses of equivalent mean diameter from= 0 1mm to = 300mm, (where that diameter was sometimes calculated from alluvial gravel

with relative lengths of the three axes 1:2:3). For the numerical coef cient he obtained a value of4 75 2 6 7, giving his expression

St =6 7

1 6(2.16)

The expression (2.15) here has been obtained by a quite different route, and the agreement betweenthe two expressions, one based on sand grains glued to the inside of a circular pipe carrying air, isencouraging. Of course, for river engineering purposes, Strickler’s result (2.16) is to be preferred.

We call the ow formula in terms of ( )2 3 and coef cients written simply as 1 or St

the Gauckler-Manning (GM) formula, however with the expression (2.16) for St, we call it theGauckler-Manning-Strickler (GMS) formula.

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Sensitivity to boundary particle sizeAs earlier, we examine the effect of uncertainty or variability in the size of the boundary particles(and any perceived ambiguity between s and ), using a power series expansion

St

St=

μ1 +

¶ 1 6

1 =1

6+

and so a fractional change in boundary particle size gives a relative change of 1 6 of that amountin . Again for this form the exact particle size is actually not so important.

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Test of logarithmic and GMS formulae

0.0

0.5

1.0

1.5

2.0

2.5

3.0

0.0 0.5 1.0 1.5 2.0

Boards, rectangularMasonry, trapezoidal - nearly rectangular

Boards, rectangular

Cement, semi-circularCement-sand, semi-circ.Boards, semi-circ.Lined tunnel

Reuss River, Seedorf

Rhine River, St Margrethen

(m/s)

(m)

Logarithmic formula, eqn (2.8)

Gauckler-Manning-Strickler

Figure 2.7: Strickler’s results approximated by two ow formulae

To test the accuracy of the GMS for-mula compared with the logarithmicformula we obtained from experiment,equation (2.11), we consider the re-sults of Strickler (1923, Beilage 4),which have been interpreted as thejusti cation for the exponent 2 3 inthe GMS formula, and leading to the“S” in that name. Strickler consideredresults from nine very different chan-nels. For each the lecturer calculatedthe equivalent s or , constant foreach channel, by least-squares ttingof the appropriate ow formula to thepoints, with results shown in the g-ure. The Gauckler-Manning-Stricklerformula gives agreement generally as good as our logarithmic formula obtained from uid me-chanics experiments, and there seems to be no need to replace it. Using the dimensionless relativeroughness = ( ) to determine resistance seems to have advantages, as set out in thefollowing section.

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Summary: the Gauckler-Manning-Strickler formulaIn the rest of this course, we sometimes write the Gauckler-Manning formula conventionally as

= =1μ ¶2 3

= St

μ ¶2 3but if we use the Strickler expression St = 6 7

1 6 we write it in the dimensionless formwhich we have called the Gauckler-Manning-Strickler formula:

= =

rwhere ( ) =

6 71 6, and = (2.17)

• We no longer have the problem that St or have dif cult units ( : L 1 3T).• The characterisation of the resistance has been reduced to that of the dimensionless relativeroughness = ( ). There are no problems with confusion of Imperial/SI units in varioussimilar formulae which group terms differently and contain terms like ( )2 3 and 1 6.

• To use the formula, we do not have to imagine a value of or St, which have no simplephysical signi cance. We do not have to look at pictures of rivers in standard references suchas Chow (1959, §5-9&10). We do not have to ring a friend to see what they used for a similarstream 20 km distant some years ago. Instead, we can always use an estimate of .

• Of course, resistance includes also that due to bed forms and vegetation, but expressing it interms of is a good basis, with an understandable quantity = ( ).

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2.3 Boundary stress in compound channels and unsteady non-uniform owsNow we relate our results to show how they relate to the computation of boundary resistancein more complicated situations. While doing this we obtain a well-known alternative resistanceformula, which is based on a simpler approximation to our above results.

