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River mechanics This textbook offers a thorough mechanical analysis of rivers from upland areas to oceans. It scrutinizes stateoftheart methods, underlining both theory and engineering applications. Each chapter includes a presentation of fundamental principles, followed with an engineering analysis and instructive problems, examples, and case studies illustrating engineering design. The emphasis is on river equilibrium, river dynamics, bank stabilization, and river engineering. Channel stability and river dynamics are examined in terms of river morphology, lateral migration, aggradation, and degradation. The text provides a detailed treatment of riverbank stabilization and engineering methods. Separate chapters cover physical and mathematical models of rivers. This textbook also contains essential reading for understanding the mechanics behind the formation and propagation of devastating ﬂoods, and offers knowledge crucial to the design of appropriate countermeasures to reduce ﬂood impact, prevent bank erosion, improve navigation, increase water supply, and maintain suitable aquatic habitat. More than 100 exercises (including computer problems) and nearly 20 case studies enhance graduatestudent learning, while researchers and practitioners seeking broad technical expertise will ﬁnd it a valuable reference. Pierre Y. Julien is Professor of Civil Engineering at Colorado State University.
River mechanics P I E R R E Y. J U L I E N
Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo Cambridge University Press The Edinburgh Building, Cambridge , United Kingdom Published in the United States of America by Cambridge University Press, New York www.cambridge.org Information on this title: www.cambridge.org/9780521562843 © Cambridge University Press 2002 This book is in copyright. Subject to statutory exception and to the provision of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published in print format 2002  
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Cambridge University Press has no responsibility for the persistence or accuracy of s for external or thirdparty internet websites referred to in this book, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate.
Dedicated to Helga and Patrick
Contents
Preface Notation
page xi xiii
1 Introduction to river mechanics
1
2 Physical properties and equations 2.1. Dimensions and units 2.2. Properties of water 2.3. Properties of sediment 2.4. River ﬂow kinematics 2.5. Conservation of mass 2.6. Equations of motion 2.7. Hydraulic and energy grade lines
9 9 10 13 22 24 25 27
3 River basins 3.1. Riverbasin characteristics 3.2. Rainfall precipitation 3.3. Interception and inﬁltration 3.4. Excess rainfall 3.5. Surface runoff 3.6. Uplanderosion losses 3.7. Sediment source and yield
31 31 35 46 51 53 63 72
4 Steady ﬂow in rivers 4.1. Steady river ﬂow 4.2. Steadynonuniform river ﬂow 4.3. Sediment transport in rivers
79 79 97 111
5 Unsteady ﬂow in rivers 5.1. River continuity equation 5.2. River momentum equations 5.3. River ﬂoodwaves
122 122 126 130
vii
viii
Contents 5.4. Looprating curves 5.5. River ﬂood routing 5.6. River ﬂow and sedimentduration curves
136 139 142
6 River equilibrium 6.1. Particle stability 6.2. Channel stability 6.3. Regime relationships 6.4. Equilibrium in river bends 6.5. Downstream hydraulic geometry 6.6. Bars in alluvial rivers 6.7. River meandering 6.8. Lateral river migration
158 158 163 165 166 170 177 179 184
7 River dynamics 7.1. River dynamics 7.2. Riverbed degradation 7.3. Riverbed aggradation 7.4. River conﬂuences and branches 7.5. River databases
199 199 204 214 221 226
8 River stabilization 8.1. Riverbank stability 8.2. Riverbank riprap revetment 8.3. Riverbank protection 8.4. River ﬂowcontrol structures 8.5. Riverbank engineering
234 234 238 253 267 282
9 River engineering 9.1. River ﬂood control 9.2. River closure 9.3. Canal headworks 9.4. Bridge scour 9.5. Navigation waterways 9.6. Dredging
286 286 297 306 310 316 325
10 Physical river models 10.1. Hydraulic similitude 10.2. Rigidbed model 10.3. Mobilebed river models
334 334 337 342
Contents
ix
11 Mathematical river models 11.1. Finitedifference approximations 11.2. Onedimensional river models 11.3. Multidimensional river models
352 353 365 376
12 Waves and tides in river estuaries 12.1. Surface waves 12.2. Tides in river estuaries 12.3. Saline wedges in river estuaries
385 385 396 401
Bibliography Index
409 427
Preface
Rivers have fascinated humanity for centuries. Most prosperous cities around the world have been founded along rivers. Today, river engineers are still designing structures to draw beneﬁts from the ﬂuvial system for developing societies. It is clear that river engineering is not based solely on a simple understanding of local hydrodynamic forces, but also on an encompassing knowledge of the watershed that supplies water and sediment to dynamic river systems. Expertise in river mechanics combines knowledge of watershed climatology, geomorphology, and hydrology, with a deep understanding of hydrodynamic forces governing the motion of water and sediment in complex river systems. Stateoftheart teaching of river mechanics clearly requires study material that emphasizes both theoretical concepts and practical engineering technology. Ideally, scientists should develop new concepts that may be applicable to engineering design, and practitioners should understand why certain structures work and why others fail. This textbook has been prepared for engineers and scientists seeking broadbased technical expertise in river mechanics. It has been speciﬁcally designed for graduate students, for scholars actively pursuing scientiﬁc research, and for practitioners keeping up with developments in river mechanics. The prerequisites simply include basic knowledge of undergraduate ﬂuid mechanics and partial differential equations. The textbook Erosion and Sedimentation, by the same author and Cambridge University Press, serves as prerequisite material for the graduate course on river mechanics taught at Colorado State University. Rather than a voluminous encyclopedia, this textbook scrutinizes a selected number of methods to meet pedagogical objectives underlining both theory and engineering applications. This text has been designed to be covered within a regular 45lecturehour graduatelevel course. The chapters of this book contain, besides theory and lecture material, various exercises, general problems, data sets, computer problems, examples, and case studies. They illustrate speciﬁc aspects of the profession from theoretical derivations through exercises, to practical solutions to real problems through xi
xii
Preface
the analysis of case studies. Most problems can be solved with a few algebraic equations; others require the use of computers. The computer problems offer students the opportunity to develop skills for solving physical problems with computers. No speciﬁc computer code or language is required. Instead of using existing software packages, I stimulate student creativity and originality in developing the students’ own computer programs. Throughout, a solid diamond () denotes equations and problems of particular signiﬁcance. Problems with a double diamond () are considered most important. I am grateful to D. B. Simons, E. V. Richardson, H. W. Shen, H. Rouse, M. Frenette, J. L. Verrette, Y. Ouellet, C. F. Nordin, S. Schumm, S. R. Abt, and J. Ruff, who inﬂuenced my teaching over the years. I am also particularly thankful to P. G. Combs and D. C. Baird for sharing their practical expertise in river engineering. This book also beneﬁts from numerous suggestions formulated by a generation of graduate students at Colorado State University. They helped me tailor this textbook to meet their needs under the constraints of quality, concision, and affordability. Jenifer Davis diligently typed successive draft versions of the manuscript, and Jean Parent prepared the ﬁgures. Finally, it has been a renewed pleasure to collaborate with Florence Padgett, Ellen Carlin, Zach Dorsey, and the Cambridge University Press production staff.
Notation
Symbols a a a acent acor ai ai+1 at aθ a˜ a, b a, b A Aa Asb At ˜ B˜ A, br c cG C Ca Cs Cﬂ Ck C0i Cr Cu
acceleration reference elevation pier width centrifugal acceleration Coriolis acceleration incremental crosssection area coefﬁcient of h i+1 partial watershed area particlestability coefﬁcient wave amplitude coefﬁcients of the resistance equation transform coefﬁcients for duration curves surface area amplitude factor surface area of a settling basin watershed drainage area wave coefﬁcients riverbend coefﬁcient wave celerity group velocity Ch´ezy coefﬁcient reference concentration Courant number Courant–Friedrich–Levy coefﬁcient grid dispersion number upstream sediment concentration runoff coefﬁcient velocity Courant number xiii
xiv
Notation
Cv , Cw , Cppm , Cmg/l d10 , d50 dm ds d∗ D Dd Dp e E E tons E˜ E() E f fl f (t) F F˜ FB Fc FD Fg Fh Fi FL FM Fp Fs FS Fw FW F() F(t) Fa (t) F p (t) Fr g G Gr
sediment concentration particle size distribution, % ﬁner by weight effective riprap size particle size dimensionless particle diameter pipe diameter degreedays drop height of a gradecontrol structure void ratio speciﬁc energy expected soil loss in tons total energy of a wave exceedance probability speciﬁc energy lost in a hydraulic jump Darcy–Weisbach friction factor Lacey silt factor inﬁltration rate force fetch length of wind waves buoyancy force centrifugal force drag force gravitational force hydrodynamic force inertial force lift force momentum force pressure force shear force in a bend submerged weight of a particle weight of water weight of a particle nonexceedance probability cumulative inﬁltration actual cumulative inﬁltration potential cumulative inﬁltration Froude number gravitational acceleration speciﬁc gravity of sediment gradation coefﬁcient
Notation Gu h hc hd hn hp hr hs ht hu hw h H H Hw Hs = 2a˜ i ib ie if j k k0 ks ks kt k˜ K K1, K2 Kc Kd K num Kp K sj l1 to l4 lc , ld La Lf Lp Lr
universal gravitation constant ﬂow depth critical ﬂow depth downstream ﬂow depth normal ﬂow depth pressure head at the wetting front rainfall depth cumulative snowmelt tailwater depth upstream ﬂow depth partial elevation drop on a watershed local change in ﬂow depth energy loss over a meander wavelength Bernoulli sum elevation drop on a watershed wave height rainfall intensity riverbed inﬁltration rate excess rainfall intensity snowmelt rate space index time index resistance parameter for laminar overland ﬂow surface roughness grain roughness total resistance to laminar overland ﬂow wave number saturated hydraulic conductivity coefﬁcients of the pier scour equation riprap coefﬁcient dispersion coefﬁcient numerical dispersion coefﬁcient plunging jet coefﬁcient submerged jet coefﬁcient moment arms moment arms in radial stability of river bends abutment length depth of the wetting front pier length river length
xv
xvi
Notation
Lr L sb LM LR LS LW L m ms mE mM M M M1 , M2 M, N n n˜ N N O() p p() p0 p0e p0i p0r pc pi p0 P P˜ P P() P0 P q qbv ∗ qbv = qbv /ω0 ds ql qm
length ratio settlingbasin length runoffmodel gridcell size grid size of rainfall precipitation correlation length of a storm length scale of a watershed length of arrested saline wedge exponent of the resistance equation sediment mass eroded from a single storm mass of the Earth mass of the Moon mass speciﬁc momentum ﬁrst and second moments of a distribution particlestability coefﬁcients Manning coefﬁcient n wave number index number of points per wavelength number of storms order of an approximation pressure probability density function porosity effective porosity initial water content residual water content fraction of material coarser than dsc sediment size fraction change in water content at the wetting front wetted perimeter total power of a wave power loss in a hydraulic jump probability power loss grid Peclet number unit discharge unit sediment discharge by volume dimensionless unit sediment discharge lateral unit discharge maximum unit discharge
Notation unit sediment discharge total unit sediment discharge river discharge bed sediment discharge watershed size correction factor peak discharge sediment discharge radial coordinate dimensionless radius of curvature cylindrical coordinate system θ downstream, r lateral, and z upward R radius of curvature of a river excess rainfall Re hydraulic radius Rh minimum radius of curvature of a channel Rm radius of the Earth RE Re Reynolds number Re∗ grain shear Reynolds number Re = V h/ν Reynolds number Re∗ = u ∗ ds /ν grain shear Reynolds number Ro = ω/κu ∗ Rouse number S slope effective saturation Se bed, friction, and watersurface slope S0 , S f , Sw radial watersurface slope So , S f , Sw dimensionless radial slope Sr∗ radial watersurface slope Sr , Swr sediment delivery ratio SDR t time t time increment time increment for sediment ts cumulative time with positive air temperature ta time to equilibrium te cumulative duration of snowmelt tf rainfall duration tr normalized storm duration tr∗ = tr /t¯r transversal mixing time tt tv vertical mixing time T period of return of extreme events T wave period qs qt Q Q bv Qe Qp Qs r r∗ r, θ, z
xvii
xviii
Notation
TE Ts u, v u¯ u∗ Uw vs v x , v y , vz V Vc V Vθ W Wm Wo x, y, z xr , yr , zr x X Xc Xe Y zb zw z∗ z
trap efﬁciency windstorm duration velocity along a vertical proﬁle average ﬂow velocity shear velocity wind velocity velocity against the stone velocity components crosssection averaged velocity critical velocity densimetric velocity downstream velocity in cylindrical coordinates channel width meander width overland plane width coordinates usually x downstream, y lateral, and z upward length ratios of hydraulic models grid spacing runoff length reach length equilibrium runoff length sediment yield bed elevation watersurface elevation dimensionless depth scour depth
Greek Symbols α αb αe β β βm γ γm γmd γs (x)
coefﬁcient of the stage–discharge relationship deﬂection angle of barges energy correction factor exponent of the stage–discharge relationship bed particle motion angle momentum correction factor speciﬁc weight of water speciﬁc weight of a water–sediment mixture dry speciﬁc weight of a water–sediment mixture speciﬁc weight of sediment gamma function
Greek Symbols δ ξ ξ˜ η η˜ λ λ λf λr = tr /te µ ν φ φ φ ρ ρm ρmd ρs ρsea = ln[−ln E(x)] ω ωE R θ θ θj θm θp θr θ0 , 0 1 = (t − tr )/te σ σd σg σ˜ τ
angle between streamline and particle direction ratio of exceedance probabilities displacement in the x direction sideslope stability number displacement in the ydirection streamline deviation angle wavelength snowmelt intensity hydrograph equilibrium number meander wavelength dynamic viscosity of water kinematic viscosity of water angle of repose of bed material latitude dimensionless soil mass eroded from a single storm velocity potential mass density of water mass density of a water–sediment mixture dry mass density of a water–sediment mixture mass density of sediment mass density of seawater double logarithm of exceedance probability settling velocity angular velocity of the Earth sinuosity coefﬁcient of secondary ﬂows in bends downstream orientation of channel ﬂow angular coordinate jet angle measured from the horizontal maximum orientation of channel ﬂow ﬂow orientation angle against a pier raindrop angle downstream bed angle sideslope angle dimensionless time stress components standard deviation of dispersed material gradation coefﬁcient angular frequency of surface waves shear stress
xix
xx
Notation
τ0 , τb τ0x , τ0y τbn τc τr τr∗ τs τsc τw τzx
bed shear stress downstream and lateral bed shear stress bed shear stress at a normal depth critical shear stress radial shear stress dimensionless radial shear stress side shear stress critical shear stress on a sideslope wind shear stress shear stress applied in the x direction in a plane perpendicular to z Shields parameter τ∗ critical value of the Shields parameter τ∗c ψ = q/i e L dimensionless discharge ψ, θ weighting coefﬁcients reduced variable Fourier coefﬁcient ζnk Superscripts and Diacriticals aˆ n˜ Cˆ e¯ hk
coefﬁcient of the logarithm resistance equation wave properties parameters of the universal soilloss equation average value time index k Subscripts
ar , aθ a x , az τc h j+1 L m , Qm L p, Q p L r , Qr ρ m , γm ρ md , γmd ρ s , γs
cylindrical coordinate components Cartesian components critical shear stress space index at j + 1 model value prototype value similitude scaling ratio properties of a water–sediment mixture properties of a dry water–sediment mixture sediment properties
1 Introduction to river mechanics
It has long been understood that water ﬂows downhill. This maybe the only statement to be remembered until a river dries out and crops wilt. Droughts unfortunately threaten humanity with the constraint that, without water, life cannot be sustained. On the other hand, the devastating consequences of excess water through ﬂoods stem from the fact that humanity, crops, and cattle are not well adapted to submerged life. Although nomadic tribes coped with the continuously changing pulses of ﬂuvial systems, sedentary conditions forced humanity to protect against ﬂoods and droughts. In arid lands, perennial streams with regulated ﬂow and a yearround supply of water are so much more valuable to humanity and wildlife than are natural sequences of short ﬂoods that succeed long droughts in dry ephemeral streams. River engineers are facing the daunting challenge of optimizing the urban and environmental resources pertaining to rivers while minimizing the damages caused by ﬂoods and droughts. Perhaps the origin of river engineering started with Yu (4000 B.C.) who was selected to be emperor of China on the premise of longlasting dikes for the protection of fertile Chinese plains against ﬂoods. For centuries, Chinese emperors were classiﬁed into “good dynasties” or “bad dynasties” depending on whether or not they succeeded in their struggle to harness large rivers. At approximately the same time, in Mesopotamia, an extensive irrigation system was developed between the Tigris and the Euphrates Rivers. Floodcontrol levees were constructed to protect fertile lands from destructive inundations. In these early periods of civilization, humanity’s cultural development was dominated by fear of thunder, lightning, rain, ﬂoods, storms, cyclones, and earthquakes. The lack of understanding of causeandeffect relationships to explain natural phenomena such as ﬂoods has characterized earlier civilizations. Nonetheless, humanity was compelled to develop hydraulic engineering and tame rivers in order to prevent famine and to survive. Today, several hydraulic structures from past civilizations serve as landmarks of excellence. Natural philosophy emerged during Greek antiquity. Thales of Miletus (circa 600 B.C.) explored natural laws through philosophical meditation in replacing 1
2
Introduction to river mechanics
mythological traditions with logical thoughts and reﬂections based on observations of nature. Basic principles underlying natural processes were deduced by rational approaches, including reﬂection and speculation. Hypotheses and assumptions were formulated by natural philosophers. For instance, Thales believed that “water was the origin of all things” and “the earth rested on water.” Plato (428–348 B.C.) speculated on matters of physics and metaphysics alike, without interest in possible discrepancies between theory and reality. Democritus of Thrace (465 B.C.) believed everything to be inherently mechanical in nature and admitted nothing fortuitous or providential. In opposition to Plato’s speculative ideas, the philosophy of Aristotle (384– 322 B.C.) contemplated nature through facts, and his writings on logic, physics, biology, metaphysics, and ethics promote continuous advances and evolution of knowledge in each ﬁeld. He recognized two types of action: a motivation, to which the speed of movement is directly proportional, and a resistance, to which motion is inversely proportional. He also believed that the motivation was proportional to the density of the body and that the resistance was proportional to the density of the medium through which it moved. These statements essentially describe the concepts of linear momentum and resistance to motion. Archimedes (287–212 B.C.) was the greatest mathematician of antiquity, with chief interest in geometry, centers of gravity, hydrostatics, a theory of ﬂoating bodies, and anticipated foundations for differential and integral calculus. Another important milestone, achieved by Hippocrates of Cos (460–380? B.C.), is worth mention. He proved the existence of evaporation by weighing a vessel ﬁlled with water over a long period of time. The most splendid achievements are twofold: (1) The concept of studying nature through experiments was born; and (2) the concept that quantitative information could be gathered from measurements was developed. It took nearly 2,000 years after Hippocrates before experiments and measurements supplanted speculations and hypotheses. The real treasures of natural philosophy found between 600 and 300 B.C. were highly reputed, but they had few practical effects, and society could hardly beneﬁt from this new knowledge. Humanity had to wait until the Renaissance, circa 1500 A.D., to appreciate the longlasting values of the Greek civilization. As much as Greeks were interested in pure rational knowledge, Romans were the true pragmatic engineers of antiquity. Marcus Vitruvius Pollio (ﬁrst century B.C.) and Sextus Julius Frontinus (40–103 A.D.) were concerned with the construction of the aqueducts that supplied water to Rome. There is still debate as to whether water conduits had been calculated, but colossal aqueduct and waterdistribution systems were designed on the basis of only experience and rough estimates. It is intriguing that the simple concept of conservation of mass was not understood at the time of the design. Frontinus could measure
Introduction to river mechanics
3
ﬂow depth, and he keenly observed that a steeper slope results in higher ﬂow velocity. However, he incorrectly considered that the discharge corresponds to the cross section given by the measured canal width and ﬂow depth. He observed that velocity increases the discharge, but he could not recognize the quantitative proportionality between velocity and discharge. Correct understanding of the relationship between discharge Q, crosssectional area A, and velocity V in terms of Q = AV is due to Hero of Alexandria (ﬁrst century A.D.). Almost 1,500 years elapsed until the discharge relationship was correctly rediscovered by Leonardo da Vinci (1452–1519) and Benedetto Castelli (1577– 1644?). The Renaissance period marks the rebirth of civilization after the Middle Ages. The development of printing contributed to rapid dissemination of knowledge. Leonardo da Vinci understood the principles of experimental science and advocated the necessity of observation: “I will treat of such a subject. But ﬁrst of all I shall make a few experiments and then demonstrate why bodies are forced to act in this manner. This is the method that one has to pursue in the investigation of phenomena of nature. It is true that nature begins by reasoning and ends by experience; but, nevertheless, we must take the opposite route: as I have said, we must begin with experiment and try through it to discover the reason.” Flow kinematics became better understood under Benedetto Castelli, who wrote “Sections of the same river discharge equal quantities of water in equal times, even if the sections themselves are unequal. Given two sections of a river, the ratio of the quantity of water which passes the ﬁrst section to that which passes the second is in proportion to the ratio of the ﬁrst to the second sections and to that of the ﬁrst and second velocities. Given two unequal sections by which pass equal quantities of water, the sections are reciprocally proportional to the velocities.” The seventeenth century brought remarkable advances in mechanics and mathematics. Dynamic concepts of inertia and momentum became clear under Ren´e Descartes (1596–1650), who wrote “I assume that the movement which is once impressed upon a given body is permanently retained, if it is not removed by some other course; that is, whatever has commenced to move in a vacuum will continue to move indeﬁnitely at the same velocity.” Pressure concepts in ﬂuids at rest were described by Blaise Pascal (1623– 1662), who postulated that pressure was transmitted equally in all directions. Christian Huygens (1629–1695) deﬁned the principle of centrifugal force and is sometimes credited with the principle of conservation of energy. Isaac Newton (1642–1727) clearly formulated three laws of motion as a concise synthesis of concepts explicitly formulated by Descartes, Wallis, Huygens, and Wren. His contribution is a concise deﬁnition of mass, momentum, inertia, and force.
4
Introduction to river mechanics
He also studied ﬂuid resistance between a ﬂuid and a solid to conclude that resistance is proportional to the relative velocity of adjacent zones. Energy concepts and presentday calculus evolved around the contribution of Gottfried Wilhelm von Leibniz (1646–1716). He introduced the concept of a live force that is proportional to the second power of velocity, now known as kinetic energy, and raised a lively debate between kinetic energy and momentum proportional to the ﬁrst power of velocity. Hydrodynamics can be attributed to outstanding mathematical developments in the eighteenth century. Daniel Bernoulli (1700–1782) dealt with ﬂuid statics and dynamics. It is nevertheless Leonard Euler (1707–1783) who rigorously derived the Bernoulli equation and the differential forms of the equations of continuity and acceleration in frictionless ﬂuids. Resistance to ﬂow remained obscure until the nineteenth century when experi ments on ﬂow in small pipes resulted from the studies of Gotthilf Heinrich Ludwig Hagen (1797–1884), JeanLouis Poiseuille (1799–1869), Julius Weisbach (1806–1871), Henry Phillibert Gaspard Darcy (1803–1858), Wilhelm Rudolf Kutter (1818–1888), Emile Oscar Ganguillet (1818–1894), and Robert Manning (1816–1897). The Navier–Stokes equations for the analysis of viscous ﬂuid motion became possible from the contributions of JeanClaude Barr´e de SaintVenant (1797–1886), Louis Marie Henri Navier (1785–1836), Baron Augustin Louis de Cauchy (1789–1857), Simeon Denis Poisson (1781–1840) and George Gabriel Stokes (1819–1903). Turbulence challenged generations of scientists including Joseph Boussinesq (1842–1929), Osborne Reynolds (1842– 1912), Ludwig Prandtl (1875–1953), and Theodor von K´arm´an (1881–1963), who contributed to unveil part of its inherent complexity. In comparison with those of hydraulics, the advances in sediment transport, which is essential to understanding river mechanics, have been extremely slow. Two contributions before the twentieth century are noteworthy: (1) the contributions of Albert Brahm on the relationship between the bedﬂow velocity and the 1/6 power of the immersed weight of bed material and (2) the concept of tractive force by Paul Francois Dominique du Boys (1847–1924) and his relationship to bedsediment transport. Today, inclass discussions can emerge from a simple question such as, Why do rivers form? It is interesting to note the required physical processes that lead to river formation. The concept of a gravitationalforce component should ﬁrst come to mind. The need for erodible material or alluvium emanates from the discussion. The concept of an alluvial river usually is gradually becoming clear. However, all of this does not explain why rivers form. Do we understand the mechanics of formation of alluvial rivers? The effects of ﬂow convergence and divergence allude to the concepts of continuity of water. Aggradation and
Introduction to river mechanics Sediment (q s = a q b )
Water (Q1 = Q 2) Q1 = q 2 W
1
Q2 = q2 W
2
1
q2 = 2 q 1
5
W
W
W
Q s1= q s1 2 W = 2 W a q 1b
Q s2 = W q s2 = W a q b 2
Q s2 > Q s1 if b > 1
Figure 1.1. Water and sediment balance for converging ﬂow.
degradation results from conservation of sediment. Does converging ﬂow tend to cause aggradation or degradation? We can formulate an intuitive understanding by using a simple sediment rating curve of the type qs = aq b , where, qs is the unit sediment discharge and q is the unit discharge; see Fig. 1.1. The results of converging ﬂow are to cause degradation when b > 1 and aggradation when b < 1. Is there any reason from our understanding of erosion and sedimentation that supports that b > 1? If so, I guess we have answered our question. In a simpliﬁed form, rivers form because sediment concentration increases with unit discharge. Flow convergence thus causes scour, and this clearly illustrates that river mechanics stems from an understanding of hydrodynamics and sediment transport. On Earth, the study of the water and sediment discharge to the oceans from the rivers around the world shows that the annual suspended sediment discharge is 13.5 × 109 metric tons per year. Some important rivers are listed in Table 1.1. Approximately half of the sediment discharge to the oceans originates from rivers in Southeast Asia. In comparison, the total freshwater ﬂow to the oceans from all rivers of the world combines to 1.2 × 106 m3 /s. The average sediment concentration of ﬂows to oceans is ∼360 mg/l. The proposed physical analysis of river mechanics is based on the concept of water and sediment transport down the rivers under the action of gravity from the upland areas to the oceans. The surface area of the land that drains into a particular river delineates the watershed, also termed the drainage basin or catchment. Chapter 2 outlines the physical properties of water and sediment and the governing equations of motion. Chapter 3 reviews the sources and the yields of water and sediment at the watershed scale. Chapter 4 treats the steadyﬂow conditions in canals and rivers. Chapter 5 delves into the mechanics of unsteady ﬂows in rivers. Chapter 6 describes the downstream hydraulic
Amazon Mississippi Congo La Plata/Parana Ob Nile Yenisei Lena Amur Yangtse Kiang Wolga Missouri Zambesi St. Lawrence Niger MurrayDarling Ganges
River
Mouth Mouth Mouth Mouth Delta Mouth Mouth Mouth Mouth Mouth Mouth Mouth Mouth Mouth Mouth Mouth Delta
Station 7.0 3.9 3.7 3.0 3.0 2.9 2.6 2.4 2.1 1.8 1.5 1.4 1.3 1.3 1.1 1.1 1.0
Catchment area (106 km2 ) 100,000 18,000 44,000 19,000 12,000 3,000 17,000 16,000 11,000 22,000 8,400 2,000 16,000 14,000 5,700 400 14,000
m3 s−1
Water
450 150 370 200 130 30 210 210 160 390 180 50 390 340 160 10 440
mm yr−1
Table 1.1. Water and sediment loads of selected rivers (after Jansen et al., 1979)
Sediment
900 300 70 90 16 80 11 12 52 500 25 200 100 10 40 30 1,500
106 tons yr−1
Discharge
90 55 15 20 4 15 3 4 15 200 10 100 50 6 25 20 1,000
µm yr−1
290 530 50 150 40 630 20 25 150 1,400 100 3,200 200 20 220 2,500 3,600
Sediment as ppm of dischargea (mg 1−1 )
a
ppm: parts in 106
Indus Orinoco Orange River Danube Mekong Hwang Ho Brahmaputra Dnieper Irrawaddi Rhine Magdelena (Colombia) Vistula (Poland) Kura (USSR) Chao Phya (Thailand) Oder (Germany/Poland) Rhone (France) Po (Italy) Ishikari (Japan) Tiber (Italy) Tone (Japan) Waipapa (New Zealand)
Mouth Mouth Mouth Mouth Mouth Mouth Bahadurabad Mouth Mouth Delta Calamar Mouth Mouth Mouth Mouth Mouth Mouth Mouth Mouth Mouth Kanakanala
0.96 0.95 0.83 0.82 0.80 0.77 0.64 0.46 0.41 0.36 0.28 0.19 0.18 0.16 0.11 0.096 0.070 0.016 0.013 0.012 0.0016
6,400 25,000 2,900 6,400 15,000 4,000 19,000 1,600 13,000 2,200 7,000 1,000 580 960 530 1,700 1,500 230 420 480 46
210 830 110 250 590 160 940 110 1,000 190 790 160 100 190 150 560 670 450 1,000 1,250 900
400 90 150 67 80 1,900 730 1.2 300 0.72 220 1.5 37 11 0.13 10 15 6 1.8 3 11
300 65 130 60 70 1,750 800 2 500 1 550 5 150 50 1 75 150 270 100 180 5,000
2,000 110 1,600 330 170 15,000 1,200 25 750 10 1,000 50 2,000 350 10 200 300 850 140 200 7,500
8
Introduction to river mechanics
geometry and equilibrium in alluvial rivers. Chapter 7 discusses the concepts of river dynamics and response to perturbations from equilibrium conditions. Chapter 8 particularly deals with river stability and presents methods to stabilize river banks. Chapter 9 presents several river engineering techniques from ﬂood control to bridge crossings and waterways. Chapter 10 focuses on physical modeling techniques, with a particular analysis on the underlying theoretical concepts. Chapter 11 introduces the reader to numerical methods used to solve river engineering problems. Finally, Chapter 12 summarizes theory and applications of river engineering problems associated with waves and tides, usually observed in river estuaries.
2 Physical properties and equations
As a natural science, the variability of river processes must be examined through the measurement of physical parameters. This chapter ﬁrst describes dimensions and units (Section 2.1), physical properties of water (Section 2.2), and sediment (Section 2.3). The equations governing the motion of water and sediment from upland areas to oceans include kinematics of ﬂow (Section 2.4), the equation of continuity (Section 2.5), the equation of motion (Section 2.6), and the concept of hydraulic and energy grade lines (Section 2.7). In this chapter and in the rest of the book, a solid diamond () denotes equations and problems of particular signiﬁcance. Problems with a double diamond () are considered most important. 2.1
Dimensions and units
Physical properties are usually expressed in terms of the following fundamental dimensions: mass (M), length (L), time (T ), and temperature (T ◦ ). Throughout the text, the unit of mass is preferred to the corresponding unit of force. The fundamental dimensions are measurable parameters that can be quantiﬁed in fundamental units. In the SI system of units, the fundamental units of mass, length, time, and temperature are the kilogram (kg), the meter (m), the second (s), and degrees Kelvin (◦ K). Alternatively, the Celsius scale (◦ C) is commonly preferred because it refers to the freezing point of water at 0◦ C and the boiling point at 100◦ C. A Newton (N) is deﬁned as the force required for accelerating 1 kg at 1 m/s2 . Knowing that the acceleration that is due to gravity at the Earth’s surface, g, is 9.81 m/s2 , we obtain the weight of a kilogram from Newton’s second law: F = mass × g = 1 kg × 9.81 m/s2 = 9.81 N. The unit of work (or energy) is the joule (J), which equals the product of 1 N × 1 m. The unit of power is a watt (W), which is 1 J/s. Preﬁxes are used in the SI system to indicate multiples or fractions of units by powers of 10: 9
10
Physical properties and equations µ (micro) = 10−6 , k (kilo) = 103 , m (milli) = 10−3 , M (mega) = 106 , c (centi) = 10−2 , G (giga) = 109 .
For example, sand particles are coarser than 62.5 micrometers, or µm; gravels are coarser than 2 millimeters, abbreviated 2 mm, and one megawatt (MW) equals one million watts (1,000,000 or 106 W). In the English system of units, the time unit is a second, the fundamental units of length and mass are, respectively, the foot (ft), equal to 30.48 cm, and the slug, equal to 14.59 kg. The force required for accelerating a mass of one slug at 1 ft/s2 is a pound force (lb). Throughout this text, a pound refers to a force, not a mass. The temperature, in degrees Celsius, TC ◦ , is converted to the temperature in degrees Fahrenheit, TF◦ , by TF◦ = 32.2 ◦ F + 1.8 TC◦ . Variables are classiﬁed as geometric, kinematic, dynamic, and dimensionless variables, as shown in Table 2.1. Geometric variables involve length dimensions only and describe the geometry of a system through length, area, and volume. Kinematic variables describe the motion of ﬂuid and solid particles, and these variables can be depicted by only two fundamental dimensions, namely L and T . Dynamic variables involve mass terms in the fundamental dimensions. Force, pressure, shear stress, work, energy, power, mass density, speciﬁc weight, and dynamic viscosity are common examples of dynamic variables. Several conversion factors are listed in Table 2.2.
2.2
Properties of water
The physical properties of a nearly incompressible ﬂuid such as water are sketched in Fig. 2.1. Mass density of water, ρ. The mass of water per unit volume is referred to as the mass density ρ. The maximum mass density of water at 4 ◦ C is 1,000 kg/m3 and varies slightly with temperature, as shown in Table 2.3. In comparison, the mass density of sea water is 1,025 kg/m3 and, at sea level, the mass density of air is 1.29 kg/m3 at 0 ◦ C. The conversion factor is 1 slug/ft3 = 515.4 kg/m3 . Speciﬁc weight of water, γ . The gravitational force per unit volume of ﬂuid, or simply the ﬂuid weight per unit volume, deﬁnes the speciﬁc weight γ . At 10 ◦ C, water has a speciﬁc weight, γ = 9,810 N/m3 or 62.4 lb/ft3 (1 lb/ft3 = 157.09 N/m3 ). Speciﬁc weight varies slightly with temperature, as given in Table 2.3. Mathematically, the speciﬁc weight γ equals the product of the mass
Properties of water
11
Table 2.1. Geometric, kinematic, dynamic, and dimensionless variables
Variable
Symbol
Fundamental dimensions
SI Units
Geometric (L) Length Area Volume
L, x, h, ds A ∀
L L2 L3
m m2 m3
Kinematic (L, T) Velocity Acceleration Kinematic viscosity Unit discharge Discharge
v x , u, u ∗ a, ax , g ν q Q
L T −1 L T −2 L 2 T −1 L 2 T −1 L 3 T −1
m/s m/s2 m2 /s m2 /s m3 /s
Dynamic (M, L, T) Mass Force Pressure Shear stress Work or energy Mass density Speciﬁc weight Dynamic viscosity
m F = ma, mg p = F/A τ x y , τ0 , τc E = F· d ρ, ρs γ , γs = ρ s g µ = ρν
M M L T −2 M L −1 T −2 M L −1 T −2 M L 2 T −2 M L −3 M L −2 T −2 M L −1 T −1
1 kg 1 kg m/s2 =1 N 1 N/m2 = 1 Pa 1 N/m2 = 1 Pa 1Nm=1J kg/m3 N/m3 1 kg/m s = 1 N s/m2 = 1 Pa s
S0 , S f G = γs /γ Re = uh/ν
— — —
— — —
Re∗ = u ∗ ds /ν √ Fr = u/ gh τ∗ = τ/(γs − γ )ds Cv , Cw
— — — —
— — — —
Dimensionless Slope Speciﬁc gravity Reynolds number Grainshear Reynolds number Froude number Shields parameter Concentration
Note: Pa stands for pascal.
density ρ times the gravitational acceleration g = 32.2 ft/s2 = 9.81 m/s2 : γ = ρg.
(2.1)
Dynamic viscosity µ. As a ﬂuid is brought into deformation, the velocity of the ﬂuid at any boundary equals the velocity of the boundary. The ensuing rate of ﬂuid deformation causes a shear stress τzx that is proportional to the dynamic viscosity µ and the rate of deformation of the ﬂuid, dv x /dz: τzx = µ
dv x . dz
(2.2)
Table 2.2. Conversion of units Unit 1 acre 1 acrefoot (acreft) 1 atmosphere (atm) 1 bar 1 barrel (U.S., dry) (bbl) 1 British thermal unit (Btu) = 778 lb ft 1 cubic foot per second (ft3 /s) 1 day 1 degree Celsius (◦ C) = (TF ◦ − 32◦ ) 5/9 1 degree Fahrenheit (◦ F) = 32 + 1.8 TC ◦ 1 drop 1 dyne (dyn) 1 dyne per square centimeter (dyn/cm2 ) 1 fathom (fath) 1 foot (ft) 1 gallon (U.S., liquid) (gal) 1 horsepower (hp) = 550 lb ft/s 1 inch (in.) 1 inch of mercury (in. Hg) 1 inch of water 1 joule (J) 1 kip 1 knot 1 liter (l) 1 micrometer (µm) 1 mile (nautical) 1 mile (statute) 1 million gallons per day (mgd) = 1.55 ft3 /s 1 Newton (N) 1 ounce (avoirdupois) (oz) 1 ﬂuid ounce (U.S.) 1 pascal (Pa) 1 pint (U.S., liquid) (pt) 1 poise (P) 1 poundfoot (lbft) 1 pound per square foot (lb/ft2 or psf) 1 pound per square inch (lb/in.2 or psi) 1 poundforce (lb) 1 poundforce per cubic foot (lb/ft3 ) 1 quart (U.S., liquid) (qt) 1 slug 1 slug per cubic foot (slug/ft3 ) 1 stoke (S) = 1 cm2 /s 1 ton (U.K., long) 1 ton (metric) (t) 1 ton (short) = 2,000 lb 1 watt (W) 1 yard (yd) 1 year (yr)
kg, m, s 4,046.87 m2 1,233.5 m3 101,325 kg/m s2 100,000 kg/m s2 0.1156 m3 1,055 kg m2 /s2 0.0283 m3 /s 86,400 s 1 degree Kelvin (K) 0.555556 degree Kelvin 61.6 mm3 0.00001 kg m/s2 0.1 kg/m s2 1.8288 m 0.3048 m 0.0037854 m3 745.70 kg m2 /s3 0.0254 m 3,386.39 kg/m s2 248.84 kg/m s2 1 kg m2 /s2 4,448.22 kg m/s2 0.5144 m/s 0.001 m3 1 × 10 −6 m 1,852 m 1609.34 m 0.04382 m3 /s 1 kg m/s2 0.02835 kg 2.957 × 10−5 m3 1 kg/m s2 0.0004732 m3 0.1 kg/m s 1.356 kg m2 /s2 47.88 kg/m s2 6,894.76 kg/m s2 4.448 kg m/s2 157.09 kg/m2 s2 0.00094635 m3 14.59 kg 515.4 kg/m3 0.0001 m2 /s 1,016.05 kg 1,000 kg 8,900 kg m/s2 1 kg m2 /s2 0.9144 m 31,536,000 s
Note: Those units for which no abbreviations are given are spelled out.
N, Pa, W
101.3 kPa 100 kPa 1,055 N m
1 × 10−5 N 0.1 Pa
745.7 W 3,386.39 Pa 248.84 Pa 1Nm = 1J 4,448.22 N
1 N/m2 0.1 Pa s 1.356 N m 47.88 Pa 6,894.76 Pa 4.448 N 157.09 N/m3
8.9 kN
Properties of sediment Surface area A Force F Shear stress
τ zx
Velocity vx
Mass density ρ Specific weight γ
Distance z
Dynamic viscosity µ = ρυ Kinematic viscosity υ
x F
dv
dv
τ zx = A = µ d x = ρυ d x z z
Figure 2.1. Newtonian ﬂuid properties.
13 The fundamental dimension of the dynamic viscosity µ is M/L T , which is a dynamic variable. As indicated in Table 2.3, the dynamic viscosity of water decreases with temperature. Fluids without yield stress for which the dynamic viscosity remains constant regardless of the rate of deformation are called Newtonian ﬂuids. The dynamic viscosity of clear water at 20 ◦ C is 1 centipoise: 1 cP = 0.01 P = 0.001 N s/m2 = 0.001 Pa s. The conversion factor is 1 lb s/ft2 = 47.88 N s/m2 = 47.88 Pa s.
Kinematic viscosity ν. When the dynamic viscosity of a ﬂuid µ is divided by the mass density ρ of the same ﬂuid, the mass terms cancel out. This results in kinematic viscosity ν with dimensions L 2 /T , which is also shown in Table 2.3, decreasing with temperature. The viscosity of clear water at 20 ◦ C is 1 centistokes = 1 cS = 0.01 cm2 /s = 1 × 10−6 m2 /s. The conversion factor is 1 ft2 /s = 0.0929 m2 /s. The change in kinematic viscosity of water, ν, with temperature T ◦ in degrees Celsius can be roughly estimated from ν=
µ = [1.14 − 0.031 (T ◦ − 15) + 0.00068 (T ◦ − 15)2 ] × 10−6 m2 /s. ρ
(2.3)
It is important to remember that both the density and the viscosity of water decrease with temperature. The maximum water density is found at 4 ◦ C, and water either colder or warmer than 4 ◦ C will be found near the surface. The density of ice increases as the temperature decreases. This causes the ice cover to crack during cold nights and expand to apply large forces on the banks of lakes, reservoirs, and wide rivers during warm days.
2.3
Properties of sediment
The physical properties of sediment are classiﬁed into single particles (Subsection 2.3.1), sediment mixture (Subsection 2.3.2), and sediment suspension (Subsection 2.3.3).
14
Physical properties and equations
Table 2.3. Approximate physical properties of clear water and ice at atmospheric pressure
Temperature ◦C
Density ρ kg/m3
Speciﬁc weight γ N/m3
Dynamic viscosity µ N s/m2
Kinematic viscosity ν m2 /s
−30◦ C −20◦ C −10◦ C 0◦ C 4◦ C 5◦ C 10◦ C 15◦ C 20◦ C 25◦ C 30◦ C 35◦ C 40◦ C 50◦ C 60◦ C 70◦ C 80◦ C 90◦ C 100◦ C
921 919 918 999.9 1,000 999.9 999.7 999 998 997 996 994 992 988 983 978 972 965 958
9,035 9,015 9,005 9,809 9,810 9,809 9,807 9,800 9,790 9,781 9,771 9,751 9,732 9,693 9,643 9,594 9,535 9,467 9,398
Ice Ice Ice 1.79 × 10−3 1.56 × 10−3 1.51 × 10−3 1.31 × 10−3 1.14 × 10−3 1.00 × 10−3 8.91 × 10−4 7.97 × 10−4 7.20 × 10−4 6.53 × 10−4 5.47 × 10−4 4.66 × 10−4 4.04 × 10−4 3.54 × 10−4 3.15 × 10−4 2.82 × 10−4
Ice Ice Ice 1.79 × 10−6 1.56 × 10−6 1.51 × 10−6 1.31 × 10−6 1.14 × 10−6 1.00 × 10−6 8.94 × 10−7 8.00 × 10−7 7.25 × 10−7 6.58 × 10−7 5.53 × 10−7 4.74 × 10−7 4.13 × 10−7 3.64 × 10−7 3.26 × 10−7 2.94 × 10−7
◦F
slug/ft3
lb/ft3
lb s/ft2
ft2 /s
0◦ F 10◦ F 20◦ F 30◦ F 32◦ F 40◦ F 50◦ F 60◦ F 70◦ F 80◦ F 100◦ F 120◦ F 140◦ F 160◦ F 180◦ F 200◦ F 212◦ F
1.78 1.78 1.78 1.77 1.931 1.938 1.938 1.936 1.935 1.93 1.93 1.92 1.91 1.90 1.88 1.87 1.86
57.40 57.34 57.31 57.25 62.40 62.43 62.40 62.37 62.30 62.22 62.00 61.72 61.38 61.00 60.58 60.12 59.83
Ice Ice Ice Ice 3.75 × 10−5 3.23 × 10−5 2.73 × 10−5 2.36 × 10−5 2.05 × 10−5 1.80 × 10−5 1.42 × 10−5 1.17 × 10−5 0.981 × 10−5 0.838 × 10−5 0.726 × 10−5 0.637 × 10−5 0.593 × 10−5
Ice Ice Ice Ice 1.93 × 10−5 1.66 × 10−5 1.41 × 10−5 1.22 × 10−5 1.06 × 10−5 0.930 × 10−5 0.739 × 10−5 0.609 × 10−5 0.514 × 10−5 0.442 × 10−5 0.385 × 10−5 0.341 × 10−5 0.319 × 10−5
Properties of sediment Size d s
15 2.3.1
Single particle
The physical properties of a single solid particle of volume ∀s are sketched in Fig. 2.2. The mass density of a solid partiMass density ρs cle, ρs , describes the solid mass per unit Specific weight γs = ρs g = G γ volume. The mass density of quartz parSpecific gravity G ticles, 2,650 kg/m3 (1 slug/ft3 = 515.4 kg/m3 ), does not vary signiﬁcantly with temperature and is assumed conFigure 2.2. Physical properties of a stant in most calculations. It must be kept single particle. in mind, however, that heavy minerals such as iron, copper, etc., have much larger values of mass density. Volume Vs
Speciﬁc weight of solid particles, γs . The particle speciﬁc weight γs corresponds to the solid weight per unit volume of solid. Typical values of γs are 26.0 kN/m3 or 165.4 lb/ft3 . The conversion factor is 1 lb/ft3 = 157.09 N/m3 . The speciﬁc weight of a solid, γs , also equals the product of the mass density of a solid particle, ρs , times the gravitational acceleration g; thus γs = ρs g.
(2.4)
Speciﬁc gravity G. The ratio of the speciﬁc weight of a solid particle to the speciﬁc weight of ﬂuid at a standard reference temperature deﬁnes the speciﬁc gravity G. With common reference to water at 4 ◦ C, the speciﬁc gravity of quartz particles is G=
ρs γs = = 2.65. γ ρ
(2.5)
The speciﬁc gravity is a dimensionless ratio of speciﬁc weights, and thus its value remains independent of the system of units. Submerged speciﬁc weight of a particle, γ˜s . Owing to the Archimedes principle, the speciﬁc weight of a solid particle, γs , submerged in a ﬂuid of speciﬁc weight γ equals the difference between the two speciﬁc weights; thus γ˜s = γs − γ = (G − 1)γ .
(2.6)
Sediment size ds . The most important physical property of a sediment particle is its size. Table 2.4 shows the grade scale commonly used in
16
Physical properties and equations
Table 2.4. Sediment grade scale and approximate properties Angle of repose φ (deg)
Critical shear stress τc (N/m2 )
>2,048 >1,024 >512 >256
42 42 42 42
1790 895 447 223
1.33 0.94 0.67 0.47
5,430 3,839 2,715 1,919
>128 >64
42 41
111 53
0.33 0.23
1,357 959
Gravel Very coarse Coarse Medium Fine Very ﬁne
>32 >16 >8 >4 >2
40 38 36 35 33
0.16 0.11 0.074 0.052 0.036
678 479 338 237 164
Sand Very coarse Coarse Medium Fine Very ﬁne
>1.000 >0.500 >0.250 >0.125 >0.062
32 31 30 30 30
0.47 0.27 0.194 0.145 0.110
0.0216 0.0164 0.0139 0.0120 0.0105
Silt Coarse Medium Fine Very ﬁne
>0.031 >0.016 >0.008 >0.004
30 30
0.083 0.065 Cohesive material
0.0091 0.0080
Clay Coarse Medium Fine Very ﬁne
>0.0020 >0.0010 >0.0005 >0.00024
Class name Boulder Very large Large Medium Small Cobble Large Small
a
Particle diameter ds (mm)
26 12 5.7 2.71 1.26
Critical shear velocity u ∗c (m/s)
Settling velocity ω0 (mm/s)
109 66.4 31.3 10.1 2.66 0.67a 0.167a 0.042a 0.010a 2.6 × 10−3 a 6.5 × 10−4 a −4 1.63 × 10 a 4.1 × 10−5 a
Possible ﬂocculation
sedimentation. Note that the size scales are arranged in geometric series with a ratio of two units (1 in. = 25.4 mm). The size of particles can be determined in a number of ways: The nominal diameter refers to the diameter of a sphere with the same volume as that of the particle, usually measured by the displaced volume of a submerged particle, the sieve diameter is the minimum length of the square sieve opening through which a particle will fall, and the fall diameter is the diameter of an equivalent sphere of speciﬁc gravity G = 2.65 having the same terminal settling velocity in water at 24 ◦ C.
Properties of sediment
17
A wetsieve method keeps the sieve screen and sand completely submerged. The sediment is washed onto the wet sieve and agitated somewhat vigorously in several directions until all particles smaller than the sieve openings have a chance to fall through the sieve. Material passing through the sieve with its wash water is then poured onto the nextsmallersize sieve. Particles retained on each sieve and those passing through the 0.062mm sieve are transferred to containers that are suitable for drying the material and for obtaining the net weight of each fraction. The drysieve method is less laborious than the wetsieve method because a mechanical shaker can be used with a nest of sieves for simultaneous separation of all sizes of interest. It requires only that the dry sand be poured over the coarsest sieve and that the nest of sieves be shaken for 10 min on a shaker that has both lateral and vertical movements. 2.3.2
Sediment mixture
The properties of a sediment mixture are sketched in Fig. 2.3. The total volume ∀t is the total of the volume of solids ∀s and the volume of voids ∀v . Particlesize distribution. An example of particlesize distribution in Fig. 2.4 shows the percentage by weight of material ﬁner than a given sediment size. The sediment size d50 for which 50% by weight of the material is ﬁner is called the median grain size. Likewise d90 and d10 are values of grain size for which 90% and 10% of the material are ﬁner, respectively. Gradation coefﬁcients σg and Gr. The gradation of the sediment mixture is a measure of nonuniformity of sediment mixtures. It can be described by 1/2 d84 σg = (2.7a) d16 d16
d50
d 84 Vv Vs
φ Figure 2.3. Properties of a sediment mixture.
or by the gradation coefﬁcient d84 d50 + Gr = 21 . d50 d16
(2.7b)
Both gradation coefﬁcients reduce to unity for uniform sediment mixtures, i.e., when d84 = d50 = d16 . The gradation coefﬁcient increases with nonuniformity, and highgradation coefﬁcients describe wellgraded mixtures.
18
Physical properties and equations
100
v.f.
Sand fine med.
Gravel v.c.
c.
fine med.
v.f.
Cobbles v.c. small large
c.
90 80
% by weight finer than
70 60 50 40 30 20 10 0 0.0625 0.125 0.25
0.5 d
10
1
2
4
8
d 50 Diameter (mm)
16
64
32 d
128 256
90
Figure 2.4. Particlesize distribution: v.f.,very ﬁne; c., coarse; v.c., very coarse.
Angle of repose φ. Typical values of the angle of repose φ of granular material are shown in Fig. 2.5. The angle of repose varies with grain size and angularity of the material. Typical values of the angle of repose are also given in Table 2.4 for material coarser than medium silt. Critical shear stress τc and shear velocity u ∗c . Approximate values of critical shear stress τc for noncohesive particles can be obtained from the extended Shields diagram. The values from Julien (1995) are given in Table 2.4 as approximate reference values. The corresponding critical shear velocity u ∗c is √ deﬁned as u ∗c = τc /ρ. Note that both τc and u ∗c do not change signiﬁcantly for sands and silts. To get crude approximations, a shearstress value of τ = 0.1 Pa is sufﬁcient to move silts but not sands, and τc = 1 Pa is sufﬁcient to move sands but not gravels. 2.3.3
Sediment suspension
The properties of a sediment suspension are sketched in Fig. 2.6, with the volume of void ∀v equal to the volume of water ∀w .
Properties of sediment
180
180
φ = 3 = 60 φ = = 45 4
3 cylinders
φ = 30
19
180
(a)
4 spheres
5 spheres
φ = 35.26
ds (mm) 10
=0
102
rock dge e l d lar she ngu a Cru ry Ve
1:1
0.9 1:1 41
n
ou
r ery
d de
V
Rounded Rounded & angular Angular
25 20
1:1 12 1:1 34
0.8
1:2 1:2 41 1:2 12 1:2 3
0.7
tan φ
35 30
1.0
Sideslope
40
180
(c)
1 45
φ=
φ = 19.46
(b)
φ (degree)
180 = 30 6
φ=
φ = 5 = 36
0.6 0.5 0.4
4
10 2
10 1
1
10
ds (inches)
Figure 2.5. Angle of repose of granular material (after Simons, 1957).
Volumetric sediment concentration Cv . The volumetric sediment concentration Cv is deﬁned as the volume of solids ∀s over the total volume ∀t . When the voids are completely ﬁlled with water, ∀v = ∀w , we obtain Cv =
∀s . ∀s + ∀w
(2.8a)
The most common unit for sediment concentration is milligrams per liter, which describes the ratio of the mass of sediment particles to the volume of the water–sediment mixture. Other units include kilograms per cubic meter (1 mg/l = 1 g/m3 ), the volumetric sediment concentration Cv , the concentration
20
Cv =
Physical properties and equations
Vs Vs + Vw
in parts in 106 (ppm) Cppm , and the concentration by weight Cw . We can easily demonstrate the following identities:
Vs
Cw =
Vv = Vw
= Figure 2.6. Properties of a suspension.
sediment weight total weight Cv G 1 + (G − 1) Cv
(2.8b)
in which G = γs /γ and Cppm = 106 Cw .
(2.8c)
Note that the percentage by weight Cppm is given by 1,000,000 times the weight of sediment over the weight of the water–sediment mixture. The corresponding concentration in milligrams per liter is then given by Cmg/l =
1 mg/l G Cppm = ρ GCv = 106 mg/l GCv . G + (1 − G) 10−6 Cppm
(2.8d)
The conversion factors from Cppm to Cmg/l are given in Table 2.5. Note that there is less than a 10% difference between Cppm and Cmg/l at concentrations of Cppm < 145,000. Table 2.5. Equivalent concentrations for Cv , Cw , Cppm , Cmg/l , P0 , and e Cv
Cw
Cppm
Suspension 0.0001 0.001 0.0025 0.005 0.0075 0.01 0.025
0.00026 0.00264 0.00659 0.01314 0.01963 0.02607 0.06363
265 2,645 6,598 13,141 19,632 26,069 63,625
265 2650 6625 13250 19875 26500 66250
Hyperconcentration 0.05 0.075 0.1 0.25 0.5 0.75
0.12240 0.17686 0.22747 0.46903 0.72603 0.88827
122,401 176,863 227,467 469,027 726,027 888,268
132,500 198,750 265,000 662,500 1,325,000 1,987,500
Note: Calculations based on G = 2.65.
Cmg/l
P0
0.95 0.925 0.9 0.75 0.50 0.25
e
19 12.3 9 3 1 0.33
Properties of sediment
21
Settling velocity ω0 . The settling velocity ω0 of sediment particles in clear water at 10 ◦ C is calculated from (G − 1)g 3 0.5 8ν 1+ ds −1 , (2.9) ω0 = ds 72ν 2 where ds is the particle diameter, ν is the kinematic viscosity, G is the speciﬁc gravity, and g is the gravitational acceleration. The values of the settling velocity in clear water are given in Table 2.4. Speciﬁc weight of a mixture, γm . The speciﬁc weight of a submerged mixture is the total weight of solid and water in the voids per unit total volume. The speciﬁc weight of a mixture, γm , is a function of the volumetric concentration Cv as γm =
γs ∀s + γ ∀v = γs (Cv ) + γ (1 − Cv ). ∀s + ∀v
(2.10)
The speciﬁc mass ρm of a submerged mixture is the total mass of solid and water in the voids per unit total volume. The speciﬁc mass of a mixture is given by ρm = γm /g. Porosity p0 . The porosity p0 is a measure of the volume of void ∀v per total volume ∀t = ∀v + ∀s . The volume of solid particles ∀s = (1 − p0 )∀t is thus p0 =
∀v e , = ∀t 1+e
(2.11)
where the void ratio e is the ratio of the volume of void ∀v to the volume of solid ∀s . The values of porosity and void ratios at various hyperconcentrations are listed in Table 2.5. Dry speciﬁc weight of a mixture, γmd . The dry speciﬁc weight of a mixture is the weight of solid per unit total volume, including the volume of solids and voids. The dry speciﬁc weight of a mixture, γmd , is a function of porosity p0 as γmd = γs (1 − p0 ) = γ G(Cv ).
(2.12)
The dry speciﬁc weight of sand deposits is approximately ∼14.75 kN/m3 or 93 lb/ft3 . The dry speciﬁc mass of a mixture is the mass of solid per unit total volume. The dry speciﬁc mass of a mixture is ρmd = γmd /g.
22
Physical properties and equations
2.4
River ﬂow kinematics
Flow kinematics describes ﬂuid motion in terms of velocity and acceleration. In rivers, two orthogonal coordinate systems are common: (1) global righthand Cartesian (x,y,z) systems, with x in the main downstream direction, y in the lateral direction to the left bank, and z upward; and (2) local cylindrical (r , θ , z) systems, in which r is the river radius of curvature in a horizontal plane, as shown in Fig. 2.7. The rate of change in the position of a ﬂuid element is a measure of its velocity. Velocity is deﬁned as the ratio between the displacement ds and the corresponding increment of time dt. Velocity is a vector quantity v that varies in both space (x,y,z) and time t. Its magnitude v at a given time equals the square root of the sum of squares of its orthogonal components: v=
v x2 + v 2y + v z2 ,
where v x = dx/dt, v y = dy/dt, and v z = dz/dt. A line tangent to the velocity vector at every point at a given instant is known as a streamline. The path line of a ﬂuid element is the locus of the element through time, e.g., the path followed by a single buoy on a river. A streak line is deﬁned as the line connecting all ﬂuid elements that have passed successively at a given point in space, e.g., instantaneous position of all buoys released over time from a single point on a river.
Vector normal to the surface z
Flow
n Surface dA
(x, y, z ) vr v
v
θ
z x x
θ r Radius of curvature
y
Axis z vertical upward
y (a)
Figure 2.7. Cartesian and cylindrical coordinates.
(b)
River ﬂow kinematics
23
The differential velocity components over an inﬁnitesimal distance ds (dx,dy,dz) and time increment dt at a point (x,y,z) are dv x =
∂v x ∂v x ∂v x ∂v x dt + dx + dy + dz, ∂t ∂x ∂y ∂z
(2.13a)
dv y =
∂v y ∂v y ∂v y ∂v y dt + dx + dy + dz, ∂t ∂x ∂y ∂z
(2.13b)
dv z =
∂v z ∂v z ∂v z ∂v z dt + dx + dy + dz. ∂t ∂x ∂y ∂z
(2.13c)
 local  
convective

The Cartesian acceleration components are obtained directly after the velocity equations are divided by the time increment dt, ax =
∂v x ∂v x ∂v x ∂v x dv x = + vx + vy + vz , dt ∂t ∂x ∂y ∂z
(2.14a)
ay =
∂v y ∂v y ∂v y ∂v y dv y = + vx + vy + vz , dt ∂t ∂x ∂y ∂z
(2.14b)
az =
∂v z ∂v z ∂v z ∂v z dv z = + vx + vy + vz , dt ∂t ∂x ∂y ∂z
(2.14c)
 local  
convective

in cylindrical coordinates, with vr = dr /dt, v θ = r dθ/dt, v z = dz/dt, and the properties of curvilinear vectors give additional convective terms in centrifugal acceleration v 2 /r in Eq. (2.15a) and the Coriolis acceleration (vr v θ )/r in Eq. (2.15b): ∂vr v θ ∂vr v2 ∂vr ∂vr dvr = + vr + − θ + vz , dt ∂t ∂r r ∂θ r ∂z ∂v θ v θ ∂v θ vr v θ ∂v θ ∂v θ dv θ = + vr + + + vz , aθ = dt ∂t ∂r r ∂θ r ∂z ∂v z v θ ∂v z ∂v z ∂v z dv z az = = + vr + + vz . dt ∂t ∂r r ∂θ ∂z ar =

local
 
convective
(2.15a) (2.15b) (2.15c)

It is shown in Eqs. (2.14) and (2.15) that the total acceleration can be separated into local and convectiveacceleration terms. Flows in which localacceleration terms vanish at any point are called steady. Flows are uniform when all convectiveacceleration terms vanish. Steadyuniform ﬂows describe the particular case of motion without any acceleration component.
24
Physical properties and equations
2.5
Conservation of mass
The equation of continuity, or law of conservation of mass, states that mass cannot be created nor destroyed. The continuity equation can be written in either differential form, discussed in this chapter, or integral form, to be discussed in subsequent chapters (e.g., Section 5.1). In differential form, consider the inﬁnitesimal control volume in Fig. 2.8 ﬁlled with a ﬂuid and a homogeneous concentration of sediment. The difference between the mass ﬂuxes entering and leaving the differential control volume equal the rate of increase of internal mass. For instance, in the x direction, the net mass ﬂux leaving the control volume is [(∂ρm v x )/(∂ x)] dx times the area dydz. The change in internal mass is (∂ρm /∂t) dxdydz. Repeating the procedure in the y and the z directions yields the following differential relationships: Cartesian coordinates (x, y, z): ∂ ∂ ∂ ∂ρm + (ρm v x ) + (ρm v y ) + (ρm v z ) = 0. ∂t ∂x ∂y ∂z
(2.16a)
ρ ρm vz + ∂ m vz dz ∂z
(up) z
ρm vx
dz ρ ρm v y + ∂ m vy dy
ρm vy
∂y
ρ ρ v + ∂ m vx dx m x
dx
∂x
dy
ρm vz
x (downstream)
Figure 2.8. Inﬁnitesimal element of a ﬂuid.
y (lateral)
Equations of motion
25
Cylindrical coordinates (r , θ , z): 1 ∂ ∂ρm 1 ∂ ∂ + (ρm r vr ) + (ρm v θ ) + (ρm v z ) = 0. ∂t r ∂r r ∂θ ∂z
(2.16b)
For the particular case in which sediment diffusion is not signiﬁcant, the conservation of solid mass is also deﬁned after ρm is replaced with Cv obtained after Eq. (2.10) is substituted into Eq. (2.16). For sedimenttransport problems in which turbulent diffusion and dispersion are signiﬁcant, sedimentcontinuity equation (10.5) in Julien (1995), including turbulentmixing coefﬁcients, should be used. For homogeneous incompressible suspensions without settling, the mass density is independent of space and time (ρs , ρ, ρm = const); consequently ∂ρm /∂t = 0 and the divergence of the velocity vector in Cartesian coordinates must be zero, i.e., ∂v y ∂v z ∂v x + + = 0. ∂x ∂y ∂z
(2.17)
When dealing with openchannel ﬂows at low sediment concentrations, we can neglect compressibility effects, and we ﬁnd that Eq. (2.17) is applicable.
2.6
Equations of motion
The analysis of ﬂuid motion results from the application of forces on a ﬁxed control volume. Given that the force F equals the product of mass m and acceleration a, the approach for ﬂuids of mass density ρ = (m/ ∀) stems from a = (F/m) = (F/ρ∀ ). The forces acting on a Cartesian element of ﬂuid and sediment (dx,dy,dz) are classiﬁed as either internal forces or external forces. The internal accelerations, or body forces per unit mass, acting at the center of mass of the element are denoted by gx , g y , and gz . The external forces per unit area applied on each face of the element are subdivided into normal and tangentialstress components. The normal stresses σx , σ y , and σz are designated as positive for tension. Six shear stresses, τx y , τ yx , τx z , τzx , τ yz , and τzy , with two orthogonal components are applied on each face, as shown in Fig. 2.9. The ﬁrst subscript indicates the direction normal to the face, and the second subscript designates the direction in which the stress is applied. The identities τx y = τ yx , τx z = τzx , and τ yz = τzy result from the sum of moments of shear stresses around the centroid. The cubic element in Fig. 2.9 is considered in equilibrium when the sum of the forces per unit mass in each direction, x, y, and z, equals the corresponding
26
Physical properties and equations ∂ τ zx dz ∂z ∂σ z dz σz + ∂z τ τzy + ∂ zy dz ∂z
τzx +
z
σy ∂ τxz dx ∂x ∂ σx dx σx + ∂x ∂ τxy dx τxy + ∂x
τxz +
τyx
gz
τxy
τxz
gx dz
gy
τyz
τzx
τzy dy
∂ τyz dy ∂y ∂σ σ y + y dy ∂y ∂ τyx dy + τyx ∂y
τyz +
σx
dx
σz y
x
Figure 2.9. Surface stresses on a ﬂuid element.
Cartesian acceleration components ax , a y , and az : ax = gx +
1 ∂τ yx 1 ∂τzx 1 ∂σx + + , ρm ∂ x ρm ∂ y ρm ∂z
(2.18a)
ay = gy +
1 ∂τx y 1 ∂τzy 1 ∂σ y + + , ρm ∂ y ρm ∂ x ρm ∂z
(2.18b)
a z = gz +
1 ∂τx z 1 ∂τ yz 1 ∂σz + + . ρm ∂z ρm ∂ x ρm ∂ y
(2.18c)
These equations of motion are general without any restriction as to compressibility, viscous shear, turbulence, or other effects. The normal stresses can be rewritten as a function of the pressure p and the additional normal stresses, τx x , τ yy , and τzz , accompanying deformation: σx = − p + τx x ,
(2.19a)
σ y = − p + τ yy ,
(2.19b)
σz = − p + τzz .
(2.19c)
Hydraulic and energy grade lines
27
Table 2.6. Equations of motion Cartesian coordinates x component ∂τ yx ∂v x 1 ∂p ∂v x ∂ yx ∂v x 1 ∂τx x ∂τzx ax = + vx + vy + vz = gx − + + + ∂t ∂x ∂y ∂z ρm ∂ x ρm ∂x ∂y ∂z
(2.20a)
y component ∂v y ∂v y ∂v y ∂v y ∂τ yy ∂τzy 1 ∂p 1 ∂τx y ay = + vx + vy + vz = gy − + + + ∂t ∂x ∂y ∂z ρm ∂ y ρm ∂x ∂y ∂z
(2.20b)
z component ∂τ yz ∂v z ∂v z ∂v z ∂v z 1 ∂p 1 ∂τx z ∂τzz + vx + vy + vz = gz − + + + ∂t ∂x ∂y ∂z ρm ∂z ρm ∂x ∂y ∂z
(2.20c)
az =
Cylindrical coordinates r component v2 1 ∂p ∂vr ∂vr ∂vr v θ ∂vr + vr + − θ + vz = gr − ∂t ∂r r ∂θ r ∂z ρm ∂r 1 ∂τθr 1 1 ∂ τθ θ ∂τzr + (r τrr ) + − + ρm r ∂r r ∂θ r ∂z θ component ∂v θ 1 ∂p v θ ∂v θ vr v θ ∂v θ ∂v θ + vr + + + vz = gθ − ∂t ∂r r ∂θ r ∂z ρ r ∂θ m 1 1 ∂ 2 1 ∂τθθ ∂τzθ + (r τr θ ) + + ρm r 2 ∂r r ∂θ ∂z z component ∂v z 1 ∂p ∂v z ∂v z v θ ∂v z 1 1 ∂(τr z ) 1 ∂τθ z ∂τzz + vr + + vz = gz − + + + ∂t ∂r r ∂θ ∂z ρm ∂z ρm r ∂r r ∂θ ∂z
(2.21a)
(2.21b)
(2.21c)
After the acceleration components ax , a y , and az from Eqs. (2.14) are considered, the equations of motion in Cartesian and cylindrical coordinates can be written as shown in Table 2.6.
2.7
Hydraulic and energy grade lines
Let us consider a small element of ﬂuid in a widerectangular channel at a bed slope S0 , as shown in Fig. 2.10. The ﬂow is one dimensional (1D) in the x direction; thus v = v x and v y = v z = 0. The shear stress that is due to element stretching is τx x = 0. The effects of bank shear τ yx in a wide channel can be neglected, τ yx = 0, but the bed shear stress is signiﬁcant, τzx = 0. At small bedslope angles, sin θ tan θ and gx = g sin θ ∼ = gS0 , Eq. (2.20a) thus reduces to 1 ∂τzx ∂v x ∼ 1 ∂p ∂v x + vx + . = gS0 − ∂t ∂x ρ ∂x ρ ∂z
(2.22)
28
Physical properties and equations
gx = g sin θ
θ g
τ zx z
h
z
x
p
So θ
1
τzx = ρg (h z) S
f
p = ρg (h  z
)
Figure 2.10. Pressure and shearstress distributions in a ﬂuid column.
We obtain the pressure distribution by integrating Eq. (2.20c), given that az = 0 and that shearstress variations are small. The resulting hydrostaticpressure approximation that is deﬁned from gz = g cos θ −g from the bed elevation is 0 h dp = ρ −gdz or p = ρg(h − z) (2.23) p
z
at the bed, z = 0, the pressure is p0 = ρgh and the relative pressure vanishes at the free surface, p = 0 at z = h. The bed shear stress τ0 is obtained from the deﬁnition of the friction slope S f for steadyuniform ﬂow as τ0 = ρgh S f . Like pressure, the shearstress vanishes at the free surface and varies linearly over the depth. The shearstress distribution is thus τzx = ρg(h − z)S f .
(2.24)
Relation (2.22) is greatly reduced after Eqs. (2.23) and (2.24) are substituted into it, owing to (∂z/∂ x) = 0 and (∂ S f /∂z) = (∂h/∂z) = 0; thus ∂v x ∼ ∂h ∂v x + vx − gS f . = gS0 − g ∂t ∂x ∂x
(2.25)
We can solve equation of motion (2.25) in dimensionless form for S f after dividing by g: v x ∂v x ∂v x ∂h − − . Sf ∼ = S0 − ∂x g∂ x g∂t
(2.26)
This very important formulation is usually attributed to SaintVenant (1871). The practical signiﬁcance is that point velocities v x can be replaced with the mean ﬂow velocities V , considering that the momentum correction factor is
Problems
29 1
So  ∂h ∂x
∂V So  ∂h  V ∂x g ∂x EGL HGL
1
So
1 Datum
Figure 2.11. Energy grade line (EGL) and hydraulic grade line (HGL).
close to unity: V ∂V 1 ∂V ∂h − − . Sf ∼ = S0 − ∂x g ∂x g ∂t  bed slope   
freesurface slope

energy slope
(2.27)

