SEDIMENT TRANSPORT Zhou Liu 3. udgave. januar 2001 Laboratoriet for Hydraulik og Havnebygning Instituttet for Vand, Jord og Miljoteknik Aalborg Universitet Preface Flow and sediment transport are important in relation to several engineering topics, e. g.

erosion around structures, back? lling of dredged channels and nearshore morphological change. The purpose of the present book is to describe both the basic hydrodynamics and the basic sediment transport mechanism. The reader’s background should be a basic course in wave theory and ? uid mechanics. Chapter 1 deals with fundamentals in ? id mechanics with emphasis on bed shear stress by currents, while Chapter 3 discusses wave boundary layer theory. They are both written with a view to sediment transport.

Sediment transport in rivers, cross-shore and longshore are dealt with in Chapters 2, 4 and 5, respectively. It is not the intention of the book to give a broad review of the literature on this very wide topic. The book tries to pick up information which is of engineering importance.

An obstacle to the study of sedimentation is the scale e? ect in model tests. Whenever small-scale tests, large-scale tests and ? ld investigations are available, it is always the result from ? eld investigations which is referred to. i CONTENTS Contents 1 Steady uniform ? ow in open channels 1. 1 1.

2 1. 3 1. 4 1. 5 1. 6 1. 7 1. 8 Types of ? ow .

. . . . . .

. . .

. . . . . . .

. . . . . .

. . . . .

. . . . .

. Prandtl’s mixing length theory . . . . . . .

. . . . . . .

. . . .

. . . . .

. . Fluid shear stress and friction velocity . . . . .

. . . .

. . .

. . . . . .

. . Classi? cation of ? ow layer . . .

. . . . . . .

. . . . . . .

. . . . . . .

. . . Velocity distribution . . . .

. . . . . .

. . . . . .

. . . . .

. . .

. . . .

. . 1 1 4 5 7 9Ch? zy coe? cient . .

. . . . . . . .

. . . .

. . . . . .

. . . . . . . .

. . .

. 12 e Drag coe? cient, lift coe? cient and friction coe? cient . . . . . . .

. . .

. . 14 Exercise . . .

. . . . . . . .

. . . . . .

. . . . .

. . . . . . . .

. . . . . .

. 16 17 2 Sediment transport in open channels 2. 1 2. 2 2. 3 2. 4 2. 5 2. 6 2.

7 2. 8 Sediment properties . . . . .

. . . . . . .

. . . . .

. . . . . . .

. . . .

. . . 17 Threshold of sediment . . . . .

. . . . . . . .

. . . .

. . . .

. . . . . . .

. 19 Bedforms, bed roughness and e? ective shear stress . . . . . . .

. . . . . . . 22 Transport modes .

. . . . . . .

. . . . . . .

. . . .

. . . .

. . . . . . . .

25 Bed-load transport formulae . . . .

. . . .

. . . . . .

. . . . . . .

. . . . . 26 Suspended load . .

. . . . .

. . . . . . . .

. . . .

. . . .

. . . . . . . .

. . 28 Total sediment transport . .

. . . .

. . . . . . .

. . . . .

. . . . . . .

. . . 32 Exercise .

. . .

. . . . . . . .

. . .

. . . . .

. . .

. . . .

. . . . . .

. . . . 33 34 3 Wave boundary layer 3.

1 3. 2 3. 3 3.

4 3. 5 3. 6 3. 7 3. 8 3. 9 Concept of wave boundary layer . .

. . . . . .

. . . . . . . .

. . . . .

. . . 34 Laminar wave boundary layer on smooth bed .

. . . . . . . .

. . . . .

. 35 Wave boundary layer thickness . . . . . .

. . . . . .

. . .

. . .

. . . . .

. 37 Wave friction coe? cient . . . .

. . . . . .