The Chézy-Weisbach ow formulaWriting shear stress in terms of the result obtained from the Darcy-Weisbach formulation of owresistance in pipes,

= 18

2 (2.18)

where is the Weisbach dimensionless resistance coef cient, expressing the relationship betweenvelocity and stress. The factor of 1 8 is necessary to agree with the Darcy-Weisbach energyformulation of pipe ow theory in circular pipes where = Diameter 4. From our simpleforce balance we already have equation (2.1): =

p( ) . Eliminating between

equation (2.18) and this gives the Chézy-Weisbach ow formula

= =

r8

=

r(2.19)

where =p8 is the Chézy coef cient, named after the French military engineer who rst

presented such an open channel ow formula in 1775. Comparing our GMS formulation equation

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(2.17) we see that it is in the same form, such that

=

r8

(2.20)

so that our is clearly related to the resistance coef cient , but in the GMS expression, was afunction of = ( ).

Generalised resistance coef cientIt is often more useful for us to introduce and use the resistance coef cient such that

=8=12

(2.21)

where we do not necessarily consider them constant, but where ( ) in general is as given byequation (2.17). In this case, the boundary stress is given by

= 2 (2.22)

Non-uniform and unsteady owsWe will be considering ows which are not uniform (vary with position ) and those which areneither uniform nor steady (vary also with time ). As the length scale of river ows is much longerin space than the cross-sectional dimensions and the time scale of disturbances is much longer thanthat of local turbulence, we will assume that the boundary stress at each place and at each timeis given by the local immediate ow conditions of velocity, in terms of discharge and area .

28

From equations (2.22) and (2.20) we have

= 2 =

μ( )

( )

¶2=

12( )

μ( )

( )

¶2(2.23)

Compound cross-sections

2

1

3

Figure 2.8: Compound cross-section

We consider compound cross-sections such as shownin Figure 2.8. Using equation (2.23) we can obtain anexpression for the boundary shear force per unit lengthof channel in each is, generalising equation (2.22) andmultiplying by the perimeter of each part:

= ( )2

2 for = 1 2

Note that the internal shear forces across faces 1-2 and 1-3 cancel. However, we do not know theindividual discharges of the three sections. An approximation would be to neglect interfacialshear and obtain the discharges for each component from the GMS equation.

The total gravitational component, force per unit length isX3

=1

where we have assumed that the slope is the same for each part.

To solve the steady ow problem then, if we might write = , where is the total29

ow.summing the components due to boundary force and equating to the gravitational componentwe obtain

2X3

=1

( ) 2

2 =X3

=1

that we could use to calculate the total ow or more complicated deductions.

Be very sceptical of the apparently simple formulae for combining St or Manning’s appearingin many books, as exempli ed by the list of 17 different compound or composite section formulaein Yen (2002, table 3), or Cowan’s formula (see page 36 of Yen’s paper) = (

P) , where

the are different contributions from surface roughness, shape and size of channel cross-section,etc., which is irrational nonsense. The proper way to proceed is by a linear sum of the forces as wehave done here.

30

2.4 General situations

Presence of different resistance elementsIn many cases the conditions in the river are more complicated than just a layer of uniform regularparticles. For example:• Irregular and variable nature of the bed particle arrangement.

The variability of resistance in real streams is often much greater than has been realised, for thearrangement of the bed “grains” or “particles” (even if they are 30 cm boulders) is very impor-tant, and can change continuously, depending on the ow history. Nikuradse’s experiments werefor sand grains levelled so that their tops were co-planar, and hence most of the particles wereshielded from the ow and resistance was small. That is what a bed looks like after a long periodof constant ow, when any individually-projecting grains have been removed, because the forceon them was larger. The bed is said to have been “armoured” – not only is the resistance small,but individual grains are hard to remove. After a sudden increase of ow, particles are morelikely to have been dislodged, moved, and deposited, leaving a random surface, where those

31

most projecting exert larger force on the uid and resistance is greater.• Bed forms – ripples, dunes, anti-dunes etc. •

Typical bedforms (after Richardson and Simons)The bed-forms which can develop if the bed is mobile will also contribute to variable resistance.

• Particle movement – if the grains are actually moving, then the force required to move the grainsappears to the water as an additional stress, whether they are moving along the bed, rolling,jumping, or carried suspended in the ow.

• Vegetation – trees (standing and/or fallen), grasses, reeds etcIt can be seen that the problem of the current instantaneous resistance in the stream is actually avery dif cult and uncertain one ...