Finally, the graphical representation in Fig. 2.11 after the consideration that S0 = −(∂z/∂ x). Thus the ﬁrst term on the righthand side of relation (2.27) physically describes the bed slope. The ﬁrst two terms on the righthand side of relation (2.27) describe the freesurface slope. The ﬁrst three terms on the righthand side of relation (2.27) describe the slope of the energy grade line. Problem 2.1 Determine the mass density, speciﬁc weight, dynamic viscosity, and kinematic viscosity of clear water at 20 ◦ C (a) in SI units and (b) in the English system of units. Answers: (a) ρ = 998 kg/m3 , γ = 9790 N/m3 , µ = 1.0 × 10−3 N s/m2, ν = 1 × 10−6 m2 /s, (b) ρ = 1.94 slug/ft3 , γ = 62.3 lb/ft3 , µ = 2.1 × 10−5 lb s/ft2 , ν = 1.1 × 10−5 ft2 /s. Problem 2.2
Determine the sediment size, mass density, speciﬁc weight, submerged speciﬁc weight, and angle of repose of small quartz cobbles (a) in SI units and (b) in the English system of units.
30
Physical properties and equations Problem 2.3
The volumetric sediment concentration of a sample is Cv = 0.05. Determine the corresponding (a) concentration by weight, Cw , (b) concentration in parts in 106 , Cppm , (c) concentration in milligrams per liter, Cmg/l , (d) porosity p0 , and (e) void ratio e. Answers: The answers to Problem 2.3 are in Table 2.5. Problem 2.4
The porosity of a sandy loam is 0.45. Determine the corresponding soil properties: (a) volumetric concentration, (b) void ratio e, (c) speciﬁc weight γm , (d) speciﬁc mass ρm , (e) dry speciﬁc weight γmd , and (f) dry speciﬁc mass ρmd . Problem 2.5
Calculate the gradation coefﬁcients σg and Gr from the particlesize distribution shown in Fig. 2.4. Answer: σg =
1 32 8 32 = 8, Gr ≈ + = 10, 0.5 2 8 0.5
i.e., a wellgraded mixture. Problem 2.6
Apply the continuity equation to 1D runoff at a unit discharge q and ﬂow depth h for rainfall intensity i e on an impervious plane. Demonstrate that (∂h/∂t) + (∂q/∂ x) = i e , where x denotes the downstream direction and t denotes time. Problem 2.7 Demonstrate Eq. (2.8b) from the deﬁnition that Cw is the sediment weight/total weight.
3 River basins
This chapter covers the characteristic of river basins (also called watersheds) that affect surface runoff and sediment yield. The main watershed characteristics are illustrated in Section 3.1, followed by rainfall precipitation (Section 3.2), interception and inﬁltration (Section 3.3), excess rainfall (Section 3.4), and surface runoff (Section 3.5). Soil eroded from upland areas is usually the source of most sediments that are transported by rivers to reservoirs and estuaries. Methods are presented to calculate upland erosion (Section 3.6) and to estimate sediment yield from watersheds (Section 3.7). 3.1
Riverbasin characteristics
The hydrologic cycle describes processes that contribute to the source and the yield of water and sediment from upland areas to the ﬂuvial system. Figure 3.1 depicts a portion of a watershed during a precipitation event. Shown in this ﬁgure are the processes of condensation, precipitation, interception, evaporation, transpiration, inﬁltration, subsurface ﬂow, exﬁltration, deep percolation, groundwater ﬂow, surface ﬂow, surfacedetention storage, channel precipitation, evaporation, and streamﬂow. All of these processes play a role in hydrology; however, precipitation, inﬁltration, overland ﬂow, and streamﬂow are most important in surface runoff, upland erosion, and river mechanics. River basins or watersheds deﬁne areas of the Earth’s surface where rainwater drains into a particular stream. The terms basin and catchment are synonymously used in the literature. Watershed characteristics can often be described in geographical terms, including physiography and topography, geology and pedology, and climatology and forestry. Watershed boundaries are delineated by drainage divides, usually located at high points, that separate different drainage areas. Watershed physiography is described primarily by topographic maps. The elevation relates to the type of precipitation in terms of rain and snow. The surface slope indicates the rate at which potential energy (PE) is transformed 31
32
River basins
Condensation Surfacerunoff divide
Surfacedetention storage
Precipitation Evaporation
Snow Surface runoff
Infiltration
Transpiration
Deep percolation
Interception
Subsurface flow
Exfiltration Evaporation Channel precipitation
Groundwater divide
Groundwater flow
Figure 3.1. Hydrologic cycle.
into the kinetic energy (KE) of surface waters. The surface slope is also a dominant parameter in the calculation of soil erosion and sediment transport. The topography of small watersheds can nowadays be depicted with a digital elevation model with data available at 30m resolution. Geographic information systems (GISs) provide standard procedures for slope calculations from the mean elevation at each pixel. Limitations lie in the size of the data ﬁles and the computing power of computers performing hydrologic calculations. Digital elevation models provide the following physiographical characteristics of a watershed: (1) drainage area, (2) extreme elevations, (3) hypsometric curves of the percentage of the drainage area at or below a given elevation (see Fig. 3.2), (4) average value and distribution of the terrain slope, (5) percentage of drainage area in lakes and reservoirs, and (6) drainage length from the physically remotest point on the watershed to the outlet. The mainstream length of a river varies with drainage area, and the relationship shown in Fig. 3.3 indicates that mainstream length increases approximately with the square root of the drainage area. Geologic information indicates the overall erodibility of large watersheds. For instance, the Yellow River in China carries a large sediment load from the Loess plateau, and the St. Lawrence River drains the granitic Laurentian shield
Riverbasin characteristics
33 1.0
Hw
hw Base plane
Div
Summit
ide
Mouth
Contour
Relative height h w /H w
Summit plane
0.8 0.6 0.4 0.2 0
Area at
Area At (entire watershed)
0
0.2
0.4 0.6 0.8 Relative area a t /At
1.0
Figure 3.2. Hypsometric curve of a watershed.
A t (km 2) 10
1
3
10
10
2
10
3
10
4
10
10
2
Lr
10
3
2
0.50
L r (km)
Mainstream length L r (miles)
At
L r = 1.73 A t 10
10
1 1 10
1
1
2
10 10 Drainage area A t (sq. miles)
10
3
10
4
Figure 3.3. Mainstream lengths of watersheds.
at an average sediment concentration of 20 mg/l. In large rivers, the location of faults can be useful in detecting changes in bed elevation through tectonic activity and also in tracking possible lateral shifting of rivers through ancient times. Paleohydrology provides broad guidance to ﬂuvial geomorphologists as to what might have been ancient ﬂuvial conditions. Without quantitative measurements of ancient tectonic activities, such information, however, often remains quite speculative.
34
River basins
Figure 3.4. Niger River: (a) climate, (b) rainfall precipitation, and (c) potential evapotranspiration.
Rainfall precipitation
35
Potential evapotranspiration > 2000 mm > 1500 mm > 1000 mm
22
00
2100
Ni
Tombouctou
ge
Malanville
rR
2000
Tchad Lake
.
Ban i R.
Koulikoro Bamako
Mopti
Sokoto R. 1900 1800
1700
160
1400
Niamey
Kaduna R.
1500
Lakoja 1500
Guinea Gulff 0
.
eR
nu Be
0
140
0
[mm]
200 400 km
(c)
Figure 3.4. (cont.)
Soil types are usually classiﬁed according to agronomic standards, such as the Soil Conservation Service (SCS) classiﬁcation. The digitized information at some ﬁne resolutions enables the estimation of inﬁltration characteristics and surfaceroughness parameters for the calculation of surface runoff. Climatic conditions are also known to change over sufﬁciently large watersheds, and many change over time, e.g., desertiﬁcation. For instance, the climate of the Niger river watershed ranges from arid to humid [Fig. 3.4(a)], and the vegetation changes from that of a desert to that of a rain forest. Qualitative changes in vegetative cover can often be corroborated with mean annual rainfall precipitation and mean annual potential evapotranspiration, as shown in Figs. 3.4(b) and 3.4(c).
3.2
Rainfall precipitation
In the United States, the mean annual rainfall precipitation increases southward from 50 mm in the north to over 4,000 mm near the Mississippi River delta. Such spatial variability in climate and vegetation is typical of very large rivers (over 1 × 106 km2 of drainage area). This contrasts with the relative homogeneity in climate and vegetation of most small watersheds. In terms of large events,
36
River basins
three hour  100 year storms Canada 45
tic O c
ean
40
25
< 2 inches < 4 inches < 6 inches < 8 inches
Pacific Ocean
30
Atlan
35
Mexico 110
115
105
Gulf of Mexico 95
100
90
80
85
75
Figure 3.5. 3h storm with 100yr period of return for the United States. 4
10
Rainfall depth h r (cm)
10 3
h r = 6 t r 0.5
10 2
10 1 hour 1 1
10
1 day
10 3 10 4 10 2 Rainfall duration tr (min)
1 month
1 year
10 5
10 6
Figure 3.6. Maximum precipitation as a function of rainfall duration.
Fig. 3.5 shows the distribution of the rainfall precipitation with a duration of 3 h and a period of return of 100 yr in the continental United States. Note the spatial variability that ranges from less than 2 in. in Nevada to more than 7 in. in Louisiana. Figure 3.6 shows the world’s largest precipitation events in terms of rainfall depth as a function of rainfall duration.
Rainfall precipitation
37
Storm
Watershed Physical scales
LS
L
R
L
W
Measurement scales LM
Figure 3.7. Characteristics of rainstorms and watersheds.
Consider a convective rainstorm passing over a watershed, as in Fig. 3.7. There are four length parameters associated with rainstorms and watersheds in the context of rasterbased hydrologic models: (1) the correlation length of the rainstorm cell, L S ; (2) the grid size of the rainfall precipitation data, L R ; (3) the characteristic length of the watershed, L W ; and (4) the runoffmodel gridcell size, L M . The radar and the model resolution parameters, L R and L M , are deﬁned by the modeler, whereas the basin and the stormsize parameters, L W and L S , are determined by the ﬁeld site and the natural storm size in the study area. In addition, it is important to note that the correlation length for watershed characteristics is of the order of tens to hundreds of meters, whereas the spatial correlation length for rainfall, L S , is hundreds to thousands of meters. An example of a rainfall event measured by a network of 46 rain gauges with an average spacing of 10.6 km is that in the Denver area, as shown in Fig. 3.8. This small mesoscale rainstorm has a diameter of 20–30 km, and the spatial cross correlation between 2min rainfall data in a frame of reference moving with the rainstorm is shown in the same ﬁgure. The correlation coefﬁcient becomes close to zero at a raingauge separation distance of ∼16 km. This means that, for this particular storm, 2min rainfall measurements with an average spacing of 15 km would be essentially uncorrelated. May and Julien (1998) demonstrated the importance of using a system of coordinates that moves with
38
River basins 30 km > 0.2 inch/hour > 0.4 inch/hour > 0.8 inch/hour
Period 2
Period 6
Period 10
Period 14
Correlation coefficient
1.0 0.8 0.6 0.4 0.2 0 0.2 0
2
4
6 8 10 12 Raingauge separation distance (km)
14
16
Figure 3.8. Rainfall correlation coefﬁcients vs. raingauge distance (after May and Julien, 1998).
the center of mass of the storm cell in order to have physically representative correlation coefﬁcients. The reason for this is that some rainstorms can move at average velocities exceeding 100 km/h. Conceptually, there are three conditions required for appropriate distributed hydrologic modeling. The ﬁrst is L M L W , so that the basin is subdivided into sufﬁciently small grid elements to describe the spatial variability of basin characteristics. The second condition is L R L S , which ensures preservation of the spatial gradients of rainfall, particularly for smaller, convective rainstorms. The third requirement is that L R ≤ L M , such that rainfall is placed in the correct watershed grid cell. The third requirement is most difﬁcult to satisfy with either rain gauges or radar. Stateoftheart rainfall precipitation estimates combine radar measurements for spatial variability and rain gauges for groundtruth calibration of radar measurements. Recent polarimetric weather radar techniques offer an opportunity to record spatially distributed rainfall events with unprecedented resolution. The
Rainfall precipitation
39
50 40
Distance (km)
30 20 10 0 10 20
(a) 12:48:00 MDT (b) 13:56:00 MDT 30 40 30 20 10 0 10 20 30 40 30 20 10 0 10 20 30 40 Distance (km) Distance (km) CSUCHILL Rainfall (mm/h); June 3, 1991; Elev. angle 1. 6°, Contour interval 10 mm/h
Figure 3.9. Colorado rainfall rates measured by the CSUCHILL radar (after Ogden and Julien, 1994).
weather radar data shown in Fig. 3.9 was recorded at the Colorado State University (CSUCHILL) radar facility, near Greeley, Colorado. The CSUCHILL radar is a dual linearly polarized coherent radar that operates at 2.75 GHz, which corresponds to a wavelength of 10.7 cm in the S band. The beam width of the 8mdiameter antenna is approximately 1.0◦ . Attenuation does not affect Sband weather radars, even in the heaviest rainfall, making them ideal for weather observation. The radar transmits an average power of 1 kW, with peak power during transmit pulses of 1 MW, and can measure Doppler velocities. Ground clutter, high reﬂectivity, ice and graupel phase, representative precipitation values at different elevations (or CAPPIs), and ground calibration remain active research areas to improve the accuracy of radar measurements. Figure 3.9 shows multiparameter rainfall rates within the 80 km × 80 km data domain at 12:48 MDT and 13:56 MDT. The radar was located at the origin of the plots. The rainfall ﬁelds were converted from radar conical coordinates to rectangular coordinates by interpolation to a 1 km × 1 km grid size. The scale of individual storm cells was approximately 4–10 km, with an average storm size near 8 km. The rainfall data from 27 radar scans were processed, and the correlation length L S was determined to be 2.3 km by both covariance models. Note that the precipitation patterns of convective rainstorms change very rapidly in both time and in space. Discretized rainfall precipitation can serve as input to rasterbased hydrologic models such as the CASC2D (Julien et al., 1995).
40
River basins
Point rainfall precipitation can be described as a random time series of discrete storm events, each having i ﬁnite duration and constant intensity. 3 i1 The principal variables of rainfall i2 precipitation sketched on Fig. 3.10 are tr tr t (1) the storm duration tr , (2) the storm tr 1 3 2 intensity i, and (3) the time arrival between successive storms. The time of Figure 3.10. Deﬁnition sketch of hyarrival of successive storms has often etographs. been described as a Poisson process. The emphasis is on the storm duration tr , average intensity i, and the rainstorm depth h r obtained from h r = itr . Considering that the average storm duration is i
t¯r =
N 1
tr , N 1
where N is the number of storms, the normalized storm duration tr∗ = tr /t¯r is distributed exponentially. The particular characteristic of exponential distribution is that the probability density function p( ) and the exceedance probability function E( ) are identical: ∗
E(tr∗ ) = p(tr∗ ) = e−tr .
(3.1)
The corresponding cumulative distribution function F(tr ) or nonexceedance probability, is obtained from F(tr ) = F(tr∗ ) = 1 − E(tr∗ ).
(3.2)
An example of the distribution of rainfall precipitation data is provided in Fig. 3.11. Similarly, the storm intensity i can be normalized after being divided by the average intensity N 1
i i¯ = N 1
over N storms. The normalized intensity i ∗ = i/i¯ is exponentially distributed with a probability density function p( ) equal to the exceedance probability E( ) as ∗
E(i ∗ ) = p(i ∗ ) = e−i , ∗
(3.3) ∗
F(i) = F(i ) = 1 − E(i ).
(3.4)
Rainfall precipitation
41
1.0 Rainfall duration 0.8
F (t r )
0.632 0.6 June Observed Fitted 58 events
0.4 4
t r ~ 1.5 x 10 s 0.2 0
0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
t r (x 10 4 s)
Figure 3.11. Cumulative distribution function of rainfall duration.
1.0 Rainfall intensity 0.8
F (i)
0.6 June
0.4
Observed Fitted 58 events
6
i c ~ 1 x 10 m/s 0.2 0
0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
6
i (x 10 m/s)
Figure 3.12. Cumulative distribution function of rainfall intensity.
For instance, the observed and the ﬁtted exponential distribution functions for rainstorm intensity, F(i ∗ ), are shown in Fig. 3.12. For the example shown in Figs. 3.11 and 3.12 with 6 yrs of rainfall data, monthly periods were found to be sufﬁciently long to assume constant values of average rainfall duration and intensity for each month. Detailed statistical analyses can show whether rainstorm duration and intensity are nearly independent for the period considered. Example 3.1 illustrates how to determine the observed and the ﬁtted exponential distribution functions.
42
River basins
Nonexceedance probability
Example 3.1 Application to distribution of rainfall duration. The rainfall durations of N = 10 rainstorms measured during one month are 50, 5, 25, 210, 320, 150, 45, 40, 25, and 40 min. Determine the distribution of these events and compare with the exponential distribution. Note that the procedure should normally be applied to large samples. In Table E.3.1.1, the events are listed in column 1 and ranked in decreasing order in column 2. Their increasing rank in column 3 is divided by N + 1 = 11 in column 4 to determine the meaRainfall duration sured nonexceedance probability Fm ( ). 1.0 The measured exceedance probability 0.8 E m ( ) = 1 − F( ) is given in column 5. Sample The rainfall duration tr is normalized 0.6 Theoretical as tr∗ = tr /t¯r after being divided by the 0.4 average rainfall duration t¯r = 91 min. 0.2 The calculated nonexceedance proba∗ 0 bility E c ( ) = 1 − e−tr . The comparison 4 0 2 3 1 between measured and calculated nonexNormalized duration t ∗r = t r / t r ceedance probability curves is shown in Fig. E.3.1.1. Note that this Figure E.3.1.1. Distribution of rainfall sample is very small in this case. duration. Case Study 3.1 The Big Thompson River flood, United States. The two reports of Grozier et al. (1976) and McCain et al. (1979) document the Big Thompson River ﬂood of July 31–August 1, 1976. As much as 12 in. (305 mm) Table E.3.1.1. Nonexceedance probability curves for rainfall duration
Duration tr (min)
Ranked duration
50 5 25 210 320 150 45 40 25 40
5 25 25 40 40 45 50 156 210 320
Rank
Measured nonexceedance probability Fm ( )
Measured exceedance probability Em ( )
1 2 3 4 5 6 7 8 9 10
0.09 0.18 0.27 0.36 0.45 0.54 0.64 0.73 0.82 0.91
0.91 0.82 0.73 0.64 0.55 0.46 0.36 0.27 0.18 0.09
Average t¯r = 91 min, N = 10
tr∗ =
tr t¯r
0.05 0.27 0.27 0.44 0.44 0.49 0.55 1.65 2.31 3.51
Calculated nonexceedance probability Fc ( ) 0.05 0.24 0.24 0.35 0.35 0.39 0.42 0.81 0.90 0.97
Rainfall precipitation
43
of rain fell on the Big Thompson River basin during the evening of July 31, 1976, causing a devastating ﬂood on the Big Thompson River and its tributaries between Estes Park and Loveland, Colorado. Larimer County ofﬁcials reported 139 lives lost and property damage of $16.5 million. During the evening hours of July 31, a series of violent thunderstorms (Fig. CS.3.1.1) released large volumes of rain along a path several miles wide from Estes Park to the Wyoming border, as sketched in Fig. CS.3.1.2. The Big Thompson River basin west of Drake was severely hit by the storm, and devastating ﬂooding occurred along the Big Thompson River between Estes Park Distance from Storm Mountain (miles) 10 10 0
20
20
60
18 55
25 35 45
40
15
12
35
10 30 25°C
8
25 20
6
15
0°C 4
10 Storm Mountain Big Thompson River
2
0 40
30
15
0°
20 10 0 10 20 Distance from Storm Mountain (km)
5
30
Schematic lines of airflow Schematic area of rainfall Radar reflectivity observed at Grover, Colorado Dashed where approximately located. Interval 10 dBZ Line of equal air temperature, in degrees Celsius Dashed within the cloud
Figure CS.3.1.1. Thunderstorms of the Big Thompson ﬂood.
40
0
Altitude above mean sea level (thousand of feet)
45
14
45
Altitude above mean sea level (thousand of meters)
50
15 25
16
River basins 10
10
8
8 Ft. Collins
6
8
Glen Haven
4
Mouth of Big Thompson Canyon Ba sin bo un da Loveland r
2 Masonville
Big Th
ompso
Loveland powerplant
10 8
Glen 2 Comfort
0 2 4 6 8 10 12
y
Drake Estes Park
4
Rainfall precipitation (inches)
2 4 6
44
n Rive
r
6
2
e att
s
omp
e Th
Littl
6
uth
5
0 0
5
r
ve
r
ive on R
Pl
So
10
Ri
10 miles 15 km
Figure CS.3.1.2. Rainfall precipitation of the Big Thompson ﬂood.
July 31 August 1
Estimated cumulative rainfall (in)
and Loveland and along the North Fork of the Big Thompson River from Glen Haven to its mouth at Drake. The isohyetal map of the total precipitation from July 31 to August 2, 1976, is shown in Fig. CS.3.1.2. Eastern Colorado was under conditions favorable for heavy rain on July 31, 1976, for a number of reasons. The surface map of that morning showed a slowly moving cold front in the state. Such fronts display lines of convergence that lift air to form thunderstorms. Also favorable was the easterly wind just north of the front, moving air upslope and aiding the frontal lifting. The lowlevel air was very moist, well above the seasonal normals, and the moisture aloft was also unusually high. Thunderstorms move with the 12 speed and the direction of the winds 10 Glen Comfort aloft, and the 500mbar (millibar) level 8 is usually adequate for judging such Glen Haven 6 movement. The 500mbar wind was 4 only ∼5 knots and was not expected to change much during the day. This was 2 the case with the thunderstorms near 0 1900 2000 2100 2200 2300 2400 Estes Park. They moved very slowly Time (h (MDT)) while putting out large amounts of water over a period of several hours, as Figure CS.3.1.3. Cumulative rainfall precipitation. shown in Fig. CS.3.1.3.
Rainfall precipitation
45
Rainfall began at approximately 18:30 MDT on July 31, 1976, and ended at approximately 23:30 MDT that evening. Additional rainfall was observed on August 1 and 2. Precipitation totals were as much as 10 in. (254 mm) between Estes Park and Drake and more than 12 in. (305 mm) in the Glen Haven area. Very little rainfall contributed to the ﬂood east of Drake and west of Estes Park. Flood runoff in the Big Thompson basin derived from an area of approximately 60 square miles (155 km2 ) centered on the Big Thompson River from Lake Estes to Drake. The topography of the area is characterized by steep northand southfacing slopes with rugged rock faces along the ridges and a thin soil mantle at lower elevations that supports a moderate stand of coniferous trees. Because of the steep slopes and small storage capacity of the soils, the storm runoff quickly reached nearby surface channels. The ﬂood lasted only a few hours. The reported peak stages on the Big Thompson River occurred as follows: 20:00 at Glen Comfort, 21:00 at Drake, 21:30 at the Loveland power plant, and approximately 23:00 at the mouth of the canyon ∼8 miles (13 km) west of Loveland. The relative timing of the peak stages was such that the peak on the Big Thompson River just downstream from Drake occurred before the peak from the North Fork arrived at Drake. The ﬂood peak moved through the 7.3mile (11.7km) length of channel between Drake and the canyon mouth in ∼2 h with no apparent reduction in discharge. 6
3
Peak discharge (ft /s)
10
5
10
4
10
1 m3/s = 35.8 ft 3/s 1 mi2 = 2.59 km 2 3
10 1 10
1
2
10 10 Drainage area (square miles)
10
3
Figure CS.3.1.4. Big Thompson peak discharge vs. drainage area.
4
10
46
River basins
East of the canyon mouth, the Big Thompson River valley widens rapidly and the ﬂood discharge was quickly reduced by valley storage and overﬂow to numerous reservoirs. The peak discharge at the mouth of the Big Thompson River near LaSalle was ∼2,500 ft3 /s (70.8 m3 /s), occurring at noon on August 1, as compared with 31,200 ft3 /s (883 m3 /s) ∼35 miles (56 km) upstream at the mouth of the canyon. The peak discharges at various locations are shown as a function of drainage area in Fig. CS.3.1.4.
3.3
Interception and inﬁltration
The excess rainfall volume available for surface runoff is the volume of precipitation in excess of the rainfall losses. Losses include interception, evapotranspiration, surface detention, storage, and inﬁltration. Some rainfall is intercepted by vegetation before it reaches the ground. The amount of interception varies with the type, density, and stage of growth of the vegetation, intensity of the rainfall, and wind speed. A dense forest canopy may intercept as much as 25% of the annual rainfall in climates with frequent, light rainfalls, low wind speeds, and evergreen vegetation. On a single storm basis, the interception storage is generally a small percentage of the total rainfall event, except in dense forests. Evapotranspiration is the combination of evaporation and transpiration. Evaporation refers to the phase change of water from liquid to vapor from wet surfaces. Water evaporation from plant surfaces is termed transpiration. All evaporation from a leaf surface is not transpiration as intercepted water is also evaporated. On an annual basis, evapotranspiration generally involves a large fraction of the total precipitation, e.g., refer to Figs. 3.4 for a comparison between arid and humid climates. In spite of the high losses on an annual basis, evapotranspiration is usually neglected in cases in which severe rainstorms are considered. Detention storage is the volume of water required for ﬁlling land depressions before surface runoff begins. Detention storage is the depth of water required for initiating surface runoff. Actual measurements of surface storage and detention are practically nonexistent. Detention storage is estimated at 0.2–0.6 in. (0.5– 1.5 cm) in pervious areas such as open ﬁelds, woodlands, and lawn grass. Values of 0.05–0.3 in. (0.13–7.6 cm) relate to paved surfaces and roofs in urban areas. Inﬁltration is the process of water’s permeating into soil pores. In general, the inﬁltration rate is dependent on soil physical properties, vegetative cover, antecedent soilmoisture conditions, rainfall intensity, and the slope of the inﬁltrating surface. Bare soils tend to have lower inﬁltration rates than soils protected by
Interception and inﬁltration 100% Clay
47
0 10
90
20
80
30
70
cla
y
40
ent Pe rc
Silty clay
Sandy clay
40
60 Clay loam
30
Sandy clay loam
20 10
50
lt
50
si nt rce Pe
Clay
60
Silty clay loam
70 80
Loam
Lo
Sandy loam
am
Sand 0 90 100% Sand
Silt loam
90
ys
and 80
Silt 70
60
50 40 Percent sand
30
20
10
100% Silt
0
Figure 3.13. Triangular soil classiﬁcation.
a vegetative cover. The impact of falling raindrops breaks down soil aggregates, and small particles are carried into the soil pores, thus sealing the surface and reducing the inﬁltration rate. The antecedent moisture condition also alters the inﬁltration rate, and wet soils have lower inﬁltration rates than dry soils. On steep slopes, the water tends to run off rapidly and there is less inﬁltration. The soil classiﬁcation in Fig. 3.13 depends on the percentage of sand, silt, and clay in the soil. For instance, a soil with 60% sand, 30% silt, and 10% clay is a sandy loam. Green and Ampt (1911) developed an approximate inﬁltration model based on Darcy’s law. They assumed vertical ﬂow in a column initially at a uniform water content p0i and saturation at the surface p0e . The saturated wetting front moves gradually into the soil zone unaffected by inﬁltration, as sketched in Fig. 3.14. The inﬁltration rate f (t) varies with time as the pistontype wetting front advances into the soil: f (t) =
K (h + L f + h p ) , Lf
(3.5)
48
River basins
where K is the hydraulic conductivity of the wetted soil part of the soil proﬁle, L f is the depth of the wetting Lf front, h p is the pressure head for wetting at the wetting front, and h is the Wetting front depth of ponding of water on the soil surface. p oi ∆po In general, the ﬂow depth h is small poe compared with the length of the wetting front L f and may be neglected in Eq. (3.5). The change in water content Figure 3.14. Pistontype inﬁltration across the wetting front, p0 , depends approximation. on the initial water content p0i , the thoroughly drained (or residual) water content, p0r , the effective saturation Se , and the total porosity p0 . The relationship is p0 = p0e − p0i = p0e − Se p0e = (1 − Se ) p0e , where the effective porosity is given by p0e = p0 − p0r and Se = p0i / p0e . By noting that F(t) = L f p0 or L f = F(t)/p0 and taking h as zero, we ﬁnd that the inﬁltration rate f (t) varies with time as po
h
h p p0 f (t) = K +1 , F(t)
(3.6)
where F(t) is the cumulative inﬁltration depth at time t. The cumulative inﬁltration is found by integration of Eq. (3.6) as F (t) F (t) = K t + h p p0 ln 1 + h p p0
(3.7)
Equations (3.6) and (3.7) apply for the case in which the ponded depth is negligible. If this is not the case, h p should be replaced with h p + h. Rawls et al. (1983) present values for p0 , p0e , h p , and K as functions of soil type in Table 3.1. In practice, because Eq. (3.7) cannot be solved explicitly for F(t), an iterative procedure is required. A trial value for F(t) is substituted into the righthand side of the equation, which is then compared with the lefthand side. This process is repeated until agreement between the two values is obtained. A good ﬁrst estimate is K t for F(t). Possibly an easier way to calculate f (t) from Eq. (3.6) is to solve Eq. (3.7) for t at various values of F(t). The cumulative inﬁltration F(t) can then be used in Eq. (3.6) to determine f (t) at corresponding time.
Interception and inﬁltration
49
Table 3.1. Green–Ampt inﬁltration parameters (after Rawls et al. 1983)a
Soil texture
p0
p0e
hp (cm)
K (cm/h)
Sand Loamy sand Sandy loam Silt loam Loam Sandy clay loam Clay loam Silty clay loam Sandy clay Silty clay Clay
0.437 0.437 0.453 0.501 0.463 0.398 0.464 0.471 0.430 0.479 0.475
0.417 0.401 0.412 0.486 0.434 0.330 0.309 0.432 0.321 0.423 0.385
4.95 6.13 11.01 16.68 8.89 21.85 20.88 27.30 23.90 29.22 31.63
11.78 2.99 1.09 0.65 0.34 0.15 0.10 0.10 0.06 0.05 0.03
a
Rawls et al. (1983) contains more information on these parameters including their standard deviations and values for various soil horizons.
Example 3.2 illustrates how to calculate inﬁltration by use of the Green–Ampt method. Typical potential inﬁltration curves for initially dry soils in terms of f (t), F(t), and f (t)/K for cumulative rainfall inﬁltration up to 5 cm are shown in Figs. 3.15(a), 3.15(b), and 3.15(c), respectively. Example 3.2 Calculation of the inﬁltration potential. Calculate the inﬁltration curve for a silt loam at 30% effective saturation. Solution: Given Se = 0.3, Table 3.1 provides the following characteristics for a silt loam as p0e = 0.486, h p = 16.7 cm, and K = 0.65 cm/h. We then calculate p0 = (1 − Se ) p0e = (1 − 0.3) 0.486 = 0.34. For instance, we compute the cumulative inﬁltration F(t) = 1 cm by rearranging Eq. (3.7): F(t) F (t) − h p p0 ln 1 + h p p0 t= K 1 1 − 16.7(0.342)ln 1 + 16.7(0.34) = 0.12 h. = 0.65 From Eq. (3.6), the inﬁltration rate at t = 0.12 h is h p p0 16.7(0.34) f (0.12 h) = K + 1 = 0.65 + 1 = 4.36 cm/h. F(t) 1
River basins
Infiltration rate f (t) (cm/h)
10 2
Sand 10 Silty clay Clay
Loamy sand Sandy loam Silt loam Loam Sandy clay loam Clay loam Silty clay loam Sandy clay
1
101
(a) Infiltration rates for dry soils 10 Silt loam Sandy loam
Loam
F (t) (cm)
Loamy sand Sand
Sandy clay loam Clay loam
Silty clay loam Sandy clay
1
Silty clay Clay
101 (b) Cumulative infiltration rates for dry soils Sandy clay loam
Clay loam
10 2
Silty clay loam
Loam
Sandy clay
f (t)/K
50
Silty clay
10 Sand Loamy sand Sandy loam 1 10 2
Clay Silt loam
10
1
10 2
10 10 Time (min) (c) Relative infiltration rates for dry soils 1
Figure 3.15. Inﬁltration characteristics for dry soils.
3
10
4
Excess rainfall
51
Table E.3.2.1. Green–Ampt inﬁltration calculations F(t)a (cm)
tb (min)
f (t)c (cm/h)
0.0 0.25 0.5 0.75 1.0 1.5 2.0 3.0 4.0 5.0
0.0 0.5 1.9 4.2 7.3 15.6 26.4 54.6 90.0 130.0
∞ 15.4 8.0 5.6 4.4 3.4 2.5 1.9 1.6 1.4
a
Assumed. From Eq. (3.7) with Se = 0.3, p0e = 0.486, h p = 16.7 cm, K = 0.65 cm/h, and p0 = 0.34 c From Eq. (3.6). b
Table E.3.2.1 illustrates the values of inﬁltration calculated from the assumed values of F. A plot of f versus t from columns 3 and 2 yields the inﬁltration curve.
3.4
Excess rainfall
Excess rainfall represents the supply of water to the surfacerunoff process. Excess rainfall represents the amount of rainfall in excess of interception, evapotranspiration, and inﬁltration. When the rainfall rate exceeds the inﬁltration rate, detention storage begins to ﬁll. Runoff will begin where the detention storage is ﬁlled. When the rainfall rate drops below the inﬁltration rate, water in surface storage is gradually depleted until surface runoff ceases. The water in detention storage then inﬁltrates. Example 3.3 provides the detailed calculation of excess rainfall by use of the Green–Ampt inﬁltration equation.
Example 3.3 Calculation of excess rainfall. Use the Green–Ampt equation to calculate the excess rainfall for the rainstorm in the ﬁrst two columns of Table E.3.3.1. The maximum detention storage is 0.75 cm, and the soil is a silt loam soil with 30% effective saturation.
52
River basins
Table E.3.3.1. Excess rainfall calculations
ta (min)
Rainfalla it (cm)
f p (t)b (cm/h)
0 15 30 45 60 75 90 105 120 135 150 165 180
0.00 0.25 0.30 0.43 0.66 1.55 4.85 0.91 0.51 0.36 0.28 0.23 0.20
Very large 15.40 7.36 4.42 2.90 2.21 1.91 1.74 1.61 1.52 1.45 1.39
Totals
10.53
F p (t)c (cm)
Inﬁltrationd Fa (t) (cm)
Very large 3.85 1.84 1.10 0.73 0.55 0.48 0.43 0.40 0.38 0.36 0.35
0.25 0.30 0.43 0.66 0.73 0.55 0.48 0.43 0.40 0.38 0.36 0.35
=
F p (t)e (cm)
Detention storage f S (cm)
Excess rainfallg Re (cm)
0.00 0.25 0.55 0.98 1.64 2.37 2.92 3.40 3.83 4.23 4.61 4.97 5.32
0.00 0.00 0.00 0.00 0.00 0.75 0.75 0.75 0.75 0.71 0.61 0.48 0.33
0.00 0.00 0.00 0.00 0.00 0.07 4.30 0.43 0.08 0.00 0.00 0.00 0.00
5.32
0.33
+ 4.88
hyetograph is in the ﬁrst two columns; rain starts at t = 15 min. inﬁltration rate from F p at the previous time increment. c Potential inﬁltration volume F = f t. p p d The actual inﬁltration volume is the smaller of column 4 and column 2 + column 7. e F (t) = F (t − 1) + F . p p a f Surface storage. g Excess rainfall. a The
b Potential
Solution: The parameters for the Green–Ampt equation from Table 3.1 are p0e = 0.486, h p = 16.68 cm, and K = 0.65 cm/h. The change in water content as a result of the passing of the wetting front is calculated as p0 = (1 − Se ) p0e = (1 − 0.3)0.486 = 0.34. The inﬁltration rate f (t) and the cumulative inﬁltration F(t) are related by Eq. (3.7) as f (t) = K
h p p0 cm 16.68(0.34) + 1 = 0.65 +1 . F(t) h F(t)
The rainfall hyetograph is deﬁned in the ﬁrst two columns of Table E.3.3.1. Note that the rainfall starts at t = 15 min. Column 3 represents the potential inﬁltration rate f p (t) calculated from the preceding equation with F p from the previous time increment. Column 4 gives the potential inﬁltration volume F p (t) calculated as f p (t)t. The actual inﬁltration volume Fa (t) for the
Surface runoff
53
time increment is the smallest of columns 4 vs. column 2 plus column 7 for the increment. Early in the storm, i.e., during the ﬁrst 60 min, the potential inﬁltration rate exceeds the rainfall rate and the actual inﬁltration volume is limited to the rainfall volume. The cumulative inﬁltration F p (t) at a particular time is equal to F p (t) from the previous time increment plus Fa (t) for the current time increment. In the time increment from 60 to 75 min, the potential inﬁltration rate falls below the rainfall rate and some detention storage and rainfall excess are generated. Because at this time the surface storage is empty, the rainfall ﬁrst satisﬁes inﬁltration with 0.73 cm, then 0.75 cm to detention storage, and 0.07 cm in surface storage or rainfall excess ready for surface runoff. This process is continued until the end of the storm. It can be seen that, for the time interval from 120 to 135 min, F(t) is 0.40 cm whereas the rainfall is only 0.36 cm. Thus 0.04 cm of water is taken from the detention storage. The actual inﬁltration Fa (t) cannot exceed the rainfall depth for the time increment f p t plus the depth of water in surface storage. Mass balance must be preserved at all times. The cumulative sum of excess rainfall, column 8 of Table E.3.3.1, plus the incremental value of cumulative inﬁltration F p (t) and detention storage S must equal the cumulative rainfall precipitation in column 2.
3.5
Surface runoff
Runoff refers to the surface ﬂow occurring during and immediately after precipitation events. Base ﬂow refers to seepage and groundwater ﬂow between precipitation events. Surface runoff is added to the base ﬂow to determine the total ﬂow. The base ﬂow from small watersheds can often be neglected in the computation of surface runoff from large rainstorms. Excess rainfall generates surface runoff as overland ﬂow and channel ﬂow. To capture the essential features of the rainfall–runoff relationship, consider a rectangular excess hyetograph of constant intensity i and duration tr , as sketched in Fig. 3.10. This section aims at deﬁning the corresponding runoff hydrograph, which requires knowledge of resistance to overland ﬂow (Subsection 3.5.1) and stage–discharge relationships (Subsection 3.5.2). Surface runoff is then calculated for overland ﬂow (Subsection 3.5.3) and snowmelt runoff (Subsection 3.5.4).
3.5.1
Resistance to overland ﬂow
Resistance to ﬂow deﬁnes the relationship between ﬂow depth h and depth¯ Resistance to ﬂow can be written in terms of the averaged ﬂowvelocity u.
54
River basins
Table 3.2. Resistance to overland ﬂow Turbulent Flow Surface Concrete or asphalt Bare sand Graveled surface Bare clay–loam soil (eroded) Sparse vegetation Short grass prairie Bluegrass sod
Laminar ﬂow k0
Manning n
Ch´ezy C (ft1/2 /s)
Darcy–Weisbach f
24–108 30–120 90–400 100–500
0.01–0.013 0.01–0.016 0.012–0.03 0.012–0.033
78–38 65–33 38–18 36–16
0.03–0.4 0.04–0.5 — —
0.053–0.13 0.10–0.20 0.17–0.48
11–5 6.5–3.6 4.2–1.8
1,000–4,000 3,000–10,000 7,000–100,000
0.1–1000 0.5–13,000 1–10,000
Darcy–Weisbach friction factor f , Manning coefﬁcient n, or Ch´ezy coefﬁcient C. The corresponding deﬁnitions of C, n, and f are, respectively, u¯ = C h 1/2 S 1/2 , 1 2/3 1/2 h S (in S.I. units), n 8g 1/2 1/2 h S . u¯ = f u¯ =
(3.8a) (3.8b)
(3.8c)
Note that only the Darcy–Weisbach coefﬁcient f is dimensionless; the Manning coefﬁcient n has the dimensions T /L 1/3 and the numerator is replaced with 1.49 in the English system of units. The Ch´ezy coefﬁcient C has the dimensions L 1/2 /T . Table 3.2 lists typical values of the Darcy–Weisbach friction factor f , the √ Ch´ezy coefﬁcient C, given the identity C = 8g/ f , the Manning coefﬁcient n, and the laminar resistance coefﬁcient k0 for various overlandﬂow conditions. Woolhiser (1975) presented a method to evaluate resistance coefﬁcients for overland ﬂow. For smooth impervious surfaces, Fig. 3.16 shows the Darcy–Weisbach fric¯ tion factor f as a function of the Reynolds number Re = [(uh)/ν] = (q/ν) given the ﬂow depth h, the mean ﬂow velocity u¯ or unit discharge q and kinematic viscosity ν. It is important to note that two ﬂow regimes are observed in this ﬁgure: (1) laminar ﬂow when Re < 1,000, characterized by a constant value of k0 ; and (2) turbulent ﬂow when Re > 1,000, approximated by the Blasius equation f ∼ = 0.223/Re0.25 . It is also interesting that rainfall intensity increases the value of k0 in the laminarﬂow regime. Raindrop impact thus only increases
Surface runoff 1 15
= = i= i= i i
Rainfall intensity (in/h) 1 in/h = 7 m/s
.0 .75 .25 0.5 3 1
Smooth surface i ~ k o + Ai f= Re
b
i=
q
0
DarcyWeisbach friction coefficient, f
55
Re = q/υ 10
1
f = 24 Re
f = 0.223 Re 0.25
TURBULENT
LAMINAR 2
10
10
2
3
10 Reynolds number, Re
10
4
Figure 3.16. Resistance to overland ﬂow (after Li and Shen, 1973).
resistance to ﬂow in the laminarﬂow regime. Raindrop effects are negligible in turbulent ﬂows, i.e., when Re > 1,000. Laminar ﬂows with raindrop impact can be described by the Darcy–Weisbach equation in which the friction factor f relates to (1) the Reynolds number Re, (2) the friction coefﬁcient k0 , and (3) the empirical coefﬁcients A and b for raindrop impact. The friction coefﬁcient kt includes both effects that are due to surface roughness and rainfall intensity. The following relationship is generally used: f =
k0 + Ai b kt = . Re Re
(3.9)
The values of k0 have been tabulated by Woolhiser for various surface characteristics, and the value k0 = 24 is representative of a smooth surface. For rainfall intensity given in feet per second, A ∼ = 1.42 × 106 s/m = 4.32 × 105 s/ft (A ∼ ∼ for i in meters per second) and b = 1. The Darcy–Weisbach friction factor f is shown in Fig. 3.17 for the case of vegetated upland surfaces. It is found that resistance to ﬂow on vegetated
56
River basins
Figure 3.17. Resistance to overland ﬂow on vegetated surfaces (after Chen, 1976).
surfaces is much higher than that for smooth impervious surfaces. The laminar regime extends up to Re < 105 , and turbulent ﬂows are found when Re > 105 . The values of k0 in Table 3.2 indicate that k0 can be several orders of magnitude larger than for a smooth surface. The effects of raindrop impact become negligible compared with the effects of vegetation.
3.5.2
Stage–discharge relationship
In general, resistance to ﬂow can be written as a stage–discharge relationship, q = αh β ,
(3.10)
Surface runoff
57
Table 3.3. Resistance relationships q = αh β for overland ﬂow
Flow Type Laminar
Resistance Coefﬁcient kt = constant
Turbulent
α
β
8gS kt ν
3
te
kt ν L 8gSi 2
Darcy–Weisbach
f = constant
8gS f
1.5
Ch´ezy
C = constant
C S 1/2
1.5
Manning (S.I. units)
n = constant
S 1/2 /n
1.67
f L2 8gSi
L2 C 2 Si nL
1/3
1/3
1/3
0.6
S 1/2 i 0.667
where the unit discharge q is a power function of ﬂow depth h. The resistance coefﬁcient α and exponent β for overland ﬂow on rectangular planes are given in Table 3.3 for laminar and turbulent ﬂows as functions of kt , f , C, and n in Table 3.2. In Table 3.3, g is the gravitational acceleration, S is the surface slope, ν is the kinematic viscosity, and i is the rainfall intensity. ¯ bed shear stress τ0 , and Froude Similarly, ﬂow depth h, ﬂow velocity u, number Fr can be determined as functions of discharge from q β1 , (3.11a) h= α β−1 q β , (3.11b) u¯ = α α q 1/β , (3.11c) τ0 = γ h S = γ S α 2β−3 u¯ = α 3/2β q 2β g −1/2 Fr = √ (3.11d) gh It is important to understand that constant values of n do not correspond to constant values of kt , C, or f . In fact, if kt is constant as a function of discharge q or Reynolds number Re, this implies that the Manning coefﬁcient n will change with discharge. For instance, in the laminarﬂow regime with constant kt , the Manning coefﬁcient n depends on the Reynolds number; after combining Eqs. (3.8b), (3.8c), and (3.9), we obtain 5/9 1/9 kt ν Re−4/9 . (3.12a) n= 8g S 1/18 We ﬁnd that, for overland ﬂows in upland areas, the Manning coefﬁcient n is inversely proportional to the Reynolds number. At very low ﬂow rates on ﬂat slopes in upland areas, the Manning coefﬁcient n can become extremely high, even greater than 1.
58
River basins
Similarly, we can combine Eq. (3.8b) with Eq. (3.8c) to ﬁnd the following relationship between the Manning coefﬁcient n, the Darcy–Weisbach factor f , and the Reynolds number Re: f =
8gS 0.1 n 1.8 . ν 0.2 Re0.2
(3.12b)
We ﬁnd that at a constant slope S and ﬂuid viscosity ν, a line of constant Manning coefﬁcient n is described by f ∼ Re−0.2 , which is approximately equivalent to the Blasius relationship in Fig. 3.16 for turbulent ﬂows at Re > 1,000. 3.5.3
Overlandﬂow hydrographs
Analytical expressions for overlandﬂow hydrographs are derived for a planerectangular surface of length L and width W at a constant slope S0 under a constant excessrainfall intensity i e . It is assumed that the overlandﬂow plane is initially dry (h = 0 and q = 0) before the beginning of precipitation at time t = 0. The ﬂow depth increases linearly with time h = i e t during the rising limb until the ﬂow depth conveys the equilibrium discharge (qm = i e L). As sketched in Fig. 3.18, at a given time t, the upstream portion of the plane X < X e reached complete equilibrium. In the completeequilibrium domain, X < X e , the ﬂow is steady and nonuniform and the unit discharge q at any point x increases as q = i e x but does not change with time. In the partialequilibrium domain, X > X e , the ﬂow is unsteady and uniform and the unit discharge at any point is constant in space at q = i e X e but changes with time given the corresponding ﬂow depth h = i e t. Excess rainfall ie = i  f
h
Complete equilibrium q = ieX h = f(x)
Partial equilibrium q = ieXe h = ie t
X < Xe
X > Xe
q = ie X X L
Xe
Figure 3.18. Sketch of overlandﬂow depth.
h = ie t e h = ie t h = 0.8 i et h = 0.6 i et etc.
Surface runoff
59
The continuity relationship (see Problem 2.6) describing the conservation of ﬂuid mass is applied to widerectangular ﬂow: ∂h ∂q + = ie . ∂t ∂x
(3.13)
Runoff discharge
Excess rainfall intensity
Note that this continuity relationship describes a steady ﬂow in the completeequilibrium domain (X < X e ) when ∂h/∂t = 0 and a nonuniform ﬂow for partial equilibrium results from ∂q/∂ x = i e . As rainfall duration increases, the length X e increases and the completeequilibrium domain becomes larger with time. From combining h = it, q = i X e , and q = αh β , we obtain the position X e as a function of time as α (3.14) X e = (i e t)β . ie The time to equilibrium te is obtained when X e reaches the downstream end of the plane; the entire plane thus beie comes under steadyﬂow conditions. At that time, t = te and the equilibrium tr Time discharge qm = i e L = α(i e te )β is solved (a) Hyetograph for te to give q ψ= 1/β ie L L te = i e[(1/β)−1)] Rising Equiliα limb brium Falling limb 1 θ= 1 ψ 1 i e L 1/β β ψ 1 1 β = . (3.15) ie α ψ = (θ +λ r ) β 1 λ r 0
θ = (t  t r )/te
t (b) Complete hydrograph ( λ r = t r > 1 ) e
Runoff discharge
ψ=
1 λ rβ
q ie L
Partial equilibrium Rising limb Falling limb
θ=
1 ψ β ψ 1 1 β
ψ = (θ + λ r ) β 0 θ = (t  t r)/t e 1 λ r 1  λ r β β λ r β 1 tr (c) Partial hydrograph (λ r = t < 1 ) e
Figure 3.19. Overlandﬂow hydrographs.
We can determine the time to equilibrium of upland areas from the excessrainfall intensity i e , the length of the plane L, and α and β from Table 3.3. Values of te for different resistance relationships are presented in Table 3.3. Completeequilibrium hydrographs are those for which the rainfall duration tr exceeds the time to equilibrium te ; hence the hydrograph dimensionless time λr = tr /te is greater than unity. The surfacerunoff hydrograph can be subdivided into three parts: the rising limb, equilibrium, and falling limb. The
60
River basins
rising limb of runoff hydrographs is characterized by q = α(i e t)β and the complete equilibrium by q = i e L; the falling limb is rather complex. In general terms, surface runoff over a rectangular plane of length L can be written in dimensionless form as ψ = [q/(i e L)] as a function of the dimensionless time = [(t − tr )/te ], where te is the time to equilibrium. Figure 3.19 illustrates the shape of surfacerunoff hydrographs on rectangular planes for any ﬂowresistance relationship. The rising limb of the completeequilibrium hydrograph is given by ψ = ( + λr )β. The equilibrium discharge simply equals ψ = 1, and the falling limb is given by =
1−ψ βψ
β−1 β
,
(3.16)
as shown in Fig. 3.19(b). For partialequilibrium hydrographs, the salient features of the hydrographs are shown in Fig. 3.19(c). Example 3.4 Calculation of surfacerunoff hydrographs. Calculate the surfacerunoff hydrograph for a 1h rainstorm of constant excessrainfall intensity of 1 in./h on a rectangular plane, 100 ft wide, 400 ft long, on a 20% slope. Also calculate the maximum ﬂow depth, velocity, and shear stress at the corresponding Froude number. The plane is sparsely vegetated. In S.I. units, i e = 7 × 10−6 m/s, L = 122 m, and tr = 3,600 s. Considering the kinematic viscosity ν = 1 × 10−6 m2 /s, the maximum Reynolds number is Remax = i e L/ν =
7 × 10−6 × 122 m s = 860. s 1 × 10−6 m2
Thus the ﬂow is laminar. For a sparsely vegetated ﬁeld, a value of k0 ∼ = 2,000 is selected from Table 3.2. From Table 3.3, β = 3 for laminar ﬂow and α = 8gS/kν =
7848 8 × 9.81 m × 0.2 s . = s2 2,000 × 1 × 10−6 m2 ms
The time to equilibrium is calculated from 1/3 kt ν L 1/3 2,000 × 1 × 10−6 × 122 m3 s2 s2 te = = = 682 s. 8gSi e2 8 × 9.81 m × 0.2 × s (7 × 10−6 )2 m2 The completeequilibrium hydrograph tr > te is calculated in three parts. First, the rising limb is calculated from ψ = ( + λr )β or q = i e L(t/te )3 =
7 × 10−6 × 122 m2 3 t = 2.69 × 10−12 t 3 . (682)3 s4
Surface runoff
61
Second, after 682 s, q = 8.54 × 10−4 m2 /s and remains constant until t = 3,600 s. Third, the unit discharge decreases with time after t > 3,600 s according to = (1 − )/β (β−1)/β or t = tr + te
(i e L − q) 682 s (8.54 × 10−4 − q) = 3,600 s + . 3(i e L)1/3 q 2/3 3(8.54 × 10−4 )1/3 q 2/3
The falling limb of the hydrograph is thus obtained from substituting values of discharge, 0 < q < 8.54 × 10−4 , into this equation to calculate the time t at which the discharge will occur. During complete equilibrium, the maximum ﬂow depth, velocity, shear stress, and Froude number are obtained from Eqs. (3.11): h=
q 1/β
u¯ = α
α
q β−1 β α
1/3 8.54 × 10−4 = 4.8 mm, 7,848 2/3 8.54 × 10−4 = 7,848 = 0.18 m/s, 7,848
=
9,810 N × 4.8 × 10−3 m × 0.2 = 9.4 Pa, m3 u¯ 0.18 = 0.83. Fr = √ =√ gh 9.81 × 4.8 × 10−3 τ0 = γ h S =
3.5.4
Snowmelt runoff
Snowmelt runoff is a complex topic that we can analyze by using a complete radiation budget. Approximations like the degreeday method are often used for simplicity. As an example, hourly snowmeltrunoff discharge data from a small experimental plot in Canada were available for comparisons with climatological data (Julien and Frenette, 1986). Cumulative snowmelt h s , in meters, of equivalent water content was successfully correlated to three factors (see Fig. 3.20): (1) the cumulative number of degree days Dd in degrees Celsius times days, (2) cumulative time ta in hours when the air temperature is above 0 ◦ C, and (3) cumulative time of snowmelt, t f , in hours measured from the experimental plot h p = 7.29 × 10−3 Dd1.2 ,
(3.16a)
h s = 4.44 × 10−9 ta3.11 ,
(3.16b)
h s = 6.15 × 10−8 t 2.7 f .
(3.16c)
River basins
x 29 7. =
1
10
10
10 8 tf 2.7
(c)
hs = 6.15 x
x 10 9 ta 3.11
(b)
10 3 D 1. d 2
(a)
hs
Cumulative snowmelt (m)
1
hs = 4.44
62
2
1
10 D d (°Cd) Degreedays
10
2
10
Time T° > 0 °C
2
t f (h)
t a (h)
Snowmelt time
Figure 3.20. Cumulative snowmelt (after Julien, 1982).
1.0 t < 150 h 50 h < t < 150 h 0.8
t > 150 h
F (i f )
0.6
6 F (i f ) = 1  e 3.21 x 10 i f 7 F (i f ) = 1  e 1.05 x 10 i f
0.4
F (i f ) = 1  e 0.2 F (i f ) = 1  e 0
0
1
8.64 x 10 6 i f
1.94 x 10 7 i f
2 3 4 Snowmelt intensity i f (10 7 m/s)
5
6
Figure 3.21. Cumulative distribution function of snowmelt rates (after Julien, 1982).
Equation (3.16c) should be given physical preference, but meteorological records may be available to use only Eqs. (3.16a) and (3.16b). The ﬁrst derivative of Eq. (3.16c) shows that the mean runoff discharge increases with the snowmelt time. The cumulative distribution functions F(q f ) of the hourly runoff intensity q f measured from this experimental plot are shown in Fig. 3.21. The ﬁeld measurements ﬁt an exponential probability density function reasonably well,
Uplanderosion losses
63
which can be written as if p(i f )di f = F(i f ) = 0
if
λ f e −λ f i f di f = 1 − e −λ f i f .
(3.17)
0
The snowmelt intensity i¯ f = 1/λ f increases as the melting period progresses. The unit discharge from snowmelt runoff can be calculated from q f = i f L. The average snowmelt rate i¯ f can be estimated by dh s /dt f from Eq. (3.16c). 3.6
Uplanderosion losses
2
Shear stress τ (N/m )
Water is the most widespread agent of erosion. Upland erosion by water can be classiﬁed into sheet erosion and rill erosion. Sheet erosion is the detachment of land surface material by raindrop impact and thawing of frozen grounds and its subsequent removal by sheet ﬂow. The surfaceerosion process begins when raindrops hit the ground and detach soil particles by splash. The KE released by raindrop impact on the ground is sufﬁciently large to break the bonds between soil particles. The character10 istics of raindrop splash depend on Shear stress raindrop size and sheetﬂow depth; Measured 1 a crownshaped crater forms a few milliseconds after impact. The impact Model 1 shear stress can be as large as 100 10 0 6 12 18 24 30 times the base shear stress from shalTime (millisecond) low sheet ﬂow. For the example Figure 3.22. Shear stress under raindrop shown in Fig. 3.22, Hartley and Julien impact (after Hartley and Julien, 1992). (1992) measured shear stress in excess of 10 Pa. This far exceeds the critical shear stress of 2.5 N/m2 for cohesive soils. In general, the effect of raindrop impact can be neglected when the sheetﬂow depth is larger than three times the raindrop size. The transport capacity of shallow overland ﬂow, usually called sheet ﬂow, increases with ﬁeld slope and unitﬂow discharge. As sheet ﬂow is concentrated and the unit discharge increases, the increased sedimenttransport capacity scours microchannels called rills. Rill erosion is the removal of soil by concentrated sheet ﬂow. Rills are small enough to be removed by normal tillage. Soil particles detached by raindrop impact are transported downstream by runoff, and the unit sediment discharge on bare soils is a power function of surface slope S, unit discharge q, and rainfall intensity i. Julien and Simons (1985) provided a quantitative evaluation of the exponents of several sedimenttransport equations for sheet erosion. For sandy soils, it was found that the
64
River basins
following equation can be used: qs ∼ = 25,500 S 1.66 q 2.035 ,
(3.18)
where qs is the unit sediment discharge from sheet and rill erosion in metric ton/m s, S is the slope, and q is the unit discharge in square meters per second. It is appropriate to note that the exponent of discharge is greater than unity and therefore rills are likely to form owing to the discussion followed in the ﬁrst chapter of this book. The foregoing analysis of upland erosion considers the analysis of a single storm (Subsection 3.6.1), followed by the expected value of soil erosion from a single storm (Subsection 3.6.2) and the universal soilloss equation (Subsection 3.6.3). 3.6.1
Soil loss from a single event
Runoff
Intensity
Upland erosion by overland ﬂow during a single event of constant excessrainfall intensity i e and duration tr is sketched in Fig. 3.23. Because the sediment discharge is a power function of the runoff discharge, the characteristics of the sedie iment discharge follow the shape of the runoff hydrograph. We obtain, in dimensionless form, the total soil mass eroded te tr Time (a) Hyetograph per unit width, φ, by integrating the sediment discharge, qs /ρν, over the dimenRising Decreasing limb Equilibrium limb sionless runoff period t ∗ = t/t¯r : q
υ
φs = 0
te
tr
∞
qs d(t ∗ ) = ρν
Time
1 ρν t¯r
∞
qs dt. 0
(3.19)
(b) Hydrograph Sediment discharge
The eroded soil mass from a single storm m s in metric tons is given by ∞ qs dt = W0 ρν t¯r φs , (3.20) m s = W0
qs ρ υ
φs
0
t∗ Dimensionless time (c) Sediment graph
Figure 3.23. Sketches of (a) hyetograph, (b) hydrograph, and (c) sediment graph.
where W0 is the plane width in meters, ρ is the mass density of water in kilograms per cubic meter, ν is the kinematic viscosity of water in square meters per second, tr is the average rainfall duration in seconds, and φs is a dimensionless term,
Uplanderosion losses
65
determined in the next paragraph, that depends on the type of hydrograph, i.e., partial or complete hydrograph. For a given constant excessrainfall intensity i e over a plane surface of slope S, Julien (1982) integrated Eq. (3.19) and obtained exact analytical solutions as functions of λr = tr /te for a complete hydrograph φc when λr > 1 and for a partialequilibrium hydrograph φ p when λr < 1, as follows: φc =
9 25,500 1.66 2 2 , S L i e tr 1 − ρν t¯r 14λr
φp =
25,500 1.66 2 2 S L i e tr ρν t¯r
2λ6 λr3 − r 2 7
(3.21)
(3.22)
Simpliﬁcations arise from the fact that the erosion rates in Eq. (3.22) become very small for short storms (λr < 1). Also, i∗ φs the erosion rates for completeequilibrium hydrographs [Eq. (3.21)] become independent of λr and increase linearly with rainfall duration tr when λ is large. t∗ r
(a) Erosion during one storm event p ( t r∗ , i ∗)
3.6.2
i∗
t r∗ (b) Joint probability density function of the duration intensity parameters of several storms
φ s x p ( t r∗ , i ∗)
1
i∗
t r∗
(c) Dimensionless function of erosion
Expected soil loss
The expected value of soil erosion from one storm of unknown duration and intensity is determined with the exponential probability density functions of rainfall duration and intensity. From the exponential distributions of rainfall duration and intensity described in Eqs. (3.1) and (3.3), the expected value of soil erosion by one storm, φ¯ s , is given as sketched in Fig. 3.24: ∞ ∞ ∗ ∗ φe −tr e −i d(tr∗ ) d(i ∗ ), φ¯ s = 0
Figure 3.24. Sketch of expected values of soil erosion (after Julien, 1982).
0
(3.23)
where t¯r is the average storm duration, i¯ is the average storm intensity, tr∗ = tr /t¯r , ∗ ¯ and i = i/i. This integral must be divided into two parts, as indicated by Eqs. (3.21) and (3.22), which are valid when tr > te and tr ≤ te , respectively.
66
River basins
After deﬁning te∗ = te /t¯r , we obtain ∞ te∗ −tr∗ −i ∗ ∗ ∗ ¯φs = φpe e d(tr )d(i ) + 0
0
0
∞
∞ te∗
∗
∗
φc e −tr e−i d(tr∗ )d(i ∗ ). (3.24)
The functions φ p and φc are given by Eqs. (3.21) and (3.22). The exact analytical integration of Eq. (3.23) was obtained by Julien (1982) and Julien and Frenette (1985). The following relation is an approximate solution: 3.4 × 105 1.66 2 ¯2 ˆ ˆ ˆ S L i e K C P, φ¯ s ∼ = ρν
(3.25)
ˆ and Pˆ are the universal soilloss equation (USLE) coefﬁcients where Kˆ , C, deﬁned in Subsection 3.6.3. In regions where inﬁltration is signiﬁcant, the expected soilerosion loss from a single storm m¯ s is multiplied by the runoff coefﬁcient Cr , and the excessrainfall intensity i e is approximated by i¯ out, or ˆ m¯ s = W0 ρ ν t¯r φ¯s ∼ = W0 3.4 × 105 S 1.66 L 2 Cr i¯2 t¯r Kˆ Cˆ P.
(3.26)
The expected amount of soil eroded during a given period of time, e.g., 1 month, is equal to the product of the expected value of soil erosion for one rainfall event m¯ s and the mean number of events ν¯ during that period. A practical approximation for the expected value of the soil loss in metric tons during a period with an average of ν¯ storms is thus ˆ E tons ∼ = 3.4 × 105 ν¯ W0 S 1.66 Cr (i¯ L)2 t¯r Kˆ Cˆ P,
(3.27)
where E tons is the expected soil loss in metric tons on a rectangular plane of width W in meters and length L in meters, the average rainfall duration t¯r is in seconds and the average rainfall intensity i¯ is in meters per second, the number of storms is ν¯ , the slope S is in meters per meter, the runoff coefﬁcient is Cr , ˆ and P, ˆ which are discussed in the next and the USLE parameters are Kˆ , C, subsection. The advantage of this formulation is that the unit discharge q = i¯ L can be replaced with snowmeltrunoff discharge for the simulation of erosion losses in cold regions. 3.6.3
Universal soilloss equation
The Universal Soil Loss Equation (USLE) was designed to predict the annual average soilerosion losses from ﬁeld areas under speciﬁed cropping and management systems. The USLE computes the soil loss Eˆ at a given site as a
Uplanderosion losses
67
product of six major factors: ˆ Eˆ = Rˆ Kˆ Lˆ SˆCˆ P,
(3.28)
where Eˆ is the soil loss per unit area normally in tons per acre, Rˆ is the rainfallerosivity factor, Kˆ is the soilerodibility factor, usually in tons per acre, Lˆ is the ﬁeldlength factor normalized to a plot length of 72.6 ft, Sˆ is the ﬁeldslope factor normalized to a ﬁeld slope of 9%, Cˆ is the cropping–management factor normalized to a tilled area with continuous fallow, and Pˆ is the conservationpractice factor normalized to straightrow farming up and down the slope. The rainfallerodibility factor Rˆ can be evaluated for each storm (summed over hours) from Rˆ = 0.01 i(916 + 331 log i), where the summation is performed over the time increments of the storm, and i is the rainfall intensity in inches per hour. Soilerosion losses from single storms strongly correlate with the maximum 30min rainfall intensity. The annual rainfallerosion index ranges from 0 to 600 in the United States. The soilerodibility factor Kˆ describes the inherent erodibility of the soil expressed in the same units as those of the annual erosion losses, tons per acre. Numerous factors control the erodibility of cohesive soils such as grainsize distribution, texture, permeability, and organic content. Typical values of the factor Kˆ relate to the general triangular soil classiﬁcation shown in Fig. 3.13. For each soil type, approximate values of Kˆ can be found in Table 3.4, given the soil type and the percentage of organic matter. The slopelengthsteepness factor Lˆ Sˆ is a topographic factor relating erosion losses from a ﬁeld of given slope and length when compared with soil losses Table 3.4. Soil erodibility factor Kˆ in tons/acre (after Schwab et al., 1981) Organic Matter Content (%) Textural Class
0.5
2
Fine sand Very ﬁne sand Loamy sand Loamy very ﬁne sand Sandy loam Very ﬁne sandy loam Silt loam Clay loam Silty clay loam Silty clay
0.16 0.42 0.12 0.44 0.27 0.47 0.48 0.28 0.37 0.25
0.14 0.36 0.10 0.38 0.24 0.41 0.42 0.25 0.32 0.23
River basins
% 18
Slope 5
16 ^^ Slope length steepness (LS)
200
%
6
Slope length L (m) 150 100
50
0
20
68
% 14%
4
12%
3
10% 8%
2
6%
1
0
4% 2% 0
100
200
300 400 500 Slope length L (ft)
600
700
800
Figure 3.25. Slopelengthsteepness factor of the USLE (after Wischmeier and Smith, 1978).
of a standard plot 72.6 ft long inclined at a 9% slope. The values of Lˆ Sˆ are plotted in Fig. 3.25, with the length and the ﬁeld slope given. The cropping–management factor Cˆ for bare soils is taken as a standard value equal to unity. The factor Cˆ accounts for soils under different cropping and management combinations such as different vegetation, canopy during growth stage, before and after harvesting, crop residues, mulching, fertilizing, and crop sequence. Typical values of Cˆ are given in Tables 3.5 for forest (Table 3.5a), pasture, rangeland, and idle land (Table 3.5b), cropland (Table 3.5c), and construction slopes (Table 3.5d). Areaaveraged values of Cˆ can be used when several vegetation types cover a given area. Table 3.5a. Cropping–management factor Cˆ for forest (after Wischmeier and Smith, 1978) Percentage of area covered by canopy of trees and undergrowth
Percentage of area covered by duff at least 2 in. deep
Factor Cˆ
100–75 70–45 40–20
100–90 85–75 70–40
0.0001–0.001 0.002–0.004 0.003–0.009
G W G W G W
Tall weeds or short brush with average drop fall height of 20 in.
Appreciable brush or bushes, with average drop fall height of 6 1/2 ft
Trees, but no appreciable low brush. Average drop fall height of 13 ft
0.36–0.42 0.36–0.42
0.28–0.40 0.28–0.40
0.17–0.36 0.17–0.36
0.45 0.45
0
0.17–0.19 0.20–0.23
0.14–0.18 0.17–0.22
0.10–0.17 0.12–0.20
0.20 0.24
20
0.09–0.10 0.13–0.14
0.08–0.09 0.12–0.14
0.06–0.09 0.09–0.13
0.10 0.15
40
0.039–0.041 0.084–0.089
0.036–0.040 0.078–0.087
0.032–0.038 0.068–0.083
0.042 0.091
60
0.012–0.013 0.041–0.042
0.012–0.013 0.040–0.042
0.011–0.013 0.038–0.041
0.013 0.043
80
0.003 0.011
0.003 0.011
0.003 0.011
0.003 0.011
95+
b
a
The listed Cˆ values assume that the vegetation and the mulch are randomly distributed over the entire area. Canopy height is measured as the average fall height of water drops falling from the canopy to the ground. The canopy effect is inversely proportional to the drop fall height and is negligible if the fall height exceeds 33 ft. c G, Cover at surface is grass, grasslike plants, decaying compacted duff, or litter at least 2 in. deep; W, cover at surface is mostly broadleaf herbaceous plants (such as weeds with little lateralroot network near the surface) or undecayed residues or both.
G W
Typec
No appreciable canopy
Vegetative canopy (type and heightb )
Cover that is in contact with the soil surface Percentage of ground cover
Table 3.5b. Cropping–management factor Cˆ for pasture, rangeland, and idle landa (modiﬁed after Wischmeier and Smith, 1978)
70
River basins Table 3.5c. Cropping–management factor Cˆ for cropland (approximate values from Wischmeier and Smith, 1978) Tilled continuous fallow Rough fallow Conventional seed bed No tillage Full canopy Residues left on the ﬁeld
1.0 0.30–0.80 0.50–0.90 0.05–0.25 0.10–0.20 0.10–0.50
Table 3.5d. Cropping–management factor Cˆ for construction slopes (modiﬁed after Wischmeier and Smith, 1978)
Type of mulch Straw Crushed stone 1/4 to 1.5 in.
Wood chips
Mulch rate (tons/acre)
Factor Cˆ
1.0–2.0 135 240 7 12 25
0.06–0.20 0.05 0.02 0.08 0.05 0.02
Table 3.6. Conservationpractice factor Pˆ for contouring, stripcropping, and terracing (after Wischmeier and Smith, 1978) P value Terracing Land slope (percent) 2–7 8–12 13–18 19–24 a b
Farming on contour
Contour stripcrop
a
b
0.50 0.60 0.80 0.90
0.25 0.30 0.40 0.45
0.50 0.60 0.80 0.90
0.10 0.12 0.16 0.18
For erosioncontrol planning on farmland. For prediction of contribution to offﬁeld sediment load.
Uplanderosion losses
71
The conservationpractice factor P equals one for downslope rows and typical values for contouring, stripcropping, and terracing are given in Table 3.6. Contour practices are most effective on slopes of less than 12%, in which case Pˆ can be as low as 0.5. Contouring does not reduce erosion losses at slopes exceeding 24%. Stripcropping and terracing reduce erosion signiﬁcantly on slopes of less than 12%. Example 3.5 Application to soil losses on a small watershed. Consider a watershed area covering 600 acres (1 acre = 4,047 m2 ) above a proposed ﬂoodwaterretarding structure. Compute the annual soilerosion loss given the conditions shown in Fig. E.3.5.1 in an area where the rainfall erosivity is Rˆ = 185. Cropland covers 280 acres of continuous corn with residues left on the ﬁeld, Pasture cultivated up and down an 8% slope that 170 acres Silt loam is 200 ft long on a silt loam. The rainfall 200 ft long at 8% erosivity Rˆ = 185 and the soilerodibility factor Kˆ ∼ = 0.46 for a silt loam in Cropland Table 3.4 the topography factor Lˆ Sˆ = 1.4 280 acres Silt loam from Fig. 3.25 with a length of 200 ft Forest 200 ft long at 8% 150 acres at an 8% slope, the cropping–manageSilt loam 100 ft long at 12% ment factor Cˆ ∼ = 0.4 for continuous cultivation and Pˆ = 1 for upslope and 1 hectare = 2.5 acres downslope practices. The annual soil loss 1 m = 3.28 ft is Eˆ = 185 × 0.46 × 1.4 × 0.4 = 47 tons Figure E.3.5.1. Example of soilloss per acre over 280 cropland acres. calculation. Pastureland covers 170 acres, half of which has a canopy cover of short brush (0.5m fall height), and 80% of the surface is covered by grass and grasslike plants, the soil is a silt loam, and the slopes at 8% are 200 ft long. The parameters are Rˆ = 185, Kˆ = 0.46, Lˆ Sˆ = 1.4, Cˆ = 0.012 from Table 3.5b, and Pˆ = 1. The annual soil loss is 1.4 tons per acre for 170 acres. Finally, forestland covers 150 acres, 30% of the area has tree canopy, and 50% of the surface is covered by litter. Silt loam slopes are 100 ft long at a 12% slope. The parameters are Rˆ = 185, Kˆ = 0.46, Lˆ Sˆ = 1.8 from Fig. 3.25, and Cˆ = 0.009 from Table 3.5a. The annual soil loss Eˆ = 185 × 0.46 × 1.8 × 0.009 = 1.37 tons/acre for these 150 acres. The total soilerosion loss on this watershed equals 13,600 tons/yr (47 tons/acre × 280 acres + 1.4 tons/acre × 170 acres + 1.37 tons/acre × 150 acres).
72
River basins
3.7
Sediment source and yield
This section on sediment source and sediment yield from watersheds ﬁrst covers the soilerosion losses (Subsection 3.7.1) followed by a method to estimate the sediment yield from watersheds (Subsection 3.7.2). 3.7.1
Soilerosion losses from large watersheds
Annual erosion losses can be calculated on watersheds. It is generally found that the topographic factors Sˆ and the landuse factor Cˆ determine most of the spatial variability in soilerosion losses. Comparatively, the rainfallerodibility and the soilerosivity parameters remain relatively constant at the basin scale. The spatial variability in the factors Sˆ and Cˆ thus requires particular attention in investigations of soilerosion mapping. When using Geographical Information System (GISs) to delineate the surface slope from rasterbased data, the gridcell size inﬂuences the slope calculations. In general, the surface slope decreases as the gridcell size increases (Molnar and Julien, 1998). The inﬂuence of the gridcell size on the computation of soilerosion rates has been studied to determine whether or not average values of the USLE parameters can be used instead of values for each pixel. Julien and Frenette (1987) deﬁned a relative gridsize factor Q ∗e for each matrix from Q ∗e
N Rˆ Kˆ Lˆ Sˆ Cˆ Pˆ = N , Rˆ i Kˆ i Lˆ i Sˆi Cˆ i Pˆ i
(3.29)
i=1
where Rˆ i denotes the value of parameter Rˆ of the USLE on pixel i and Rˆ is the average value of Rˆ i over a large area. The mean value of the relative correction factor Q¯ ∗e shown in Fig. 3.26 gradually decreases as the number of pixels, N , increases. This can be attributed to the fact that calculated slopes decrease as grid size increases. The use of average values of USLE parameters tends to underestimate soilerosion losses when applied to large areas. When applied to watersheds, the values of the correction factor have been shown to vary primarily with drainage areas, as in Fig. 3.27. As a result, the average value of the watershedsize correction factor Q¯ e can be written as a function of the drainage area At in square kilometers as Q¯ e = 0.8 At−0.137 ,
At > 0.125 km2 .
(3.30)
Sediment source and yield
73
1.2
1.0
Q e∗
0.8
0.6 A o = 0.028 km2 Data set A A o = 0.25 km2 Data set B A o = 4 km2 Data set C Regression curves
0.4
0.2
0 1
2
10
3
10
10
Number of cells N
Figure 3.26. Erosion ratio vs. number of cells (after Julien, 1979). Watershed drainage area A t (mi2 ) 10
10
2
10
1
1
10
10 3
10 2
Qe
Q e from Julien and Frenette, 1987 Confidence intervals at 95% Conca de Tremp and Chaudi re data
1
1
10
10
2
10
1
1 10 2 10 Watershed drainage area A t (km2 )
10 3
Figure 3.27. Erosion coefﬁcient vs. watershed area (after Julien and del Tanago, 1991).
74
River basins
The correction factor Q¯ e remains constant ( Q¯ e = Q¯ es ) when At < 0.125 km2 , which indicates that the soilloss equation can be applied to small areas without bias. As the drainage area increases beyond At > 0.125 km2 , the correction factor Q¯ e decreases gradually, as shown in Fig. 3.27. Consequently, the annual erosion losses from a large watershed can be estimated from the erosion Eˆ calculated with watershedaveraged values of the USLE coefﬁcients divided by the correction factor Q¯ e : Eˆ Eˆ ∼ . = Qe
(3.31)
This relationship is quite practical because quick estimates of the gross erosion on large watersheds can be obtained from the average characteristics of the watershed. Calculations can be based on either the USLE [Relation (3.28)] or on the surfacerunoff parameters in Eq. 3.26. In this case, the watershed √ slope is calculated from S¯ = H/1,000 At , where H is the elevation difference between the highest and the lowest elevation on the watershed and At is the watershed drainage area in square kilometers. The runoff width W0 = 106 At /L, with At in square kilometers and the average runoff length L in meters. We may also consider the mean precipitation h¯ = ν¯ t¯r i¯ during the given period to reduce Eq. (3.27) to the following relation for annual erosion losses on large watersheds: 3.4 × 1011 ˆ Eˆ tons At L¯ S¯1.66 Cr h¯ i¯ Kˆ Cˆ P, Q¯ e
(3.32)
where Eˆ is the erosion loss in tons, At is the drainage area in square kilometers, L¯ is the average runoff length in upland areas in meters (∼100 m), √ S¯ = H/1,000 At , H is the elevation difference in meters, h¯ is the average precipitation in meters during the period considered, Cr is the runoff coefﬁcient, ˆ and Pˆ are the i¯ is the average rainfall intensity in meters per second, and Kˆ , C, watershedaveraged values of the USLE parameters. As shown in Fig. 3.27, the variability around the mean value is approximately a factor of 2. The erosion loss is expected to range between 50% and 200% of the value estimated from relation (3.31). 3.7.2
Sediment yield from large watersheds
The sum of upland and channel erosion in a watershed amounts to the gross erosion. All eroded particles in a watershed, however, do no reach the watershed
Sediment source and yield
75 Drainage area A t (km2)
1
SDR
10
10
1
10
10
10 2
10
3
10 4
1
10 2 2 10
(A t in mi2 ) S DR = 0.31 A0.3 t SDR = 0.41 At0.3 (A t in km2) 1
10
10 2 10 Drainage area A t (mi2 )
1
10
3
10 4
Figure 3.28. Sediment–delivery ratio (modiﬁed after Boyce, 1975).
outlet. Particles detached from bare upland areas are trapped in vegetated areas farther downstream. Some material carried in natural streams is deposited in the channels to cause channel aggradation. Some material deposits on the ﬂoodplain during major ﬂoods and large amounts are permanently trapped in lakes and reservoirs. The total amount of sediment that is delivered to the outlet of the watershed is known as the sediment yield. The sediment–delivery ratio is deﬁned as the ratio between the sediment yield and the gross erosion on a watershed. The ability of a channel network to convey eroded material to the outlet depends on drainage area, watershed slope, drainage density, and runoff. It is found from Fig. 3.28 that the sediment– delivery ratio decreases primarily with the size of the drainage area. The sediment yield Y from a large watershed can thus be estimated from the following procedure: At ˆ Y ∼ = SDR ¯ 3.4 × 1011 L¯ S¯1.66 Cr h¯ r i Kˆ Cˆ P, Qe
(3.33)
where SDR is the sediment–delivery ratio from Fig. 3.28, At is the drainage area in square kilometers, Q¯ e is the watershedsize correction factor from Fig. 3.27, √ S¯ = H/1,000 At with H as the elevation difference in meters between the highest and the lowest elevation on the watershed, h¯ r in meters is the average precipitation during the period considered, L¯ is the average runoff length in meters, Cr is the runoff coefﬁcient, i¯ is the average rainfall intensity in meters ¯ and P¯ are the watershed average values of the USLE per second, and K¯ , C, parameters. Finally, the speciﬁc degradation of a watershed is obtained from the sediment yield divided by the drainage area. A calculation example is presented in Case Study 3.2.
76
River basins
Cumulative sed. load (kt)
Sediment load (kt)
Case Study 3.2 Soil losses of the Chaudi`ere watershed, Canada. The Chaudi`ere watershed covers At = 5,830 km2 , as measured from topographic maps (1:250,000). The average cropping–management factor, Cˆ ∼ = 0.35, is determined from topographic maps (1:50,000) and from forest and agricultural maps (1:250,000). The average soilerodibility parameters are Kˆ ∼ = 0.17 ∼ ˆ and P = 1.0. The characteristic slope is computed from the relationship S¯ = √ H/1, 000 At = 0.0156, in which H = 1190 m is the elevation difference in meters between the highest and the lowest point on the watershed. The runoff length L¯ is the length of sheet and rill ﬂow on upland areas. Based on topographical maps and ﬁeld observations, this length is estimated at 300 ft, or L¯ = 91.4 m, and is assumed constant over the watershed. The data from 22 meteorological stations on the Chaudi`ere watershed and from hourly data summaries (1943–1970) were analyzed to conclude that the mean annual rainfall precipitation is ∼770 mm and the runoff coefﬁcient is ∼0.7. Assume that the Figs. 3.11 and 3.12 are representative of rainfall condition, i¯ = 1 × 10−6 m/s. The average number of rainstorms is ν¯s = 51/yr, and the average storm duration is t¯r = 1.5 × 104 s. We can estimate the expected value of annual erosion loss from this watershed from the average watershed characteristics by using relation (3.31) with Q¯ e ∼ = 0.24 from Fig. 3.27 with At = 5,830 km2 , E = ( E¯ tons /Q e ) ∼ = [(3.4 × 1011 )/ 0.24] × 5,830 × 91.4 m × (0.0156)1.66 × 0.7 × 0.77 m × 1 × 10−6 m/s × 0.17 × 0.35 × 1 2.4 × 106 tons/yr. The mean annual sediment yield Chaudi re River from this watershed is estimated with 400 200 Observed the sediment–delivery ratio SDR ∼ = 150 300 0.03 from Fig. 3.28: Y ∼ = E × SDR = Computed 2.4 × 106 × 0.03 = 720 ktons/yr. The 200 100 mean annual sediment yield meas100 50 ured from 1968–1976 is ∼363 ktons/ 0 yr. The reader is referred to Julien and 0 J M M J S N Frenette (1986) for a detailed analTime (months) ysis of sediment yield from snowFigure CS.3.2.1. Sediment yield of the melt on this watershed. Comparison Chaudi`ere watershed (after Julien and with the monthly sediment yield Frenette, 1985). in Fig. CS.3.2.1 shows reasonable agreement between measured and calculated sediment yields on a monthly basis.
Problems
77
Exercise 3.1
Combine the equations h = ite , q = i L , and q = αh β to derive the expressions for the time to equilibrium in Table 3.3. Exercise 3.2
Combine Eqs. (3.8b) and (3.8c) to derive the relationship between the Darcy– Weisbach friction factor f and the Reynolds number Re in Eq. (3.12b). Computer Problem 3.1
Find a sample of rainfall duration, intensity, or rainfall depth and compare with the exponential distribution as per the method shown in Example 3.1. Problem 3.1
Repeat the inﬁltration calculations of Example 3.3 for a silty clay. Plot the hyetograph, inﬁltration rate, detention storage, and excess rainfall as functions of time from Table E.3.3.1. Compare the results for silty loam vs. silty clay. Problem 3.2
Consider a 1h storm at 1in./h rainfall intensity on a 100mlong plot at a 5% slope. Determine the maximum ﬂow depth, ﬂow velocity, Froude number, and shear stress if the surface is rough and impervious. (Hint: Determine the appropriate resistance equation.) Also calculate and plot the surfacerunoff hydrograph. Problem 3.3 Estimate the time to equilibrium of the largest possible 3h storm on a rectangular farmland covering 35 acres. Estimate the inﬁltration on bare soil for silty clay loam. Problem 3.4 Calculate the average rate of snowmelt in meters per second, from the derivative of Eq. (3.16c) after a snowmelt period of 100 h. Calculate λ2 f and compare with the results shown in Fig. 3.21.
78
River basins Problem 3.5
From the data in Table 1.1, plot on a log–log scale the water discharge, sediment discharge, and concentration as functions of drainage area. Problem 3.6
From the expression for sediment yield [Relation (3.33)] and the mean annual ¯ determine the expression for mean sediment concentration. ﬂow, Cr At h,
4 Steady ﬂow in rivers
Steady ﬂow refers to ﬂow conditions that do not change with time. Steady ﬂows can be either uniform when the conditions do not change with space or nonuniform when ﬂow conditions change with space. Steady ﬂow in rivers (Section 4.1) includes description of atastation hydraulic geometry, followed by a description of steadyuniform ﬂow and resistance to ﬂow. Steadynonuniform ﬂows (Section 4.2) include an analysis of the momentum equations followed by rapidly varied ﬂow and gradually varied ﬂow. Sediment transport in rivers (Section 4.3) includes a simple description for sediment transport in steadyuniform ﬂow followed with calculations of aggradation and degradation in river reaches. 4.1
Steady river ﬂow
Drainage networks have been studied by geomorphologists and topologists. In general, topologists search mathematical ranking and order among subwatersheds without speciﬁc reference to physical entities. The results of several years of experimental studies from the Rainfall Erosion Facility at Colorado State University by Schumm et al. (1987) attempted to relate basin morphology and sediment yield. Although considerable sedimentyield variability is assumed to result from climatic ﬂuctuations and landuse changes, the experiments show that sediment yield is highly variable under steady rainfall conditions. The complex response of channel network evolution seems to be characterized by an exponential decrease in sediment yield as the channel network develops. Rivers follow the low points along the watershed topographic proﬁles. With very few exceptions in arid areas, the lowest point of a watershed is located at the river outlet. A watershed map usually shows the hydrographic area and drainage network with lakes, reservoirs, and perhaps also gauging stations, counties, states, and/or countries. The drainage network of a watershed implies a discontinuous increase in drainage area and discharge at river conﬂuences. Except 79
80
Steady ﬂow in rivers
Figure 4.1. Longitudinal proﬁle of the Mississippi River.
for river captures and artiﬁcial changes like transmountain water diversions, the drainage network of natural channels does not change signiﬁcantly with time. The river length is usually measured in the downstream direction from the uppermost river elevation. Left and right banks are usually referenced to a downstreamlooking direction. A longitudinal proﬁle is quite useful to detect bedrock controls, headcuts, nickpoints, and alluvial reaches. The approximate slope of alluvial reaches can be estimated from gradual changes in elevation over long distances. The valley slope corresponds to the ﬂoodplain elevation drop over the valley length. The channel slope corresponds to the watersurface elevation drop over the channel length. Channel and valley slopes are deﬁned as positive although the elevation decreases in the positivedownstream direction. The river sinuosity is then deﬁned as the ratio of the channel length to the valley length between two points located on the river. Examples of river reaches and longitudinal proﬁles are shown in Fig. 4.1 for an alluvial river, and Figs. 4.2 and 4.3 for semialluvial rivers. The longitudinal proﬁle of an alluvial river reach is gradual. Bedrock control in semialluvial river reaches causes discontinuities in longitudinal proﬁles, bedmaterial sizes, and ﬂow conditions. At engineering time scales, it can be considered that the channel characteristics can change signiﬁcantly during extreme ﬂood events or after active tectonic and volcanic periods. River reaches can be examined one by one to determine their physical characteristics, such as length, sinuosity, width, depth, crosssectional geometry, surface roughness, and hydraulicresistance factor. The straightening of river reaches through meander cutoffs can change the
Steady river ﬂow
81
Figure 4.2. Longitudinal proﬁle of the Matamek River (after Frenette and Julien, 1980).
length of rivers and their corresponding sinuosity and slope. River mileage is often measured downstream from a major river conﬂuence or upstream from the mouth of a river. It is important to consider that the exact location of these reference points usually changes over the years. The following discussion focuses on atastation hydraulic geometry (Subsection 4.1.1) and on steadyuniform ﬂow in rivers (Subsection 4.1.2). 4.1.1
Atastation hydraulic geometry
At a given station along the river, a crosssection proﬁle can be drawn in the direction perpendicular to the main ﬂow direction. Cross sections that are not measured perpendicular to the ﬂow direction will appear wider than reality. The crosssection proﬁle shows the ﬂowdepth distribution across the river as well as the elevation of the banks including the ﬂoodplain. It is important to reference the elevation to a benchmark that provides an exact horizontalreferenceplane elevation such as the mean sea level. Elevation scales are usually linked to the geodetic reference elevation [above mean sea level (ASL)] or the National Geodetic Vertical Distance (NGVD), which are both absolute vertical elevations. In some cases, the elevation is given with respect to a lowwater
82
Steady ﬂow in rivers (a)
(b)
Gravel Velocity 2 ft/s
Depth Velocity
Sand Silt
Scale 0
50 100
200 ft
Figure 4.3. Morphology and bed material of the Matamek River (after Frenette and Julien, 1980).
reference plane (LWRP), which is the watersurface elevation in the river that is exceeded 97% of the time. Note that the LWRP is not horizontal, but slopes with the river. An example of a crosssectional proﬁle is given in Fig. 4.4. In this particular case, the downstream bedrock control retains water in the channel when the discharge reduces to zero. Indications of the substrate material are given and the crosssectional geometry extends onto the ﬂoodplain, where sketchy information on vegetation in terms of deciduous and coniferous trees, bushes and grasses is provided. These attributes provide basic information on aquatic habitat and expected ﬂoodplain roughness for runoff simulations during ﬂoods. From this crosssectional proﬁle, the following geometrical parameters can be determined as functions of stage: (1) topchannel width W, (2) wetted perimeter P, (3) crosssectional area A, (4) mean ﬂow depth h¯ = A/W , and (5) hydraulic radius Rh = A/P. These parameters describe the geometry of a cross section. It is important to consider that the mean ﬂow depth is different from the stage. An increase/decrease in mean ﬂow depth does not necessarily correspond to the same increase/decrease in stage.
Steady river ﬂow
83
Figure 4.4. Cross section of the Matamek River (after Frenette and Julien, 1980).
Figure 4.5. Matamek River: (a) vertical velocity proﬁle and (b) transversal velocity proﬁle (after Frenette and Julien, 1980).
The concept of atastation hydraulic geometry stems from velocity measurements along a cross section. An example of a velocity proﬁle along one vertical of a cross section at a given stage provides sufﬁcient measurements for the determination of the depthaveraged ﬂow velocity v¯ [Fig. 4.5(a)].
84
Steady ﬂow in rivers
The depthaveraged velocity is normally obtained from a measured velocity proﬁle. Total discharge measurements are obtained from evenly spaced depth h i and depthaveraged velocity v¯ i measurements along the cross section. The total crosssectional area A is the sum of incremental areas ai , A = i ai = spacing between verticals of a cross section. i Wi h i , where W is the The total discharge is Q = i ai v¯ i , where v¯ i is the depthaveraged ﬂow velocity normal to the incremental area. The crosssectional average velocity is V = Q/A. Figure 4.5(b) shows a crosssectional velocity proﬁle with evenly spaced depthaveraged ﬂow velocities along the crosssectional proﬁles. At this cross section, the following hydraulic geometry is found: top width W = 280 ft, wetted perimeter P = 280.5 ft, crosssectional area A = 684 ft2 , mean ﬂow depth h¯ = 2.44 ft, and hydraulic radius Rh = A/P = 2.43 ft. The discharge is Q = 210 ft3 /s, and the crosssectional average velocity V = 0.31 ft/s = 9.4 cm/s. Note that it is common to ﬁnd that the wetted perimeter is close to the top width, ¯ P∼ = h. = W , and that the hydraulic radius is close to the mean ﬂow depth, Rh ∼ A ﬂowrating curve, or stage–discharge relationship, displays the change in stage with discharge. In channels with bedrock control, the stage–discharge relationship is unique and well deﬁned, as shown in Fig. 4.6(a). In large alluvial rivers, the stage–discharge relationship may shift over time because of a combination of processes including (1) bed aggradation or degradation, (2) changes in bedform conﬁguration, or (3) looprating effects that are due to dynamic ﬂood routing. The example of the Mississippi River is shown in Fig. 4.6(b). Speciﬁcgauge records indicate the water level at a given ﬂow discharge. The analysis of speciﬁcgauge records can determine longterm aggradation or degradation trends of a river. For instance, Fig. 4.7(a) shows the speciﬁcgauge records of the Atchafalaya River between Simmesport and Morgan City from 1950 to 1997. It clearly shows that the river has been gradually degrading between Simmesport and Chicot Pass and aggrading downstream of Chicot Pass. In some alluvial rivers, water temperature can affect the speciﬁcgauge records. For the example of the Mississippi River, in Fig. 4.7(b), the water temperature at a ﬁxed discharge of 106 ft3 /s can alter the stage by as much as 5 ft. Higher stages correspond to warm water temperature. The analysis of hydraulicgeometry relationships becomes simple once the ﬂowrating curve is known. At a given discharge, the mean ﬂow depth is obtained from the ﬂowrating curve and the corresponding width, hydraulic radius, and crosssectional area are obtained from the crosssectional proﬁles. This process can be repeated at various stages or discharges to deﬁne the atastation hydraulicgeometry relationships. The rate at which the
Steady river ﬂow
85
Figure 4.6. Stage–discharge relationship: (a) Matamek River and (b) Atchafalaya River.
hydraulicgeometry parameters change with discharge is quite important in deﬁning atastation hydraulic geometry relationships. The example of the Matamek River is illustrated in Figs. 4.8. The analysis of 10 cross sections over a 1.4km reach displays the local variability in hydraulicgeometry relationships of a nearly straight river. The wetted perimeter, or river width, varies by a factor of 2 within this reach; the rate of increase with discharge is fairly constant. When the hydraulicgeometry parameters are plotted vs. discharge on a log–log scale, the gradient of the proﬁles gives the exponent b of the relationship P = a Q b . For instance, the wetted perimeter in Fig. 4.8(a) can be approximated by P∼ = 126 Q 0.1 , with Q in cubic feet per second and P in feet. Note that the
86
Steady ﬂow in rivers
Figure 4.7. Speciﬁcgauge records: (a) Atchafalaya River and (b) temperature effect (U.S. Army Corps of Engineers, 1999).
exponent is very small, which means that both the wetted perimeter and the riversurface width do not vary signiﬁcantly with discharge. In fact, for many practical applications, the use of a rectangular cross section is a fairly good approximation. Beyond bankfull ﬂows, the channel width suddenly increases, which alters calculations of mean ﬂow depth and hydraulic radius. Figure 4.8(b) shows the crosssection average ﬂow depth vs. discharge at several cross sections of the same reach. Except at very low ﬂows, the slope of the lines on this diagram indicates that the mean ﬂow depth increases approximately as h¯ = 0.35 Q 0.36 with Q in cubic feet per second and h¯ in feet.
Figure 4.8. Atastation hydraulic geometry of the Matamek River (after Frenette and Julien, 1980).
87
88
Steady ﬂow in rivers
Figure 4.8. (cont.)
The crosssectional averaged velocity can be plotted against discharge on a log–log plot, as shown in Fig. 4.8(c) for the same reach. At a station, the velocity varies largely with discharge, in this case, V ∼ = 0.022 Q 0.54 . The deﬁnition of volumetric ﬂux requires that the product of width, depth, and velocity be equal to the discharge. Hence the relationships must satisfy ¯ ∼ Q = W hV = 126 Q 0.1 × 0.36 Q 0.36 × 0.022 Q 0.54 ; thus the product of coefﬁcients is 126 × 0.36 × 0.022 = 1 and the sum of exponents is 0.1 + 0.36 + 0.54 = 1. Typical atastation hydraulicgeometry relationships show that changes in discharge affect primarily the ﬂow velocity and, to a lesser extent, the ﬂow depth. Unless the discharge exceeds bankfull conditions, the river surface width is fairly constant except at very low discharges. In some cases, the watersurface slope of a river may also change with discharge. The example shown in Fig. 4.8(d) illustrates a slight increase in slope with discharge.
4.1.2
Steadyuniform river ﬂow
Steadyuniform ﬂow implies no change in hydraulic conditions with either space or time. We easily obtain that ∂ Q/∂ x = 0, or Q = AV = constant and ∂ A/∂t = 0, or A is constant in space and time. Equation of motion (2.27) reduces to S f = S0 = tan θ ∼ = sin θ because of steady ﬂow [(∂ V /∂t) = 0] and uniform ﬂow [(∂h/∂ x) = 0 and (∂ V /∂ x) = 0]. The shearstress relationship τ0 for uniform ﬂow reduces to τ0 = γ Rh S0 .
(4.1)
For steadyuniform ﬂow, shearstress and resistance relationships can be directly written as a function of bed slope. With reference to Fig. 4.9, with identical pressure distributions at the upstream and the downstream cross sections, the weight component in the downstream x direction is balanced by only the
Steady river ﬂow
89
Figure 4.9. Equilibrium sketch for a river reach.
bed shear force Fw sin θ = τ0 W d x, which, for widerectangular channels with Fw = γ W hd x, and sin θ ∼ = tan θ = S0 at small angles, simply reduces to τ0 = γ h S0 because S f = S0 . Shear velocity u ∗ deﬁnes a kinematic substitute for the dynamic bed shear stress τ0 . The identity stems from τ0 = ρu 2∗ or τ0 = g Rh S f . (4.2) u∗ = ρ The shear velocity is not a measurable quantity but serves as a scaling parameter for kinematicvelocity proﬁles in turbulent boundary layers. The shear velocity is also often used in sedimenttransport studies. Resistance to ﬂow is evaluated from steadyuniform ﬂow conditions, S f = S0 . Because no shear exists under hydrostatic conditions, the shear stress τ0 is assumed to vary with ﬂuid velocity and, after dimensional considerations, a dimensionless friction factor f has been deﬁned after the work of Darcy and Weisbach as 8g Rh S f 8τ0 = . (4.3) f = ρV 2 V2 Resistance to ﬂow can thus be described from the Darcy–Weisbach friction factor f. For widerectangular channels, Rh = h, we also easily demonstrate √ that S f = f Fr2 /8. We may also consider that 8/ f = V /u ∗ . The depthaveraged ﬂow velocity is then simply obtained as 1/ 1/ 8g 1/2 1/2 2 2 Rh S f = C Rh S f , (4.4) V = f where C is the Ch´ezy coefﬁcient. The identity C 2 = 8g/ f is always valid, and it is important to note that f describes ﬂow resistance whereas C describes ﬂow conveyance. The Ch´ezy coefﬁcient C is a constant as long as f is a constant.
90
Steady ﬂow in rivers
It has been observed over the years that the Ch´ezy coefﬁcient C or the Darcy– Weisbach factor f varies with relative submergence and the Manning equation is a convenient approximation. Total resistance to ﬂow can be described in terms of the Ch´ezy coefﬁcient C, the Darcy–Weisbach friction factor f, or the Manning coefﬁcient n. The following identity among these factors has been established: √ 1/6 1/6 C ≡ (8g/ f ) ≡ (Rh /n) (in S.I. units) ≡ (1.49/n) Rh (in English units). For steadyuniform ﬂow, S f = S0 , in a widerectangular channel, Rh = h, the normal depth h n is obtained after q = V h is substituted into Eq. (4.3): 1/3 f q2 (4.5a) hn = 8gS0 or nq 3/5 hn = in S.I. (4.5b) 1/2 S0 The normal depth thus increases with discharge and friction factor but decreases with increasing slope. Evaluation of the Darcy–Weisbach friction factor for pipe ﬂows yields three ﬂow regimes clearly shown in Fig. 4.10. The ﬂow is laminar on smooth surfaces when the Reynolds number Re < 2000 and the Darcy–Weisbach friction factor is inversely proportional to the Reynolds number, f = (kt /Re). The ﬂow becomes turbulent when Re > 2000 and the surface boundary roughness exerts
Figure 4.10. Resistancetoﬂow diagram.
Steady river ﬂow
91
an inﬂuence on resistance to ﬂow. When the ﬂow is hydraulically smooth, the Darcy–Weisbach friction factor f gradually decreases with the Reynolds number. When the ﬂow is hydraulically rough, resistance to ﬂow depends on only the ratio of pipe diameter to surface roughness. For turbulent ﬂows in rough pipes, the Darcy–Weisbach friction factor f and the Ch´ezy coefﬁcient C remain constant at any value of the Reynolds number. The evaluation of the Darcy–Weisbach friction factor in rivers is a complex matter. The friction factor f varies with ﬂuid viscosity, ﬂow depth, grain size, bedform conﬁguration, and vegetation. Exact values can be obtained experimentally for ﬁeld conditions or laboratory conditions representative thereof. River ﬂows are turbulent, and we essentially need to differentiate between hydraulically rough versus hydraulically smooth boundaries. We can do this by comparing the bedmaterial grain size d50 with the laminar sublayer thickness δ = 11.6 ν/u ∗ , where ν is the kinematic viscosity of water and u ∗ is the shear velocity from Eq. (4.2). The order of magnitude for δ corresponds to tenths of millimeters, that corresponds to sand sizes. When δ > 3 ds , the ﬂow is hydraulically smooth; the ﬂow is hydraulically rough when δ < ds /6. Resistance to ﬂow in plane bed rivers with bed material ﬁner than sand is said to be hydraulically smooth: C 3.67u ∗ h 8 V = √ = 5.75 log , (4.6) = u∗ f g ν √ where u ∗ = τ0 /ρ is the shear velocity and ν is the kinematic viscosity of the ﬂuid. For hydraulically rough boundaries without bedforms, resistance to ﬂow in rivers with bed material coarser than sand can be approximated by 12.2Rh 8 C V = 5.75 log =√ = , (4.7) u∗ g f ks where ks ∼ = 3 d90 or ks ∼ = 6.8 d50 can serve as ﬁrst approximations. Three equivalent ﬂowresistance formulations remain commonly used in river engineering practice: (1) the Ch´ezy coefﬁcient C, (2) the Manning coefﬁcient n, and (3) the Darcy–Weisbach friction factor f. Both Ch´ezy and Manning coefﬁcients are dimensional, and the equivalence among the three is simply stated as 1/6 1/6 R 1.49 Rh 8g ≡ h (S.I. units) ≡ (English units). (4.8) C≡ f n n Typical values for grain resistance are shown in Table 4.1. Both f and n increase with surface roughness. The Ch´ezy coefﬁcient describes ﬂow conveyance and decreases with surface roughness.
92
Steady ﬂow in rivers
Table 4.1. Grain resistance and velocity formulations for turbulent ﬂow over hydraulically rough plane boundaries (C = C and f = f ) Formulation
Range
Velocitya
Resistance parameter
Ch´ezy
h/ds → ∞
Manning
h/ds > 100
Logarithmic
C √ = g
8g 1/2 1/2 constant V = C Rh S f f 1/6 1/6 Rh C Rh 1 2/3 1/2 ∼ (S.I.) (S.I.) V = Rh S f =a = √ ∼ g ds n n 1/6 ∼ n = 0.062 d50 (d50 in meters) 1/6 n∼ = 0.046 d75 (d75 in meters)
C=
8 12.2 Rh = 5.75 log f ks
1/6 n∼ = 0.038 d90 (d90 in meters) 12.2 Rh V = 5.75 log g Rh S f ks ∼ ks = 3 d90 ks ∼ = 3.5 d84 ks ∼ = 5.2 d65 ∼ 6.8 d50 k = s
The hydraulic radius Rh = A/P is used, where A is the crosssectional area, and P is the wetted perimeter; the friction slope S f is the slope of the energy grade line.
a
For practical purposes, it is clear from Fig. 4.11 that the logarithmic equation applies over a wide range of h/d50 . As a ﬁrst approximation, the following can be used: 2h 8g = 5.75 log . (4.9) f d50 Manning’s equation is also found to be applicable. For instance, the rela1/6 tionship in which n = 0.064 d50 , with d50 in meters, should be in reasonable agreement with the ﬁeld measurement when h/ds >100 and h/ds < 10,000. It is concluded that Manning’s equation may not applicable in shallow mountain streams (h < 10 ds ) and in very deep sandbed rivers (h > 10,000 ds ). The logarithmic form of grain resistance in Eq. (4.9) can be transformed into an equivalentpower form in which the exponent m varies with relative submergence h/ds : m ˆ h 8 V bh ˆ = = a ≡ a ln (4.10) u∗ f ds ds under the transformation that imposes the constraint that the value and the ﬁrst
Steady river ﬂow
93
Table 4.2. Typical values of resistance coefﬁcients
Boundary type
DarcyWeisbach f
Manning n
Ch´ezy C (m1/2 /s)
Smooth Plane sand bed Sand antidunes Ripples Sand dunes Gravel bed Cobble bed Boulder bed Vegetation
0.0056 0.0046–0.0078 0.0078–0.015 0.015–0.042 0.018–0.076 0.011–0.042 0.018–0.057 0.029–0.076 0.042–0.24
0.01 0.010–0.013 0.013–0.018 0.018–0.030 0.020–0.040 0.015–0.030 0.020–0.035 0.025–0.04 0.03–0.07
118 100–130 72–100 43–72 32–65 43–86 37–65 32–52 18–43
Figure 4.11. Resistance to ﬂow for hydraulically rough rivers.
derivative be identical: aˆ ds m a= , m h m=
1 ˆ ln(bh/d s)
.
(4.11a) (4.11b)
Vegetation increases resistance to ﬂow, and only crude empirical resistance coefﬁcients can be obtained. Tables 4.2 provides typical values of resistance coefﬁcients for various conditions. Ranges of values for the Manning coefﬁcient n are summarized in Tables 4.2 and 4.3. In the case of hydraulically smooth channels, values of 0.01< n < 0.02
94
Steady ﬂow in rivers
Table 4.3. Typical bedform characteristics
Manning n
Sediment concentration (mg/l)
Dominant type of roughness
Bedform surface proﬁles
Lower ﬂow regime Plane bed Ripples Dunes Washedout dunes
0.014 0.018–0.028 0.020–0.040 0.014–0.025
0 10–200 200–3000 1000–4000
Grain Form Form Variable
— — Out of phase Out of phase
Upper ﬂow regime Plane bed Antidunes Chutes and pools
0.010–0.013 0.010–0.020 0.018–0.035
2000–4000 2000–5000 5000–50000
Grain Grain Variable
— In phase In phase
Bedform
are appropriate. In sandbed channels, the presence of bedforms increases resistance and values of the Manning coefﬁcient n can be as high as 0.05. In gravel1/6 bed and cobblebed streams, grain resistance is predominant, with n ∼ ds , as given in Tables 4.1 and 4.2. Other values of the Manning coefﬁcient n are given in Table 4.3. Example 4.1 explains how to calculate the normal depth and shear stress in a river. Example 4.2 illustrates how to determine the main characteristics of channel ﬂows, including shear stress, resistance parameters, and friction slope, from ﬁeld measurements of a velocity proﬁle.
Example 4.1 Application to smooth plane bed. Consider a steadyuniform ﬂow in a 10mwide smooth rectangular channel; if the discharge is 10 m3 /s and the slope is 26 cm/km, estimate the friction factor in terms of f and equivalent C and n and calculate the normal ﬂow depth and the corresponding applied bed shear stress. For a very smooth surface, we can consider f ∼ = 0.01 from Table 4.1, the calculated normal ﬂow depth given the unit discharge q=
10 m3 1 m2 Q = = W s 10 m s
is hn =
0.01 × 1 m4 s2 2 s × 8 × 9.81 m × 26 × 10−5
1/3 = 0.788 m.
The equivalent resistance parameters C and n are calculated, with the hydraulic
Steady river ﬂow
95
radius given in S.I. units, as A 10 m × 0.788 m W hn = = 0.681 m, = P W + 2h n 10 m + 2 × 0.788 m 8 × 9.81 m = 88 m1/2 /s, C= s 2 × 0.01
Rh =
1/6
Rh (0.681 m)1/6 = = 0.01 s/m1/3 . C 88 m1/2 s The applied bed shear stress for steadyuniform ﬂow S f = S0 is obtained from Eq. (4.1): n=
9810 N × 0.681 m × 26 × 10−5 m3 1.73 N = = 1.73 Pa. m2
τ0 = γ Rh S f = γ Rh S0 =
Example 4.2 Application to a turbulentvelocity proﬁle. Consider the given measured velocity proﬁle in a 200ftwide river in Fig. 4.5(a). Consider two points, 1 and 2, near the bed, z1 u∗ ln , v1 = κ ks v2 =
z2 u∗ ln , κ ks
and estimate the following parameters: (a) shear velocity: u∗ =
κ (v 2 − v 1 ) 0.4(0.85 − 0.55)ft = = 0.11 ft/s = 0.0335 m/s. z2 1.5 ln ln s z1 0.5
(b) boundary shear stress: ft2 1.92 slug lb (0.11)2 2 = 0.023 2 = 1.1 Pa. 3 s ft ft (c) depthaveraged ﬂow velocity: τ0 = ρ u 2∗ =
V ∼ = 0.85 ft/s = 0.26 m/s. (d) unit discharge: q = V h = 0.85 ft/s × 3.7 ft = 3.1 ft2 /s = 0.292 m2 /s.
96
Steady ﬂow in rivers (e) hydraulic radius: Rh =
Wh 200 × 3.7 ft2 A = = = 3.56 ft = 1.09 m; P W + 2h 200 + 2 × 3.7 ft
the hydraulic radius Rh = 3.56 ft is close to ﬂow depth 3.7 ft. (f) Froude number: Fr = √
V ∼ V 0.85 ft/s = = 0.078. =√ 2 g Rh gh 32.2 × 3.7 fts2
(g) friction slope from Eq. 4.1: 0.023 lb ft3 τ0 ∼ τ0 10 cm = 2 . = 9.96 × 10−5 ∼ = = γ Rh γh km ft 62.4 lb × 3.7 ft
Sf =
(h) Darcy–Weisbach friction factor from Eq. 4.3: f =
8 Sf 8 × 9.96 × 10−5 = = 0.13. 0.0782 Fr2
(i) Manning coefﬁcient from Table 4.1: 1.49 ft1/3 s 1.49 2/3 1/2 Rh S f = (3.7 ft)2/3 (9.96 × 10−5 )1/2 V 0.85 ft m1/3 = 0.042 s/m1/3 .
n=
Note that, because of the conversion factor 1.49 ft1/3 /m1/3 , the value of n is the same in both S.I. and English units. (j) Ch´ezy coefﬁcient from Eq. 4.8: 8g 8 × 32.2 = = 44.5 ft1/2 /s = 24.6 m1/2 /s. C= f 0.13 Note that the metric value of C is commonly used. (k) momentum correction factor [Eq. (4.17)] in Subsection 4.2.1: 1 1 2 v x2 dA ∼ v dh i , βm = = 2 A Vx A hVx2 i xi Note that dh i = 1 ft except for the uppermost velocity measurement s2 ft3 1 2 2 2 2 [0.55 + 0.85 + 1.0 + (1.1 × 0.7)] 2 3.7 ft (0.85)2 ft s2 = 1.074.
βm ∼ =
Steadynonuniform river ﬂow
97
(l) energy correction factor [Eq. (4.23b)] in Subsection 4.2.2:
1 ∼ 1 αe = v x3 dA = v 3 dh i , 3 3 AVx A hVx i xi s3 ft4 1 3 3 3 3 [0.55 + 0.85 + 1.0 + (1.1 × 0.7)] 3.7 ft (0.85)3 ft3 s3 = 1.194.
αe ∼ =
Note that αe > βm and that both are greater than unity.
4.2
Steadynonuniform river ﬂow
Steadynonuniform ﬂow in rivers implies that the total discharge does not change with time but can vary in the downstream direction. Mathematically, steady ﬂow implies that (∂h/∂t) = 0, (∂ V /∂t) = 0, (∂ W/∂t) = 0, and (∂ Q/∂t) = 0. In a 1D channel without rainfall, inﬁltration, and lateral inﬂow, discharge also remains constant in the downstream direction. Nonuniform ﬂow is possible under steady discharge when the mean ﬂow velocity, channel width, and ﬂow depth change in the downstream direction, or (∂ V /∂ x) = 0, (∂ W/∂ x) = 0, and (∂h/∂ x) = 0. In the following discussion, momentum equations are derived (Subsection 4.2.1), followed by rapidly varied converging ﬂow in rivers (Subsection 4.2.2) and gradually varied ﬂow (Subsection 4.2.3). 4.2.1
Momentum equations for steady ﬂow
Momentum equations deﬁne the hydrodynamic forces exerted by surface ﬂows. After the equations of motion [Eqs. (2.18)] are multiplied by the mass density of the water ρ, the terms on the lefthand side of the equations represent the rate of momentum change per unit volume and the rate of impulse per unit volume is found on the righthand side. Integration over the total volume ∀ shows that the rateofmomentum change equals the impulse per unit time. For example, the x component in the Cartesian coordinates for steady ﬂow is ∂v x ∂v x ∂v x + vy + vz d∀ ρ vx ∂x ∂y ∂z ∀ τ yx ∂τzx ∂τx x ∂p d∀ + + + = ρgx d∀ − d∀. (4.12) ∂x ∂y ∂z ∀ ∀ ∂x ∀ The integrand on the lefthand side can be rewritten as ∂ρv x v y ∂ρv y ∂ρv x ∂ρv x v z ∂ρv z ∂ρv x2 + + − vx + + . ∂x ∂y ∂z ∂x ∂y ∂z
(4.13)
98
Steady ﬂow in rivers
By virtue of the continuity Eq. (2.17), the terms in parentheses of expression (4.13) can be dropped. The volume integral of the remaining momentum and stress terms can be transformed into surface integrals by means of the divergence theorem Eq. 5.2. The result is the general impulse–momentum relationship. x component ∂x ∂y ∂z ∂x + vy + vz ρv x v x p dA dA = ρgx d∀ − ∂n ∂n ∂n A ∀ A ∂n ∂x ∂y ∂z + τ yx + τzx dA. (4.14a) + τx x ∂n ∂n ∂n A y component ∂x ∂y ∂z ∂y + vy + vz ρv y v x p dA dA = ρg y d∀ − ∂n ∂n ∂n A ∀ A ∂n ∂x ∂y ∂z + τ yy + τzy dA. (4.14b) τx y + ∂n ∂n ∂n A z component ∂x ∂y ∂z ∂z + vy + vz ρv z v x p dA dA = ρgz d∀ − ∂n ∂n ∂n A ∀ A ∂n ∂x ∂y ∂z + τ yz + τzz dA. (4.14c) τx z + ∂n ∂n ∂n A It is observed that momentum is a vector quantity, the momentum change that as due to convection is embodied in the surface integral on the lefthand side of Eqs. (4.14), and all the stresses are expressed in terms of surface integrals. Consider a detailed application of the momentum equations to openchannel ﬂows. With reference to the rectangular channel sketched in Fig. 4.12, the momentum relationship [Eqs. (4.14)] in the downstream x direction is applied to this channel, now subjected to rainfall at an angle θr , velocity Vr over an area Ar , wind shear τw , bank shear τs = τ yx , and bed shear τb = τ0 = τzx : ∂x ∂y ∂z ∂x + vy + vz ρv x v x p dA dA = ρgx d∀ − ∂n ∂n ∂n A ∀ A ∂n ∂x ∂y ∂z + τ yx + τzx + dA. (4.15) τx x ∂n ∂n ∂n A Several integrals vanish for 1D ﬂow in impervious channels, v y = v z = τx x = 0, leaving ∂x ∂z ∂y ∂x ρv x2 dA + p dA = ρgx d∀ + τzx dA + τ yx dA ∂n ∂n ∂n A A ∂n ∀ A A (4.16)
Steadynonuniform river ﬂow
99
Figure 4.12. Forces applied on a river reach.
Consider an incompressible homogeneous ﬂuid, ρm = ct, and deﬁne the momentum correction factor βm , given the crosssectional averaged velocity Vx : 1 v 2 dA. (4.17) βm = AVx2 A x The value of βm is generally close to unity; the reader can refer to Example 4.2 for a detailed calculation from a measured velocity proﬁle. With average values of pressure p, velocity V, and area A at upstream cross section 1 and downstream cross section 2, the integration of the momentum equation for this control volume ∀ of length X c , width W , and height h yields βm ρ A2 V22 + p2 A2 − βm ρ A1 V12 − p1 A1 − ρ Ar Vr2 sin(θ + θr ) = γ ∀ sin θ − τb W X c − τs 2h X c + τw W X c .
(4.18)
Assuming that the boundary shear stress τ0 equals the bank shear stress τs and the bed shear stress τb , the equation with negligible rainfall, Ar → 0, without wind shear, τw → 0, can be rewritten when the channel inclination θ is small (sin θ ∼ = S0 , the bed slope) as A1 + A2 2 2 +γ X c S0 − τ0 (W + 2h)X c . p2 A2 + βm ρ A2 V2 = p1 A1 + βm ρ A1 V1 2  downstream force   upstream force     shear force  weight force (4.19)
100
Steady ﬂow in rivers
This equation indicates equilibrium of forces in the downstream direction. The downstream and the upstream forces equal the sum of a pressure force and a momentum force. Further reduction is possible when the weightforce component cancels the shear force. In the case of steady ﬂow, Q 1 = Q 2 , in a rectangular channel of width W, the crosssection area is A = W h and the average pressure is p = 0.5 ρ gh. Further assuming that the Boussinesq correction factor βm ∼ = 1, we ﬁnd that the hydrodynamicforce component Fh reduces to Fh = p A + βm ρ AV = 0.5ρghW h + ρW h 2
Q Wh
2 .
(4.20)
The speciﬁcmomentum function M is obtained after the hydrodynamic force Fh is divided by the channel width W and the speciﬁc weight of the ﬂuid ρg; thus M=
h2 q2 Fh = + , ρgW 2 gh
(4.21)
where the unit discharge q = Q/W . This speciﬁcmomentum function M can be plotted as a function of ﬂow depth at a given unit discharge q. For instance, the speciﬁcmomentum functions for q = 1 m2 /s are shown in Fig. 4.13. Two ﬂow depths with identical speciﬁc momenta are called conjugate depths. The lowest value of the speciﬁcmomentum function is obtained when ∂ M/∂h = 0, which corresponds to the critical ﬂow depth h c = (q 2 /g)1/3 . Given the property that q = V h, the critical ﬂow depth corresponds to (q 2 /gh 3c ) = (V 2 /gh c ) = Figure 4.13. Speciﬁcmomentum dia2 Fr c = 1. A critical value of the Froude gram. number Fr = 1 relates to the minimum value of the speciﬁcmomentum function. Flow is supercritical when Fr > 1 or h < h c and subcritical when Fr < 1 or h > h c . Example 4.3 Application of momentum to hydraulic jump. Steady ﬂow, Q = 10 m3 /s, in a rectangular channel, W = 10 m, is such that the upstream ﬂow velocity V = 4 m/s is rapidly reduced over a short distance to form a
Steadynonuniform river ﬂow
101
hydraulic jump. If the channel slope counterbalances the shear force, determine the ﬂow depth, velocity, and forces downstream of the hydraulic jump. The increased turbulence in the aerated portion of the ﬂow near the surface is not accounted for by the bedresistance equations. For this reason, energy will be dissipated at a faster rate than it would through bedresistance. Because ﬂow depth changes over a short distance, it can be shown that the bed friction force on a smooth surface without bafﬂe blocks is very small compared with pressure forces. We can therefore consider that equilibrium will be approximately dictated by the balance of pressure and hydrodynamic forces upstream and downstream of the hydraulic jump. The answer can be found with the use of conjugate depths on the speciﬁcmomentum diagram, Fig. 4.13. Exact values of the conjugate ﬂow depth can be calculated from M1 = M2 in the form q2 g
1 1 − h1 h2
=
1 2 h − h 21 . 2 2
(E.4.3.1)
The relationship Fr21 = (q 2 /gh 31 ) allows to rewrite Eq. E.4.3.1 as Fr21 =
1 h2 2 h1
h2 +1 . h1
(E.4.3.2)
The solution of the quadratic equation in h 2 / h 1 is called the Belanger equation: 1 h2 = h1 2
1+
8 Fr21
−1 .
(E.4.3.3)
The conjugate downstream ﬂow depth h 2 is calculated directly from the upstream ﬂow depth h 1 and the upstream Froude number Fr1 . For the case in point, Q = 10 m3 /s and W = 10 m, or q = Q/W = 1 m2 /s, the upstream ﬂow depth is (q/V1 ) = [(1 m2 s])/(s × 4 m)] = 0.25 m and the upstream Froude number is Fr21 =
q2 1 m 4 s2 = 2 = 6.52. 3 s × 9.81 m × (0.25)3 m3 gh 1
The corresponding downstream conditions are calculated from the Belanger √ equation ﬂow depth h 2 = (0.25/2)( 1 + 8 × 6.52 − 1) = 0.788 m and downstream velocity V2 = (q/h 2 ) = (1.27 m/s) (see Fig. E.4.3.1).
102
Steady ﬂow in rivers
Figure E.4.3.1. Hydraulic jump.
Hydrodynamic forces are calculated by multiplying the speciﬁcmomentum value by ρgW . For instance, the downstream pressure force is F p2 = ρg W h 2 /2 =
9.81 kN (0.788 m)2 = 30.4 kN. × 10 m × m3 2
Simpliﬁed solutions are possible for two types of nonuniform ﬂow: (1) rapidly varied converging ﬂow and (2) gradually varied ﬂow. In both cases, convective changes in ﬂow depth, width, and velocity head are signiﬁcant and cannot be neglected. Rapidly varied ﬂows are usually induced by structures and other perturbations to the ﬂow. Because ﬂow conditions change over a short distance, it can generally be assumed that the energy loss and the change in bed elevation are small compared with the convective terms of the St. Venant equation. We can assume as a ﬁrst approximation that V2 d h+ = 0, (4.22a) dx 2g which can be integrated over x to yield conservation of speciﬁc energy over this short reach. The speciﬁcenergy diagram hence becomes extremely useful in the analysis of rapidly varied steady ﬂows. Gradually varied ﬂows are those in which changes in width, depth, and velocity take place over a reasonably long distance; the convectiveacceleration terms can be neglected but friction losses remain important over a long reach. Gradually varied ﬂows refer to changing ﬂow depth in the downstream direction such that, with a change in ﬂow depth h in the downstream x direction, all four
Steadynonuniform river ﬂow
103
terms of the St.Venant equation are rearranged in the following manner: dE ∼ d V2 (4.22b) h+ = S0 − S f . = dx dx 2g 4.2.2
Rapidly varied converging river ﬂow
Rapidly varied ﬂow conditions refer to large spatial derivatives over a short channel reach. In converging ﬂows, it is often considered that the energy level will be maintained constant because the bed friction does not take place over a long distance. In widerectangular channels, the gradient of the integral form of the Bernoulli sum can be rewritten as d −dz b p V2 d E˜ = = − S f = S0 − S f ∼ + αe = 0; (4.23a) dx dx γm 2g dx this implies that the integral form of speciﬁc energy E˜ = ( p/γm ) + [αe (V 2 /2g )] remains constant. The energy correction factor αe is deﬁned as 1 v 3 dA. (4.23b) αe = AVx3 A x After considering that αe ∼ = 1, hydrostatic distribution p = γ h, and that the unit discharge q = V h, we ﬁnd that the speciﬁc energy E corresponds to the sum of pressure and velocity head above the channelbed elevation: E =h+
q2 . 2gh 2
Figure 4.14. Speciﬁcenergy diagram.
(4.24) The speciﬁcenergy function E can be plotted as a function of ﬂow depth at a given unit discharge q, e.g., q = 1 m2 /s in Fig. 4.14. Two ﬂow depths with identical speciﬁc energies are called alternate depths. The lowest value of the speciﬁc energy corresponds to (∂ E/∂h) = 1 − (q 2 /gh 3c ) = 0, which deﬁnes the critical ﬂow depth when the Froude number Fr = 1: 2 1/3 q . (4.25) hc = g
104
Steady ﬂow in rivers
We also easily demonstrate that the minimum value of speciﬁc energy E min = 1.5 h c . Once the critical ﬂow depth is known, the Froude number can be directly calculated from q 2 = gh 3c as q = Fr = √ h gh
hc h
3/2 .
(4.26)
Applications of rapidly varied ﬂows are shown for river ﬂow contraction in Example 4.4 and for ﬂow under a sluice gate in Example 4.5. Example 4.4 Application to openchannel ﬂow contractions. Consider steady ﬂow at Q = 10 m3 /s in a 10m widerectangular channel from Example 4.3. Determine (a) the maximum possible elevation of a sill z max in section A that will not cause backwater and (b) the maximum lateral contraction Wmax of the channel in section A that will not cause backwater. In rapidly varied ﬂow, it is expected that the energy losses through friction are equivalent to friction losses for steadyuniform ﬂow. The accelerating ﬂow is analyzed with the speciﬁcenergy diagram, Fig. E.4.4.1.
Figure E.4.4.1. Sill and river contraction.
Steadynonuniform river ﬂow
105
Solution: (a) The maximum elevation of the sill z max in section A is such that the ﬂow will be critical on top of the sill and z max + E min = E 1 or z max = E 1 − E min . In this example, the unit discharge q = (Q/W ) = [(10 m3 /s)/10 m] = 1 m2 /s, the critical ﬂow depth 1/3 1 m4 s4 q2 = 2 = 0.467 m, hc = 3 g s × 9.81 m and the minimum speciﬁc energy E min = 3/2 h c = 0.70 m. The approaching steadyuniform ﬂow depth from Example 4.3 is h 1 = 0.788 m. The corresponding velocity is V1 = (q/h 1 ) = [1 m2 /(s × 0.788 m)] = (1.27 m/s). The speciﬁcenergy level E 1 = h 1 + V12 /2g = 0.788 m + (1.27 m/s)2 2 × 9.81 m = 0.870 m. The maximum sill elevation z max = E 1 − E min = 0.870 m − 0.70 m = 0.17 m. (b) The minimum channel width W2 in section A that does not cause backwater is such that the total discharge remains constant, Q = W1 q1 = W2 q2 , and the ﬂow is critical in the contracted section A2 , or h c2 = 0.67 E 1 = 0.67 E min2 , with E 1 = 0.87 m and 1 = q22 /gh 3c2 , or W2 =
Q g(0.67E 1 )3
.
In this example, 10 m3 s = 7.22 m. W2 = s 9.81 m (0.667 × 0.87 m)3 The maximum lateral contraction Wmax = W1 − W2 = 10 m − 7.22 m = 2.78 m. Example 4.5 Application to rapidly varied ﬂow under a sluice gate. A sluice gate controls the 10m3 /s ﬂow discharge in a 10m widerectangular channel. If the ﬂow depth downstream of the sluice gate is at h 1 = 0.25 m and rapidly increases to the normal ﬂow depth h 2 = 0.788 m in a hydraulic jump located at the toe of the sluice gate, determine the following: (a) What is the water level upstream of the sluice gate? (b) What is the force applied onto the sluice gate? (c) How much energy is lost in the hydraulic jump? Solution: (a) The energy losses through friction along the sluice gate can be neglected because the ﬂow is rapidly varied and converging. It is assumed that the speciﬁcenergy level on both sides of the gate are identical. At a unit
106
Steady ﬂow in rivers
ﬂow discharge q = (Q/W ) = 1 m2 /s, the speciﬁcenergy diagram in Fig. 4.14 shows that the downstream speciﬁc energy corresponding to a downstream ﬂow depth h 1 = 0.25 m is E1 = h1 +
q2 1 m4 s2 = 0.25 m + = 1.06 m. s2 × 2 × 9.81 m (0.25 m)2 2gh 21
The alternate depth at the same speciﬁcenergy level corresponds to the ﬂow depth h 2 ∼ = 1 m upstream of the gate. (b) The force applied on the gate can be calculated from the speciﬁcmomentum function plotted in Fig. 4.13. We obtain graphically, at an upstream ﬂow depth of 1 m, M = 0.6 m2 , and, at the downstream ﬂow depth of 0.25 m, M∼ = 0.45 m2 . The corresponding force applied to the gate corresponds to the difference in speciﬁcmomentum times γ W , or F = γ W M = (9,810 N /m3 ) × 10 m × (0.6 − 0.45) m2 = 14.7 kN. (c) The hydraulic jump with q = 1 m2 /s and upstream ﬂow depth h 1 = 0.25 m was previously examined in Example 4.3. The speciﬁcenergy level at the upstream end is given by E1 = h1 +
q2 1 m4 s2 = 0.25 m + 2 = 1.06 m, 2 s × 2 × 9.81 m × (0.25 m)2 2gh 1
as shown in Fig. 4.14. At the downstream end of the hydraulic jump, the previously calculated conjugate depth h 2 = 0.788 m corresponds to the speciﬁc energy q2 1 m4 s2 = 0.788 m + 2 2 s × 2 × 9.81 m × (0.788 m)2 2gh 2 = 0.87 m.
E2 = h2 +
The speciﬁc energy lost in the jump is E = 1.06 m − 0.87 m = 0.19 m. The speciﬁc energy lost in a hydraulic jump can be calculated directly after the following transformation: q2 q2 + − h E = h 2 + 1 2gh 22 2gh 21 q2 1 1 − , (E.4.5.1) = h2 − h1 + 2g h 22 h 21 given the constraint of conjugate depths 1 1 q2 1 = h 22 − h 21 − g h1 h2 2 from Eq. (E.4.3.1).
Steadynonuniform river ﬂow
107
After q 2 /g is substituted with a function of h 1 and h 2 , E in Eq. (E.4.5.1) can be solved as a function of h 1 and h 2 after algebraic manipulations as E =
(h 2 − h 1 )3 . 4h 1 h 2
(E.4.5.2)
In our practical example, E =
(0.788 − 0.25)3 = 0.197 m. 4 × 0.788 × 0.25
The power loss P in the hydraulic jump is then calculated from P = γ QE =
9,810 N 10 m3 × 0.197 m = 19.3 kW. × m3 s
The larger the Froude number upstream of the jump, the larger the power lost through turbulence in the hydraulic jump.
4.2.3
Gradually varied river ﬂow
The term graduallyvaried ﬂow refers to ﬂow conditions that change over long distances. Simpliﬁcations of the momentum equations are possible for 1D ﬂows. As a ﬁrst approximation, it is often assumed that resistance to ﬂow in gradually varied ﬂow can be calculated as for steadyuniform ﬂow. In widerectangular channels, h = Rh with steadyuniform ﬂow, S f = S0 , the normal ﬂow depth h n from Eq. (4.5a) compares with the critical ﬂow depth h c from Eqs. (4.25) and (4.26) according to hn = hc
f 8S0
1/3 =
1 Fr
2/3 ·
(4.27)
at normal depth
Graduallyvaried watersurface elevation proﬁles are commonly called backwater curves. Their analysis results from a direct application of the quasisteady dynamic wave of the St. Venant equation. For a steady 1D ﬂow, in widerectangular channels, relation 2.27 can be rewritten as dE dh V ∂V dh ∂h dE = = S0 − S f = + = (1 − Fr2 ). dx dh dx ∂x g ∂x dx
(4.28)
Note that, in this derivation, it is assumed that V = q/ h and Fr2 = q 2 /gh 3 and an equation for resistance to ﬂow is not required because q is considered constant.
108
Steady ﬂow in rivers
The relationship describing watersurface elevation for a steady 1D ﬂow of an incompressible sedimentladen ﬂuid is S0 − S f dh = . dx 1 − Fr2
(4.29)
Using the properties of critical ﬂow depth h c from Eq. (4.25) and normal depth h n from Eq. (4.5), in widerectangular channels, Rh = h, the governing equation for steady ﬂow, with constant q and f , becomes 3 S0 1 − hhn dh = 3 . dx 1 − hhc
(4.30)
Note that dh/dx → 0 as the ﬂow depth h approaches the normal depth h n . Also, dh/dx → ∞ near critical ﬂow as h → h c . The sign of dh/dx depends on the relative magnitude of h, h n , and h c . Five types of backwater proﬁles are possible: 1. 2. 3. 4. 5.
H proﬁles for horizontal surfaces with h n → ∞, M proﬁles for mild slopes when h n > h c or S0 < f /8, C proﬁles for critical slopes when h n = h c or S0 = f /8, S proﬁles for steep slopes when h n < h c , or S0 > f /8, A proﬁles or adverse slopes when S0 < 0.
A subcritical normal depth h n > h c corresponds to f > 8 S0 and a supercritical normal depth h n < h c is obtained when f < 8 S0 . Accordingly, a stream slope is said to be mild when S0 < f /8 and steep when S0 > f /8. The ratio of friction slope S f to bed slope S0 can be obtained from S f = ( f q 2 /8 gh 3 ) and S0 = ( f q 2 /8 gh 3n ); thus Sf = S0
hn h
3 ,
(4.31)
and, ﬁnally, the ratio of applied bed shear stress τb to the bed shear stress applied at normal depth τbn for widerectangular channels with constant q and f is 2 τb ∼ γ h S f h hn 3 hn = = . (4.32) = τbn γ h n S0 hn h h Typical watersurface proﬁles in open channels are shown in Fig. 4.15.
Steadynonuniform river ﬂow
109
4.15 Backwater proﬁles for mild and steep slopes.
Numerical calculations can be initiated from a given ﬂow depth h 1 , and the distance increment x at which h 2 = h 1 ± h is approximated by 3 h 1 − hh c 1 x ∼ (4.33a) = 3 S0 1 − hh n 1
or
3 S0 x 1 − hhn h ∼ = 3 . 1 − hhc
(4.33b)
Note that it is also possible to iterate on h until a predetermined value of x is obtained. The friction slope S f in gradually varied ﬂows with constant q and f can be approximated by Eq. (4.31), which shows that S f < S0 when the ﬂow depth exceeds the normal depth and increases very rapidly when h < h n . This shows that the bed shear stress increases (τ > τn ) at ﬂow depths less than normal depth (h < h n ). Bed shearstress distributions for 1D mild M and steep S backwater curves are sketched in Figs. 4.15 and 4.16. Shear stress
110
Steady ﬂow in rivers increases in the downstream direction for converging ﬂows (M2 and S2 backwater curves), and decreases for diverging ﬂows (M1, M3, S1, and S3 backwater curves). Consequently, bed sediment transport is expected to increase in the downstream direction for converging ﬂows and decrease for diverging ﬂows.
Example 4.6 Application to gradually varied ﬂow. Consider a 10mwide canal discharging 10 m3 /s at a slope of 26 cm/km along the reach sketched in Fig. E.4.6.1. Given the sluicegate opening at 0.2 m, a sill height at 0.17 m, and free downstream overfall conditions, draw the watersurface proﬁle, also called the hydraulic grade line (HGL), and the energy grade line (EGL). The ﬂow depth at C is 0.2 m and the bed surface is smooth (f ∼ = 0.01). Identify the backwater curves and sketch the shearstress distribution along the reach. Given the speciﬁcenergy diagram for the same unit discharge q = 1 m2 /s in Fig. 2.10, the alternate depth to h = 0.2 m is approx4.16 Typical backwater proimately 1.5 m upstream of the gate at B. ﬁles. Critical ﬂow depth, h c = (q 2 /g)1/3 = (1/9.81)1/3 = 0.467 m, controls the ﬂow depth on top of the sill at G and near the free overfall at J. The sill controls subcritical ﬂow upstream of G and the reach H–I is sufﬁciently long to sustain steadyuniform ﬂow between EF and HI at a normal depth h n = 0.788 m, from Example 4.1. Rapidly varied ﬂow is obtained between BC and FH, as is the hydraulic jump DE previously calculated in Example 4.3. Graduallyvaried ﬂow is observed among AB, CD, and IJ. The bed slope S0 = 26 cm/km corresponds to a mild slope because S0 < f /8 = 125 cm/km. In the reach AB, h > h n > h c and the backwater type is classiﬁed as M1. The ﬂow is accelerating between IJ with h n > h > h c to form an M2 backwater curve, and the reach CD, with h n > h c > h, corresponds to an M3 backwater curve. The EGL is drawn with conservation of energy in rapidlyvaried ﬂow between BC and FH taken into consideration. The speciﬁc energy for steadyuniform
Sediment transport in rivers
111
E.4.6.1 Sketch of HGL and EGL.
ﬂow was previously calculated in Example 4.5 at 0.87 m, which remains constant between E and I. From Example 4.5, the energy lost in the hydraulic jump is E = 0.197 m and, from Eq. (4.47), the friction slope S f is much greater than S0 when h < h n , as observed between CD and IJ. Finally, the velocity head is very small in the reservoir and the EGL practically corresponds to the free surface. Bed shearstress distribution is also sketched along the reach as calculated from Eq. 4.32, shear stress increases largely when h < h n and should be largest downstream of the sluice gate in the M3 backwater curve. Energy losses in the hydraulic jump are caused by nearsurface turbulence. Bed shear stress should be highly variable but larger than that for steadyuniform ﬂow. Around the sill, bed shear stress increases in converging ﬂow between F and G and decreases in diverging ﬂow between G and H. Bed shear stress increases again near the free overfall as h < h n .
4.3
Sediment transport in rivers
This section presents a brief summary of sediment transport in rivers. Equilibrium conditions are covered in Subsection 4.3.1 and aggradation–degradation characteristics are discussed in Subsection 4.3.2. More speciﬁc details are available in Julien (1995).
4.3.1
Equilibrium sediment transport
The dimensionless particle diameter d∗ is deﬁned from the speciﬁc gravity G of sediment, the kinematic viscosity of the ﬂuid ν, and the gravitational
112
Steady ﬂow in rivers
acceleration g as d ∗ = ds
(G − 1)g ν2
1/3 .
(4.34)
The settling velocity ω of a sediment particle in still water is deﬁned as 8ν ω= ds
0.5 d∗3 1+ −1 . 72
(4.35)
The ratio of shear force to bed particle weight deﬁnes the Shields parameter τ∗ as τ∗ =
τ0 u 2∗ = , (γs − γ ) ds (G − 1) g ds
(4.36)
where τ0 is the bed boundary shear stress, u ∗ is the shear velocity, γs is the speciﬁc weight of a sediment particle, γm is the speciﬁc weight of water, ds is the particle size, and g is the gravitational acceleration. The critical value of the Shields parameter τ∗c corresponding to the beginning of motion (τ0 = τc ) depends on d∗ , as shown in Fig. 4.17. Critical values of the Shields parameter τ∗c and shear stress τc for different particle sizes are listed in Table 2.4.
4.17 Particlemotion diagram (after Julien, 1995).
Sediment transport in rivers
113 The unit sediment discharge by volume can be deﬁned from the Meyer– Peter and the M¨uller formulas as qbv = 8 (τ∗ − τ∗c )1.5 0.5 × (G − 1)g ds3 .
(4.37a)
An alternative formulation based on the Shields parameter and the settling velocity is shown in Fig. 4.18. Also, a very crude approximation for sands, where 0.1 < τ ∗ < 1, is √ (4.37b) qbv ≈ 18 g ds3/2 τ∗2 · Sediment transport can be subdivided into three zones that describe the dominant mode of transport: bedload, mixed load, and suspended load. 4.18 Sedimenttransport diagram (after It is interesting that, for turbulent ﬂow Julien, 1995). over rough boundaries, incipient motion corresponds to u ∗ /ω ∼ = 0.2. Figure 4.19 shows the ratio of suspended to total load as a function of u ∗ /ω and h/ds . In most rivers, bedload is dominant at values of u ∗ /ω less than ∼0.4. A transition zone, called a mixed load, is found
4.19 Ratio of suspended to total sediment load (after Julien, 1995).
114
Steady ﬂow in rivers
4.20 Sediment concentration proﬁles (after Woo et al., 1988).
where 0.4 < u ∗ /ω < 2.5 in which both the bedload and the suspended load contribute to the total load. The sediment concentration C at an elevation z above the bed for the suspended load can be calculated from the Rouse equation as C = Ca
h−z z
a h−a
κuω
∗
,
(4.38)
where Ca is the concentration at an elevation a above the bed, h is the ﬂow depth, and κ is the von K´arm´an constant (κ ∼ = 0.4). An example of a concentration proﬁle is shown in Fig. 4.20.
4.3.2
Riverbed aggradation and degradation
Owing to continuity of sediment, part of the total load deposits on the channel bed as the sedimenttransport capacity decreases in the downstream direction. The sediment continuity relationship for advective ﬂuxes is ∂qt y ∂qt x ∂qt z ∂Cv + + + = 0, ∂t ∂x ∂y ∂z
(4.39)
where the mass ﬂuxes qt x , qt y , and qt z account for the total unit sediment discharge by volume in the x, y, and z directions, respectively.
Sediment transport in rivers
115
Assuming a steady supply of sediment (∂Cv /∂t = 0), Eq. (4.39) for a wide channel without lateral sediment inﬂow (∂qt y /∂ y = 0) reduces to ∂qt z ∂qt x + = 0, ∂x ∂z
(4.40)
which reduces further after it is assumed that the advective ﬂuxes qt x = v x Cv and qt z = −ωCv are dominant: ∂ωCv ∂v x Cv − = 0. ∂x ∂z
(4.41)
A practical approximation is obtained for gradually varied ﬂow (∂v x /∂ x → 0), constant fall velocity ω, and ∂Cv /∂z ∼ = −Cv / h; thus ωCv v x ∂Cv + = 0. ∂x h
(4.42)
The solution for grain sizes of a given fraction i (constant fall velocity) at a constant unit discharge q = V h, given v x = V , is a function of the upstream sediment concentration C0i of fraction i at x = 0: X ωi
Ci = C0i e− hV .
(4.43)
This shows that the concentration left in suspension is negligible (Ci /C0i = 0.01) at a distance X Ci : X Ci = 4.6
hV . ωi
(4.44)
This relationship is very useful in the design of settling basins. For the analysis of reservoir sedimentation, the percentage of sediment fraction i that settles within a given distance X deﬁnes the trap efﬁciency TEi as: TE i =
−W X ωi X ωi C0i − Ci = 1 − e− hV = 1 − e Q . C0i
(4.45)
It is interesting to note that the trap efﬁciency for particles of given settling velocity ωi , basin length x, and discharge Q = W hV increases with basin width W . We thus conclude that, for a given discharge and sediment discharge Q t x , increasing the channel width induces aggradation. When the trap efﬁciency of silt and clay particles is calculated, careful consideration must also be given to density currents and possible ﬂocculation, in which case the ﬂocculated settling velocity must be used instead of ωi : TEi
W ∂z i ∂ Q t xi + (1 − p0 ) =0 ∂x ∂t
(4.46a)
116
Steady ﬂow in rivers
or TEi ∂ Q t xi ∂z i =− . ∂t (1 − p0 ) W ∂ x
(4.46b)
Values of porosity p0 depend of the speciﬁc weight of sediment deposits. Dry speciﬁc weight of sediment deposits varies with the proportion of sand, silt, and clay and changes with time. The conversion of the incoming weight of sediment to volume necessitates knowledge of the average dry speciﬁc weight of a mixture γmd , deﬁned in Chap. 2 as the dry weight of sediment per unit total volume including voids. For material coarser than 0.1 mm, the dry speciﬁc weight of the mixture remains practically constant around γmd = 14.75 kN/m3 or 93 lb/ft3 . As a rough ﬁrst approximation, 1 ton of sediment corresponds to 20 ft3 of bed material. The corresponding dry mass density of the mixture, ρmd , is given by ρmd = γmd /g = 1500 kg/m3 or 2.9 slug/ft3 . The porosity p0 of sand material is then obtained from p0 = 1 − γmd /γs = 0.43. The volumetric sediment concentration is Cv = 1 − p0 and the void ratio is e = p0 /(1 − p0 ). For distances separating successive cross sections X larger than X C , the trap efﬁciency is essentially unity and aggradation responds directly to changes in the sedimenttransport capacity of the stream. For X smaller than X C , only part of the sediment load in suspension will settle within the given reach. The sediment load at the downstream end will then exceed the sedimenttransport capacity of the stream. The reader is referred to Julien (1995) for a detailed analysis of erosion and sedimentation. Exercise 4.1 Demonstrate the conjugate ﬂowdepth relationship, Eq. (E.4.3.3), from the speciﬁcmomentum identity M1 = M2 for ﬂow in a widerectangular channel. Exercise 4.2 Carry out the algebraic transformations from Eq. (E.4.5.1) to demonstrate that the speciﬁcenergy loss in a hydraulic jump is E =
(h 2 − h 1 )3 . 4h 1 h 2
Exercise 4.3 Apply the law of conservation of volume to an incompressible ﬂuid ﬂowing A through a 1D control volume of discharge Q = 0 v x dA given the crosssectional area A and top width W . Consider the lateral inﬂow of unit discharge
Problems
117 ql , rainfall intensity ir , and inﬁltration rate i b leaving through the wetted perimeter P (see Fig. Ex.4.3.1). [Hint: calculate the volume ﬂuxes entering the control volume ∀ = Adx and compare to the rate of change in control volume (∂∀/∂t = (∂ A/∂t)dx.] Exercise 4.4
Ex.4.3.1 Conservation of mass.
Demonstrate the 1D formulation of the A equation of motion (4.15). Consider the momentum ﬂux F = βm ρ AVx2 = ρ 0 v x2 dA passing through a crosssectional area A and top width W . Neglect momentum contributions from rainfall, lateral inﬂow, and inﬁltration. The shear stress τ0 is applied over the wetted perimeter P and the bed slope is S0 = tan θ (see Fig. Ex.4.4.1).
Problem 4.1
The crosssectional and depthaveraged ﬂowvelocity proﬁles are given in digital form with the orientation reference along the cross section. Complete Table P.4.1.1 to calculate the cross section area A = i ai and the ﬂow discharge Q = i Q i = ai v i sin θ. Note that a large eddy formed near the right bank and reverse ﬂow is measured. Draw the cross section proﬁle and the velocity proﬁle as per Fig. 4.5(b).
Ex.4.4.1 Conservation of momentum.
Problem 4.2
Consider the cross section of a 10m widerectangular minor channel with a bankfull depth of 2 m. Given a ﬂoodplain width that extends 50 m on each side of the minor channel bounded with a nearvertical escarpment (see Fig. P.4.2.1). Calculate the following parameters as functions of ﬂow depth h up to 5 m: (1) top channel width W , (2) wetter perimeter P, (3) crosssectional area
118
Steady ﬂow in rivers
Table P.4.1.1. Crosssection data Distance from left bank (ft)
Flow depth (ft)
Depthaveraged ﬂow velocity (ft/s)
Deviation angle (◦ )
0 10 20 30 40 60 80 100 120 140 160
0 1.3 1.7 1.8 2.1 2.4 2.9 3.6 4.1 2.2 0
0 +0.6 +0.2 0.4 1.1 1.4 2.1 2.6 1.9 0.8 0
— 260 280 110 90 89 86 85 88 90 —
ai (ft2 )
Qi (ft3 /s)
0 −7.68 17 6.76 31.5 67.19 58 186.48 82 35.2 0
Answer: A = 383.5 ft2 , Q = 596.5 ft3 /s.
¯ and (5) hyA, (4) near ﬂow depth h, draulic radius Rh . Discuss the effect of the width discontinuity at h = 2 m. Problem 4.3
The cross section in Fig. P.4.3.1 is located 1.25 km upstream of the control section on the Matamek River. Given the watersurface slope [Fig. 4.8(d)] and the stage–discharge relationship P.4.2.1 Cross section.
P.4.3.1 Cross section.
Problems
119
[Fig. 4.6(a)], determine the following: (a) A pool is formed because of downstream bedrock control. Draw the water surface elevation when Q = 0. (b) At the given discharge, Q = 465 ft3 /s, calculate surface width, wetted perimeter, crosssection area, mean ﬂow depth, mean ﬂow velocity, and hydraulic radius. Plot these points on Figs. 4.8. (c) Estimate the bankfull discharge at that cross section, assuming a constant Manning coefﬁcient n at all stages. Problem 4.4
Consider a steadyuniform ﬂow discharge of 15 m3 /s in a 70mwide 2–4mm gravelbed stream. Given the bed slope of 5 cm/km, determine the following: (a) (b) (c) (d) (e) (f)
the critical ﬂow depth; the resistance coefﬁcients f , n, and C; the normal ﬂow depth and compare it with the hydraulic radius; the Froude number at normal ﬂow depth; the type of slope for backwater proﬁle calculations; the applied bed shear stress at normal ﬂow depth.
Problem 4.5
Draw the position of the free surface, the HGL, and the EGL in a 10m widerectangular channel discharging 10 m3 /s in the smooth canal at a bed slope of 26 cm/km (see Fig. P.4.5.1). Answer: Combine the information from Examples 4.1 to 4.6. Energy is conserved between A–B and D–E. Friction losses occur between C and D, to
P.4.5.1 Deﬁnition sketch.
120
Steady ﬂow in rivers
P.4.6.1 Concentration proﬁle of the Missouri River (after Bondurant, 1963).
which turbulence losses are added between B and C. Momentum is conserved between B and D, and friction forces can be neglected between B and C, but are dominant between C and D. Added forces are provided by the sill and the sluice gate, which can be calculated as per Example 4.3. Problem 4.6
From the cross section in Fig. P.4.6.1, do the following: (a) Plot the surface width vs. stage and crosssectional areas vs. stage. (b) Estimate the Manning coefﬁcient n of the river if the slope is 1.5 × 10−4 . Problem 4.7
Use the values of aˆ and bˆ from Eqs. (4.10) and use the transforms for Eqs. (4.11) to deﬁne a and m for a power relationship applicable when h ∼ = 5 d50 . Once a and m are deﬁned, plot the straightline resistance relationship on Fig. 4.11 and compare with the ﬁeld measurements. Determine what the range of applicability is of this equation. Answer: From aˆ = 2.5, bˆ = 2, and h/d50 = 5, we obtain m = 0.434 √ and a = 2.86. Plotting (8/ f ) = 2.86 (h/d50 )0.434 in Fig. 4.11 is in very good agreement with the ﬁeld measurements when h/d50 < 50.
Problems
121
Computer Problem 4.1 Backwater behind a reservoir Consider steady ﬂow (q = 3.72 m2 /s) in the impervious rigid boundary channel sketched in Fig. CP.4.1.1. Assume CP.4.1.1 Deﬁnition sketch. that the channel width remains large and constant regardless of ﬂow depth and that f = 0.03. Determine the distribution of the following parameters along the 25km reach of the channel when the watersurface elevation at the dam is 10 m above the bed elevation: (a) ﬂow depth in meters, (b) mean ﬂow velocity in meters per second, and (c) bed shear stress in newtons per square meter.
5 Unsteady ﬂow in rivers
This chapter is concerned with unsteady ﬂow in rivers. The governing equations describing unsteady ﬂow in open channels include continuity in Section 5.1 and momentum in Section 5.2 for 1D ﬂow. Governing equations are followed by the concept of ﬂoodwave propagation (Section 5.3), looprating effects (Section 5.4), ﬂood routing (Section 5.5), and ﬂow and sedimentduration curves in Section 5.6.
5.1
River continuity equation
The continuity equations simply express conservation of mass. In most rivers, it is assumed that the mass density ρ is constant, and the continuity equations imply conservation of volume. A general formulation is presented in Subsection 5.1.1, followed by the commonly used twodimensional (2D) form in Subsection 5.1.2. A simpliﬁed 1D form is then given in Subsection 5.1.3.
5.1.1
General continuity formulation
The threedimensional (3D) form of the continuity equation is simply the integral over a control volume ∀ of the differential form [Eq. 2.16a]: ∂ρv y ∂ρv z ∂ρv x ∂ρ d∀ + + + d∀ = 0. (5.1) ∂x ∂y ∂z ∀ ∂t ∀ This volume integral of partial derivatives can be transformed into surface integrals owing to the divergence theorem, e.g., application to a vector F, ∂F ∂x d∀ = dA, (5.2) F ∂n ∀ ∂x A in which ∂ x/∂n is the cosine of the angle between the coordinate x and the normal vector n pointing outside the control volume. The integral form of the 122
River continuity equation
123
continuity equation simply states that the difference between inﬂow and outﬂow results in volumetric storage, as shown in Example 5.1.
Example 5.1 Application of the continuity equation. Consider the fairly straight river reach of length X c , constant width W, and ﬂow depth h, as sketched in Fig. E.5.1.1. The ﬁrst integral of (5.1) is zero in this case because the ﬂuid is incompressible and the control volume ∀ = W h X c is constant. The divergence theorem (Eq. 5.2) is applied to the second term: ∂x ∂y ∂z + ρv y + ρv z dA = 0. (E.5.1.1) ρv x ∂n ∂n ∂n A The values of ∂ x/∂n, ∂ y/∂n, and ∂z/∂n are the cosines of the angle between the vector normal to the surface n pointing outside the control volume and the Cartesian coordinates x, y, and z, respectively. Figure E.5.1.1 illustrates the direction cosines on the downstream face. In this example, the direction cosine ∂ x/∂n vanishes on all faces except the upstream (∂ x/∂n = −1) and downstream (∂ x/∂n = +1) cross sections. Similarly, the direction cosine ∂ y/∂n vanishes except on the left bank (∂ y/∂n = +1) and right river bank (∂ y/∂n = −1). The direction cosine ∂z/∂n vanishes except at the free surface (∂z/∂n = +1) and the riverbed (∂z/∂n = −1). Moreover, for an incompressible homogeneous suspension ρm1 = ρm2 = ρm with the vertical velocity, v z = dh/dt at the free surface, the inﬁltration rate through the banks v y = i, and v z = −i at the bed. The surface integral of the ﬁrst term in Eq. (E.5.1.1) yields the downstream ﬂux A2 V2 ∂ x/∂n (note that ∂ x/∂n = +1) and the upstream ﬂux −A1 V1 (note that V1 is positive in the x direction and ∂ x/∂n = −1 upstream).
Figure E.5.1.1. Deﬁnition sketch.
124
Unsteady ﬂow in rivers
The second term gives the inﬁltration through the channel banks 2i h X c (note that on the right bank the ﬂux is in the negative y direction and ∂ y/∂n = −1). The third term gives the ﬂux through the of freesurface area W X c dh/dt and the inﬁltration through the bed i W X c . The sum of all components results in A2 V2 − A1 V1 
outﬂow
 