. . . . . . . . . . . . . . . . . . 39 Mechanism of sediment transport in coastal regions . . . . . . . . . . . . . 41 Boundary layer of irregular waves . . . . . . . . . . . . . . . . . . . . . . . 42 Boundary layer of wave and current: Fredse’s model . . . . . . . . . . . . . 42 Boundary layer of wave and current: Bijker’s model . . . . . . . . . . . . . 47 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 50 4 Cross-shore sediment transport and beach pro? e 4. 1 Sediment size and its sorting on beaches . . . . . . . . . . . . . . . . . . . 50 ii CONTENTS 4. 2 4. 3 4. 4 4. 5 4. 6 4. 7 4. 8 4. 9 Threshold of sediment under wave actions . . . . . . . . . . . . . . . . . . 51 Depth of closure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 Bed form and bed roughness . . . . . . . . . . . . . . . . . . . . . . . . . . 54 Beach classi? cation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 Berms and longshore bars . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 Equilibrium beach pro? le (x2/3 ) . . . . . . . . . . . . . . . . . . . . . . . 59 Erosion and accretion predictors . . . . . . . . . . . . . . . . . . . . . . . . 60 Shoreline retreat due to sea level rise . . . . . . . . . . . . . . . . . . . . . 61 4. 10 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 5 Longshore sediment transport 5. 1 5. 2 63 CERC-formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 Bijker’s method: Wave + current . . . . . . . . . . . . . . . . . . . . . . . 66 71 6 References iii 1 STEADY UNIFORM FLOW IN OPEN CHANNELS 1 Steady uniform ? ow in open channels This chapter is written with a view to sediment transport.The main outcome is the current friction coe? cient. The coordinate system applied in this chapter is shown in Fig. 1. Fig. 1. Coordinate system for the ? ow in open channels. 1. 1 Types of ? ow Description of various types of ? ow are given in the following. Laminar versus turbulent Laminar ? ow occurs at relatively low ? uid velocity. The ? ow is visualized as layers which slide smoothly over each other without macroscopic mixing of ? uid particles. The shear stress in laminar ? ow is given by Newton’s law of viscosity ?? = ? ? du dz (1-1) where ? is density of water and ? kinematic viscosity ( ? 10? 6 m2 /s at 200 C ). Most ? ows in nature are turbulent. Turbulence is generated by instability in the ? ow, which trigger vortices. However, a thin layer exists near the boundary where the ? uid motion is still laminar. A typical phenomenon of turbulent ? ow is the ? uctuation of velocity U = u + u where U u W w W = w + w (1-2) instantaneous velocity, in x and z directions respectively time-averaged velocity, in x and z directions respectively instantaneous velocity ? uctuation, in x and z directions respectively u w Turbulent ? ow is often given as the mean ? ow, described by u and w. 1 1STEADY UNIFORM FLOW IN OPEN CHANNELS In turbulent ? ow the water particles move in very irregular paths, causing an exchange of momentum from one portion of ? uid to another, and hence, the turbulent shear stress (Reynolds stress). The turbulent shear stress, given by time-averaging of the NavierStokes equation, is ? t = ? ? u w (1-3) Note that u w is always negative. In turbulent ? ow both viscosity and turbulence contribute to shear stress. The total shear stress is ? = ? ? + ? t = ? ? du + (?? u w ) dz (1-4) Steady versus unsteady A ? ow is steady when the ? ow properties (e. g. density, velocity, pressure etc. at any point are constant with respect to time. However, these properties may vary from point to point. In mathematical language, ? (any ? ow property) = 0 ? t (1-5) In the case of turbulent ? ow, steady ? ow means that the statistical parameters (mean and standard deviation) of the ? ow do not change with respect to time. If the ? ow is not steady, it is unsteady. Uniform versus non-uniform A ? ow is uniform when the ? ow velocity does not change along the ? ow direction, cf. Fig. 2. Otherwise it is non-uniform ? ow. Fig. 2. Steady uniform ? ow in a open channel. 2 1 STEADY UNIFORM FLOW IN OPEN CHANNELSBoundary layer ? ow Prandtl developed the concept of the boundary layer. It provides an important link between ideal-? uid ? ow and real-? uid ? ow. Here is the original description. For ? uids having small viscosity, the e? ect of internal friction in the ? ow is appreciable only in a thin layer surrounding the ? ow boundaries. However, we will demonstrate that the boundary layer ful? l the whole ? ow in open channels. The boundary layer thickness (? ) is de? ned as the distance from the boundary surface to the point where u = 0. 995 U . The boundary layer development can be expressed as laminar ? w turbulent ? ow ? x ? x = 5 = 0. 4 U x ? ?0. 5 when Rex = when Rex = U x ? U x ? < 5 ? 105 > 5 ? 105 U x ? ?0. 2 Fig. 3. Development of the boundary layer. Example 1 Given Wanted Solution Development of the boundary layer ? ow. ?ow velocity U = 1m/s and water depth h = 10m 1) x value where the boundary layer ? ow starts to ful? l the whole depth 2) type of the boundary layer ? ow Based on the expression for turbulent boundary layer ? ow (U/? )0. 