32

Results for resistance coef cients in real riversHere we attempt to obtain understanding and a formula for the resistance coef cient using resultsfrom a number of eld measurements. To compare with several experimental works, we use theChézy-Weisbach formulation. We considered the results of Hicks & Mason (1991), a catalogueof 558 stream-gaugings from 78 river and canal reaches in New Zealand, of which 55 were siteswith grading curves for boundary material, so that particle sizes were known. Neither vegetationnor bed-form resistance can be isolated. Hicks & Mason based their approach on Barnes (1967),who provided values of Manning’s resistance coef cient = 1 St for a single ow at each of 50separate river sites in the United States of America, of which boundary material details were givenfor 14. We also include those results here.

From both catalogues we took the values of 84, the boundary particle size for which 84%of the material was ner, and from the values of , calculated the relative roughness84 = 84 ( ), and used the measured values of Chézy’s to calculate values of( =

p8 ). Results are shown on the next page. We have plotted them for a parameter

= 8, which it will be more convenient for us to use later.

33

10 3

10 2

10 1

100

10 3 10 2 10 1 100

= 0 0 — co-planar bed

= 0 5 — irregular

= 1 0 — fully exposed grains

= 2 0 — moving

=8

Relative roughness 84

Barnes (1967)Hicks & Mason (1991) — stable bedHicks & Mason (1991) — moving bedEq. (2.24)Aberle & Smart (2003, eqn 9)Pagliara et al. (2008, eqn 15), = 0, = 0

Ditto, but = 0 2

Yen (2003), Eq. (2.25)Strickler, Eq. (??)

34

• Many of the results from each study are for large bed material 84 0 1, possibly a re ection ofthe hilly and mountainous nature of New Zealand and Paci c North-West of the United Statesof America (and which applies to Austria ...).

• There is a wide scatter of results. But not all that very wide if we consider that the streams rangefrom large slow-moving rivers with extremely small grains to mountain torrents with 30 cmboulders. Most of the results, unless the grains are moving, fall between 0 005 and 0 02.

• There is, as we have seen, slow variation with relative roughness: am increase in by a factorof 10 leads to an increase in of about 2, as we have already seen.

• The points, we believe, have a tendency to group around three of the curves shown and the restto be bounded below by the fourth (upper) curve shown. The curves have been drawn using theexpression, found by trial and error:

=0 06 + 0 06

(1 0 0 6 ln 84)2 (2.24)

with values of = 0 0 5, 1, and 2. The parameter is an arbitrary one that we use to identifythe state of the particles making up the bed. This will now be explained.

• The rst grouping of points comprises those around the bottom curve. We hypothesise that thesepoints, having the lowest resistance, are those forming beds where the particles are relativelyco-planar such that the bed is armoured. We assigned = 0 to this state, and used that inequation (2.24) to plot the curve.

• The next grouping of points is around the second curve from the bottom, which can be seen35

to substantially coincide with the a curve corresponding to exposed boulders on top of the bedoccupying 0.2 of the surface area. Of course, with a number of these grains thus exposed, theresistance is greater. We assigned a value of = 0 5 to this intermediate state.

• Substituting = 1 in equation (2.24) gives the third curve on the gure, passing throughwhat we believe is the third grouping of particles. This is probably the state for the maximumresistance for a stable bed corresponding to exposed grains occupying something like 50% ofthe surface area. Any more such grains will cause shielding of particles, the bed will start toresemble the co-planar case, and resistance will actually be reduced.

• Further evidence supporting our assertions is obtained from the expression proposed by Yen(2002, eqn 19), who considered results from a number of experimental studies using xedimpermeable beds. We used his formula, converted to = 8, used an in nite Reynoldsnumber, and converted his equivalent sand roughness = 2 84. It can be seen that the curvepasses (left to right) from our curve = 1 for small particles, which are unlikely to have thetops levelled so that particles are exposed, to the second curve for larger particles, more likely tobe levelled in the laboratory experiments, with = 0. Yen obtained the approximation for :

=1

4

μlog10

μ112

s+1 95

R0 9

¶¶ 2

(2.25)

where R is the channel Reynolds number R = ( ) , in which is the kinematic viscosity.• The logarithmic formula we obtained above, equation (2.8), leads to, if we use = 2 84, and

36

as =p8 = 1 ,

=

μ1ln11¶ 2

= (6 0 2 5 ln (2 84))2 (2.26)

giving results quite similar to those from Yen’s formula.• For points above the third curve almost all experimental points had shear stresses greater thanthe critical one necessary for movement. If particles move, not only do many particles protrudeabove others, increasing the stress, but there is the additional force required to maintain thesliding and rolling and jostling of all the particles. Hence, the resistance is greater. And, if thereis a need to maintain particles in suspension, that will contribute also to resistance. We haveshown the fourth curve as drawn for = 2.