inﬂow
∂h = 0. ∂t  storage 
+ 2i h X c + i W X c + W X c  

inﬁltration
(E.5.1.2)
In general, continuity simply formulates that the storage is the inﬂow minus the outﬂow. In the case of an impervious channel, i = 0, the integral form of the continuity equation reduces to ∂h = 0. ∂t  storage 
A2 V2 − A1 V1 + W X c 
outﬂow

inﬂow

(E.5.1.3)
This form is equivalent to 1D equation (5.8) written as −W ∂h A2 V2 − A1 V1 . = Xc ∂t
5.1.2
(E.5.1.4)
Twodimensional continuity for rivers
Depthintegrated formulations of the governing equations are useful in describing most river systems with typical variability in planform geometry such as changes in lakes, reservoirs, and estuaries. The system of coordinates typically sets the x axis in the main downstream direction at a small slope S¯ 0x = tan θ from the horizontal, a horizontal y axis toward the left bank, and the upward z axis deviating at a slope S¯ 0x from the vertical. The depthintegrated continuity relationship is obtained for homogeneous suspensions (constant ρ) from the integration of Eq. (2.16a) along the upward z axis. The surface integration is carried out over a control volume of ﬁxed grid dx and dy with bed elevation z b and watersurface elevation z w that vary in space, as sketched in Fig. 5.1. Given the ﬂow depth h = z w − z b , the depthintegrated velocity components Vx and Vy are Vx =
1 h
zw
v x dz, and zb
Vy =
1 h
zw
v y dz. zb
(5.3)
River continuity equation
125 The net volume ﬂux leaving the control volume in the x direction is ∂ (hVx ) dx − dyhVx dy hVx + ∂x = dx dy
∂ (hVx ) . ∂x
(5.4)
Repeating the process in the y direction gives dx dy ∂(hVy )/∂ y and, considering the inﬁltration ﬂux i b dx dy and the entering rainfall ﬂux i dx dy, the net volume change within the control volume is dx dy
∂hVy ∂hVx ∂h + dx dy + dx dy ∂t ∂x ∂y
+ (i b − i) dx dy = 0.
(5.5)
The depthintegrated form of the continuity equation is obtained after Eq. (5.5) is divided by dx dy: Figure 5.1. Depthintegrated control volume.
∂hVy ∂hVx ∂h + + + (i b − i) = 0. ∂t ∂x ∂y (5.6)
The rate of change in ﬂow depth, dh/dt, relates to the net downstream ﬂux ∂(hVx )/∂ x, the net lateral ﬂux ∂(hVy )/∂ y, the vertical inﬁltration rate i b deﬁned as positive outward, and the vertical rainfall intensity i deﬁned as positive inward. In this form, h = z w − z b , where both watersurface elevation z w and bed elevation z b can vary and Vx and Vy are depthintegrated velocities. All parameters, h, Vx , Vy , i b , and i, can vary in space x − y and time. Equation (5.6) is applicable to homogeneous suspensions (constant ρm ) and to the orthogonal system of coordinates in which the z axis is near vertical (small angle θ0 ).
5.1.3
Onedimensional continuity for rivers
The deﬁnition sketch in Fig. 5.2 describes a river reach with a top width W, crosssection area A, wetted perimeter P, hydraulic radius Rh = A/P, and mean ﬂow depth h = A/W . The total discharge Q is given from the product of the mean
126
Unsteady ﬂow in rivers
Figure 5.2. Continuity for a river reach.
ﬂow velocity V and area A; the unit discharge of the lateral ﬂow is ql . The rainfall intensity is i, and the inﬁltration rate through the wetted perimeter is i b . The net volumetric ﬂux leaving the control volume is (∂ Q/∂ x)dx + i b Pdx. The net volumetric ﬂux entering the control volume is ql dx + i W dx. The difference between entering and leaving volumetric ﬂuxes corresponds to volumetric storage ∂ Adx = ∂(W h)dx per unit time ∂t. After dividing by dx, we easily demonstrate that
∂Q ∂A + + i b P − i W − ql = 0, ∂t ∂x
(5.7)
where i b is the rate of inﬁltration through the wetted perimeter P, i is the rainfall intensity through the reachaveraged river width W , A is the reachaveraged crosssectional area, and ql is the unit discharge of lateral inﬂow. For an impervious channel (i b = 0) without rainfall (i = 0) and without lateral inﬂow (ql = 0), the 1D equation of continuity simply reduces to ∂A ∂Q + = 0. ∂x ∂t
(5.8)
This simple differential equation that expresses conservation of mass is widely used in the analysis of ﬂoodwave propagation.
5.2
River momentum equations
In this section, the momentum equations are presented for 2D ﬂows in Subsection 5.2.1, followed by a 1D formulation in Subsection 5.2.2. 5.2.1
Twodimensional momentum for rivers
For most rivers, depthintegrated formulations of the equation of motion are sufﬁciently accurate as long as the acceleration in the nearvertical z direction can be neglected. The resulting depthintegrated formulation is 2D in x and y
River momentum equations
127 along with a hydrostatic pressure distribution. For homogeneous suspensions (constant ρm ), the integration over depth is applied to a control volume, as shown in Fig. 5.3. It is assumed that the wind and the rainfall forces applied through the free surface are negligible. The bed shear stress τzx = τ0x and the pressure is hydrostatic such that the depthaveraged pressure p is given by p = 0.5 ρgh. The sum of forces in the x direction gives ρax hdxdy = ρghdxdy sin θ −
ρg ∂h 2 dxdy 2 ∂x
+
∂τ yx h dxdy ∂y
− τ0x dxdy. Figure 5.3. Stresses on a depthintegrated volume.
ax =
(5.9)
Dividing through by the mass ρhdxdy with small θ0 , such that sin θ0 ∼ = tan θ0 = S¯ 0x (and sin θ = S¯ 0x − ∂z b /∂ x), and using Eq. (2.14a), results in
∂ Vx ∂ Vx ∂ Vx ∂ Vx + Vx + Vy + Vz ∂t ∂x ∂y ∂z  local  
convective

∂τ yx ∂h τ0x ∂z b − g − + . ∂x ∂x ρh ρ∂ y  gravity  bed elevation  pressure  bed shear  side shear  = g S¯ 0x − g
(5.10)
It is assumed that the momentum correction factors are close to unity, i.e., that the ﬂow velocity is fairly uniform over the depth. It can be further assumed that Vz is very small and that the variability of riverbank shear stress is negligible compared with that of the bed shear stress. The following approximation for
128
Unsteady ﬂow in rivers
the depthintegrated equation of motion is ∂ Vx ∂ Vx ∂ Vx ∂h τ0x + Vx + Vy = g S¯ 0x + gS0x − g − . ∂t ∂x ∂y ∂x ρh
(5.11)
With a similar analysis applied in the y direction we obtain τ0y ∂ Vy ∂ Vy ∂ Vy ∂h + Vx + Vy = gS0y − g − , ∂t ∂x ∂y ∂y ρh
(5.12)
where S¯ 0x is the average downstream bed slope and the average lateral slope is S¯ 0y = 0. The local bed slopes are S0x = − (∂z b /∂ x) and S0y = − (∂z b /∂ y), respectively. These simpliﬁed equations are applicable to 2D rasterbased models of rivers with large width–depth ratios. The main difﬁculty lies in the evaluation of the bed and the side shearstress components. For ﬂow in bends, the relative magnitude of radial acceleration terms in cylindrical coordinates indicates that the centrifugal acceleration is counterbalanced by pressure gradient and radial shear stress, as suggested by Rozovskii (1957): 1 ∂τr Vθ2 = gSwr − , r ρ ∂z
(5.13)
where the local downstream velocity Vθ , the radial shear stress τr , and the radial watersurface slope Swr vary with the vertical elevation z and/or the radius of curvature r . 5.2.2
Onedimensional momentum for rivers
When the channel width remains fairly constant, the lateral velocity component Vy from Eq. (5.12) can be neglected. The velocity reduces to V = Vx , the bed slope to S = S0x , and the bed shear stress to τ0 = τ0x : ∂ Vx Vx ∂ Vx τ0x ∂h + = g S¯ 0x + gS0x − g − , ∂t ∂x ∂x ρm h
(5.14)
where S¯ 0x is the reachaveraged positive slope and S0x = −∂z b /∂ x. It is left to the reader in Exercise 5.1 to demonstrate with the continuity equation [Eq. (5.8)] that the following formulation reduces to Eq. (5.14) when A is constant and βm = 1: ∂βm QV τ0 P A∂h ∂Q + = g AS0 − g − . (5.15) ∂t ∂x ∂x ρ Assume a hydrostatic pressure distribution, p = ρgh, and assume that the bed shear stress τ0 in widerectangular channels (Rh ∼ = h and P W ) approximately equals τ0 = ρgh S f and that gx is a function of bed slope gx = gS0 ,
River momentum equations
129
where S0 = S¯ 0x + S0x . The 1D equation of motion [Eq. (5.14)] with a single bed slope S0 simply reduces to St. Venant equation (2.27) following dimensionless form after being rearranged and divided throughout by gravitational acceleration g and A: V ∂V 1 ∂V ∂h − − . Sf ∼ = S0 − ∂x g∂ x g ∂t (1) (2) (3) (4) (5)  kinematic    
diffusive
(5.16)

quasisteady dynamic
 
The terms of the dimensionless equation of motion are sketched in Fig. 5.4. They describe (1) the friction slope or the slope of the EGL, (2) the bed slope of the channel, (3) the pressure gradient or downstream change in ﬂow depth that is due to backwater, (4) the velocity head gradient or downstream change in velocity head that is due to backwater and/or changes in channel width, and (5) the localacceleration term for unsteady ﬂow. The full use of Eq. (5.16) is referred to as the dynamicwave approximation of the St. Venant equation. In many applications, however, the last term (5) is very small and can be neglected. For all steady ﬂows, as well as unsteady Figure 5.4. Slope deﬁnition sketch. ﬂows in which (5) is negligible, the quasisteady dynamicwave approximation includes the ﬁrst four terms of Eq. (5.16); this corresponds to the backwater equations for gradually varied ﬂows used in Chap. 4. In most rivers, the ﬂow is subcritical, Fr < 1, and the velocity head gradient (4) can be neglected before the pressure gradient (3). The diffusivewave approximation, S f = S0 − ∂h/∂ x, is commonly used in river mechanics. Finally, the kinematicwave approximation is obtained in which all but the ﬁrst two terms vanish. This is the case when ∂h/∂ x S0 . Example 5.2 illustrates how the continuity and the momentum equations can be combined to determine the celerity of surface waves.
130
Unsteady ﬂow in rivers
Example 5.2 Application to surface perturbations. Consider a solitary wave traveling in a frictionless channel without change of shape or velocity. For instance, a solitary wave can be produced by a sudden horizontal displacement of a vertical gate in a laboratory ﬂume. As sketched in Fig. E.5.2.1(a), the solitary wave travels to the right with celerity c in a stationary ﬂuid. An observer moving along the wave crest at a velocity equal to the celerity will perceive steady ﬂow [Fig. E.5.2.1(b)] in which the wave appears to stand still while the ﬂow moves at a velocity equal to c in magnitude. When continuity relationship (5.8) is Figure E.5.2.1. Propagation of a small applied to the steadyﬂow case, conperturbation. stant discharge implies that the relative velocity under the crest is ch/(h + h). It is interesting to note that in relative motion [Fig. E.5.2.1(b)], the relative velocity decreases with ﬂow depth. The equation of motion is then applied to the steady relative motion. When friction is neglected and a small slope is assumed, the energy equation between the normal section of the ﬂow and the section at the wave crest simply describes conservation of speciﬁc energy, or 2 c2 h c2 = h + h + . (E.5.2.1) h + 2g 2g h + h Solving for c, we obtain 2g (h + h)2 . c= 2h + h
(E.5.2.2)
This is commonly known as the Lagrangian celerity equation for 1D propagation of small waves in still water. For waves of moderate amplitude, the wave celerity increases with wave height.
5.3
River ﬂoodwaves
This section treats the subject of ﬂoodwave propagation in 1D channels. Unsteady ﬂow describes changes in ﬂow discharge Q with time t, and downstream
River ﬂoodwaves
131
distance x. Therefore the exact differential is dQ =
∂Qdt ∂Qdx + . ∂x ∂t
(5.17)
By deﬁnition, the celerity c = dx/dt deﬁnes the location where the ﬂow is steady, i.e., dQ = 0. The celerity c at which space–time changes take place is simply given by the solution of Eq. (5.17) for dx/dt, or −∂ Q dx ≡ ∂t . c≡ ∂Q dt ∂x
(5.18)
When considering conservation of mass in a onedimensional impervious channel, we obtain from Eq. (5.8) that (∂ Q/∂ x) = (−∂ A/∂t). Substituting this into Eq. (5.18) gives c=
∂Q . ∂A
(5.19)
This relationship for ﬂoodwave celerity is referred to as the Kleitz–Seddon law. It is interesting to point out that it is essentially the result of the conservation of mass with the restrictions pertaining to Eq. (5.8). For a widerectangular channel, the unit discharge q varies with depth h according to resistance relationships, such as q = V h = αh β . The equation of wave propagation and the celerity relationship in ﬂowing water then reduce to ∂h ∂h +c = 0, ∂t ∂x ∂q c= = βαh β−1 = βV. ∂h
(5.20) (5.21)
The ﬂoodwave celerity c is always faster than the ﬂow velocity when β > 1. For instance, β = 3 for laminar ﬂow and β = 1.5 for turbulent ﬂow with constant Darcy–Weisbach factor f (or Ch´ezy coefﬁcient C); see Table 3.3. It is of foremost importance to understand that the celerity of ﬂoodwaves increases with ﬂow depth. It implies that larger ﬂoodwaves (larger ﬂow depth) propagate faster than small ﬂoodwaves. This causes nonlinearity in the downstream propagation of ﬂoodwaves, and linear techniques based on superposition, such as the unit hydrograph, fail to adequately simulate ﬂoodwave propagation in channels. It also indicates that the method of isochrons used in hydrology is not applicable to both small and large ﬂoodwaves. The following treatment of equation of motion (5.16) considers the resistance and the continuity relationships. The resistance relationship V = αh β−1 yields the following derivatives ∂ V = (β−1) (V / h)∂h. The reader can demonstrate
132
Unsteady ﬂow in rivers
in Exercise 5.2 that equation of motion (5.16) can be rewritten as S f = S0 − [1 + (β − 1)Fr2 ]
1 ∂V ∂h − . ∂x g ∂t
(5.22)
Note that for steady ﬂow this equation is slightly different from Eq. (4.28) because resistance to ﬂow is considered and q is not constant. A complete list of conversion factors considering q = V h = αh β , ∂h/∂t = −∂q/∂ x, and c=βV is presented in Table 5.1. Continuity relationship (5.8) then yields ∂h/∂t = −∂q/∂ x = −V ∂h/∂ x − h∂ V /∂ x, which can be combined with ∂ V = [(β − 1)V / h]∂h to reduce Eq. (5.22) to S f = S0 − [1 − (β − 1)2 Fr2 ] 
ﬂoodwave diffusivity
∂h , ∂x
(5.23a)