Hopefully the gure and approximating curves have given us an idea of the magnitudes andvariation of the quantities, and maybe even some results for use in practice.

37

2.5 Computation of normal owAt last we turn to a common practical problem in River Engineering. "Normal ow" is the namegiven to a uniform ow, and the depth is called the normal depth. If the discharge , slope ,resistance coef cient St = 1 , and the relationship between area and depth and perimeter anddepth are known, the GMS formula becomes a transcendental equation for the normal depth . Tosolve this is a common problem in river engineering

A numerical methodAny method for the numerical solution of transcendental equations can be used, such as Newton’smethod. Here we develop a simple method based on direct iteration, where we develop a trick,giving us rapid convergence.

In the case of wide channels, (i.e. channels rather wider than they are deep, a common case), thewetted perimeter does not vary much with depth . Similarly the width does not vary much with. Consider the GM formula in the conventional form, written now

= St

5 3

2 3

we divide both sides by 5 3, and showing functional dependence of and on :

5 3= St

( ( ) )5 3

2 3( )

The term ( ) is approximately the width of the channel, which for many channels varies little38

with , as does the perimeter ( ). So, the right side of the equation varies slowly with , soby isolating the 5 3 term and taking the 3 5 power of both sides of the equation, we obtain theequation in a form suitable for direct iteration

=

μSt

¶3 5×

2 5( )

( )(2.27)

where the rst term on the right is a constant for any particular problem, and the second termvaries slowly with depth – a primary requirement that the direct iteration scheme be convergentand indeed be quickly convergent.

For an initial estimate we suggest making a rough estimate of the approximate width 0 and so,making a wide channel approximation, setting ( ) 0 and ( ) 0, in the generalscheme of (2.27) gives

0 =

μSt 0

¶3 5(2.28)

Experience with typical trapezoidal sections shows that the method works well and is quicklyconvergent.

39

Trapezoidal section

1

Most canals are excavated to a trapezoidal section,and this is often used as a convenient approximationto river cross-sections too. In many of the problemsin this course we will consider the case of trapezoidalsections. Consider the quantities shown in the gure:the bottom width is , the depth is , the top width is , and the batter slope, de ned to be theratio of H:V dimensions is . Geometrically, = + 2 , area = ( + ), wettedperimeter = + 2 1 + 2 .

Example 1 Calculate the normal depth in a trapezoidal channel of slope 0.001, St = 25, bot-tom width = 10m, with batter slopes = 2, carrying a ow of 20 m3s 1. We have =(10 + 2 ), = 10 + 4 472 . For 0 we use = 10m. Equation (2.28) gives

0 =

μSt 0

¶3 5=

μ20

25× 10 0 001

¶3 5= 1 745m

Then, equation (2.27) gives

+1 =

μSt 0

¶3 5× (10 + 4 472 )2 5

10 + 2= 6 948× (10 + 4 472 )2 5

10 + 2

With 0 = 1 745, 1 = 1 629, 2 = 1 639, 3 = 1 638m, and the method has converged.

40

3. Froude numberWilliam Froude (1810-1879, pronounced as in "food") was a naval architect who proposedsimilarity rules for free-surface ows. A Froude number is a dimensionless number from a velocityscale and a length scale , F = In the original de nition, of a ship in deep water, theonly length scale was , the length of the ship. In river engineering it is not obvious what thelength scale is. Might it be the wetted perimeter , might it be the geometric mean depth ,where is cross-sectional area and is surface width?