and, for Manning’s equation, 4 Q 0.2 S 0.9 ∂h S f = S0 − 1 − . 9 gn 1.8 W 0.2 ∂ x
(5.23b)
The ﬂoodwavediffusivity term depends on the value of β and Fr. For instance, Manning’s equation is applicable in most rivers (β = 5/3), and, for Fr < 1.5, the ﬂoodwavediffusivity term is positive. Conversely, the term is negative when Fr > 1.5. In rivers with low Froude number, the ﬂoodwave diffusivity becomes close to unity. It is also quite important to examine the ratio of the second to the ﬁrst term of the St. Venant equation. From Eq. (5.22), the ratio (−∂h/S0 ∂ x) is a measure of the ﬂoodwave attenuation at low values of the Froude number. From Table 5.1, this diffusivity ratio can be written (see Exercise 5.5) as 1 ∂q 1 W 1−2/β ∂ Q −1 ∂h = = . S0 ∂ x S0 c2 ∂t β 2 S0 α 2/β Q 2−2/β ∂t
(5.24)
1/2
In the case of the Manning equation, α = (S0 /n), β = 5/3 , the diffusivity ratio reduces to 9 n 1.2 ∂Q −1 ∂h = . 1.6 0.2 0.8 S0 ∂ x 25 W S0 Q ∂t
(5.25)
The reader will note that rapid changes in discharge (large ∂ Q/∂t) increase ﬂoodwave diffusivity. For a given ﬂoodwave, a given Q, and ∂ Q/∂t, diffusivity increases with Manning coefﬁcient n and decreases with channel slope. It is
∂V ∂x
−β 2 q ∂ V (β − 1) ∂ x
βq ∂V (β − 1)V ∂ x
−βV
∂V ∂x
−βh ∂ V (β − 1) ∂ x
h ∂V (β − 1)V ∂ x
∂V ∂x
βq ∂V (β − 1)V ∂t
∂V −q (β − 1)V 2 ∂t
∂V ∂t
−1 ∂ V βV ∂t
h ∂V (β − 1)V ∂t
∂V −h β(β − 1)V 2 ∂t
∂V ∂t
Note: Calculated from q = hV = αh β , (∂h/∂t) = (−∂q/∂ x), and c = (dx/dt) = βV .
βq ∂h h ∂t
∂q −β 2 q 2 ∂h = ∂t h2 ∂ x
V ∂h h ∂t
−∂h ∂t
(β − 1)
−(β − 1) ∂h βh ∂t
∂h ∂t
−1 ∂h βV ∂t
∂h ∂t
∂q βq ∂h = ∂x h ∂x
∂V V 2 ∂h = −β(β − 1) ∂t h ∂x
∂V V ∂h = (β − 1) ∂x h ∂x
∂h ∂h = −βV ∂t ∂x
∂h ∂h = ∂x ∂x
∂h ∂x
Table 5.1. Conversion factors for ﬂoodwave propagation
−βV
∂q ∂x
−(β − 1)
∂q ∂x
V 2 ∂q q ∂x
(β − 1) V ∂q β q ∂x
−∂q ∂x
h ∂q βq ∂ x
∂q ∂x
∂q ∂t
−1 ∂q βV ∂t
(β − 1) V ∂q β q ∂t
−(β − 1) ∂q β2q ∂t
h ∂q βq ∂t
−h 2 ∂q β 2 q 2 ∂t
∂q ∂t
134
Unsteady ﬂow in rivers
therefore clear that, in many cases, channel straightening (higher S0 ) and channel lining (lower n) decrease ﬂoodwave diffusivity of natural channels. The following relationships are important in determining the ﬂow velocity V and the ﬂoodwave celerity c, assuming that the Manning equation is applicable: ∂h 1/2 1 2/3 S0 − [1 − (β − 1)2 Fr2 ] , (5.26) V = Rh n ∂x 1/2 β 2/3 2 2 ∂h . c = Rh S0 − [1 − (β − 1) Fr ] n ∂x
(5.27)
The ﬂoodwavediffusivity term plays a dominant role in the alteration of ﬂoodwaves. This process is illustrated with the following example. Consider a triangular ﬂoodwave propagating in a widerectangular channel, as sketched in Fig. 5.5. Two points at equal ﬂow depth are located upstream (A) and downstream (B) of the wave crest. The reference discharge Q r corresponds to the kinematicwave approximation, where, S0 = S f , thus assuming Manning’s 2/3 1/2 equation Q r = (A/n)Rh S0 . Three types of ﬂoodwaves are recognized: (1) a dynamic wave when Fr > 1/(β − 1); (2) a kinematic wave when Fr = 1/(β − 1), and (3) a diffusive wave when Fr < 1/(β − 1). Dynamic waves in steep channels thus tend to form pulsating ﬂows or surges, also called roll waves. In laminar ﬂow (β = 3), roll waves can theoretically form when Fr > 0.5. Measurements in sheet ﬂows are possible for subcritical ﬂow at Fr > 0.7 (Julien and Hartley, 1985). In turbulent ﬂows (β = 5/3), roll waves develop on very steep smooth channels under supercritical ﬂows (Fr > 1.5). Roll waves and supercritical ﬂows should be avoided when open channels are being designed because of surface instabilities and cross waves incurred by any perturbation of the bank and/or the bed. This can best be achieved by an increase in boundary roughness to the extent that the ﬂow will remain subcritical. Kinematic waves are obtained when Fr = 1/(β −1). As sketched in Fig. 5.5(b), the bed and the friction slopes are identical. This implies that wave celerity and discharge increase solely with ﬂow depth. The wave celerities and discharges at points (A) and (B) are therefore identical, and the distance separating (A) and (B) remains constant as the ﬂoodwave travels downstream. The celerity of the wave crest at (C) is nevertheless larger than that of (A) or (B), and the crest gradually moves to form a welldeﬁned wave front, also referred to as kinematic shock. In most rivers, the ﬂow is subcritical and ﬂood routing is adequately described by the diffusivewave approximation of the St. Venant equation. In such cases, the wave celerity and discharge do not vary solely with ﬂow depth but also depend on the gradient of ﬂow depth in the downstream direction. For instance,
River ﬂoodwaves
135
Figure 5.5. Dynamic, kinematic, and diffusive waves.
as sketched on Fig. 5.5(c), the initially triangular ﬂoodwave shows ∂h/∂ x < 0 on the downstream side and ∂h/∂ x > 0 on the upstream side of the crest. This implies that the two points, (A) and (B), at identical ﬂow depth will propagate at different celerities where the downstream point (B) moves faster downstream than point (A). This will result in ﬂoodwave attenuation, or stretching of the distance separating (A) and (B), as the ﬂood propagates in the downstream direction. The ﬂoodwave will elongate and the peak discharge decrease as the
136
Unsteady ﬂow in rivers
wave travels downstream. Floodwave attenuation is most effective when the Froude number is low, and ∂∂hx is large compared to S0 . 5.4
Looprating curves
Solutions to unsteady ﬂow problems in most rivers require the use of the diffusivewave formulation of the St. Venant equation. Based on the frictionslope relationship, total discharge Q in a river can be written as Q=
1/2 ∂h A 2/3 Rh S0 − . [1 − (β − 1)2 Fr2 ] n ∂x
(5.28)
At a given ﬂow depth, the friction slope S f at low Froude numbers is slightly larger than the bed slope during the rising stage (∂h/∂ x) < 0. Unless the kinematicwave approximation is applicable, river ﬂoods will induce loops in the rating curve relationship because S f = S0 during the rising and the falling levels of ﬂood hydrographs. Typically, at a given ﬂow depth, the rising limb of a ﬂoodwave will display a higher discharge than the falling limb. A counterclockwise loop is obtained for stage–discharge relationship (h versus Q) of most channels. The example of the Mississippi River Figure 5.6. Looprating curve of the Mississippi River at Vicksburg (after is shown in Fig. 5.6. The maximum Combs, 1994). discharge is reached before the maximum ﬂow depth. The looprating curve also has an impact on shear stress and sedimenttransport calculations. Bed shearstress τ0 calculations based on the friction slope are given by ∂h . (5.29) τ0 = γ Rh S f = γ Rh S0 − [1 − (β − 1)2 Fr2 ] ∂x Shearstress calculations deviate from kinematicwave considerations. It is particularly shown that at a given ﬂow depth, shear stress is larger on the rising limb than on the falling limb. Bedload sediment transport also increases with bed shear stress, thus resulting in larger sediment transport during the rising limb at a given ﬂow depth than for the falling limb. For instance, the Meyer–Peter Muller formula shows that qs ∼ (τ0 − τc )1.5 .
Looprating curves
137
Figure 5.7. Loop of the sedimentrating curves: (a) Bell River and (b) Yampa River.
Sediment transport is described by the sedimentrating curve (sediment discharge Q s versus water discharge Q). Although both Q s and Q increase during the rising stages, Q s increases faster than Q. The sedimentrating curve is characterized by a clockwise looprating effect, as shown in Fig. 5.7. This effect is predominant for streams transporting mostly bedload. Streams carrying predominantly washload may not respond to changes in bed shear stress. In summary, Fig. 5.8 sketches the effects of the dynamic, kinematic, and diffusive waves on discharge and bedload sediment transport. The dynamic wave usually found in upland areas tends to cause riverbed degradation as the ﬂoodwave ampliﬁes downstream. Conversely, the diffusive wave typically found in most subcritical rivers causes ﬂoodwave attenuation and riverbed aggradation. Example 5.3 Application of the St. Venant equation. An observer measures the ﬂow depth in a 50m widerectangular channel inclined at S0 = 3 × 10−3 with Manning coefﬁcient n = 0.03. Initially, the ﬂow depth is 1.0 m and the water level rises at a rate of 1 m/h. Calculate (1) the initial discharge at a distance of 1 km downstream, (2) calculate the relative magnitude of the acceleration terms in the St. Venant equation, and (3) determine whether the ﬂoodwave attenuates as it propagates downstream.
138
Unsteady ﬂow in rivers
Figure 5.8. Dynamic, kinematic, and diffusive waves.
Step 1: The initial upstream discharge is W 5/3 1/2 50 5/3 h u S0 = 1 3 × 10−3 = 91.3 m3 /s. Qu = n 0.03 From continuity equation (5.8), Q =
1 −W hX = −50 × × 1,000 m3 = −13.9 m3 /s. t 3,600 s
The downstream discharge Q d is 91.3 − 13.9 = 77.4 m3 /s. Step 2: The downstream ﬂow depth is 0.6 n Q 3/5 0.03 × 77.4 = = 0.906 m. hd = √ W S 1/2 50 × 3 × 10−3
River ﬂood routing
139
The upstream and the downstream velocities are, respectively, Vu =
Qu 91.3 Qd 77.4 = 1.826 m/s, Vd = = 1.708 m/s. = = W hu 50 × 1 W hd 50 × 0.906
Over a distance of 1 km, the ﬂow depth changes by 0.094 m and V = 0.118 m/s. The terms of the St. Venant equation (Eq. 5.16) are (2) S0 = 0.003 = 3 × 10−3 , (3)
−0.094 m ∂h = = −9.4 × 10−5 , ∂x 1,000 m
(4)
1.77 (−0.118) V ∂V = = −2.13 × 10−5 . g ∂x 9.81 1,000
Given the ﬂoodwave celerity c = βV = 5/3 × 1.77 m/s = 2.95 m/s, the 1km distance is traveled in 390 s. (5)
s2 0.118 m 1 ∂V = = 3.53 × 10−5 . g ∂t 9.81 m 340 s2
The friction slope is S f = S0 − (∂h/∂ x) − (V ∂ V /g∂ x) − (1/g)(∂ V /∂t) = 3.08 × 10−3 . Note that the Froude number, Vu 1.826 Fr = √ =√ = 0.58 gh u 9.81 is less than unity and term (4) is smaller than term (3). Step 3: We can use Eq. (5.22) to calculate the ﬂoodwave diffusivity term as 1 − (β − 1)2 Fr2 = 1 − 49 (0.58)2 = 0.85. We determine that the ﬂoodwave is diffusive and attenuates as it propagates downstream.
5.5
River ﬂood routing
We can calculate the propagation of ﬂoodwaves in rivers by solving the equations of conservation of mass and momentum. The basic approach for the numerical simulation of ﬂoodwave routing stems from the concept of celerity, which is sketched for a single widerectangular channel in Fig. 5.9. Any change in ﬂow depth corresponds to a change in discharge, and the perturbation propagates at a celerity c. Because c = dx/dt, numerical models should attempt to set x and t such that c ∼ = x/t. The stability of several numerical
140
Unsteady ﬂow in rivers
Figure 5.9. Sketch of stable and unstable numerical schemes.
schemes depends on the Courant–Friedrich–Levy condition, which can be written as ct , x
C fl =
(5.30)
where C f l is the Courant number. When C f l ≤ 1, the numerical scheme is stable; the scheme is usually unstable when C f l > 1 m, as sketched in Fig. 5.9. Because the ﬂoodwave celerity is maximum near the peak discharge of the ﬂoodwave (c = βV ), the time and space increments t and x, respectively, can be related as t =
x C f l C f l x = β−1 , βV βα 1/β q β
(5.31)
and, for Manning’s equation in S.I. units, we easily demonstrate in Exercise 5.6 that, for stability, t
1 for a single time step. The Froude number is Fr =
(3,000)0.1 (1.5 × 10−4 )0.45 Q 0.1 S 0.45 = = 0.34. g 0.5 n 0.9 W 0.1 (9.81)0.5 (0.015)0.9 (260)0.1
The natural ﬂoodwave should attenuate because Fr < 1.5. The Froude number is sufﬁciently small and the diffusivity term from Eq. (5.23), 1 − (β − 1)2 Fr2 = 1 − 49 (0.34)2 = 0.95, remains close to unity. The diffusivewave approximation is therefore recommended over the fulldynamicwave approximation. The maximum diffusivity ratio from Eq. (5.25) is 9 n 1.2 Q max −1 ∂h = 1.6 0.2 0.8 S0 ∂ x 25 W S0 Q t =
(0.015)1.2 1,000 9 = 0.019. 25 (260)0.2 (1.5 × 10−4 )1.6 (3,000)0.8 86,400
The diffusivity is so small that the solution should be very close to kinematic routing. The initial watersurface elevation at Q = 1,000 m3 /s is h = 2.534 m. The equation of continuity for an impervious channel without rainfall and lateral inﬂow is rewritten as h i =
(Q in − Q out ) t . W x
(E.5.4.1)
142
Unsteady ﬂow in rivers
Table E.5.4.1. Floodwave routing calculation
Days (i)
Q in (m3 /s)
hu (m)
h i (m)
hd a (m)
Kinematic Q out (m3 /s)
1 2 3 4 5 6 7
1,000 2,000 3,000 2,500 2,000 1,500 1,000
2.53 3.84 4.90 4.39 3.84 3.23 2.53
0 1.02 1.28 −0.43 −0.51 −0.54 −0.59
2.53 2.53 3.56 4.82 4.39 3.88 3.33
1,000 1,000 1,759 2,925 2,499 2,033 1,579
C fl
Diffusive Q out b (m3 /s)
hd c (m)
0.67 0.67 0.84 1.03 0.97 0.89 0.81
1,000 1,015 1,786 2,912 2,485 2,021 1,567
2.53 2.55 3.65 4.71 4.42 3.90 3.33
= h di + h i 1/2 W ∂h 1/2 ∼ h i . h d 5/3 S0 1/2 1 − = 212.3 h d 5/3 1 + out = n S0 ∂ x S0 C f l x c Floodstage calculated with x = 32.5 km and t = 2 h.
ah
di+1
bQ
On a daily basis (t = 86,400 s), the discharge Q out at the downstream station is then updated from the new downstream ﬂow depth h d i+1 = h d i + h i , and the downstream discharge is calculated with the diffusivewave approximation in S f = S0 − ∂h/∂ x. The procedure is illustrated in Table E.5.4.1 and the results are shown in Fig. E.5.4.1. For the sake of comFigure E.5.4.1. Calculated ﬂood parison, the ﬂoodstage calculated on a ﬁner routing. grid, x = 32.5 km and t = 2 h, is shown in the last column of Table E.5.4.1. The results are quite similar to those of the diffusivewave formulation.
5.6
River ﬂow and sedimentduration curves
Duration curves describe the percentage of time that a certain water or sediment discharge is exceeded. The curves are usually based on daily records and are useful in estimating how many days per year an event will be exceeded. Let the uppercase letter X denote a random variable, and let the lowercase letter x denote a possible value of X . For a random variable X , its cumulative distribution function (cdf ), denoted as F(x), is the probability P that the random variable X will be less than or equal to x: F(x) = P(X ≤ x).
(5.34)
River ﬂow and sedimentduration curves
143
F(x) is also called the nonexceedance probability for the value x. The exceedance probability E(x) = 1 − F(x). The probability density function (pdf), p(x), is the derivative of the cdf: p(x) =
dF(x) . dx
(5.35)
Exceedance probability transforms are useful in the analysis of duration curves. Properties of an exponential pdf (epdf) of variable x are such that after variable ¯ is deﬁned as ¯ = x/x¯ . x is divided by the mean value x¯ , a reduced variable ¯ ¯ The properties of the epdf p() and the exceedance probability E(), deﬁned ¯ d, ¯ are ¯ = ¯∞ p() as E() ¯ = p() ¯ = e− . E() ¯
(5.36)
It is interesting to note that both the epdf and the exceedance probability are identical. The purpose of the transform is to determine whether the pdf of a runoff ¯ after the following or sediment variable x reduces to an epdf of variable transform: ¯ = ax b ,
(5.36a)
where a and b are, respectively, the transform coefﬁcient and exponent, hereby referred to as transform parameters. When successful, this transforms the unknown exceedance probability of variable x into a simple epdf of the reduced ¯ variable . The inverse transform is simply deﬁned from Eq. (5.36a) as b1 1 ¯ bˆ , ¯ b1 = aˆ x = a
(5.36b)
where aˆ and bˆ are the inverse transform parameters simply calculated from a ˆ and b as aˆ = (1/a)1/b and bˆ = (1/b). Useful conversions are a = aˆ −1/b = aˆ −b ˆ The inverse transform exponent bˆ is important in determining and b = (1/b). the degree of nonlinearity of the variable x and can be related to deterministic relationships. An example of range of values of bˆ for various processes is summarized in Table 5.2. In increasing order of nonlinearity, we ﬁnd point rainfall, surface runoff, chemical and sediment concentration, and chemical and sediment yield. The properties of the transform are such that once the transform parameters a and b are known, the exceedance probability of variable x is calculated directly
144
Unsteady ﬂow in rivers Table 5.2. Typical values of bˆ for different variables Process
bˆ
Point rainfall Upland snowmelt Upland chemical concentration Upland runoff River ﬂow discharge River sediment concentration Upland chemical yield Upland sediment yield River sediment discharge
0.85–1.28 1.15–1.37 0.46–1.40 1.46–1.62 1.25–1.77 1.13–2.14 1.48–2.33 2.20–2.35 1.70–2.84
from Eqs. (5.35) and (5.36a). The pdf p(x) is p(x) = abx
b−1 −ax b
e
¯ b−1 b −¯ = ab e . a
(5.37)
¯ ¯ and E(x) = E(). ¯ We can also demonstrate that p(x)d x = p()d The procedures to evaluate the parameters of this distribution are presented in Subsection 5.6.1, followed by practical considerations in Subsection 5.6.2.
5.6.1
Parameter evaluation
Two procedures are examined for the transform parameter evaluation: (1) a graphical method and (2) the method of moments. Graphical method. The graphical method capitalizes on the properties of the epdf and the exceedance probability in Eq. (5.36). The natural logarithm of Eq. (5.36) is combined with Eq. (5.36a) to give ¯ = ¯ = ax b . −ln E()
(5.38)
The transform parameters a and b are evaluated graphically after the natural logarithm of Eq. (5.38) is taken in the form ¯ = ln ¯ = ln a + b ln x, = ln[−ln E()]
(5.39)
where the shorthand exceedance probability designates the double logarithm of the exceedance probability. Of course, this transform requires that all values of x be positive.
River ﬂow and sedimentduration curves
145
A straight line is ﬁtted on the graphical presentation of as a function of ln x to yield estimates for a and b. It becomes clear from the properties of Eq. (5.39) that ln a corresponds to the value of when x = 1. Likewise, if the points on the graph vs. ln x assemble on a straight line, the slope of the line gives directly the transform exponent b. The graphical method is subjective but provides the user a qualitative appreciation of the goodness of ﬁt. Ideally, all points should plot on a straight line for the power transform in Eq. (5.38) to be exactly applicable. Method of moments. Parametric estimation from the method of moments takes advantage of the information contained in the ﬁrst and the second moments of the sample. The transform parameters a and b can be evaluated after being equated with the ﬁrst and the second moments of the transformed variables. Speciﬁcally, the ﬁrst moment, or mean value, of the sample x¯ is equated 1/b ¯ and to the ﬁrst moment M1 of the transformed distribution, given x = /a ¯ : ¯ p(x) dx = p()d b1 ∞ 1 ¯ ¯ b1 e−¯ d M1 = x p (x) dx = a 0 0 b1 1 1 ˆ bˆ = x¯ . 1+ = = a! a b
∞
(5.40)
Accordingly, the expected value x¯ can be simply evaluated from a simple gamma ˆ where ! designates ˆ b, function of the inverse transform parameters as x¯ = a! ˆ the factorial function of the argument b. Likewise, the second moment of the sample x 2 is equated to the second moment M2 of the transformed distribution: b2 ∞ ∞ 1 ¯ ¯ b2 e−¯ d x 2 p(x) dx = M2 = a 0 0 b2 2 1 ˆ = x 2. = 1+ (5.41) = aˆ 2 !(2b) a b The evaluation of the transform parameters a and b follows after the transform coefﬁcient a is eliminated from the ratio of Eq. (5.41) to the square of Eq. (5.40); thus 1 + b2 x2 = 2. (5.42) 2 x¯ 1+ 1 b
From the calculated values of x¯ and x 2 of the sample, the value of b on the lefthand side of Eq. (5.42) can best be evaluated numerically. For instance, an interpolation procedure is based on the numerical values given in Table 5.3
146
Unsteady ﬂow in rivers Table 5.3. Useful values of the transform parameter b b 0.1 0.15 0.2 0.25 0.3
x2 / x2
(b)
!(b)
184,756 2,213 252 70 30.2
9.51 6.22 4.59 3.62 2.99
0.951 0.933 0.918 0.906 0.897
0.35 0.4 0.45 0.5 0.55
16.77 10.86 7.79 6 4.861
2.54 2.21 1.97 1.77 1.61
0.891 0.887 0.886 0.886 0.888
0.6 0.65 0.7 0.75 0.8
4.090 3.543 3.138 2.830 2.588
1.49 1.38 1.30 1.22 1.16
0.893 0.900 0.908 0.919 0.931
0.85 0.9 0.95 1 1.05
2.395 2.238 2.108 2 1.907
1.11 1.07 1.02 1.00 0.973
0.945 0.962 0.980 1.00 1.02
1.1 1.2 1.3 1.4 1.5
1.828 1.700 1.601 1.523 1.460
0.951 0.918 0.897 0.887 0.886
1.046 1.10 1.17 1.24 1.33
1.6 1.7 1.8 1.9 2
1.409 1.366 1.330 1.299 1.273
0.893 0.909 0.931 0.962 1.00
1.43 1.54 1.68 1.83 2.0
2.5 3 3.5 4 4.5
1.183 1.132 1.100 1.078 1.063
1.33 2.0 3.32 6.0 11.6
3.32 6.0 11.6 24.0 52.3
proves to be sufﬁciently accurate. The value of the transform coefﬁcient a then ˆ bˆ = x¯ in Eq. (5.40) as follows from a! a =
b 1 + b1 . x¯
(5.43)
River ﬂow and sedimentduration curves
147
Parameter estimation from the method of moments is direct and not subjective but lacks the visual information inherent in the graphical method. The evaluation of the transform parameters with both the graphical method and the method of moments is illustrated in Example E.5.5.
Example 5.5 Application of the duration curve procedure. Consider the following sample of an unknown variable: x¯ = 4.5, 1.0, 7.0, 2.0, 9.0, 0.5, 6.0, 11.0, 3.5. The ﬁrst step consists of ranking the n = 9 numbers in decreasing order of x, as shown in the second column of Table E.5.5.1; the values of x are squared in the third column. The second step consists of calculating the exceedance probability by use of the Weibull plotting position. Accordingly, the numbers in decreasing order are ranked from 1 to n as shown in the fourth column, 1 being the largest and n being the smallest number. After the rank is divided by 1 + n, the plotting position corresponds to the exceedance probability ¯ or E(x) in the ﬁfth column. The values of ln x and = ln [−ln E()] ¯ E() are tabulated in the sixth and the seventh columns, respectively, for the plot shown in Fig. E.5.5.1. Graphically, the parameter estimation gives a ≈ 0.1 and b ≈ 1.3. The line is usually ﬁt through the higher values of ln x because, for sedimenttransport parameters, the large values of x are usually those contributing to most of the sediment load. With the method of moments, the average value, x¯ = 4.94, is calculated at the bottom of the second column of Table E.5.5.1. The average value of x 2 = 36.1 is compiled at the bottom of the third column. From Table E.5.5.1, the ratio (x 2 /x¯ 2 ) = [36.1/(4.94)2 ] = 1.479 in the second column corresponds to a value of b ≈ 1.45 from interpolation with the values given in the ﬁrst column of Table 5.3. The value of a ≈ 0.085 is thereafter calculated from Eq. (5.43). Table E.5.5.1. Example of transformation procedure Sample x˜ 4.5 1.0 7.0 2.0 9.0 0.5 6.0 11.0 3.5 Average:
Ranked x 11.0 9.0 7.0 6.0 4.5 3.5 2.0 1.0 0.5 x¯ = 4.94
x2 121.00 81.00 49.00 36.00 20.25 12.25 4.00 1.00 0.25 x 2 = 36.1
Rank
E(x)
1 2 3 4 5 6 7 8 9
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
ln (x) 2.3979 2.1972 1.9459 1.7918 1.5041 1.2528 0.6931 0.0000 −0.6931
ln [− ln E(x)] 0.8340 0.4759 0.1856 −0.0874 −0.3665 −0.6717 −1.0309 −1.4999 −2.2504
148
Unsteady ﬂow in rivers
Figure E.5.5.1. Flow duration example.
The gamma function can also be approximated by Stirling’s asymptotic series as √ 1 139 1 + − · · · , !(x) = (x + 1) = 2π x x x e−x 1 + 12x 288x 2 51,840x 3 (E.5.5.1) which is readily available on most scientiﬁc calculators.
5.6.2
Practical considerations
The transforms provide useful information in the analysis of exceedance probability and ﬂow duration curves. Indeed, the exceedance probability E(x) of a variable x can be directly calculated from E(x) = e−ax , b
(5.44)
given the transform parameters. For instance, the ﬂow discharge in the Chaudi`ere river (x = Q) is given by φ = 0.048 Q 0.66 , or a = 0.048 and b = 0.66. The exceedance probability E(1,000 m3 /s) of a daily discharge Q = 1,000 m3 /s is calculated from E(1,000) ∼ = e−0.048×(1,000)0.66 = 0.01. The transforms also enable the user to estimate what value of a parameter will be exceeded a certain fraction of the time directly from 1/b 1 ˆ = aˆ [− ln E(x)]b . (5.45) x = [− ln E(x)]1/b a For instance, we calculate the daily sediment discharge of the Colorado River at Lee’s Ferry that is exceeded 1% of the time, or 3.65 days a year, from Eq. (5.45)
River ﬂow and sedimentduration curves
149
after solving for x = Q s . For the Colorado River, aˆ = 120,230 and bˆ = 1.7, a daily sediment discharge exceeded 1% of the time, E(Q s ) = 0.01, is simply given by Q s = 120,230(− ln 0.01)1.7 = 1.6 × 106 tons/day. Once the magnitude x1 and exceedance probability E 1 of an event is known one can determine the unknown magnitude x2 of another event of exceedance ˆ At probability E 2 simply as a function of the inverse transform exponent b. ˆ a given b we deﬁne ξ = E 2 /E 1 to obtain the unknown η = x2 /x1 . From b b Eq. (5.44), we obtain E 1 = e−ax1 and ξ E 1 = e−a(ηx1 ) to be solved for ξ as a function of η and b as b b (η ξ = e−ax1 (η −1) = E 1
b
−1)
,
ˆ or conversely for η as a function of ξ and b, ˆ ˆ ln E 2 b ln ξ b = , η = 1+ ln E 1 ln E 1
(5.46)
(5.47)
which shows that at any given value of ξ and E 1 , the magnitude of η increases ˆ with the exponent b. As a ﬁnal result, the transforms enable the user to calculate the magnitude of infrequent events from the mean value of a variable and the inverse transform ˆ From Equations 5.40 and 5.45, we demonstrate that the exceedance parameter b. probability of the mean value is only a function of bˆ as 1/bˆ . (5.48) −ln E(x¯ ) = !bˆ The value of x that has an exceedance probability E(x) is then directly calculated from x¯ and bˆ as ˆ ˆ ln E(x) b [− ln E(x)]b x¯ = x¯ . (5.49) x= ln E(x¯ ) !bˆ Values of the multiplication coefﬁcient in the braces of Eq. (5.49) are given in Table 5.4. One of the practical applications of this method is the determination of the twoyear ﬂood, which usually corresponds to the dominant discharge of alluvial rivers. Extreme rainfall events and the resulting ﬂoods can claim thousands of lives and cause billions of dollars in damage. Floodplain management and the design of ﬂoodcontrol works, reservoirs, bridges, and other investigations need to reﬂect the likelihood or probability of such events. Figure 5.10 shows extreme speciﬁcdischarge conditions as functions of drainage area. A frequency distribution is precisely determined when a sufﬁciently long record is available, e.g., ﬂood discharges, rainfall precipitation, or pollutant loadings. Frequency analyses are required when available data from past
150
Unsteady ﬂow in rivers
Table 5.4. Multiplication factor k for infrequent events x = kx Inverse transform coefﬁcient bˆ rainfall Exceeded
E(x)
5% of the time 1% of the time 1 day per year 1 day per 1.5 years 1 day per 2 years 1 day per 5 years 1 day per 10 years
0.05 0.01 2.738 × 10−3 1.825 × 10−3 1.37 × 10−3 5.47 × 10−4 2.73 × 10−4
runoff
sediment
1
1.5
2
2.5
3
3.0 4.6 5.9 6.3 6.6 7.51 8.2
3.9 7.4 10.7 11.9 12.7 15.5 17.6
4.49 10.6 17.4 19.9 21.7 28.2 33.6
4.68 13.7 25.5 30.1 33.6 46.6 58.0
4.48 16.3 34.2 41.8 47.8 70.6 92
Figure 5.10. Speciﬁc discharge vs. drainage area (modiﬁed after Creager et al., 1945).
measurements are insufﬁcient to deﬁne precisely the risk of large ﬂoods, extreme rainstorms, and pollutant loadings. Engineers typically have to work with a sample of 10 to 100 observations to estimate events with exceedance probabilities of 1 in 100, 1 in 1,000, and even the ﬂood ﬂows for spillway design exceeded with a chance of 1 in 10,000. The return period (sometimes called the recurrence interval) is often speciﬁed rather than the exceedance probability. A return period T can be understood as follows: In a ﬁxed Tyear period the expected number of exceedances of the Tyear event is exactly one if the distribution of ﬂoods does not change over
River ﬂow and sedimentduration curves
151
that period. Thus an average of one ﬂood greater than the Tyear ﬂood level occurs in a Tyear period. A return period has been incorrectly understood to mean that one and only one Tyear event can occur every T years. Actually, the probability of the Tyear ﬂood’s being exceeded is 1/T every year; thus the 1% exceedance event can be described as a value with a 1 in 100 chance of being exceeded each year.
Case Study 5.1 Daily sediment discharge of the Colorado River, United States. The duration curve of 5 years of daily sedimentdischarge measurements of the Colorado River at Taylor’s Ferry before the construction of the dams is shown in Fig. CS.5.1.1. The magnitude of infrequent events can be esˆ timated from the mean values x¯ and b. For instance, given the mean daily sediment discharge of 185,700 tons/day in the Colorado River and bˆ = 1.7, estimate ¯ s) ˆ and E( Q (1) the parameters a, b, a, and (2) the magnitude of the daily sediment discharge x2 exceeded 1 day per Figure CS.5.1.1. Sediment discharge year (P2 = 0.00274). of the Colorado River (after Julien, First, b = 1/bˆ = 0.588, !(1.7) = 1.54 1996). from Table 5.3, aˆ = x¯ /1.54 = 120,230 from Eq. (5.40), and a = aˆ −b = 1.028 × 10−3 from Eq. (5.36b). The exceedance probability of the mean daily sediment discharge of the Colorado River, ¯ This gives Q¯ s = 185,700 tons/day, is calculated from Eq. (5.44) with X = x. −3 0.588 E 1 = E( Q¯ s ) = e−1.028×10 ×185,700 = 0.275. Second, the daily sediment discharge exceeded 1 day/yr is calculated from Eq. (5.49) as Q s = [(−ln 0.00274)1.7/1.54]185,700 = 2.46 million tons per day. As a last practical example, estimate the daily sediment discharge with a period of return of 10 yr from this 5yr record. Using Eq. (5.49), we obtain Q s10 yr = [(− ln 2.74 × 10−4 )1.7/1.54] 185,700 = 4.31 million tons per day, which is 23 times the mean daily sediment discharge. Case Study 5.2 Fluvial data set for the Rhine River, The Netherlands. This case study presents a summary of the 1998 Rhine River ﬂood data in Table CS.5.2.1, followed with stage measurements along the RhineWaal River in The Netherlands in Table CS.5.2.2. Selected particlesize distributions of the bed material are presented in Table CS.5.2.3 and selected velocity and
152
Unsteady ﬂow in rivers
Table CS.5.2.1. Flood of the Rhine river near the Pannerdens Canal Large
Dune
Small
Dune
Date in Discharge Stage Depth Velocity Slope (m) (m) (m/s) (cm/km) 1998 (m3 /s)
length (m)
height (m)
length (m)
height (m)
29 Oct. 30 31 1 Nov. 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
7.43 8.31 11.12
0.34 0.41 0.48
0 0 0
16.24 19.78 22.57 23.90 26.02 28.84
0.74 0.87 0.97 1.08 1.13 1.15
0 0 0 0 0 0
32.42
0.93
0
34.78 36.63
0.79 0.74
6.94 6.72
0.27 0.25
4,077 4,783 6,180 8,119 9,045 9,464 9,149 8,267 7,273
11.71 12.33 13.42 13.96 14.48 14.88 15.01 14.91
8.51 9.13 10.22 10.76 11.28 11.52 11.49
1.82 1.75 1.71 1.71
10.42 11.77 12.35 13.12 13.70 13.32 11.58 10.61 9.84 9.07 8.49 7.91
5,122 4,850
11.17 14.14 10.94 13.76 10.56 13.46 10.05 13.11 12.85 9.82 12.69 9.49 12.46 9.25
4,522
12.18
9.1
40.01
0.66
6.61
0.28
4,527
12.17
9.1
42.28
0.56
7.95
0.26
5,640
1.63 1.53
sedimentconcentration proﬁles are shown in Table CS.5.2.4. The navigable width of the Rhine River is maintained at 260 m. The assistance of G. Klaassen at Delft Hydraulics and W. ten Brinke at the Rijkswaterstaat, is gratefully acknowledged. Exercise 5.1 Demonstrate that the lefthand side of Eq. (5.15) reduces to Eq. (5.14) when βm = 1 in wide rivers. Hint: use continuity equation (5.8). Exercise 5.2
Demonstrate from the derivatives of the resistance relationship V = αh β−1 that equation of motion (5.16) reduces to Eq. (5.22). Answer: h∂ V = (β − 1) V ∂h, or (V /g) (∂ V /∂ x) = (β − 1)Fr 2 (∂h/∂ x).
River ﬂow and sedimentduration curves
153
Table CS.5.2.2. Stages along the RhineWaal River Station
Pannerdense Nijmegen Lobith Kop Haven Tiel Zaltbommel Vuren kilometer section 862.18 867.22 884.87 913.25 934.78 951.78 Distance 4,998.1 10,180.858 27,225.159 55,528.35 76,652.202 93,204.14 Date Time cm cm cm cm cm cm Oct. 1 Nov. 1 2 3 4 4 4 4 4 4 5 5 5 5 5 6 7 8 9 10 11
09:00 12:00 12:00 12:00 11:00 16:00 17:00 18:00 19:00 20:00 07:00 08:00 09:00 11:00 12:00 12:00 12:00 13:00 11:00 10:00 13:00
1372 1457 1514 1554 1573 1574 1573 1573 1572 1572 1563 1562 1561 1559 1558 1515 1468 1427 1392 1358 1327
1318 1395 1450 1486 1502 1502 1502 1502 1502
1118 1188 1231 1266 1283 1284 1284 1284 1284 1283 1277 1278
1494 1494 1493 1490 1489 1455 1413 1376 1346 1314 1285
1276 1274 1248 1215 1189 1165 1137 1110
776 840 890 939 955 958 958 958 958 959 958 957 957 956 956 933
467 536 585 646 670 673 673 674 674 675 675 675 674 673 673 657 626 593 568 541 514
871 852 826 799
224 277 315 377 409 409 412 414 415 416 418 419 418 145 414 402 370 337 321 300 276
Table CS.5.2.3. Particlesize distribution of the bed material Position (m from axis)
d90 (mm)
d84 (mm)
d75 (mm)
d65 (mm)
d50 (mm)
d35 (mm)
d25 (mm)
d16 (mm)
d10 (mm)
−33 −33 −33 0 0 0 33 33 33
7.810 11.480 7.894 11.506 12.190 12.000 11.789 11.767 5.642
5.524 8.768 5.770 8.809 9.905 9.600 9.263 9.227 3.717
3.000 5.296 3.040 5.767 6.400 4.000 6.800 6.109 2.622
1.864 2.539 1.474 3.347 3.040 1.851 4.800 3.232 1.925
0.948 0.980 0.808 1.901 1.182 0.942 2.686 0.901 1.435
0.609 0.651 0.556 1.306 0.710 0.698 1.109 0.472 0.967
0.453 0.469 0.448 0.952 0.495 0.536 0.684 0.397 0.775
0.372 0.378 0.378 0.762 0.401 0.414 0.459 0.334 0.602
0.315 0.319 0.325 0.635 0.338 0.339 0.396 0.297 0.486
Average (mm)
10.231
7.843
4.782
2.675
1.310
0.786
0.579
0.456
0.383
154
Unsteady ﬂow in rivers
Table CS.5.2.4. Velocity and concentration proﬁles Height z (m) October 31, 1998 0.2 0.2 0.2 0.3 0.3 0.4 1.1 1.1 1.2 2.5 3.0 4.4 6.5 7.3 7.3 8.3 8.2 Surface 8.9 November 3, 1998 0.3 0.3 0.3 0.5 0.8 0.4 1.2 0.9 1.3 2.2 3.5 4.1 4.0 5.3 6.0 7.3 8.0 9.0 Surface 9.9 November 5, 1998 0.2 0.2 0.3 0.5 0.5 0.5 0.8 0.9 0.9
Concentration c Velocity v (mg/1) (m/s) 233 246 159 83 75 97 39 38 9
0.97 1.02 1.00 1.28 1.23 1.25 1.50 1.48 1.45
4 5 4 3 2 3 2 2
1.73 1.71 1.72 1.90 1.87 1.85 1.98 1.96
494 488 498 398 293 432 132 185 134 83
0.74 0.81 0.72 0.84 1.22 0.47 1.38 1.34 1.47 1.63
49 43 43 35 33 26 25 23
1.92 1.86 1.85 1.99 1.98 2.08 2.04 1.90
216 212 250 227 218 230 157 170 160
1.27 1.33 1.30 1.47 1.36 1.47 1.41 1.46 1.41
Height z (m) 1.9 3.1 3.7 5.1 5.5 5.9 6.8 7.9 9.0 Surface 10.5 November 7, 1998 0.2 0.2 0.2 0.7 0.6 0.2 1.1 1.3 1.3 2.1 2.5 3.3 4.7 5.1 5.9 6.8 7.9 9.0 Surface 9.4 November 10, 1998 0.3 0.3 0.3 0.4 0.2 0.4
Concentration c Velocity v (mg/1) (m/s) 71 40 41 29 23 28 20 18 8
1.66 1.81 1.82 2.06 2.00 1.94 1.99 2.03 2.07
108 140 99 53 96 303 31 24
0.84 0.89 0.85 1.02 1.11 0.93 1.35 1.46
26 20 23 21 11 8 8 7 5 4
1.36 1.45 1.46 1.43 1.76 1.77 1.79 1.79 1.77 1.84
143 153 103 124 181 87
0.85 0.95 0.91 0.95 0.73 0.96
1.3
15
1.30
0.9 1.1 2.0 3.2 3.8 4.5 5.1 6.3 6.9 8.1 Surface 9.1
42 28 11 7 8 7 7 6 5 4
1.11 1.24 1.38 1.51 1.53 1.44 1.52 1.62 1.70 1.74
Problems
155
Exercise 5.3
Demonstrate Eq. (5.23a) from the derivatives of the resistance relationship V = αh β−1 combined with the continuity relationship [Eq. (5.8)]. Hint: V ∂h ∂h ∂q V ∂h h∂ V 1 ∂V = (β − 1) , =− =− + . g ∂t gh ∂t ∂t ∂x ∂x ∂x Exercise 5.4
As per Exercises 5.2 and 5.3, demonstrate that, for ﬂoodwave propagation, term (5) of St. Venant equation (5.16) is always −β times larger than term (4). Exercise 5.5
Demonstrate Eqs. (5.23a) and (5.23b) from Eq. 5.16, Table 5.1, c = βV, Q = W q, and q = αh β . For Manning’s equation, substitute α = S 1/2 /n and β = 5/3. Exercise 5.6 Demonstrate Eqs. (5.24) and (5.25) from q = αh β , α = S 1/2 /n and β = 5/3 for Manning’s equation. Problem 5.1
Determine the celerity of the ﬂoodwave in Example 5.3. What is the time required for the upstream perturbation to reach the downstream station? Answer: c = βV = 5 × 1.82/3 = 3.03 m/s, 1,000 m = 5.5 min. T = c Problem 5.2
Repeat the calculations from Example 5.4 for different values of x. Describe and explain the inﬂuence of the Courant number on ﬂood routing calculations. Problem 5.3 Estimate the peak discharge of a 3km2 watershed under extreme rainfall conditions. What is the equivalent rainfall intensity?
156
Unsteady ﬂow in rivers Problem 5.4
With reference to Case Study CS.5.1, for the daily sediment discharge of the Colorado River, bˆ = 1.7, the daily sediment discharge of x1 = 1 × 106 tons/day is exceeded 10 days/yr (E 1 = 0.0274). Calculate the exceedance probability E 2 of a daily sediment discharge x2 = 2 × 106 tons/day. Answer: In this case, ξ is calculated, given η = x2 /x1 = 2 and b = 0.59 1/bˆ = 0.59 for the Colorado River; thus ξ = (0.0274)(2 −1) = 0.162, or E 2 = ξ E 1 = 4.45 × 10−3 or 1.62 days/yr. Problem 5.5
From the cross section of the Missouri River (Fig. P.4.6.1) determine the following: (a) If the Manning coefﬁcient n is constant at any stage, determine the celerity relationship from the relation between the crosssection area and discharge. (b) When the discharge is Q = 31,600 ft3 /s and the water level rises at a rate of 1 ft per day, what is the ﬂow discharge at a cross section located 10 miles upstream (assume that the cross section is unchanged and without lateral inﬂow along the reach). Problem 5.6
Consider that the Missouri River ﬂow in Fig. P.4.6.1 is controlled by clearwater releases from a reservoir and determine the following: (a) Determine the celerity of the wave generated from suddenly increasing the discharge from a steady 10,000 ft3 /s to a discharge ﬁxed at 20,000 ft3 /s. (b) Assume simple approximations for partial derivatives over a reach on a daily basis and estimate the relative magnitude of the various terms of the St. Venant equation when the discharge increases from 20,000 to 30,000 ft3 /s in 1 day. What approximation of the St. Venant equation is best suited for ﬂoodwave routing? Answer: (a) c 2.5 m/s, (b) diffusivewave approximation.
Problems
157
Problem 5.7
Take a sample, at least 100 values, containing daily measurements of either discharge, sediment concentration, sediment discharge, or nutrient or chemical concentration. Determine the following: (a) exceedance probability curve (b) inverse transform parameters aˆ and bˆ (c) average value of the sample and value exceeded 1 day/yr Problem 5.8
From the data section in Case Study 5.2, determine the ﬂoodwave celerity. Determine which approximation of the St. Venant equation is best suited for ﬂood routing.
6 River equilibrium
In a strict sense, a channel is stable when all particles along the wetted perimeter are not moving. This implies that, without transport of bed material, a crosssectional geometry cannot change with time. The geometry of stable channels, also termed nonalluvial channels, depends on rock outcrops and artiﬁcial riprap and does not depend on sediment transport. Many rivers, however, ﬂow in their own deposits, called alluvium, to form alluvial channels. Equilibrium of alluvial channels implies a balance between incoming and outgoing water discharge and sediment load. Poised and graded streams are synonyms describing equilibrium conditions. Whenever a balance is obtained between incoming and outgoing sediment discharges, the crosssectional geometry may locally change as long as the deposition volume with a river reach is equal to the erosion volume. For instance, river bends may reach equilibrium condition between the rate of erosion on the outside bank and the rate of sedimentation on the point bar. In a broad sense, the crosssectional geometry of a meandering channel is in equilibrium. However, lateral migration of the bend implies that the planform geometry of the stream is not stable. This chapter deals with stable and equilibrium river conditions. Section 6.1 details particle stability in a strict sense, and the concept of stable channel is used in Section 6.2 to deﬁne the ideal atastation crosssectional geometry of a straight channel. Empirical regime relationships in Section 6.3 are followed by ﬂow conditions in river bends in Section 6.4. The downstream hydraulic geometry of alluvial channels is derived in Section 6.5. Planform geometry and bars are discussed in Section 6.6, followed by a discussion of river meandering in Section 6.7. Lateral migration rates are discussed in Section 6.8, followed by a case study of a meandering river. 6.1
Particle stability
Figure 6.1 illustrates the forces acting on a cohesionless particle resting on an embankment inclined at a sideslope angle 1 and a downstream bedslope 158
Particle stability
159
Figure 6.1. Particlestability analysis.
angle 0 . These are the lift force FL , the drag force FD , the buoyancy force FB , and the weight of the particle FW . As long as the watersurface slope in the downstream direction is small, the buoyancy force can be subtracted from the particle weight to give the submerged weight, FS = FW − FB . The lift force is deﬁned as the ﬂuid force normal to the embankment plane whereas the drag force is acting along the plane in the same direction as that of the velocity ﬁeld surrounding the particle. For notational convenience, we deﬁne two geometrical parameters, a and tan , from the sideslope angle 1 and the downstream bedslope angle 0 . These two parameters describe the projection of the submerged weight vector along the embankment plane. The angle is obtained from the ratio of the two projection components of FS in the embankment plane as tan = [(cos 1 sin 0 )/(cos 0 sin 1 )], which can be approximated by tan ∼ = (sin 0 / sin 1 ) as long as both angles are fairly small (less than ∼20◦ ). The
160
River equilibrium
fraction of the submerged weight that is normal to the embankment plane is given by a = 1 − cos2 0 sin2 1 − cos2 1 sin2 0 , which is approximated by a ∼ = cos2 1 − sin2 0 when both angles are small (less than ∼20◦ ). As a realistic approximation, the submerged weight has one sideslope component, FS sin 1 , one downslope component, FS sin 0 , and a component normal to the plane, FS a , as shown in Fig. 6.1. The streamline deviates from the downstream direction at an angle λ along the embankment plane (λ is deﬁned positive downward). When unstable, a particle moves at an angle β from the direction of steepest descent. A particle moves along the horizontal when β = 90◦ and moves in the downstream direction when β + = 90◦ . When β + > 90◦ , a particle moves up the sideslope toward the free surface. Conversely, when β + < 90◦ , a particle moves down the sideslope toward the thalweg. In most streams, the downstream slope 0 will be sufﬁciently small to consider that ∼ = cos 1 . The downstream direction thus = 0 and a ∼ practically corresponds to β = 90◦ , values of β < 90◦ indicate particles moving toward the thalweg, and β > 90◦ indicates motion toward the free surface. Stability against rotation of a particle determines incipient motion when the equilibrium of moments about the point of rotation is satisﬁed. The deviation angle δ is measured between the particle direction and the streamline. Considering the angles δ and β and the moment arms l1, l2 , l3 , and l4 that are shown in Fig. 6.1, we ﬁnd that stability about point 0 corresponds to 2 l2 FS a = l1 FS 1 − a cos β + l3 FD cos δ + l4 FL . (6.1) The lefthand side of Eq. (6.1) deﬁnes the stabilizing moment that is due to the particle weight. Clearly the last term on the righthand side denotes the lift moment, which always destabilizes the particle. The ﬁrst two terms on the righthand side determine about which pivot point particle P is to rotate. In most cases, their net sum is positive, and moments about 0 are considered. Should their net sum be negative, when λ < 0, the particle will then rotate about point 0 instead of about 0. The stability factor SF0 , for rotation about point 0, is deﬁned as the ratio of the resisting moments to the moment generating motion. In the case shown in Fig. 6.1, in which both cos δ and cos β are positive, the stability factor SF0 is the ratio of the sum of counterclockwise moments about 0 to the sum of clockwise moments about 0; thus SF0 =
l2 FS a . 2 l1 FS 1 − a cos β + l3 FD cos δ + l4 FL
(6.2)
Note that each term in Eq. (6.2) must be positive; otherwise the formulation is
Particle stability
161
changed to express the ratio of positive stabilizing moments to positive destabilizing moments. Because the stability factor SF0 equals unity when the angle 0 or 1 equals the angle of repose φ under static ﬂuid conditions (FD = FL = 0), it is found that tan φ = l2 /l1 . Dividing both the numerator and the denominator by l1 FS transforms Eq. (6.2) into SF0 =
a tan φ , 2 η1 tan φ + 1 − a cos β
(6.3)
in which η1 = M + N cos δ after we deﬁne M=
l4 FL , l2 FS
N=
l 3 FD . l2 FS
(6.4)
The parameter M/N represents the ratio of lift to drag moments of force. The case of no lift is given by M/N = 0 and equal moments are described by M = N or (M + N )/N = 2. The variable η1 is called the sideslope stability number for the particle on the embankment. The variable η1 relates to the stability number η0 = M + N for particles on a planehorizontal surface (0 = 1 = δ = 0) after we consider λ + δ + β + = 90◦ : (M/N ) + sin(λ + β + ) , 1 + (M/N ) τ0 τ0 η0 = = . τc (G − 1)ρ g ds τ∗c
η1 = η0
(6.5) (6.6)
The stability number η0 is calculated from the applied shear stress τ0 , the critical shear stress on a planehorizontal surface τc , the particle diameter ds , the mass density of the particle ρs = G ρ , the mass density of the ﬂuid ρ, the gravitational acceleration g, and the critical value of the Shields parameter τ∗c . This normalized form of the Shields parameter shows that η0 = 1 describes the incipient motion of particles on a planehorizontal bed. When the ﬂow is fully turbulent over a hydraulically rough horizontal surface, incipient motion approximately corresponds to τ∗c = 0.047 and SF0 = η0 = 1. The second equilibrium condition indicates the direction of a moving particle from equilibrium conditions along the section normal to AA in Fig. 6.1: 2 sin β. l3 FD sin δ = l1 FS 1 − a
(6.7)
162
River equilibrium
After δ is written as a function of λ, , and β, solving for β gives −1
β = tan
cos(λ + ) . 2 (M + N ) 1 − a + sin(λ + ) N η0 tan φ
(6.8)
The angle β determines the direction of motion of a sediment particle in contact with the inclined plane. The particleorientation angle depends on (1) surface topography by means and a , (2) streamﬂow direction at an angle λ, and (3) particle characteristics such as angle of repose φ and excess shear η0 = τ0 /τc . A complete calculation example is given in Example 6.1 . Example 6.1 Application to particlestability analysis. The following example details the calculations of particle stability for typical riverbend conditions by use of Eqs. (6.3), (6.5), (6.6), and (6.8). The overall agreement with ﬁeld data being slightly better with M/N = 0, this value was given preference for all calculations. A 16mm quartz particle stands on the bed of a channel; if the downstream channelslope angle is 0 = 0.074◦ and the sideslope angle is 1 = 15◦ , calculate the particledirection angle under an applied bed shear τ0 = 12 Pa when the streamlines are deﬂected upward at λ = −10◦ . The calculation procedure is as follows: (a) The particle size is ds = 16 mm. (b) The angle of repose is approximately φ = 37◦ , and the speciﬁc gravity G = 2.65 is assumed. (c) The sideslope angle is 1 = 15◦ . (d) The downstream slope angle is 0 = 0.074◦ . ◦ (e) The angle = tan−1 (sin 0 /sin 1 ) = 0.28 . 2 2 (f) The geometric factor a = cos 1 − sin 0 = 0.965. (g) The applied bed shear stress is τ0 = 12 Pa; this corresponds to a ﬂow depth of ∼1 m. (h) The streamline deviation angle λ = −10◦ means that the downstream shearstress component is τ0 cos λ = 11.8 Pa and the transverse shear stress is τ0 sin λ = −2.08 Pa toward the free surface (negative). (i) From Eq. (6.6) and τ∗c = 0.047, the planebedstability number is η0 =
12 Nm3 s2 = 0.985. m2 (1.65) 1,000 kg × 9.81 m × 0.016 m × 0.047
Channel stability
163
( j ) From Eq. (6.8), assuming M/N = 0, the 16mm particledirection angle is cos(−10◦ + 0.28◦ ) = 79.4◦ . β = tan−1 1 − (0.965)2 ◦ ◦ + sin(−10 + 0.28 ) 0.985 tan 37◦ (k) The 16mm particle moves toward the thalweg because β + < 90◦ . ( l ) The stability factor is calculated from Eq. (6.3). SF◦ =
0.965 tan 37◦
0.985 sin (−10◦ + 79.4 + 0.28◦ ) tan 37◦ + = 0.97.
√
1 − 0.9652 cos 79.4◦
The particle is unstable. We easily repeat the calculations with a 32m particle to ﬁnd out that it is stable under the given hydraulic conditions.
6.2
Channel stability
Consider the crosssectional geometry of a straight channel for which all particles of weight Fs along the wetted perimeter are at incipient motion. The critical shear stress τsc on a sideslope θ1 is determined in analogy to the critical shear stress τc that corresponds to the angle of repose φ. A very simpliﬁed but elegant equilibrium relationship was proposed by Lane (1953) who, after assuming that τc ∼ Fs tan φ and the resultant on the sideslope R/Fs cos θ1 = tan φ, obtained sin2 θ1 tan2 θ1 τsc . (6.9) = 1− = cos θ1 1 − 2 τc tan2 φ sin φ A detailed derivation can also be found in Julien (1995). As sketched in Fig. 6.2, the thalweg shear stress is τc = γ h 0 S and τ0 = γ h S cos θ1 on the sideslope. Isolating γ S gives the following identity: τ0 h 0 = τc h cos θ1 .
(6.10)
The differential equation for the ideal crosssection geometry is obtained from tan θ1 = −dh/dy after τ0 /τc is canceled from Eqs. (6.9) and (6.10), thus 2 2 dh h + tan2 φ = tan2 φ, (6.11) dy h0
164
River equilibrium 50° Flow
φ= θ1
Fs
Fs cos θ 1 F s sin θ 1
45°
40°
τ sc
40°
1.5 H : 1 V
Sid
esl
θ1
35°
30°
R
ope
Bed
30°
2 H :1 V
25°
3H:1V
20°
20°
10°
0°
0.2
0
0.4
τ sc τ c
0.6
0.8
1.0
φ
φ ho
τ o = τc
h
θ 1  dh dy Critical shear stress τ sc = γ h S cos θ 1
Figure 6.2. Ideal crosssection geometry.
where h 0 is the ﬂow depth at the centerline of the channel. The ideal crosssection geometry in which all particles are at incipient motion has a cosinusoidal shape, with y measured laterally from the centerline: h = cos h0
y tan φ . h0
(6.12)
The ﬂow depth h 0 corresponds to incipient motion on a plane surface h 0 ≈ 0.047 (G − 1)ds /S. The surface width, h 0 π/tan φ, is obtained from the values of y in Eq. (6.12) where h = 0. The width–depth ratio is constant and equal to π cot φ. The crosssection area is A = 2 h 20 cot φ, and the hydraulic
Regime relationships
165
radius is Rh =
8h 0 cos φ . π(4 − sin2 φ)
The mean ﬂow velocity V is then obtained from a resistance relationship, e.g., the Manning equation. The ﬂow discharge is then obtained from A and V (see Problem 6.2). In practice, only coarse bed channels may reach incipient motion at high discharge. Sediment transport in all sandbed channels cannot be ignored, and the width–depth ratio of most alluvial streams far exceeds the ideal geometry conditions.
6.3
Regime relationships
The construction of irrigation canals in India and Pakistan fostered investigations of the design of canals under regime conditions, i.e., a canal that is nonsilting and nonscouring. Empirical relationships have been proposed by Kennedy (1895), Lacey (1929), and Blench (1969), among many others. The Lacey silt factor f e was deﬁned to designate the properties of “Kennedy’s standard silt” of the Upper Bari Doab canal. The geometry of canals with different bed material were compared in terms of different values of f e . The silt factor was shown 1/2 to increase with grain size; the approximate relationship fl = 1.59 dmm can be used. The key relationships from Lacey determined the mean ﬂow velocity V in feet per second, the hydraulic radius Rh in feet, the crosssection area A in square feet, the wetted perimeter P in feet, and the dimensionless slope as a function of the design discharge Q in cubic feet per second, and the Lacey silt factor: 1/3
V = 0.794 Q 1/6 fl R = 0.47 A = 1.26
,
−1/3 , Q fl −1/3 , Q 5/6 fl
P = 2.66 Q
1/3
(6.14) (6.15)
,
(6.16)
5/3 fl Q −1/6 .
(6.17)
1/2
S = 0.00053
(6.13)
In wideshallow channels, the hydraulic radius is approximately equal to the ﬂow depth, and the channel width can often be approximated by the wetted perimeter. The width–depth ratio thus increases slightly with discharge and decreases slightly with grain size. Another characteristic of the regime equation is that once the discharge and grain size are determined a unique value of bed slope is calculated from Eq. (6.17). Note that the method does not allow
166
River equilibrium
Flood plain
W1 A1 V1
h1 = A1/W1
Width W 2 A2
P1 Q1 = A1 V1
Flood plain Mean flow depth h2 = A2/W2
Velocity V2 Q2 = A2 V2
Wetted perimeter P2
Figure 6.3. Downstream hydraulic geometry.
changes in slope at a given discharge and grain size. Consequently, the velocity in Eq. (6.13) does not depend on slope. The regime equations refer to downstream channel geometry at a given bankfull discharge. As shown in Fig. 6.3, different discharges in Eqs. (6.13)– (6.17) correspond to bankfull conditions in different channels. Because these channels can be located at different positions in the same ﬂuvial system, channel properties are referred to as downstream hydraulic geometry. Downstream hydraulic geometry describes bankfull conditions for different cross sections, as opposed to the atastation hydraulic geometry that describes channel properties at different discharges at a given cross section.
6.4
Equilibrium in river bends
Secondary circulation in curved channels is generated through a change in downstream channel orientation. The streamlines near the surface are deﬂected toward the outer bank whereas those near the bed are deviated toward the inner bank. The nearbed velocity, the tangential bed shear stress, and the drag on the bed particles are commonly directed toward the inner bank. Flow in bends is analyzed in cylindrical coordinates. The relative magnitude of radialacceleration terms indicates that the centrifugal acceleration is counterbalanced by pressure gradient and radial shear stress, as suggested by Rozovskii (1957): 1 ∂τr v2 = gSr − , r ρ ∂z
(6.18)
where the local downstream velocity v, the radial shear stress τr , and the radial watersurface slope Sr vary with the vertical elevation z and/or the radius of curvature r. In Fig. 6.4, the transverse boundary shear stress τr R at point R A , the
Equilibrium in river bends
167
Figure 6.4. Flow in river bends.
radial watersurface slope Sr A at point A, the radius of curvature R at the same point, the average ﬂow depth h, and the top channel width W serve as scaling factors. These scaling factors deﬁne dimensionless parameters for channel width w ∗ = w/W , ﬂow depth z ∗ = z/ h, radius of curvature r ∗ = r/R, velocity v ∗ = v/V¯ , radial shear stress τr∗ = τr /τr R , and the radial surface slope Sr∗ = Sr /Sr A . The element of ﬂuid volume d∀ = dx dy dz for a reach of given length dx = Rdθ is reduced to a dimensionless volume d∀∗ = d∀/W Rh. Radial equation of motion (6.18) is multiplied by ρ and d∀, reduced in dimensionless form, and then integrated over the dimensionless volume ∀∗ of the reach. The resulting dimensionless momentum equation in the radial direction is ρW h V¯ 2
∀∗
v ∗2 ∗ d∀ = ρg RW h Sr A r∗
∀∗
Sr∗ d∀∗ − RW τr R
∀∗
∂τr∗ ∗ d∀ . ∂z ∗
(6.19)
The corresponding force diagram is sketched in Fig. 6.4, after the centrifugal force is denoted on the lefthand side of Eq. (6.19) by Fc , the pressure force F p describes the ﬁrst term on the righthand side, and the last integral represents the shear force Fs . The pressure force is found to balance the sum of the centrifugal force exerted at a distance lc above point A whereas the shear force is exerted at a distance ld below point A. Moment equilibrium around point A
168
River equilibrium
gives
∂τr∗ ∗ d∀ ∗ ρh V¯ ∗ ∂z R = = ∀ ∗2 . Rτr R v ∗ lc d∀ ∗ ∀∗ r 2
ld
(6.20)
The dimensionless parameter R denotes the ratio of the centrifugal force generating secondary motion to the shear force abating the motion and dissipating energy. The resulting ratio of radial shear stress τr R to the downstream bed shear stress τθ deﬁnes the deviation angle λ of the streamlines near the bed. Therefore, combining Eqs. (6.20) and (4.10), we obtain a 2 h 2m h τr R . (6.21) = tan λ = τθ R ds R Rozovskii (1957) found that the value of the term in brackets of Eq. (6.21) is approximately equal to 11. Slightly different values have been proposed by Engelund (1974), de Vriend (1977), Odgaard (1981), and Hussein and Smith (1986). We conclude that deviation angle λ depends primarily on the ratio of ﬂow depth to radius of curvature. Sharp bends will exhibit stronger secondary ﬂows. In curved channels, secondaryﬂow effects on particle stability are examined through the inﬂuence of the deviation angle λ on the values of the stability factor SF0 . Stability calculations based on Eqs. (6.6), (6.8), (6.5), and (6.2) with λ = 0 demonstrate that slight downward deviations decrease particle stability, whereas small upward deviations increase particle stability. On the other hand, large upward deviations will also decrease particle stability. Quantitative dimensionless results are summarized in Fig. 6.5 for typical 2D ﬂow conditions in a curved alluvial channel. The local crosssectional shape is described as the ratio of the embankment slope θ to the angle of repose φ. This ﬁgure illustrates the relative particlestability ratio λ = SF0 (λ = 0)/ SF0 (λ = 0) of the stability factor with secondary circulation (λ = 0) over the stability factor without secondary circulation (λ = 0). It is shown in Fig. 6.5 that, when the deviation angle λ is relatively small, for example less than 15◦ , the downward deﬂected streamlines near the outer bank (θ/φ > 0) induce particle motion, as expected from λ < 1. Conversely, opposite effects are observed near the inner bank (θ/φ < 0); the upward deviations of the streamlines increase the stability of particles, as shown by the values of λ > 1. The conditions induced by secondary circulation at angles λ < 15◦ decrease particle stability near the outer bank and increase particle stability near the inner bank of curved alluvial channels.
Equilibrium in river bends
169
Figure 6.5. Relative particle stability in curved channels.
When the strength of secondary circulation increases, 15◦ < λ < 55◦ , asymmetry in the particlestability curves shown in Fig. 6.5 develop and a larger proportion of the channel becomes unstable. For extreme conditions (λ > 55◦ ), the entire cross section becomes unstable and scour occurs for all particles on the wetted perimeter of the alluvial channel, thus widening the channel. This analysis of secondarycirculation effects on particle mobility in curved alluvial channels highlights a continuum of conditions between the following two extremes: (1) at small deviation angles, around λ < 15◦ , equilibrium prevails between outerbank erosion and innerbank deposition; and (2) at large deviation angles, typically when λ > 55◦ , the particle stability decreases over the entire cross section and should result in widening of the alluvial channel. Secondarycirculation effects being primarily felt near the outer bank, the deviation angle λ thus relates to the widening process in alluvial channels. The result of downward lateral forces is to destabilize the bed particles whereas upward lateral forces stabilize bed particles [Fig. 6.6(b)]. Under signiﬁcant secondary ﬂows, crosssectional geometry becomes asymmetric, with the thalweg moving toward the outer bank, as sketched in Fig. 6.6. Stability can be maintained only through a heavier particle weight to counterbalance the higher erosive forces. Conversely, ﬁner particles can deposit to form a point bar near the inner bank.
170
River equilibrium
Figure 6.6. Stability and equilibrium in river bends.
To reach stability in graded bed material in which armoring is possible, coarser grains can be found near the outer bank and ﬁner grains near the inner bank [Fig. 6.6(c)]. In the case of uniform erodible bed material, secondary ﬂows will induce scour at the toe of the outer bank, leading to bank caving and lateral migration. Equilibrium is possible only when bank caving balances inside deposition on the point bar [Fig. 6.6(d)]. It is also important to note that the change in crosssectional geometry depends on the magnitude of the streamline deviation angle λ. From Eq. (6.21) we obtain that the angle λ increases with ﬂow depth h. At high ﬂows, we thus expect to ﬁnd the strongest secondary ﬂows with the maximum scour potential near the outer bank. The asymmetry in the crosssectional geometry should be observed during ﬂoods. At low ﬂows, low values of the angle λ will tend to bring the thalweg position back closer to the channel centerline and the crosssectional geometry should become more symmetrical.
6.5
Downstream hydraulic geometry
Flow discharge varies with time, and sediment transport is most active during ﬂoods. The morphology of alluvial channels should therefore depend on a high discharge value that contributes to the channel formation. This value should be high enough such that sediment transport is quite active, yet it should
Downstream hydraulic geometry
171
Figure 6.7. Natural levees.
reoccur a sufﬁcient number of times to maintain a channel free of vegetation. A bankfull discharge is determined by the discharge that a channel can convey when reaching the ﬂoodplain level. Natural levees result from the deposition of coarser fractions of the suspended load on the ﬂoodplain adjacent to the channel. As sketched in Fig. 6.7, levees can form near the outer bank of the channel or along both sides of alluvial channels. The reoccurrence frequency of bankfull ﬂows is quite variable (Williams, 1978). Periods of return of 1.5 and 2 yrs are most often encountered in the literature, but mean annual ﬂoods, mean annual ﬂows, and 5yr ﬂoods have also been cited. The dominant discharge is of sufﬁcient magnitude and frequency to determine the dimension and geometry of an alluvial channel. For hydraulicgeometry relationships, it is taken to be the bankfull discharge, which has a period of return of approximately 1.5 yr in many natural channels. It is important to remember that the concept of dominant discharge cannot be precisely quantiﬁed. The dominant discharge of a river remains subjective and variable. The downstream hydraulicgeometry relationships developed in this section can provide only approximative values of width, depth, velocity, and slope. The downstream hydraulic geometry of noncohesive alluvial channels can be analytically determined from the following four relationships. First, under steadyuniform bankfull ﬂow conditions, the dominant discharge Q is Q = W hV,
(6.22)
where the mean velocity vector V is taken normal to the crosssectional area, h is the ﬂow depth, and W is the bankfull width. For channels with large width–depth ratios, the hydraulic radius Rh becomes equal to the ﬂow depth h. Second, the power form of the resistance equation from Eq. (4.10) is √ V =a g
h ds
m h 1/2 S 1/2 ,
(6.23)
where g is the gravitational acceleration, ds is the grain diameter, S is the
172
River equilibrium
slope, and the exponent m = 1/ln(12.2 h/ds ). Note that the value of m = 1/6 corresponds to the Manning–Strickler resistance equation. Third, the stability of noncohesive particles in straight alluvial channels is described by the relative magnitude of the downstream shear force and the weight of the particle. The ratio of these two forces deﬁnes the Shields number τ∗ from Eq. 4.36: τ∗ =
hS , (G − 1) ds
(6.24)
where G is the speciﬁc gravity of sediment particles. The critical value of the Shields number, τ∗ ≈ 0.047, identiﬁes the beginning of motion of noncohesive particles in turbulent ﬂows over rough boundaries. For values of the Shields number below the critical value (τ∗ ≤ τ∗c ), the particles on the wetted perimeter of the alluvial channel are stable. Beyond this threshold (τ∗ > τ∗c ), the particles enter motion and the rate of sediment transport increases with the Shields number. Two signiﬁcant concepts are associated with the Shields number: (1) the threshold concept described by τ∗c for the beginning of motion of noncohesive particles and (2) the concept that beyond the threshold value, the sedimenttransport rate increases with the Shields number. Because the Shields number depends primarily on ﬂow depth, it is thus associated with the vertical processes of aggradation and degradation in alluvial channels. Fourth, for ﬂow in bends, with the radius of curvature R being proportional to the channel width W , Eq. (6.21) shows that tan λ = br
h ds
2 m
h , W
(6.21a)
where the value of br = [(a 2 W )/( R R)] is assumed to be constant. Equations (6.21)–(6.24) contain 13 variables, namely W , h, V , S, Q, ds , τ∗ , tan λ, g, a, br , G, and m. Four relationships enable the deﬁnition of the four dependent variables, W , h, V , and S, as functions of the others. Julien (1988) showed that Q, ds , and τ∗ were the primary independent variables. The variability in the other parameters is considered to be relatively small. Julien and Wargadalam (1995) determined approximative empirical values for the remaining parameters from a large data set, including data from 835 rivers and canals. The downstream hydraulic geometry for noncohesive alluvial channels for hydraulically rough turbulent ﬂows were derived for ﬂow depth h in meters, surface width W in meters, average ﬂow velocity V in meters per second, and
Downstream hydraulic geometry
173
friction slope S as −1
6 m−1
1
h = 0.133 Q 3 m+2 ds6 m+4 τ∗6 m+4 , −4 m−1
2 m+1
−2 m−1
W = 0.512 Q 3 m+2 ds 6 m+4 τ∗ 6 m+4 , V = 14.7 Q
m 3 m+2 −1
2−2 m 6 m+4
ds
5 6 m+4
S = 12.4 Q 3 m+2 ds
(6.25a)
2 m+2 6 m+4
τ∗
,
6 m+5 6 m+4
τ∗
(6.25b) (6.25c) (6.25d)
from the equilibrium or dominant ﬂow discharge Q in cubic meters per second, the median grain size ds = d50 in meters, and the Shields parameter τ∗ = γ h S/(γs − γ )d50 , given the resistance exponent m calculated from m = 1/ ln(12.2 h/d50 ). An example of the agreement among predicted and measured width, depth, velocity, and slope from Julien and Wargadalam (1995) is shown in Fig. 6.8. The recommended calculation procedure for the downstream hydraulic geometry starts with the user selection of three independent variables. To include the effects of sediment transport, the user may want to calculate four dependent variables of average ﬂow depth h in meters, surface width W in meters, average ﬂow velocity V in meters per second, equilibrium slope S as a function of three known independent variables in discharge Q in cubic meters per second, median grain size ds in meters, and dimensionless Shields number τ ∗ ; Eqs. (6.25a)–(6.25d) are solved with the ﬁvestep procedure outlined below in Example 6.2. When Manning’s equation is applicable, m = 1/6, a simpliﬁed form of Eqs. (6.25) is obtained as h∼ = 0.133 Q 0.4 τ∗−0.2 ,
(6.26a)
W ∼ = 0.512 Q 0.53 ds−0.33 τ∗−0.27 ,
(6.26b)
V ∼ = 14.7 Q
(6.26c)
0.07
ds
0.33
τ∗0.47 ,
S∼ = 12.4 Q −0.4 ds τ∗1.2 .
(6.26d)
The hydraulic geometry of stable channels is obtained from Eqs. (6.26) when τ∗ ∼ = 0.047. Higher sediment transport implies higher velocity and slope and reduced width and depth. This system of equations is sufﬁciently ﬂexible to allow the user to use a different set of known independent variables. For instance, geomorphologists may prefer to calculate ﬂow depth h, width W, mean velocity V, and Shields number τ∗ as explicit functions of discharge Q in cubic meters per second,
River equilibrium 10
2
(a) Depth (m)
Predicted
10 1 1
10
2
10
10
Predicted
10 3
10
2
10
1
1
10
2
2
10
1 10
Predicted
10
(b) Width (m)
1
2
10
3
10
10
(c) Velocity (m/s)
1
10
1
10 1 1
1
1
10
(d) Slope
10 Predicted
174
2
10 3
10
4
10
5
10
10
5
4
10
3
2
10 10 Measured
10
1
1
Figure 6.8. Downstream hydraulicgeometry method of Julien and Wargadalam (1995).
Downstream hydraulic geometry
175
median grain size ds in meters, and channel slope S. The empirically recalibrated equations of Julien and Wargadalam (1995) are 6m
2
−1
h = 0.2 Q 5+6 m ds5+6 m S 5+6 m , W = 1.33 Q V = 3.76 Q
2+4 m 5+6 m
1+2 m 5+6 m
−4 m 5+6 m
ds
−2 m 5+6 m
ds
S S
−5 5+6 m
2
τ∗ = 0.121 Q 5+6 m ds
(6.27a)
−1−2 m 5+6 m
2+2 m 5+6 m
S
,
(6.27b)
,
4+6 m 5+6 m
(6.27c) ,
(6.27d)
where m = 1/ln(12.2 h/ds ). The equations are solved with the procedure given in Example 6.2. For the particular case in which Manning’s equation is acceptable, m = 1/6, the downstream hydraulicgeometry relationships simplify to h = 0.2Q 0.33 ds0.17 S −0.17 , W = 1.33Q
0.44
ds−0.11
S
−0.22
(6.28a) ,
(6.28b)
V = 3.76Q 0.22 ds−0.05 S 0.39 ,
(6.28c)
τ∗ = 0.121Q 0.33 ds−0.83 S 0.83 .
(6.28d)
It is observed that the channel width varies primarily with discharge whereas other parameters also affect ﬂow depth and ﬂow velocity. A plot of channel width vs. discharge is shown in Fig. 6.9.
10
Channel width W (m)
10
5 4 3
10
2
10 10 1
1
10 4 10
3
10
2
10
1
10
2
1 10 10 10 Discharge Q (m3 /s)
3
10
4
10
5
10
6
10
7
Figure 6.9. Channel width vs. discharge (after Kellerhals and Church, 1989).
176
River equilibrium
Equation (6.28d) relates sediment transport by means of τ∗ to the discharge, grain size, and slope. For instance, if we consider the following crude approximation for the unit bedsediment discharge qs in square meters per second as qs ∼ g ds3 τ∗2 , the solution for the bedmaterial discharge Q s = W qs 18 = is obtained from Eqs. (6.28b) and (6.28d): Q s ds0.28 Q 1.11 S 1.44 ,
(6.29)
where Q s is the bedmaterial load in cubic meters per second, ds is the bedmaterial size in meters, Q is the bankfull discharge in cubic meters per second, and S is the bed slope. This approximation relates to downstream conditions and does not apply to atastation conditions. This crude relationship shows the equilibrium between, on the right side, the product of hydraulic parameters (discharge and slope) and, on the left side, the product of sediment characteristics (sediment discharge and grain size). Lane (1955a) had proposed a similar form in which Q S ∼ Q s ds . Example 6.2 illustrates how the hydraulic geometry of a channel can be estimated from the discharge, grain size, and slope.
Example 6.2 Application to stable channel geometry. Calculate the downstream hydraulic geometry given Q = 104 m3 /s, d50 = 0.056 m, and τθ∗ = 0.047 at the beginning of motion. Step 1: Roughly estimate the ﬂow depth, e.g., h = 1 m. Step 2: From the ﬂow depth and grain size calculate m from m=
1
= 0.186. 12.2 h ln ds
Step 3: Calculate the exponents for ﬂow depth from Eq. (6.25a), given m = 0.186: h = a Q b dsc τθ∗d = 0.133(104)0.39 (0.056)0.023 (0.047)−0.195 = 1.38 m. Step 4: Repeat steps 2 and 3 with the calculated ﬂow depth in step 3 until convergence: m = 0.175 gives h = 1.49 m,
and m = 0.172 gives h = 1.51 m.
Step 5: Calculate the channel width W, ﬂow velocity V , and slope S by using the last value of m and the exponents of Q, ds , and τθ∗ in Eqs. (6.25), e.g., with
Bars in alluvial rivers
177
m = 0.172: W = 0.512(104)0.534 (0.056)−0.335 (0.047)−0.267 = 36.4 m, V = 14.7(104)0.068 (0.056)0.329 (0.047)0.466 = 1.87 m/s, S = 12.4(104)−0.397 (0.056)0.994 (0.047)1.199 = 2.86 × 10−3 .
6.6
Bars in alluvial rivers
Channelbed conﬁgurations include bedforms and bars. Bedforms include ripples, dunes, and antidunes and remain submerged, except during droughts or in ephemeral streams. The reader is referred to Julien (1995) regarding bedform classiﬁcation and formation. Bars refer to large bedform conﬁgurations that are often exposed during low ﬂows. They are usually submerged at least once a year in order to prevent vegetation growth. Bars can be viewed as alluvial bed deposits that can be transported under high ﬂow conditions. When bars do not get submerged approximately every year, vegetation grows and stabilizes the bars to form islands or to reduce the active channel width of a river. Alternate bars form in straight channels with deposits alternating from the right bank to the left bank. As illustrated in Fig. 6.10(a), the wavelength of alternate bars is proportional to the channel width W, and 2π W. Alternate bars tend to form in channels where the Froude number is high and the Shields parameter is close to incipient motion. The height of alternate bars can reach the ﬂow depth. The thalweg is said to wander or weave between both banks. Additional information regarding alternate bars can be found in Fujita and Muramoto (1982, 1985) and in Ikeda (1984). Point bars are sketched in Fig. 6.10(b). Their formation process is akin to that of the secondary ﬂows discussed in Section 6.4, and they usually are found near the inner bank of river bends. The increased particle stability near the inner bank induces sedimentation and ﬁning of the bed deposits. The process of pointbar sedimentation is usually associated with the erosion of the outer bank and lateral migration of the river. During major ﬂoods, pointbar deposits can be remobilized to form chute cutoffs, as sketched in Fig. 6.10(c). At low ﬂows, chute cutoffs may induce sedimentation in both river branches. The truncated point bar is also referred to as a middle bar. Tributary bars form in the main channel near the conﬂuence, with tributaries carrying a signiﬁcant sediment load. As shown in Fig. 6.10(d) the tributary bar contributes to streamlining of the conﬂuence of both streams. The
178
River equilibrium
Figure 6.10. Bar formations in rivers.
tributary bars can be reworked, depending on the magnitude, sediment load, and timing of the ﬂoods in both channels. Figure 6.10(e) shows submeanders, which are deﬁned as small meanders conﬁned within the banks of a perennial stream channel. Submeanders are caused by very low ﬂows compared with ﬂood discharges.
River meandering 6.7
179
River meandering
River meandering is characterized by a succession of alternating meander loops. A meander loop is the channel reach between two inﬂection points. A meander consists of a pair of loops in opposite directions. Numerous hypotheses have been suggested to explain the origin of meandering, including secondary ﬂows, perturbation theory, and extremal hypotheses. Meandering is basically a ﬂuid mechanics problem in which vorticity plays a leading role. The motion of ﬂuid in a curved channel is generally based on the equations of motion. Different approaches were proposed by Einstein (1926), Rozovskii (1957), Yen (1970), DeVriend (1977), Odgaard (1981), Nelson and Smith (1989), and others. Several attempts focus on the hydrodynamic stability of straight alluvial channels. A perturbation technique is used to determine whether small oscillations amplify or decay. Examples can be found in Callander (1969), Anderson (1967), Engelund and Skovgaard (1973), Parker (1976), and Ikeda et al. (1981). Extremal hypotheses include the principle of minimum variance ﬁrst proposed by Langbein and Leopold (1966). The minimization involves the adjustment of the planimetric geometry and the hydraulic factors of depth, velocity, and local slope. Yang (1976) stated that the time rate of energy expenditure explains the formation of meandering streams. Other studies by Maddock (1970) and Chang (1980) use the principle of minimum stream power. Chang (1979a) concluded that a meandering river is more stable than a straight one as it expends less stream power per unit channel length for the system. Julien (1985) treated meandering as a variational problem in which the energy integral corresponds to the functional of a variational problem, the solution of which is the sinegenerated curve. Consider the reach of a meandering alluvial river, as sketched in Fig. 6.11. Two systems of coordinates are deﬁned: one rectilinear and one curvilinear. The downvalley axis X deﬁnes the rectilinear system along the centerline of the meandering pattern downstream of the valley slope. In the curvilinear system, the sinuous axis x follows the centerline of the meandering river path. The angle θ separates the directions x and X along the ﬂow path. Measured from the river centerline, the radius of curvature R in the transversal direction y remains orthogonal to the downstream axis x. Both the magnitude and the direction of the radius of curvature R vary along the path of the channel width W and mean ﬂow velocity V . The radius of curvature is minimum, Rmin , at the apex and is maximum, R = ∞, at the crossing. A complete meander loop is denoted between O and M. At M, the river length is L and the meander length is . The amplitude of the meander belt, or meander width, is Wm .
180
River equilibrium
Figure 6.11. Deﬁnition sketch of a meandering river.
When plotting the orientation angle θ as a function of downstream distance x, Langbein and Leopold (1966) found that θ is a function of the maximum angle θm set at the origin, the downstream distance x and the river length L: θ = θm cos
2π x . L
(6.30)
This sinegenerated curve is compared with an observed meandering pattern in Fig. 6.12. The meander length is computed from the following relationship: = 0
L
cos θ dx = 0
L
2π x cos θm cos L
dx.
(6.31)
The sinuosity , deﬁned as = L/, increases gradually with θm in radians,
River meandering
181
N
F
A
E Reference axis
B O Flood plain
G
W = 100 ft
Rm ~ 260 ft D
0
C
400 ft
200
H (from planetable map by Leopold and Worman, August 1953)
90
θ m = 1.35 rad = 77°
Channel direction θ (°)
Popo Agie River, Wyoming
Rm ~ Λ~ L ~ Ω~ Wm ~
45 0
260 ft 1380 ft 2050 ft 1.5 800 ft
45 90
O
A
B
C
D
E
F
G
H
Figure 6.12. Meandering planform geometry (after Langbein and Leopold, 1966).
as illustrated in Fig. 6.13: L
L = ≡
L
∼ =1+
cos θ dx
2 θm π
5 = 1 + 0.1 θm5 .
(6.32)
0
The radius of curvature R can be obtained from dx = Rdθ; thus combining this with Eq. (6.30) gives 2 πx L csc . (6.33) R= 2 π θm L The minimum radius of curvature Rm at the apex is obtained when csc [(2π x)/L] = 1 or Rm =
L . 2π θm
(6.34)
For a given meander length , the minimum radius of curvature (which obviously corresponds to the maximum value of /Rm ) varies with θm as 2π θm 2 π θm = = . Rm L
(6.35)
182
River equilibrium
Figure 6.13. Properties of meandering rivers.
We easily obtain Rm = [L/(2π θm )] = [( )/(2π θm )]. The ratio /Rm varies with θm as shown in Fig. 6.13 and reveals that the minimum radius of curvature for a given meander wavelength corresponds to the maximum angle θm = 75◦ = 1.3 rad. Consequently, the increase in radius of curvature beyond this point constitutes an extremely important feature because the radius of curvature controls the magnitude of the centrifugal force in bends. Leopold et al. (1960) empirically observed that the meander length is ∼10 times the channel width W , as shown in Fig. 6.14(a). The properties of sinegenerated curves are also supported by ﬁeld evidence. For instance, the ratio of wavelength to minimum radius of curvature /Rm for meandering streams in Fig. 6.13 varies between 3 and 5. Field measurements from Leopold and Wolman (1960) indicate an average ratio of 4.7, as shown in Fig. 6.14(b). Also, the mean radius of curvature R¯ m 2.3 W is obtained from 10 W 4.7 R¯ m in Figs. 6.14(a) and 6.14(b). From 10 W and the above, we obtain 5 Rm . W π θm
(6.36)
River meandering 6
(a)
1 m = 3.28 ft
5
5
Meander length Λ (ft)
10
10
5
10
(b)
4
4
10
10
4
10 10
3
10 10
10
3
2
10
10
2
10 10 1 Channel width W (ft)
2
1
3
3
2
Meander length Λ (ft)
10
183
10
4
10
10 10 10 10 Mean radius of curvature Rm (ft)
5
Bankfull discharge Q (m 3/s) 10
5
Meander length Λ (ft)
10
2
10
1
1
10
10
2
10
3
10
4
(c)
4
10
10
10
3
2
1 m 3/s = 35.3 ft 3/s 10 10
1
1
10
2
3
10 10 10 Bankfull discharge Q (ft 3/s)
4
10
5
10
6
Figure 6.14. Empirical geometry of meanders (after Leopold et al., 1960 and Ackers and Charlton, 1970).
The meander width Wm as deﬁned L/4in Fig. 6.11 is evaluated analytically by the following integral: Wm = 2 0 sin θ dx. The ratio of meander width Wm to the wavelength is a dimensionless measure of the amplitude:
x dx sin θm cos 2π Wm L . = 0L x cos θm cos 2 π dx L 0 L/4
2
(6.37)
184
River equilibrium 5
10
Wm ~ 4.5 W 4
10
10
6
3
10
5
2
10
4
10
10
1
10
1
10
2
10
3
1
10
10 10
3
2
Meander length Λ (m)
Meander width Wm (m)
10
1
10
10
3
10
2
1
10
2
1 10 10 Channel width W (m)
10
3
4
10
10
2
5
10
Figure 6.15. Empirical geometry of meanders (after Zeller, 1967a).
Equation (6.37) has been integrated numerically as a function of θm , and the ratio Wm / is plotted in Fig. 6.13. The meander width increases rapidly when θm exceeds 90◦ and reaches the value 3.25 at the meander cutoff (θm = 125◦ ). Field results in Fig. 6.15 give empirical values Wm = 4.5 W ∼ = 0.45 .
6.8
Lateral river migration
The effects of channel meandering on hydraulic geometry and sediment transport can be examined through the effect of sinuosity on channel slope. The energy gradient of the valley, S f 0 , is given by the ratio of the energy loss H over a meander wavelength . The friction slope S f of a meandering stream, however, corresponds to the energy loss H over the sinuous
Lateral river migration
185
stream length L: Sf =
Sf0 H H = = . L L
(6.38)
The inﬂuence of the sinuosity of the channel width, depth, velocity, and Shields parameter can be examined through Eqs. (6.28). For instance, for the Manning– Strickler equation with m = 1/6, the ﬂow depth h ∼ 1/6 , W ∼ 0.22 , V ∼ −0.39 , and τ∗ ∼ −0.83 . As a numerical example, when compared with a straight channel, a meandering channel with sinuosity = 2 would have a ﬂow depth 12% larger and a channel width 16% larger. The ﬂow velocity would decrease by 24%, and the shear stress would decrease by 44%. Sinuosity thus increases the crosssection area and decreases ﬂow velocity and sediment transport. Because sinuosity varies with θm [Eq. (6.32)], the ratio of the Shields parameter for a meandering channel τ∗ is shown in Fig. 6.16 as a function of θm . Sediment transport
High
1.0
Low
Longitudinal shear stress
Relative shear stresses
0.8
Transversal shear stress
0.6
Lateral mobility High
Low
Low
0.4
0.2
0
0
1.03
1.13
20
40
0
60
1.76
2.69
80
100
1.0
0.5
5
Sinuosity Ω 1.34
1.5
θm
3
2 R m /W
2
(radians)
6.00 (degrees) 2.0
2.5
Figure 6.16. Relative longitudinal and transversal shear stresses.
5
186
River equilibrium
When θm > 90◦ , the Shields parameter of a meandering channel is less than half that of a straight channel. The decrease in the longitudinal Shields parameter as the sinuosity increases demonstrates that sinuous channels have a reduced ability to transport sediments compared with straight channels at any given valley slope. The transversal shear stress has been shown to be proportional to tan λ and the square of the velocity. For instance, from Eq. (6.20), we obtain τt0 ∼
hV 2 , R
(6.39)
where τt0 is the bed shear stress in the transversal direction. The maximum transversal shear stress τt max corresponds to the minimum value of the radius of curvature Rm = /2π θm from Eq. (6.35). Combining Eq. (6.39) with Eqs. (6.28a) and (6.28c) for a given discharge and grain size results in τt max ∼
2 π θm −1/6 ∼ θm −1.61 . 0.77 (6.40)
Figure 6.17. Examples of lateral migration.
The value of τt max from Eq. (6.40) is shown in Fig. 6.16 as a function of θm in radians, Eq. (6.40). High values of the transversal Shields parameter are found when 40◦ < θm < 80◦ . From Fig. 6.13, this corresponds to values of ∼ 4 Rm . With 10 W in Fig. 6.15, this leads us to the important conclusion that the lateral shear stress should be high when Rm ∼ 2.5 W . Lateral migration in meandering rivers results from the erosion of the outer bank combined with equivalent sedimentation near the inner bank. This process is illustrated in Fig. 6.17(a). In natural rivers, erosion rates are variable and also depend on bankmaterial strength, cohesion, armoring, and vegetation. Typical patterns are
Lateral river migration
187
Figure 6.18. Types of meander scroll formation (after Richardson et al., 1990).
also sketched in Fig. 6.17(b). Widening is the result of erosion on one bank in excess of sedimentation near the opposite bank. Narrowing is the result of sedimentation near one bank in excess of the erosion rate of the opposite bank. The alluvium deposits near the inner bank sometimes form small ridges or terraces called scrolls caused by the reworking of pointbar deposits at low and high ﬂows. Scrolls mark the successive positions of former meander loops, are often visible on aerial photographs, and are indicative of past innerbank locations. Figure 6.18 illustrates several types of scrolls. The radius of curvature to width ratio can be a useful tool for predicting erosion rates in river systems. Biedenharn et al. (1989) studied the effects of Rm /W and bank material on the erosion rates of 160 bends along the Red River in Louisiana and Arkansas. As indicated in Fig. 6.19, at Rm /W values greater than ∼5, the erosion rates were generally low at 10–35 ft/yr (3–10 m/yr) with no discernible increasing or decreasing trend. The maximum erosion rates were observed in the Rm /W range of 2 to 4. The relative migration rate, deﬁned as the annual migration rate divided by the channel width, can be developed as a function of Rm /W . Nanson and Hickin (1986) studied the relative migration rate of 18 rivers in western Canada, including the Beatton River, British Columbia. As shown in Fig. 6.20, their study indicated that the migration rate of meanders is maximized at 2.0 < Rm /W < 4.0. The relative migration rate can vary by 1 order of magnitude at a given value of Rm /W . The considerable scatter in the data suggests that the migration rate of meanders is extremely complex and may be a function of factors other than Rm /W . The lateral stability of different stream reaches can be compared by means of a dimensionless erosion index. The erosion index is the product of its median bank erosion rate expressed in channel widths per year, multiplied by the percentage of reach along with erosion occurred, multiplied by 1,000.
200
60 Red River, Louisiana
175 50
40
125
100
30
75 20 50 10 25
0
0 0
2
4
6 R m /W
8
10
12
Figure 6.19. Lateral migration rates (after Biedenharn et al., 1989).
Envelope
0.10
curve
Relative migration rate (channel width per year)
0.12
18 rivers in Canada 0.08
0.06
0.04
0.02
Beatton River 0
0
2
4
6 R m /W
8
10
12
Figure 6.20. Relative migration rates (after Nanson and Hickin, 1986).
188
Average erosion rate (m/year)
Average erosion rate (ft/year)
150
Lateral river migration
189
36 32 28
Erosion index
24 HIGH 20 Meandering Braided
16 12 8 4
LOW 0 1.0
1.2
1.4
1.6
1.8
2.0 2.2 Sinuosity
2.4
2.6
2.8
3.0
Figure 6.21. Erosion index (after Brice, 1984a).
Erosion indexes for 41 streams in the United States are plotted against sinuosity in Fig. 6.21. The length of most of these reaches is 25 to 100 times the channel width. The highest erosionindex values are for reaches with sinuosity of less than 2. It is clear from Fig. 6.16 that this corresponds to high lateral mobility. We can thus infer that lateral erosion of meandering channels is correlated to the value of the transversal shear stress. The erosion index value of 5 was suggested by Brice (1984a) as a boundary between stable and unstable reaches. Reaches having erosionindex values of less than 5 are unlikely to cause lateral erosion problems at bridges. A neck cutoff is the natural result of lateral migration of a meandering pattern over a long period of time. Indeed, channel sinuosity increases as lateral migration of the outer bend progresses. When the channel sinuosity becomes very large, e.g., > 3, the downstream transport capacity, channel slope, and ﬂow velocity are reduced. As a consequence, the risk of ﬂooding is largely increased. Neck cutoffs form when the valley slope far exceeds the channel slope and the sedimenttransport capacity is sufﬁcient to cut through the neck of the long meander loops. Once the cutoff has occurred, silting at both ends of the
190
River equilibrium
Oxbow lake
Sedimentation Clay plugs
Lateral migration
Neck cutoff
Figure 6.22. Oxbowlake formation process.
meander loop results in clay plugs that isolate the oxbow lake. The oxbowlake formation process through neck cutoffs is illustrated in Fig. 6.22. Neck cutoffs result in decreased sinuosity, increased slope, velocity, and sediment transport to be conveyed through or deposited in downstream bends. There may also be local incision and head cutting upstream of the neck cutoff. Over time, meander loops readjust their slope sinuosity and sediment transport to reach a renewed state of equilibrium.
Case Study 6.1 Sediment transport in the Fall River, Colorado, United States. Fall River is a meandering stream ﬂowing through the Horseshoe Park area of Rocky Mountain National Park, Colorado (Fig. CS.6.1.1). During the late spring snowmelt, the daily discharge exceeds bankfull ﬂow for several weeks through early summer. The bankfull ﬂow of Fall River is ∼7 m3 /s, and its winter lows are less than 0.5 m3 /s. The failure of the Lawn Lake Dam, located at the headwaters of Roaring River on July 15, 1982, is at the origin of a large alluvial fan, shown in Fig. CS.6.1.1. The alluvial fan supplies coarse sediment for bedload transport to the Fall River. The Fall River slope is gentle at 0.0013. Sediment loads as high as 0.6 kg/m s have been measured with sediment fractions ranging from 0.125 to 32 mm moving as bedload. During ﬁeld seasons 1986 and 1987, measurements were taken along two consecutive bends at 22 cross sections (Anthony, 1992) to determine the patterns of internal crosssectional adjustments from high to low ﬂows (Anthony and Harvey, 1991). Measurements included bed and freesurface topography, velocity proﬁles, and sediment transport by size fraction. Duplicated measurements at 1m intervals across the channel included at least three vertical velocity
Lateral river migration
191
a Ro
HORSESHOE PARK STUDY AREA
rin r ive gR
N
Fall Rive r
Debris Fan
Rd.
To Estes Park
d. idge R Trail R r ive Stage recorder and ll R a gauging station F Study reach
Flood Lake
0
500 m
Figure CS.6.1.1. Location of the Fall River reach.
proﬁles (for both longitudinal and transverse ﬂow) and six 1min bedload measurements. Topographic measurements included bed and watersurface elevations at 1m intervals across the channel for each of the 22 cross sections identiﬁed in Fig. CS.6.1.2. At these high discharges, the crosssectional geometry of this meandering channel alternates between a relatively rectangular shape at crossings to the most asymmetrical form near the bend apex. Near the apex, the crosssectional geometry features (1) a deep thalweg along the outside of the bends, (2) a pointbar planform along the inside of the bends, and (3) a pointbar slope connecting the two. At each crosssection, 2D velocity proﬁles were measured with a Marsh– McBirney current meter at 1m intervals across the channel. Both transverse and longitudinal velocities were measured, starting from the bottom (with the current meter resting on the bottom) and then in approximately 10cm intervals to the water surface. The data set of 1986 near bankfull discharge describes the ﬂow pattern for the entire study reach. At each point in the proﬁle, a 30s sampling duration was allowed for measuring each velocity component. This sampling duration was extended to 1 min when turbulence was signiﬁcant. The repeated sampling done in 1987 provided averageﬂow vectors for closely spaced points at each cross section. The streamline deviation angle λ was obtained from the 2D velocity measurements made near the channel bed. Both the velocity measurements close to the bed and those 10 cm above the bed were available for analysis. The measurements made 10 cm above the bed showed a more consistent pattern of ﬂow direction and magnitude and were used to determine the streamline deviation angle λ.
192
River equilibrium 6
5
7
8
(a) d95 = 8.00 mm
9 10
4
11
3
22 12
2
21 13
1
20 14
Scale 0
19
10 m
5
18
Bedload percentage 0
15
17
20%
5
16
6
7
8 (b) d10 = 0.25 mm
9 10
4
11
3
22 12
2
21 13
1
20 Scale
14
0
5
19
10 m
Bedload percentage 0
20%
18 15
17 16
Figure CS.6.1.2. Bedload sediment transport in bends (after Julien and Anthony, in press.)
Lateral river migration
193
Sediment transport in the layer covering 3 in. (7.67 cm) above the bed was measured with a Helley–Smith sampler. Two 1min bedload samples were collected at each vertical of each cross section, and the measurements were repeated three times. The bedload measured at each sampling location was later sieved to determine the particlesize distribution of moving material. Grainsize distributions from duplicate measurements were quite similar. Measured bedload movement at each sampling location was divided into weights for each size fraction. At each cross section, the bedload movement for each size fraction was then summed over the entire cross section. For a given grain size, the percentage of the crosssection total material transported was then calculated for each sampling point. After this process was repeated at each cross section, a bedload percentage map over the entire reach was obtained for that particular size fraction. Bedload percentage maps were then produced for each size fraction, i.e., 16, 8, 4, 2, 1, 0.5, 0.25 and 0.125 mm. Typical bedload percentage maps are shown in Fig. CS.6.1.2 to represent coarse grains in transport (d95 ∼ = 8 mm) and ﬁne grains in transport (d10 ∼ = 0.25 mm). The values of bedload percentages at each cross section were used to calculate the position of the center of mass of bedload transport for each size fraction. It is interesting to note in Fig. CS.6.1.2 that the location of the center of mass for ﬁne grains is different from that of coarse grains. The lines linking the successive positions of the center of mass for different grain sizes are shown in Fig. CS.6.1.3. Near the crossing, the bedload centerofmass curves for each
Scale
Thalweg
0
5
10 m
Point bar. Apex Crossings Crossings Apex Center of mass location Point bar. 8 mm (~d 95) 1 mm (~d50 ) 0.25 mm (~d 10 )
Thalweg
Figure CS.6.1.3. Mean bedload trajectory in Fall River (after Julien and Anthony, in press.)
194
River equilibrium
size fraction are fairly parallel and oriented in the downstream direction. Near the apex, the bedload centerofmass curves are shifting across the channel. The angle difference between coarse and ﬁne grains can be calculated from the angle βˆ of the particlestability analysis in Example 6.1. The particlestability analysis determines the mean particledirection angle β, given the particle grain size, sideslope angle, downstream slope angle, and the shear stress. The results of calculations with the method in Example 6.1 with different size fractions for the Fall River bend are shown in Fig. CS.6.1.4. Finally, the crosssection geometry of the Fall River bends also depends on discharge. As shown in Fig. CS.6.1.5, increasing discharge causes erosion of the thalweg and sedimentation on the point bar. The larger ﬂow depth increases the deviation angle λ according to Eq. (6.21). Conversely, as the
120
110
Measured β
100
Toward point bar Particle size 0.125 mm 0.25 mm 0.5 mm 1 mm 2 mm 4 mm 8 mm 16 mm
90
80
Toward thalweg
70 Toward point bar
Toward thalweg 60 60
70
80
90
100
110
120
Average calculated β Figure CS.6.1.4. Particledirection angle in Fall River (after Julien and Anthony, in press.)
Lateral river migration
195
(a) Increasing discharge 6/7/86 6/1/86
0
Depth (m)
0.5 Point bar sedimentation 1.0 6/1/86 6/7/86
1.5 Cross section 9
2.0 2.5
Pool erosion 0
2
4 6 8 Distance across channel (m)
10
12
Depth (m)
(b) Decreasing discharge 0
6/7/86
0.5
7/17/86 3/30/86
1.0
Point bar erosion
1.5 Cross section 9 2.0 2.5
Pool sedimentation 0
2
4 6 8 Distance across channel (m)
6/7/86 7/17/86 3/30/86 10
12
Figure CS.6.1.5. Changes in crosssection geometry: (a) increasing discharge and (b) decreasing discharge (after Anthony, 1992).
discharge decreases, the angle λ decreases and the cross section gradually becomes more rectangular through erosion of the point bar and sedimentation in the pools.
Exercise 6.1
With reference to Fig. 6.2, derive Eq. (6.9) from τc ∼ Fs tan φ and R/(Fs cos θ1 ) = tan φ.
Exercise 6.2
Derive Eq. (6.11) from Eqs. (6.9) and (6.10); then substitute Eq. (6.12) into Eq. (6.11) to demonstrate that it is indeed the solution.
196
River equilibrium Exercise 6.3
Derive the expression for the radius of curvature in Eq. (6.33) from dx = Rdθ and the sinegenerated curve, Eq. (6.30). Exercise 6.4
Examine the ﬁeld observation in Figs. 6.9 and 6.14(c). Determine empirical ratios for the meander length as a function of channel width. Problem 6.1
Calculate the particledirection angle under an applied shear stress τ0 = 10 Pa, where the streamlines are deﬂected upward at λ = −10◦ . The downstream bedslope angle is θ0 = 0.05◦ and the sideslope angle is θ1 = 10◦ . Consider a particle of size 10 mm and compare the result with a 1mm particle under identical ﬂow conditions. Problem 6.2 Deﬁne the ideal crosssection geometry for a 100mm cobblebed canal with all particles at beginning of motion. The slope of the channel is 0.01. Also estimate the ﬂow discharge in this canal. (Assume that φ = 40◦ , n = 0.03). Answers: h 0 = 0.78 m, W = 2.9 m, A = 1.45 m2 , Rh = 0.42 m, V = 1.87 m/s, Q = 2.71 m3 /s, τ∗ = (G −h S1) ds = 0.047, incipient motion Problem 6.3
The Cache la Poudre River near Rustic, Colorado, has a bankfull discharge of 17.6 m3 /s, a width of 12.8 m, a depth of 0.65 m, and a slope of 0.0048 with a grain size of 150 mm. Compare the actual geometry with the regime equations and downstream hydraulicgeometry relationships. Problem 6.4
Use the regime relationships to calculate the hydraulic geometry of an irrigation canal that conveys 5000 ft3 /s in a very ﬁne gravelbed channel. Compare with the hydraulic geometry for stable channels. (Hint: τ∗ = 0.047 for stable channels.) Answers: V = 4.6 ft/s, R = 5.7 ft, P = 188 ft, S = 6.9 × 10−4 from the regime equations; V = 0.72 m/s, h = 1.78 m, W = 109 m, S = 1.3 × 10−4 from Eqs. (6.26) with ds = 3 mm.
Problems Distance (miles) 0
x
197
15
0 1 2 miles
x x x
x
5
x
30
10 20
Axis x
25
θ (degrees)
100 50 0
50
100 0
5
10 15 20 25 Distance (miles)
30
Figure P.6.5.1. Mississippi River near Greenville (after Langbein and Leopold, 1966).
Problem 6.5
From Fig. P.6.5.1, determine the orientation angle of the river at positions equally spaced by 1 mile and plot the values to complete the diagram. Fit a sine function and determine the values of θm and the origin of the downriver axis x. Determine L and , measure the sinuosity, and compare with calculations based on θm and Eq. (6.32). Measure the minimum radius of curvature Rm and compare with values calculated from L and θm . Compare the measured meander width with the calculations, assuming the sinegenerated curve. Problem 6.6
From the information presented in Case Study 6.1, determine the following: downstream angle θ0 , sideslope angle θ1 at high ﬂow, bed shearstress estimate τ0 , radius of curvature, and streamline deviation angle λ. Also calculate the downstream hydraulic geometry from the relationships of Section 6.5 and compare with ﬁeld observations. Determine the river length L, the meander length , the sinuosity, maximum planform deviation angle θm and meander width Wm , and compare with relationships in Section 6.7. Plot the ﬁeld observations on Fig. 6.9, 6.13, 6.14 and 6.15, and determine whether lateral mobility should be high or low. Answers: θ0 = 0.0745, θ1 15◦ from Fig. CS.6.1.5, τ0 ∼ =γhS∼ = −1 ◦ 12 Pa, Rm 15 m, λ tan , (11 h/R) = 36 at high ﬂow and λ 10◦ at low ﬂow. From Q = 7 m3 /s, ds = 8 mm and S = 0.0013, Eqs. (6.28) give h 0.52 m, W = 22 m, V = 0.55 m/s, and τ∗ = 0.05. The actual crosssection geometry is deeper and narrower than calculated. Planform geometry, L 103 m, 50 m, 2, θm 95◦ = 1.66 rad, and Wm 45 m. This gives Wm 4.5 W, = 5 W 3.5 Rm , Rm 1.5 W , which is very low. Field measurements compare well with the ﬁgures. Based on Fig. 6.16, the lateral mobility should be fairly low.
198
River equilibrium Problem 6.7
Determine the sinuosity of the channel in Fig. P.6.7.1. Locate the inﬂection points and locate the clockwise and counterclockwise loops. Determine the number of loops and the number of meanders. Determine the average river wavelength, meander length, meander width, and length of the meander belt. Compare meander width with that of Fig. 6.13. Anticipate where neck cutoffs might occur and determine where the oxbow lakes would be. Discuss the impact of possible neck cutoffs on hy0 100 m draulic geometry and sediment transport. Do you expect the lateral mobility Figure P.6.7.1. Example of meandering channel. of this river to be high or low?
7 River dynamics
Deviations from equilibrium conditions will trigger a dynamic response from the alluvial river system to restore the balance between inﬂowing and outﬂowing water and sediment discharges. Section 7.1 of this chapter deals with the dynamics of stream response to changes in water and sediment discharges. Sections 7.2 and 7.3 describe the dynamic response of alluvial systems to degradation and aggradation, respectively, particularly the effects on hydraulic geometry and channel morphology. Section 7.4 focuses on river conﬂuences and branches. Finally, Section 7.5 provides guidelines on river databases, data sources, and ﬁeld surveys. 7.1
River dynamics
Conceptually, the ﬂuvial system of the watershed sketched in Fig. 7.1 can be divided into three main zones: (1) an erosional zone of runoff production and sediment source; (2) a transport zone of water and sediment conveyance; and (3) a depositional zone of runoff delivery and sedimentation. The second zone is characterized by nearequilibrium conditions between the inﬂow and the outﬂow of water and sediment. The bed elevation in this equilibrium zone is fairly constant and the hydraulic geometry is described in Chap. 6 and this section. The upper zone is characterized by net erosion of bed material and channel degradation. The dynamic response of degrading ﬂuvial systems is discussed in Section 7.2. The lower zone is characterized by net sedimentation and channel aggradation. The dynamic response of aggrading ﬂuvial systems is discussed in Section 7.3. Conceptually, the hydraulic geometry of alluvial channels is related to stream characteristics that vary with time. As sketched in Fig. 7.2, the concept of dominant discharge used in Chap. 6 refers to bankfull discharge conditions or ﬂood discharge with a period of return of ∼1.5 yr. The corresponding sediment discharge is divided into washload for size fractions of bed material ﬁner than d10 . The bedmaterial discharge Q bv corresponds to size fractions of 199
200
River dynamics
Figure 7.1. Erosion, transport, and sedimentation in a ﬂuvial system.
Q
Qs
Bankfull discharge
Dominant discharge Q Period of return ~ 1.5 year
t Washload Bed d s < d10 material Bed material load discharge by volume d s > d10 Q bv t
% finer
100 Bed material Particle size distribution
50 10 0
d10 d s = d50
ds
Figure 7.2. Time changes in discharge and sediment transport.
bed material coarser than d10 . The bedmaterial discharge is most important because it relates to changes in bed conditions and thus to changes in hydraulic geometry. Section 6.4 dealt with equilibrium downstream hydraulic geometry. Accordingly, the relationship of Eqs. (6.26) among channel slope S, dominant discharge Q, grain size ds , and Shields parameter τ∗ can be rewritten as 1
5
4+6 m
S Q 2+3 m = 1.24ds4+6 m τ∗5+6 m .
(7.1)
We remember that the unit of the bankfull discharge Q is cubic meters per second and that of the median grain diameter ds is meters, whereas
River dynamics
201
the slope S and the Shields parameter τ∗ are dimensionless. After considering that the sediment discharge Q s is proportional to the Shields number τ∗ , we ﬁnd that the product of slope and discharge on the lefthand side of Eq. (7.1) must be balanced by the product of grain size and sediment discharge on the righthand side of Eq. (7.1), which is known as Lane’s (1955) relationship, written as Q S ∼ Q s ds .
(7.2)
This qualitative relationship states that equilibrium conditions exist between hydraulic conditions on the lefthand side of relation (7.2) and sediment conditions on the righthand side of relation (7.2). Perturbations to one or several parameters in relation (7.2) will be balanced by a change in one or several of the remaining parameters. A quantitative relationship between hydraulic and sediment variables is possible after deﬁning a sediment transport relationship. It is important to use a bedsediment discharge because riverbed changes are induced by erosion and/or sedimentation of bed material. The following empirical relationship for sand transport is used as a ﬁrst approximation: √ qbv 18 gds3/2 τ∗2 .
(7.3)
This relationship for the unit bedload discharge by volume qbv in square meters per second is approximately valid for 0.1 < τ∗ < 1, where the grain diameter ds is in meters. Accordingly, the bedload discharge by volume is Q bv = qbv W , and from relations (6.27) and (7.3) we obtain 6+4 m
−2.5+5 m 5+6 m
Q bv 0.77Q 5+6 m ds
7+10 m
S 5+6 m .
(7.4a)
Speciﬁcally for m = 1/6, this reduces to Q bv d S0.28 0.77Q 1.11 S 1.44 .
(7.4b)
We thus should apply this approximation only as a downstream hydraulic relationship among dominant discharge Q in cubic meters per second, the bedmaterial discharge by volume Q bv in cubic meters per second, and the grain diameter ds in meters. Relations (7.4) should not be used to deﬁne atastation properties such as sedimentrating curves. The sediment concentration in milligrams per liter, Cmg/l = 106 GCv = 6 10 G(Q bv /Q) can be obtained directly from relations (7.4) as Cmg/l 2.65 × 106 Q 0.11 ds−0.28 S 1.44 .
(7.5)
This relationship deserves testing and should be accurate within an order of magnitude as long as the sedimenttransport relationship and the approximation
202
River dynamics
m = 1/6 are applicable. Nevertheless, it clearly shows the dominant role played by the channel slope S in the conveyance of sediment in alluvial channels. We can assess the effects of sediment discharge Q bv on the downstream hydraulic geometry of sandbed streams after substituting S into Eqs. (6.27) with the function of the bedmaterial discharge Q bv in cubic meters per second, ds in meters, and Q in cubic meters per second from relation (7.4b). We obtain the solution for the bankfull ﬂow depth in meters, channel width W in meters, ﬂow velocity V in meters per second, and Shields parameter τ∗ after assuming that m = 1/6 and substituting S from relation (7.4b) into Eqs. (6.28): , h 0.19Q 0.46 ds0.13 Q −0.12 bv
(7.6a)
W 1.3Q 0.62 ds−0.15 Q −0.15 , bv
(7.6b)
V 4Q −0.08 ds0.02 Q 0.27 bv ,
(7.6c)
S 1.2Q −0.77 ds0.19 Q 0.69 bv ,
(7.6d)
τ∗ 0.14Q −0.31 ds−0.67 Q 0.57 bv .
(7.6e)
Alternatively, the effects of bedmaterial sediment concentration Cmg/l that correspond to the dominant discharge can be assessed after Q bv = 3.8 × 10−7 Cmg/l Q is substituted into relations (7.6a)–(7.6e). We simply obtain downstream hydraulicgeometry relationships as functions of bankfull bedmaterial sediment concentration in milligrams per liter, ds in meters, and Q in cubic meters per second: −0.12 , h 1.1Q 0.34 ds0.13 Cmg/l
(7.7a)
−0.15 W 12Q 0.47 ds−0.15 Cmg/l ,
(7.7b)
0.27 V 0.075Q 0.19 ds0.02 Cmg/l ,
(7.7c)
0.69 S 4.4 × 10−5 Q −0.08 ds0.19 Cmg/l ,
(7.7d)
0.57 τ∗ 3 × 10−5 Q 0.26 ds−0.67 Cmg/l .
(7.7e)
At sediment concentrations of less than 1,000 ppm, sediment concentrations in milligrams per liter Cppm can be used instead of Cmg/l . The difference between Cmg/l and Cppm is less than 10% at concentrations of less than 145,000 ppm.
River dynamics
203
Problem 7.2 illustrates how relations (7.7) can be used to estimate the characteristics of an alluvial channel, based on discharge, grain size, and bedmaterial concentration. We can thus infer from relations (7.6) that an increase in dominant discharge Q + is expected to cause a signiﬁcant increase in bankfull width W + and in depth h + , a signiﬁcant decrease in slope S − , and a lesspronounced decrease in Shields parameter τ∗− . An increase in dominant sediment discharge Q + bv corresponds to a signiﬁcant increase in slope S + and Shields parameter τ∗+ , a lesspronounced increase in velocity V + , and slight decreases in channel width W − and ﬂow depth h − . The effects of increases in grain size ds+ are comparatively less signiﬁcant, except for a decrease in Shields parameter τ∗− . In summary, we can expect the following dynamic responses of alluvial systems to perturbations in water and sediment discharges: Q + → W + h + S − τ∗− ,
(7.8a)
+ + + Q+ bv → S τ∗ V ,
(7.8b)
ds+ → τ∗− .
(7.8c)
We thus note that the downstream hydraulic geometry in width and depth primarily depends on discharge. Increases in water and sediment discharges exert counterbalancing effects on the channel slope and the Shields parameter. Grainsize effects are comparatively small. The opposite effects are of course obtained for decreasing values of the parameters. As sketched in Fig. 7.3, changes in hydraulic geometry take place through reworking of the alluvium in which rivers ﬂow. Flow depth increases as the bed degrades, and ﬂow depth decreases through aggradation. Channel widening will occur through bank erosion. Channel narrowing will probably take place through shoaling and the formation of bars and islands, followed by incision of the main channel. In general, shoals are submerged sandbars at low ﬂows. Sandbars and gravel bars are submerged at high ﬂow, and islands are covered with vegetation. Changes in ﬂow velocity are quite naturally linked to bedforms and the ability of a stream to transport sediment. Changes in bed slope, on the other hand, are quite problematic in the sense that they require adjustment of bankfull conditions and changes in the ﬂoodplain. As an example, consider an alluvial ﬁne sandbed channel that has a bankfull width W1 = 300 ft, a ﬂow depth h 1 = 10 ft, slope S1 = 8 ft/mile, and a ﬂow velocity V1 = 10 ft/s. Determine the expected change in downstream hydraulic geometry if the dominant ﬂow discharge is decreased by 50%. The bedmaterial size and the sediment concentration are expected to remain the
204
River dynamics
Figure 7.3. Adjustments in river width, depth, and ﬂow velocity.
same. The changes in hydraulicgeometry characteristics with Q 2 /Q 1 = 0.5 are calculated from relations (7.7): h 2 / h 1 ∼ = (0.5)0.34 = 0.80, W2 /W1 ∼ = (0.5)0.47= 0.19 −0.08 ∼ = 0.87, and S /S = 1.05. (0.5) (0.5) 0.72, V2 /V1 ∼ = 2 1 = Accordingly, the ﬂow depth is expected to decrease to ∼8 ft, the bankfull width should decrease to ∼210 ft, the ﬂow velocity should decrease to ∼9 ft/s, and the bed slope should slightly increase to ∼8.5 ft/mile for the new equilibrium condition. These expected changes only can be used as a ﬁrst approximation.
7.2
Riverbed degradation
Channel degradation refers to the general lowering of the bed elevation that is due to erosion. In some cases, the bed material is ﬁne and degradation will result in channel incision, which is covered in Subsection 7.2.1. In other cases, the material is sufﬁciently coarse to form an armor layer that prevents further degradation, as discussed in Subsection 7.2.2.
Riverbed degradation 7.2.1
205
Incised rivers
Slope adjustments refer to streams that would require either a steeper or a milder slope for reaching equilibrium between incoming and outgoing water and sediment discharges. Stated in simple terms, when the outgoing exceeds the inﬂowing sediment load, alluvial streams will scour bed material and degrade. Degradation results in channel incision and milder slopes, as sketched in Fig. 7.4.
Figure 7.4. Schematic of riverbed degradation.
206
River dynamics
Figure 7.5. Effects of dams downstream of alluvial reaches.
Riverbed degradation
207
Incised channels tend to be narrow and deep compared with equilibrium conditions. Channel incision will take place until equilibrium condition is reached. Incised channels are typical of upland areas whereby the sedimenttransport capacity increases in the downstream direction. Rills are smallscale channels found in upland areas. Gullies are largerscale features also found in upland areas. Conventionally, rills can be crossed by farm machinery whereas gullies cannot. In rivers, channel incision is found in arroyos and canyons. Arroyos are ephemeral channels in arid areas with ﬂashy hydrographs that carry large sediment loads during short periods of time. Many arroyos dry out in the downstream direction as a result of inﬁltration and evaporation. The sediment load eventually deposits on the channel bed downstream of arroyos to form wideshallow streams. Canyons are usually deeply entrenched in vertical bedrock walls. Incised channels typically are narrower and deeper then equilibrium channels and are characterized by a shortage of sediment. Channel degradation also causes the banks to become unstable and subject to failure. Gullylike incised channels become very unstable, and bank erosion may become a signiﬁcant source of sediment to the channel. Incised channels can often be found where the stream slope increases in the downstream direction. Knickpoints indicate points with a sudden change in bed slope. Headcuts usually refer to sudden drops in bed elevation. Headcuts usually start downstream, and their upstream migration is a characteristic feature of incised channels. As sketched in Fig. 7.5, artiﬁcial structures such as dams alter the equilibrium between the ﬂow of water and sediment in alluvial channels. Reservoirs tend to decrease the magnitude of ﬂood ﬂows and increase low ﬂows. The clearwater release from the dam also causes the reach below the dam to degrade in the form of a wedge starting below the dam. The magnitude and the extent of the degradation below dams depend on the reservoir size and operation and on the size and availability of alluvium below the dam. Incision tends to be deepest in sandy materials and is subject to armoring in gravels. Degradation of the main river stem at river conﬂuences causes headcutting and degradation in the tributaries. As sketched in Fig. 7.6, the headcut propagates upstream from the conﬂuence and can cause severe stability problems in structures on shallow foundations such as bridges and some gradecontrol structures. The ensuing gullying in a tributary can cause signiﬁcant bank inFigure 7.6. Schematic of headcut mistabilities and channel widening. gration.
River dynamics
50 Simmesport Stage (ft)
40
Q = 400,000 ft 3/s
30
15
10
Q = 200,000 ft 3/s
20
5 10
3
3
1 m /s = 35.5 ft /s 0 1936
1944
Stage (m)
208
1952
1960
1968 Year
1976
1984
1992
0 2000
Figure 7.7. Speciﬁcgauge record of the Atchafalaya River at Simmesport (after U.S. Army Corps of Engineers, 1999).
Speciﬁcgauge records are often used to determine whether a stream tends to aggrade or degrade over time. A speciﬁcgauge record is the watersurface elevation that corresponds to a given discharge. When gauge records are available for a long period of time, plotting the gauge elevation at given discharges can detect longterm river trends. For instance, the speciﬁcgauge record of the Atchafalaya River is shown in Fig. 7.7. Both discharges show a gradual lowering of the watersurface elevation with time. Over a period of 50 yrs, the watersurface elevation is ∼15 feet lower than in the mid1950s. When the procedure is repeated at several stream gauges along a river, longitudinal watersurface proﬁles at a speciﬁc discharge can be obtained. The example in Fig. 4.7(a) illustrates the watersurface lowering in the Atchafalaya River from Simmesport to Chicot Pass and the rising in watersurface elevation between Chicot Pass and Morgan City. It should be noted that speciﬁcgauge records do not provide information on bed elevation, and, per se, aggradation or degradation changes cannot be determined from speciﬁcgauge records. Changes in bedforms, bed material, resistance to ﬂow, looprating effects, and channel widening can alter the trends observed with speciﬁc gauges. It is possible to observe opposite trends at different discharges; for instance, speciﬁc gauges may indicate a decreasing trend at a low discharge and an increasing trend at a high discharge. It is thus recommended to compare cross sections over time in order to conﬁrm any trend detected with speciﬁc gauges.
7.2.2
Riverbed armoring
Armoring of the bed layer refers to coarsening of the bedmaterial size as a result of degradation of wellgraded sediment mixtures. The selective erosion of ﬁner particles of the bed material leaves the coarser fractions of the mixture on the
Riverbed degradation
209
bed to induce coarsening of the bed material. When the applied bed shear stress is sufﬁciently large to mobilize the larger bed particles, degradation continues; when the applied bed shear stress cannot mobilize the coarse bed particles, an armor layer forms on the bed surface. The armor layer becomes coarser and thicker as the bed degrades until it is sufﬁciently thick to prevent any further degradation. The armor layer is representative of stable bed conditions and can be mobilized only during large ﬂoods. A riverbed is sometimes said to be paved when the armor layer can be mobilized only during exceptional ﬂoods. Three conditions need to be satisﬁed to form armor layers: (1) the stream must be degrading, (2) the bed material must be sufﬁciently coarse, and (3) there must be a sufﬁcient quantity of coarse bed material. Relative to the ﬁrst condition, the sedimenttransport capacity must exceed the sediment supply such that the stream attempts to scour the bed. The second condition can be quantiﬁed as follows from the Shields diagram. The incipient condition of motion with τ∗c 0.05 can be rewritten in terms of minimum grain size at the beginning of motion. dsc 10 h S,
(7.9a)
where dsc is the minimum grain diameter, h is the ﬂow depth during ﬂoods, and S is the channel slope. The units of grain size are the same as those of the ﬂow depth. Alternatively, we can estimate the ﬂow depth corresponding to the beginning of motion by h dsc /10 S.
(7.9b)
The third condition refers to the fraction of material pc coarser than dsc available in the bed material. When this percentage is large, the armor layer will form rapidly and the extent of degradation will be minimal. When this percentage is low, a large volume of bed material will be scoured before the armor layer can form. The effect of the armor layer in this case will be limited. Quantitatively, we can consider that an armor layer of approximately twice the grain size will stabilize the bed. The scour depth z that will form an armor layer equal to 2dsc can be estimated from 1 −1 . (7.10) z = 2dsc pc The scour depth becomes very large when pc is small, and it is therefore important to have a particlesize distribution that is representative of the sublayer, including clay, sand, and gravel layers. Example 7.1 illustrates some characteristics of river degradation and armoring.
210
River dynamics
Once an armor layer has formed, it plays a very important role in channel stability and morphology. Indeed, the riverbed is stable except under large ﬂoods, and the armor layer protects the bed against further degradation. Its removal through natural and/or artiﬁcial activities can cause signiﬁcant streaminstability problems. For instance, gravel mining operations remove the coarse armor layer from bed streams at low ﬂows. The damage can often be seen during subsequent ﬂoods in which large discharges cause severe degradation in the channel. Headcut development and upstream migration can also cause the failure of upstream structures such as bridges. Case study 7.1 illustrates the complex effects of dams, landuse changes, and gravel mining on riverchannel morphology. Example 7.1 Application to riverbed degradation. The Meuse River in The Netherlands is 250 km long. The mean annual discharge is 230 m3 /s near Maastricht and exceeds 3,000 m3 /s during major ﬂoods. The Meuse is more or less controlled by weirs to enable river navigation, and the lower stretches of the Meuse have dikes. From Maastricht to Maasbracht, the Meuse meanders over shallow gravel banks, and barge trafﬁc is possible only in a parallel canal. The river is easily navigable between Maasbracht and Lith. Near Maasbracht, the river shows a sharp transition from a gravelbed to a sandbed river with d50 decreasing from ∼16 to 3 mm and the bed slope decreasing from 48 to 10 cm/km. Figure E.7.1.1 from MurrilloMu˜noz (1998) shows several characteristics of the bed material of the Meuse River. At a ﬂow depth of 3 m, the critical grain size at the beginning of motion is calculated from the slope and relations (7.9). In this case, dsc 12 × 3 m × 48 × 10−5 = 17 mm for the upper reach and dsc 12 × 3 m × 10 × 10−5 = 3 mm for the power reach. It is interesting to note that these size fractions approximately correspond to d50 of the bed material. To examine whether armoring is possible during ﬂoods, the critical grain sizes at a ﬂood ﬂow depth h = 6 m are 25 and 7 mm. Considering the particlesize distribution of the lower reach, particle sizes exceeding 35 mm cannot be found in large quantities and degradation can be expected during ﬂoods. Considering that only ∼3% of material, pc 0.03, is coarser than 35 mm in the lower reach, the extent of degradation calculated from Eq. (7.10) is z 2.3 m, which is comparable with the degradation measurements shown in Fig. E.7.1.1. Note that Eq. (7.10) is very sensitive to low values of pc and an inﬁnite degradation depth is obtained when pc → 0. It is therefore very important to carefully determine the particlesize distribution of coarse bed fractions. When several meters of degradation are expected, borings are required for examining the substrate and looking for possible gravel and cobble layers that can limit the extent of
Riverbed degradation 10
211
2
d 50 (mm)
Gravel 10 1 1
Sand
2
Silt
10
10
0
10
20
30
100
40
50
60 70 80 90 100 110 120 130 140 150 River chainage (km)
Sand
Gravel
% Passing
80 60 km 145
40
km 88
km 110
km 46.9
20
Thalweg elevation (m + NAP)
10 15
Thalweg elevation (m + NAP)
0 1 10
14 12 10 8 6 4 2 0 70
1
Grain diameter (mm)
2
10
10
40 S ~ 48 cm/km
35 30
1972
25
19091916
20 15
1995 Rijkswaterstaat surveys 20
25
30
35 40 45 50 River chainage (km)
1909 1916
55
60
65
70
S ~ 10 cm/km
1972 1995
75
80
85
90
95 100 105 110 115 120 125 130 135 140 145 River chainage (km)
Figure E.7.1.1. Meuse River in The Netherlands (after MurilloMu˜noz, 1998).
212
River dynamics
degradation. The presence of bedrock outcrops along the river proﬁle can be most useful when severe degradation is expected. In the ﬁeld, quick estimates of the ﬂow depth required for beginning of motion of the bed material are obtained by h m 10 ds /Scm/km ,
(E.7.1.1)
where the ﬂow depth h is in meters, the median grain diameter is in millimeters, and the slope is in centimeters per kilometers. In this example, the ﬂow depth for the upper reach with ds 16 mm and S = 48 cm/km is approximately 3 m. The ﬂow depth for the lower reach with ds 3 mm and S = 10 cm/km is also approximately 3 m. Downstream ﬁning in this river seems to follow the empirical rule that ds /S is constant. Case Study 7.1 Dynamic changes of Dry Creek, United States. Dry Creek is a major tributary to the Russian River just south of Healdsburg, California (Fig. CS.7.1.1). The Russian River Basin drains approximately 235 km2 and Dry Creek has a drainage area of 34 km2 . (b)
sian
ry cr ee
Rus
D k
Lambert bridge
Elevation (ft above msl)
Rive
r
Warm Springs Dam
Westside Bridge
Healdsburg Dam Sand and gravel mining
80
60
1972
40 30 22
24
15
10 34
26 28 30 32 Russian River miles
Maximum
800
(m) 200
700 Width (ft)
3 ft (~ 1 m)
Lambert bridge
Westside Bridge
(d) 900 Dry creek mouth
Russian River
1940
50
(Not to scale)
(c)
(m) 20
70 Dry creek Healdsburg Dam
(a)
600 Average
500 400
Minimum
300 18 ft (5.5 m)
Rock outcrop (Not to scale)
150
200 100 1940
1973
1950
1960 Year
1970
100 50
1980
Figure CS.7.1.1. Russian River in California (after Simons and Julien, 1983).
Riverbed degradation
213
Healdsburg Dam serves as a control on the Russian River, and Dry Creek will be even more signiﬁcantly altered when the Warm Springs Dam is completed and put in operation. Signiﬁcant landuse changes are also evident from the aerial photographs. The Dry Creek valley is agricultural, with citrus fruits as the major crop. The upland areas are rugged, consisting primarily of hilly and mountainous terrain. Recently this condition has been aggravated by record storms and wildﬁre, both contributing to abnormal runoff volumes. Other landuse changes include some urbanization and the conversion of forest lands to agricultural and grazing lands. Dry Creek is located in a valley of high relief (>300 m) on a welldeﬁned wide ﬂoodplain averaging 1.2 km in width (Fig. CS.7.1.1). The lowﬂow channel is incised within a widerﬂow channel bordered by low scarps. The lowﬂow channel has a sinuosity of approximately 1.20, whereas the wider ﬂood ﬂow channel is straight (sinuosity approximately 1.05). Aerial photographs indicate that Dry Creek is extremely active. This corroborates the ﬁndings in Fig. 6.16. Although the overall channel sinuosity remains approximately the same from year to year, a comparison of the location of meander bends and crossovers reveals that their positions change with time. Channel boundaries are alluvial, being composed of sand and gravel. The lowﬂow channel is locally braided and locally anabranched. Both the lowﬂow and the ﬂoodﬂow channels are equiwidth with the development of wide point bars prevalent along the lowﬂow channel. Tree cover is generally less than 50% of the bankline, and cut banks are evident. Bank material is generally noncohesive silt, sand, gravel, and cobbles. Dry Creek is a braided stream that is overloaded with sands and gravel. The deposition of sediments from the overloaded condition creates gravelbar formation and aggradation of the channel bed. The presence of the gravel bars enhances lateral migration of the channel. These conditions create severe bankerosion problems because of the highly erodible bank. Bankerosion problems were detected more than 30 km upstream of the conﬂuence with the Russian River and have been documented from records and photographs as early as 1940. In addition, the similarity of degradation and lateral migration tendencies between the areas upstream and downstream of the Lambert Bridge (control) indicates the signiﬁcance of natural channel instability. Gravel mining in the Russian River and other tributaries such as Dry Creek has been an important industry in Sonoma County since the early 1900s. In recent years, such activities have continued in Dry Creek and in the Russian River near the mouth of Dry Creek. There is evidence of gravel mining near the conﬂuence of Dry Creek and the Russian River, resulting in a general deepening of the Russian River. This is due to the closing of the
214
River dynamics
Healdsburg Dam in 1952 and the Coyote Dam in 1958. The lowering of the base level in the Russian River has induced a general lowering of the base of Dry Creek. The channel of Dry Creek downgraded signiﬁcantly in the 1950s and 1960s because of the drop in the base level of the Russian River. This drop was the result of instream gravel mining and the construction and operation of the Healdsburg and Coyote Dams. A total streambed elevation drop of 7 m has been recorded in a 0.6km length near the Healdsburg Dam since 1940. The current Healdsburg Dam structure, completed in 1952, has maintained the past upstream bed elevation and is acting as a control that prevents upstream headcuts in the Russian River. However, the baselevel drop in the Russian River initiated a headcut in Dry Creek that has propagated a total distance of 13 km upstream from the mouth. The headcut was controlled when rock just upstream of the Lambert Bridge was exposed in 1972–1977. Calculations indicate that the headcut traveled at an estimated rate of 0.6 km/yr, requiring 23 yrs to travel the entire 13 km. With the deeper channel system and with controlled ﬂooding, higher banks became exposed to attack by the ﬂowing water. Abnormal ﬂooding and ﬁre sequence produced record runoff and sediment that has caused the deeply incised lower ends of Dry Creek to begin to widen again from extensive bank erosion. Analysis of aerial photographs indicates a signiﬁcant increase in channel width. Qualitative analysis reveals that the erosion problems along Dry Creek are complicated. The accelerated channel degradation has been primarily caused by (1) lowering of the base level that is due to the construction of dams on the Russian River; (2) increased runoff resulting from record storms, complicated by landuse changes and wildﬁres; and (3) gravel mining activity on the Russian River and Dry Creek.