In fact, the answer will usually depend on the problem. If we consider the mean total head of achannel ow at a section, where is mean velocity and is surface elevation:

= +2

2

it does not appear explicitly. Neither does it appear in the momentum ux at a section

=¡¯ + 2

¢Here, we are going simply to de ne it in terms of a vertical length scale, the depth scale :

F2 =2

=2

3(3.1)

where = is discharge. For many years the lecturer was more speci c, using terms ofkinetic and potential energy, now all he says is that F2 is a measure of dynamic effects relative togravitational, the latter measured by mean depth.

41

Even if we were to consider rather more complicated problems such as the unsteady propagation ofwaves and oods, and to non-dimensionalise the equations, we would nd that the Froude numberF itself never appears in the equations, but always as F2 or F2, depending on whether energy ormomentum considerations are being used.

Flows which are fast and shallow have large Froude numbers, and those which are slow and deephave small Froude numbers. Generally F2 is an expression of the wave-making ability of a ow,and in conversation we usually use “high/ low Froude number” as an expression of how fast a owis. For example, consider a river or canal which is 2 m deep owing at 0 5 ms 1 (make some effortto imagine it - we can well believe that it would be able to ow with little surface disturbance!).We have

F =0 5

10× 2 = 0 11 and F2 = 0 012

and we can imagine that the wavemaking effects are small. Now consider ow in a street gutterafter rain. The velocity might also be 0 5 ms 1, while the depth might be as little as 2 cm. TheFroude number is

F =0 5

10× 0 02 = 1 1 and F2 = 1 2

and we can easily imagine it to have many waves and disturbances on it due to irregularities in thegutter.

42

Near-constancy of Froude number in a streamIt is interesting to calculate the Froude number F of a steady uniform ow given by the Chézy-Weisbach formula for discharge:

=

r3

Immediately this gives

F2 =2

3=

and as for wide channels, we see that the square of the Froude number is approximatelyequal to the ratio of bed slope to resistance coef cient , giving some signi cance and physicalfeeling for . This means that for a particular reach of river, where slope is effectivelyindependent of ow, where also does not vary much with the ow and often does not varymuch, the Froude number F does not change much with ow. While a ood might look moredramatic than a more-common low ow, because it is faster and higher, the Froude number isroughly the same for both.

43

4. The effect of obstructions on streams – an approximatemethod

River Traun, Bad Ischl, Oberösterreich

Structures such as weirs can almost completely blocka river, but there are also other types of obstaclesthat are only a partial blockage, such as the piersof a bridge, blocks on the bed, Iowa vanes, etc. orpossibly more importantly, the effects of trees placedin rivers (”Large Woody Debris”), used in theirenvironmental rehabilitation. It might be importantto know what the forces on the obstacles are, or,more importantly for us, what effects the obstacleshave on the river. Here we set up the problem inconventional open channel theoretical terms. Then,however, we obtain an analytical solution by makingthe approximation that the effects of the obstacle aresmall. This gives us more insight into the nature andimportance of the problem.

44

The physical problem and its idealisationSurface if no obstacle: slowly-varying owSurface along axis and sides of obstacleMean of surface elevation across channel

1 2

1 2

12

(a) The physical problem, longitudinal section showing backwaterat obstacle decaying upstream to zero

(b) The idealised problem, uniform channel with no friction or slope

Figure 4.1: A typical physical problem of ow past a bridge pier, and its idealisation for hydraulic purposes

45

Momentum ux in a channelThe momentum ux across a section is de ned to be the sum of the pressure force, plus the massrate of transport multiplied by the velocity. For a vertical section, the mass rate of transport isd , so the momentum ux is

=

Z ¡+ 2

¢d

Substituting the hydrostatic pressure distribution, = ( ), where is the free surfaceelevation, we obtain

=

Z ¡( ) + 2

¢(4.1)

• The integral R ( ) is simply the rst moment of area about a transverse horizontal axisat the surface, we can write it as R

( ) = ¯ (4.2)

where ¯ is the depth of the centroid of the section below the surface.• The uid inertia contribution

R2 : although it has not been written explicitly, it is

understood that equation (4.1) is evaluated in a time mean sense. In equation (1.2) we saw thatif a ow is turbulent, then 2 = ¯2 + 02, such that the time mean of the square of the velocityis greater than the square of the mean velocity. In this way, we should include the effects of

46

turbulence in the inertial momentum ux by writing the integral on the right of equation (4.1)Z2 =

Z ³¯2 + 02

´(4.3)

Usually we do not know the nature of the turbulence structure, or even the actual velocitydistribution across the ow, so that we approximate this in a simple sense such that we write forthe integral in space of the time mean of the squared velocities:Z

2 =

Z ³¯2 + 02

´¯ 2 =

μ ¶2=

2

(4.4)

The coef cient is called a Boussinesq coef cient, after the French engineer who introducedit to allow for the spatial variation of velocity. Allowing for the effects of time variation,turbulence, has been a recent addition. has typical values of 1 05 to something like 1 5 or morein channels of irregular cross-section. Almost all textbooks introduce this quantity for openchannel ow (without turbulence) but then assume it is equal to 1. In this course we consider itimportant and will include it.

In equation (4.1), collecting contributions (4.2) and (4.4), we have the expression for the momentumux at a section

=

μ¯ +

2¶

(4.5)

47

Momentum conservationConsider the momentum conservation equation if a force is applied in a negative direction to aow between two sections 1 and 2:

=

μ¯ +

2¶1

μ¯ +

2¶2

(4.6)

Usually one wants to calculate the effect of the obstacle on water levels. The effects of drag can beestimated by knowing the area of the object measured transverse to the ow, , the drag coef cientD, and , the mean uid speed past the object:

=1

2D

2 (4.7)

and so, substituting into equation (4.6) gives, after dividing by density,1

2D

2 =

μ¯ +

2¶1

μ¯ +

2¶2

(4.8)

We consider the velocity on the obstacle as being proportional to the upstream velocity, such thatwe write

2 =

μ1

¶2(4.9)

where is a coef cient which recognises that the velocity which impinges on the object isgenerally not equal to the mean velocity in the ow. For a small object near the bed, could bequite small; for an object near the surface it will be slightly greater than 1; for objects of a vertical

48

scale that of the whole depth, 1. Equation (4.8) becomes1

2D

2

21

=

μ¯ +

2¶1

μ¯ +

2¶2

(4.10)

A typical problem is where the downstream water level is given (sub-critical ow, so that thecontrol is downstream), and we want to know by how much the water level will be raised upstreamif an obstacle is installed. As both 1 and 1 are functions of 1, so that we would need to know indetail the geometry of the stream, and then to solve the transcendental equation for 1. However,by linearising the problem, solving it approximately, we obtain a simple explicit solution that tellsus rather more.

2¯2

¯1

2

2

1 1

2

1

Figure 4.2: Cross-section showing dimensions for water levels at 1 and 2

49

Consider the stream cross-section shown in Fig. 4.2, with a small change in water level1 = 2+ . We now use geometry to obtain approximate expressions for quantities at 1 in termsof those at 2. It is easily shown that

1 = 2 + 2 +³( )2

´and

¡¯¢1=¡¯¢2+ 2 +

³( )2

´and similarly we write for the blockage area 1 = 2+ 2 +

³( )2

´, where 2 is the surface

width of the obstacle (which for a submerged obstacle would be zero). We have actually been usingTaylor series expansions, but the physical interpretations seem simpler than the mathematical!

The momentum equation (4.10) gives us1

2D

2

212 = 2 +

2

2 + 2

2

2

Now we use a power series expansion in to simplify the term in the denominator:1

2 + 2=

1

2 (1 + 2 2)=1

2(1 + 2 2)

1 1

2(1 2 2)

neglecting terms like ( )2 (see equation A-1). The momentum equation becomes1

2D

2

212 2

μ1

2232

¶As 1 = 2 + ( ) we replace 1 by 2 and introduce F22 = 2

232, the square of

the Froude number of the downstream ow. The equation is easily solved to give an explicit50

approximation for the dimensionless drop across the obstacle ( 2 2), where 2 2 is themean downstream depth:

2 2=

12 D F221 F22

2

2(4.11)

This explicit approximate solution has revealed the important quantities of the problem to us andhow they affect the result: downstream Froude number F22 = 2

232 and the relative blockage

area 2 2. For subcritical ow F22 1 the denominator in (2.27) is positive, and so is , sothat the surface drops from 1 to 2, as we expect. If the ow is supercritical, F22 1, we ndnegative, and the surface rises between 1 and 2. If the ow is near critical F22 1, the change indepth will be large, which is made explicit, and the theory will not be valid.