7.3
Riverbed aggradation
Channel aggradation refers to a gradual bedelevation increase that is due to bedload sedimentation. This section considers braiding in Subsection 7.3.1 and alluvial fans and deltas in Subsection 7.3.2.
7.3.1
Braided rivers
When the inﬂowing sediment discharge exceeds the outgoing sediment capacity, alluvial channels tend to deposit their sediment load throughout the reach. Streams carrying mostly washload will not change their morphology because
Riverbed aggradation
215
the sediment overload will be carried downstream to settle in lakes, reservoirs, or estuaries. Streams carrying most of their sediment load in suspension change their morphology gradually as the excess sediment load settles in the downstream direction. The riverbed material size becomes gradually ﬁner in the downstream direction. From Lane’s relationship, downstream ﬁning is usually accompanied by a downstream decrease in bed slope. On the other hand, streams that carry predominantly bedload material will respond quite rapidly to a change in sedimenttransport capacity. A decrease in transport capacity induces direct settling on the bed of alluvial channels. As sketched in Fig. 7.8, the settling of bedload forces aggrading channels out of the bankfull conditions. The ﬂow spreads on the ﬂoodplain with accumulation of the bedsediment load to form natural levees on a wide ﬂoodplain. There is a tendency for the stream to widen and become very shallow with bars subjected to rapid changes in morphology. At high ﬂows, braided streams have a low sinuosity and often appear to be straight. At low ﬂows, numerous small channels weave through the exposed bars. These streams are known to braid as the bed slope increases through aggradation. The ﬂow velocity of braided streams is high, and the bed material can be easily mobilized. Braided streams are rather unstable in that they are prone to severe lateral migration, frequent shifts, and changes in crosssection geometry. The bars of braided streams are generally submerged once a year and are devoid of vegetation. Islands are different from bars in that they are stabilized by vegetation and rivers with multiple islands are anastomosed. Anastomosed rivers are usually more stable than braided channels because vegetation straightens the banks and stable islands control the ﬂow between the branches. During ﬂoods, vegetated islands trap sediment and aggrade. Because braided channels require large bedload transport, most braided rivers are steep, and therefore, at a given discharge, braided rivers should be steeper than meandering rivers. Several criteria based on bankfull discharge and slope are shown in Fig. 7.9. The range of slope variability, however, is quite extensive, and it remains difﬁcult to separate braiding from meandering channels solely on the basis of bankfull discharge and slope. An alternative approach is based on the width–depth ratio. In general, braided channels have a width–depth ratio in excess of 100. Case Study CS.7.3 illustrates several features of a large braided river. A single channel with given dominant discharge is thus thought to meander on mild slopes and braid on steep slopes. This concept has been expanded by Lane (1957) who proposed a slopedischarge relation for sandbed channels. Empirically, braided channels were observed when S Q 1/4 > 0.01 and channels
216
River dynamics
Figure 7.8. Schematic features of riverbed aggradation.
were meandering when S Q 1/4 < 0.0017, given the slope S and the dominant discharge Q in cubic feet per second. Unfortunately, this criterion is not always valid. Lane’s diagram is considered only as an index describing the morphological pattern.
Riverbed aggradation
217
2
(a)
10
Lane
Leo
pol
d&
Slope (m/m)
3
10
Braided Wo lma n SQ 0.25
Lane
Transitional 4
= 0.0
SQ 0. 44
041
=0 .01 25 SQ 0.25 = Meandering 0.000 70
10
5
10
2
Sinuosity
(b)
4
3
5
10 10 10 10 Mean annual discharge (m 3/s)
1
Schumm
10
Meandering thalweg channel
Thalweg sinuosity
Point bar Straight
Meandering Combination Braided thalweg of meandering channel channel and braided
Thalweg
Slope Channel type Mixed load
Schumm
1
Widthdepth ratio Gradient High High
Channel pattern meandering
2
Braided
Bed load
3a 3b Legend 4 Channel boundary Flow Bars 5 High Relative stability (3% >) Low Small Small Low Low
Bed load total load ratio Sediment size Sediment load Flow velocity Stream power
Meander shift Meander shift chute cutoff neck cutoff
Low Low
Straight
Suspended load
High Low Relative stability Channel shift Alternate bar avulsion shift
(c)
Low High (> 11%) Large Large High High
Figure 7.9. Braiding and meandering: (a) predictors of Lane (1957) and Leopold and Wolman (1960), (b) sinuosity diagram of Schumm (1977), and (c) relative stability diagram of Schumm (1977).
River dynamics (a) 100
(b)
% by weight finer
Sand 80
Gravel
Bed load average 75% 197781
60
50% 40 Bed material average 197981
20
0 1 2 4 8 16 32 64 0.062 0.25 0.125 0.5 Sieve size (mm) (c)
Bed material d 75 (mm)
218
50 40 30 20 10 0 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Slope (ft per 1000 ft)
600 Braided pattern sandy gravel bed
Anastomosing pattern gravelly sand bed
550
Elevation (ft)
S ~ 0.0013 500
Transitional length of rapid change in slope, channel pattern and bed material S ~ 0.001 d 75 = 40 mm
450 S ~ 0.0005 400
350 0
d 75 = 20 mm
S ~ 0.0003 d 75 = 10 mm
10 20 30 40 50 Length along middle of active alluvial system (miles)
60
Figure 7.10. Bed characteristics of the Tanana River in Alaska (after Buska et al., 1984).
Direct measurements of bedload transport and comparisons with transport capacity are too uncertain to provide any good indication of a stream tendency to braid. One useful indicator is that braiding often occurs near a sudden decrease in slope, provided that the stream carries excess bedload material. An example of a sudden decrease in bed slope with downstream ﬁning and braiding is shown in Fig. 7.10 for the Tanana River in Alaska. In this case, from Buska et al. (1984), channel aggradation takes place with selective deposition of coarser material upstream. Note that the bed slope is proportional to the grain size. The magnitude of slope and grain size is quite comparable with those of the degrading Meuse River in Case Study 7.1.
Riverbed aggradation 7.3.2
219
Alluvial fans and deltas
Alluvial fans are found where steep mountain channels reach valley ﬂoors, as sketched in Fig. 7.11. The sudden break in bed slope causes the bed material transported by the river to deposit. The accumulation of debris usually takes a conical shape. The volume of material in the alluvial fan is indicative of the sedimenttransport capacity of the Apex stream through geologic times. The Alluvial fan aggradation takes place on the riverbed and on natural levees between the apex of the alluvial fan and the valley ﬂoor. Valley floor An example of an alluvial fan is discussed in Case Study 7.2. Figure 7.11. Sketch of an alluvial fan. Through aggradation and natural levee formation, a river raises its own bed elevation above the surrounding ﬂoodplain to form a perched river. Perched rivers are stable as long as they cannot breach their levees. Perched rivers are prone to avulsion in which rivers select a new ﬂow path that can be located up to hundreds of kilometers away from their original river courses. Old channels of perched rivers rapidly dry out, and the process of aggradation and natural levee formation starts at the new river location. To some extent, tectonic activities exert a similar inﬂuence on alluvial river morphology as aggradation and degradation (sketched in Fig. 7.12). Uplift and subsidence on the side of a river may result in river perching and may result in river avulsions. Uplift along the water course should cause aggradation and possible braiding upstream and degradation and possible incision downFigure 7.12. Sketch of possible tecstream. Subsidence along a river course tonic effects.
220
River dynamics
should cause the opposite effects, with degradation upstream and aggradation downstream. Deltas are observed when rivers reach large lakes, reservoirs and oceans. The sediment deposits extend in a deltaic form into the water, and the aggradation in the upstream river causes the deposits to spread laterally in the lower reaches of the river. The features of deltas and alluvial fans are quite similar where the valley ﬂoor is replaced with the water level and deltas are much ﬂatter than alluvial fans. The delta deposits are usually ﬁne (ﬁner than sand) as opposed to alluvial fans (coarser than sand). Rivers transporting large quantities of washload may remain sinuous and meander to the river mouth. Rivers transporting large quantities of bed material tend to braid. Case Study 7.2 Lawn Lake Dam failure, Colorado, United States. Jarrett and Costa (1986) documented the failure of Lawn Lake Dam in Rocky Mountain National Park, Colorado, on the morning of July 15, 1982. The dam released 674 acreft (8.3 × 10−5 m3 ) of water at an estimated peak discharge of 18,000 ft3 /s (500 m3 /s) down the Roaring River valley. Three people were killed and damages totaled $31 million. The probable cause of failure was deterioration of the lead caulking used for the connection between the outlet pipe and the gate valve. The resulting leak eroded the earthﬁll, and progressive piping led to failure of the embankment. Floodwaters from Lawn Lake Dam overtopped a second dam, Cascade Lake Dam, located 6.7 miles downstream, as shown in Fig. CS.7.2.1. Cascade Lake Dam, a 17fthigh, concrete gravity, 12.1acreft capacity dam, failed by toppling with 4.2 ft of water ﬂowing over its crest. The ﬂood continued down the Fall River and caused extensive damage from overbank ﬂow to the city of Estes Park. Peak discharges were determined with a variety of indirect methods including a dambreak model. Peak discharges for the ﬂood were estimated at (1) 18,000 ft3 /s from Lawn Lake Dam; (2) 12,000 ft3 /s at Horseshoe Falls where Roaring River joins the Fall River; (3) 7,210 ft3 /s into Cascade Lake Dam at the east end of Horseshoe Park; (4) 16,000 ft3 /s after the failure of Cascade Lake Dam; (5) 13,100 ft3 /s ∼1 mile downstream from Cascade Lake Dam; and (6) 8,520 ft3 /s just upstream from Estes Park. Maximum depths ranged from 6.4 to 23.8 ft, maximum widths ranged from 97 to 1,112 ft, and mean velocities ranged from 3.3 to 12.6 ft/s. Travel times of the ﬂood were determined from eyewitness accounts. The leading ﬂoodwave took 3.28 h to travel 12.5 miles (average 3.8 miles/h). Flood peaks were 2.1 to 20 times the 500yr ﬂood for selected locations along the ﬂood path. It probably was the largest ﬂood that these basins have experienced since the retreat of the glaciers several thousands years ago.
River conﬂuences and branches
N
on
Cr ee k Rive
r
Roar i
sil
ng R iver
Yp
Lawn Lake Lawn Lake Dam
Fall
Horseshoe Falls Alluvial fan Olympus iver on R s Dam p m Tho Big Estes Park
ve Ri
Horseshoe Park Cascade Lake and dam
ll Fa
Thousand Falls
221
r
er
1
0 0
1
n pso
2 miles 2 km
Riv
Lake Estes
om
Big
Th
Figure CS.7.2.1. Lawn Lake Dam failure.
In the Roaring River valley, alternate river reaches were either scoured or ﬁlled, depending on valley slope. Channels were widened tens of feet and scoured from 5 to 50 ft locally. Generally, reaches steeper than 7% were scoured and reaches less than 7% were ﬁlled. In the Roaring River, 56% of the channel reach was scoured, some by as much as 50 ft, and 44% was ﬁlled with coarse sediments, 2 to 8 ft thick. An alluvial fan extending 42.3 acres contained 364,600 yd3 of material deposited at the mouth of the Roaring River. The fan had a maximum thickness of 44 ft and an average thickness of 5.3 ft. A 452ton boulder measuring 14 × 17.5 × 21 ft moved with the ﬂood and was deposited on the alluvial van. Down the ﬂow axis, average particle sizes changed from 7.5ft boulders at the fan apex to ﬁne sand and silt within a distance of 1,900 ft. The alluvial fan dammed the Fall River, forming a 17acre lake upstream from the fan.
7.4
River conﬂuences and branches
The effects of channel narrowing and widening are quite important in alluvial rivers. Indeed, a widening stream tends to aggrade and a narrowing tends to degrade. The combination of these two mechanisms stabilizes the river width, and this is the primary reason why rivers tend to have fairly constant widths over long river reaches.
222
River dynamics
River conﬂuences in equilibrium have to convey the water and sediment discharges in the downstream direction. When the sediment concentration and particle size are the same, relations (7.7) can be used to determine the expected changes in hydraulic geometry from an increase in discharge while keeping ds and Cppm constant. We thus obtain that river conﬂuences will cause a signiﬁcant increase in bankfull width, an increase in ﬂow depth and shear stress, and a slight increase in ﬂow velocity. The channel slope should also slightly decrease. The example of the conﬂuence of the Ganges and Jamuna Rivers ﬂowing into the Padma River is shown in Case Study 7.3. In particular, Table CS.7.3.1 conﬁrms the expected changes in width, depth, velocity, and slope. River branches under equilibrium condition have to convey water and sediment discharges in the downstream direction. From relations (7.7), with constant grain size and sediment concentration, the opposite changes are expected from river branching: (1) signiﬁcant decrease in river width, (2) moderate decrease in ﬂow depth and shear stress, (3) slight decrease in ﬂow velocity, and (4) slight increase in riverbed slope. Deviations from equilibrium conditions are possible from river captures that result from a new conﬂuence as a result of avulsion of a tributary. At a new river capture, we can expect degradation and channel widening downstream of the new conﬂuence. Likewise, ﬂow diversions out of alluvial channels will trigger sedimentation below the diversion. Channel aggradation, narrowing, reduced ﬂow velocity, and gradual steepening of the river are the expected changes in channel branches. As an example, consider the sediment diversion of the Mississippi River into the Atchafalaya River. Approximately 22% of the ﬂow discharge of the Mississippi River is diverted into the Atchafalaya River. The sand concentration in the Mississippi River varies largely with discharge, as shown in Fig. 7.13(a). Sand concentrations remain less than approximately 200 ppm at all discharges less than 1,000,000 ft3 /s. Sand concentrations increase approximately linearly with discharge. The effect of a ﬂow diversion should be to decrease the sand concentration downstream of the diversion. An analysis of sediment transport by size fraction and a sediment budget are shown in Fig. 7.13(b). The very ﬁne sand and ﬁne sand fractions are in reasonable equilibrium, and the inﬂowing sediment load equals the outﬂowing sediment load. For the medium sand and coarse sand fractions, the inﬂowing sand load of ∼19 million yd3 /yr far exceeds the outﬂowing of 2.25 million yd3 /yr. We thus expect the sedimentation of ∼17 million yd3 of medium and coarse sand, 0.25 mm < ds < 1 mm. In a river reach that is 262 miles long and a river
River conﬂuences and branches 5
Suspended sand concentration (ppm)
(a)
10
223 10
15
3
30 x 10 m 3/s
20
2
10
1 10
6
5
(b)
Fine and very fine sand 0.0625 < ds < 0.25 mm 40
ippi
siss
6.5
(million cubic yards per year)
Medium and coarse sand 0.25 < ds < 1.0 mm
Vicksburg R. mile 437
19
0.7
r
Rive
Atchafalaya River
Mis
Diversion to the
10
Discharge (cfs)
36
Donaldsonville R. mile 175
1.5
Figure 7.13. Sediment diversion at the Old River Control Complex (after U.S. Army Corps of Engineers, 1999). (cfs, cubic feet per second.)
that is ∼2,000 ft wide, the accumulation of sediment represents an average accumulation of sand of 0.16 ft/yr. The tendency toward braiding is possible, but the Mississippi River has a low width–depth ratio (∼2,000 ft wide and 50 ft deep). At this rate, it would take 100 yr to raise the bed by 16 ft. Case Study 7.3 Alluvial changes of the Jamuna River, Bangladesh. The Jamuna River is the lowest reach of the Brahmaputra River in Bangladesh. It drains an area of 550,000 km2 , and the mean annual discharge is 20,000 m3 /s. It is a large braided sandbed river, and the number of braids at low ﬂows
224
River dynamics
Table CS.7.3.1. Characteristics of the Jamuna, Ganges, and Padma Rivers
River
Drainage area (km2 )
Mean annual discharge (m3 /s)
Bankfull discharge (m3 /s)
Slope (cm/km)
ds (mm)
Width (m)
Depth (m)
Velocity (m/s)
Jamuna Ganges Padma
550,000 1,000,000 1,550,000
20,000 11,000 28,000
48,000 43,000 75,000
7.5 5 4.5
0.20 0.14 0.10
4,200 3,700 5,200
6.6 6.5 7.5
1.70 1.78 1.93
Jamua
Old Brahmaputra Bhairab Bazar
Hardinge Bridge Gang
es
Gorai Offtake
Tilly Dha Aricha lesw ari (Teota) Baruria Dhaka
Pa
dm
Arial Khan Offtake
0
20
40 km
Upper Meghna
a Mawa
Lower Meghna
Figure CS.7.3.1. Large river conﬂuences and branching in Bangladesh.
varies between 2 and 3, as shown in Fig. CS.7.3.1. The total width of the braided channel pattern varies between 5 and 17 km. At the conﬂuence with the Ganges, the average annual ﬂood is ∼60,000 m3 /s and lowﬂow discharges vary between 4,000 and 12,000 m3 /s. The maximum discharge recorded in 1988 reached 100,000 m3 /s. The watersurface slope decreases from 10 to 6 cm/km and the bed material is quite uniform with d50 ∼ = 0.25 mm near the Indian boarder and 0.16 mm at the conﬂuence with the Ganges River (See Table CS.7.3.1). Klaassen et al. (1993) report that the Jamuna River is quite active, with frequent channel shifts and lateral migration rates E frequently exceeding 500 m/yr. The shifting rate of the ﬁrstorder channel is 75 to 150 m. Bankerosion rates of secondorder channels of 250 to 300 m are common. Lateral
River conﬂuences and branches
225
migration rates exceeding the channel width W in 1 yr have been measured, and, in general, E/W decreases with R/W , where R is the channel radius of curvature. An example of lateral shift is shown in Fig. CS.7.3.2. Signiﬁcant changes in crosssection geometry can take place within a few years. Examples of braid conﬂuences and branching are shown in Fig. CS.7.3.3. The conﬂuence shown in Fig. CS.7.3.3 (a) did not move signiﬁcantly upstream or downstream. In general, slackwater zones are behind wide bars and prone to quick deposition within a year or so. Conﬂuences thus tend to be streamlined. The branching shown in the same ﬁgure migrates upstream at a rate of ∼900 m/year. In general, symmetrical branches move upstream. Asymmetrical bifurcations are characterized by one dominant channel in the main downstream direction. The small channel tends to bifurcate at a right angle from the main ﬂow direction. These smaller channels tend to get smaller in size and eventually disappear. The characteristics of the Ganges and the Padma Rivers are also shown in Table CS.7.3.1 for comparison. The reader will note that the sediment size is ﬁner in the Padma River, and the slope is accordingly reduced. The width, depth, ﬂow velocity, and coarse sediment transport of the Jamuna River are shown in Fig. CS.7.3.4. The ﬁgures indicate a signiﬁcant variability in measurements that are typical of measurements in alluvial rivers. As much as there is debate as to what the dominant discharge of a stream really is, the reader will note that, at a given discharge near 50,000 m3 /s, the variability in width and depth is approximately a factor of 2. This illustrates the fact that the dominant discharge and downstream Figure CS.7.3.2. Lateral migration of hydraulicgeometry relationships are the Jamuna River (after Klaassen et al., 1993). not exact features of alluvial rivers.
226
River dynamics (a)
N 1977
(b) 1977
1978 1984 Reference point Crosssection Previous situation Bare sand
1986
1978
1986
1984
0
1987 5 km 1987
N
Figure CS.7.3.3. Conﬂuences and branching of the Jamuna River (after Klaassen et al., 1993).
7.5
River databases
River databases for the analysis of channel stability include (1) historical developments, (2) maps and photos, (3) streamﬂow data, (4) sediment data, and (5) ﬁeld surveys. Historical information affecting channel morphology and stability should be reviewed. Upstream basin information includes landuse changes, ﬂow diversions, and artiﬁcial structures for ﬂood control, irrigation, and navigation. Topographic maps at various scales, depending on the stream size, indicate the nature of the drainage area and the ﬂuvial system. The stream planform geometry, longitudinal proﬁle, and estimates of channel slope can be obtained from topographic maps with contours. Aerial photographs are most useful in examining sediment deposits. The comparison of several sets of aerial photographs enable the evaluation of the lateral migration rates of alluvial channels. Geographic information systems (GIS) are useful in examining topography, soil types, and landuse data. The analysis of lateral migration in very large rivers, like the Jamuna River, is sometimes possible with GIS data. Satellite imagery is sometimes useful in the analysis of watershed data and some turbidity and ﬂow patterns in lakes and estuaries.
River databases
227
4
Width (m)
10
W ~9 Q
0.57
3 (a)
Coarse sediment (million tons/day) Cross section average velocity (m/s)
Depth (m)
10 10
h ~ 0.4 Q
0.26
(b)
1 10
V ~ 0.27 Q 0.17 1
1 (c)
10 10 1
1
10
2
10
3 (d)
10
10 3
4
10 Discharge (m 3/s)
10
5
Figure CS.7.3.4. Hydraulic geometry of the Jamuna River (after FAP24, 1996).
Streamﬂow data include discharge data on a daily basis for the entire period of record. The entire ﬂowdischarge record can be used to determine the ﬂowduration curves and for ﬂoodfrequency analysis. Hydrometric gauges are useful in determining the stage–discharge relationship. Looprating effects that are due to aggradation/degradation, bedform changes, and/or hydrodynamic effects can be examined from stage–discharge relationships. It is also useful to examine speciﬁcgauge records to detect aggradation/degradation trends over long periods. The bankfull discharge should normally ﬁt within the range of
228
River dynamics
1 to 5 yr in the ﬂoodfrequency analysis. Extrapolations to periods of return far exceeding the length of record can be misleading. Sediment data include bed material and sediment transport; the particlesize distribution of bed material should be determined as accurately as possible. The variability in bed material in alluvial rivers can be high. Several samples at different locations are often desirable. In degrading channels, careful attention should be paid to the coarse fractions of the surface material and underlying deposits. Sieve analyses are best suited to ﬁnegrained streams, and gravelbed and cobblebed streams require the examination of large volumes of bed material to determine the median grain size. Suspended sediment records should indicate the ﬂuxaverage sediment concentration and the sediment load. Sediment load by size fractions is most valuable to separate washload from bedmaterial load. Sediment concentration proﬁles enable evaluation of the Rouse parameter. Sediment budgets by size fractions are sometimes most useful in determining the different patterns of sediment transport for washload and bedmaterial load. Field surveys are most effective after a review of maps and photos. It is very important to gather additional information on the crosssectional geometry of the river, including bankfull conditions and ﬂoodplain elevation, land use, and vegetation. Field notes should include indications of actual upstream basin conditions, and recent changes in vegetation, land use, and sediment sources should be reviewed. Riverbed, banks, proﬁle, and planform should be examined for particlesize distribution of bed material, sediment deposits, aggradation/degradation, headcutting, and bedrock control. Bank stability and mode of failure, stratigraphy and seepage, lateral migration, and vegetation should be considered. River hydraulics includes velocities, highwater marks on bridge piers, structures, and the ﬂoodplain; river choking, debris, ice cover and ice jams, and ﬂow controls provide important information for watersurface calculations with hydraulic models. Flights over the river are quite informative on the overall planform stability of rivers during ﬂoods. Case Study 7.4 illustrates how river databases can be used to examine the dynamics of river systems by means of the analysis of mass curves. Case Study 7.4 Bernardo reach of the Rio Grande, New Mexico, United States. The 10milelong Bernardo reach of the Middle Rio Grande in New Mexico is included in the habitat designation for two federally listed endangered species: the silvery minnow and the southwestern willow ﬂycatcher. A complete database of ﬂow and sediment transport in the Rio Grande below Cochiti Dam has been assembled at Colorado State University by G. Richard,
River databases
229
6
Cumulative discharge (10 acrefeet)
45 40
km3 50
San Acacia Bernardo
35
1.09E+06 af/yr
43 37
30 1.09E+06 af/yr (Bernardo))
25
30 24
20 15
18
0.60E+06 af/yr
1978
10
12
0.54E+06 af/yr (Bernardo))
5
6 0 2001
1996
1991
1986
1981
1976
1971
1966
1961
1956
1951
1946
0 Time (years)
400 350
3.2E+06 tons/yr San Acacia Bernardo
1986 5.97E+06 tons/yr 1973
4
3
1991
300 250 2
9.44E+06 tons/yr
2.99E+06 tons/yr
8
200
(x 10 metric tons)
450
150 1
100
0
2002
1998
1994
1990
1986
1982
1978
1974
1970
1966
1962
1958
1954
0
1950
50 1946
6
Cumulative suspended sediment load (10 tons)
Figure CS.7.4.1. Mass curve for the discharge of the Rio Grande (after Richard et al., 2000). (af/yr, acreft per year.)
Time (years)
Figure CS.7.4.2. Mass curve for the suspended sediment of the Rio Grande (after Richard et al., 2000).
C. Leon, and T. Bauer in collaboration with D. Baird at the U.S. Bureau of Reclamation. Besides the complete data reports (Leon et al., 1999, and Bauer et al., 2000), the geomorphic analysis of Richard et al. (2000) shows a 500ftwide river with a ﬁne sand bed at d50 0.3 mm. The channel width varies from 150 to 1,200 ft within a 10milelong reach. A discharge mass curve at Bernardo and San Acacia in Fig. CS.7.4.1 represents the cumulative runoff
River dynamics 3
300
6
250
19781995 2.290 mg/L
19681977 10.777 mg/L
2 200
19821997 2.227 mg/L
19561966 7.250 mg/L
8
Cumulative sediment load (10 tons)
350
19631978 8.004 mg/L
150
1
100 19521956 31.145 mg/L
50 0 0
(x 10 metric tons)
230
San Acacia Bernardo + Rio Puerco
19471952 7.376 mg/L
4
8
12
16
20
24
28
32
0 36
6
Cumulative discharge (10 acrefeet)
Figure CS.7.4.3. Double mass curve of the Rio Grande (after Richard et al., 2000).
Figure CS.7.4.4. Sediment accumulation in the Bernardo reach of the Rio Grande (after Richard et al., 2000).
volume as a function of time. During the same period, the sediment mass curves in Fig. CS.7.4.2 describe the cumulative sediment yield as a function of time. The inﬂuence of Cochiti Dam is clearly identiﬁed after 1975. Double mass curves represent the cumulative sediment load as a function of the cumulative runoff water volume. The case of Bernardo reach of the Rio Grande is illustrated in Fig. CS.7.4.3. The slope of double mass curves deﬁnes average sediment concentrations in suspension. In the case of the Rio Grande, the slope reduction reﬂects the effects of Cochiti Dam in terms of reduced sediment concentration after 1975.
Problems
231
Mass difference curves show the difference between sediment inﬂow and sediment outﬂow to a given reach. It can be determined from the sum and the differences between the sediment curves of all tributaries and branches of a given river reach. Mass difference curves provide information on the net sediment balance on a river reach over time and indicate whether a river is aggrading (+) or degrading (−). The case of the Bernardo reach of the Rio Grande is illustrated in Fig. CS.7.4.4, and the negative values indicates degradation of the river reach, which has been conﬁrmed with ﬁeld measurements of the crosssectional geometry. It is important to include the contribution of all tributaries in the analysis of mass difference curves. Exercise 7.1
Derive relations (7.4a) and (7.4b) from Relations (6.27) and (7.3). Exercise 7.2 Derive hydraulicgeometry relations (7.6a)–(7.6e) from combining Relations (6.28) and (7.4). Exercise 7.3
Derive hydraulicgeometry relations (7.7a)–(7.7e) from Relations (7.6) and the deﬁnition of Cppm . Exercise 7.4
Calculate the annual rate of bed aggradation in the Mississippi River from the data provided in the river branching section, Section 7.4. Problem 7.1 Determine the combined effects of a 50% decrease in water discharge Q − and a 200% increase in sediment discharge Q + bv on channel width, ﬂow depth, ﬂow velocity, slope, and Shields parameter. Answers: New ﬂow depth ∼67% of initial ﬂow depth h 0.67 h 0 , W 58%W0 , V 1.27V0 , S 2.75S0 , and τ∗ 1.84τ∗0 . Problem 7.2
Estimate the hydraulic geometry of an alluvial stream at a bankfull discharge of 4,500 ft3 /s with d50 = 0.5 mm and a bedmaterial concentration of 150 ppm.
232
River dynamics
Answers: With Q = 127 m3 /s, ds = 0.0005 m, and Cmg/l 150, we obtain from relations (7.7a)–(7.7e) h 1.2 m, W 170 m, V = 0.63 m/s, S 2.2 × 10−4 , and τ∗ 0.3. Problem 7.3 Varmint Creek drains a 320squaremile watershed (see also Problem 7.2). The mean annual rainfall is 45 in. The largest known peak discharge in a 45yr record was 26,000 ft3 /s in 1929. The 10yr suspended sediment record indicates a mean annual yield of 48,000 tons, mostly a washload of silt and clay. The channel slope is ∼2.5 ft/mile. The 2yr ﬂood peak is 4,500 ft3 /s. The corresponding bedsediment concentration is estimated at 150 ppm. Compare the measured peak discharge and sediment yield with those of comparable watersheds of the same drainage area. Problem 7.4
The Jamuna River is a large braided river with a median grain size of 0.2 mm. The river conveys ∼48,000 m3 /s at bankfull conditions and the corresponding bedmaterial discharge is approximately 2.6 million tons per day. Estimate the downstream hydraulic geometry of the river. Answers: Calculate by using Relations (7.6a)–(7.6e) with Q = 48,000 m3 /s, ds = 0.0002 m, and Q bv = 11.6 m3 /s to give h 6.7 m, W 2,500 m, V 2.8 m/s, S 3.2 × 10−4 and τ∗ 6. Field measurements in Table CS.7.3.1 indicate h 6.6 m, W 4,200 m, V 1.7 m/s, S 7.5 × 10−5 , and τ∗ 15. The calculated slope far exceeds the measured slope, and the stream may be aggrading and braiding. Problem 7.5
From the information in Case Study 7.1 and Fig. CS.7.1.1, estimate the minimum grain size at the beginning of motion during a ﬂood of the Meuse River at 100,000 ft3 /s. Answers: The ﬂow depth is ∼25 ft during ﬂoods, or h 7.6 m, and the slope is ∼4 ft/mile, or S 7.6 × 10−4 . We obtain from Relations (7.9) dsc 12 h S = 0.069 m or 69 mm. This indicates that all sand and gravel sizes are in motion during ﬂoods. Only cobbles are stable and can armor the riverbed if available in sufﬁciently large quantity.
Problems
233
Problem 7.6
With reference to the Lawn Lake Dam failure in Case Study 7.2, determine the following: (1) the ﬂow depth required for mobilizing a 2m boulder in Roaring River and (2) the volumetric sediment concentration of the Roaring River ﬂow. Answers: (1) From Relation (7.9b) with S = 0.07 and ds = 2 m, we obtain a ﬂow depth of ∼2.8 m; (2) a volumetric sediment concentration of ∼33% is obtained from the alluvial fan volume and the Lawn Lake water release. It can be classiﬁed as hyperconcentrated ﬂow. Problem 7.7 With reference to the Jamuna River in Case Study 7.3, determine the planform geometry from the planform predictors based on discharge and slope. Also use the width–depth ratio as an indicator. Problem 7.8
Anticipate the effects of a 30% ﬂow reduction and 30% sediment diversion on the hydraulic geometry (W, h, V, S, τ∗ ) of a large meandering river. Problem 7.9
With reference to the Jamuna, Ganges, and Padma Rivers’ conﬂuences in Case Study 7.3, apply the hydraulicgeometry relationships and compare with ﬁeld measurements of W, h, V, and S. Compare the slopes upstream and downstream of the conﬂuence. Would a river conﬂuence in equilibrium require an increase or decrease in slope in the downstream direction? Problem 7.10
The Bernardo reach of the Rio Grande in New Mexico features a channel width ranging from 150 to 1,200 ft with an average of ∼500 ft. The average reach slope is 80 cm/km, a median grain size of 0.3 mm, and a sandload up to 10,000 tons per day at discharges of ∼5,000 ft3 /s. Compare with the range of channel width and slope calculated at a sediment concentration of sand varying from 500 to 2,000 mg/l. Answers: From Relations (7.7) and C =500 mg/l, W =163 m and S = 4.6 × 10−4 .
8 River stabilization
Riverstabilization structures are designed to protect the riverbanks and prevent lateral migration of alluvial channels through bank erosion. Riverstabilization methods can be classiﬁed according to two different approaches: (1) strengthening the banks and (2) reducing hydrodynamic forces. This chapter ﬁrst examines the bank stability of alluvial streams in Section 8.1. Bankprotection methods through strengthening the banks with riprap are discussed in Section 8.2, and other bankstrengthening methods are covered in Section 8.3. Flowcontrol structures offer an alternative approach by reducing the hydrodynamic forces applied against the riverbanks. The ﬂowcontrol structures covered in Section 8.4 aim at gaining control over the ﬂow depth and the direction and magnitude of ﬂow velocity near the river banks. Some engineering considerations are discussed in Section 8.5.
8.1
Riverbank stability
Bank stability is examined in this section. First, the processes are reviewed in Subsection 8.1.1, followed by conceptual solutions for slope reduction in Subsection 8.1.2 and subsurface drainage in Subsection 8.1.3.
8.1.1
Bankerosion processes
Processes of bank erosion are directly linked to the lateral migration of alluvial channels. Bank erosion is the result of ﬂowing water that applies active forces met by the passive forces of the bank material to resist motion. As discussed in Chap. 6, the hydrodynamic forces in river bends induce secondary ﬂow where the freesurface streamlines are deﬂected toward the outer bank and the nearbed streamlines are deﬂected toward the inner bank. Along a cross section, the streamlines are deﬂected downward near the outer bank and deﬂected upward on the point bar. The resulting effect is to decrease the 234
Riverbank stability
235
Figure 8.1. Bankfailure types.
stability of sediment particles and cause degradation near the outer bank. On the other hand, the particlestability increases and aggradation is expected near the point bar. The scour at the toe of the outer bank shifts the thalweg to the outside of river bends and causes steepening of the outer bank. Increased steepening of the outerbank material causes bank failure. Three modes of failure are typical of alluvial rivers, as sketched in Fig. 8.1. With noncohesive granular material, grain removal at the toe of the outerbank induces sliding of the granular material as soon as the bank angle exceeds the angle of repose of the material [Fig. 8.1(a)]. In the case of cohesive bank material, rotational failure is typical and the presence of tension cracks may accelerate the bankerosion process [Fig. 8.1(b)]. In alluvial streams ﬂowing in stratiﬁed deposits, the underlying noncohesive material is mobilized, thus leaving the overlying cohesive material unsupported and subject to tension cracks and cantilever failure [Fig. 8.1(c)]. In general, the most erosive banks are sandy and silty, whereas the least erosive are clayey and gravelly.
236
River stabilization
Factors affecting streambank failure include hydraulic parameters that control the active forces such as discharge magnitude and duration, velocity, and applied shearstress magnitude and orientation. Additional active forces that are due to seepage, piping, surface waves, and ice can also contribute to bank erosion. Passive forces relate to bankmaterial size, gradation, and cohesion. Biological factors such as vegetation can play a signiﬁcant stabilizing role. Artiﬁciallyinduced activities such as urbanization, drainage, ﬂoodplain farming and development, boating and commercial navigation, and waterlevel ﬂuctuations from hydropower generation can have detrimental effects on bank stability. Changes in channel geometry through bank erosion are particularly signiﬁcant during ﬂoods. The sedimenttransport capacity is often several orders of magnitude greater during ﬂoods than at intermediate or low ﬂows. Most cases of riverbank instability in alluvial rivers take place during the small percentage of time when the dominant discharge is exceeded. The analysis of ﬂowduration curves and sedimentduration curves in Section 5.6 is useful in the analysis of bank stability. Landslides refer to the downslope movement of earth and organic materials [Fig. 8.1(b)]. Active forces are involved in mass wasting. These forces are associated with the downslope gravity component of the slope mass. Resisting these downslope forces are the shear strength of the Earth’s materials and any additional contributions from vegetation by means of root strength or human slopereinforcement activities. When a slope is acted on by a stream or river, an additional set of forces is added. These forces are associated with removal of material from the toe of the slope, ﬂuctuations in groundwater levels, and vibration of the slope. A slope may fail if stable material is removed from the toe. When the toe of a slope is removed, the slope loses more resistance by buttressing than it does by downslope gravitational forces. The slope materials may then tend to move downward into the void in order to establish a new balance of forces or equilibrium. The presence of water in riverbanks and its movement toward or away from the river affects bank stability and bank erosion in various ways. The outﬂow of water from the river into the adjacent banks stabilizes the riverbanks. Rivers that continuously seep water into the banks tend to have smaller widths and larger depths for a particular discharge. The converse is true of rivers that continuously gain water by an inﬂow through their banks. The inﬂow destabilizes riverbanks. Piping is another phenomenon common to the alluvial banks of rivers. With stratiﬁed banks [e.g., Fig. 8.1(c)], ﬂow is induced in more permeable layers
Riverbank stability
237
by changes in river stage and by wind and boatgenerated waves. If the ﬂow through the permeable lenses is capable of dislodging and transporting ﬁne particles from the permeable lenses, the material is slowly removed, undermining portions of the bank. Without this foundation material to support the overlying layers, a block of bank material drops down and results in the development of tension cracks, as sketched in Fig. 8.1(c). These cracks allow surface ﬂows to enter, further reducing the stability of the affected block of bank material. Mass wasting is an alternative form of bank erosion. If the bank becomes saturated and possibly undercut by ﬂowing water, blocks of the bank may slump or slide into the channel. Mass wasting may be further aggravated by construction of homes on riverbanks, operation of equipment on the ﬂoodplain adjacent to the banks, added gravitational force resulting from tree weight, saturation of banks, and increased inﬁltration into the ﬂoodplain. 8.1.2
Figure 8.2. Slopereduction methods.
Slope reduction and benching
The weight of soil at the toe of the slope counterbalances the weight of soil in the upper part of the slope and aids the shear strength of the soil in resisting against failure. Direct slope reduction is an excavation method whereby soil is removed from the slope to ﬂatten the slope angle, as shown in Fig. 8.2(a). This method is usually preferred when adequate space is available on the ﬂoodplain. Where there is insufﬁcient space, the slope may be ﬂattened by use of ﬁll material, as shown in Fig. 8.2(b), or by cutandﬁll operations, as shown in Fig. 8.2(c). Slope, or buttress, benching is also an indirect method of slope reduction. It differs
238
River stabilization
from the method described above only in the ﬁnal shape of the slope section. The benching method produces a benched or a series of stepped sections, as shown by Fig. 8.2(d). The end effect of this method is a gross reduction in the angle of the slope. 8.1.3
Subsurface drainage
The control of groundwater within a slope is often a feasible means of stabilizing the slope. The control of groundwater may be achieved by two methods: (1) the prevention of inﬁltration of surface water into soils, and (2) the provision for subsurface drainage to remove the water from the soil mass. The ﬁrst method is generally accomplished by merely providing adequate surface drainage. The second method uses various subsurfacedrainage techniques. Subsurface drains are most effective when only a small quantity of water is required to be removed to affect stabilization. A thin conﬁned aquifer can be intercepted to reduce the artesian pressure. Horizontal drains consist of a slotted pipe to remove excessive water from cut slopes experiencing stability problems. Figure 8.3(a) shows an idealized concept for the application of horizontal drains to streambank stability. Stabilization of highway cut slopes frequently involves the use of a combination of horizontal and vertical drainage systems. Vertical drains can be utilized in riverbank stabilization, as sketched in Fig. 8.3(b). The main function of a drainwell installation is to intercept groundwater moving toward the bank and thus to relieve pore pressures that would otherwise develop in a saturated ﬁnegrained bank material. Usually this highcost concept is applicable only in cases in which high groundwater levels exist above maximum river elevations. A drainage trench is shown conceptually in Fig. 8.3(c). Its function is similar to that of the vertical drain to intercept any groundwater before it reaches the bank slope. A slotted or perforated pipe is placed at the bottom of the trench to drain the water intercepted by the trench. As with vertical drains, this method would be effective for bank stabilization only in cases in which high groundwater levels occur above maximum river elevations. 8.2
Riverbank riprap revetment
Several engineering methods can be used to strengthen riverbanks against erosion. Methods commonly encountered include the use of riprap or large stones that are not easily removed from the banks. Besides riprap in this section, the other methods considered in Section 8.3 are vegetation, gabions, blocks and
Riverbank riprap revetment
239
Figure 8.3. Riverbank drainage methods.
rocks, rockﬁll trenches, windrow revetment, soil cement, fences, bulkheads, and mattresses. When economically available in sufﬁcient size and quantity, rock riprap is usually the most widely used material for bank protection. A rock riprap blanket is ﬂexible and is neither impaired nor weakened by slight movement of
240
River stabilization
the bank resulting from settlement or other minor adjustments. Local damage or loss is easily repaired by the placement of more rock. Construction is not complicated, and special equipment and construction practice are not necessary. Riprap is usually durable and recoverable and may be stockpiled for future use. Locally available riprap usually provides a costeffective alternative to many other types of bank protection. The appearance of rock riprap is natural, and after a period of time vegetation will grow between the rocks. Finally, wave runup on rock slopes is usually less than on other types of structures. The important factors to be considered in designing rock riprap blanket protection are: (1) the velocity (both magnitude and direction) of the ﬂow or shear stress in the vicinity of the rock; (2) the sideslope of the bankline being protected; (3) the density of the rock; (4) the angle of repose for the rock, which depends on stone shape and angularity; (5) the durability of the rock; (6) the riprap blanket thickness; (7) the ﬁlter needed between the bank and the blanket to allow seepage but to prevent erosion of bank soil through the blankets; (8) the blanket stabilization at the toe of the bank; and (9) the blanket must be tied into the bank at its upstream and downstream ends. Two methods to determine the size of rock riprap required for streambank stabilization are presented: (1) shearstress method in Subsection 8.2.1 and (2) ﬂowvelocity method in Subsection 8.2.2. Speciﬁcations are given for riprap gradation in Subsection 8.2.3 and for ﬁlters in Subsection 8.2.4. Protection against riprap failure is then considered in Subsection 8.2.5. 8.2.1
Shearstress method
Riprap stability on a sideslope is a function of (1) the magnitude and the direction of the ﬂow velocity or shear stress in the vicinity of the streambank, (2) the sideslope angle, and (3) the properties of the rock including size, density, and angularity. The functional relationship that determines the stability factor of a particular stone under given hydrodynamic forces has been analyzed previously in Chap. 6. The stability factor SF can be determined from Eqs. (6.3), (6.5), (6.6), and (6.8), as detailed in Example 6.1. On the outer bank of natural meandering channels, the streamlines are deﬂected downward at an angle λ that can be estimated from tan λ ∼ = 11 h/R, where h is the mean ﬂow depth and R is the bend radius of curvature. A simpliﬁed approach to the calculation of the rock size that corresponds to the beginning of motion on a sideslope θ1 is possible when the following approximations are acceptable: (1) ﬂow in the downstream direction, i.e., in the absence of secondary ﬂows, or λ ∼ = 0; = 0; (2) the downstream bed slope is negligible, θ0 ∼ (3) the speciﬁc gravity of the rock is close to 2.65; and (4) the viscous drag is small compared with the lift force.
Riverbank riprap revetment
241
The effective rock size dm required for stabilizing a riverbank under applied shear stress τ0 is estimated from Lane’s relationship as dm =
τ0
τ∗c (γs − γ ) cos θ1
tan2 θ1 1− tan2 φ
(8.1a)
or τ0
dm = τ∗c γ (G − 1)
sin2 θ1 1− sin2 φ
,
(8.1b)
where τ0 is the applied shear stress, γs and γ are the speciﬁc weight for the rock and water, respectively, θ1 is the sideslope angle, φ is the angle of repose of the rock riprap, and τ∗c is the critical value of the Shields number. Examples 8.1 and 8.2 illustrate how to determine riprap size and particle stability. It must be considered that the term in brackets of Eqs. (8.1) becomes very small when the sideslope angle becomes approximately equal to the angle of repose of the material. This tends to require extremely large material on steep banks. Example 8.1 Calculation of riprap size in straight channels. Determine the rock riprap size required for stabilizing the banks of a straight river given the river width W = 300 m, a ﬂow depth of h = 7 m, and a channel slope of S = 60 cm/km. The bank slope is θ1 = 30◦ , the rock density G is 2.7, and the angle of repose is φ = 40◦ . Step 1: The shear stress applied on the particle is τ0 = γ h S = 9,810 N/m3 × 7 m × 60 × 10−5 = 41 Pa. Step 2: The term in brackets of Eqs. (8.1) is calculated as 0.5 tan θ1 2 tan 30 2 η = cos θ1 1 − = cos 30 1 − = 0.63. tan φ tan 40 Note that very large riprap sizes may be obtained when the term η becomes very small. Step 3: The effective riprap size is obtained from Eqs. (8.1) assuming τ∗c = 0.047 as dm ∼ = or ∼3.3 in.
τ0 41 Pa m3 = 0.085 m, = ητ∗c (γs − γ ) 0.63 × 0.047 (2.65 − 1) 9,810 N
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River stabilization
Example 8.2 Application to riprap stability in a bend. The mean ﬂow velocity in a river bend reaches 18 ft/s (5.5 m/s) during ﬂoods at a Manning coefﬁcient of n = 0.03, the river is 300 ft wide (91 m) and 10 ft deep (3.0 m) with a radius of curvature of 700 ft (215 m). Determine the stability factor of 6in. crushed rock riprap at a sideslope of 1V :2.25H , or θ1 = 24◦ . The angle of repose of the material is φ = 40◦ and the speciﬁc gravity G = 2.65. Step 1: The slope is calculated from Manning’s formula 2 nV 0.03 × 18 2 = = 6.1 × 10−3 S = tan θ0 = 1.49 h 2/3 1.49 × 102/3 or 610 cm/km, or θ0 = 0.35◦ . Step 2: The shear stress is obtained from τ0 = γ h S = 62.4 × 10 × 6.1 × 10−3 = 3.8 lb/ft2 or 182 Pa. Step 3: From the method in Chap. 6, the angle θ = tan−1 (sin θ0 / sin θ1 ) = 0.86◦ . Step 4: The factor aθ = cos2 θ1 − sin2 θ0 = cos2 24 − sin2 0.35 = 0.91. Step 5: The streamline deviation angle is λ tan−1 (11h/R) = tan−1 [(11 × 10)/700] = 8.9◦ . Step 6: η0 =
21τ0 21 × 3.8 lbft3 = 2 = 1.55. (G − 1)γ ds ft (1.65) 62.4 lb × 0.5 ft
Step 7: The angle β, assuming M = N , is cos(λ + θ ) −1 β = tan 2 1 − aθ2 + sin(λ + θ) η tan φ 0
= tan−1
cos(8.9 + 0.86) √ = 50.7◦ . 2 1 − 0.91 2 + sin(8.9 + 0.86) 1.55 tan 40
Riverbank riprap revetment Step 8:
243
1 + sin (λ + β + θ ) 2 1 + sin (8.9 + 50.7 + 0.86) = 1.55 = 1.45. 2
η1 = η0
Step 9: The stability factor is aθ tan φ SF = η1 tan φ + 1 − aθ2 cos β =
0.91 tan 40◦ = 0.52 √ 1.45 tan 40◦ + 1 − 0.912 cos 50.7◦
The stone is unstable because SF < 1. As a ﬁrst approximation, assuming λ = 0 and θ0 = 0, the stone size at incipient motion is estimated from Eqs. (8.1) as τ0 dm ∼ = sin2 θ1 1− 0.047 (γs − γ ) sin2 φ =
ft
2
3.8 lbft3
sin2 24◦ 1− 0.047 × 1.65 × 62.4 lb sin2 40◦
1.0 ft or 30 cm.
Calculation steps 6–9 account for secondary ﬂow and can be repeated for different stone sizes until SF = 1. Based on ﬂow velocity, the ﬁrst estimate of riprap size can be obtained from Eqs. (8.4) in Subsection 8.2.4 solved for ds as ds = =
sin φ Vc2 K c2 2(G − 1)g(sin2 φ − sin2 θ1 )1/2 (18)2 ft2 s2 sin 40 = 2.7 ft. s2 (1.2)2 2 × 1.65 × 32.2 ft sin2 40 − sin2 24
In this case, this value can also be obtained from Fig. 8.4 in Subsection 8.2.2.
8.2.2
Velocity method
The stone size needed to protect a streambank from erosion by a current that is moving parallel to the embankment can also be determined as a function of ﬂow velocity. The diameter d50 is that of a spherical stone that would have the same weight as the 50% size of the stone. For stone riprap, the velocity at the
244
River stabilization
top of the stone, called velocity against the stone, v s , is related to the shear velocity u ∗ as v s = 5.75 u ∗ . From the incipient motion condition of material with angle of repose φ, for hydraulically rough ﬂow conditions, τ∗ c =
u 2∗c (G − 1)gds
= 0.06 tan φ,
we obtain the critical velocity against the stone v sc as v sc = 5.75 u ∗c = 2(G − 1)gds tan φ .
(8.2)
The velocity against the stone v s also relates to the mean ﬂow velocity V as a function of the ﬂow depth h and the stone diameter ds as V = v s log (4 h/ds ). After combining this with Eq. (8.2), we obtain the critical mean ﬂow velocity Vc as Vc = K c 2(G − 1)gds , (8.3a) where K c = log
4h ds
tan φ .
(8.3b)
The critical mean ﬂow velocity thus depends on relative submergence h/ds √ and angle of repose φ. It is noticed that, for riprap, tan φ 1 and K c 1.2, when h/ds 5. The critical mean ﬂow velocity for representative riprap design conditions (h 5 ds or K c = 1.2) is shown in Fig. 8.4. This graphic should be used when h < 10 ds . At larger ﬂow depths (h > 10 ds ), the shearstress method is preferable. The reader is referred to Maynord (1992) and Abt (Colorado State University, 2001, pers. comm.) for recent developments on riprap design procedures based on ﬂow velocity. On a sideslope without secondary ﬂows, the critical mean ﬂow velocity Vc can be approximated according to Lane’s approach by 1/4 sin2 θ1 . (8.4) Vc K c 2(G − 1)gds 1 − sin2 φ Typical curves at an angle of repose φ = 40◦ are shown in Fig. 8.4 for sideslope angles up to 33◦ . It is observed that sideslope effects become very signiﬁcant when θ1 > 20◦ . Gradations calculated with the U.S. Army Corps of Engineers (1981) method can be obtained from Table 8.1.
Riverbank riprap revetment
245
Figure 8.4. Particlestability diagram.
8.2.3
Riprap gradation
The concept of a representative grain size for riprap is fairly simple. A uniformly graded riprap with a median size d50 scours to a greater depth than a wellgraded mixture with the same median size. The uniformly distributed riprap scours to a depth at which the velocity is less than that required for the transportation of d50 size rock. The wellgraded riprap, on the other hand, develops an armor plate; that is, some of the ﬁner materials, including sizes up to d50 and larger, are transported by the high velocities, leaving a layer of large rock sizes that can not be transported under the given ﬂow conditions. Thus the size of rock representative of the stability of the riprap is determined by the
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River stabilization
Table 8.1. Riprap weight from the U.S. Army Corps of Engineers (1981) Stone weight (lb) for percent lighter by weighta 100
50
15
d100 max (in.)
max
min
max
min
max
min
d30 max (ft)
d90 max (ft)
12 15 18 21 24
86 169 292 463 691
35 67 117 185 276
26 50 86 137 205
17 34 58 93 138
13 25 43 69 102
5 11 18 29 43
0.48 0.61 0.73 0.85 0.97
0.70 0.88 1.06 1.23 1.40
G = 2.65, or γs = 165 lb/ft3 = 26 kN/m3 , 1 ft = 12 in. = 30.5 cm. The relationship between diameter and weight is based on a spherical shape.
a Assuming
larger sizes of rock. The representative grain size dm for riprap is larger than the median rock size d50 . The effective size dm approximately corresponds to d65 , and we can use dm ∼ = 1.25 d50 . The computations of the representative grain size dm for the recommended gradation are illustrated in Fig. 8.5 in terms of d50 . Recommended gradations Figure 8.5. Suggested riprap grafrom the U.S. Army Corps of Engineers are dation curve. presented in Tables 8.1 and 8.2. Riprap consisting of angular stones is more suitable than that consisting of rounded stones. Control of the gradation of the riprap is almost always made by visual inspection. When necessary, poor gradations of rock can be used as Table 8.2. Suggested riprap size gradation Percent ﬁner by weight
Sieve diameter (×d50 )
Stone diameter (×d50 )
0 10 20 30 40 50 60 70 90 100
0.25 0.35 0.50 0.65 0.80 1.00 1.20 1.60 1.80 2.00
— 0.28 0.43 0.57 0.72 0.90 1.10 1.50 1.70 1.90
Riverbank riprap revetment
247
riprap, provided that a ﬁlter is placed between the riprap and the bank or bed material. Considering the practical problems of quarry production, a gradation band is usually speciﬁed by the U.S. Army Corps of Engineers (1981) rather than a single gradation curve, and any stone gradation within the limits is acceptable. The Corps criteria for establishing gradation limits for riprap are as follows: (1) the lower limit of d50 stone should not be less than the size of stone required for withstanding the design shear forces; (2) the upper limit of a d50 stone should not exceed ﬁve times the lower limit of a d50 stone, the size that can be obtained economically from the quarry or the size that satisﬁes layerthickness requirements; (3) the lower limit of d100 stone should not be less than two times the lower limit of d50 stone; (4) the upper limit of d100 stone should not exceed ﬁve times the lower limit of d50 stone, the size that can be obtained economically from the quarry or the size that satisﬁes layerthickness requirements; (5) the lower limit of d15 stone should not be less than 1/16 the upper limit of d100 stone; and (6) the bulk volume of stone lighter than the d15 stone should not exceed the volume of voids in the structure without this lighter stone. The riprap thickness should not be less than (1) 12 in. (30 cm) for practical placement, (2) less than the diameter of the upper limit of d100 stone, or (3) less than 1.5 times the diameter of the upper limit d50 stone, whichever is greater. If riprap is placed under water, the thickness should be increased by 50%, and if it is subject to attack by large ﬂoating debris or wave action it should be increased 6–12 in. (15–30 cm). Riprap placement is usually accomplished by dumping directly from trucks. If riprap is placed during construction of the embankment, rocks can be dumped directly from trucks from the top of the embankment. Rock should never be placed by dropping down the slope in a chute or pushed downhill with a bulldozer. These methods result in segregation of sizes. With dumped riprap there is a minimum of expensive handwork. Draglines with orange peel buckets, backhoes, and other power equipment can also be used advantageously to place riprap. 8.2.4
Filters
Filters are used under the riprap revetment to allow water to drain easily from the bank without carrying out soil particles. Filters are required when the d15 of the riprap gradation exceeds ﬁve times the d85 of the bank material. Filter blankets must meet two basic requirements: stability and permeability. The ﬁlter material must be ﬁne enough to prevent the base material from escaping through the ﬁlter, but it must be more permeable than the base material. There is
248
River stabilization
no standard ﬁlter that can be used in all cases. Two types of ﬁlters are commonly used: gravel ﬁlters and synthetic ﬁlter cloths. Gravel ﬁlters consist of a layer, or blanket, of wellgraded gravel placed over the embankment or riverbank before riprap placement. Sizes of gravel in the ﬁlter blanket should be from 3/16 in. (5 mm) to an upper limit, depending on the gradation of the riprap, with maximum sizes of approximately 3–31/2 in. (90 mm). The ﬁlter thickness should not be less than 6–9 in. (20 cm). Filters that are onehalf the thickness of the riprap are quite satisfactory. Suggested speciﬁcations for gradation are as follows: d50 (ﬁlter) < 40, d50 (base) d15 (ﬁlter) < 40, 5< d15 (base) d15 (ﬁlter) < 5. d85 (base)
(8.5a) (8.5b) (8.5c)
If the base material is a ﬁnegrained cohesive soil, such as fat or lean clay, these requirements are not applicable, and the stability criterion is that the d15 size of the ﬁlter cannot exceed 0.4 mm. Multiple ﬁlters may be used when the base material is very ﬁne. In such a case, each layer must satisfy the stability and permeability requirements relative to the underlying layer. A detailed ﬁlter design calculation example is presented in Example 8.3. Synthetic ﬁlter cloths (plastic cloth and woven plastic materials) are also used as ﬁlters, replacing a component of a graded ﬁlter. Numerous plastic ﬁlter fabrics exist with a wide variation in size of number of openings and in strength and durability of material. Opening areas of 25%–30% appear desirable to minimize the possibility of clogging and to reduce head loss. It is often desirable to place a protective blanket of sand or gravel on the ﬁlter or to take care in placing the rock to that the ﬁlter fabric is not punctured. The sides and the toe of the ﬁlter fabric must be sealed or trenched so that the base material does not leach out around the ﬁlter cloth. Care is also required in joining adjacent section soft ﬁlter fabric together. Example 8.3 Application to ﬁlter design. The following ﬁlter design example involves the properties of the base material and the riprap given in Table E.8.3.1. The riprap does not contain sufﬁcient ﬁnes to act as the ﬁlter because 100 d15 (riprap) = = 67, d85 (base) 1.5
Riverbank riprap revetment
249
Table E.8.3.1. Sizes of materials Base material
Riprap
Sand
Rock
d85 = 1.5 mm d50 = 0.5 mm d15 = 0.17 mm
d85 = 400 mm d50 = 200 mm d15 = 100 mm
which is much greater than 5, the recommended upper limit [requirement (8.5c)]. Also d15 (riprap) 100 = = 600, d15 (base) 0.17 which is much greater than 40, the recommended upper limit [requirement (8.5b)]. The properties of the ﬁlter to be placed adjacent to the base, from requirement (8.5c), are as follows: d50 (ﬁlter) < d50 (base) d15 (ﬁlter) > (b) d15 (base) d15 (ﬁlter) < (c) d15 (base) d15 (ﬁlter) < (d) d85 (base) (a)
40
so d50 (ﬁlter) < (40)(0.5) = 20 mm.
5
so d15 (ﬁlter) > (5)(0.17) = 0.85 mm.
40
so d15 (ﬁlter) < (40)(0.17) = 6.8 mm.
5
so d15 (ﬁlter) < (5)(1.5) = 7.5 mm.
Thus, with respect to the base, 0.85 mm < d15 (ﬁlter) < 6.8 mm, d50 (ﬁlter) < 20 mm. The properties of the ﬁlter to be placed adjacent to the riprap are as follows: d50 (riprap) < d50 (ﬁlter) d15 (riprap) > (b) d15 (ﬁlter) d15 (riprap) < (c) d15 (ﬁlter) d15 (riprap) < (d) d85 (ﬁlter) (a)
40
so d50 (ﬁlter) > 200/40 = 5 mm.
5
so
40
so d15 (ﬁlter) > 100/40 = 2.5 mm.
5
so
d15 (ﬁlter) < 100/5 = 20 mm.
d85 (ﬁlter) > 100/5 = 20 mm.
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River stabilization
Figure E.8.3.1. Example of ﬁlter design.
Therefore, with respect to the riprap, the ﬁlter must satisfy these requirements: 2.5 mm < d15 (ﬁlter) < 20 mm, d50 (ﬁlter) > 5 mm, d85 (ﬁlter) > 20 mm. These riprap ﬁlter requirements along with those for the base material are shown in Fig. E.8.3.1. Any ﬁlter having sizes represented by the double crosshatched area is satisfactory. For example, a good ﬁlter could have these sizes: d85 = 40 or 20 mm, d50 = 10 mm, d15 = 4 mm.
8.2.5
Preventing riprap failure
A major shortcoming in riprap design techniques is the assumption that riprap failure is due only to particle erosion. Riprap failure modes are identiﬁed as follows: (1) particle erosion; (2) translational slide; (3) slump; and (4) sideslope failure, as illustrated in Fig. 8.6. Particle erosion by ﬂowing water is the most commonly considered erosion mechanism [Fig. 8.6(a)]. Particle erosion can be initiated by abrasion,
Riverbank riprap revetment
251
impingement of ﬂowing water, eddy action/reverse ﬂow, local ﬂow acceleration, freeze/thaw action, ice, or toe erosion. Probable causes of particle erosion include: (1) stone size that is not large enough or is reduced by abrasion; (2) individual stones that are removed by impact; (3) a bank sideslope that is so steep that the angle of repose of the riprap material is easily exceeded; and (4) riprap gradation that is too uniform. The solution requires coarser riprap material and careful consideration of riprap gradation and angularity of the material. A translational slide is a riprap failure caused by the downslope mass movement of stones. The initial phases of a translational slide are indicated by cracks in the upper part of the riprap blanket that extend parallel to the channel. Slides are usually initiated by channelbed degradation that undermines the toe of the riprap blanket. It has been suggested that the presence of a ﬁlter blanket may provide a potential failure plane for Figure 8.6. Riprapfailure types. translational slides. The causes of translational slides include: (1) a bank sideslope that is too steep; (2) the presence of excess hydrostatic pore pressure; and (3) loss of material at the toe of the riprap blanket. The solution requires strengthening the toe of the riprap blanket and possible use of larger stones near the channel bed. Modiﬁed slump failure of riprap [Fig. 8.6(c)] is the mass movement of material within only the riprap blanket. Probable causes of modiﬁed slump are: (1) a bank sideslope that is too steep; and (2) lack of toe support. The solution requires adding coarse material at the toe of the embankment and reducing the sideslope angle in the upper part of the embankment. A sideslope failure is a rotation–gravitational movement of material along a surface of rupture that has a concave upward curve [Fig. 8.6(d)]. The cause
252
River stabilization
Figure 8.7. Riprap revetments.
of slump failures is related to shear failure of the underlying base material that supports the riprap. Probable causes of slump failures are: (1) excess pore pressure in the base material; and (2) sideslopes that are too steep. The solution requires reduction of the embankment slope or possibly draining the base material. Riprap should not be used at slopes steeper than 1V:1.5H. The upstream and the downstream ends of the blanket should be tied into the bank to prevent stream currents from unraveling the blanket. The most common method to tie into the bank is to dig a trench at both ends of the blanket (Fig. 8.7). The depth of a trench should be twice the blanket thickness, and the bottom width of the trench should be three times the thickness. The most effective method to prevent undermining is the launching apron sketched in Fig. 8.7(b). A ﬂexible “launching apron” is laid horizontally on the bed at the foot of the revetment, so that when scour occurs the materials will settle and cover the side of the scour hole on a natural slope. This method is recommended for cohesionless channel beds in which deep scour is expected. In cohesive channel beds, bank revetment should be continued down to the expected worst scour level and the excavation then reﬁlled. Alternatives to the launching apron include: (1) excavating and continuing the revetment down to a nonerodible material or to below the expected scour level; and (2) driving a “cutoff wall” of sheet piling from the toe of the revetment down to a nonerodible material or to below the expected scour level.
Riverbank protection 8.3
253
Riverbank protection
This section discusses bankprotection measures other than riprap: vegetation (Subsection 8.3.1), windrows and trenches (Subsection 8.3.2), sacks and blocks (Subsection 8.3.3), gabions and mattresses (Subsection 8.3.4), articulated concrete mattresses (Subsection 8.3.5), soil cement (Subsection 8.3.6), and retaining walls (Subsection 8.3.7). 8.3.1
Vegetation
Vegetation is probably the most natural method for protecting streambanks. It is less expensive than most structural methods and it improves environmental conditions for wildlife. The presence of vegetation below the water surface can effectively protect a bank in two ways. First, the root system helps to hold the soil together and increases overall bank stability by forming a binding network. Vegetation takes water from the soil, providing additional capacity for inﬁltration, and may improve bank stability by water withdrawal. Second, the exposed stalks, stems, branches, and foliage provide resistance to the streamﬂow, causing the ﬂow to lose energy by deforming the plants rather than by removing soil particles. Dense vegetation reduces ﬂow velocities and induces deposition. Above the water surface, vegetation prevents surface erosion by absorbing the impact of falling raindrops. Vegetation is generally divided into two broad categories: grasses and woody plants (trees and shrubs). The grasses are less costly to plant on an eroding bank and require a shorter period of time to become established. Woody plants offer greater protection against erosion because of their more extensive root systems; however, under some conditions the weight of the plant will offset the advantage of the root system. On very high banks, tree roots do not always penetrate to the toe of the bank. If the toe becomes eroded, the weight of the tree and its root mass may cause a bank failure. Using planted vegetation for streambank erosion control also has its limitations. These may include the following: (1) their failure to grow; (2) they are subject to undermining; (3) they may not withstand alternate periods of wetting and drying for varied durations; (4) they may be uprooted by freezing and thawing of ice; and (5) they may suffer wildlife or livestock damage. Native plants should normally be used because they have become adapted to the climate, soils, and other ecological characteristics of the area. Exotic plants, in contrast, are often met with local opposition. Plants chosen should have some tolerance to ﬂooding. A mixture of grasses, herbs, shrubs, and trees should be used to provide a diversity of wildlife habitats. Some nitrogenﬁxing plants may be required in poor soils and difﬁcult climates.
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River stabilization
Streambank zones depend on the ability of certain plants to tolerate various durations of ﬂooding and their attributes of dissipating wave and current energies. The splash zone located between normal high water and normal low water is the zone of highest stress. The splash zone is exposed frequently to wave wash, erosive river currents, ice and debris movement, wet–dry cycles, and freezingthawing cycles. This section of the bank would be inundated throughout most of the year (at least 6 months/yr). The bank zone lies above the normal highwater level; yet this site is exposed periodically to wave wash, erosive river currents, ice and debris movement, and trafﬁc by animals or humans. The site is inundated for at least a 60day duration once every 2–3 yrs. The water table in this zone frequently is close to the soil surface because of its closeness to the normal river level. The terrace zone, inland from the bank zone, is usually not subjected to erosive action of the river except during occasional ﬂooding. For the splash zone, only herbaceous semiaquatic plants, such as reeds, rushes, and sedges, are suggested for planting. These types of plants can tolerate considerable ﬂooding and are more likely to live in this zone. Reeds also protect streambanks in various ways. With their roots, rhizomes, and shoots, they bind the soil under the water, sometimes even above the water. In the reed zone along the riverbank, they form a permeable underwater obstacle that slows down the current and waves by friction, thereby reducing their impact on the soil. In the bank zone, both herbaceous and woody plants are used. These should still be quite ﬂood tolerant and able to withstand partial to complete submergence for up to several weeks. Various willows can be used in this zone. In periods of high water, the upper branches of such shrubs reduce the speed of the current and increase friction and thereby decrease the erosive force of the water. The branches of such shrubs have a great resilience, springing back after currents subside. The terrace zone is less signiﬁcant for bank protection because it is less often ﬂooded, and thus less easily eroded. The terrace zone contains native grasses, herbs, shrubs, and trees that are slightly less ﬂood tolerant than those in the bank zone. The tree species also become taller and more massive. Trees are noted for their value in stabilizing banks of streams and rivers. The banks of some rivers have not been eroded for durations of 100–200 yrs because heavy tree roots bind the alluvium of ﬂoodplains. A combination of trees, shrubs, and grasses in this zone will not only serve as an integrated plant community for erosion control, but will improve wildlife habitat diversity and aesthetic appeal. Grasses can be planted by hand seeding, sodding (transplanting clumps of grass or herbaceous plants, sprigging (planting plant stems or rhizomes), or by mechanical spreading mulches consisting of seed, fertilizer, and other organic mixtures. Several commercial manufacturers now market erosioncontrol
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255
matting that will hold the seed and soil in place until new vegetation can become established. The matting is generally installed by hand and secured to the bank where plantings have been made to prevent erosion. A fence should be placed along the top bank if livestock requires access to the stream. We construct reed rolls by combining sections of sod, rhizomes, and shoots, and enclosing them within a wire net, and placing all components in a trench. Usually the sod must be held in place with wire netting or stakes. Shrublike willow, dogwood, and alder transplants or 1yrold rooted cuttings are effectively used in this zone and can augment the sodding practice. Hydroseeding can be a useful and effective means of direct seeding, particularly on steep slopes. Often barges with hydroseeders mounted on them can be ﬂoated on the stream and used adjacent to the site. Seeds should be blown on ﬁrst in a water slurry and then mulches applied following seeding to reduce soil moisture loss. The mulch also will tend to tie down and cover the seeds and reduce immediate surface soil erosion by wind and water. Case Study 8.1 illustrates the effectiveness of vegetation protection on the neck of Thompson Bend, Mississippi River. Case Study 8.1 Thompson Bend on the Mississippi River, United States. Thompson Bend is located on the right descending bank of the Mississippi River between river miles 30 and 45, above the conﬂuence of the Mississippi and Ohio Rivers; see Fig. CS.8.1.1. The river ﬂows in a gooseneck encompassing approximately 10,000 acres (40 km2 ) of valuable agricultural land. At the throat
Figure CS.8.1.1. Thompson Bend of the Mississippi River.
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of the bend, the overland distance is approximately 2 miles. The river distance along the thalweg is approximately 14 miles (22.5 km). During large ﬂoods the river naturally tries to ﬂow straight across Thompson Bend. The water surface drop along the thalweg is 7 ft (2.1 m) for a slope 0.5 ft/mile (9.5 cm/km) along the river and 3.5 ft/mile (66 cm/km) across the neck. In the early 1980s, severe erosion of the upper bankline began along the right descending bank in the upper reaches of the bendway. In addition, localized surface erosion reached an estimated rate of 40,000 (tons/acre)/event [16,000 (tons/hectare)/event]. Continued erosion could have allowed for development of a chute cutoff across the bend. This would have impaired navigation to the steep slopes and resulting high velocities in the new channel across the bend. It would also have destroyed thousands of acres of valuable farmland and changed the river regime for miles upstream and downstream. Thompson Bend clearly illustrates the vital importance that vegetation exerts on controlling overbank scour. The revegetation process began in 1985 and early 1986. The results were immediately evident during the fall ﬂood of 1986. Very little erosion was observed. The area was tested again in the ﬂood of 1990, when very little erosion occurred. The Thompson Bend also suffered very little visible damage during the Great Flood of 1993, when record high stages occurred, and the duration of the overland ﬂows reached an unprecedented 130 days. However, the ﬂood took its toll on the vegetation. Numerous trees that were inundated for most of the 130 days died. The ﬂood of 1994 did not allow any signiﬁcant revegetation to occur. Subsequently, the ﬂood of 1995, the second highest ever at Cape Girardeau, produced some visible scour for the ﬁrst time since the revegetation was initiated. However, the erosion was very minor compared with the massive amounts of scour that occurred in the early 1980s.
8.3.2
Windrows and trenches
A windrow revetment consists of piling a sufﬁcient supply of erosionresistant material on the existing land surface along the bank. Trenches are similar except that the material is buried as sketched in Fig. 8.8(b). Windrows and trenches permit the area between the natural riverbank and the windrow to erode through natural processes until the erosion reaches and undercuts the supply of riprap. As the rock supply is undercut, it falls onto the eroding area, thus giving protection against further undercutting and eventually halting further landward movement. High banks tend to produce a nonuniform revetment alignment and have a tendency for large segments of the bank to break loose and rotate slightly. Comparatively, low banks simply slough into the stream.
Riverbank protection
Figure 8.8. Windrows and trenches.
8.3.3
257 The velocity and the stream characteristics dictate the size of stone forming a windrow revetment. The size of stone must be large enough to resist being transported by the stream. An important design parameter is the ratio of the relative thickness of the ﬁnal revetment to the stone diameter. Large stone sizes will require more material than smaller stone sizes to produce the same relative thickness. A wellgraded stone is important to ensure that the revetment does not fail from leaching of the underlying bank material. The stream velocity was found to have a strong inﬂuence on the ultimate sideslope of the revetment. In general, the greater the velocity, the steeper the sideslope of the ﬁnal revetment.
Sacks and blocks
Burlap sacks ﬁlled with soil or sand–cement mixtures have long been used for emergency work along levees and streambanks during ﬂoods (Fig. 8.9). The sacks protect streambanks for which riprap of suitable size and quality is either not available or must be transported over long distances. Although most types of sacks are easily damaged and will eventually deteriorate, sacks ﬁlled with sand–cement mixtures can provide longterm protection if the mixture has set up properly. Precast cellular blocks can be manufactured with locally available sand, cement, and aggregates or can be obtained from commercial sources. Blocks are durable, exceeding riprap in freeze/thaw resistance and are less likely to be encapsulated and lifted off the bank by ice. Also, most designs provide easy pedestrian access to the water’s edge, and may be more aesthetic than competing materials. Channel boundary roughness is less than with many other techniques, and wave runup is less than that for smooth concrete surfaces. Cellular blocks are cast with openings to provide for drainage and to allow vegetation to grow through the blocks, thus permitting the root structure to strengthen the bank. Fabric or a gravel blanket can be used as a ﬁlter under the blocks if there is any danger that the bank soil will be eroded through the block
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openings by streamﬂow or seepage. Hand placement is frequent instead of the use of specialized equipment. After the blocks have been placed, the revetment offers sufﬁcient ﬂexibility to conform to minor changes in bank shape. H V Weepholes should be included in the revetment design to allow drainage of groundwater and prevent pressure buildup that could cause revetment failure. V H Sacks do have certain advantages over stone riprap, as follows: (1) they allow possible placement on steep slopes; (2) they use locally available materials; (3) they result in a smooth boundary, if channel conveyance is a major consideration; and (4) the cobV H blestone effect may be considered to be more aesthetic. Sack revetments also have disadvantages compared with stone riprap: (1) they are highly labor intensive, and Figure 8.9. Sacks and blocks. thus are generally more costly than stone; (2) they have a tendency to act monolithically and therefore do not have as much ﬂexibility as riprap and are more susceptible to excess hydrostatic pressure; (3) uniformly sized sacks are not as effective against erosion and leaching as wellgraded stones, and therefore a sack revetment is more likely to require a ﬁlter material; and (4) synthetic bags may be vulnerable to environmental hazards such as ﬁre, ice, vandalism, and livestock trafﬁc. The preferred placement technique is sketched in Fig. 8.9. A rule of thumb is to consider ﬂat placement only if the bank slope is ﬂatter than 1V :2.5H . On slopes of 1V :2H , the bags should be overlapped by being placed with the long dimension pointing toward the bank, whereas on slopes steeper than 1V :2H , the bags should be overlapped with the short dimension pointed toward the bank. The maximum slope should be 1V :1H . V
8.3.4
H
Gabions and mattresses
Gabions are patented rectangular wire boxes (or baskets) ﬁlled with relatively smallsized stone, usually less than 8 in. (20 cm) in diameter. Where ﬂow
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velocities are such that small stones would not be stable if used in a riprap blanket, the wire boxes provide an effective restraint. Limiting recommended maximum velocity for use of gabions ranges from 8 to 15 ft/s (2 to 5 m/s), depending on the manufacturer. The baskets are commercially available in a range of standard sizes and are made of heavy galvanized steel wire with a polyvinyl chloride (PVC) coating when used in a corrosive environment. They are supplied at a job site folded ﬂat and are assembled manually, by use of noncorrosive wire. The baskets are normally 0.5 m deep × 1 m × 2 m and are set on a graded bank for revetments. A ﬁlter blanket or ﬁlter cloth is used where required to prevent leaching of base material and undermining of the baskets. Box gabions are normally stacked on relatively steep slopes to form a massive structure capable of resisting the forces of both river ﬂows and also unstable bankline materials. The ﬂexibility of their mesh and ﬁller stone allows them to maintain their structural integrity even after some degree of displacement, undercutting, or settlement. Box gabion structures generally are aligned either along the streambank toe to form a retaining wall for the bank materials or out from the bank to form dikes for diverting ﬂows away from the bank. Examples are shown in Fig. 8.10. Mattress gabions are shaped into shallow, broad baskets and are tied together side by side to form a continuous blanket of protection. They are normally placed on a smoothly graded riverbank slope. Gabions and mattresses are among the more expensive methods of streambank erosion protection. However, their record of satisfactory performance is making them more and more popular. Unit costs of bankline length protected vary widely, depending on the complexity of the protection design.
Figure 8.10. Gabions and mattresses.
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Gabion structures need to be periodically inspected, and proper maintenance ensures reliable performance. Wire mesh is subject to damage from ﬂoating debris, water pollutants, corrosion, wear from highvelocity sediments, and vandalism. Any cracks or breaks in the PVC coating will expose the wire to corrosive elements, thereby providing negligible advantage over standard galvanized gabions. Freezing temperatures have caused the coating to lose its ductility. Fabricating, ﬁlling, and lacing gabions under these conditions have caused an unsatisfactory number of coating cracks and splits. 8.3.5
Articulated concrete mattresses
In large rivers, precast concrete blocks held together by steel rods or cables can be used to form ﬂexible articulated mats, sketched in Fig. 8.11. Block sizes may vary to suit the contour of the bank. It is particularly difﬁcult to make a continuous mattress of uniformly sized blocks to ﬁt sharp curves. The open spacing between blocks permits removal of bank material unless a ﬁlter blanket of gravel or plastic ﬁlter cloth is placed underneath. For embankments that are subjected to only occasional ﬂood ﬂows, the spaces between blocks may be ﬁlled with earth and vegetation can be established. The use of articulated concrete mattresses (ACMs) has been limited primarily to the Mississippi River. This is due to the large cost of the plant required for the placement of the mattress beneath the water surface. The present ACM originated from experiments started in 1915 to develop a ﬂexible and permanent underwater willow mattress. After many failures and discouragements, the concrete mattress sketched in Fig. 8.11 was developed. The basic unit of this mattress is 4 ft wide (1.3 m) × 25 ft long (7.6 m) and 3 in. (7.5 cm) thick.
Figure 8.11. Articulated concrete mattresses.
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The articulated concrete is ﬂexible, strong, and durable, and ensures complete coverage of the riverbank when properly placed. However, ∼8% of its surface area is open, which permits ﬁnes to pass through. The open areas are undesirable but necessary to facilitate placement of the mattress in swift deep water, to relieve hydrostatic pressure, and to provide the required ﬂexibility. Yearly maintenance runs approximately 2% of inplace work. This type of revetment has an excellent service record and is considered the standard for the Lower Mississippi River. 8.3.6
Soil–cement
The use of concrete, soil–cement, or rollercompacted concrete generally comes into play when the design and/or site conditions preclude more conventional ﬂexible techniques. Although concrete generally implies permanency, the material itself can be manufactured and placed to respond and withstand changing ﬁeld conditions. The use of concreterelated materials offers a costeffective alternative that has proven to be effective under a wide variety of ﬁeld conditions. The key to success, however, is proper recognition of the rigid nature of the material in a ﬂexible boundary channel. In areas where riprap is scarce, the combination of onsite soil with cement provides a practical alternative. Figure 8.12 sketches a typical soil–cement construction for bank protection. For use in soil–cement, soils should be easily pulverized and contain at least 5%, but not more than 35%, silt and clay (material passing through the No. 200 sieve). Finer soils usually are difﬁcult to pulverize and require more cement as do 100% granular soils that have no material passing through the No. 200 sieve. Soil–cement can be placed and compacted on slopes as steep as 2H :1V . Best results have been achieved on slopes no steeper than 1V :3:H . A stairstep construction is recommended on channel embankments with relatively steep slopes. Placement of 6–9ft (2–3 ms)wide horizontal layers of soil–cement (6in. or 15cmthick layers) can progress more rapidly than a
Figure 8.12. Soil–cement.
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large quantity of ﬁll material. Special care should be exercised to prevent raw soil seams between successive layers of soil–cement. If uncompleted embankments are left at the end of the day, a sheepsfoot roller should be used on the last layer to provide an interlock for the next layer. The completed soil–cement installation must be protected from drying out for a 7day hydration period. After completion, the material has sufﬁcient strength to serve as a roadway along the embankment. When velocities exceed 6–8 ft/s and the ﬂow carries sufﬁcient bedload to be abrasive, the aggregates should contain at least 30% gravel particles retained on a No. 4 (4.75mm) sieve. It should be emphasized that soil–cement provides rigid bank protection. The depth of the bank protection should be sufﬁcient to protect the installation from the anticipated total scour. A soil–cement blanket with 8%–15% cement may be an economical and effective streambank protection method for use in areas where vegetation is difﬁcult to establish in sandy bank material. However, soil–cement has three major disadvantages; impermeability, low strength, and susceptibility to temperature variations. If the bank behind the blanket becomes saturated and cannot drain, failure may occur. Also, because a soil–cement blanket is relatively brittle, very little, if any, trafﬁc (vehicular, pedestrian, or livestock) can be sustained without cracking the thin protection veneer. In northern climates the blanket can easily break up during freeze–thaw cycles. 8.3.7
Retaining walls
Retaining walls are nearvertical structures designed to prevent streambank erosion or failure. Vertical retaining walls provide a substantial increase in waterfront land area and often improve the access to water. Although some soils may be relatively stable in a vertical embankment under dry conditions, wet soils are unstable in a vertical embankment. Retaining walls, have been classiﬁed into three distinct types, discussed below: (1) gravity walls; (2) cantilever walls; and (3) sheetpiling walls. Gravity walls are massive walls that rely on their mass to restrain the movement of soil. The walls are constructed to such proportions that any developed soil or hydrostatic pressures that would tend to cause movement or failure of the wall are resisted by the weight of the wall and resultant shear forces that develop at the base of the wall. Examples of gravity walls are shown in Fig. 8.13. The concrete walls in Fig. 8.13(a) are usually designed such that no tension is developed within the concrete and no reinforcing steel is required for their construction. Variations of this type of wall may consist of stone masonry, as
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Figure 8.13. Gravity walls.
shown in Fig. 8.13(b). This type of wall has been successfully adapted to the construction of very large walls. Retaining walls constructed of stoneﬁlled gabions in Fig. 8.13(c) are another type of gravity wall. As discussed in Subsection 8.3.4, gabions are baskets made of metal mesh or geotextiles that are ﬁlled with stone and have wide usage in hydraulic structures. To form a wall, the gabions are simply stacked atop one another in such a number as to provide sufﬁcient mass to retain the soil within a streambank.
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The timber crib walls in Fig. 8.13(d) are similar to common fence construction except that all wood should be treated with a preservative to minimize deterioration that is due to repetitive wetting and drying or insect activity and the toe of the crib should always be protected with riprap. The most common cause of failure is scour around the pilings, followed by the structure tipping over because of the pressure of the bank behind the bulkhead. The ﬁll material should drain freely. Filter fabric or gravel can be placed as a ﬁlter behind openings in the fence to prevent ﬁne soils from leaching through. The bulkhead should be tied into the bank at the upstream and the downstream ends of the structure to prevent ﬂanking from ﬂow behind the bulkhead. The crib walls in Fig. 8.13(d) are constructed of interlocking structural members of treated wood, precast reinforced concrete, or metal, usually galvanized steel or aluminum. The structural components of a crib wall interlock to form “boxes” that are ﬁlled with stone, blocks, sand, or rubble to provide the mass necessary for stability. Caissons, or large “boxes” of reinforced concrete in Fig. 8.13(e) are constructed on land and then ﬂoated to the site and ﬂooded and sunk to form a continuous wall. The caisson is then ﬁlled with concrete or compacted sand to provide the mass required for stability. A cantilever wall refers to a reinforced concrete base with a stem wall cantilevered upward from the base. The stem is designed to resist the lateral earth and hydrostatic forces. The soil above the base provides mass to resist movement. Reinforced concrete is required for strengthening cantilevered walls. The wall stem may be supported or stiffened by buttresses fronting the wall or by counterforts behind the wall, as shown in Fig. 8.14. The simplest form of cantilever wall would be wooden posts driven into the streambed with wooden lagging or timbers placed behind the posts or nailed or bolted to the posts in order to retain the soil within the banks. This type of construction is restricted to small walls, although larger walls have been constructed in a similar manner by use of steel H piling rather than posts and with timber lagging or reinforced concrete panels placed between the ﬂanges of the piling in order to retain the soil. Sheetpiling walls are sometimes referred to as ﬂexible walls or ﬂexible bulkheads. As shown in Fig. 8.15, the walls are normally constructed by driving of the sheet piling and then excavation of the earth fronting the wall. Although wooden and reinforced concrete sheet piling have been used, steel sheet piling is most commonly used. Because of the limited stiffness of the sheet piling and the resulting large deﬂections, the height of these walls is somewhat restricted. Anchors increase the allowable height of ﬂexible cantilevered walls. Earth anchors are constructed to provide a horizontal restraining force to the wall. Walls of this sort are often referred to as “anchored” or “tiedback” walls, as
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Figure 8.14. Reinforced cantilevered walls.
Figure 8.15. Sheetpiling walls.
shown in Fig. 8.15(b). Anchors may be constructed by the drilling of horizontal or angled holes through the wall and into the retained soil a distance beyond any zone of potential active movement of the soil. Threaded steel rods are then pressure grouted into the holes and bolts are torqued against the wall to tension the rods and to apply a restraining force on the wall. Anchors may also
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Figure 8.16. Diagrams of forces that act on a retaining wall.
be provided when a structural member, or “deadman,” is buried beyond the zone of active movement within the soil and a rod connected to that member is tensioned against the wall. In some cases, when the walls serve as components of a hydraulic structure or a cofferdam, it may be possible to brace against the face of the wall and restrain movement. Retainingwall stability. The external forces that act on a retaining wall are sketched in Fig. 8.16. The lateral earth pressures that act behind the structure are generally referred to as active. The earth pressures that resist movement in front of the wall are referred to as passive. Hydrostatic pressures result from groundwater behind the wall and water within the channel. For the development of active earth pressure, the assumption is made that the wall moves or deﬂects enough to allow for the development of shearing resistance under the structure. The pressure exerted under the wall is necessary to maintain equilibrium of the soil mass. The active and the passive pressures represent the limits of a broad spectrum of stress conditions. When a wall is restrained and not free to move, “atrest” earthpressure conditions are assumed to exist. These conditions lie between the active and passive limits. For design purposes, the limiting conditions of active and passive earth pressures are normally selected for freestanding and cantilevered walls. For braced or restrained walls, an atrest condition may be assumed, often in concert with the limiting conditions. The active and the passive earth pressures are usually computed based on methods originally proposed by Rankine and modiﬁed by others. The atrest earth pressures are usually computed by use of factors developed from physical
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tests or empirical means. Lateral earth pressures resulting from granular soils are usually more accurately predicted than those resulting from cohesive soils. For this reason, and for purposes of drainage, granular material is usually selected as backﬁll for retaining walls. With the rise and fall of stream stages, a difference in water elevation across the retaining wall may develop. The differences in water elevation will then result in an imbalance in hydrostatic forces that act on the wall. Further, hydrostatic forces acting upward on the base of the wall will lessen the effective weight of the structure, thereby lessening its ability to retain the soil. The stability of freestanding walls is a relatively simple problem of statics, once the forces acting on a wall have been deﬁned. Groundwater control is required for preventing imbalances in water elevations across a retaining wall. Whenever conditions allow, the backﬁll for a wall should be a clean, freedraining granular material. Drains or weepholes through the wall should be provided to ensure a prompt lowering of the groundwater behind the wall with the lowering of the stage within the stream. Erosion control is critical to the stability of all retaining walls. Under submerged conditions, there is always the potential for the loss of material at the toe of the retaining wall.
8.4
River ﬂowcontrol structures
Flowcontrol structures are designed to reduce hydrodynamic forces against streambanks by controlling the direction, velocity, or depth of ﬂowing water. Among the most important properties of a ﬂowcontrol structure is its degree of permeability. As used here, the term “permeable” means that a structure has deﬁnite openings through which water is intended to pass. An impermeable structure may deﬂect a current entirely, whereas a permeable structure may serve mainly to reduce the ﬂow velocity. Structures made of riprap, or ﬁlled with riprap, have some degree of permeability, but these are classed as impermeable because they act essentially as impermeable barriers to a rapidly moving current of water. Types of ﬂowcontrol structures include hardpoints (Subsection 8.4.1), spurs (Subsection 8.4.2), guidebanks (Subsection 8.4.3), retards (Subsection 8.4.4), dikes (Subsection 8.4.5), jetties (Subsection 8.4.6), fences (Subsection 8.4.7), vanes (Subsection 8.4.8), bendway weirs (Subsection 8.4.9), and drop structures (Subsection 8.4.10).
8.4.1
Hardpoints
Hardpoints consist of stone ﬁlls spaced along an eroding bank line (Fig. 8.17). The structures protrude only short distances into the river channel and are supplemented with a root section extending landward into the bank to preclude
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Figure 8.17. Hardpoint.
ﬂanking, should excessive erosion persist. The majority of the structure cannot be seen as the lower part consists of rock placed underwater, and the upper part is covered with topsoil and seeded with native vegetation. The structures are especially adaptable in long, straight reaches not subject to direct attack.
8.4.2
Spurs or groynes
A spur, also called a groyne, is a structure or embankment projected a fair distance from the bank into a stream to deﬂect ﬂowing water away from the bank. Spurs prevent erosion of the bank and establish a more desirable width and channel alignment. By deﬂecting the current away from the bank and causing sediment deposits, a spur or a series of spurs may protect the streambank more effectively and at a lower cost than riprap. Also, if the location of any scour is moved away from the bank, failure of the riprap on the spur can often be repaired before damage is done to structures along and across the rivers. Very short spurs are similar to hardpoints. Long spurs or groynes may also be called spur dikes, and very long spurs can be referred to as dikes and jetties (Subsections 8.4.4 and 8.4.5). Spurs are also used to channelize a wide, poorly deﬁned stream into a welldeﬁned channel that neither aggrades nor degrades, thus maintaining its location from year to year. Spurs on streams with suspended sediment discharge induce sedimentation to establish and maintain the new alignment.
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Figure 8.18. Spurs of groynes.
As the spur length increases, the scour depth at the spur tip increases, the severity of ﬂow deﬂection increases, and the length of channel bank protection increases. The projected length of impermeable spurs should be held to less than 15% of the channel width at bankfull stage. The projected length of permeable spurs should be held to less than 25% of the channel width. The spacing of spurs in a bankprotection scheme is a function of the spur’s length, angle, and permeability, as well as the channel bend’s degree of curvature (see Fig. 8.18). As a rule of thumb, the spacing should be 3–5 times the projected length. Reducing the spacing between individual spurs results in a reduction of the local scour at the spur tip and causes the ﬂow thalweg to stabilize farther away from the concave bank toward the center of the channel. Spurs angled downstream produce a less severe constriction of ﬂows than those angled upstream or normal to the ﬂow. Retardant spurs should be designed perpendicular to the primary ﬂow direction. The spur height should be sufﬁcient to protect the regions of the channel bank affected by the erosion processes active at the particular site. If the design ﬂow stage is lower than the channel bank height, spurs should be designed to a height
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no more than 3 ft (1 m) lower than the design ﬂow stage. If the design ﬂow stage is higher than the channel bank height, spurs should be designed to bank height. Permeable spurs should be designed to a height that will permit passage of heavy debris over the spur crest and not cause structural damage. When possible, impermeable spurs should be designed to be submerged by approximately 3 ft (1 m) under their worst design ﬂow condition, thus minimizing the impacts of local scour and ﬂow concentration at the spur tip and the magnitude of ﬂow deﬂection. Permeable spurs should be designed with level crests unless bank height or other special conditions dictate the use of a sloping crest design. Impermeable spurs should be designed with a slight fall toward the spur head to allow different amounts of ﬂow constriction with stage (particularly important in narrow channels). A simple straight spur head form is recommended. The spur head or tip should be as smooth and rounded as possible. Smooth, wellrounded spur tips help minimize local scour, ﬂow concentration, and ﬂow deﬂection.
8.4.3
Guidebanks
Guidebanks are placed at bridge crossings near the ends of approach embankments to guide the stream through the bridge opening, as shown in Fig. 8.19. Constructed properly, ﬂow disturbances, such as eddies and cross ﬂow, will be eliminated to make a more efﬁcient waterway under the bridge. They are also used to protect the highway embankment and reduce or eliminate local scour at the embankment and adjacent piers. The recommended shape of a guidebank is
Figure 8.19. Guidebank.
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a quarter ellipse with a major to minor axis ratio of 2.5. The major axis should be approximately parallel to the main ﬂow direction. For some ﬂow conditions, a short curved guidebank on one side and a long straight bank on the other may be the best solution. The crest elevation should be 1 ft (30 cm) higher than the elevation of the design ﬂood, with the effect of the contraction of the ﬂow taken into consideration; this is because the design ﬂow should not overtop the guidebank. Besides improved erosion protection, guidebanks provide a more efﬁcient water ﬂow (less head loss) through a bridge opening. They also decrease scour depth and move the scour hole away from the abutments.
8.4.4
Retards
A retard is a low permeable structure located near the toe of the bank slope parallel to the streamﬂow, as shown in Fig. 8.20. The function of a retard is to decrease velocity behind the structure and eliminate erosive secondary currents, thereby inducing deposition and growth of vegetation. Retards are most successful on streams that carry a large bed material load. Occasional tieback connections to the bank are desirable and are mandatory where the retard is located any distance away from the bank. A satisfactory structure height is usually 1/3 to 2/3 of the streambank height. A top elevation
Figure 8.20. Retard.
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that is less than the bank height allows drift during high ﬂows to pass over and lodge against the bank rather than against the structure. Retards have several advantages over the basic technique of sloping the bank and paving with stone, as follows: (1) they can be adapted to a wide range of conditions; (2) the channel alignment can be improved; (3) they are usually less costly; and (4) little if any bank grading is required, which simpliﬁes rightsofway acquisition and material disposal problems. Retards are less desirable than stone paving in the following respects: (1) they offer no direct immediate improvement in bankslope stability; (2) they offer no direct immediate prevention of erosion by overbank drainage and return ﬂows; (3) they are subject to damage by ice, drift, ﬁre, vandalism, and deterioration from the elements; (4) they may reduce channel capacity, particularly after vegetation is established; (5) they interfere with local access to the stream channel; and (6) they may not be aesthetically pleasing. Pile retards can be made of concrete, steel, or timber. They may be used in combination with bankprotection works such as riprap. The retard then serves to reduce the velocities sufﬁciently so that either smaller riprap can be used, or riprap can be eliminated. The design of timber pile retards is essentially the same as timber pile dikes (Subsection 8.4.5). 8.4.5
Dikes
There are two principal types of dikes: (1) stone ﬁll dikes; and (2) timber pile dikes. Stone ﬁll dikes are very long spur dikes or groynes. Except for their extended length, their characteristics are described in Subsections 8.4.2 and 8.4.3. The design approach for dikes and retards requires a comparison of recent and old aerial photographs to identify bankcaving rates and to become familiar with the overall streammeander pattern. A preliminary determination of upstream and downstream termination points and general structural alignment can then be made. A site inspection then conﬁrms the general stream characteristics, survey of cross sections, bed and bank material, soil boring if piles are to be driven, and drift carried by the stream. The usual alignment problem is to stabilize the outside of a bend. It is not mandatory to have a constant radius of curvature, but the alignment should be smooth without abrupt changes in radius, as sketched in Fig. 8.21(a). Unless a retard is located in a zone of very mild attack, toe protection is highly desirable [Fig. 8.21(b)]. Toe deepening can cause structural failure and/or loss of ﬁller material from the retard. Toe protection is more effective than excess penetration of piles or posts. Toe protection should be heavier at and downstream of the point of maximum attack determined from aerial photographs and site inspection and survey. Extending the heavier toe protection downstream of this
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273
Figure 8.21. Bankstabilization sketch.
point is desired because impingement points usually tend to migrate downstream with time. The simplest method is to windrow aggraded stone along the channel side of the retard at a uniform rate per linear foot, varying the rate along the bend if appropriate. A rule of thumb is that 1 ton of stone per foot will accommodate 3 ft (1 m) of scour. Mistakes in selecting upstream and downstream termination points are a common reason for retard failure because points of attack migrate down valley with time, which can be detected from comparative aerial photographs. Therefore, the downstream termination should be downstream of the existing knickpoint of bank stability. Conversely, the upstream terminating point should correspond to the downstream end of the adjacent pointbar deposition zone. The usual termination method is to turn the structure azimuth back into the bank at an angle. The top elevation can be kept constant.
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Dikes and retards allow the channel alignment to be modiﬁed if that is a project requirement. They are well suited to the incremental construction approach. Because dikes and retards extend into the channel, they are subject to severe attack. Therefore inspection and maintenance are essential. Channel capacity at high ﬂow is decreased initially. The channel will usually adjust by forming a deeper, though narrower, cross section, and the ultimate effect may even be an increase in capacity. Because conservative assumptions on future deposition and vegetative growth would be necessary, dikes or retards must be approached with caution on ﬂoodsensitive projects. Timber pile dikes may consist of closely spaced single, double, or multiple rows of timber piles, as shown in Fig. 8.22. Wire fence may be used in conjunction with pile dikes to collect debris and thereby cause effective reduction of velocity. Double rows of timber piles can be placed together to form timber
Figure 8.22. Timber pile dikes.
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275
cribs, and rocks may be used to ﬁll the space between the piles. Timber pile dikes are vulnerable to failure through scour. This can be overcome if the piles can be driven to a large depth to achieve safety from scour, or the base of the piles can be protected from scour with dumped rock in sufﬁcient quantities. The arrangement of timber piles depends on the velocity of ﬂow, quantity of suspended sediment transport, and depth and width of the river. If the velocity of ﬂow is large, timber pile dikes are not likely to be very effective. On the other hand, they are quite effective in moderate ﬂow velocities with high concentrations of suspended sediments. The deposition of suspended sediments in the pile dike ﬁeld is a necessary consequence of reduced velocities. If there is not sufﬁcient material in suspension, or if the velocities in the dike ﬁelds are too large for deposition, the permeable timber pile dikes will be only partially effective in training the river and protecting the bends. The length of each dike depends on channel width, position relative to other dikes, ﬂow depth, and available pile lengths. Generally, pile dikes are not used in large and deep rivers. On the other hand, banks of wide shallow rivers can be successfully protected with dikes. The spacing between dikes varies from 3 to 20 times the length of the upstream dike, with closer spacing favored for best results. 8.4.6
Jetties
Jetty ﬁelds add roughness to a channel or overbank area to train the mainstream along a selected path. The added roughness along the bank reduces the velocity and protects the bank from erosion. Jetty ﬁelds are usually made up of steel jacks tied together with cables. Both lateral and longitudinal rows of jacks are used to make up the jetty ﬁeld, as shown in Fig. 8.23. The lateral rows are usually angled approximately 45◦ to 70◦ downstream from the bank. The spacing varies, depending on the debris and sediment content in the stream, and may be 50 to 250 ft (15 to 80 m) apart. Jetty ﬁelds are effective only if there is a signiﬁcant amount of debris carried by the stream and the suspended sediment concentration is high. When jetty ﬁelds are used to stabilize meandering rivers, it may be necessary to use jetty ﬁelds on both sides of the river channel because in ﬂood stage the river may otherwise develop a chute channel across the point bar. Steel jacks are devices with basic triangular frames tied together to form a stable unit. The resulting framework is called a tetrahedron. The tetrahedrons are placed parallel to the embankment and cabled together with the ends of the cables anchored to the bank. Wire fencing may be placed along the row of tetrahedrons. The aesthetical value of steel jacks is minimal, particularly in rivers used for boating, ﬁshing, rafting, and kayaking.
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Figure 8.23. Jetties.
8.4.7
Fences
Fencing can be used as a lowcost bankprotection technique on small to medium size streams that are usually wide and shallow (see Fig. 8.24). Special structural design considerations are required in areas subject to ice and ﬂoating debris. Both longitudinal (parallel to stream) fence retards and transverse (perpendicular to stream) fences have been used in the prototype with varying degrees of success: (1) the channel gradient must be stable and not be steep (subcritical ﬂow); (2) toe scour protection can be provided by extending the support posts well below the maximum scour expected or by placing loose rock Figure 8.24. Fences. at the base of the fence to launch
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Figure 8.25. Vane dikes.
downward if scour occurs at the toe; (3) tiebacks to the bank are important to prevent ﬂanking of the fence and to promote deposition behind the fence; (4) fence retards generally reduce attack on the bank so that vegetation can become established; and (5) metal or concrete fences are preferred because of ice damage and ﬁre loss of wooden fences.
8.4.8
Vanes
Vane dikes are structures designed to guide the ﬂow away from an eroding bank line (see Fig. 8.25). The structures can be constructed of rock or other erosionresistant material, the tops of which are constructed below the design watersurface elevation and would not connect to the high bank. Water is free to pass over or around the structure, with the main thread of ﬂow near the surface directed away from the eroding bank. The ﬁndings from model investigations of these structures include the effects of various vane dike orientation, vane dike length, and gap length. The ends of the dikes are subjected to local scour, and appropriate allowance should be made for loss of dike material into the scour hole. The rock size to be used for the dike depends on availability of material. Large rocks are generally used to cover the internal section constructed with smaller rocks or earthﬁll. Side slopes of 1V:5H and 1V:2H are common.
8.4.9
Bendway weirs
A bendway weir is a low sill located in a navigation channel bend. Bendway weirs are usually angled 20◦ to 30◦ upstream with approximately equal length and spacing. The level crest is at an elevation low enough to allow normal
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Figure 8.26. Bendway weirs.
river trafﬁc to pass over unimpeded (Fig. 8.26). The weir should be high and long enough to intercept a large percentage of ﬂow at the river cross section. Weirs are typically built in sets (4 to 14 weirs per bend) and are designed to control nearbed ﬂow velocities and current directions through the bend and well into the downstream crossing. Secondary ﬂows are reduced, and water ﬂowing over the weir is redirected at an angle perpendicular to the weir. When the weirs are angled upstream, water is directed away from the outer bank and toward the inner part of the bend. Prototype and model results indicate that construction of a series of bendway weirs in a navigation channel bend results in: (1) improved navigation through the bend; (2) deposition at the toe of the revetment on the outside of the bend and thus an increase in bank stability; (3) more uniform surfacewater velocities across any cross section; (4) ﬂow patterns in the bends that are generally parallel with the banks; and (5) the thalweg of the channel being moved away from the toe of the outerbank revetment.
Case Study 8.2 Bendway weirs of the Mississippi River, United States. A physical movablebed model study of a 20mile section of the middle Mississippi River was built at Waterways Experiment Station and operated for the St. Louis District of the Corps of Engineers (Pokrefke and Combs, WES, 1998, personal communication). The model was designed to study and improve bendway navigation and address environmental concerns: (1) highmaintenance dredging costs; (2) need to protect least tern nesting areas; (3) constricted navigation channel; (4) high velocities; (5) detrimental highﬂow current patterns; and (6) inadequate navigation channel widths in the crossing downstream of the bend. With this model, the bendway weir concept, developed in January 1988, was reﬁned for Price’s Landing and Brown’s Bends. Bendway weirs have been
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Figure CS.8.2.1. Bendway weirs of the Mississippi River.
tested in 11 models at WES, e.g., Fig. CS.8.2.1, to improve both deep and shallowdraft navigation, align currents through highway bridges, divert sediment, and protect docking facilities. From 1989 to 1995, over 120 weirs have been built in 13 bends of the Mississippi River. Analysis of the ﬁve oldest weir installations show that from 1990 to 1995 dredging was reduced by 80%, saving
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U.S. $3 million. In addition, towboat accidents were reduced, tow delay times at bends were reduced, sediment and ice management was improved, least tern nesting areas were undisturbed, aquatic habitat area was increased, and ﬁsh size and density in the weir ﬁelds increased (ﬁvefold in some areas).
8.4.10
Drop structures
Gradecontrol structures, or drop structures, are used to reduce the slope of a channel. The purpose of gradecontrol structures is to stabilize the banks and bed of a channel by reducing stream slope and ﬂow velocity. The efﬁciency of gradecontrol structures decreases with increasing stream size. Vertical concrete, timbers and sheet pile weirs, riprap sloping sills, and soil–cement or gabion drop structures can be designed, as sketched in Fig. 8.27, after considering the stability of the structure and the depth of the scour hole at the toe of the structure. Drop structures must be designed to contain the effects of energy dissipation while protecting both the bed and the banks upstream and downstream of the structure. Protection of the approach sections is usually ensured with riprap. Rock protection is efﬁcient in breaking up the remaining turbulence in the stilling basin and in protecting bed and bank materials from direct attack. Filters must also be designed when appropriate. Log and timber drop structures are appropriate in very small streams and ﬁeld gullies [Fig. 8.27(a)]. Corrugated pipe gradecontrol structures consist of a culvert driven or buried vertically into the stream bed, depending on channelbed width. Riprap is placed within the culvert and for a short distance downstream to prevent scour below the structure. The downstream sidewall of the culvert section is trimmed ﬂush with the bottom. The structures are effective for small drops, and a series of them may be used to control grade in a relatively steep channel over a long distance. Gabions may be useful in constructing weirs or drop structures in areas where an adequate ﬁller material is available. These baskets are versatile in that they may be used in series or singularly to provide various drop heights. They may also be used to protect the streambanks in the approaches and to form a stilling basin. Concrete, because of its durability, is probably the most frequently used material in the construction of larger drop structures. These structures vary in design, particularly in the stilling basin portion of the structure. Weirs are probably the most widely used form of gradecontrol structure. This is due to their relative ease of design and construction, low cost in many instances, and versatility and adaptability for other purposes. The simplest form
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281
Figure 8.27. Drop structures, weirs, and sills.
is probably that made of sheet piles, the piling serving as both control section and cutoff wall. To form sheetpile weirs, sheet piles are driven to a depth two to three times the anticipated depth of maximum scour, or to refusal, and are trimmed to a cross section approximately that of the original channel in that area [Fig. 8.27(b)]. If high banks are present, the sheetpile weir should be tied
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into the bank at both ends. Riprap protection is suggested both upstream and downstream of this type of weir. Sheet piles are particularly useful in wide channels where the required length of concrete or rock structures could be too costly. The piles may be trimmed at or above the natural channel grade. By trimming the piles above the channel bottom, the channel will aggrade for a short distance upstream, resulting in a ﬂatter EGL in that reach. In a stream requiring reduction of energy over a longer reach, several weirs may be used in sequence. Broadcrested weirs, sloping sills, or rockﬁll weirs are structures formed when a pile of stone is placed across the bottom of a stream [Fig. 8.27(c)]. The amount of scour that could be expected to occur just downstream of the weir, because of turbulence or a migrating headcut, would determine the volume of rock needed. Speciﬁc methods to determine the local scour depth downstream of hydraulic structures are presented in Section 9.2. 8.5
Riverbank engineering
The essence of a successful bankstabilization project is that it should be both effective and environmentally sound before the economics are considered. The engineering challenge is to determine the most suitable technique to solve a speciﬁc problem. Environmental and economic factors can then be integrated into the ﬁnal selection. The required project life is short for emergency stabilization during an extreme ﬂow event. Immediate action is required under conditions not permitting the design and construction of a permanent solution. An intermediate project life refers to temporary solutions. For instance, an eroding channel can be stabilized with the expectation that the stream channel will be relocated in the future. The most common situation refers to a very long project life that typically exceeds 100 yrs. Operation and maintenance costs can sometimes be traded against initial construction costs. If the project sponsor has the capability to monitor the condition of the work and maintain it as required, a less durable technique may be preferable to a “bombproof” method requiring a higher initial investment. Climatic conditions affect project durability through: (1) the effect of freezing and thawing on stone; (2) the effect of ice ﬂows on protection structures; (3) the effect of heaving on slope armoring; (4) the effect of wetting and drying, with the accompanying damage by bacterial growth and insects, on wooden components; (5) the effect of sunlight on synthetic materials; and (6) the effect of corrosion on wire meshes for gabions, mattresses, jacks, and fences. Corrosion and abrasion
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can greatly reduce the durability of techniques that rely on metallic components for longterm structural integrity. The critical factors are water chemistry, air quality, and concentration and velocity of coarse sediment impinging on the metallic components. Debris carried by the ﬂow, usually in the form of uprooted trees, can cause such extensive damage on some streams as to rule out some techniques and may prohibitively raise the cost of other structures designed to withstand debris loads. Debris can make structures more effective in reducing nearbank velocities and accumulating sediment. Debris can also fail structures that accumulate large debris in zones of high velocity. Vandalism and theft can be reduced by selection of a technique that minimizes temptation. Some materials that are obvious targets for vandals and thieves are posts, boards, concrete blocks and stones of attractive size and shape, small cables and wire, and easily removable fasteners. It is worth considering an increase in the size and the weight of components to make their removal or destruction more tedious. Peeling threaded fasteners and thinly grouting the surface of vulnerable mattresses may sufﬁce to keep the work functional. Animals can cause problems in a stabilization method. Beavers have a remarkable talent for girdling, felling, or eating vegetation of all sizes and species. This can be disconcerting if the success of the work depends on quickly establishing a strong vegetative cover. Cows, deer, rabbits, and other animals may also ﬁnd tender young vegetation on new stabilization plantings to their liking. Insect damage can be a problem for wooden components or vegetation. Preservative treatment for wooden components is common practice, and chemical treatment of vegetation at vulnerable stages may be feasible. Water quality and environmental considerations may rule out these options for some projects. Adjustment to scour or settlement remain important. A stabilization method that has the ability to adjust to scour or bank subsidence has a signiﬁcant advantage over those that do not. Completely rigid methods must be carefully designed and constructed and perhaps even then supplemented by ﬂexible materials at critical points. The property of ﬂexibility reaches its ultimate application in the design of toe protection. The methods that have this property are stone and other adjustable armor, ﬂexible mattress, and a few types of dikes and retards. Streambank limitations also need to be considered. It is sometimes expensive or impractical to grade the bank for geotechnical stability or to allow placement of armor protection because of adjacent structures, restricted rights of way, or restrictions on disposing of the excavated material. In such cases, a technique must be selected that leaves at least part of the existing bank intact.
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Figure P.8.2.1. Revetment example.
Problem 8.1
With reference to Example 8.2, determine the rock size that is stable and determine an appropriate gradation curve for the riprap blanket. Answer: d50 1.25 ft = 40 cm, d100 = 2.5 ft 80 cm, d20 = 0.625 ft 20 cm. Problem 8.2
Use the channel sketched in Fig. P.8.2.1 and the following properties: (a) (b) (c) (d) (e) (f) (g) (h)
speciﬁc weight of stone = 165 lb/ft3 (26 kN/m3 ), local depth at toe of outer bank = 25 ft (7.6 m), local depth at 20% upslope from toe = 20 ft (6.1 m), channel sideslope = 1V:2H, channel downstream slope = 2 ft/mile (3.8 cm/km), minimum centerline bend radius = 1700 ft (520 m), average velocity = 7.2 ft/s (2.2 m/s), watersurface width = 500 ft (150 m)
Determine the size, thickness, and geometry of the riprap blanket required for the stabilization of the outer bank of this channel. If the riprap is placed over uniform 0.5mm sand, determine the characteristics of the ﬁlter required for preventing leaching.
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285
Table P.8.3.1. Riprap gradation Weighta (lb)
% Lighter by weight
Maximum
Minimum
100 50 15
292 86 43
117 58 18
a Unit
weight of stone is 165 lb/ft3 (26 kN/m3 ).
Problem 8.3
Table P.8.3.1 shows the results of a riprap sample that is supposed to meet the requirements for riprap gradation. Determine whether or not this gradation meets the requirements for stability. If d50 corresponds to incipient motion, determine what gradation is required.
9 River engineering
This chapter presents a discussion of various types of river engineering structures. The aim of this chapter is to familiarize the reader with solutions to a wide spectrum of river engineering problems. This chapter covers an overview of the following types of river engineering structures: (1) ﬂood control in Section 9.1; (2) river closure and local jet scour in Section 9.2; (3) canal headworks in Section 9.3; (4) bridge crossings in Section 9.4; (5) navigation and waterways in Section 9.5; and (6) dredging in Section 9.6.