We could immediately estimate how important this is. We see that, for small Froude numberF22 ¿ 1, such that 1 F22 1, the relative change of depth is equal to 12 times 1 (for abody extending the whole depth), times D 1 for cylinders etc , multiplied by F22, usually small,multiplied by the blockage ratio 2 2, which is also probably small. So, the relative result isusually small. However, the absolute value might still be nite compared with resistance losses, aswill be seen below.

Another bene t of the approximate analytical solution is that it shows that such an obstacle formsa control in the channel, so that the nite sudden change in surface elevation is a function of2, or a function of , in a manner analogous to a weir. In numerical river models it should

ideally be included as an internal boundary condition between different reaches as if it were a typeof xed control.

51

The mathematical step of linearising has revealed much to us about the nature of the problem thatthe original momentum equation did not.

Example 2 It is proposed to build a bridge, where the bridge piers occupy about 10% of the "wettedarea" of a river with Froude number 0 5 (which is quite large). How much effect will this have onthe river level upstream?

As the bridge piers occupy all the depth, we have = 1. A typical drag coef cient is D 1. Wewill use = 1 (this is an estimate!). So we nd, using equation (2.27):

2 2=

12 D F221 F22

2

2

12 × 1× 1×

0 52

1 0 52× 0 1

= 0 017,

about 2% of the mean depth. This seems small, but if the river were 2 m deep, there is a 4 cm dropacross the bridge. If the slope of the river were = 10 4, this would correspond to the surfacelevel change in a length of 400 m, which can hardly be neglected.

52

5. Reservoir routing

Surface Area

= +

=

( )

( ( ) )

Surface Area +

Figure 5.1: Reservoir or tank, showing surface level varyingwith in ow, determining the rate of out ow

Consider the problem shown in gure 5.1,where a generally unsteady in ow rate ( )enters a reservoir or a storage tank, and wehave to calculate what the out ow rate ( )is, as a function of time . The action of thereservoir is usually to store water, and torelease it more slowly, so that the out owis delayed and the maximum value is lessthan the maximum in ow. Some reservoirs,notably in urban areas, are installed justfor this purpose, and are called detention

reservoirs or storages. The procedure of solving the problem is also called Level-pool Routing.

Time

In ow

Out ow

Figure 5.2:

The process is shown in gure 5.2. When a ood comes down theriver, in ow increases, the water level rises in the reservoir until atthe point O when the out ow over the spillway now balances thein ow. At this point, out ow and surface elevation in the reservoirhave a maximum. After this, the in ow might reduce quickly, butit still takes some time for the extra volume of water to leave thereservoir.

53

Detention reservoir in a public parkin Melbourne, Australia

It is simple and obvious to write down the relation-ship stating that the rate of surface rise d d isequal to the net rate of volume increase divided bysurface area:

d

d=

( ) ( )

( )(5.1)

where is the free surface elevation, and ( ) isthe surface area, possibly given from planimetricinformation from contour maps, and ( ) is thevolume rate of out ow, which is usually a simplefunction of the surface elevation , from a weir or

gate formula, usually involving terms like ( outlet)1 2 and/or ( crest)

3 2, where outlet is theelevation of the pipe or tailrace outlet to atmosphere and crest is the elevation of the spillway crest.There might be extra dependence on time if the out ow device is opened or closed. This is adifferential equation for the surface elevation itself. The procedure of solving it is called Level-poolRouting.

The traditional method of solving the problem, described in almost all books on hydrology, is touse an unnecessarily complicated method called the “Modi ed Puls” method of routing, whichsolves a transcendental equation for a single unknown quantity, the volume in the reservoir, ateach time step. It is simpler and more fundamental to treat the problem as a differential equation(Fenton 1992). :-)

54

Numerical solution of the differential equation by Euler’s methodEuler’s method is the simplest (but least-accurate) of all methods, being of rst-order accuracyonly. For river engineering purposes it is usually quite good enough. However there is a goodmethod for making it more accurate, which we will use. Euler’s method is to approximate thederivative in a differential equation at a time step by a forward difference expression in terms of atime step , here applying it to equation (5.1):

d

d

¯̄̄̄+1 =

( ) ( )

( )

giving the scheme to calculate the value of at +1 as

+1 = +( ) ( )

( )+

¡2¢

(5.2)

where we use the notation for the solution at time step . We have shown that the error of thisapproximation is proportional to 2. It is necessary to take small enough that this is small.