9.1
River ﬂood control
This section discusses river engineering solutions to ﬂood control. The methods include ﬂow regulation with reservoirs in Subsection 9.1.1, ﬂoodways in Subsection 9.1.2, channel conveyance in Subsection 9.1.3, and levees in Subsection 9.1.4.
9.1.1
Reservoirs
The most direct method of ﬂood control is the storage of surface runoff. Flood control with reservoirs redistributes ﬂoodwaters and attenuates ﬂoodwaves. A reservoir ﬁlls up when the inﬂow exceeds the outﬂow, and water storage acts as a buffer to decrease the peak discharges and increase the low discharges. An increase in low discharges is beneﬁcial to hydropower, navigation, and irrigation. It is best to keep the reservoir as full as possible in order to maintain sufﬁcient storage capacity to increase the low discharge reserve in periods of drought. For ﬂood control, however, the reservoir should be kept as low as possible to store unexpected large ﬂoods and decrease ﬂood peaks. In upper watersheds, detention basins are simply equipped with an uncontrolled conduit and a spillway. As long as the spillway is not overtopped, the ﬂood volume is stored, resulting in a considerable attenuation of the ﬂoodwave, 286
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287
Figure 9.1. Flood storage.
as shown in Fig. 9.1. When the spillway is overtopped, some storage above the spillway crest reduces the outﬂow discharge. Floodcontrol reservoirs are most effective when the volume of a reservoir is of the same order of magnitude as that of the upstream ﬂoodwave volume.
9.1.2
Floodways
Floodwave attenuation is sometimes possible if some of the ﬂoodwater is diverted away from the river. Floodways are used to divert ﬂoodwaters into a topographic depression near the river, or into a lake, river, or sea. Floodway outlets consist of spillways or control gates, which are usually located on or near the ﬂoodplain to regulate the overbank ﬂow discharge. Normally ﬂoodways are not used for long periods of time. It is therefore important to operate the facilities periodically to ensure proper usage in case of emergency. Floodwaters usually carry a signiﬁcant suspended sediment load, and the possibility of scour around the structures or ﬁlling in the ﬂoodway needs to be considered. In some cases, sedimentation can be expected downstream of the diversion but can be removed after the ﬂood. Erosion and sedimentation problems may become more serious in the case of possible capture of the river by the distributary, e.g., the possible capture of the Mississippi River by the Atchafalaya River near the Old River control complex.
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9.1.3
Channel conveyance
Lowering the ﬂood levels in a river reach by increasing the discharge capacity of the river channel is sometimes possible: (1) by reducing the roughness of the river bed; (2) by enlarging the conveying cross section; and/or (3) by shortening the river channel and thus steepening the channel slope. Each of these improvements results in lowering of the speciﬁc stage in the river reach considered. Effective improvements are restricted to clearing the banks and the ﬂoodplain from vegetation and other obstacles, eliminating sandbars and islands, and smoothing the banks with revetments. In cold regions, explosives or ice breakers can break up ice jams. In streams in which resistance to ﬂow depends on bedforms, smoothing the bed by removal of sand dunes is rarely effective because bedforms will shortly reappear. Enlarging the conveying cross section can sometimes be done by deepening or widening the river channel and by lowering or widening the ﬂoodplain. This approach is successful only if the sediment load is small and the bed is stable. In most cases, deepening the lowwater bed by dredging will result in a temporary improvement at a high cost that is likely to reoccur. In many cases, an enlarged cross section will gradually ﬁll up with sediment until the original bed level and slope have been restored. Deepening of the channel is attractive only if the sediment load is small and the bed is stable. If not, a permanent improvement can be achieved only by continuous dredging, ultimately resulting in degradation of the river bed. Such a solution might be economically justiﬁed if the dredged material can be used elsewhere. However, the stability of hydraulic engineering structures might be endangered by headcuts and bed degradation. Shortening the river channel can be achieved by meander cutoffs, but such a channel rectiﬁcation should be carried out with great care. If it is not ﬁxed by embankments, the river might start meandering again. Moreover, the locally steep slope will increase the sediment load, causing erosion upstream and sedimentation downstream. In most cases, the new river channel will gradually increase its sinuosity and regain its original slope. Also, ﬂood protection problems are not solved with an increase in stream conveyance; they are merely passed on downstream.
9.1.4
Levees
A levee is an earth embankment constructed along a stream to protect land on the ﬂoodplain from being ﬂooded (Fig. 9.2). A ﬂoodwall is a concrete structure that serves the same purpose and is found in urban areas where insufﬁcient space prohibits building a levee. For thousands of years, river levees have been built for the
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289
protection of people and their property against ﬂooding. It still is the most expedient method for ﬂood control. Levees reduce the storage capacity of the ﬂoodplain and thus restrict the ﬂow conveyance of the river. These reductions will induce higher water levels and limit ﬂoodwave attenuation. Because of the lateral migration of meandering streams, the levees should be placed at a fair distance from migrating channels, preferably outside the meander belt. The elevation of levees is primarily determined by the stage of the design ﬂood. River engineers perform a beneﬁt–cost analysis to compare the ﬂoodprotection gain against the construcFigure 9.2. Levees and ﬂoodwalls. tion cost of raising the levee above the design level. In general, they measure the beneﬁt of additional protection by considering the exceedance and the cost of ﬂood damages at different water levels. The cost of possible levee breaching and the corresponding extent of ﬂooding need to be considered. Moreover, if ﬂood control is regarded as a purely economical problem, it is seldom justiﬁed. Public, environmental, political, and military pressure may justify the large sums needed to protect against ﬂooding. If the increased value of landside property, with improved protection against ﬂooding, is taken into account, the ratio between costs and beneﬁts may become more favorable. Other factors affecting the levee height include tides and wind waves in coastal areas and wide rivers. The design of the levee must prevent breaching as a result of seepage, piping, sliding, slope failure, and revetment erosion. Levees should provide safety until their crests are overtopped. The choice of elevation of the levees may also depend on anticipated aggradation of the riverbed in coming years. River sedimentation within the levees causes the ﬂowrating curves to shift upward, thus necessitating the levees’ being raised beyond the initial level, e.g., Yellow River and Rio Grande. As sketched in Fig. 9.2(d), the volume of material required for levee protection increases with the square of the levee height, and the cost can become prohibitive in the case of perched rivers. For rivers for which the ﬂood hydrographs rise or fall quickly, the levee section should be analyzed for instability following
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rapid drawdown. Also, the design of internal ﬁlters is particularly critical because of the frequent need to use less than desirable material for embankment construction. Levee crown elevation is based on the design ﬂood proﬁle plus an allowance for settlement and freeboard. Settlement of either the foundation or the levee embankment, or both, may result in a loss of freeboard, which provides additional levee height for factors that cannot be rationally accounted for in ﬂoodproﬁle computations. Minimum freeboard allowances are usually 2 ft (60 cm) for agricultural levees and 3 ft (1 m) for urban levees. Additional freeboard may be needed at the upstream end of a levee, near drainage structures, bridges, and other constrictions, and for wave action. Levee crown width is primarily dependent on dimensions for the patrol road on the levee crown. Minimum widths of 10 to 12 ft (3 to 9 m) are commonly used with occasional wider turning or passing areas. Where public roads are located on the levee crown, a wider crown width is required. Compacted levees with sheep foot or rubbertired rollers require strong foundations of low compressibility and water content of borrow material close to speciﬁed range. Semicompacted levees are not commonly used, and uncompacted levees with ﬁll dumped in place with little spreading and compaction is used only for emergency work. Borrow areas should be on the riverside of the levee, and long, shallow borrow areas along the levee alignment are favored because there are fewer potential problems (Fig. 9.3). Riverside borrow areas should be designed to ﬁll slowly on rising stages and drain fully on fall stages, and the bottom of the excavated area should slope away from the levee. Underground seepage in pervious foundations beneath levees may result: (1) in the buildup of excessive hydrostatic pressure beneath the impervious top layer on the landside; and (2) in sand boils or in piping beneath the levee itself unless seepage control measures are provided. Principal control measures include riverside impervious blankets, landside seepage berms, pervious toe trenches, pressurerelief wells, cutFigure 9.3. Leveed ﬂoodplain. off trenches, and sheet piling.
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291
Agricultural use on the landside requires the control of the groundwater table. The efﬁciency of the drainage system improved by the construction of drains, ditches, and canals may be affected by aggradation or degradation of the riverbed and the ﬂoodplain. Local surface drainage of landside agricultural ﬁelds through the levees is usually possible with: (1) levees along the tributary channels; (2) culverts with ﬂood gates; or (3) pumping plants. For major tributaries, levees are constructed along the tributaries, tying to the main channel levee and extending upstream to the limits of backwater inﬂuence. Culverts with ﬂoodgates are effective only when the main river is at low ﬂow. The culverts can be equipped with control devices or with ﬂap gates that automatically open to permit outﬂow when the water level on the landside is higher than on the riverside of the levee. Floodgates automatically close to prevent entry of backwater from the main river when the head differential is reversed in ponding areas, and pumping stations are usually required for land drainage when the main river is at high stage. Pumping plants are most frequently effective where a limited amount of temporary ponding is available. Pumping plants may be combined with culverts with ﬂoodgates in some situations.
Case Study 9.1 Flood Protection of the Mississippi River, United States. The Mississippi River has threatened the valley for a long time. In 1543, la Vega, in his history of the expedition by DeSoto, described the ﬁrst recorded ﬂood of the Mississippi River. The severe ﬂood began on approximately March 10, 1543 and crested 40 days later. The river had returned to its banks by the end of May, having been in ﬂood for ∼80 days. The Mississippi River discharges an average of 520 km3 of water each year past the cities of Vicksburg and Natchez, Mississippi. Not all parts of the Mississippi River drainage basin contribute water in equal measure; see Fig. CS.9.1.1(a). Onehalf of the water discharged to the Gulf of Mexico is contributed by the Ohio River and its tributaries (including the Tennessee River) whose combined drainage areas constitute only onesixth of the area drained by the Mississippi River. By contrast, the Missouri River drains 43% of the area but contributes only 12% of the total water. The Mississippi River now discharges an average of about 200 million metric tons of suspended sediment per year past Vicksburg and eventually to the Gulf of Mexico. The suspended sediment load carried by the Mississippi River to the Gulf of Mexico has decreased by onehalf in the past 200 yrs. The decrease has happened mostly since the 1950s, as the largest natural sources of sediment in the drainage basin were cut off from the Mississippi River main stem by the construction of large reservoirs on the Missouri and Arkansas Rivers. This large decrease in
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Figure CS.9.1.1. Regime of the Mississippi River.
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293
sediment load from the western tributaries was counterbalanced somewhat by a ﬁvefoldtotenfold increase in sediment load in the Ohio River as a result of deforestation and rowcrop farming. By the year 1879, the need for improvement of the Mississippi River had become widely recognized. The necessity for coordination of engineering operations through a centralized organization had ﬁnally been accepted. Accordingly, in that year, the Congress established the Mississippi River Commission (MRC) with the assignment and the duty to protect the banks of the Mississippi River, improve the navigation thereof, prevent destructive ﬂoods, promote and facilitate commerce, trade, and postal service. Great ﬂoods in 1882 and subsequent years plagued the valley as levees were overtopped or crevassed. These disasters and the rising ﬂood heights between the levees caused many to question the total reliance on building levees to contain the river’s ﬂoodwaters. Other approaches to improving ﬂood protection – reforestation of the ﬂoodplain, cutoffs to speed up the river’s ﬂow, reservoirs to hold back ﬂoodwaters, and ﬂoodways to divert ﬂows away from the main channel – were suggested but always rejected by the MRC in favor of a “leveeonly” policy. The role of the MRC grew with each ﬂood, ﬁnally culminating in the Flood Control Act of 1917, which authorized the MRC to construct an extensive program of ﬂood protection with cost sharing by states and local interests. The program maintained the leveeonly approach and included new levee construction and strengthening of existing levees to standards set 3 ft above the high water of 1912. By the end of 1926, the improved levee system had successfully passed several major highwater events. These successes convinced the MRC and the public that the ﬂoodcontrol problem was nearly solved. The false sense of security in the Lower Mississippi Valley vanished in the ﬂood of 1927, a natural disaster of great proportions. This tremendous ﬂood extended over nearly 26,000 miles2 , killed more than 500 people, and drove more than 700,000 people from their homes. Thirteen crevasses in the main Mississippi River levees occurred, demonstrating that even the largest and strongest levees would not alone protect from ﬂooding. To prevent a recurrence of the 1927 ﬂood, Congress authorized the Mississippi River and Tributaries (MR&T) project in the Flood Control Act of 1928. The leveeonly policy of the past was discarded and the U.S. Army Corps of Engineers adopted a new approach based on improved levees plus ﬂoodways, including a spillway to divert water at Bonnet Carr´e into Lake Pontchartrain above New Orleans. The four major elements of the MR&T project are: (1) levees for containing ﬂood ﬂows; (2) ﬂoodways for the passage of excess ﬂows past critical reaches of
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Figure CS.9.1.2. Design of ﬂoodcontrol plan of the Mississippi River.
the Mississippi; (3) channel improvement and stabilization in order to provide an efﬁcient navigation alignment, to increase the ﬂoodcarrying capacity of the river, and for protection of the levee system; and (4) tributary basin improvements for major drainage and for control, such as dams and reservoirs, pumping plants, and auxiliary channels. The ﬂoodcontrol plan sketched in Fig. CS.9.1.2 is designed to control the “project ﬂood.” It is a ﬂood larger than the record ﬂood of 1927. At Cairo, the project ﬂood is estimated at 2,360,000 ft3 /s or 66,820 m3 /s. The project ﬂood is 11% greater than the ﬂood of 1927 at the mouth of the Arkansas River and 29% greater at the latitude of Red River Landing, amounting to 3,030,000 ft3 /s (85,800 m3 /s) at the location ∼120 miles (200 km) below Vicksburg. The Atchafalaya River. When the ﬁrst European settlers arrived, they found the Red River emptying into the Mississippi at Turnbull’s Bend and the Atchafalaya River a welldeﬁned distributary ﬂowing out of Turnbull’s Bend a few miles to the South [Fig. CS.9.1.3(a)]. In 1931, Captain Shreve dug a canal across the narrow neck of Turnbull’s Bend. The river accepted the shortcut
River ﬂood control
295
Figure CS.9.1.3. Old River Control.
and abandoned its old channel, the upper part of which eventually silted up, leaving the lower section open, which became known as Old River [Fig. CS.9.1.3(b)]. The Red River no longer ﬂowed into the Mississippi River, but into the Atchafalaya River. Old River connected them to the Mississippi. The current usually ﬂowed west from the Mississippi through Old River into the Atchafalaya; however, during high water on the Red River, the ﬂow sometimes reversed. For years the head of the Atchafalaya River was blocked by a massive “raft” – a 30milelong (50km) log jam – that deﬁed efforts of settlers to remove it. In 1839, the State of Louisiana began to dislodge the raft to open up the river as a freeﬂowing and navigable stream. The removal of the log jam provided an opportunity for the Atchafalaya to enlarge, become deeper and wider, and carry more and more of the ﬂow from the Mississippi. The Atchafalaya offered the Mississippi River a shorter outlet to the Gulf of Mexico – 142 miles (227 km)
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Figure CS.9.1.4. Locks and dams of the Upper Mississippi River.
compared with 315 miles (504 km) – and by 1951 it was apparent that, unless something was done soon, the Mississippi River would take the course of the Atchafalaya River. The Old River control structures were designed and operated to maintain the distribution of ﬂow and sediments between the Lower Mississippi River and the Atchafalaya River in approximately the some proportions as occurred naturally in 1950. That distribution was determined to be approximately 30% of the total latitude ﬂow (combined ﬂow in the Red River and the Mississippi River above the control structures) passing down the Atchafalaya River on an annual basis [Fig. CS.9.1.3(c)]. The Mississippi River is the main stem of a network of inland navigable waterways that form a system of ∼12,350 miles (19,760 km) in length, not including the Gulf Intracoastal Water of 1,173 miles (1,877 km). Waterborne commerce on the Mississippi rose from 30 million tons in 1940 to almost 400 million tons in 1984. In Fig. CS.9.1.4, the entire 1080km reach of the Upper Mississippi River between Minneapolis, Minnesota, and St. Louis, Missouri, is controlled for barge navigation by a series of 29 lockanddam structures. The ﬁrst structure, completed in 1913 at Keokuk, Iowa, was built to impound water to generate hydroelectric power. The other 28 structures were built, mostly during the 1930s, to maintain a minimum river depth of 9 ft (2.7 m) for barge navigation. Before the dams were built, navigation during lowwater periods was extremely hazardous, if not impossible, across rapids such as those at Keokuk and Rock Island, and it was difﬁcult in many other reaches of the upper river.
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297
Figure 9.4. Dam construction beside the river.
9.2
River closure
The construction of dams requires closing off a river reach and diverting streamﬂow around a dry construction site. There are essentially three types of river closures: (1) dam construction beside the river; (2) complete closure and ﬂow diversion; and (3) partial closure and ﬂow contraction. The ﬁrst type is illustrated in Fig. 9.4, in which a dry construction site can be found next to the river in the ﬁrst stage. When the construction is complete, the river is then diverted on the structure in the second stage. This is obviously possible for only low structures such as locks and dams. The complete closure and ﬂow diversion is discussed in Subsection 9.2.1, along with the methods to calculate local scour from permanent structures such as sills, drop structures, sluice gates, and culverts in Subsection 9.2.2. The partial closure and ﬂow contraction approach by use of cofferdams is covered in Subsection 9.2.3. 9.2.1
River closure and diversion
For the construction of earth dams and high concrete dams in deep, narrow canyons, the entire river channel is generally closed by building upstream and downstream earth and rockﬁll cofferdams. For large dams on major rivers, the streamﬂow must be conveyed around the work site through tunnels, as sketched in Fig. 9.5. Such tunnels may serve a dual purpose: (1) ﬂow diversion during construction; and (2) regulated outlet works later. On small streams, a temporary ﬂume or pipeline may be adequate to divert Figure 9.5. River diversion. streamﬂow around the construction site.
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The size of a diversion tunnel depends primarily on the diversion design ﬂood and on the upstream cofferdam height. The tunnel inlet must be low enough for ﬂow through the tunnel as soon as the river is closed off. A log boom or trash rack upstream of the inlet may be required for preventing partial blockage of a diversion tunnel by debris. Diversion tunnels are designed to be closed temporarily, usually by gates or stoplogs, at the upstream portal to permit ﬁlling of the reservoir. If the tunnel is used solely for diversion, permanent tunnel closure is ensured with a concrete plug. The construction sequence for channel diversions is as generally follows: (1) construct diversion tunnel; (2) construct temporary upstream cofferdam; (3) construct temporary downstream cofferdam; (4) dewater construction site within the cofferdams; (5) construct permanent upstream cofferdam; and (6) construct main dam. Initial closure is usually made with temporary cofferdams across the main river channel upstream and downstream of the construction site. The water in the work area between the temporary structures is pumped out for the construction of a permanent cofferdam in the dewatered area. The permanent upstream cofferdam can be constructed as an extension of the temporary upstream cofferdam or as a part of the main dam embankment section. River closure can be effected in various ways, but the two primary procedures are to: (1) successively narrow and eventually close off the channel by end dumping from one or both banks [Fig. 9.6(a)]; and/or (2) gradually raise a low sill across the channel by dumping uniformly across the gap from a construction bridge or barge or by use of a dragline or moveable dredge disposal line [Fig. 9.6(b)]. During a river closure, the upstream water level is raised from the backwater effect of choking the ﬂow. The difference in water levels upstream and downstream of the closure causes an increase in ﬂow velocity and a greater potential for scour in the closure section. In end dumping, the ﬂow Figure 9.6. River closure. velocity increases. The ﬁnest material in the ﬁll will move downstream rather than deposit in the gap. As velocity increases, all the ﬁll material being dumped into the closure section will be transported downstream and will not contribute to closing the channel. Just before complete closure, velocities through the gap may be so high that only
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299
large stones and prefabricated concrete blocks may be required for making the ﬁnal closure. In gradually raising the crown of a low sill across the opening, a mound is built up uniformly across the channel by dumping from a bridge or suspension cable. As material is added, the ﬂow is choked and forces backwater effects upstream of the sill, as described in Example 4.4. The increasing waterlevel difference upstream and downstream of the closure section increases the ﬂow velocity, tractive force, and sedimenttransport capacity. As ﬁll material continues to be added and the tractive force becomes sufﬁciently high to transport largersize fractions of the ﬁll, the crest lengthens in the downstream direction. Obstruction of the channel by the sill raises the upstream water surface and diverts part of the streamﬂow through the diversion tunnel. The reduced discharge over the sill eventually permits the sill to be raised to the water surface for complete closure. To determine the stability of a rounded stone deposited in ﬂowing water, Isbash’s equation has been modiﬁed for river closures and stilling basins. As per Eqs. (8.3), the relationship between critical ﬂow velocity Vc for beginning of motion and diameter of a stone ds is Vc = K c [2g(G − 1)]1/2 ds1/2 ,
(9.1)
where K c is a coefﬁcient equal to 1.2 for river closure and 0.86 for stilling basins, g is the gravitational acceleration, and G is the speciﬁc gravity of the stone. The stone diameter ds that can withstand an average ﬂow velocity Vc is calculated from 2 Vc 1 . (9.2) ds = 2g(G − 1) K c The stone weight FW corresponding to a spherical stone of diameter ds is calculated from πds3 , (9.3) 6 where γs is the speciﬁc weight of the stone. A sample calculation for a river closure is presented in Example 9.1. FW = γs
Example 9.1 Application to a river closure. During the closure of a river, the watersurface elevation drops approximately h = 6 ft (1.82 m) within a short distance. Estimate the stone diameter ds at a speciﬁc weight γs of 165 lb/ft3 or G = 2.65 that would be stable. Assuming conservation of energy within this short reach, the velocity corresponding to a 6ft drop is calculated as √ Vc = 2gh = 2 × 32.2 × 6 = 19 ft/s (or 5.8 m/s). (E.9.1.1)
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In the case of a shallow opening, the velocity may correspond to the critical ﬂow depth, given the unit discharge q as h c = (q 2 /g)1/3 . The required stone diameter calculated from Eq. (9.2) is 19 2 1 = 2.35 ft (or 0.71m), ds = 2 × 32.2 × 1.65 1.2 and the corresponding stone weight from Eq. (9.3) is π(2.35)3 = 1,121 lb (or 5 kN). 6 These results are comparable with those previously presented in Fig. 8.4. We can substitute Eq. (E.9.1.1) into Eq. (9.2) to ﬁnd that the required stone diameter increases linearly with h. The appropriate stone size is approximately oneﬁfth of the drop height. FW = 165 ×
9.2.2
Jet scour
The construction of large dams is associated with the need to release water periodically downstream. The bed scour caused by plunging jets or submerged jets needs to be considered in the stability analysis of hydraulic structures. With reference to the four cases illustrated in Fig. 9.7, the scour depth z below plunging jets in Fig. 9.7(a) can be estimated from the empirical equation of Fahlbusch (1994) as a function of unit discharge q, the jet velocity V1 entering the tailwater depth h t at an angle θ j measured from the horizontal at the water surface, and the gravitational acceleration g: q V1 sin θ j − h t . (9.4) z = K p g The coefﬁcient for plunging jet Kp depends on grain size with Kp 20 for silts, 5 < Kp < 20 for sand, and 3 < Kp < 5 for gravel. Submerged jets discharge entirely under the free surface, as sketched in Fig. 9.7(b). The example of ﬂow under a sluice gate downstream of a hydraulic structure has a considerable scour potential. Hoffmans and Verheij (1997) applied Newton’s second law to a control volume and found the equilibrium scour depth z from V2 , (9.5) z = K s j yj 1 − V1 where V2 is the outﬂow velocity, V1 is the inﬂow velocity, y j is the inﬂow jet thickness, and K s j is a scour coefﬁcient for submerged jets. The value
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301
Figure 9.7. Jet scour.
of K s j depends on the particle size and varies from K s j 50 for silts, to 20 < K s j < 50 for sand, and to 7 < K s j < 20 for gravel. As sketched in Fig. 9.7(c), scour below gradecontrol structures and also sills and drop structures can be estimated from the method of Bormann and Julien (1991) as
sin φ z = 1.8 sin(θ j + φ)
0.8
q 0.6 V1 sin θ j [(G − 1) g]0.8 ds0.4
− Dp,
(9.6)
where z is the scour depth below the gradecontrol structure, D p is the drop height of the gradecontrol structure, q is the unit discharge, V1 is the approach velocity, ds is the particle size, g is the gravitational acceleration, G is the speciﬁc gravity of bed material, φ is the angle of repose of the bed material, and θ j is the jet angle measured from the horizontal. Examples 9.2 and 9.3 illustrate how these empirical relationships can be used to estimate the scour depth below hydraulic structures. As sketched in Fig. 9.7(d), local scour below circular culvert outlets has been studied by Ruff et al. (1982). The scour depth z can be predicted as ( z = 2.07 D
Q g D5
)0.45 ,
(9.7)
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River engineering
where Q is the discharge, D is the culvert diameter, and g is the gravitational acceleration. The local scour depths predicted from the empirical relationships are subjected to improvements as more ﬁeld and laboratory data become available. A recent review of different methods showing the analysis of different data sets has been presented by Hoffmans and Verheij (1997).
Example 9.2 Application to scour depth below a sluice gate. A sluice gate, as sketched in Fig. E.9.2.1 is operated at an opening of 0.34 m. The upstream water level is 10 m and the downstream water level is 5 m. The unit discharge in this wide opening is 2 m2 /s. The bed material consists of fairly uniform gravel with d50 = 5 mm and d90 = 7 mm. Estimate the scour depth: Figure E.9.2.1. Local scour below a sluice gate.
Step 1: The ﬂow velocity in the vena contracta is V1 =
q 2 = 5.9 m/s . = yj 0.34
Step 2: The outﬂow velocity is V2 =
0.4 m q 2 . = = ht 5 s
Step 3: The scour depth obtained from Eq. (9.5) with K s j = 15 is 0.4 z = 15 × 0.34 1 − 4.8 m. 5.9 Example 9.3 Application to scour depth below a gradecontrol structure. A broadcrested weir is built across a 50mwide river. The drop height is 2.25 m, and the face angle of the structure is 60◦ . The scour slope is approximately 1V:2H in noncohesive material with d50 = 2 mm and d90 = 2.5 mm (see Fig. E.9.3.1). Estimate the scour depth when the river discharge is 160 m3 /s: Step 1: Determine q = (Q/W ) = (160/50) = 3.2 m2 /s, D p = 2.25 m, ds = 0.002 m, θ j = tan−1 1/2 = 26◦ , and g = 9.81 m/s2 .
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303
Figure E.9.3.1. Local scour below a drop structure.
Step 2: Estimate φ = 40◦ and G = 2.65. Step 3: Assume that the critical ﬂow condition is obtained at the sill crest. The critical ﬂow depth h c is obtained from h c = (q 2 /g)1/3 = (3.22 /9.81)1/3 = 1 m. Step 4: The approach ﬂow velocity is assumed to be critical, or V1 =
q 3.2 = 3.2 m/s. = hc 1
Step 5: The scour depth estimated from Eq. (9.6) is sin (40◦ ) 0.8 3.20.6 3.2 sin 26◦ z = 1.8 − 2.25 m = 2.7 m. sin (66◦ ) (1.65 × 9.81)0.8 (0.002)0.4
9.2.3
Cofferdams
In large rivers in which the entire ﬂow cannot be diverted from the construction site, a partial river closure with ﬂow contraction can be considered. Cellular cofferdams can enclose a part of the streambed that can be dewatered by pumping to provide a dry area for construction. The twostage construction of a lock and dam is sketched in Fig. 9.8. Although cofferdams are temporary structures, each stage may be in place for several years, and they should be as cheap as possible while providing the optimum degree of protection. The cofferdam cost is generally one of the major cost items in river construction, particularly when ﬂood stages are frequent or ﬂashy in nature and the construction extends through several years. For preliminary studies, ﬂoods with the following recurrence intervals have been
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Figure 9.8. Flow contraction with cofferdams.
used for cofferdam design (Petersen, 1986): (1) a 3–5yr period of return for river construction not exceeding 2 yr; or (2) a 5–25yr ﬂood when the construction period exceeds 2yr. Where long streamﬂow records are available, elevation of the second or the third largest historical ﬂood can be used to determine the cofferdam crest elevation including freeboard. Other important considerations include backwater effects and property damage behind large dams with high cofferdams and potential downstream effects of water release from a sudden cofferdam failure. For ﬁnal design, the tradeoff between risk and potential damage that is due to overtopping and the cost of a higher cofferdam can be assessed. If the risk of loss of life and property damage downstream is high, the standard project ﬂood may be adopted for cofferdam design. With stage construction on alluvial streams, considerable scour can be expected in the constricted channel cross section at ﬂood stages. Large contraction scour and abutment scour can be calculated from the methods discussed in Subsections 9.4.2 and 9.4.3. If studies indicate that excessive scour might endanger the cofferdam, it may be possible to change the cofferdam conﬁguration to reduce local scour or to armor the bed where maximum scour is expected. In general, the scour model tests from Franco and McKellar (1968) indicated that: (1) the scour is local rather than general and the point of maximum scour is usually near the upstream corner of the cofferdam; (2) the depth, location, and area of maximum scour can be affected by the alignment of the upper arm of the cofferdam; (3) the extent of scour along the cofferdam is increased as the angle of the upstream arm is increased with respect to the ﬂow direction; and (4) the area of maximum scour can be moved away from the cofferdam with a spur dike or guidebank at the upstream corner of the cofferdam. Cellular cofferdams are predominately designed either with circular or diaphragm cells (Fig. 9.9). As a rule of thumb, the cell width should
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305
approximately equal the cell height. The circular cells are connected by cells of circular arcs. Circular cells are stable so that each cell can be ﬁlled immediately on completion, making it possible for equipment to work from one cell to the next. The diaphragm type consists of two walls of circular arcs connected by straight diaphragm sections, but individual cells are not stable. A template is used for guidance in driving the piles, and when Figure 9.9. Cofferdam cells. a cell is completed the template is removed and reused for the next cell. According to Petersen (1986), templates are somewhat easier to set for diaphragm cells than for circular cells. Several diaphragm cells are driven and then carefully ﬁlled by keeping the differences in ﬁll elevation in adjacent cells within approximately 4–5 ft (1.2–1.5 m) to avoid distortion of the diaphragms. Circular cells usually require less steel for high structures. However, less steel is usually required for diaphragm cells for low cofferdams. The failure of a circular cell is usually local, with damage limited to one cell, but diaphragm cell failure may extend to adjacent cells. Sheetpile cells are usually ﬁlled hydraulically with readily available local material such as sand, gravel, rock, or earth for stability. It is particularly important to ﬁll the lower half of each cell with pervious material to facilitate drainage and avoid high hydrostatic forces on the cell walls. A concrete cap (∼6 in. or 15 cm) is poured to protect the top of the cell ﬁll against scour in the event the cofferdam is overtopped, to prevent inﬁltration of rainwater into the ﬁll, and to provide a working surface for the contractor’s equipment. The concrete cap also provides a base for sandbagging to provide limited additional height and protection against overtopping beyond the cofferdam design ﬂood elevation and extends the time required for water to saturate the ﬁll in the event of overtopping. Cellular cofferdams are constructed by driving a wall of interlocking steel sheets through water and saturated pervious material into underlying more impervious clays or rock. Cofferdams are driven to rock whenever possible, and if the material is soft shale, piles are driven 6 in.–1 ft (15–30 cm) into the shale. In pervious materials, the piles are driven as deep as feasible to increase the seepage path and decrease seepage ﬂow into the work area and thus reduce
306
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pumping costs. Following the completion of cofferdam construction, the work area inside the cofferdam is dewatered at a limited drawdown rate (usually ∼5 ft in 24 h or 1.5 m per day), depending on riverbank stability and the rate of cell drainage. 9.3
Canal headworks
The regular supply of water to irrigation canals demands adequate control of the water level and sediment transport at the canal intake. Canal headworks control the water level with the construction of a weir to artiﬁcially raise the water level at the irrigation canal intake, as sketched in Fig. 9.10. The weir located downstream of the headworks raises the upstream water level, thus artiﬁcially decreasing the channel slope and ﬂow velocity near the canal headworks. The artiﬁcial slope is therefore less than the equilibrium slope of the channel under natural conditions. Finally, sedimentation problems upstream of the weir and local scour downstream of the weir need to be addressed (e.g., Section 9.2). There are a number of problems inherent to ﬁxed weirs in streams with variable discharge. Raising the water level upstream of the weir is required for waFigure 9.10. Weirs and gradecontrol structures. ter supply during periods of low river discharge. In periods of high discharge, however, the raised water level may increase the risk of ﬂooding. The erosive forces near the weir at high discharges may also necessitate expensive bed and bank stabilization. Canal headworks also control the sediment intake into the canal. Fixed weirs usually reduce the sediment concentration at the intake for a period of time. The decreased sedimenttransport capacity upstream of the weir results in aggradation that will eventually reach the canal intake. Canal headworks may be located on the outside of a stable river bend. In river bends, the surface water (low sediment concentration) seeks the outside of the bend whereas the sedimentladen streamlines near the bed are deﬂected away from the canal headworks (e.g., refer to Fig. 6.6). Therefore, headworks on the concave bank beneﬁt from the natural river ﬂow curvature and reduce the necessity for extensive sediment exclusion and sediment ejection at the canal headworks, as sketched in Fig. 9.11. The concave bank approach is possible
Canal headworks
307 when the river is stable, without lateral migration, and when the river does not carry a signiﬁcant load of ﬂoating debris. Subsection 9.3.1 focuses on sediment exclusion followed by sediment ejection in Subsection 9.3.2. 9.3.1
Sediment exclusion
Figure 9.11. Sediment exclusion and ejection.
Sediment exclusion is intended to prevent sediment from entering the canal by deﬂecting sediment away from the canal headworks, as sketched in Fig. 9.11. Natural ﬂow curvature causes the heavily sediment laden water to ﬂow away from the canal intake, as shown in Fig. 9.12(a). Guide walls can be designed on both small and large systems to create a favorable ﬂow path, as shown in Fig. 9.12(b). Guide walls increase exclusion efﬁciency and are effective in continuous sluicing operations. Guide vanes simply produce favorable secondary currents for sediment removal. Bottom and surface vanes, shown in Fig. 9.13, induce secondary currents that divert the bottom streamlines containing a heavy sediment load away from the canal headworks, and surface water containing a relatively light sediment load can be diverted through the canal headworks. Under steady ﬂow, guide
Figure 9.12. Canal headworks with river curvature.
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River engineering
Figure 9.13. Canal headworks with vanes.
vanes can exclude practically the entire bedload. Conversely, they should be avoided when the ﬂow is unsteady. 9.3.2
Sediment ejection
As sketched in Fig. 9.11, sediment ejectors remove sediment that should not be transported through the canal system. Ejection methods do not preclude other sediment control methods, but provide a factor of safety should the exclusion devices fail to perform according to design. The sediment ejector sketched in Fig. 9.14 should be close to the head regulator where settling is predominant; otherwise the sediment deposited in the canal may not reach the ejector. The tunnel ejector shown in Fig. 9.14(a) can eject coarse bedload from straight canals. The bed of the canal forms a ramp sloping down toward the ejector. The ramp height may sometimes be level with the roof of the ejector. The incoming bed material rolls over the ramp and is ejected through the tunnels. The roof over the guide vanes prevents coarse material from passing over the vanes because
Canal headworks
309
of the turbulence resulting from the interaction of the bottom guide vanes and large canal discharges. For small discharges the roof can be omitted and the guide vanes alone will deﬂect the bedload. The vortex tube in Fig. 9.14(b) is extremely efﬁcient at removing coarse bedloadlike gravels in small canals. The main feature of the vortex tube is a pipe with a slit opening along the top side. As water ﬂows over the tube, the shearing action sets up a vortex motion that catches the bedload as it passes over the lip of the opening. The sediment is carried to the outlet at the downstream end of the vortex tube. Settling basins sketched in Fig. 9.14(c) remain popular for the removal of both bedload and suspended load at canal headworks. The underlying principle is simply to provide a section wide and Figure 9.14. Sediment ejectors. long enough to reduce ﬂow velocity and allow sediment to settle out. The basic relation for the design of settling basins is the trap efﬁciency TE , which represents the ratio of the weight of sediment settling in the basin to the weight of sediment entering the basin: TE = 1 − e
−ωW L sb Q
,
(9.8)
where W is the settlingbasin width, ω is the settling velocity of a particle, L sb is the length of settling basin, and Q is the ﬂow discharge. At a given discharge and particle size, the efﬁciency depends on the surface area of the settling basin. A calculation example is detailed in Example 9.4.
Example 9.4 Calculation of a settlingbasin length. Determine the length of a 30mwide settling basin designed to trap 90% of ds = 0.05mm silt entering a canal at a discharge of Q = 3 m3 /s. The unit discharge q = Q/W = 3/30 = 0.1 m2 /s. The settling velocity of 0.05mm particles is approximately 2 mm/s.
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River engineering
The settlingbasin length is calculated from Eq. (9.8), solved for L as L sb =
0.1 s m2 q [−ln (1 − TE )] = [−ln (1 − 0.9)] = 115 m. ω 0.002 m s
Repeating the application to different basin widths shows that identical results are obtained as long as the settlingbasin area Asb = L sb W = 3,450 m2 .
9.4
Bridge scour
River engineers are concerned with bridge crossings in the following regards: (1) careful selection of the bridge site to minimize the total bridge costs; and (2) protection against possible structural failure from scour undermining the embankments and piers. The depth of scour during a ﬂood has to be determined in order to design stable foundations for the bridge. The river crossing should preferably be located in straight river reaches or in stable bends without lateral migration. Sites with narrow ﬂoodplain width, rock outcrops, or high bluffs are good locations for bridge crossings. Straight river reaches are often selected in order to avoid problems with lateral migration and deep scour holes in bends. Protecting embankments and piers against scour requires consideration of the following items: (1) general scour (Subsection 9.4.1); (2) contraction scour (Subsection 9.4.2); (3) abutment scour (Subsection 9.4.3); and (4) pier scour (Subsection 9.4.4). The total scour depth is obtained from the sum of all components. Livebed scour occurs when there is transport of upstream bed material into the scour hole. Conversely, clearwater scour is without upstream bed sediment transport. Selected methods are presented below, and a detailed treatment of scour at bridges can be found in Richardson and Davis (1995). It should be remembered that all scour estimation procedures serve as approximations and engineering judgement should be exercised.
9.4.1
General scour
Progressive degradation or aggradation can be associated with changes in the river regime caused by natural processes or human activities on the stream or watershed. Factors that affect longterm bed changes are: (1) dams and reservoirs both upstream and downstream of the bridge site; (2) changes in watershed land use like urbanization, deforestation, etc.; (3) channel stabilization and rectiﬁcation; (4) natural or artiﬁcial cutoff of a meander bend; (5) changes in the downstream base level of the bridge reach, including headcuts; (6) gravel
Bridge scour
311
mining from the streambed; (7) diversion of water into or out of the stream; (8) lateral migration of a river bend; and (9) thalweg shifting in a braided stream. The engineer must assess the present state of the stream and watershed and consider planned future changes in the river system. Rivers are free to change crosssection geometry, and the location of the thalweg may also shift considerably in time, e.g., Fig. CS.7.3.2. The bed level may still vary considerably as a consequence of traveling of sand waves, bedforms, rifﬂes, and pools in relatively straight river reaches and changes in pools and crossings during Figure 9.15. Bridge crossing. ﬂood conditions. If bedlevel measurements are available from previous years, an analysis of speciﬁcgauge data may yield some insight into the trend to be expected in the future, e.g., see Figs. 4.7 and 7.7. When general scour is expected to be signiﬁcant, as sketched in Fig. 9.15, it is often advisable to construct a gradecontrol structure downstream of the bridge crossing. The purpose of the structure is to control the bed elevation at the bridge site and protect the abutment and piers.
9.4.2
Contraction scour
Contraction scour results from ﬂow acceleration in river contractions. The approach ﬂow depth h 1 and average approach ﬂow velocity V1 result in the sediment transport rate q S1 . The total transport rate to the contraction is W1 q S1 , in which W1 is the width of the approach. If the water ﬂow rate Q 1 = W1 q1 in the upstream channel is equal to the ﬂow rate at the contracted section, then, by continuity, q2 =
W1 q1 , W2
(9.9)
where q1 = h 1 V1 , q2 = h 2 V2 , and the subscript 2 refers to conditions in the contracted section. The sedimenttransport rate at the contracted section after
312
River engineering equilibrium is established must be q S2 =
W1 qS . W2 1
(9.10)
With reference to Fig. 9.16, the depth of scour z that is due to the contraction is then z = h 2 − h 1 .
(9.11)
Here h 1 , V1 , and W1 are the depth, velocity, and width of the approach ﬂow, respectively, and h 2 is the contracted ﬂow depth at the bridge. The contracted Figure 9.16. Contraction scour (after Nordin, 1971). scour depth represents an average over the channel width, and symmetry is assumed in the calculation. When only one side is being contracted, the abutment scour equation of the following subsection should be considered.
9.4.3
Abutment scour
Abutments, as well as spur dykes, can have different shapes, and they can be set at various angles to the ﬂow. As sketched in Fig. 9.17, the tip of earth and
Figure 9.17. Abutment scour (after Richardson et al., 1990).
Bridge scour
313
rockﬁll abutments will generally have a spillthrough shape when compared with sheet piles standing as vertical walls. Local scour at abutments depends on the amount of ﬂow intercepted by the bridge abutments. The equilibrium scour depth for local livebed scour in sand near a spillthrough abutment under subcritical ﬂow is 0.4 La z = 1.1 Fr0.33 (9.12a) 1 , h1 h1 where z is the equilibrium depth of abutment scour, h 1 is the average upstream ﬂow depth in the main channel, L a is the abutment and embankment length measured at the top of the water surface and normal to the side of the channel from where the top of the design ﬂood hits the bank to the outer edge of the abutment, and Fr1 = {V1 /[(gh 1 )0.5 ]} is the upstream Froude number. If the abutment terminates at a vertical wall and the wall on the upstream side is also vertical, then the scour hole in sand calculated by Eq. (9.12a) nearly doubles (Liu et al., 1961, and Gill, 1972): 0.4 La z = 2.15 Fr0.33 (9.12b) 1 . h1 h1 Field data for scour at the end of rock dikes on the Mississippi River indicate that the equilibrium scour depth for spillthrough abutment scour when L a /h 1 > 25 can be estimated by z = 4 Fr0.33 1 . h1
(9.13)
This method can be used for very long abutments when ﬂood waters extend onto very wide ﬂoodplains. Example 9.5 illustrates abutment scour calculations. 9.4.4
Pier scour
Pier scour is caused by the horseshoe vortex induced by secondary ﬂow at the pier base [Fig. 9.18(a)]. The horseshoe vortex removes bed material away from the base region in front of and along the side of the pier. The strength of the vortex decreases as the depth of scour is increased. Equilibrium scour is reached when the transport rates entering and leaving the scour hole are equal. The Colorado State University (CSU) equation calculates the pier scour as 0.65 a z = 2.0 K 1 K 2 Fr0.43 (9.14) 1 , h1 h1
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Figure 9.18. Pier scour (after Richardson and Davis, 1995).
where z is the scour depth, h 1 is the ﬂow depth just upstream of the pier, K 1 is the correction for pier shape from Table 9.1 corresponding to Fig. 9.18(b); K 2 is the correction for the ﬂow angle of attack from Table 9.2, a is the pier width and Fr1 = V1 /(gh 1 )0.5 is the upstream Froude number. The extent to which a pier footing or pile cap affects local scour at a pier is not clearly determined. Under some circumstances the footing may serve to impede the horseshoe vortex and reduce the depth of scour. In other cases in which the footing extends above the streambed into the ﬂow, it may increase the effective width of the pier, thereby increasing the local pier scour. In the calculations, the pier width can be used if the top of the pier footing is slightly Table 9.1. Piertype correction factor K 1 (after Richardson et al., 1990) Pier type
K1
Square nose Round nose Circular cylinder Sharp nose Group of cylinders
1.1 1.0 1.0 0.9 1.0
Note: See Fig. 9.18(b).
Bridge scour
315
Table 9.2. Flowangle correction factor K 2 (after Richardson et al., 1990) Correction factor K 2 Attack Angle θ p (deg)
L p /a = 4
L p /a = 8
L p /a = 12
0 15 30 45 90
1.0 1.5 2.0 2.3 2.5
1.0 2.0 2.5 3.3 3.9
1.0 2.5 3.5 4.3 5.0
Note: θ p = skew angle of ﬂow, L p = pier length, a = pier width.
above or below the streambed elevation. The footing width is used when the pier footing projects well above the streambed or when general scour is expected in the river reach. As a rule of thumb, the scour depth is approximately two times the pier width. In debrisladen streams or when ice jams can be expected at the bridge site, the actual local scour may be considerably larger than that determined by the equation. The extra scour resulting from debris or ice accumulation against the pier must be given due consideration. Ice and debris can both produce static and dynamic pressure up to 200 Pa on bridge piers. These forces are over and above the static and dynamic forces caused by the trafﬁc on the bridge and the forces exerted on the piers by the ﬂowing water. When bedforms travel along the riverbed, the maximum scour depth will be onehalf the dune height, greater than the mean scour depth. The CSU equation has been slightly modiﬁed to account for dune bedforms and for scour reduction from bed armoring (Richardson and Davis, 1995). Scour may also be reduced if riprap is placed around bridge piers. A riprap layer twice the design diameter thick should extend between 1.5 and 6 times the pier width. Example 9.5 illustrates how to calculate the local scour depth around a bridge pier. Example 9.5 Application to abutment and pier scour. A 200mlong bridge is to be constructed over a sandbed channel with 300mlong spillthrough abutments 1V :2H . Six rectangular bridge piers measuring 1.5 m thick and 12 m long are aligned with the ﬂow. At a design 100yr ﬂow discharge of 850 m3 /s, the upstream ﬂow velocity is 3.75 m/s and the ﬂow depth is 2.8 m upstream of the piers. Estimate (a) the abutment scour depth and (b) the pier scour depth.
316
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(a) Abutment scour Step 1: The approach Froude number is V1 3.75 Fr1 = √ =√ = 0.71. gh 1 9.81 × 2.8 Step 2: The ratio of abutment length L a to ﬂow depth h 1 is (L a / h 1 ) = (300/2.8) = 107. Step 3: The abutment scour depth is calculated with Eq. (9.13) because L a > 25h : = 4 × 2.8 × 0.710.33 = 10 m. z = 4 h 1 Fr0.33 1 Guidebanks in Subsection 8.4.3 could be designed to reduce abutment scour. (b) Pier scour Step 1: The Froude number is Fr1 = 0.71. Step 2: The rectangular pier shape corresponds to K 1 = 1.1 from Table 9.1. Step 3: The pier length/width ratio (L/a) = (12/1.5) = 8 and K 2 = 1.0 from Table 9.2 as long as the pier is aligned with the ﬂow, θ p = 0. Step 4: The pier scour depth calculated from the CSU equation (9.14) is z = 2 h 1 K 1 K 2
a h1
0.65 Fr0.43 = 2 × 2.8 × 1.1 × 1 × 1
1.5 2.8
0.65 (0.71)0.43
= 3.5 m.
9.5
Navigation waterways
Channel improvement for navigation is justiﬁed on the basis of savings in commercial shipping costs. The reduced cost of moving commodities by means of waterways instead of other modes of transportation is compared with the costs of construction, operation, and maintenance. Navigation requirements are discussed in Subsection 9.5.1, followed by waterway alignment and cutoffs in Subsection 9.5.2, and locks and dams in Subsection 9.5.3.
Navigation waterways 9.5.1
317
Navigation requirements
General requirements for channel depth, channel width, and lock dimensions for commercial navigation are governed by a number of factors, including type and volume of probable future tonnage, types and sizes of vessels, and tows in general use on connecting waterways. Tows on the Lower Mississippi frequently have more than 40 barges and transport 50,000 to 60,000 tons (450–540 MN) of cargo (Fig. 9.19). Higherpowered towboats have an average of ∼3,000 hp (2.2 MW), with three 10ft (3m) diameter, ﬁvebladed stainless steel propellers housed in Kort nozzles. Three barge types are common: (1) openhopper barges for transporting coal, sand and gravel, and sulfur; (2) coveredhopper barges for grain and mixed cargo; and (3) tank barges for petroleum and chemicals (Fig. 9.19). Barge sizes vary around a standard 35 ft (10.7 m) in width and 195 ft (60 m) in length. Pilots navigate towboats at the stern of the tow with control of the engine thrust and direction of the rudder. The navigation width depends on channel alignment, size of tow, and whether oneway or twoway trafﬁc is planned. Oneway trafﬁc may be adequate when the trafﬁc is light if the reach is relatively straight with good visibility and if passing lanes are provided. Twoway trafﬁc permits heavy trafﬁc to move faster except when tows are meeting or passing. Figure 9.20(a) shows the recommended channel widths for commercial navigation in straight channels. Wider navigation channels are required in bends because tows take an oblique position [shown in Fig. 9.20(b)]. The drift angle α varies with the radius of
Figure 9.19. Barge types (after Petersen, 1986).
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River engineering
Figure 9.20. Navigation characteristics.
curvature of the channel; the speed, power, and design of the craft, and wind forces; whether the tow is empty or loaded; whether the trafﬁc is going up or down the river; and the ﬂow pattern. The navigation channel width is a direct function of the drift angle, which is larger for downbound tows than for upbound tows. For example, the drift angle for downbound tows of various sizes in a bend with a 3,000ft (914m) radius is shown in Fig. 9.20(b).
9.5.2
Waterway alignment and cutoffs
Navigation is preferable in fairly straight channels or bends with long radii of curvature. In wide channels, the waterway alignment can be controlled with revetments, spur dykes, and longitudinal dykes. Once the radius of curvature has been determined, the design of river alignment structures is based on the methods discussed in Chap. 8, considering the cost and the effectiveness of different structures. Navigation in meandering channels usually requires straightening the river. Cutoffs across the neck of long meander bends improve the alignment of the river, reduce sinuosity, and shorten the river length, thus increasing the channel
Navigation waterways
319
Figure 9.21. Artiﬁcial cutoff.
slope. For instance, the natural cutoffs of the Mississippi River shortened the river by ∼50 miles (80 km). Artiﬁcial cutoffs were constructed to improve river alignment for navigation and for ﬂood control. Indeed, straight river reaches are easier to navigate than long meandering rivers, e.g., Fig. 9.21. Also, steeper gradients reduce ﬂood stages. Historically, structural cutoffs of the Mississippi River were prevented because they (1) disrupted the river regime, (2) aggravated bank erosion upstream, (3) increased downstream shoaling stages, and (4) produced currents difﬁcult to navigate. Pilot channels are excavated from the downstream end for the construction of artiﬁcial cutoffs. Pilot channels have 1V :3H sideslopes with bottom widths from 50 to 200 ft (15 to 60 m) and bottom elevations from 6 to 12 ft (2 to 4 m) below the lowwater reference plane. When the river length is long compared with the pilot channel, the slope ratio if favorable for natural enlargement of the cut and narrow pilot channels are adequate. Wider and deeper pilot channels are initiated when there is little length or slope advantage. A parallel trench is excavated for the placement of riprap revetment. Earth plugs near the upstream end of the excavated pilot channels are left in place to block low ﬂows. Plugs are designed to be overtopped and washed out during ﬂoods after the revetment has been completed. An example of river alignment is shown in Fig. 9.22. A radius of curvature of 8,000 ft (2.7 km) is maintained throughout the reach. The construction of riprap revetments, spur dykes, and longitudinal dykes provides ﬂow control as long as dyke overlaps and tiebacks prevent ﬂanking of the structures. Note the increase in slope that is due to the reduced reach length. Case Study 9.2 illustrates the changes in the Rhine and the Waal Rivers in The Netherlands.
320
River engineering
Figure 9.22. River control for navigation.
Case Study 9.2 Control of the Rhine and the Waal Rivers, The Netherlands. According to RIZA (1999), people living near the Rhine River in past centuries had to cope with ﬂood disasters. Levees failed repeatedly during ﬂoods. In winter, drifting ice often led to the formation of ice jams that raised
Navigation waterways
321
Figure CS.9.2.1. Waal River near Tiel in The Netherlands (after Jansen et al., 1979).
the water levels and increased the risk of levee overtopping. Levees, built on the ﬂoodplain and groynes or spur dykes, have been built since the eighteenth century to channelize the river, as shown in Fig. CS.9.2.1. The width of the main channel has been reduced from more than 500 to 260 m. Islands and sandbanks were removed, and the banks were protected from erosion with a series of groynes spaced approximately every 200 m on both sides of the river. As a result, the main channel of the Rhine River has eroded considerably over the past 100 years. In the past 60 years, the Rhine’s channel at Lobith has degraded an average of 1–2 cm each year. This in turn threatens the stability of the groynes and the banks. Approximately onethird of the sediment transported by the Rhine reaches the North Sea. The rest is deposited in the river’s lower reaches. During ﬂoods, material deposits between the groynes and on the ﬂoodplain, particularly behind
322
River engineering
the summer levees, which is the primary reason for the doublelevee system. Up to 10 cm of sediment can deposit along the riverbanks during extended periods of ﬂooding. The main navigable channel of the Waal River is currently maintained at 150 m wide with at least 2.5 m of water 95% of the time. The river engineering challenge arises from increasing the navigation width and depth without detrimental effects on the environment. Measures include the use of bottom vanes and bendway weirs to widen bends and extension of groynes and construction of longitudinal dykes to increase the ﬂow depth locally. Stream rehabilitation projects do not necessarily imply restoring the stream to natural conditions that existed before humanity appeared. Natural conditions that existed several thousands of years ago can be looked at as reference conditions. In many instances, stream restoration to ancient conditions is not a feasible proposition. Stream rehabilitation offers a compromise between reference conditions and the present situation. The present policy for stream rehabilitation may include the vital components for mankind development and offer adequate habitat for aquatic and riparian species as well as ensure survival of endangered species. Figure CS.9.2.2 shows the present policy reﬂects a compromise solution between present and referenced ﬂoodplain situations.
Figure CS.9.2.2. Floodplain management of the RhineWaal in The Netherlands (after RIZA, 1999).
Navigation waterways
323
Target situations are often idealistic, and the completion of a stream rehabilitation project does not guarantee that the designed habitat is ﬁt for the intended species. Efforts to restore aquatic habitat and develop wetlands have become an integral part of river engineering practice.
9.5.3
Locks and dams
Dams are required to sustain an adequate navigation depth, and locks allow ships to navigate in a river steepened by artiﬁcial cutoffs. Locks and dams must be spaced to maintain the navigation depth throughout the pools. Spacing the dams farther apart may eliminate dams and reduce the initial cost of construction. However, maintenance dredging and channel contraction with dikes are also required for maintaining the sedimenttransport capacity of the natural channel. Straight reaches are desirable locations for locks because they are easier to navigate than bends. However, straight reaches tend to be unstable with inadequate ﬂow depth in the downstream approach channel. Wide straight reaches are also prone to sedimentation. Adverse cross currents from spillway discharges may also affect trafﬁc in the lock approaches. In river navigation projects the lock is usually located near the bank at one end of the dam to minimize the adverse effects of spillway discharge on trafﬁc. A typical lock and dam layout is illustrated in Fig. 9.23. A lock is a structure designed to enable vessels to gain access to lower or higher water levels on either side of the dam. It is an open chamber with gates at both the upper and the lower pool. Locks admit water to ﬁll the chamber from the upper pool and discharge to the lower pool to empty the lock, as illustrated schematically in Fig. 9.24. A downstreambound tow approaches a lock, the emptying valves and lower lock gates are closed, and the water surface in the lock chamber is brought to the same elevation as that of the upper pool by opening the ﬁlling valves. The upper lock gates are then opened for the tow to move into the lock chamber. The upper lock gates and ﬁlling valves are then closed, and the emptying valves are opened to bring the water surface in the lock down to the level of the lower pool. The lower lock gates are then opened, and the tow moves out of the lock chamber into the lower pool. The procedure is reversed for a tow moving upstream. Locks are designed to provide adequate depth for navigation. Undersized locks may force large tows to be very slow and cautious when entering a lock chamber. Large locks are not only more expensive, but they require longer ﬁlling and emptying times. Filling and emptying times for a lock are designed to be as short as possible without excessive turbulence,
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River engineering
Figure 9.23. Lock and dam (after Petersen, 1986).
surges, or crosscurrents in the lock chamber that might damage the tow or cause the tow to damage the lock. Locks ﬁll and empty through culverts and ports in or on the chamber ﬂoor or at the base of the lock walls. Deep locks provide a cushion of water to dampen turbulence so that tows are not damaged and stresses in the hawsers (the cables that secure tows to the lock walls) are within acceptable limits. In the lock chamber, shown in Fig. 9.25, ﬂow from the wall culverts pass into a crossover culvert across the center of the lock chamber. The splitter wall at the crossover culvert distributes ﬂow equally into two longitudinal ﬂoor culverts with ports. Ports in the upper guard wall reduce crosscurrents by permitting the ﬂow intercepted by the lock to pass through the wall to the spillway. The total crosssectional area of port openings in the guard wall should be equivalent to the crosssectional area of the approach channel. Upper guard wall ports should be designed as low as possible to increase bottom velocities and reduce shoaling. Currents and velocities from a lockemptying system can be dangerous to tows approaching the lower lock. The lock emptying outlet should be outside the lower lock approach. However, when the discharge outlet is located outside the lock approach, the watersurface elevation at the outlet sometimes may be higher than that in the lower lock approach, resulting in difﬁculty in opening the lower lock gates. The outlet discharge manifold in the lower lock approach can induce turbulence to reduce shoaling.
Dredging
325
Figure 9.24. Navigation through a lock (after Petersen, 1986).
9.6
Dredging
Dredging is the process of removing material from the bed or the banks of a waterway for the purpose of deepening or widening navigation channels or to obtain ﬁll material for land development. Dredging is a very costly operation (in excess of U.S. $1.00 per cubic yard) that requires heavy equipment and long pipelines. Dredging equipment can be classiﬁed as either mechanical or hydraulic. Mechanical dredges (Fig. 9.26) lift the dredged material by means of diggers or buckets. Hydraulic dredges pick up the dredged material by means of suction pipes and pumps. Mechanical dredges remove bed material by a dipper or bucket, and the excavated material is dumped into disposal barges for unloading at the disposal site. Mechanical dredges are usually not selfpropelled and must be towed to the work site. Dipper dredges have considerable digging power to excavate hard compacted material and blastedrock fragments. It can operate with very little maneuvering
326
River engineering
Figure 9.25. Flow pattern through a lock (after Petersen, 1986).
space and can be accurately controlled in the vicinity of bridges and other structures. Bucket dredges use interchangeable buckets (clamshell, orange peel, dragline) for different operational purposes. An open bucket digs into 12 yd3 (9 m3 ) of bed material, and then closes to be raised and emptied. A modiﬁed system in which buckets are ﬁxed on a conveyor belt has also been used in Europe. Considerable ﬁne material is lost from the bucket as it is raised from depths up to 100 ft (30 m), and the maximum concentration of the suspended turbidity plume is typically less than 1,000 ppm. Hydraulic suction dredges, shown in Fig. 9.27, can be categorized by means of picking up the dredged material (cutterhead, plain suction, and dustpan dredges) and by means of disposal of the dredged material (hopper, pipeline, and sidecasting dredges). Hopper dredges, sketched in Fig. 9.27(a), are selfpropelled seagoing vessels used primarily for maintenance dredging and progressive deepening by successive passes. Hopper dredges draw concentrated material in contact with the channel bottom through suction pipes and store it in hoppers in the
Dredging
327
Figure 9.26. Mechanical dredges (after U.S. Army Corps of Engineers, 1983).
dredge. Sediment resuspension occurs near the suction lines at concentrations of a few parts per thousand, and up to several tens of parts per thousand when there is overﬂow from the hoppers. When fully loaded, the dredge moves to the disposal area under its own power. Hopper dredges are emptied by opening of the bottom doors and dumping of the entire contents in a few seconds. A welldeﬁned plume of dredged material entrains water as it settles to the bottom. Most of the material forms a mound on the bottom, and some spreads horizontally. Sidecasting dredges are selfpropelled seagoing vessels designed to remove material from shallow coastal harbors. A sidecasting dredge picks up bottom material through two suction pipes and pumps it directly overboard beside the dredge. The dredge operates back and forth across the bar, successively deepening the channel on each pass. Sidecasting dredges are not suitable for dredging contaminated material. Hydraulic pipeline dredges loosen the bottom material with a cutterhead or with water jets (dustpan dredges). The slurry is then pumped through a ﬂoating discharge line to the disposal site. Dustpan dredges [Fig. 9.27(b)] are selfpropelled vessels suitable for dredging only noncohesive material in waters without signiﬁcant wave action. Dustpan dredges are equipped with: (1) pressure water jets that loosen the bottom material; and (2) a wideﬂared and ﬂat
328
River engineering
Figure 9.27. Suction dredges (after U.S. Army Corps of Engineers, 1983).
suction line intake for sediment removal. It normally discharges into open water through a relatively short pipeline, up to 100 ft (30 m) long; a longer disposal line requires a booster pump. Cutterhead dredges are the most efﬁcient and versatile, and thus the most widely used. The cutterhead dredge shown in Fig. 9.27(c) has a rotating cutter around the suction pipe intake and can dig and pump alluvial material including
Dredging
329
compacted clays and hardpans. Suction pipe diameters range from 8 to 30 in. (20 to 90 cm). The cutterhead dredge sketched in Fig. 9.27(c) consists generally of a cutterhead, baskets, ladder, suction line, Aframe, Hframe, pumps, gantry, spuds, and a pipeline up to 30 in. (90 cm) in diameter. A cutterhead dredge operates by circling about one anchored spud with the cutterhead describing an arc. As the swing is completed, the second spud is anchored and the ﬁrst one raised to cirFigure 9.28. Cutterhead dredge with cle in the opposite direction and move pipeline (after U.S. Army Corps of Engineers, 1983). forward, as illustrated in Fig. 9.28. The ﬂoating discharge line is made up of 30–50ft (9–16m) long pipe sections, each supported by pontoons. They are connected together by ﬂanges, ball joints, or rubber sleeves to give the dredge some ﬂexibility in moving. Land pipeline disposal can range up to 3 miles (5 km) without additional sections of shore pipe. For longer transport distances, booster pumps are required on the discharge line. Slurries of 10%–20% sediment concentration by weight are typically pumped in pipelines up to 30 in. (90 cm) in diameter at velocities ranging from 15 to 20 ft/s (5 to 8 m/s). According to Petersen (1986), the disposal of dredged material has received considerable environmental consideration since the 1960s. Not all dredged material is contaminated, and only a small percentage of the sites are highly contaminated. Clay content and organic matter in dredged material tend to retain many contaminants, and sands easily release contaminants through mixing, resuspension, and transport. Adverse water quality effects are unlikely unless the dredged material is highly contaminated. The physical impacts of openwater disposal are likely to be of greater potential consequence than the chemical and the biological impacts. Adverse biological effects are often unlikely because many organisms usually recolonize disposal sites. Except during ﬁsh migration and spawning, turbidity in river waters is more likely to be an aesthetic problem than a biological problem. Conﬁned land disposal is sometimes the only possible alternative for disposing of some highly polluted sediments. The dredge material is ponded until the suspended solids have settled out. The longterm storage capacity of land disposal sites depends on consolidation of the dredged material, compressibility of foundation soils, effectiveness in dewatering the dredged material, and
330
River engineering
management of the site. Case Study 9.3 illustrates how recurring dredging problems can be alleviated with ﬂowcontrol structures.
Case Study 9.3 Dredging at Choctaw Bar of the Mississippi River, United States. In this river engineering case study from Phil Combs (USACE WES, 1998), Choctaw Bar is an important location along the Mississippi River. The Choctaw Bar area is located between river mile 557–565 AHP (above head of passes) on the Mississippi River. It is located ∼17 river miles upstream of Greenville, Mississippi. The small town of Arkansas City, Arkansas, is located adjacent to Choctaw Bar and is protected by the Mississippi River Mainline levee. Choctaw Bar is located on the right descending bank of the river, and the EutawMounds revetment is located on the left descending bank. Stabilization of the left bank began as early as 1947 and has continued until the present, costing more than $10 million. Stabilization of the river in this reach was critical as levee crevasses had occurred many times in the past and the mainline levee is relatively close to the river. In addition to the stabilization of the left bank, dikes have been constructed on the right bank. Dikes in the vicinity of Chicot Landing were constructed in 1968 for approximately $1.2 million. Stabilization of the river is for ﬂood protection and to ensure a 9ftdeep navigable channel. Channel maintenance for navigation has become a signiﬁcant problem within the reach. Table CS.9.3.1 lists the channel maintenance dredging for the years 1971–1973. Review of the 1972 hydrographic surveys indicates that a bar in the vicinity of mile 559 AHP essentially blocked the navigation channel. Table CS.9.3.1 of maintenance dredging reﬂects the attempt to maintain the channel by mechanical means. The 1972 hydrographic surveys also indicate that there is a considerable chute cutoff development on the right bank, as shown in Fig. CS.9.3.1(a). Regular discharge measurements have been taken to establish the ﬂow distribution in the channel and the right chute. The ﬂow in the chute has increased to the point that, in 1972, ∼50% of the total river ﬂow was passing through the chute. Table CS.9.3.1. Dredging data (after Combs, USACE WES, 1998) Year
Days
Cubic Yards
1971 1972 1973
15 27 71
404,000 1,531,000 4,121,000
Dredging
331
Figure CS.9.3.1. Dredging at Choctaw bar of the Mississippi River (after Combs, USACE, WES, 1998).
Additional information includes the facts that: (1) the left bank has been stabilized; (2) rock dikes are located generally on the righthand side of the channel; and (3) contours on plan sheets refer to elevations in feet relative to the low water reference plane (LWRP). The LWRP is the watersurface elevation plane corresponding to the discharge exceeded 97% of the time on the ﬂow duration curve. The elevation of the LWRP shown on the plan sheets is varying from 97–100 ft NGVD (National Geodetic Vertical Datum). NGVD elevations are identical to above meansealevel (MSL) elevations.
332
River engineering
The assignment has been to develop an engineering plan to provide dependable navigation through the reach. The plan should utilize and enhance the existing stabilization infrastructure in the reach, if at all possible. The plan should consider that the aquatic ecosystem in the right chute is a valuable resource. Additionally, the plan should not conﬂict with the overall ﬂood carrying capacity of the river. If the existing infrastructure cannot be used, sufﬁcient engineering and environmental justiﬁcation has to be provided to alter the plan. Figure. CS.9.3.1(b) shows the structural solution proposed by the U.S. Army Corps of Engineers and the topography at Choctaw Bar in 1992. Note the effect of reduced channel width on ﬂow depth and ease of navigation. Problem 9.1
During a river closure by the enddumping method, the watersurface elevation drops ∼15 ft, or 4.5 m, within a short distance. Estimate the stone diameter and the stone weight required for closure. Answer: ds 3 ft (90 cm) and FW 2,200 lb (10 kN). Problem 9.2
Estimate the scour depth below a plunging jet at a velocity of 18 ft/s (5.5 m/s) at an angle of 60◦ from the horizontal. The jet thickness is 1 ft (30 cm) and the tailwater depth is 6 ft (1.8 m). Sand covers a thick layer of gravel ∼5 ft (1.5 m) below the surface. Answer: z 5.8 ft (1.8 m). Problem 9.3
Estimate the scour depth below the sluice gates of Example 9.2 when the opening is 1 m high. Determine the ﬂow velocity, assuming conservation of energy on both sides of the gate. Answer: V1
√
2gh = 9.9 m/s, q = V h = 9.9 m2 /s,
V2 2 m/s, z 12 m. Problem 9.4
Estimate the scour depth below the gradecontrol structure in Example 9.3 when
Problems
333
(a) the discharge is 320 m3 /s (b) for scour in medium sand (Q = 160 m3 /s). Problem 9.5
Estimate the scour depth below a 2ftdiameter culvert ﬂowing full at a velocity of 15 ft/s. Answer: The discharge Q = 47 ft3 /s (1.33 m3 /s) and z 5 ft (1.5 m). Problem 9.6 Repeat the length calculations for the settling basin in Example 9.4 for a width of 60 m and show that the basin area is 3,450 m2 . Also determine the trap efﬁciency of a basin area of 2,000 m2 . Answer: TE = 74%. Problem 9.7
Calculate the local scour depth at the end of a 100mlong spillthrough abutment in a river ﬂowing at a velocity of 3 m/s and a ﬂow depth of 4 m. If the water spreads 200 m farther out the ﬂoodplain during ﬂoods, estimate the scour depth, considering that the Froude number does not change signiﬁcantly. Answer: The scour depth for Eq. (9.12a) is 12.5 m and does not change during ﬂoods because L a > 25 h 1 . Problem 9.8
Estimate the scour depth around a rectangular pier that is 1 m wide and 5 m long in a river that is 4 m deep. The ﬂow velocity is 3 m/s at an angle of 30◦ from the pier alignment. Problem 9.9
Reevaluate the pier scour depth from Example 9.5 should the ﬂow alignment against the pier change to 20◦ in the coming years. Answer: At L p /a = 8, θ p = 20◦ , and K 2 2.2 from Table 9.2, the scour depth would more than double at z = 7.7 m.
10 Physical river models
Physical models of rivers have existed at least since 1875, when Louis Jerome Fargue built a model of the Garonne River at Bordeaux. Physical models are usually built to test various river engineering structures and to carry out experiments under controlled laboratory conditions as opposed to costly ﬁeld programs. The main purposes of physical models include: (1) a smallscale laboratory duplication of a ﬂow phenomenon observed in a river; (2) the examination of the performance of various hydraulic structures or alternative countermeasures to be considered in the ﬁnal design; and (3) investigation of the model performance under different hydraulic and sediment conditions. This chapter ﬁrst describes hydraulic similitude in Section 10.1 in terms of geometric, kinematic, and dynamic similitude. Hydraulic models can be classiﬁed into two categories: (1) rigidbed models, discussed in Section 10.2 and (2) mobilebed models, discussed in Section 10.3. The analysis leads to the deﬁnition of modelscale ratios, and several examples and case studies are presented. 10.1
Hydraulic similitude
The prototype conditions, denoted by the subscript p, refer to the fullscale ﬁeld conditions for which a hydraulic model, subscript m, is to be built in the laboratory. Model scales, subscript r, refer to the ratio of prototype to model conditions. For instance, the gravitational acceleration in the prototype is g p , the gravitational acceleration in the model is gm , and the scale ratio for gravitational acceleration is deﬁned as gr = g p /gm . Hydraulic models usually have the same gravitational acceleration in the model and the prototype; thus gr = 1. For all scale models, the following considerations are relevant: (1) the model length must be large enough to ensure the accuracy of the measurements, e.g., a ﬂowdepth measurement error of 1 mm in a model at a scale of 1:100, or zr = 100, represents an error of 10 cm in the prototype ﬂowdepth measurement; (2) we must consider the physical limitations on space, water discharge, and 334
Hydraulic similitude
335
instrumentation accuracy, e.g., we cannot realistically model the Mississippi River in a 100mlong hydraulics laboratory; and (3) we must appropriately simulate the boundary conditions, e.g., the stage and the discharge of inﬂow tributaries, and the possible tidal effects at the downstream end must be properly accounted for. Hydraulic models use water and require that the scales of mass density ρr and kinematic viscosity νr be unity. Because the scale ratio for gravitational acceleration gr = 1, the scale for speciﬁc weight γr and dynamic viscosity µr are also unity; thus ρr = gr = νr = γr = µr = 1 in hydraulic models. Geometric similitude describes the relative size of two Cartesian systems of coordinates (x, y, z). The vertical zr length scale is deﬁned as the ratio of the prototype vertical length z p to the model vertical length z m such that zr = z p /z m . For instance, a length scale zr = 100 indicates that a model length of 1 m corresponds to a prototype length of 100 m. The horizontal length scales are deﬁned in a corresponding manner in the downstream x and lateral y directions as xr and yr . Exact geometric similitude is obtained when the vertical and the horizontal length scales are identical, i.e., L r = xr = yr = zr . The corresponding area and volume scales are respectively Ar = L r2 and Volr = L r3 . If accurate quantitative data are to be obtained from a model study, there must be exact geometric similitude in every linear dimension. Model distortion implies that the vertical zr and the lateral yr scales are not identical. The distortion factor is obtained from the ratio of yr /zr . Model tilting results from different vertical zr and downstream horizontal xr scales. The downstream model slope Sr = zr /xr is effectively tilted when the horizontal length scale is different from the vertical scale. In a distorted model, the surface area scales for horizontal and crosssectional surfaces are respectively Ar = xr yr and A xr = zr yr . The volume scale for a distorted model corresponds to Volr = xr yr zr . Kinematic similitude refers to parameters involving length and time, e.g., velocity V, acceleration a, kinematic viscosity ν, etc. For instance, the velocity scale Vr is deﬁned as the ratio of prototype to model velocities as Vr = V p /Vm . The time scale tr = t p /tm appropriately describes kinematic similitude when ﬂuid motion in the model and the prototype are similar. With the kinematic relationships that L = V t and V = at, time can be canceled from these two relationships to obtain V 2 = a L. Any experiment in which the gravitational acceleration is the same in the model and the prototype requires that ar = gr = 1. When applied to the model and the prototype, this relationship yields one of the most important kinematic relationships in physical modeling: Vr = 1. zr0.5
(10.1)
336
Physical river models
This is known as the Froude similitude criterion. Accordingly, the time scale and the velocity scales for exact kinematic similitude are identical, tr = Vr = zr0.5 . It is important to consider that the time scale for distorted and tilted models varies with direction. Distorted and tilted models are restricted to simulate 1D ﬂows and time scales for ﬂow velocities in the y and the z directions are irrelevant. Distorted and tilted models therefore cannot appropriately account for 2D and 3D convective and turbulent accelerations and should not be used to model vorticity, diffusion, turbulent mixing, and dispersion. Dynamic similitude implies a similarity in the dynamic behavior of ﬂuids. Dynamic similitude refers to parameters involving mass, e.g., mass density ρ, speciﬁc weight γ , and dynamic viscosity µ. For instance, the massdensity scale ρ r is deﬁned as the ratio of prototype to model mass densities as ρr = ρ p /ρm . The mass scale Mr = M p /Mm appropriately describes dynamic similitude, besides the readily deﬁned length and time scales. The basic concept of dynamic similitude is that individual forces acting on corresponding ﬂuid elements must have the same force ratio in both systems. Individual forces acting on a ﬂuid element may be due either to a body force such as weight in a gravitational ﬁeld, or surface forces resulting from pressure gradients, viscous shear, or surface tension. The resulting inertial force necessitates that the force polygon be geometrically similar. Gravitational and viscous effects are respectively described by the Froude number Frr = Vr /(gr zr )0.5 and the Reynolds number Rer = Vr zr /νr . In hydraulic models, the Froude and the Reynolds numbers, i.e., Frr = Rer = 1, can be simultaneously satisﬁed only when Vr2 /zr = Vr zr , which is the trivial full scale Vr = zr = 1. We thus conclude that exact similitude of all force ratios in hydraulic models is strictly impossible except at full scale. Of course, forces that are negligible compared with others will not affect the force polygon. Therefore the art of hydraulic modeling is to center the analysis around the force components that are dominant in the system. The art of hydraulic modeling thus consists of determining whether gravity or viscosity is the predominant physical parameter and to determine the scale parameters accordingly. This approach is reasonable as long as either gravitational or viscous terms can be neglected. In open channels, gravitational effects are typically predominant and resistance to ﬂow does not depend on viscosity as long as ﬂows are hydraulically rough. In most river models, however, the weight force ratio Fgr = ρr L r3 gr should balance the inertial or hydrodynamic force Fir = ρr L r2 Vr2 . The ratio of inertial to weight forces implies that Fir /Fgr = Vr2 /L r gr = 1, which is simply the Froude similitude criterion. The Froude similitude criterion thus satisﬁes similarity in the ratio of inertial to weight forces. We also recognize that the Froude number also properly scales the ratio of velocity head to ﬂow depth. Consequently Froude similitude
Rigidbed model
337
describes similarity conditions in speciﬁcenergy diagrams for the model and the prototype. The Froude similitude is therefore useful in describing rapidly varied ﬂow conditions. 10.2
Rigidbed model
Rigidbed models are built to simulate ﬂow around river improvement works and hydraulic structures. A rigid boundary implies that the bed is ﬁxed, i.e., no sediment transport. This is the case when the Shields parameter of the bed material is τ∗ < 0.03. Rigidbed model scales can be determined in either one of two cases: (1) exact geometric similitude, in which resistance to ﬂow can be neglected; and (2) distorted/tilted models, in which resistance to ﬂow is important. Exact geometric similitude and Froude similitude can be simultaneously maintained in rigidbed models only when resistance to ﬂow can be neglected (Subsection 10.2.1). Such models are well suited to the analysis of 3D ﬂow around hydraulic structures, in which sediment transport is not important. When long river reaches are considered and resistance to ﬂow cannot be neglected, both the Froude and the resistance similitude can be simultaneously satisﬁed in tilted/distorted models (Subsection 10.2.2). 10.2.1
Exact Froude similitude
The model scales for hydraulic models with exact geometric similitude can be determined from the Froude similitude criterion. Exact geometric similitude is required when the ﬂow is 3D and when vertical accelerations are not negligible. This type of model is particularly well suited for modeling the ﬂow near hydraulic structures. The scale ratios for hydraulic models with exact geometric similitude reduce to Vr = tr = L r0.5 and Mr = L r3 . All other scale ratios can be derived from the length, time, and mass scales by use of the fundamental dimensions of any considered variable. The scale ratio for several variables is listed in the third column of Table 10.1. The exact geometric similitude imposes constraints that are usually difﬁcult to work with in modeling river reaches. Scaling the size of roughness elements according to the length scale will maintain the same resistance parameter as long as Re∗ > 70 for both the model and the prototype. In practice, this is possible for only very coarse bed channels such as cobble and boulder bed streams. The difﬁculty with the undistorted (untilted) Froude similitude criterion is that the nearbed conditions are drastically changed. Indeed, the laminar sublayer thickness δ in hydraulic models is relatively too thick. For instance, it requires 1/2 a scale δr = (u ∗r )−1 = (L r ) . A strict geometrical similitude cannot be maintained because the length similitude requires very small particles whereas scale
dsr Wr h r
Particle diameter
Xsection area
—
Vr u ∗r ωr Qr qbr
Velocity
Shear velocity
Settling velocity
Discharge
Unit bedload discharge
—
Lr
1/2
1/2 zr −1/2 z r xr
—
3/2
yr zr
—
—
−1/2
xr z r
1/2 Lr 1/2 Lr
tbr
Time (bed)
1/2
xr yr zr
L r3 Lr
zr yr
zr
zr
zr
zr 2+5m 1+m
1
3/2
1−2m 2+2m
yr zr
1−2m 2+2m
1/2 zr
1+7m 2+2m
2+5m 1+m
2m −1 2+2m
1+4m 1+m
yr zr
yr zr
zr
zr4 xr−3
Lr L r2
zr yr
Complete General
zr
zr yr
Tilted
xr
Lr
Lr Lr
Exact
—
tr
Time (ﬂow)
Kinematic
X r Wr h r
xr
Length
Volume
hr Wr
Geometric Depth Width
Scale
Rigidbed (Froude)
Table 10.1. Scale ratios for hydraulic models
1
yr zr1.5
zr0.286
zr0.286
1/2 zr
zr2.428
zr0.928
yr zr2.43
yr zr
zr−0.286
zr1.43
zr yr
(d∗r = τ∗r = 1) m = 1/6
yr zr1.5 1+m dsr zr0.5−m
—
m zr0.5−m d sr
1/2 zr
−1−3 m zr1.5+3m dsr
−2m zr0.5+2m dsr
−2 m yr zr2+2 m dsr
yr zr
dsr
−2 m zr1+2 m dsr
zr yr
d∗r = 1
Mobilebed
1
−1−m yr zr1+m dsr
−1 dsr
−1 dsr
−1−m zrm dsr
2 zr3 dsr
3+m zr2−m dsr
2 yr zr3 dsr
zr yr
dsr
2 zr2 dsr
zr yr
Incomplete Frr = 1
—
(G − 1)r
Sediment density
—
d∗r
Dimensionless diameter
Lr
—
— zr
1
zr5 xr−3.5
3/2
Re∗r
Grain Reynolds
3 − 6m 2 + 2m
1
3/2
zr 1
Lr —
zr3/2 —
−3 m 1+m
1
3/2
Rer τ∗r
Reynolds Shields
zr
zr xr−1
−3m 1+m
2 + 5m 1+m
1 − 2m 1+m
2 + 5m 1+m
zr
1
zr
zr xr− 1
yr zr
zr
yr zr
1
Froude
1
fr Frr
Darcy–Weisbach
Sr
Slope
1
xr yr zr
L r3
Fr
Force
Dimensionless
zr2 xr−1
Lr
τr
Shear stress
zr
Lr
pr
xr yr zr
Pressure
L r3
Mr
Mass
Dynamic
zr0.857
1
1
zr1.5 1
1
zr−0.43
zr− 0.43
yr zr2.43
zr0.57
zr
yr zr2.43
zr
1−2 m 3
2m − 1 zr1 − 2 m dsr
dsr
2+2 m 3
1+m z 0.5−m dsr r
3/2
zr 1
1
2 m z −2 m dsr r
dsr2 m zr− 2 m
−2 m yr zr2+2 m dsr
2m zr1−2 m dsr
zr
−2 m yr zr2+2 m dsr
−3 dsr
1
1
− 1−m zr1+1 m dsr 1
−1−m zrm−0.5 dsr
2m zr−2 m dsr
−2 zr− 1 dsr
2 yr zr3 dsr
−2 dsr
zr
2 yr zr3 dsr
340
Physical river models
modeling produces a very large laminar sublayer thickness in the model. In the hydraulically smooth regime, resistance to ﬂow increases as the Reynolds number decreases, and resistance to ﬂow will be larger for the model than for the prototype. Exact similitude of nearbedﬂow conditions cannot be preserved when the same ﬂuid is used because the viscous effects cannot be neglected. Exact similitude in these cases would require different ﬂuids for the model and the prototype. 10.2.2
Froude similitude for tilted river models
Model distortion and tilting is viewed as a feasible practical alternative. Model distortion and tilting are acceptable only when vertical and lateral accelerations of the water can be neglected with respect to the gravitational acceleration. This practical solution allows the use of different scales for ﬂow depth and sediment size. The model is distorted when yr = zr and tilted when xr = zr , which should be appropriate for near 1D ﬂow conditions. Rigidbed hydraulic models require that resistance to ﬂow be the same for the model and the prototype. In the hydraulically rough regime, Re∗ > 70, resistance to ﬂow depends on relative submergence h/ds . The governing equation to be preserved in gradually varied ﬂow models with rigid boundaries is the resistance relationship Sr = fr Frr2 whereby tilting is required because Frr = 1 and dsr = 1. In general terms, resis√ tance to ﬂow can be deﬁned as 8/ f = a(h/ds )m where m = [1/(ln 12.2 dhs )]. The governing equation of similitude for resistance to ﬂow can be written as zr zr 2 m . (10.2) Frr2 = xr dsr As a particular case, according to Strickler’s relationship between Manning co1/6 efﬁcient n and bed roughness diameter n ∼ ds , the Manning–Strickler equation corresponds to m = 1/6. The model scales in distorted Froude models must simultaneously satisfy the Froude and the Manning–Strickler similitude criteria. The Manning–Strickler 1/2 1/2 similitude criterion in a distorted model is deﬁned as(zr /dsr)1/6 [(zr Sr )/Vr] = 1. A tilted hydraulic model Sr = zr /xr = 1 that satisﬁes the Froude similitude Frr = 1 implies that dsr = zr4 /xr3 . According to this relationship, the user has 2 degrees of freedom in selecting two of the three scale parameters, xr , zr , or dsr . During the calibration of rigidboundary models, model roughness is typically increased when disproportionately large blocks and sticks are used to reproduce a stage–discharge relationship comparable with that of the prototype. The modeling of design structures with distorted rigidboundary models thus requires the intuition and judgement of experienced engineers. The scale ratios
Rigidbed model
341
for a distorted rigidbed model are presented in column 4 of Table 10.1 and a calculation example is presented in Example 10.1. Model distortion is often encountered in engineering practice whereby the ﬂow depth is increased compared with that of exact similitude. A distorted model with different horizontal and vertical scales allows different scales for the bed material and for ﬂow depth. The practical interest in distorted models is that in increasing ﬂow depth and decreasing resistance to ﬂow, and the model user can empirically increase the size of roughness elements until the model results compare with ﬁeld measurements. The model is then said to be calibrated. Because the kinematic similarity is not exact, however, any attempt to determine kinematic properties such as streamlines and turbulent mixing cannot be properly scaled in distorted models.
Example 10.1 Application to a tiltedrigidbed model. A 2,000m gravelbed river reach has a ﬂow depth of 2 m, a width of 50 m, and a mean velocity of 0.3 m/s. If the prototype Manning coefﬁcient n = 0.025 and d50 = 5 mm, determine the model scales. The maximum length in the laboratory is 20 m. The prototype Froude number is Vp 0.3 = √ = 0.0677, Fr p = gh p 9.81 × 2 and the prototype slope is 0.0252 × 0.32 n2 V 2 = = 2.23 × 10−5 . Sp ∼ = h 4/3 2 4/3 The Shields parameter is τ∗ p =
h p Sp 2 × 2.23 × 10−5 = 0.005 < 0.03. = (G − 1) ds p 1.65 × 0.005
Therefore a rigidbed model is appropriate. The length scale xr = (x p /xm ) = (2,000/20) = 100 or a model depth of h m = h p / h r = 2/100 = 0.02 m and an exact Froude similitude gives √ Vr = xr = 10, and the model velocity Vm = V p /Vr = 0.03/10 = 0.03 m/s. The corresponding model Reynolds number is Rem = [(Vm h m )/ν] = [(0.03 × 0.02)/(1 × 10−6 )] = 600 and the model ﬂow would be laminar. Model distortion is necessary in order to increase the accuracy of ﬂowdepth and velocity measurements and to guarantee turbulent ﬂow in the model. The user has 1 degree of freedom in selecting either the vertical scale, the slope, or the size of roughness elements. The vertical scale is arbitrarily set as zr = 25
342
Physical river models
or a ﬂow depth of 8 cm. The model Reynolds number would then be Rem =
Re p Vp h p 2 × 0.3 = = = 4,800. 3/2 Rer 1 × 10−6 × 251.5 νzr
Model scales are calculated from column 4 of Table 10.1 for tilted rigid beds. For example, planform geometry is similar when √ yr = xr = 100. The model √ / z ) = 0.3/ 25 = 0.06 m/s, and the time velocity is Vm = (V p /Vr ) = (V√ p r −1/2 = (100/ 25) = 20. The model discharge is Q m = scale is tr = xr zr 3/2 Q p /Q r = V p W p h p /yr zr = [(0.3 × 50 × 2)/(100 × 253/2 )] = 0.0024 m3/s. The tilted model slope is Sm = S p /Sr = S p /zr xr−1 = 2.23 × 10−5 × 100/25 = 8.9 × 10−5 . The size of the roughness elements should be dsm = dsp /dsr = dsp xr3 /zr4 = 0.005 1003 /254 = 13 mm. In this type of model, resistance blocks or plates would be placed on a smooth surface until the ﬁeld stage–discharge conditions can be adequately duplicated. This is a trialanderror calibration procedure.
10.3
Mobilebed river models
Mobilebed models are useful when sediment transport is signiﬁcant, e.g., when τ∗ > 0.06. Typical examples include drop structures, local scour, erosion below spillways, sills, locks and dams, reservoir sedimentation, etc. The bed mobility provides 1 additional degree of freedom in selecting the mass density of sediment. Similitude in sediment transport is obtained when the Shields parameter τ∗ and the dimensionless grain diameter d∗ are similar in both systems, i.e., τ∗r = 1 and d∗r = 1. Of course, these conditions also imply that Re∗r = 1 because τ∗ d∗3 = Re∗2 . There are four similitude criteria for mobilebed models: (1) Froude similitude; (2) resistance, e.g., Manning–Strickler; (3) dimensionless grain diameter; and (4) bedmaterial entrainment or Shields parameter. These four similitude criteria must be simultaneously satisﬁed in river reaches with rapidly varied ﬂow and sediment transport. The governing criteria involve seven parameters: Vr , gr , zr , dsr , Sr , (G − 1)r , and νr . Hydraulic models with gr = νr = 1 and four equations of similitude leave only 1 degree of freedom, e.g., the model length scale zr , besides the lateral scale yr , which is not speciﬁed by the equations. The model is tilted because the equations impose Sr = zr /xr . For instance, the user may prefer a model that is undistorted but tilted, yr = zr , to a distorted and tilted model that preserves planform geometry xr = yr . The mobilebed similitude is said to be complete, with 1 degree of freedom, when the four equations of similitude are simultaneously satisﬁed (Subsection 10.3.1). When complete similitude is impossible, it is sometimes possible
Mobilebed river models
343
to sacriﬁce one of the governing equations for an additional degree of freedom (Subsection 10.3.2). 10.3.1
Complete mobilebed similitude
It is important to acknowledge that complete mobilebed similitude implies that the downstream direction is dominant and the accelerations in the lateral and the vertical directions are negligible. Complete mobilebed similitude is therefore essentially suitable for 1D sedimenttransport processes. Similitude in dimensionless particle diameter d∗r = 1 in hydraulic models imposes the following relationship between the particle diameter and the particle density: dsr3 =
1 . (G − 1)r
(10.3)
It is clear from this relationship that hydraulic models require very light sediment when large particles are used in the model. The properties of light material commonly used in practice are listed in Table 10.2. It is interesting to note that, for prototype sediment at a speciﬁc gravity of 2.65, any lightweight material corresponds to a speciﬁc scale ratio for the particle diameter. Table 10.2. Lightweight sediment properties for mobilebed models
Material
Speciﬁc gravity G
Typical size ds (mm)
Polystyrene
1.035–1.05
0.5–3
Gilsonite Nylon (polyamidic resins) Lucite PVC Perspex Acrylonitrile butadiene styrene Coal
1.04 1.16
0.1–5
1.18 1.14–1.25 1.18–1.19 1.22
1.5–4 0.3–1 2–3
Ground walnut shells Bakelite Pumice Loire sand Lytag (ﬂyash) Quartz sand
1.2–1.43 (up to 1.6) 1.33 1.38–1.49 1.4–1.7 1.5 1.7 2.65
0.3–4 0.15–0.41 0.3–4.0
0.63–2.25 1–3 0.1–1
Comment Durable but difﬁcult to wet and tends to ﬂoat
Hydrophobic Dusty Adds detergent against airbubble adherence Possible inhomogenity in speciﬁc gravity and sorting Deteriorate in 2–3 months, color water (dark brown) Porous, tends to rot, changes diameter, and ﬂoats Dusty Porous
344
Physical river models
Similitude in Shields parameter τ ∗r = 1 imposes the following relationship between the particle diameter and the slope similitude Sr : τ ∗r = zr Sr /[(G − 1)r dsr ] = 1.
(10.4)
It is clear from Eqs. (10.3) and (10.4) that in order to simultaneously satisfy 1/2 d∗r = 1, Sr = zr /xr , and τ∗r = 1, the condition dsr = xr /zr must be satisﬁed. The scale ratios are obtained from simultaneously satisfying Eqs. (10.1)–(10.4). In a nutshell, the particlediameter scale dsr is directly obtained as a result of the Manning–Strickler relationship and the Froude similitude dsr = zr4 /xr3 . The density of the sediment is then obtained from the dimensionless particle diameter such that (G − 1)r = xr9 /zr12 . Finally, by substitution into the Shields equation, the condition that satisﬁes the same Shields parameter imposes that 7/10 zr = xr . This condition further simpliﬁes previous requirements as dsr = −0.2 xr and (G − 1)r = xr0.6 . The criterion for sediment suspension based on settling velocity ω can be deﬁned from ω = 8(ν/ds )[(1 + 0.0139 d∗3 )0.5 − 1]. The settling velocity scale in water is thus ωr = 1/dsr as long as d∗r = 1. The criterion for sediment suspension is deﬁned from identical values of the ratio of shear velocity to settling 1/2 velocity, or ωr /u ∗r = 1. This leads directly to dsr = xr /zr , which is identical to the condition previously obtained from the Shields parameter. We can thus conclude that similitude in Shields parameter is equivalent to similitude in the ratio of bedload to sediment suspension, provided that d∗r = 1. This strengthens the requirement that d∗r = 1 and τ∗r = 1 for similitude in sediment transport. Similitude in bedload sediment transport can be determined from the Einstein–Brown relationship as qbv /ωds = f (τ∗ ), in which, with ωr = 1/dsr when d∗r = 1, we obtain directly qbvr = f (τ∗r ). It is interesting to note that qbvr = 1 when τ∗r = 1, which again strengthens the requirement for similitude in Shields parameter through τ∗r = 1. Bed aggradation and degradation relates to the sediment continuity relationship applied to bedload discharge written in 1D form as ∂qb /∂ x = − p0 (∂z 0 /∂ts ), where qb is the unit bedload discharge, p0 is the porosity of the bed material, z 0 is the bedelevation, and ts refers to time. The time scale for bedload motion tsr that describes bedelevation changes is then obtained as tsr = [( p0r zr xr )/qbr ]. It can be assumed that the porosity ratio p0r = 1. This time ratio refers to the sedimentation time scale that is useful in the analysis of local bedelevation changes through local scour, bedforms, and changes in bedload transport. The time scale for bedelevation changes is different from the time scale obtained from the Froude similitude criterion. In diffusion–dispersion studies, the time scale for vertical mixing can be estimated from tvr = zr /u ∗r comparatively with the time to lateral mixing given by ttr = yr2 /zr u ∗r . Of course these two scales are equivalent only as long as
Mobilebed river models
345
the model is not distorted. Also, the length scale for vertical mixing is xvr = [(zr Vr )/u ∗r ] is comparable with the length for lateral mixing given by xtr = Vr yr2 /zr u ∗r . These length scales are compatible for only undistorted models. The scale ratios for rapidly varied mobilebed models are listed in column 5 of Table 10.1. Example 10.2 illustrates how to calculate the scale ratios for mobilebed models. In practice, complete mobilebed similitude is somewhat restricted to model scales that are not too small (approximately zr < 25) because larger scale models necessitate unreasonably light material. Example 10.2 Calculation of complete mobilebed similitude. Consider the model of a large sand bed at a ﬂow depth of 8 m and a velocity of 2 m/s. The slope is 7 × 10−5 , and the discharge is 40,000 m3 /s. Determine the scale ratios for complete similitude at zr = yr = 100. The prototype Shields parameter is τ∗ p =
h p Sp 8 × 7 × 10−5 = 1.7 > 0.03. = (G − 1) p dsp 1.65 × 0.0002
× 8 −1 The bed is mobile and the value of m = [ln( 12.2 )] = 0.076. The scale 0.0002 ratios for the following parameters are obtained from column 5 of Table 10.1:
m = 0.0763 downstream distance xr = 1001.21 = 266 dsp 0.2 mm = = 1.2 mm model particle diameter dsm = dsr 100−0.393 time scale tr = 1000.71 = 26.6 time scale of bed tsr = 1002.21 = 26,300 Vp 2 m/s = 0.2 m/s model velocity Vm = = Vr 10 Qp 40,000 m3 /s model discharge Q m = = = 0.4 m3 /s Qr 1005/2 Sp 7 × 10−5 model slope Sm = = = 1.9 × 10−4 Sr 100−0.21 (G − 1) p 1.65 = 7 × 10−3 . model sediment density (G − 1)m = = 1001.18 230 G m = 1.007, which is lower than the density of polystyrene. An incomplete mobilebed model will unfortunately be indicated at this model scale.
10.3.2
Incomplete mobilebed similitude
When the conditions for complete similitude are not practically possible, one constraint can sometimes be sacriﬁced in order to beneﬁt from an additional
346
Physical river models
degree of freedom. As the model further deviates from complete similitude, there is a greater risk that the model may yield incorrect results. There are nevertheless a number of possibilities, depending on which conditions of similitude should be preserved in both the model and in the prototype. Two types of models are considered here: (1) nonFroudian similitude Frr = 1; and (2) quasi similitude in sediment transport d∗r = 1. First, nearequilibrium streams in which the ﬂow is gradually varied can be simulated with different values of the Froude number as long as the ﬂow is subcritical, i.e., ﬁnegrained alluvial rivers with low Froude numbers can be simulated with the same bed material at higher, yet subcritical, Froude numbers. Second, coarse bed material in which bedload transport is predominant can be simulated with smaller values of d∗ as long as the ﬂow is hydraulically rough. The scale values for these two cases of incomplete mobilebed similitude are presented in columns 7 and 8 of Table 10.1. In nonFroudian models, graduallyvaried ﬂow in large alluvial rivers does not impose large changes on the speciﬁcenergy diagram. In cases in which the Froude number remains fairly constant along the study reach, we may use different values of the Froude number for the model and the prototype as long as the ﬂow remains subcritical in both systems. In this case only the three governing equations, (10.2), (10.3), and (10.4), are simultaneously satisﬁed. The Froude similitude criterion would be replaced with an additional degree of freedom, e.g., yr , zr , and dsr . When the same density of sediment is used, the similitude in dimensionless particle diameter d∗ implies that (G − 1)r = 1/dsr3 ; thus the sediment must have the same density and particle diameter. The Shields similitude condition can be 1/2 rewritten as dsr = xr /zr = 1 and Manning–Strickler equation (10.2) can be rewritten in terms of the Froude number as Frr = zr−0.5+m . The Froude scale is not unity but changes slightly with model scale and the scale ratios for gradually varied alluvial ﬂow models are listed in column 8 of Table 10.1. This type of model allows the use of the same sediment in the model and the prototype. Undistorted crosssectional geometry can be simulated with yr = zr , and similitude in planform geometry is obtained when xr = yr . This type of model offers similitude in bedload and suspended sediment transport. The drawbacks of this approach are that the model does not obey the Froude similitude and should be limited to nearuniform subcritical ﬂows. This type of model should appropriately simulate sediment transport and resistance to ﬂow. However, the force diagrams and the lateral/vertical accelerations are not appropriately simulated. Models with quasi similitude in sediment transport, d∗r = 1, may be used from Table 10.1, column 7, in which bedload transport is dominant, as detailed in Example 10.3. It is noteworthy that the similitude in total sediment transport
Mobilebed river models
347
is questionable when τ∗r = 1 and d∗r = 1. Scale effects can be found in other sedimenttransport processes such as suspended load and bedform prediction. Case Study 10.1 and Problem 10.3 illustrate how the scales of a model that uses the same material for both systems can be determined. Case Study 10.2 relates physical modeling with protection against scour at highway bridges. Example 10.3 Application to bedload transport in coarsegravelbed streams. A gravelbed river with d50 ≈ 30 mm and d90 = 100 mm has a natural bed slope S = 3 × 10−3 . If the ﬂow depth reaches 4 m during ﬂoods, determine the model scales that would allow a 10km reach to be modeled within 40 m in the hydraulics laboratory. The downstream length scale xr = 10,000/40 = 250 and the Shields parameter of the prototype is τ∗ p =
h p Sp 4 m × 3 × 10−3 = 0.24 > 0.06; = (G − 1) p ds p 1.65 × 0.03 m
hence there is sediment transport. Complete mobilebed similitude would require a model sediment size of dsr = (ds p /dsm ) ∼ = zr−0.286 = 250−0.286 = 0.208 or dsm 30 mm, which is impractical. The modeler must therefore resort to incomplete mobilebed modeling. In this case, a model that does not satisfy the Froude condition would not be indicated because the prototype Froude number is certainly quite high. Because the bed material is very coarse and the ﬂow is hydraulically rough, Re∗ p =
gh p S p ds p ν
√ =
9.81 × 4 × 3 × 10−3 × 0.03 = 10,300, 1 × 10−6
it is considered that d∗r = 1 would be appropriate because ﬁner bed material would still be in the hydraulically rough regime. The scale ratios for incomplete mobilebed similitude with d∗r = 1 leave 2 degrees of freedom. From proh 4 −1 ) = 0.135, and the totype values, m = [ln(12.2 × dsp )]−1 = (ln 12.2 × 0.03 p 2 degrees of freedom in zr = 175 and dsr = 20 are arbitrarily selected, along with yr = 250. There is no similitude in planform geometry because yr = 250 and xr = zr1+2 m dsr−2 m = 314. The resulting scale ratio for grain shear Reynolds number from Table 10.1, column 7 is Re∗r = (Re∗ p /Re∗m ) = dsr1+m zr0.5−m = 201+0.135 1750.5−0.135 = 197; then Re∗m = Re∗ p /Re∗r = 10,300/197 = 52, which is considered hydraulically rough. Other model scales √ calculated 1/2 from Table 10.1 are Q r = yr zr1.5 = 5.8 × 105 and Vr = zr = 175 = 13.3. The model sediment diameter is dsm /dsr = 0.03 m/20 = 1.5 mm at a
348
Physical river models
particle density of (G −1)m = (G −1) p /(G − 1)r = 1.65/zr1−2 m dsr2 m−1 = 1.65/ 1751−2×0.135 202×0.135−1 = 0.34 or G m = 1.34, in which coal or Bakelite would be appropriate from Table 10.2. The model will also require tilting and Sm = S p /Sr = S p /ds2r m zr−2 m = 3 × 10−3 /202×0.135 175−2×0.135 = 0.0054. The scale for unit bedload discharge is calculated assuming similitude in qbr /u ∗r dsr = 1 or qbr = dsr1+m zr0.5−m = 201+0.135 1750.5−0.135 = 197. Finally, the time scale for m = 1751.5+3×0.135 bedelevation changes is obtained from tsr = zr1.5+3 m ds−1−3 r −1−3×0.135 20 = 278, which means that bed changes over 1 min of model time correspond to 4.6 h of prototype time. Suspended sediment transport is not in similitude. Therefore the similitude in total sediment discharge is questionable, but aggradation and degradation features from bedload discharge should be well represented by the model. Case Study 10.1 Mobilebed model of the Jamuna River, Bangladesh. Klaassen (1990, 1992) reported on a physical scale model of the Jamuna River in Bangladesh. The Jamuna River is a ﬂat sandbed river, with, on the average, three braids upstream of Sirajganj (near the proposed bridge site). The purpose of this movablebed scale model investigation is: (1) to study the effect of the river training works on channel patterns; (2) to identify “worst” channel patterns (during both ﬂoods and lowﬂow conditions) for the river training works; and (3) to study the upstream river shifts on the channel patterns near the bridge and the training works. For the present model investigation the limitations include: (1) model dimensions at a maximum of approximately 50 × 20 m2 because of costs; (2) model material sand, because lightweight material (which would be preferable) is too expensive in large quantities; and (3) model material size, which should be ∼200 µm because it is readily available in large quantities. The main characteristics of the related model are listed in Table CS.10.1.1. The scaling procedure resulted in a considerably distorted model. For a model discharge of ∼100 1/s, a fair reproduction of the channel pattern is probably obtained for a model slope of approximately 8 × 10−3 . If the Ch´ezy coefﬁcient Cm in the model is assumed to be ∼25 m1/2 /s, the average velocity in the channels is estimated to be ∼0.4 m/s. This implies that the Froude number in the model is of the order of 0.5 whereas in the prototype Fr ∼ = 0.2. Locally in the model even higher values of the Froude number may be found. Such high Froude numbers will certainly affect the ﬂow pattern reproduction and consequently induce scale effects. Very good morphological results were obtained with the model, and the objectives have been satisfactorily met. One of the main concerns about the model relates to local scour. Local scour is reproduced correctly in only undistorted models that satisfy the Froude condition. Because both conditions are
Mobilebed river models
349
Table CS.10.1.1. Characteristics of the Jamuna River model (after Klaassen, 1990) Parameter Particle size Density Slope Discharge Bankfull width Total width Flow depth Sediment transport Flood duration Froude number
Prototype
Scale factor
dsp = 0.2 mm G p = 2.65 S p = 7 × 10−5 Q p = 10,000 m3 /s = 90,000 m3 /s W p = 3,000 m 15,000 m h p = 5.8 m h mp = 8 m qsp = 1.4 × 10−3 m2 /s T p = 78 days Fr p = 0.1−0.2
dsr = 1 (G − 1)r = 1 Sr = 0.01 Q r = 106 yr = 1,000 yr = 1,000 zr = 200 qsr = 80 tr = 2,500 Frr = 0.25
Model dsm = 0.2 mm G m = 2.65 Sm = 7 × 10−3 Q m = 0.01 m3 /s = 0.09 m3 /s Wm = 3.3 m 15 m h m = 0.032 m h mm = 0.04 m qsm = 1.34 × 10−5 m2 /s Tm = 0.03 day Frm = 0.4 − 0.8
not fulﬁlled, local scour cannot be scaled. Another concern is the local presence of supercritical ﬂow. Through comparisons with the conditions in Example 10.2, the reader will note that complete mobilebed similitude requires excessively light material. It is also given as Problem 10.3 that the reader can verify that the model scales are quite comparable with those obtained with incomplete mobilebed similitude with zr = 100, yr = 1,000, and dsr = 1. Case Study 10.2 Bridge scour model for Schoharie Creek, United States. Richardson and Lagasse (1996) report that there are more than 575,000 bridges in the U.S. National Bridge Inventory. Approximately 84% of these bridges span over water. On April 5, 1987, the center span of the 165mlong bridge over Schoharie Creek, New York, collapsed during a nearrecord ﬂood of ∼ 1,750 m3 /s. Each rigidframe pier was supported on a spread footing bearing on a glacial till just below the streambed. The bridge designers assumed that the glacial till substrate was nonerodible. A physical model of the Schoharie Creek bridge was built at a model scale of 1:50 in a 20 ft long × 100 ft ﬂume. The following scale ratios were obtained from an exact Froude similitude with zr = 50: (1) the velocity scale was 7.07; (2) the time scale was also 7.07; and (3) the discharge scale was 17,678. The horseshoe vortex near the base of the bridge pier was at the origin of the scour underneath the pier and the spread footing. The 1987 bridge failure resulted in establishing a National (U.S.) Program that requires each state to evaluate all bridges over water for vulnerability to failure from erosion of the foundations. This evaluation is to be carried out by an interdisciplinary team consisting of hydraulic, structural, and geotechnical
350
Physical river models
engineers. This evaluation is in addition to the National Bridge Inspection Program that requires the states to inspect all bridges every two years to determine their structural integrity. As a design philosophy, bridges should be designed by use of the ﬂow of a 100yr ﬂood and to withstand the effects of scour from a ﬂow exceeding the 100yr ﬂood. Scour monitoring devices include sonic fathometers for scour data measurements that can be stored in data loggers. Also, magnetic sliding collars have been designed to slide freely over a small stainless steel pipe driven into the streambed at the expected scour location. An audible signal can be heard at the bridge deck when the magnet comes close to the sensor. Exercise 10.1
Derive the scale ratios for tiltedrigidbed models from similitude in Froude number and Manning–Strickler. Compare the results with those of column 4 in Table 10.1. Also determine the results of the particular case in which xr = yr = zr = L r . Exercise 10.2 Demonstrate that, in a hydraulic model, the scale parameters that satisfy the same Reynolds number and the same Shields parameter also satisﬁes the same dimensionless particle diameter. Exercise 10.3
In Example 10.2, show that the grain Reynolds number of the model and the prototype are quite similar despite the model distortion. Exercise 10.4
Derive the scale ratios for complete mobilebed similitude from simultaneously solving for similitude in: (1) Froude number Frr = 1; (2) resistance equation; and (3) sediment transport τ∗ p = d∗r = 1. Compare the results with those of Table 10.1, column 5, and check the values for the Manning–Strickler relationship (m = 1/6). Exercise 10.5
Derive the scale ratios for incomplete mobilebed similitude after considering similitude in: (1) the resistance equation; and (2) sediment transport with τ∗r = d∗r = 1. Compare the results with those of Table 10.1, column 8, for Frr = 1.
Problems
351
Exercise 10.6
In Case Study CS.10.1, determine whether the scale factors selected in Table CS.10.1.1 satisfy: (1) the Froude similitude; (2) the Manning–Strickler similitude; and (3) sedimenttransport similitude in d∗r = τ∗r = 1. Problem 10.1
A clearwater openchannel model is to be designed such that the maximum laboratory discharge is 2 l/s for a stream discharge of 300 m3 /s. If the laboratory space allows a maximum length of 60 m to model a river reach of 10 km, determine a suitable scaling length for the model. From this, determine the scaling ratios for time, discharge, and hydropower. Problem 10.2 The ﬁlling and emptying gates of a canal lock extend the full height of the lock chamber. When a vessel is lowered in the prototype lock, the gates at the outlet end are programmed to open at a ﬁxed rate. The waves and the currents produced by the outﬂow cause the vessel to pull at its moorings. In a 1/25 scale model of the system that uses water, the maximum tension in the moorings is 1.6 lb (0.36 N) when the gates are opened at the proper rate. Determine the maximum mooringline tension in the prototype. Answer: F p = Fm Fr = (l.6 lb)L r3 = 25,000 lb (112 kN). Problem 10.3
With reference to Case Study 10.1, demonstrate that the model scales in Table CS.10.1 are comparable with those obtained from Table 10.1, column 8, with yr = 1,000, zr = 100, dsr = 1, and m = 0.076. Answer: tr = 1002−m = 7050, Q r = yr zr1+m = 142,000, Sr = zr−1 = 0.01, and Frr = 100m−0.5 = 0.14. Problem 10.4
With reference to Example 10.3, calculate the ratio u ∗ /ω for both the model and the prototype and verify that most of the sediment transport is bedload in both cases. Answer: u ∗ p /ω p = 0.57 and u ∗m /ωm 0.5, thus predominantly bedload.
11 Mathematical river models
Numerous river engineering problems can be conveniently investigated by means of mathematical models. Mathematical models must properly describe the physical processes and provide a numerical solution to a system of differential equations that are solved together with suitable boundary conditions and empirical relationships that describe resistance to ﬂow and turbulence. The differential equations describing river mechanics problems are usually simpliﬁed forms of the equations of conservation of mass and momentum, leading to a set of partial differential equations involving two independent variables (time and space or two spatial variables). Examples that use the ﬁnitedifference method are presented in this chapter. The ﬁniteelement method also provides useful solutions to river engineering problems but is beyond the scope of this chapter. The algorithms to be used in the ﬁnitedifference method depend on the type of differential equation to be solved. Table 11.1 provides a simple classiﬁcation of river engineering problems. The information propagates at a celerity c in hyperbolic equations, and the celerity is effectively inﬁnite in parabolic equations. Once a river engineering problem has been deﬁned and a mathematical model chosen, ﬁeld data need to be gathered to describe initial and boundary conditions, geometrical similitude, material properties, and design conditions. Additional data are also required for calibration and veriﬁcation. The governing equations can be simpliﬁed to preserve the main features of the physical problem; the time and the space increments are determined at this stage. A schematization can be made of the design conditions to be investigated. Model calibration is usually necessary because empirical parameters are involved to describe resistance to ﬂow and because of the impliﬁcations to the governing equations. Parameters can be adjusted to obtain good correspondence between numerical results and continuum values. Of course, the adjustment should not be extended beyond physically acceptable values. The
352
Finitedifference approximations
353
Table 11.1. Differential equation types in river engineering
Equation type Hyperbolic
Parabolic Elliptic
Equation ∂φ ∂φ +v =0 ∂t ∂x 2 2 ∂ φ ∂ φ = c2 2 ∂t 2 ∂x ∂2φ ∂φ = Kd 2 ∂t ∂x ∂2φ ∂2φ + 2 =0 ∂x2 ∂y
River engineering problem Advection (v constant) Floodwave propagation (c2 constant) Diffusion–dispersion (K d constant) Flow net
precision of a model refers to the error margin of the numerical calculations. Model accuracy usually refers to the comparison of the model with ﬁeld measurements. For instance, a model that calculates the ﬂoodstage to the nearest centimeter but is 1 m off from the ﬁeld measurements is precise but not accurate. The method of adjusting parameters by running the model at different values until a satisfactory result is obtained is called hindcasting. It is a very useful way to determine the sensitivity of the model results to changes in the model parameters. The calibration phase should also comprise a check of the numerical accuracy by varying numerical parameters such as the time step. Model veriﬁcation involves simulation for a different set of prototype data with the coefﬁcients previously obtained during the calibration. If a model run satisfactorily reproduces the measured prototype conditions without further adjustment, a reasonable conﬁdence is gained in the application of the model to design conditions that have never occurred in the prototype. It is often possible to calibrate a model with the ﬁrst half of a ﬁeld data set and to verify the model with the second half. This chapter describes ﬁnitedifference approximations in Section 11.1, followed by some typical 1D models in Section 11.2, and a brief discussion of multidimensional models in Section 11.3.
11.1
Finitedifference approximations
Let us consider a function h(x, t) deﬁned in space x and time t. We may divide the x–t plane into a grid, as shown in Fig. 11.1(a). The grid spacing along the x axis is x and the time interval along the t axis is t.
354
Mathematical river models
The value of the variable h will use the spatial location as a subscript and the time as a superscript, e.g., h kj refers to the value of ﬂow depth at the jth spatial grid point and kth time grid point. By the known time level, we mean that the values of different dependent variables are known at the time level h k and we want to compute their values at the unknown time level h k+1 . If the computations progress from one step to the next, then the procedure is referred to as a marching procedure. Most of the phenomena described by hyperbolic partial differential equations are solved with marching procedures. The conditions speciﬁed at time t = 0 are referred to as Figure 11.1. Finitedifference grid the initial conditions. The conditions specand approximation. iﬁed at the channel ends are called the end, or boundary, conditions. Finitedifference approximations are ﬁrst introduced before a presentation on consistency and convergence (Subsection 11.1.1), a linear stability analysis (Subsection 11.1.2), higherorder approximations (Subsection 11.1.3), and boundary conditions (Subsection 11.1.4). The ﬁnitedifference method is based on a Taylor series expansion of the variable h j+1 written as a function of h j as
h kj+1
=
h kj
∂h + x ∂x
k
x 2 + 2! j
∂ 2h ∂x2
k j
x 3 + 3!
∂ 3h ∂x3
k + 0(x 4 ), (11.1) j
where the derivative (∂h/∂ x)kj is evaluated at grid point j and time level k and 0(x m ) indicates morder terms. The Taylor series could be similarly expanded k from h kj as to deﬁne h j−1 h kj−1
=
h kj
− x
−
x 3!
3
∂h ∂x
∂ h ∂x3 3
k j
k
x 2 + 2!
∂ 2h ∂x2
+ 0(x 4 ). j
k j
(11.2)
Finitedifference approximations
355
Rearranging Eqs. (11.1) and (11.2) and dividing by x gives, respectively,
∂h ∂x
k
h kj+1 − h kj
=
x
j