Accurate results with simple methods – Richardson extrapolationWe introduce a clever device for obtaining more accurate solutions from Euler’s method and others.

Consider the numerical value of any part of a computational solution for some physical quantityobtained using a time or space step , such that we write ( ). Let the computational scheme beof known th order such that the global error of the scheme at any point or time is proportional to

55

, then if (0) is the exact solution, we can write the expression in terms of the error at order :

( ) = (0) + + (5.3)

where (0) is the solution for a vanishingly small time step, so that it should be exact. The is anunknown coef cient; the neglected terms vary like +1. If we have two numerical simulations orapproximations with two different 1 and 2 giving numerical values 1 = ( 1) and 2 = ( 2)then we write (5.3) for each:

1 = (0) + 1 +

2 = (0) + 2 +

These are two linear equations in the two unknowns (0) and . Eliminating , which is notimportant, between the two equations and neglecting the terms omitted, we can solve for (0), anapproximation to the exact solution:

(0) = 2 1

1+

¡+11

+12

¢(5.4)

where = 2 1. The errors are now proportional to step size to the power + 1, so that wehave gained a higher-order scheme without having to implement any more sophisticated numericalmethods, just with a simple numerical calculation. This procedure, where is known, is calledRichardson extrapolation to the limit.1. For simple Euler time-stepping solutions of ordinary differential equations, = 1, and if we

56

perform two simulations, one with a time step and then one with 2, we have

( 0) = 2 ( 2) ( ) +¡

2¢

(5.5)

where the numerical solution at time has been shown as a function of the step. This is verysimply implemented.

2. For the evaluation of an integral by the trapezoidal rule, = 2.

Example 3 Consider a small detention reservoir, square in plan, with dimensions 100m by 100m,with water level at the crest of a sharp-crested weir of length of = 4m, where the out ow overthe sharp-crested weir can be taken to be

( ) = 0 6 3 2 (5.6)

where = 9 8 ms 2. The surrounding land has a slope (V:H) of about 1:2, so that the length of areservoir side is 100 + 2× 2× , where is the surface elevation relative to the weir crest, and

( ) = (100 + 4 )2 .

The in ow hydrograph is:

( ) = min + ( max min)

μmax

1 max

¶5(5.7)

where the event starts at = 0 with min and has a maximum max at = max. This general formof in ow hydrograph mimics a typical storm, with a sudden rise and slower fall, and will be used inother places in this course. In the present example we consider a typical sudden local storm event,

57

with min = 1 m3s 1 at max = 1800 s.

0

5

10

15

20

0 1000 2000 3000 4000 5000 6000

Discharge¡m3 s 1

¢

Time (sec)

In owOut ow — accurateEuler — step 200sEuler — step 100sRichardson extrapn

Figure 5.3: Computational results for the routing of a sudden storm through a small detention reservoir

The problem was solved with an accurate 4th-order Runge-Kutta scheme, and the results are shownas a solid blue line on gure 5.3, to provide a basis for comparison. Next, Euler’s method (equation5.2) was used with 30 steps of 200 s, with results that are barely acceptable. Halving the time stepto 100 s and taking 60 steps gave the slightly better results shown. It seems, as expected fromknowledge of the behaviour of the global error of the Euler method, that it has been halved at each

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point. Next, applying Richardson extrapolation, equation (5.5), gave the results shown by the solidpoints. They almost coincide with the accurate solution, and cross the in ow hydrograph with anapparent horizontal gradient, as required, whereas the less-accurate results do not. Overall, it seemsthat the simplest Euler method can be used, but is better together with Richardson extrapolation. Infact, there was nothing in this example that required large time steps – a simpler approach mighthave been just to take rather smaller steps.

The role of the detention reservoir in reducing the maximum ow from 20 m3s 1 to 14 7 m3s 1 isclear. If one wanted a larger reduction, it would require a larger spillway. It is possible in practicethat this problem might have been solved in an inverse sense, to determine the spillway length for agiven maximum out ow.

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