forward difference
x 2!
−
 x 
∂h ∂x
k =
h kj − h kj−1 x
j

backward difference
+
− j
x 2!
∂ 2h ∂x2
 h kj − h kj−1 x 
x 2 3!
∂ 3h ∂x3
upwind
k j

+ 0(x),
k − j
(11.3)

downwind

+ 0(x 3 )
k
truncation error
h kj+1 − h kj
+ 0(x 3 )
∂ 2h ∂x2
x 2 3!
truncation error
+ 0(x).
∂ 3h ∂x3
k j
 (11.4)

The ﬁrst approximation of the partial derivative in Eq. (11.3) is written in the form of a forward (downwind) difference and a ﬁrstorder truncation error 0(x) that will approach zero as x becomes very small. Similarly, Eq. (11.4) includes a backward (upwind) difference and a truncation error 0(x). The truncation error approaches zero as x approaches zero as long as the high derivatives remain continuous. Therefore both forward and backward ﬁnitedifferences are ﬁrstorder approximations. A central ﬁnitedifference approximation can then be obtained from taking half of the sum of Eqs. (11.3) and (11.4), or
∂h ∂x
k = j
h kj+1 − h kj−1 2x
+ 0(x 2 ).
(11.5)
The truncation error is of the order of (x)2 because the terms in x in Eqs. (11.3) and (11.4) cancel. The central difference is thus said to be a secondorder approximation. Figure 11.1(b) shows a geometrical representation of the forward, backward, and central ﬁnitedifference approximations. The real slope is the tangent of the function at B. The forward ﬁnitedifference approximation uses the slope of the secant curve line BC, the backward ﬁnitedifference approximation uses the slope of line AB, and the central ﬁnitedifference approximation uses the slope
356
Mathematical river models Table 11.2. Explicit and implicit ﬁnite differences Finite difference
Explicit
Implicit
Backward upwind
h kj − h kj−1 ∂h ∂x x
− h k+1 h k+1 ∂h j j−1 ∂x x
Forward downwind
h kj+1 − h kj ∂h ∂x x
k+1 h k+1 ∂h j+1 − h j ∂x x
Central
h kj+1 − h kj−1 ∂h ∂x 2x
k+1 h k+1 ∂h j+1 − h j−1 ∂x 2x
of the line AC, although all three approximations become exact as x goes to zero (ﬁrstorder approximation). It is clear from these ﬁgures that the central ﬁnitedifference approximation is more accurate (secondorder approximation) than the forward or the backward ﬁnitedifference approximations. Explicit formulations refer to partial derivatives at the known level k whereas implicit formulations refer to the unknown level k + 1. Table 11.2 lists some typical explicit and implicit ﬁnitedifference approximations for the spatial partial derivative, ∂h/∂ x, at the grid point ( j,k). 11.1.1
Consistency and convergence
Four properties of consistency, stability, convergence, and accuracy are important in numerical analysis. The following formulation of the advection equation, or ﬂoodwave propagation problem, is used to illustrate these properties. Hence ∂h ∂h +c = 0, (11.6) ∂t ∂x where c is the celerity, is approximated with a forward difference in time and a backward difference in space (FTBS) to give − h kj h k+1 j
+c
h kj − h kj−1
= 0. (11.7) t x Rearranging to ﬁnd the ﬂow depth at the unknown level k + 1 as a function of the ﬂow depth at the known level k, we obtain = h kj − h k+1 j
ct k h j − h kj−1 , x
h k+1 = Cc h kj−1 + (1 − Cc ) h kj , j where Cc = (ct/x) is the Courant number.
(11.8a) (11.8b)
Finitedifference approximations
357
To get started, the initial condition of ﬂow depth needs to be known for all j values at k = 0. The algorithm can then march in time given the boundary condition of ﬂow depth for all k values at j − 1 = 0. Consistency is the property of a ﬁnitedifference scheme to reduce to the partial differential equation as the truncation error disappears. In our example, the values of h kj+1 and h kj−1 from Eqs. (11.1) and (11.2) are substituted back into Eq. (11.8b) to give h kj + t +
ct x
k t 2 ∂ 2 h ct 3 h kj + 0(t ) = 1 − 2 2! ∂t x j j k k ∂h x 2 ∂ 2 h k 3 + + 0(x ) . (11.9) h j − x ∂x j 2! ∂x2 j
∂h ∂t
k
+
Rearranging the equation after canceling the terms in h kj and dividing by t results in

∂h ∂t
k
∂h +c ∂x j
original equation
k
t + 2! j

∂ 2h ∂t 2
k

cx − 2! j
∂ 2h ∂x2
truncation error
k
+ 0(t ) + 0(x ) = 0, 2
2
j

(11.10)
where the ﬁrst part is the original equation. This numerical scheme is unconditionally consistent with the partial differential equation because the truncation error vanishes regardless of how t and x approach zero. A method is said to be convergent when the difference between the solutions of the differential and difference equations tends to zero as the time step goes to zero. It has been shown that a consistent method, if stable, is also convergent and vice versa. Consequently it is generally sufﬁcient to check consistency and stability to ensure convergence. It is therefore indicated to examine the stability of numerical schemes.
11.1.2
Linear stability analysis
The stability of a difference method is concerned with the propagation of an error, introduced for example by inaccurate initial or boundary data or rounding in the numerical calculations. Such errors will be propagated by the difference method. If they do not grow, the method is called stable. The linear stability analysis, also referred to as the von Neumann procedure, examines the property of the response of the ﬁnitedifference scheme to input
358
Mathematical river models
perturbations written as a Fourier series in complex form as h kj =
N /2
ζn˜k e
i n˜ 2 π jx L
,
(11.11)
˜ n=1
√ where N is the number of points per wavelength L = N x, i = −1, and n˜ is the wavenumber index. The complex function ei α˜ j n˜ can be separated into ˜ + a real and an imaginary part according to Euler’s relation ei α˜ j n˜ = cos(α˜ j n) ˜ When viewed in the complex plane, the Fourier coefﬁcients ζn˜k i sin(α˜ j n). exhibit the amplitude and the angle α˜ = 2π/N represents a phase angle. The linear stability analysis method examines how each Fourier coefﬁcient changes ˜ Looking at any wave number (say n˜ = 1), in time for any wavenumber index n. we obtain h jk = ζ k ei α˜ j .
(11.12a)
In the following example from Abbott and Basco (1989), the stability analysis of the algorithm in Eq. (11.8) is examined after h jk+1 and h kj−1 are deﬁned from Eq. (11.12a) as h jk+1 = ζ k+1 ei α˜ j , k h j−1
k i α( ˜ j−1)
=ζ e
.
(11.12b) (11.12c)
The terms in Eqs. (11.12) are substituted back into Eq. (11.8b): ˜ j−1) ζ k+1 ei α˜ j = Cc ζ k ei α( + (1 − Cc )ζ k ei α˜ j .
(11.13)
After canceling the common term ei α˜ j , we obtain ζ k+1 =
[(1 − Cc ) + Cc e−i α˜ ]ζ k  ampliﬁcation factor Aα 
(11.14)
The term in the brackets of Eq. (11.14) is the ampliﬁcation factor Aα , which is a complex number. As sketched in Fig. 11.2(a), in the complex plane, the factor Aα is simply a circle of magnitude Cc centered at 1 − Cc , and the modulus Aα  determines whether the Fourier coefﬁcient grows (when Aα  > 1), stays constant (when Aα  = 1), or decays (when Aα  < 1) as a function of time. Of course, growth in the Fourier coefﬁcients means that perturbations grow as the calculations progress in time, even when x → 0 and t → 0. Therefore a ﬁnitedifference scheme is stable as long as Aα  ≤ 1. In this exam˜ + iCc sin α˜ and Aα  = ple, shown in Fig. 11.2, Aα = (1 − Cc + Cc cos α) √ [1 − Cc (1 − cos α)] ˜ 2 + Cc2 sin2 α. ˜ Numerical stability Aα  ≤ 1 corresponds to Cc ≤ 1, which is referred to as the Courant–Friedrich–Levy (CFL)
Finitedifference approximations (a)
Imaginary
1 for stabilty
Aα
359
Cc = 1
+1
C c = 0.5
Unstable +1
1
Real
Cc > 1 1 (b) 1.6 Cc = 2
A
1.4
α
>1 Unstable
1.2 Cc = 1
1.0
Stable
A α 0.8 0.6
C c = 0.5 A
0.4
α
1
0.2 0
0
2
4
6
8
10
12 N
14
16
18
20
22
24
Figure 11.2. Stability diagram (modiﬁed after Abbott and Basco, 1989).
condition of stability for this numerical scheme. It is interesting to note that the CFL stability condition implies that the celerity c of the analytical solution must be less than the celerity (x/t) of the numerical solution, or Cc =
c ≤ 1. x/t
(11.15a)
Accordingly, once the space size x has been determined, the time increment t that satisﬁes the stability requirement is speciﬁed as t
1, and there is numerical diffusion when Cc < 1. The convergence of this numerical scheme is examined at Cc = 0.5 and Cc = 2.0 after the space increment x = 1,000 m is halved, as shown in Table E.11.1.2. It is most instructive to observe that in the case of Cc = 0.5, halving x results in a better approximation of the peak value than that calculated in Table E.11.1.1. Speciﬁcally, after t = 2,000 s, the peak value of h max calculated with x = 1 km is 0.61 m compared with h max = 0.5 at x = 2 km from a correct value of h max = 1 m. The improvement shows that the truncation error decreases as x → 0, which is a simple example of convergence in which accuracy is gained at the expense of more extensive calculations. In comparison, the case with Cc = 2 shows that decreasing x by onehalf does not increase the accuracy of the calculation. Indeed, the maximum ﬂow depth h max calculated at t = 4,000 s is 8 m for x = 1,000 m compared with h max = 4 m at the same time when x = 2,000 m. This illustrates the fact that decreasing x and t does not necessarily improve convergence, even if the numerical scheme is consistent. A consistent numerical scheme can be convergent only when it is stable.
Routing times (s) 0 1,000 2,000 3,000 4,000
Routing times (s) 0 250 500 750 1,000 1,250 1,500 1,750 2,000
Distance (km) Initial depth (m) 0.25
0
0.5
2 0.75
3 1
4
0.25 0.125 0.0625 0.031 0.016 0.008 0.004 0.002 0.001
0.5 0.375 0.025 0.156 0.094 0.0547 0.031 0.016 0.009
0.75 0.625 0.5 0.375 0.266 0.18 0.127 0.07 0.05
1 0.875 0.75 0.625 0.5 0.383 0.28 0.20 0.145
0 0 0 0 0
0.25 −0.25 +0.25 −0.25 +0.25
0.5 0 −0.5 1 −1.5
0.75 0.25 −0.25 −0.75 2.75
1 0.5 0 −0.5 −1
= 2 h kj−1 − h kj Cc = 2.0 or t = 1000 s, h k+1 j
0 0 0 0 0 0 0 0 0
0.5 0.75 0.81 0.78 0.7 0.6 0.49 0.39 0.29
0.5
5
0.5 1.5 −0.5 0.5 −1.5
= 0.5 h kj−1 + 0.5 h kj Cc = 0.5 or t = 250 s, h k+1 j
1
0
Table E.11.1.2. Wave propagation at x = 1,000 m
0 1 2 −3 4
0 0.25 0.5 0.656 0.72 0.71 0.655 0.57 0.46
0
6
0 0 2 2 −8
0 0 0.125 0.31 0.484 0.6 0.655 0.655 0.61
0
7
0 0 0 4 0
0 0 0 0.06 0.19 0.336 0.468 0.56 0.61
0
8
0 0 0 0 8
0 0 0 0 0.03 0.11 0.22 0.344 0.45
0
9
0 0 0 0 0
0 0 0 0 0 0.01 0.06 0.14 0.24
0
10
Onedimensional river models 11.2
365
Onedimensional river models
There are numerous onedimensional models available for the simulation of steady backwater ﬂow, unsteady ﬂoodwave propagation, advection–dispersion of sediment and contaminants, and aggradation–degradation in alluvial rivers. The foregoing presentation focuses on a few numerical schemes: (1) explicit schemes in Subsection 11.2.1; (2) the Leonard scheme in Subsection 11.2.2; (3) the MacCormack scheme in Subsection 11.2.3, an explicit predictor– corrector scheme; and (4) the Preissmann scheme in Subsection 11.2.4, an implicit scheme. Subsection 11.2.5 discusses aggradation–degradation simulations in alluvial rivers.
11.2.1
Explicit scheme
Consider the combination of advection and dispersion as a mechanism for the transport of sediment, or contaminant, at a concentration φ. The substance is said to be conservative in the sense that the total mass remains constant and therefore without sedimentation or chemical reaction that would cause the decay of the substance; the 1D advection–dispersion equation is written as ∂φ ∂ 2φ ∂φ +v = Kd 2 , ∂t ∂x ∂x
(11.19)
where v is the mean ﬂow velocity in the downstream x direction and K d 250 hu ∗ is the dispersion coefﬁcient. Note that a similar approach could be used for turbulent diffusion; however, it is assumed here that the substance is well mixed and the concentration is uniform at a given cross section. It can also be noted that when K d is very small, Eq. (11.19) becomes quite similar to the ﬂoodwave propagation analysis of Section 11.1, as described by Eq. (11.6). The mathematical interest in Eq. (11.19) arises from the fact that the equation is a hybrid between a hyperbolic equation when K d = 0 and a parabolic equation when v = 0. In rivers, the ﬂow velocity is usually important and the system should be centered around a solution to hyperbolic equations, and the numerical scheme developed in Section 11.1 can serve as a basis for further analysis. The FTBS scheme can thus be adopted for the advective term and a secondorder central difference approximation used for the dispersion term. The resulting ﬁnitedifference scheme is written as − φ kj φ k+1 j t
+v
φ kj − φ kj−1 x
= Kd
φ kj+1 − 2φ kj + φ kj−1 x 2
.
(11.20)
366
Mathematical river models
The terms are rearranged in an explicit form after Cu = vt/x and Ck = K d t/x 2 , are deﬁned; thus φ k+1 = Ck φ kj+1 + (1 − Cu − 2Ck ) φ kj + (Cu + Ck )φ kj−1 . j
(11.21)
Although the dispersion term is a secondorder approximation, the advection term is approximate to only the ﬁrst order. The truncation error contains terms on a higher order than the advection term and may thus induce numerical dispersion. Indeed, with reference to the consistency analysis in Subsection 11.1.2, advection equation (11.10) can be rewritten, after φ is substituted for h and v for c, as k k 2 k k ∂φ t ∂ φ x ∂ 2 φ ∂φ 2 2 +v + −v + 0(t ) + 0(x ) = 0. ∂t j ∂x j 2! ∂t 2 j 2! ∂ x 2 j (11.22) The advection scheme can be written as ∂φ ∂φ = −v + 0(t, x). ∂t ∂x
(11.23)
After taking space and time derivatives with constant v, we obtain ∂ 2φ ∂ 2φ = −v 2 + 0(t, x), ∂t∂ x ∂x
(11.24a)
∂ 2φ ∂ 2φ + 0(t, x), = −v 2 ∂t ∂ x∂t
(11.24b)
which can be recombined to obtain the wave equation 2 ∂ 2φ 2∂ φ = v + 0(t, x). ∂t 2 ∂x2
(11.25)
Substituting Eq. (11.25) back into Eq. (11.22) gives k 2 k 2 k ∂φ −v t vx ∂φ ∂ φ + +v = + 0(t 2 , x 2 ). ∂t j ∂x j 2 2 ∂x2 j 
advection term


truncation error

(11.26)
We obtain the fact that the truncation error of the advection scheme includes a numerical diffusion term, which, with Cu x = vt, can be rearranged as K num =
−v 2 t vx vx + = (1 − Cu ). 2 2 2
(11.27)
Onedimensional river models
367
It is now becoming clear that the proposed scheme in Eq. (11.20) really solves the following equation: ∂φ ∂ 2φ v∂φ + = (K d + K num ) 2 · ∂t ∂x ∂x
(11.28)
The numerical dispersion determined in Eq. (11.27) vanishes as Cu approaches unity, hence the interest in running computer models with values of Cu close to unity. This is a challenge, as it has been shown that this numerical scheme becomes unstable as Cu > 1. This now explains why the simulations in Example 11.1 were ﬂawless when Cc = 1, whereas diffusion was observed for Cc = 0.5. The grid Peclet number P = Cu /Ck = vx/K d is a measure of the ratio between advection Cu = vt/x and dispersion Ck = K d t/x 2 . As dispersion gradually tends toward a normal distribution, the standard deviation √ σd increases with time t as σd = 2K d t, and 95% of the dispersed material is contained in a plume of length ±2σd . When Cu = 1, the advection length is simply given by x = vt, the grid Peclet number then corresponds to 2x 2 /σd2 . The physical signiﬁcance, as sketched in Fig. 11.4, is that x = σd
Figure 11.4. Advection and dispersion characteristics.
368
Mathematical river models
when P = 2Cu , or Ck = 1/2, and dispersion is dominant when P < 2 Cu . A substance cannot propagate upstream when x > 2σd , which corresponds to P = 8 Cu , or Ck = 1/8; dispersion is dominant when P > 8 Cu . The physical dispersion is sufﬁciently large to counteract local convection instabilities in regions of large velocity gradients when P ≤ 2 Cu . In practice, modelers must guard against simulations in which K num K d . For instance, consider a model with v = 1 m/s, K d = 100 m2 /s, x =2,000 m, and t = 1,000 s. This model is stable, Cu = vt/x = 0.5 < 1, and essentially advective, Ck = K d t/x 2 = 0.025 Cu . However, K num = (vx/2)(1 − Cu ) = 500 is ﬁve times larger than the physical dispersion K d , and the model is physically meaningless as far as the dispersion calculations are concerned. A better approach would be to use, for instance, half the original K d value, or K d /2 = 50 m2 /s, and set K num = Kd/2 to determine the grid spacing x and time increment t as functions of Cu and K d as x = K d /v(1 − Cu ),
(11.29a)
t = K d Cu /v (1 − Cu ).
(11.29b)
2
For instance, modeling this case with v = 1 m/s and K d = 100 m2 /s would require setting the model for Cu = 0.9 with x = 1,000 m and t = 900 s to obtain K num = 50 m2 /s. A ﬁner grid spacing x would be required for reducing K num . It is also clear from Eqs. (11.29) that both x and t become excessively small when K d decreases. This approach thus becomes quite restrictive when we are solving for diffusion problems (lower K d values). Also, when this algorithm is applied to rivers with variable ﬂow velocity, the most restrictive conditions on x and t are obtained when v is large. This practically means that explicit models are most sensitive to instabilities where and when v is the largest, i.e., near the peak discharge of a ﬂood hydrograph. Peak ﬂood conditions should therefore be used to determine the maximum velocity used in the calculations of x and t.
11.2.2
Leonard scheme
Leonard (1979) developed a thirdorder approximation to solve advection– dispersion equation (11.19). The algorithm eliminates the numerical diffusion term contained in the truncation error. Canceling higherorder space derivative terms provides a higher accuracy for the advection term. The explicit algois given, without derivation, as a function of Cu = vt/x and rithm for φ k−1 j
Onedimensional river models
369
Ck = K d t/x 2 , and values of several modes at the known level φ k : Cu 2 k C = φ + C (1 − C ) − − 3 C + 2 φ k+1 φ kj+1 k u u u j j 6 Cu 2 Cu − 2 Cu − 1 φ kj − Ck (2 − 3Cu ) − 2 (11.30) Cu 2 k Cu − Cu − 2 φ j−1 + Ck (1 − 3 Cu ) − 2 Cu 2 Cu − 1 φ kj−2 . + Ck (Cu ) + 6 The linear stability analysis shows a wide stability range, as in Fig. 11.5. For practical purposes, it is generally sufﬁcient to know that the algorithm is stable Cu 2.0
1.6
Unstable Ck = 0
1.4
P = 10
1.8
Unstable
P =5
1 < Cu < 2 1.2 1.0
=2
0.8
Stable
P
0.6
=
1
P
0.4
P
5 = 0.
0.2
P = 0 .2 0
0
0.2
0.4
0.6
0.8
1.0
1.2 Ck
Figure 11.5. Stability diagram of the Leonard scheme (modiﬁed after Leonard, 1979).
370
Mathematical river models
for Cu < 1 and Ck < 0.4. We can therefore determine x and t as functions of K d and v as t =
K d Cu2 , v 2 Ck
(11.31a)
x =
K d Cu · v Ck
(11.31b)
For given K d and v, the values of x and t are proportional to Cu and inversely proportional to Ck . For instance, when Cu = 1 and Ck = 0.1, t = 10 K d /v 2 , x = 10 K d /v, and P = 10. The coefﬁcients of Eqs. (11.30) reduce to = 0.1 φ kj + 0.8 φ kj−1 + 0.1 φ kj−2 . φ k+1 j
(11.32a)
This algorithm should be convenient as long as the changes in φ are relatively gradual. Small oscillations may otherwise be ampliﬁed because the grid Peclet number P = 10. To ensure a numerically smooth simulation without oscillations, a value of P = Cu /Ck = 2 may provide better results, for instance, Cu = 0.8 and Ck = 0.4. The corresponding grid spacing x = 2K d /v and time step t = 1.6 K d /v 2 are smaller and the coefﬁcients of Eq. (11.30) reduce to = 0.048 φ kj+1 + 0.376 φ kj + 0.304 φ kj−1 + 0.272 φ kj−2 . φ k+1 j
(11.32b)
In the case of pure advection (K d = Ck = 0), the algorithm is stable for Cu < 1, which simply imposes t < x/v regardless of x; this is quite convenient as long as v remains fairly constant within x. Example 11.2 Application to Advection–Dispersion. Consider advection and dispersion in a steep and very rough mountain channel with a slope S = 4 × 10−3 . The ﬂow depth h is 3 m, the mean ﬂow velocity is V = 2 m/s, and the dispersion coefﬁcient K d 250 hu ∗ 257 m2 /s. At the upstream end of a 30km reach, very ﬁne sediment is released at a concentration of 100,000 mg/l for 15 min with a 3min pulse at 200,000 mg/l 1 h after the initial release. Consider that the reach is fairly uniform and that the total sediment mass is preserved at all times. Determine the maximum concentration at a distance of 10 km downstream of the release. How long will the concentration exceed 10,000 mg/l at a point located 20 km downstream of the release? A simple ﬁnitedifference model can be used. For instance, with t = 180 s = 0.05 h and x = 500 m, the Courant number Cu = V t/x = 0.72
Onedimensional river models
371
and the model should be stable because Cu < 1. The dispersion number Ck = K d t/x 2 = 0.185, and the grid Peclet number is P = 3.86. The user of the Leonard method should check that the coefﬁcients a j−2 = 0.075607, a j−1 = 0.577656, a j = 0.337868, and a j+1 = 0.008869 sum up to unity and that none is negative. The initial sediment concentration can be set at 0, and the upstream boundary condition set as 1 × 105 for the ﬁrst ﬁve time steps and 2 × 105 at the 20th time step; all other values are zero. These boundary conditions are not sufﬁcient for using the Leonard scheme. An additional upstream boundary condition is required because the algorithm includes both φ j−1 and φ j−2 . Additionally, the downstream boundary condition must be speciﬁed because the algorithm requires φ j+1 . To compare the total mass, the second upstream boundary condition could satisfy the requirement that no sediment propagates upstream from the point of release. At the
Figure E.11.2.1. Advection–dispersion calculation example.
372
Mathematical river models
downstream end, advection is dominant and the concentration could be set identical to the value calculated one grid space upstream at the previous time step. The use of the numerical scheme in Subsection 11.2.1 is comparatively simpler because only one upstream boundary condition must be speciﬁed. The numerical results shown in Fig. E.11.2.1 indicate that the second sediment pulse disperses very rapidly and produces a second peak with lower sediment concentration. The maximum concentration 10 km downstream is 61,000 mg/l at a time of 1 h 27 min after release. The Leonard algorithm gives Cmax at 59,000 mg/l at 1 h 30 min. The results of both algorithms are essentially identical 20 km downstream, and the concentration exceeds 10,000 mg/l between 2 h 25 min until 4 h after release.
11.2.3
MacCormack scheme
The MacCormack scheme is an explicit, twostep predictor–corrector scheme (MacCormack, 1969) that is secondorder accurate in both space and time and is capable of capturing the shocks without isolating them. This scheme has been applied for analyzing 1D unsteady openchannel ﬂows by Fennema and Chaudhry (1986). For 1D ﬂow, two alternatives of this scheme are possible: (1) backward ﬁnite differences are used in the predictor part and forward differences are in the corrector part; or (2) forward ﬁnite differences are used in the predictor part and backward differences in the corrector part. It is possible to alternate the direction of differencing from one time step to the next. Better results are produced if the direction of differencing in the predictor step is the same as that of the movement of the wave front. The ﬁnitedifference approximations for the ﬁrst alternative are given below. The predictor algorithm is U ∗j − U kj ∂U , ∂t t k F jk − F j−1 ∂F , ∂x x
(11.33)
in which the superscript * refers to variable U computed during the predictor part for the unknown level k + 1. Substitution of these ﬁnite differences into the governing equation in the form ∂U/∂t + ∂ F/∂ x + R = 0 and simpliﬁcation of the resulting equation yield U ∗j = U kj −
t k k − R kj t. F j − F j−1 x
(11.34)
Onedimensional river models
373
In the continuity equation, the computed value of U ∗j corresponds to ﬂow depth h ∗ at the level k + 1, from which we determine the values of crosssection area A∗ , discharge Q ∗ , and velocity U ∗ at the unknown level k + 1. These values are then used to compute F ∗ and R ∗ . The corrector algorithm is implicit in space, as k U ∗∗ ∂U j − Uj = , ∂t t
(11.35)
∗ F j+1 − F j∗ ∂F = . ∂x x
(11.36)
Substituting these ﬁnite differences and R = R ∗j into the governing equation yields k U ∗∗ j =Uj −
t ∗ (F − F j∗ ) − R ∗j t, x j+1
(11.36a)
where the superscript ** refers to the values of the variables after the corrector step. The value of U k+1 at the unknown time level k + 1 is ﬁnally given j by 1 Ujk+1 = (U ∗j + U ∗∗ j ), 2
(11.36b)
from which all other parameters can be determined at the level k + 1. The MacCormack scheme is stable if the CFL condition is satisﬁed.
11.2.4
Figure 11.6. Deﬁnition sketch of the Preissmann scheme.
Preissmann scheme
The Preissmann scheme has been used since the early 1960s (Preissmann and Cunge, 1961; Liggett and Cunge, 1975). It offers the advantage that a variable spatial grid may be used; steep wave fronts may be properly simulated by varying the weighting coefﬁcients θ and ψ, and the scheme yields an exact solution of the linearized form of the governing equations for a particular value of x and t. The partial derivatives are approximated as sketched in Fig. 11.6 with the following
374
Mathematical river models
weighted fourpoint scheme k h k+1 h k+1 − h kj ∂h j+1 − h j+1 j (1 − ψ) +ψ , ∂t t t
(11.37a)
k+1 h kj+1 − h kj h k+1 ∂h j+1 − h j (1 − θ ) + θ , ∂x x x
(11.37b)
in which 0 ≤ θ ≤ 1 and 0 ≤ ψ ≤ 1 are weighting coefﬁcients. In the partial derivatives, h refers to either velocity V or ﬂow depth h. By selection of a suitable value for θ, the scheme may be made totally explicit (θ = 0) or implicit (θ = 1). The scheme is stable if 0.5 ≤ θ ≤ 1. The truncation errors are of the order of t 2 and x 2 when ψ = θ = 0.5. Taking θ = 0.5 gives a scheme that is stable without numerical diffusion. Taking values θ > 0.5 introduces truncation errors that produce numerical diffusion. Steep wave fronts are properly simulated for low values of θ, but there are oscillations behind the wave front. These oscillations are eliminated for θ close to unity; however, steep wave fronts are somewhat smeared. For typical applications, θ = 0.6–0.7 may be used. By substituting these ﬁnitedifference approximations into the St. Venant equation, we have 2N equations (N is the number of reaches on the channel). We cannot write these equations for the grid points at the downstream end. However, we have 2(N + 1) unknowns, i.e., two unknowns for each grid point. Thus, at each time step for a unique solution, we need two more equations. These are provided by two boundary conditions. Boundary conditions at the upstream and the downstream extremities of the routing reach must be speciﬁed in order to obtain solutions to the St. Venant equations. In fact, in most unsteadyﬂow applications, the unsteady disturbance is introduced at one or both of the external boundaries. At the upstream boundary, either a speciﬁed discharge or watersurface elevation time series (hydrograph) can be used. The hydrograph should not be affected by downstream ﬂow conditions. At the downstream boundary, speciﬁed discharge or watersurface elevation time series or a tabular relation between discharge and watersurface elevation (singlevalued rating curve) can be used. Another downstream boundary condition can be a looprating curve based on the Manning equation. The loop is produced with the friction slope S f rather than the channel bottom slope S0 . The Preissmann scheme produces a pentadiagonal banded matrix. The set of nonlinear algebraic equations may be solved by an iterative technique. The solution of 2N × 2N simultaneous equations requires an efﬁcient technique such as the doublesweep elimination procedure. Algorithms for subcritical and supercritical ﬂows with velocity or ﬂow depth as the boundary condition
Onedimensional river models
375
are given in Abbott and Basco (1989). The lengthy procedure is beyond the scope of this chapter.
11.2.5
Aggradation–degradation scheme
The analysis of bedelevation changes in 1D channels involves the combined effects of watersurface calculations and of changes in bed elevation through the sediment continuity relationship. The stability of explicit schemes depends not only on the hydrodynamic conditions, but also on the type of sedimenttransport relationship. To illustrate this point, consider a steady 1D ﬂow of unit discharge q in a rectangular canal. Assume a slight perturbation of the bed elevation at point j, as sketched in Fig. 11.7, and let us Figure 11.7. Sketch of riverbed. calculate the change in bed elevation z by using a backward difference of the sediment continuity relationship: z j+1 = −
TE (qs j+1 − q s j )ts . (1 − p0 ) x
(11.38)
where TE = 1 − exp[(−xω)/q] is the trap efﬁciency, p0 is the porosity of the bed material, ω is the settling velocity, q is the unit discharge, x is the grid spacing, and ts is the time increment for sediment. In practice, the grid spacing is usually sufﬁciently long that TE 1. Also, the porosity is usually assumed constant. The unit sediment discharge qs is the subject of discussion here because h j > h j+1 and V j < V j+1 . Therefore, if the sedimenttransport relationship is proportional to the ﬂow depth, qs ∼ h b , then qs j > qs j+1 and z j+1 > 0 cause aggradation at node j + 1 and the scheme is unstable. Conversely, the scheme is stable (as long as ts is reasonable) if qs ∼ V b because qs j < qs j+1 results in degradation. Algorithms based on shear stress, qs ∼ aτ b = a(h S)b , can be equivalent to the ﬂowdepth algorithm if the slope is taken as a reach average watersurface slope. Opposite results in terms of scheme stability are obtained if the backward difference z j+1 is replaced with a forward difference z j . Also, the case of supercritical ﬂow would also yield different results. The stability of aggradation–degradation algorithms depends not only on the stability of the hydrodynamic scheme but also depends largely on the sedimenttransport relationship. In practice, it is often preferable to calculate z from Eq. (11.38) and
376
Mathematical river models
split z as z j = αz and z j+1 (1 − α)z, with 0 < α < 1 as a weighting factor between forward and backward differences. The time increment for sediment ts is different than the time increment t for hydrodynamic calculations. It is clear from Eq. (11.38) that the aggradation–degradation increment z is linearly proportional to ts . When the change in bed elevation z during ts = t is small compared with the ﬂow depth, the assumption of a rigid boundary calculation for hydrodynamic calculations during t is justiﬁed. The aggradation–degradation calculations can be performed independently, and the equations are said to be uncoupled. The use of ts > t is thus possible as long as z is small compared with h. Conversely, when z is large compared with the ﬂow depth when ts = t, the sediment and the hydrodynamic equations have to be solved simultaneously, and the equations are said to be coupled. Fortunately, bedelevation changes are usually sufﬁciently small during ﬂoods to use uncoupled formulations. An example of aggradation–degradation calculations is shown in Fig. 11.8. The formulation is uncoupled, and the unit discharge is constant throughout the reach. The upstream sediment discharge depends on the local upstream slope, and the upstream bed elevation is assumed to be ﬁxed. The sandtransport capacity is used in the calculations. Over time, aggradation takes place as sediment transport decreases in the downstream direction. The reach becomes increasingly uniform with a new slope that is several times less than the initial reach slope. The sedimenttransport capacity also gradually becomes fairly uniform with slightly decreasing unit sediment discharge in the downstream direction. This numerically illustrates the fact that alluvial rivers gradually tend to become quite uniform in the downstream direction with very gradual changes in depth, velocity, slope, and sedimenttransport characteristics. 11.3
Multidimensional river models
2D depthintegrated models have been applied to predict surface runoff and sedimenttransport rates. A quasisteady approach can often be used, although some 2D unsteadyﬂow models are available. A complete description of multidimensional models is beyond the scope of this chapter. A signiﬁcant number of 2D and 3D codes are commercially available, and some are readily available in the public domain. The fast development of computers make new numerical solutions possible to river engineering problems of increasing complexity. In many cases in which the vertical variation in ﬂow velocity and turbulence are of little interest, vertically averaged horizontal 2D models can be used.
Multidimensional river models
377
Figure 11.8. Example of riverbed aggradation and degradation.
The following assumptions are usually considered in depthintegrated 2D models: (1) the radius of curvature is usually much larger than the channel width and the mildly curved channel approximation is appropriate; (2) the shallowwater approximation is appropriate, and this assumption is particularly valid when the width–depth ratio of meandering or braiding streams is large enough to neglect sidewall effects; (3) a 2D horizontal ﬂow is considered, the vertical velocity component is disregarded, and this type of model cannot account for secondary currents in bends; (4) hydrostatic pressure distribution is assumed;
378
Mathematical river models
(5) the wavelength of the bed deformations is assumed to be longer than the wavelength of ripples, dunes, or antidunes, and bedforms are considered only as roughness elements; (6) the spatial variation in the hydraulic roughness can be neglected; (7) the inﬂuence of grain sorting is insigniﬁcant, and uniform bed material is considered; and (8) the celerity of the bed disturbances is small, the model assumes a rigid boundary, and hydrodynamic calculations are uncoupled with the aggradation–degradation calculations. 2D models usually solve the continuity and the momentum equations after an empirical relationship is assumed for resistance to ﬂow (e.g., Manning or Ch´ezy). The calculations can be performed in a rasterbased or vectorbased format, depending on the GIS database available for the calculations. Finitedifference schemes are well suited to rasterbased data whereas the ﬁniteelement method seems more appropriate for vectorbased data and complex geometries. The calculations are repeated at successive time steps for unsteadyﬂow calculations. Sedimenttransport calculations can effectively yield aggradation and degradation results with uncoupled formulations. Examples of models include CASC2D for surface runoff simulation at the watershed scale (e.g., Julien et al., 1995) and FLO2D for mudﬂow simulation, typically on alluvial fans (e.g., O’Brien et al., 1993). Quasi3D ﬂow models simulate depthaveraged mathematical ﬂow models in combination with the depthintegrated velocity proﬁle. The continuity and the momentum equations are solved with empirical logarithmic velocity proﬁles. The bedload transport rate in the transverse direction is calculated with empirical formulas. The depthaveraged steadystate equation for the suspended load is described in an orthogonal, curvilinear coordinate system. Models can solve the sediment continuity equation for the 2D bed and suspended sediment transport. Once the depthaveraged 2D velocity calculations are completed, standard logarithmic velocity proﬁles are considered to determine the vertical velocity proﬁle. Because the model calculations are in two dimensions, secondary currents in river bends cannot be properly simulated with quasi3D models. 3D models are generally steadystate models used for turbulentﬂow simulations. In κε models, the state of turbulence is characterized by the energy and dissipation parameters κ and ε. 3D models typically solve the depthaveraged Reynolds approximation of the momentum equation for velocity. The depthaveraged mass conservation determines the watersurface elevation. The deviation from the depthaveraged velocity is computed for each cell by the solution of the conservation of mass equation in conjunction with a κε closure for vertical momentum diffusion. Sedimentation computations are based on 2D solid
Multidimensional river models
379
mass conservation for the channel bed and the exchange of sediment between bedload and suspended load. Data required for running multidimensional models include: (1) channel geometry with cross sections; (2) upstream/downstream boundary conditions in terms of discharge and stage as functions of time for unsteadyﬂow models and ﬂowvelocity proﬁles for 3D models; (3) particlesize distribution of the bed material; (4) upstream/downstream sediment load (some models require both bedload and suspended load); and (5) suspended sediment concentration proﬁles for 3D models. The data requirements increase with the number of dimensions in the model. Some model features may require data that are not available, and many times assumptions must be made regarding missing data. For instance, some models can calculate sediment transport by size fractions, and sediment data of the bed material and boundary conditions may not be available for each size fraction. Models may not necessarily handle the data in the most appropriate manner. For instance, some κε should account for the turbulence generated behind bedforms. Lumped values of κ and ε will be assumed in the model even if longitudinal proﬁle data showing bedforms are available. Steady 3D models are applied to estimate the initial rate of sedimentation and erosion in a given situation. The reason for this is the vast computer time required for stabilizing the models under steadystate conditions. To run the model over long time periods under different ﬂow conditions to determine aggradation and degradation would be prohibitive. The initial models provide good insight into the shortterm effects of a proposed structure (channel diversion, new harbor, closure of a channel, etc.). However, they are of limited value for longterm simulation. It is usually preferable to run longterm 1D simulations in parallel to gain basic knowledge of morphological processes and longterm changes to be expected at the site. Case Study 11.1 illustrates how various models can be used to solve complex river engineering problems.
Case Study 11.1 Lower Mississippi River Sediment Study, United States. The diversion of water and sediment from the Mississippi River into the Atchafalaya River has been closely monitored with physical and mathematical model studies (USACE, 1999). Models of the ﬂow and sediment diversion at the Old River Control Complex required extensive ﬁeld and laboratory measurements. Among the data needs at the site, the bathymetry and daily water and sedimentdischarge records over a period of 50 yr served to
380
Mathematical river models
calibrate and test 1D HEC6 models of the Mississippi and Atachafalaya Rivers. A 3D model, CH3DSED, also required more accurate ﬁeld measurements of velocity proﬁles with acoustic Doppler current proﬁlers, and sediment concentration proﬁles by size fractions, measured with a P63 at discharges of ∼1,000,000 ft3 /s.
r
ve
Ri
ss
ssi
Mi
i ipp
<  20 ft below LWRP >  20 ft below LWRP >  30 ft below LWRP
Figure CS.11.1.1. Mississippi riverbed nead the Old River Control Complex.
Multidimensional river models
381
Figure CS.11.1.2. Surfacevelocity calculation near a diversion (after U.S. Army Corps of Engineers, 1999).
Selected results are shown here to illustrate the type of information that can be gained from physical and mathematical models. Figure CS.11.1.1 shows two results from the physical model. The position of the thalweg is extremely important for navigation purposes. Comparing the results illustrates the sediment accumulation and the nonuniform thalweg depths downstream of sharp bends as opposed to the more uniform thalweg depths obtained for gently curved channels. The surface ﬂow velocities calculated with the model CH3DSED are shown in Fig. CS.11.1.2 and, as expected, maximum ﬂow velocities can be found near the concave bank of the diversion channel. The sediment concentration of ﬁne sand in suspension is shown in Fig. CS.11.1.3. The higher sediment concentrations of ﬁne sand on the point bars are the result of secondary ﬂows, as expected.
382
Mathematical river models
Figure CS.11.1.3. Calculated sediment concentration proﬁles of the Mississippi River (after U.S. Army Corps of Engineers, 1999).
Exercise 11.1
Plot the modules of the ampliﬁcation factor A of Eq. (11.14) on Fig. 11.2(b) as a function of 2 < N < 20 for values of Cr = 0.25, 0.5, 0.75, 1.0, 1.25, 1.5, and 2.0. Show that A approaches unity as N → ∞ for all values of Cr . Exercise 11.2
With reference to Example 11.1, repeat the calculations for x = 500 m and x = 250 m for values of Cr = 0.5 and 2.0, respectively. Compare the results with those of Tables E.11.1.1 and E.11.1.2. Plot the results at t = 2,000 s for Cr = 0.5 and x = 2,000, 1,000, 500, and 250 m and show that the results converge. Also, plot the corresponding results for Cr = 2.0 and show that the results diverge as x decreases even if the scheme is consistent.
Problems
383
Exercise 11.3
Reconcile the results obtained in Exercise 11.2 with the fact that A approaches unity as N is large, as obtained in Exercise 11.1. Exercise 11.4
With reference to Subsection 11.2.1, demonstrate that x = 2K d /V (1 − Cu ) and t = 2K d Cu /V 2 (1 − Cu ) from K = K num in Eq. (11.27) and Cu = V t/x. Also, calculate x and t for V = 1 m/s, K d = 50 m2 /s, and Cu = 0.9. Compare the results with those of Subsection 11.2.1. Exercise 11.5 Derive Eq. (11.32b) from a stable Leonard scheme with Cu = 0.8 without oscillations and with a grid Peclet number P = 2. Exercise 11.6 Consider the following time approximation: h k+1 − h˜ kj ∂h 1 j with h˜ kj = α h kj+1 + h kj−1 + (1 − α)h kj , ∂t t 2 where α is a weighting coefﬁcient between a central difference and the nodal value at h kj . Deﬁne the ﬁnitedifference approximation, Eq. (11.6), with a central difference in time. Show that when α = Cr , the scheme reduces to a simple = h kj − c(t/x)(h kj − h kj−1 ). FTBS scheme, or h k+1 j Computer Problem 11.1
Calculate the ﬂoodwave propagation of Example 5.4 at a grid size x = 32.5 km and t = 2 h and use the diffusivewave approximation. Answer: in Table E.5.4.1 Computer Problem 11.2
With reference to Example 11.2, what would be the maximum concentration 10 km downstream of the release if the ﬁrst pulse only lasted 9 min at a concentration of 100,000 mg/l2 . Also, compare the results with and without the second pulse.
384
Mathematical river models Computer Problem 11.3
With reference to Computer Problem 4.1, calculate sediment transport for ds = 0.3 mm over the 25km reach and use the aggradation–degradation algorithm to calculate changes in bed elevation through time. Repeat the calculations and correct for the water surface elevation changes through time. Provide graphical output of sedimenttransport capacity, bed elevation, and hydraulic grade line at three different times. (Hint: The results are somewhat similar to those of Fig. 11.8.)
12 Waves and tides in river estuaries
This chapter relates to river features observed in wide, open areas and in river estuaries. Section 12.1 presents the theory of surface waves with applications to wind waves. Section 12.2 speciﬁcally deals with tides in river estuaries. Section 12.3 presents a brief discussion of saline wedges in river estuaries.
12.1
Surface waves
As sketched in Fig. 12.1, let us consider short gravity waves propagating in a smooth canal of depth h and unit width. The ﬂuid is incompressible and the motion is irrotational. The velocity potential is deﬁned such that the velocity component in the downstream x direction is v x = (−∂/∂ x) and v z = (−∂/∂z) in the vertical z direction. The equation of continuity can be rewritten as ∂ 2 ∂ 2 + = 0. ∂x2 ∂z 2
(12.1)
To obtain a solution to the equation of continuity, the method of separation of variables is considered, and the wavelength λ is assumed to propagate in time t at a celerity c: = f (z) cos
2π (x − ct) . λ
(12.2)
Substituting Eq. (12.2) into Eq. (12.1) yields an equation of the form d2 f (z ) − f (z ) = 0. dz 2
(12.3)
The solution to this equation requires the use of hyperbolic functions. The hyperbolic sine, cosine, and tangent are deﬁned by the three relations sinh z =
e z + e−z sinh z e z − e−z e z − e−z , , cosh z = , tanh z = = z 2 2 cosh z e + e−z 385
(12.4)
386
Waves and tides in river estuaries
Table 12.1. Properties of hyperbolic functions
f (z)
Value at z → 0
Value at z → ∞
Value at z → −∞
sinh z cosh z tanh z
z 1 z
1/2 e z
−1/2 e−z
1/2 e z
1/2 e−z
1
−1
d f (z) dz cosh z sinh z
f (z)dz z cosh z sinh z
Figure 12.1. Surfacewave diagram.
respectively, which should be compared with the relations sin z =
ei z + e−i z sin z ei z − e−i z , cos z = , tan z = , 2i 2 cos z
(12.5)
√ where i = −1 for the circular functions. It follows at once from their deﬁnitions that cosh2 z − sinh2 z = 1. We note that, like their circular cousins, sinh z is an odd function and cosh z is even. In addition we have, by deﬁnition, cosech z = 1/sinh z, sech z = 1/cosh z, and coth z = 1/tanh z. It is clear from the table that sinh z and cosh z are solutions of Eq. (12.3). Other elementary properties of the hyperbolic functions which follow at once are summarized in Table 12.1 and sketched in Fig. 12.2. Therefore the complete solution for the potential function can be written as 2π z 2π 2π z (x − ct) , + B˜ sinh (12.6) cos = A˜ cosh λ λ λ where coefﬁcients A˜ and B˜ and celerity c can be determined from the boundary conditions. The ﬁrst boundary condition is that the vertical velocity is zero, or vz =
∂ = 0 at z = 0. ∂z
(12.7)
Surface waves
387
7 6
~ f (kh)
5
0.19% (both sinh and cosh)
4 3 2
~ e kh ~ f 3 (kh) = 2
cosh sinh ~ ~ f 1 (kh) = kh ~ f 2 (kh) = 1.0
1
0.4%
tanh 0
0
1
2πh ~ kh = λ
2
3
1/20
h/λ 1/2 Intermediate depth waves Shallow water waves Deep water waves (long waves) (short waves)
Figure 12.2. Properties of hyperbolic functons.
Substituting Eq. (12.6) into Eq. (12.7) imposes the condition that B˜ = 0. There are two remaining boundary conditions to deﬁne the celerity c in Subsection 12.1.1 and the coefﬁcient A˜ as a function of wave amplitude in Subsection 12.1.2. The concept of wave energy is presented in Subsection 12.1.3, followed by group velocity in Subsection 12.1.4 and wave power in Subsection 12.1.5. Finally, applications to wind waves are presented in Subsection 12.1.6. 12.1.1
Wave celerity
An expression for the wave celerity can be obtained from the boundary condition at the free surface. The expression emerges from the equation of motion without friction. For 2D ﬂow (v y = 0), the equation of motion in the vertical z direction is 1 ∂p ∂v z ∂v z ∂v z + vx + vz = −g − . ∂t ∂x ∂z ρ ∂z
(12.8)
In the case of irrotational ﬂow, (∂v z /∂ x) = (∂v x /∂z), the following is obtained from the ﬂow potential deﬁnition v z = (−∂/∂z): p ∂ −∂ 1 2 2 + v + v z + gz + = 0. (12.9) ∂z ∂t 2 x ρ Considering that the velocity of the water is small, we can neglect the term in ˜ and the equation of pressure v 2 . The potential that is due to gravity is g (z + η)
388
Waves and tides in river estuaries
becomes p ∂ = − g (z + η) ˜ ρ ∂t
(12.10a)
p 1 ∂ −z− . g ∂t ρg
(12.10b)
or η˜ =
Differentiating the reduced equation of motion with respect to time gives vz =
1 ∂ 2 ∂ η˜ 1 ∂p = . − ∂t g ∂t 2 ρg ∂t
(12.11a)
At the free surface z = h, the atmospheric pressure p is constant and the vertical velocity v z can be obtained as 1 ∂ 2 ∂ =+ . ∂z g ∂t 2  velocity potential   equation of motion  vz = −
(12.11b)
After substituting the potential function from Eq. 12.6 with B˜ = 0, into Eq. (12.11b), we obtain at z = h 2π h 1 2πc 2 2π h 2π sinh =− − cosh λ λ g λ λ (12.12a)
or c2 =
2π h gλ tanh . 2π λ
This is the relationship for wave celerity c, also known as the phase velocity, derived by Airy. There are two particular cases of interest: (1) long waves, also called shallowwater waves, where λ h; and (2) short waves, also called deepwater waves, where λ h. For shallowwater waves, the wavelength λ is very long compared with the ﬂow depth h (λ h): c2
gλ 2π h = gh. 2π λ
(12.12b)
This is the case of tidal waves, in which the velocity of propagation is independent of the wavelength. For deepwater waves, the wavelength λ is very short compared with the ﬂow depth h(λ h): c2
gλ . 2π
(12.12c)
Surface waves
389
This is the case of deep sea waves that are due to the wind. As the celerity is proportional to the square root of the wavelength, a high wind gives rise to longer waves than does a gentle breeze. The wave period T relates to the wavelength λ and the celerity c as λ = cT , and the following identities are obtained directly from Eqs. (12.12): T = λ=
2π h 2π λ coth , g λ
(12.13a)
2π h gT 2 tanh . 2π λ
(12.13b)
The relationship among celerity, period, and ﬂow depth is illustrated in Fig. 12.3. It is sometimes convenient to deﬁne the wave number k˜ = (2π/λ) ˜ = (2π/T ) to deﬁne the surface waves of and the angular frequency σ˜ = kc ˜ − σ˜ t). amplitude a˜ as η˜ = a˜ sin(kx
16 c = g λ tanh 2 π h 2π λ λ= cT
14
12
Celerity c (m/s)
14 12 11
10 9 8
7
Shallow water waves
10
Period T (s) 8 16 1
6
8
5 4
6
Deep water waves 3
4 2
2
0
0
5
10
15 20 Depth h (m)
25
Figure 12.3. Wave celerity as a function of depth and period.
30
390
Waves and tides in river estuaries
12.1.2
Displacement and velocity
The velocity components are obtained directly from the velocity potential equation (12.6) with B˜ = 0: ∂ 2π A˜ 2π z 2π ∂ ξ˜ =− = cosh sin (x − ct), ∂t ∂x λ λ λ ∂ 2π A˜ 2π z 2π ∂ η˜ =− =− sinh cos (x − ct), vz = ∂t ∂z λ λ λ
vx =
(12.14a) (12.14b)
and integrating over time gives the horizontal ξ˜ and vertical η˜ displacements; 2π z 2π A˜ cos (x − ct), ξ˜ = cosh c λ λ
(12.15a)
A˜ 2π z 2π sinh sin (x − ct). c λ λ
(12.15b)
η˜ =
 wave amplitude a˜ 
This relationship deﬁnes the constant A˜ as a function of three wave characteristics: (a) wave amplitude a˜ , (b) wavelength λ, and (c) wave celerity c. From cos2 + sin2 = 1, the path of a ﬂuid element is seen to be an ellipse: ξ˜ 2 η˜ 2 2 + ˜ 2 = 1, ˜ 2π z 2π z A A cosh sinh c λ c λ
(12.15c)
with its major axis horizontal because cosh > sinh. As the hyperbolic sine and cosine approach the same limiting value for large values of the argument, this ellipse becomes a circle near the surface of deepwater waves, as shown in Fig. 12.4.
Figure 12.4. Particle motion in water waves.
Surface waves
391
Figure 12.5. PE deﬁnition sketch.
12.1.3
Wave energy
The total energy of a wave is divided into potentialenergy (PE) and kineticenergy (KE) components generally expressed in terms of average energy over a complete wavelength per unit surface area. The PE that is due to the progressive waveform on the free surface is obtained from subtracting the PE without the wave from the PE with the wave, as sketched in Fig. 12.5(a); the incremental PE per unit width, dPEa, in a small column of water is the height of the center of gravity times the mass increment dm. In a water column h + η˜ high and dx long for water of speciﬁc weight γ , we obtain ˜ 2 /2]γ dx. The average PE dPEa = (height to center of gravity) g dm = [(h + η) per unit surface area is obtained from integration of the incremental PE over one wavelength λ and one wave period T: t+T x+λ γ PEa = (h + η) ˜ 2 dxdt. (12.16) 2λT t x ˜ − σ˜ t) from Eq. (12.15b), we ﬁnd that Eq. (12.16) becomes Using η˜ = a˜ sin(kx t+T x+λ γ ˜ − σ˜ t) + a˜ 2 sin2 (kx ˜ − σ˜ t)] dxdt PEa = [h 2 + 2a˜ sin(kx 2λT t x or simply PEa =
γ a˜ 2 γ h2 + , 2 4
which is the average PE per unit surface area of all the water above z = 0. The PE in the absence of a wave, as sketched in Fig. 12.5(b), is t+T x+λ γ γ h2 . PEb = h 2 dxdt = 2λT t 2 x The average PE per unit area that is attributable to the wave is PEa − PEb =
γ a˜ 2 . 4
(12.17)
392
Waves and tides in river estuaries From Fig. 12.6, the KE per unit width of a small element dx long and dz high with velocity components v x and v z is given by 1 2 v x + v z2 dm 2 1 2 = ρ v x + v z2 dzdx. 2
d(KE) =
Figure 12.6. KE deﬁnition sketch.
The average KE per unit surface area is then given after integration over the ﬂow depth, one wavelength, and one wave period t+T x+λ h+η˜ 2 ρ v x + v z2 dzdxdt. (12.18) KE = 2λT t x 0 Using the velocity components compatible with the progressive wave η˜ = ˜ − σ˜ t) and velocity components given by Eqs. (12.14), we ﬁnd that a˜ sin (kx Eq. (12.18), then becomes t+T x+λ h+η˜ 2 2 ˜ ˜ ρ ˜ 2 cosh 2 kz sin (kx − σ˜ t dzdxdt. k˜ 2 A KE = ˜ cos2 (kx ˜ − σ˜ t) + sinh kz 2λT t x 0 ˜ = 1 [1 + cosh 2 kz]; ˜ By using the following identities: (1) cosh2 kz = kz 2 1 2 ˜ 2 ˜ 2 ˜ ˜ (2) sinh kz = − 2 (1 − cosh2 kz); (3) cos (kx − σ˜ t) − sin (kx − σ˜ t) = cos2 ˜ − σ˜ t) + sin2 (kx ˜ − σ˜ t) = 1, the average KE den˜ − σ˜ t); and (4) cos2 (kx (kx sity becomes ˜ 2 t+T x+λ h+η˜ ρ k˜ 2 A ˜ − cos 2(kx ˜ − σ˜ t)]dzdxdt KE = [cosh 2 kz 4λT t x 0 ˜2 ρ k˜ A ˜ = sinh 2 kh. 8 ˜ = 2 sinh kh ˜ cosh kh; ˜ (2) A˜ = ac/sinh ˜ It reduces further from: (1) sinh2 kh ˜ from Eq. (12.15b); and (3) c2 = (g/k) ˜ tanh kh ˜ from relations (12.12): kh KE =
γ a˜ 2 . 4
(12.19)
The KE of smallamplitude waves [Eq. (12.19)] thus equals the PE [Eq. (12.17)]. The average total energy E˜ per unit surface area is the sum of the average potential of Eq. (12.17) and average KE of Eq. (12.19): γ a˜ 2 . E˜ = PE + KE = 2
(12.20)
Surface waves 12.1.4
393
Group velocity
Several waves can be superposed simply by the addition of the ﬂowpotential functions, which results in adding the wave amplitudes as η˜ = a˜ 1 sin σ˜ 1 t + a˜ 2 sin σ˜ 2 t,
(12.21a)
η˜ = a˜ 1 (sin σ˜ 1 t + sin σ˜ 2 t) + (a˜ 2 − a˜ 1 ) sin σ˜ 2 t.
(12.21b)
In deep water, celerity is a function of wavelength, and two sets of waves of slightly different wavelengths travel at different velocities. This causes local reinforcement and interference as one set gains on the other. The velocity with which the regions of reinforcement or interference advance is known as the group velocity. To ﬁnd this velocity, consider two trains of simple harmonic waves of the same amplitude but of slightly different wavelength and frequency, such as 2π (x − c1 t) , λ1 2π (x − c2 t) . η˜ 2 = a˜ sin λ2
η˜ 1 = a˜ sin
(12.22a) (12.22b)
The following identity is used: sin x + sin y = 2 sin 12 (x + y) cos 12 (x − y). The resultant water elevation η is obtained from Eqs. (12.21) and (12.22): 1 c1 1 c2 η˜ = η˜ 1 + η˜ 2 = 2 a˜ cos π − − x− t λ1 λ2 λ1 λ2 (12.23) c1 1 c2 1 x− t . + + sin π λ1 λ2 λ1 λ2 As the wavelengths and the velocities are only sightly different, we may replace λ1 and λ2 with the mean wavelength λ and c1 and c2 with the mean celerity c in the second trigonometrical function in braces: c 2π 1 (x − ct) . (12.24) η˜ 2a˜ cos π d x −d t sin λ λ λ  group envelope  This expression represents a wave traveling with phase velocity c with an amplitude of 2a˜ and a group envelope in braces. At a given instant the amplitude plotted against x is represented by a cosine curve, the distance from one region of maximum amplitude to the next being the reciprocal d (1/λ). Moreover, this region of maximum amplitude is progressing in the positive x direction with
394
Waves and tides in river estuaries
η
cG c
c x
Figure 12.7. Groupvelocity sketch.
the group velocity cG : c λdc − cdλ d λ λ2 cG = = 1 − dλ2 d λ λ dc = c−λ . dλ
(12.25)
Therefore we have a wave within a wave, as represented in Fig. 12.7. The fullline wave curve advances with the celerity c whereas the dashed curve advances with the group velocity cG . If c increases with λ, the phase wave travels more rapidly than the group wave. In this case the fullline waves pass through the group from rear to front, each crest disappearing at the front of the group while a new crest appears at the rear. The group velocity obtained from substituting the wave celerity from relations (12.12) into Eq. (12.25) is 2˜kh c , (12.26) 1+ cG = ˜ 2 sinh 2kh ˜ is large (deepwater waves) and which asymptotically reduces to c/2 when kh ˜ reduces to c when kh is very small (shallowwater waves). The group velocity for shallowwater waves is identical to the wave celerity because the wave celerity is independent of wavelength. 12.1.5
Wave power
The increment of power in a frictionless ﬂuid is obtained by the product of the elementary force ax dm and velocity in the same direction v x = −∂/∂ x. Considering that the channel banks and bed do not move and that the pressure is atmospheric at the free surface, the increment of power can then be integrated over the ﬂow depth during one period T to give the total power P˜ of the wave per unit surface area: t+T h+η˜ ∂ −∂ 1 dzdt. (12.27a) P˜ = ρ T ∂t ∂x t 0 Substituting the ﬂowpotential function from Eq. (12.6) into Eq. (12.27a) gives γ a˜ 2 cG . P˜ = 2
(12.27b)
Surface waves
395
It is concluded that the energy of a wave previously calculated in Eq. (12.20) propagates at a velocity corresponding to the group velocity.
12.1.6
Wind waves
The minimum wind speed to generate gravity waves is approximately 6 m/s. The characteristics of wind waves include a full spectrum of wavelengths and amplitudes. The signiﬁcative wave height Hs represents the average height of the large waves in the upper third of the spectrum. The wave height Hs is ˜ The average wavelength of these waves given as twice the wave amplitude a. determines the signiﬁcative wavelength λs . Under a constant wind speed Uw , the wave height increases with the downwind distance, called the fetch length ˜ and with time until equilibrium is obtained. In deep water, either the wind F, duration or the fetch length can limit the growth of waves. In shallow water, the ﬂow depth h can also limit the growth of wind waves. ˜ The In rivers, the wind waves are usually limited by the fetch length F. ﬂow depth may also limit the wave height during large storms when the fetch length exceeds ∼3–5 km. The signiﬁcative wave height Hs can be estimated as a function of wind speed Uw , ﬂow depth h, and fetch length F˜ from Fig. 12.8. The development of wind waves under a constant wind speed Uw over a fetch length F˜ is sketched in Fig. 12.8. To determine the wave height H˜ s , it can be assumed that the work done by the shear stress τ ∼ ρUw2 over the fetch length F˜ corresponds to the energy of the wave E˜ ∼ ρgHs2 from Eq. (12.20). After the work done and the energy are equated, the wave height H˜ s is related to the fetch length F˜ as ˜ (12.28) H˜ s 0.003 Uw F/g. We can estimate the windstorm duration t˜s required for reaching equilibrium by considering the ratio of the fetch length F˜ to the group velocity cG . In √ ˜ gh. shallow water, the storm duration can be approximated by t˜s = F/ ˜ In Waves are fetch limited as long as H˜ < 0.3 h, or Uw < 100 gh 2 / F. this case, the signiﬁcative wave height is approximated by Relation 12.28 and The the wavelength is λ 15 H˜ s . The period is obtained from Eq. (12.13a). ˜ ﬂow depth limits the growth of surface waves when h < 0.01 Uw F/g. In the case of depthlimited waves, the wave height is limited to approximately 30% to 40% of the ﬂow depth, or H˜ s 0.35 h. Fetchlimited waves are likely to cause erosion of river banks whereas depthlimited waves will likely also cause resuspension of the bed material.
396
Waves and tides in river estuaries (a)
Uw
~ Hs ~ F
(b) Flow depth h = 15 ft (4.6 m)
Wind speed Uw (mph)
100 80
1
2
3
4
8 10
6
20
30
6 5
60
50 km m/s 30
4 3
40
20
2 1.5 1.0 H~s ( ft) 0.5
30 20
10 103
10 8 6 10 miles
4
1 mile
10 ~ Fetch length F (ft)
5
10
(c) Flow depth h = 30 ft (9.2 m)
Wind speed Uw (mph)
100 80
1
2
3
4
6
8 10
20
60 4
40
5
6
7
89
30 10
50 km m/s 30 20
3
2 1.5
30 20
10 8
1.0
H~s ( ft) 0.5
10 103
1 mile
Note: 100 mph = 44.7 m/s
6 4
10 ~ Fetch length F (ft) 1.0 m = 3.28 ft
10 miles
5
10
Figure 12.8. Amplitude of wind waves.
12.2
Tides in river estuaries
Tides are complex oscillatory waves in large water masses. The most obvious tidal period corresponds to approximately a half daily cycle, more speciﬁcally 12 h 25 min. The tidal range and time of high water not only vary from place to place but also throughout the month and the year. Tidal oscillations result from massive gravitational and centrifugal force balances among Earth, Moon, and Sun.
Tides in river estuaries Earth
397
2
Considering ﬁrst the Earth alone, we ﬁnd that the gravitational attraction is Ω proportional to the universal gravitation gE FgE = 6.673 × 10−11 Nm2 /kg2 . constant G u m φ At the surface of the Earth of mass φ FcE ME RE m E = 5.976 × 1024 kg and radius R E = RE = 6375 km Fc 6,371 km, the gravitational acceleration is g = G u m E /R 2E = 9.8 m/s2 . The cenTE = 2 π = 24 h ωE trifugal acceleration is proportional to the ω E = 7.3 x 10 5 rad/s square of the angular velocity multiplied by the radius of rotation. As sketched Figure 12.9. Earth rotation sketch. in Fig. 12.9, the centrifugal acceleration that is due to the Earth’s rotation is acE = ω2E R E cos φ = 0.034 m/s2 cos φ, where ω E = 7.272 × 10−5 rad/s is the angular velocity of the Earth and φ is the latitude of the point considered. Suppose the Earth is entirely covered with water. A unit mass of water on the surface is simultaneously subjected to gravitational and centrifugal accelerations. At the poles, there is no local centrifugal action, whereas at the equator the centrifugal acceleration is of the order of 0.034 m/s2 , or 0.3% of the gravitational acceleration. This combined action varies with latitude and draws the water toward the equator. Although there is in fact a preferential accumulation of water near the equator, the daily rotation of the Earth does not cause tides but contributes to tides through Coriolis acceleration. The presentation of the Coriolis acceleration (Subsection 12.2.1) is followed by tidal acceleration (Subsection 12.2.2) and tide amplitude and propagation (Subsection 12.2.3). 12.2.1
FgE = GmME /RE FcE = mωE2 RE cos φ FcE ~~ 0.003 F
Coriolis acceleration
Moving water at the surface of the Earth is subjected to acceleration, known as the Coriolis acceleration, that is due to the rotation of the Earth. The Coriolis acceleration is a function of the angular velocity vector of the Earth’s rotation and the ﬂowvelocity vector V of moving water in a frame of reference at the center of mass of the Earth and rotating with the Earth (x to the East, y North, and z vertical up). The Coriolis acceleration that is due to the Earth’s rotation = yˆ ω E cos φ + zˆ ω E sin φ that is applied at the center of mass of the Earth and is due to the velocity vector V with v x to the East, v y to the North, and v z vertical
398
Waves and tides in river estuaries
up, is then deﬁned as acor = −2 × V = 2 ω E [xˆ (v y sin φ − v z cos φ) − yˆ v x sin φ + zˆ v x cos φ].
(12.29)
For horizontal ﬂow at the surface of the Earth (v z = 0), the Coriolis acceleration acor depends on the latitude angle φ. Fluid motion is deﬂected to the right in the northern hemisphere and to the left in the southern hemisphere. Given the angular velocity of the Earth, ω E = 7.3 × 10−5 rad/s and the latitude angle φ, the magnitude of the Coriolis acceleration is acor  2 ω E sin φV.
(12.30)
The ratio of the Coriolis acceleration to the downstream gravitational acceleration is deﬁned as 2 sin φω E V /(gS). Given ω E = 7.3 × 10−5 rad/s, this ratio is small (ratio 2 m/s. Example 12.1 illustrates one case in which the Coriolis acceleration is not negligible. The Coriolis acceleration can also be compared with the centrifugal acceleration in a river bend of radius R1 as 2 sin φ ω E R1 /V . This ratio is usually negligible in all but fairly straight large rivers. Although negligible in rivers unless the surface slope is very small, the Coriolis force plays a dominant role in large water masses and in the atmosphere, in which it generates geostrophic winds.
Example 12.1. Calculation of the Coriolis acceleration. Consider the lower Mississippi River at a latitude φ = 30◦ in a river bend at a radius of curvature R1 = 4 miles (6.4 km). The watersurface slope of the river is Sw ≈ 6 × 10−5 is applied on and the water moves horizontally to the south at 10 ft/s (3 m/s). The westward Coriolis acceleration due to the rotation of the Earth is acor  = 2 ω E sin φV = 2 × 7.3 × 10−5 × sin 30◦ × 10 = 7.3 × 10−4 ft/s2 (2.22 × 10−4 m/s2 ). The centripetal acceleration in the river bend is calculated from acent = V 2 /R1 = 100 ft2 /s2 /4 × 5,280 ft = 4.7 × 10−3 ft/s2 (1.43 × 10−3 m /s2 ). The Coriolis acceleration is ∼15% of the centripetal acceleration in this river bend. In comparison, the downstream gravitational acceleration component is gS = 32.2 ft/s2 × 6 × 10−5 = 1.9 × 10−3 ft/s2 (5.8 × 10−4 m/s2 ). The Coriolis acceleration is a signiﬁcant fraction of the downstream gravitational acceleration. Indeed, the Coriolis acceleration is ∼38% of the downstream gravitational acceleration and should not be neglected in this case.
Tides in river estuaries
399
Field velocity measurements of the Mississippi River that use the acoustic Doppler current proﬁlers at the Old River Control Complex show that the random turbulentvelocity ﬂuctuations mask the effects of the Coriolis acceleration. Winkley (1989) nevertheless concludes that the Coriolis force exerts an inﬂuence on the geometry of the Lower Mississippi River. The Coriolis effects become even more signiﬁcant for oscillations of longer periods, such as tides.
12.2.2
Tidal accelerations
The Earth and the Moon may be considered as a single system, consisting of two bodies having a common center both of mass and rotation C, as sketched in Fig. 12.10. The common center of mass is just inside the surface of the Earth – as the Earth diameter and mass are respectively approximately 3.7 and 81 times those of the Moon and their distance apart is ∼30 Earth diameters. In a frame of reference rotating in space about C with a period of ∼27.3 days, the water is subject to the gravitational accelerations of the Earth and the Moon and the centripetal acceleration. The centrifugal and gravitational accelerations exerted by the Moon are equal and opposite at the center of mass of the Earth. Locally at the Earth surface, the gravitational acceleration attracts water toward the Moon at point A while the centrifugal acceleration about point C forces the water away from the Moon toward point B. Their resultant force causes the water surface to form a spheroid that is oblate along the Earth–Moon axis. The above argument can be applied Earth
Moon
TE = 24 h 5 ωE = 7.3 x 10 rad/s ωC = 2π/TC
ω s = 2π/Ts Ts = 365 d FcC
TC = 27.3 d FgM
FgE
B
A
C
6375 km 4670 km mE = 5.98 x 10
24
kg
384 410 km mM = 7.34 x 10 22 kg
Figure 12.10. Earth–Moon inﬂuence on tides.
400
Waves and tides in river estuaries
to the effect of the Sun’s giving differential gravitation effects that are 2.17 times smaller than those for the Moon. Consequently, maximal tides should be obtained during new and full Moons. Also, the gravitational and the centripetal accelerations at the poles do not change during the day. The largest variations are obtained at the equator with, for instance, the gravitational force of the Sun near the east at sunrise and near the west at sunset. In simpliﬁed form, the differential in latitudinal accelerations on a daily basis generates ﬂow velocities that are proportional to the cosine of the latitude. The maximum differential is obtained when the Sun is in the equatorial plane, which should increase the magnitude of fall and spring tides at the times of the equinoxes. The magnitude of the Coriolis acceleration from relation (12.30) is therefore proportional to sin φe cos φe . Accordingly, tides should be small near the poles and the equator and should be maximum at a 45◦ latitude. The exact analysis of tides is further complicated by the facts that: (1) the axis of rotation of the Earth is at 66◦ 33 minutes from the plane of Earth’s rotation around the Sun; and (2) the plane of the Moon is also inclined at 5◦ 9 minute from the plane of the Earth’s rotation around the Sun. Highwater levels do not usually coincide with the maximum lunar attraction at a given locality. Displacements in harmonic systems are out of phase with accelerations. Tidal currents are not necessarily in phase with the displacements because of the ﬁnite wavepropagation speed and its dependence on depth. For reasons of boundary shape and friction effects, local systems rotate around amphidromic points where no tide elevation exists. An example is shown in Fig. 12.11.
12.2.3
Tide amplitude and propagation
The propagation of tides in estuaries with variable widths is calculated on the basis that the effects of friction and wave reﬂections remain negligibly small. Thus the law of conservation of energy provides a convenient way of expressing the effect of changes in width and depth on tide amplitude. The total energy for the entire surface of the wave from Eg. (12.20) is 1 1 W0 λ0 E˜ = W0 λ0 γ a˜ 02 = Wx λx γ a˜ x2 , 2 2
(12.31)
where W is the width of the estuary, λ is the tidal wave length and a˜ is the tide amplitude. The subscript 0 refers to the initial or reference position and the subscript x to any location. Because the celerity c of shallowwater waves √ c = gh x and because the period T = λ/c is independent of any deformation
Saline wedges in river estuaries
401
SeptIles
6
50° 5
4
7 8 9
3
10 11
Rimouski
12 13 14 1
ft
Quebec ´
2
4
Grondines
36
44 40
32 28
20
24
16
45°
Tide amplitude in feet (1 m = 3.28 ft)
10 ft
8
70°
6
5
12
65°
60°
Figure 12.11. Tide amplitude of the St. Lawrence estuary.
√ of the wave, the tide amplitude a˜ x can be determined after λx = T gh x is substituted into Eq. (12.31) to obtain W0 1/2 h 0 1/4 a˜ x = . (12.32) a˜ 0 Wx hx This is known as Green’s law, which assumes conservation of energy for shallowwater waves. In the case of intermediateor deepwater waves other than tidal waves, the more general relationship from conservation of power [Eq. 12.27(b)] is recommended. The case of tide propagation with friction is best analyzed with numerical models. From Fig. 12.11 we can examine the tidal amplitude changes as it propagates up the St. Lawrence estuary. It is clear that the increase in amplitude is caused by the reduction in channel width and ﬂow depth. 12.3
Saline wedges in river estuaries
As rivers enter seas and oceans, the density difference between clear freshwater and saltwater often causes density stratiﬁcation. A saline wedge refers to the saline layer underlying freshwater in a river communicating with a tideless sea.
402
Waves and tides in river estuaries
The still form is called an arrested saline wedge, and the length and the shape of arrested saline wedges are described below. Let ρsea be the density of saltwater (usually ρsea 1,035 kg/m3 ), and let ρm be the mass density of the river freshwater, including suspended sediment. Then a densimetric velocity V may be deﬁned as ρ gh, (12.33) V = ρm where ρ = ρsea − ρm , g is the gravitational acceleration, and h is the river depth. The physical signiﬁcance of V is linked to the propagation velocity of internal waves of large wavelength. The initial velocity of saline fronts moving in still water is also proportional to V . A saline wedge extends laterally as long as V > V1 . As sketched in Fig. 12.12, we can calculate the height of the wedge at the river mouth h s1 by deﬁning the position where the local velocity V1 equals the local densimetric velocity V1 . The densimetric velocity is applied at the river mouth: 2 = V12 = V1
ρ g (h − h s1 ) . ρm
(12.34)
If mixing at the interface is small, the continuity condition for freshwater in a river of constant width gives V1 (h − h s1 ) = V h.
(12.35)
Eliminating V1 between expressions (12.34) and (12.35) and solving for h s1 / h gives 2/3 V h s1 =1− . (12.36) h V The length L of an arrested saline wedge has been studied experimentally and the following empirical relationship is suggested: V h 1/4 V 2.5 L 6.0 . (12.37) h ν V The length of an arrested saline wedge depends primarily on the river depth h and the density difference ρ. The shape of the arrested saline wedge sketched in Fig. 12.12 is deﬁned from the height of the saline wedge h s , the height of the wedge at the river mouth h s1 , and the distance L in terms of the wedge length L . The shape of the arrested saline wedge is practically independent of seawater salinity, river
Saline wedges in river estuaries
403
Figure 12.12. Arrested saline wedges.
velocity, water depth, channel width, and viscosity, as shown in Fig. 12.12(b). Case Study 12.1 provides an example of interaction between the saline wedge and tides in river estuaries.
Case Study 12.1 Salmon capture in the Matamek River estuary, Canada. The Matamek River is a small tributary of the St. Lawrence River estuary near SeptIles. When leaving rivers to enter the sea, juvenile Atlantic salmon (salmo salar), called smolts, spend 20–30 days in the saline wedge of river estuaries to gradually adapt to saltwater. To study the migration pattern in the Matamek River, the capture–recapture method was used to estimate the population of smolts in the estuary during the migration period in June and early July. Captures were made during high tides at a ﬁxed location (point A
Waves and tides in river estuaries
Water level (m)
Average salmon capture
404
1975
1976
3 2 1
3 2 1 10
June
20
July
10
10
20 June
10 July
Figure CS.12.1.1. Salmon capture and tides of the Matamek River.
in Fig. 4.2) in the estuary for a period of 2 yrs. The daily average number of smolts captured per seine haul is shown in Fig. CS.12.1.1. The double peak in ﬁsh migration intrigued scientists. At the time, arguments were supporting two different ﬁsh population’s migrating at slightly different times. Scientiﬁc studies were pursued to determine whether there could be genetic differences in the ﬁsh population. When the moon cycle in Fig. CS.12.1.1 is plotted against the salmon capture chart, the peaks correspond to halfmoon periods. Could the Moon also exert inﬂuence on ﬁsh capture? When the water levels in the estuary were plotted as a function of time, Fig. CS.12.1.1 shows that the captures were low when tides were very high. The saline wedge is located at the capture site during halfmoons when the tide is moderately high. The saline wedge was located upstream during higher tides and downstream for lower tides, as sketched in Fig. 12.12(c). It became clear that the number of ﬁsh captured depended primarily on whether or not the seine was hauled in the saline wedge or not. A simple regression equation based on the high and the lowwater levels was set up to determine ahead of time whether or not ﬁshing would be good for any given day. So much for the inﬂuence of the Moon on ﬁsh behavior and for the two salmon populations in the Matamek River. Case Study 12.2 Tidal data for the St. Lawrence River, Canada. This case study presents tidepropagation data in the St. Lawrence River. The data presented in Fig. CS.12.2.1 and Table CS.12.2.1 illustrate tidal wave
Saline wedges in river estuaries 3
405
March 8, 2000 Sept les
2 1 0 5
Rimouski
4 3 2 1 0 6
Qu bec
Stage (m)
5 4 3 2 1 0 4
Grondines
3 2 1 0 2
TroisRivi res
1 0
0
2
4
6
8 10 12 14 16 March 8, 2000, Time (h)
18
20
22
24
Figure CS.12.2.1. Tide propagation in the estuary of the St. Lawrence River.
propagation in river estuaries. The database serves for speciﬁc calculations in Problem 12.5.
Exercise 12.1
Satisfy Eq. (12.6) with the boundary condition that the vertical velocity is zero at the bottom of the canal [Eq. (12.7)] to show that B˜ = 0.
406
Waves and tides in river estuaries
Table CS.12.2.2. Tides of the St. Lawrence River SeptIles
Rimouski
Quebec
Grondines
Date
m
Time
m
Time
m
Time
m
Time
March 2000 4
2.1 0.5 2.7 0.4
0h44 6h31 12h57 19h23
3.2 1 3.8 0.7
1h17 7h09 13h26 19h53
4.1 0.5 4.8 0.3
5h23 12h15 17h29 0h52
2 0.5 2.6 0.4
8h06 15h03 19h58 4h16
5
2.2 0.5 2.8 0.3
1h19 7h11 13h32 19h53
3.4 0.8 4 0.6
1h50 7h45 13h59 20h21
4.4 0.4 5 0.2
5h59 12h58 18h07 1h30
2.2 0.5 2.8 0.5
8h32 16h03 20h34 4h57
6 New Moon
2.4 0.2 2.8 0.2
1h52 7h49 14h06 20h19
3.6 0.6 4.1 0.5
2h22 8h19 14h32 20h43
4.7 0.3 4.8 0.2
6h34 13h39 18h45 2h07
2.4 0.6 2.5 0.6
8h59 16h57 21h14 5h36
7
2.5 0.1 2.8 0.1
2h25 8h26 14h39 20h53
3.8 0.5 4.2 0.4
2h54 8h54 15h05 21h20
4.9 0.2 5.3 0.2
7h09 14h19 19h24 2h43
2.7 0.7 3 0.7
9h29 17h47 21h45 6h14
8
2.6 0.1 2.7 0.1
2h59 9h05 15h14 21h25
3.9 0.5 4.1 0.4
3h28 9h30 15h41 21h53
5.1 0.2 5.3 0.2
7h45 14h59 20h03 3h19
2.9 0.8 3 0.8
10h04 18h33 22h23 6h51
9
2.7 0.1 2.6 0.1
3h35 9h45 15h51 21h59
4 0.5 4 0.4
4h04 10h09 16h20 22h28
5.2 0.1 5.1 0.2
8h21 15h39 20h44 3h56
3.1 0.8 3 0.8
10h42 19h18 23h05 7h27
10
2.7 0.2 2.4 0.2
4h14 10h29 16h31 22h37
4 0.6 3.8 0.6
4h43 10h53 17h03 23h07
5.2 0.2 4.9 0.3
9h00 16h21 21h28 4h33
3.1 0.9 2.8 0.9
11h25 20h03 23h51 8h04
11
2.7 0.4
4h57 11h20
3.9 0.7
5h27 11h42
5.1 0.3
9h42 17h05
3.1 0.9
12h12 20h50
Exercise 12.2
Substitute Eq. (12.6) into Eq. (12.11b) to obtain wave celerity relationships (12.12).
Exercise 12.3 From the deﬁnition of velocity in Eqs. (12.14), demonstrate that the convectiveacceleration terms in Eq. (12.9) are small compared with the local acceleration
Problems
407
for waves of small amplitude (˜a λ). [Hint: Examine the ratio (v x2 + v z2 )/ ∂/∂t.] Exercise 12.4 Demonstrate the relationship for the group velocity in Eq. (12.26) from substituting wave celerity relationships (12.12) into Eq. (12.25). Also show that cG = c/2 for deepwater waves. Exercise 12.5
Demonstrate that the energy of gravity waves propagates at the group velocity by substituting Eq. (12.6) into Eq. (12.27a). [Hint: Use the identities shown in Eq. (12.19).] Exercise 12.6
Calculate the length of the saline wedge for the clearwater of the Matamek River. The ﬂow velocity is 1 m/s and the ﬂow depth is ∼4 m. Answer: L = 1,650 m, V = 1.17 m/s, and h S1 = 0.4 m, with ρsea = 1,035 kg/m3 . Problem 12.1
Estimate the wave height, wavelength, and waveperiod in a river that is 10 m deep with at fetch length of 10 km. Consider a hurricane with a wind speed of 75–90 mph (∼40 m/s). Answer: H˜ s 3.5 m, λs 50 m, and the period T 6 s. The celerity c = λs /T = 8.3 m/s, and the waves are depth limited. Problem 12.2
Estimate the wave height, wavelength, and waveperiod in a river that is 4 m deep, with a fetch length of 1 km. Consider a storm with a wind speed of 65–75 mph (∼30 m/s). Problem 12.3
Compare the magnitude of the Coriolis acceleration in the Rhine River with the downstream gravitational acceleration and also with the centrifugal acceleration
408
Waves and tides in river estuaries
in a 2kmradius river bend near Nijmegen in The Netherlands. (Hint: Consider the ﬁeld measurements h = 10 m, V 2 m/s, and S 1 × 10−4 ). Answer: acor = 2×10−4 m/s2 , acor /gS 22%, and acor /acent 10%. Problem 12.4
Consider a branch of the Mississippi River delta with a discharge of 270,000 ft3 /s (7,640 m3 /s), a branch width of 1,500 ft (460 m), and a depth of 45 ft (13.7 m). Estimate the densimetric velocity V and the length L of the salinity wedge. Answer: Assume that ρ/ρm = 0.035, V = 4.0 ft/s = 1.2 m/s, and V = 7.11 ft/s = 2.17 m/s. The length L 16.5 miles (26 km). Problem 12.5
With reference to the tide propagation data in Case Study 12.2, examine the following characteristics: (a) Estimate the ﬂoodwave celerity from the shallowwater equation and compare with ﬁeld observations. (Hint: Assume a ﬂow depth of ∼10 m near Grondines, ∼50 m near Qu´ebec, up to 100 m near Rimouski, and up to 200 m near SeptIles.) (b) Compare the shape of tidal waves between SeptIles and TroisRivi`eres and discuss the effects of bed roughness on tides. (c) From Table CS.12.2.1, examine the days when the highest tides occur at various locations. Discuss the observations. Computer Problem 12.1
From the data in Case Study 12.2, calculate at 30min intervals the ﬂow velocity at Qu´ebec on March 8, 2000. Assume a simple prismatic approximation for the river channel 120 km long with a width of 2 km, depth 15 m, constant inﬂow discharge of 10,000 m3 /s, and constant stage near TroisRivi`eres. Why is there ﬂow reversal at Qu´ebec?
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Index
2D model, 378 3D model, 378 452ton boulder, 221 500yr ﬂood, 220 Aframe, 329 abutment, 271 abutment scour, 312–313, 315 advectiondispersion, 365, 370 adverse slopes, 108 aerial photograph, 187, 226 aggradation, 114, 199, 215, 219 aggradationdegradation scheme, 375 agricultural levee, 290 alluvial channel, 158, 171 alluvial fan, 219, 221 alluvial river bar, 177 alluvial system, 199, 203 alternate bar, 177–178 alternate depth, 103 ampliﬁcation factor, 358 anabranching, 216 anastomosed river, 215, 216, 218 anchor, 265 angle of repose, 18, 161 angular frequency, 389 annual erosion loss, 74 annual migration rate, 187 apex, 178, 180 aquatic habitat, 322 armor layer, 209, 210 arrested saline wedge, 402 artesian pressure, 238 articulated concrete mattress, 260 artiﬁcial cutoff, 319 atastation hydraulic geometry, 81, 87 Atchafalaya River, 86, 222, 295, 379 Atlantic salmon, 403 avulsion, 216, 219 backward ﬁnitedifference, 355, 361 backwater proﬁle, 107, 110
427
bafﬂe, 324 bank caving, 170 bank failure, 235 bankerosion process, 234 bank stabilization, 282 bankfull discharge, 171, 183, 199, 224 barge type, 317 basket, 259 bedform characteristics, 94 bedload percentage map, 193 bedload sediment transport, 113, 192–193, 218 bedmaterial discharge, 199 bedrock outcrop, 212 bed shear stress, 27 Belanger equation, 101 benching, 237 bend apex, 191 bendway weir, 277–279 Big Thompson ﬂood, 42, 44 blanket, 252 block, 257, 258 borrow area, 290 bottom vane, 308 boundary condition, 362 box gabion, 259 brackish water, 403 braided river, 214–218, 223 braiding and meandering, 216–217 branches, 221 bridge inspection program, 350 bridge scour, 310 broadcrested weir, 282, 302 bucket dredge, 325, 327 bulkhead, 264 buoyancy force, 159 burlap sack, 257 buttress, 237, 264 caisson, 263–264 calibration phase, 353 canal headwork, 306, 307–308
428
Index
cantilever wall, 262, 264 canyons, 207 capture, 216 capture–recapture, 403 celerity, 130–131, 389 cellular cofferdam, 303–304 centerofmass curve, 193 central ﬁnitedifference, 355, 361 centrifugal acceleration, 23, 128, 166, 398 channel conveyance, 288 channel degradation, 204 channel improvement, 288, 294, 316 channel incision, 205, 207 channel intake, 306 channel morphology, 199 channel shift, 224 channel stability, 163 channel widening, 222 Chaudi`ere watershed, 76 Ch´ezy coefﬁcient, 54, 89, 91 choking, 298 chute cutoff, 178, 330 circular cell, 305 clay plug, 190 climate, 34 cofferdam, 303–304 cohesive bank, 235 Colorado River, 151 commercial navigation, 317 compacted levee, 290 complete closure, 297 complete mobilebed similitude, 345 completeequilibrium hydrograph, 58–59, 60 concave bank, 180 concentration proﬁles, 154 concrete wall, 263 conﬁned aquifer, 238 conﬂuence, 225 conjugate depth, 100–101 conservation of mass, 24, 122 conservation practice, 70–71 consistency, 356–357, 363, 366 contaminant, 329, 365 continuity, 122 contraction scour, 311–312 control volume, 122, 125 convection instabilities, 368 convective acceleration, 23 convective rainstorm, 37, 39 convergence, 356–357, 363 conversion of units, 12 convex bank, 180 Coriolis acceleration, 23, 397–398, 400 corrector algorithm, 373 Courant number, 140–141, 356 Courant–Friedrich–Levy (CFL), 140, 358 coveredhopper barge, 317
crib wall, 263–264 critical ﬂow depth, 100, 103 critical shear stress, 18, 163 cropland, 71 croppingmanagement, 68, 69–70 crossing, 178, 180 crossover culvert, 324 crosssection geometry change, 195 cross wave, 134 culvert scour, 291, 301 cumulative distribution function, 40 cumulative inﬁltration, 48, 53 cumulative time of snowmelt, 61 cut and ﬁll, 237 cutoff, 316, 318–319 cutoff trench, 290 cutoff wall, 252 cutterhead dredge, 328 dam, 323 dam break, 220 Darcy–Weisbach friction factor, 54, 89, 91 database, 228 deadman, 266 debris, 283 decreasing discharge, 195 deepwater wave, 388–389, 401 degradation, 114, 199, 207, 210, 219 degree days, 61 delta, 220 densimetric velocity, 402 density stratiﬁcation, 401 depositional zone, 199 depthintegrated 2D model, 377 depth of ponding, 48 design ﬂood elevation, 305 detention basin, 286 detention storage, 46, 53 diaphragm cell, 304–305 differential equation types, 353 diffusive wave, 129, 134–136, 138, 141 dike, 272, 274 dipper dredge, 325, 327 Dirichlet type of boundary condition, 362 discharge duration, 206 discharge measurement, 84 dispersion number, 371 distorted model, 335, 348 divergence theorem, 122 diversion tunnel, 297–299 dominant discharge, 171, 203, 206 double levee system, 289, 322 double mass curve, 229 downbound tow, 318 downstream boundary condition, 362 downstream ﬁning, 212, 215
Index downstream hydraulic geometry 158, 166, 172–175, 202 downwind, 356 drain, 267 drain well, 238, 239 drainage network, 79 drainage trench, 238 dredged material, 329 dredging, 325 drift angle, 318 drop structure, 280, 281 Dry Creek, 212 dry speciﬁc weight, 21 dumped riprap, 247 duration curve, 142–143, 147, 151 dustpan dredge, 327–328 dynamic response, 199, 203 dynamic similitude, 336 dynamic variables, 10 dynamic viscosity, 11 dynamic wave, 135, 138 earth cofferdam, 297 earth plug, 319 effective riprap size, 241 effective saturation, 48 elliptic equations, 362 emptying, 323 enddumping, 298 endangered species, 228, 322 energy correction factor, 97, 103 energy dissipation, 280 energy grade line (EGL), 29, 110–111 equation of motion, 25, 27, 128 equilibrium in river bends, 166 equivalent concentrations, 20 erosion coefﬁcient, 73 erosion index, 187, 189 erosional zone, 199 Euler’s relation, 358 evapotranspiration, 46 exact geometric similitude, 335, 337 exceedance probability, 40, 42, 143, 147, 149, 151 excess rainfall, 51–53 expected soil loss, 65–66 explicit ﬁnite difference, 356, 365 external forces, 25 extremal hypotheses, 179 extreme rainfall, 149 extreme speciﬁc discharge, 149 Fall River, 190 fan apex, 221 fence, 276 fetch length, 395
429 ﬁeld survey, 228 ﬁlling, 323 ﬁlter, 247 ﬁlter design, 250 ﬁlter fabric, 264 ﬁnitedifference approximation, 353–354 ﬁsh population, 404 ﬂash ﬂood, 45 ﬂexibility, 283 ﬂoating debris, 276 ﬂoodcarrying capacity, 294 ﬂood control, 286, 289 ﬂood damage, 289 ﬂoodfrequency analysis, 227 ﬂoodgate, 291 ﬂoodplain, 171, 288 ﬂood protection, 288, 293 ﬂood routing, 139, 141 ﬂoodwall, 289 ﬂoodwave ampliﬁcation, 135 ﬂoodwave attenuation, 135, 137, 286–287, 289 ﬂoodwave celerity, 131, 140 ﬂoodwave diffusivity, 132, 134, 139 ﬂoodwave propagation, 133 ﬂoodway, 287, 293 ﬂow contraction, 104, 297, 304 ﬂow convergence, 4 ﬂowdischarge record, 227 ﬂow diversion, 297 ﬂowduration curve, 227 ﬂow in bends, 128 ﬂow kinematics, 22 ﬂuvial system, 199 ﬂux, 124 forestland, 71 forward ﬁnitedifference, 355, 361 Fourier series, 358 freeboard, 290, 304 freezethaw cycle, 262 Froude similitude, 336–337, 340 fundamental dimensions, 9 gabion, 258–259, 263, 280 gabion structure, 260 gabion wall, 263 Ganges River, 224–225 gantry, 329 gate pier, 324 general scour, 310–311 geographic information system (GIS), 226 geometric similitude, 335 geometric variables, 10 geometry of meanders, 183 geotextile, 263 gooseneck, 255 gradation coefﬁcients, 17 gradecontrol structure, 207, 280, 301–302
430
Index
graduallyvaried ﬂow, 102, 107, 110 grain resistance, 92 gravel ﬁlters, 248 gravel mining, 210, 213–214 gravity wave, 385 gravity wall, 262 grazing, 213 great ﬂood, 293 Green and Ampt, 47 Green’s law, 401 grid Peclet number, 367, 371 groundwater control, 267 group velocity, 393–394 groynes, 268–269, 321 guidebank, 270–271, 304 guide vane, 307 guide wall, 307 gullies, 207
inﬂection point, 180 initial water content, 48 interception, 46 interference, 393 irrigation canal, 306 island, 215
hardpoint, 267 headcut, 207, 210, 214 high tide, 403 higherorder approximation, 360, 362 hopper dredge, 325, 327–328 horizontal drain, 238–239 horseshoe vortex, 313–314 hydraulic conductivity, 48 hydraulicgeometry changes, 204 hydraulicgeometry relationship, 171 hydraulic geometry of stable channel, 173 hydraulic grade line (HGL), 29, 110–111 hydraulic jump, 101, 106 hydraulic model, 339 hydraulic modeling, 336 hydraulic pipeline dredge, 327 hydraulic similitude, 334 hydraulic suction dredge, 325 hydraulically rough/smooth, 91 hydrologic cycle, 31 hydroseeding, 255 hyetograph, 52 hyperbolic function, 385–387 hyperbolic partial differential equation, 354, 365 hypsometric curve, 33
Lacey silt factor, 165 Lagrangian celerity, 130 landslide, 236 Lane’s relationship, 201 lateral earth pressure, 267 lateral migration, 170, 184, 186, 188–189, 224 lateral mobility, 185, 189 lateral stability, 187 launching apron, 252, 257 Lawn Lake, 220 Leonard scheme, 368, 371 levee, 171, 219, 288–290, 293 liftdrag ratio, 161 lift force, 159 lightweight sediment properties, 344, 349 linear stability analysis, 357, 369 local acceleration, 23 local scour depth, 302 lock and dam, 296, 323–324 logarithmic equation, 92 logjam, 295 long wave, 388 longitudinal dyke, 319 longitudinal proﬁle, 80 longitudinal shear stress, 185 loop, 180 looprating curve, 136 low sill, 299 low tide, 403 low water reference plane (LWRP), 81, 331 lower guard wall, 324 Lower Mississippi River, 261
ice and debris movement, 236, 254, 320 ideal crosssection geometry, 163–164 impermeable spur, 270 implicit ﬁnite difference, 356 impulsemomentum relationship, 98 incipient motion, 160 incised river/channel, 205, 207 incomplete mobilebed similitude, 346–347, 349 increasing discharge, 195 inﬁltration, 46, 48–49
Jamuna River, 223–225, 348 jet scour, 300–301 jetty ﬁeld, 275–276 Kellner jack, 276 kinematic similitude, 335 kinematic variables, 10 kinematic viscosity, 13 kinematic wave, 129, 134–135, 138 Kleitz–Seddon law, 131 knickpoints, 207
MacCormack scheme, 372 magnetic sliding collar, 350 mainstream lengths, 33 maintenance, 282 maintenance dredging, 330
Index Manning coefﬁcient, 54, 90–91 ManningStrickler similitude, 340 marching procedure, 354 masonry wall, 263 mass curve, 229 mass density, 10 mass wasting, 237 Matamek River, 83, 87, 403 mathematical model, 352 matting, 255 mattress, 258–259 maximum precipitation, 36 maximum scour, 304 maximum sill elevation, 105 mean annual ﬂood, 171 mean daily sediment discharge, 151 mean radius of curvature, 183 meander belt, 180 meander cutoff, 288 meander length, 179–184 meander loop, 179, 187, 190 meander width, 179–184 meandering, 186, 189, 217 mechanical dredge, 325 method of moment, 144–145 Meuse River, 210 midchannel bar, 178 migration rate, 188 mild slopes, 108 minimum channel width, 105 minimum radius of curvature, 181–182 minimum stream power, 179 Mississippi River, 222, 255, 278, 291, 330, 379, 398 Mississippi River Commission (MRC), 293 mobilebed model, 339, 342, 344, 348 mobilebed similitude, 343 model accuracy, 353 model calibration, 352 model dimension, 348 model distortion, 340–341 model scale, 334, 342 model veriﬁcation, 353 modiﬁed slump failure, 251 momentum correction factor, 96, 99 momentum equation, 97, 126 mountainous terrain, 213 multidimensional river model, 376 National Geodetic Vertical Datum (NGVD), 81, 331 native plants, 253 navigable waterway, 296 navigation, 316–318 neck cutoff, 189–190 Neuman type of boundary condition, 362 noncohesive alluvial channel, 171
431 noncohesive bank, 235 nonexceedance probability, 40, 42, 143 nonuniform thalweg depth, 381 normal depth, 90 normal stresses, 25 numerical diffusion, 366–367 numerical stability, 141, 359, 368 numerically smooth simulation, 370 openhopper barge, 317 overland ﬂow, 57–58, 63 overtopping, 305 oxbow lake, 190 Padma River, 222, 224–225 parabolic equation, 365 partial closure, 297 partialequilibrium hydrograph, 58, 60 particle erosion, 250 particleorientation angle, 162 particlesize distribution, 17 particle stability, 158, 162, 168, 194 pasture land, 71 path line, 22 paved bed, 209 peak discharge, 220 perched river, 216, 219, 289 perforated pipe, 238 permeable spur, 270 perturbation theory, 179 pervious toe trench, 290 phase velocity, 388 physical model, 334 physical properties, 9, 14 pier footing, 314 pier scour, 313–315 pile, 273 pile cap, 314 pile cluster, 274 pile retard, 272 pilot channel, 319 pipeline, 329 piping, 236, 290 planform geometry, 181 plunging jet scour, 300–301 point bar, 169–170, 177 point bar sedimentation, 194 pointbar slope, 191 point rainfall precipitation, 40 pontoon, 329 pore pressure, 238 porosity, 21 port, 329 potential evapotranspiration, 34 potential inﬁltration, 52 precast cellular block, 257
432
Index
precision, 353 predictorcorrector scheme, 372 Preissmann scheme, 373 preservative treatment, 283 pressurerelief well, 290 probability density function, 143 project durability, 282 project ﬂood, 294 project life, 282 properties of sediment, 13 pumping plant, 291, 306 quasi similitude, 347 quasisteady dynamic wave, 129 radial acceleration, 128 radial shear stress, 128, 166–167 radial surface slope, 167 radius of curvature, 167, 172, 178–179, 181, 187 raindrop impact, 55, 63 rainfall erodibility, 67 rainfall excess, 53 rainfall precipitation, 35 rapidly varied ﬂow, 102 Red River, 295 regime equation, 165 regulated outlet work, 297 relative inﬁltration, 50 relative migration rate, 187 relative particle stability, 169 reservoir, 286 resistance to ﬂow, 93 resistance to overland ﬂow, 53–54 retaining wall, 262, 266 retard, 271–274 revegetation, 256 revetment, 252, 330 Rhine River, 151, 152, 321 rigidbed hydraulic model, 337, 339–341 rills, 63, 207 Rio Grande River, 228–229 riprap design 238, 240–242, 244–247, 250 risk, 304 river branches, 222 river closure, 297–299, 304 river conﬂuences, 221–222 river control, 320 river crossing, 310 river database, 226 river diversion, 297 river dynamics, 199 river engineering, 286 river estuaries, 385, 401 river ﬂood control, 286 river ﬂoodwave, 130
river ﬂowcontrol structure, 267 river formation, 4 river meandering, 179 river model, 365 riverstabilization structure, 234 riverbank engineering, 234, 236, 253, 282 riverbed armoring, 208 riverbed degradation, 204 rivers around the world, 5 Roaring River, 190, 220 rock dike, 331 rock durability, 240 rock ﬁll, 274 rock riprap, 240 rockﬁll abutment, 313 rockﬁll cofferdam, 297 rockﬁll weir, 282 roll wave, 134 rotational failure, 235 Rouse equation, 114 runoff, 51 Russian river, 212 sack revetment, 257–258 SaintVenant equation, 28, 129, 139 saline wedge, 401–404 salmon capture, 404 sampling duration, 191 satellite imagery, 226 scale ratio, 339 Schoharie Creek, 349 scour adjustment, 283 scour depth, 303, 310 scour monitoring, 350 scrolls, 187 secondorder approximation, 361 secondary circulation, 166, 169 sediment ejection, 306, 308 sediment exclusion, 306–307 secondary ﬂow, 168 sediment budget, 222 sediment concentration, 19 sediment continuity relationship, 114 sedimentdelivery ratio, 75 sediment discharge, 113 sediment diversion, 223 sediment grade scale, 16 sediment overload, 215 sedimentrating curve, 137 sediment record, 228 sediment source, 72, 199 sediment transport, 111 sediment yield, 72, 74 settlement, 283 settling basin, 309 settling velocity, 21, 112
Index shallowwater wave, 388 shear stresses, 25 shear velocity, 18, 89 sheepfoot roller, 262 sheet erosion, 63 sheet ﬂow, 63 sheet pile, 262, 264, 281–282, 305 Shields parameter, 112, 161, 172 short wave, 388 sidecasting dredge, 327 sideslope failure, 250–251 sieve analysis, 17 sill raising, 298 similitude criteria, 342 sinegenerated curve, 180, 182 sinuosity, 180, 184–185, 189 slackwater zone, 225 slope adjustment, 205 slope length steepness, 67 slopereduction method, 237 slope reinforcement, 236 sloping sill, 281, 282 sluice gate, 105, 300, 302 slump, 250 smolts, 403 snowmelt runoff, 61 sodding, 254 soil cement, 261–262 soil classiﬁcation, 47 soil erodibility, 67 soil loss, 64, 71 speciﬁc discharge, 150 speciﬁc energy, 110, 130 speciﬁcgauge record, 84, 86, 208 speciﬁc gravity, 15 speciﬁc momentum, 100 speciﬁc weight, 10, 15, 21 spillthrough abutment, 313 spillway crest, 287 spillway gate, 324 splash zone, 254 splitter wall, 324 sprigging, 254 spud, 329 spur, 268–270, 321 St. Lawrence River, 404 stability, 357 stability factor, 160–161, 168 stability number, 161 stability diagram, 217 stabilizing moment, 160 stable bed condition, 209 stable channel, 158 stable channel geometry, 176 stable foundation, 310 stagedischarge relationship, 56, 84, 227 starboard, 329
433 steady 3D model, 379 steadynonuniform ﬂow, 97 steadyuniform ﬂow, 88 steel jack, 275 steep slopes, 108 stilling basin, 280, 324 stone ﬁll dike, 272 stone gradation, 247 storage capacity, 286 stratiﬁed bank, 235 streak line, 22 streambank erosion control, 253 streambank failure, 236 streamline, 22 streamline deviation angle, 160, 168, 170, 191 stream rehabilitation, 322 subcritical normal depth, 108 submeander, 178 submerged jet scour, 300–301 submerged speciﬁc weight, 15 submerged weight, 159–160 subsidence, 219 subsurface drainage, 238 summer levee, 322 supercritical normal depth, 108 surface runoff, 53 surfacerunoff hydrograph, 60 surface vane, 308 surface wave, 236, 385–386 suspended load, 113 swing cable, 329 synthetic ﬁlter cloth, 248 Tanana River, 218 tank barge, 317 tectonic activity, 219 temperature effect, 86 tension crack, 235 terrace zone, 254 tetrahedron, 276 thalweg, 169–170, 178, 191 thalweg erosion, 194 thunderstorms, 44 tieback, 252, 271, 273 tide, 385, 388, 396, 399–401, 404 tilted river model, 335, 340 timber crib wall, 264 timber pile dike, 272, 274–275 timber pile fence, 274 time scale for bedload motion, 344 time to equilibrium, 59, 60 toe protection, 258, 272–273 toe scour, 235 topographic measurement, 191 towboat, 317 translational slide, 250–251
434
Index
transport zone, 199 transversal shear stress, 185–186 trap efﬁciency, 115 trench, 239, 256–257 trenchﬁll revetment, 320 tributary bar, 177–178 tributary basin improvement, 294 truncation error, 361 tunnel, 297 tunnel ejector, 308–309 turbulentvelocity proﬁle, 95 uncoupled formulation, 376 underground seepage, 290 uniform thalweg depth, 381 universal soilloss equation, 66 upland erosion, 63–64 uplift, 219 upper lock gate, 323 upstream boundary condition, 362 upwind, 356 urban levee, 290 urbanization, 213 valley slope, 186, 189 vandalism, 283 vane dike, 277 vegetation, 253 velocity against stone, 244
velocity measurement, 191 vertical drain, 238 vertical wall, 313 volumetric ﬂux, 126 von Neumann procedure, 357 vortex tube, 309 Waal River, 321–322 washload, 199, 214 water and sediment loads, 6 water quality, 329 watersheds, 31 waterway alignment, 316, 318 wave, 131, 254, 362, 364, 385, 387, 389–391, 394 weather radar, 39 weephole, 258, 267 Weibull plotting position, 147 weighting coefﬁcient, 373 wetting front, 48 wildﬁre, 213 wind wave, 395–396 windrow, 256, 257 windrow revetment, 257 windstorm duration, 395 wire fencing, 275 Yampa River, 137