25 1. 25 h 0. 40. 8 (1/10? 6 )0. 25 101. 25 0. 40. 8 x ?=h = = = 1171 m Rex = Comment 1 ? 1171 U x = = 1. 171 ? 109 > 5 ? 105 ? 10? 6 turbulentThe example demonstrates that the ? ow in open channels is always a turbulent boundary layer ? ow. 3 1 STEADY UNIFORM FLOW IN OPEN CHANNELS 1. 2 Prandtl’s mixing length theory Prandtl introduced the mixing length concept in order to calculate the turbulent shear stress. He assumed that a ? uid parcel travels over a length before its momentum is transferred. Fig. 4. Prandtl’s mixing length theory. Fig. 4 shows the time-averaged velocity pro? le. The ? uid parcel, locating in layer 1 and having the velocity u1 , moves to layer 2 due to eddy motion. There is no momentum transfer during movement, i. e. the velocity of the ? id parcel is still u1 when it just arrives at layer 2, and decreases to u2 some time later by the momentum exchange with other ? uid in layer 2. This action will speed up the ? uid in layer 2, which can be seen as a turbulent shear stress ? t acting on layer 2 trying to accelerate layer 2, cf. Fig. 4 The horizontal instantaneous velocity ? uctuation of the ? uid parcel in layer 2 is u = u1 ? u2 = du dz (1-6) Assuming the vertical instantaneous velocity ? uctuation having the same magnitude w = ? du dz (1-7) where negative sign is due to the downward movement of the ? uid parcel, the turbulent shear stress now becomes ? = ? ? u w = ? 2 du dz 2 If we de? ne kinematic eddy viscosity ? = 2 du dz (1-8) the turbulent shear stress can be expressed in a way similar to viscous shear stress ? t = ? ? du dz 4 (1-9) 1 STEADY UNIFORM FLOW IN OPEN CHANNELS 1. 3 Fluid shear stress and friction velocity Fluid shear stress The forces on a ? uid element with unit width is shown in Fig. 5. Because the ? ow is uniform (no acceleration), the force equilibrium in x-direction reads ? z ? x = ? g (h ? z) ? x sin ? For small slope we have sin ? ? tan ? = S. Therefore ? z = ? g (h ? z) S The bottom shear stress is ? b = ? z=0 = ? g h S (1-10)Fig. 5. Fluid force and bottom shear stress. Bottom shear stress In the case of arbitrary cross section, the shear stress acting on the boundary changes along the wetted perimeter, cf. Fig. 5. Then the bottom shear stress means actually the average of the shear stress along the wetted perimeter. The force equilibrium reads ? b O ? x = ? g A ? x sin ? where O is the wetted perimeter and A the area of the cross section. By applying the hydraulic radius (R = A/O) we get ? b = ? g R S (1-11) In the case of wide and shallow channel, R is approximately equal to h, eq (1-11) is identical to eq (1-10). 5 1STEADY UNIFORM FLOW IN OPEN CHANNELS Friction velocity The bottom shear stress is often represented by friction velocity, de? ned by u? = ? b ? (1-12) The term friction velocity comes from the fact that and it has something to do with friction force. Inserting eq (1-11) into eq (1-12), we get u? = gRS ?b /? has the same unit as velocity (1-13) Viscous shear stress versus turbulent shear stress Eq (1-10) states that the shear stress in ? ow increases linearly with water depth, cf. Fig. 6. Fig. 6. Shear stress components and distribution. As the shear stress is consisted of viscosity and turbulence, we have ? = ?? + ? t = ? g (h ? z) S (1-14) On the bottom surface, there is no turbulence (u = w = 0, u = w = 0), the turbulent shear stress ? t = ? ? u w = 0 Therefore, in a very thin layer above the bottom, viscous shear stress is dominant, and hence the ? ow is laminar. This thin layer is called viscous sublayer. Above the viscous sublayer, i. e. in the major part of ? ow, the turbulent shear stress dominates, cf. ?g. 6. The measurement shows the shear stress in the viscous sublayer is constant and equal to the bottom shear stress, not increasing linearly with depth as indicated by Fig. 6. 6 1STEADY UNIFORM FLOW IN OPEN CHANNELS 1. 4 Classi? cation of ? ow layer Scienti? c classi? cation Fig. 7 shows the classi? cation of ? ow layers. Starting from the bottom we have 1) Viscous sublayer: a thin layer just above the bottom. In this layer there is almost no turbulence. Measurement shows that the viscous shear stress in this layer is constant. The ? ow is laminar. Above this layer the ? ow is turbulent. 2) Transition layer: also called bu? er layer. viscosity and turbulence are equally important. 3) Turbulent logarithmic layer: viscous shear stress can be neglected in this layer.Based on measurement, it is assumed that the turbulent shear stress is constant and equal to bottom shear stress. It is in this layer where Prandtl introduced the mixing length concept and derived the logarithmic velocity pro? le. 4) Turbulent outer layer: velocities are almost constant because of the presence of large eddies which produce strong mixing of the ? ow. Fig. 7. Scienti? c classi? cation of ? ow region (Layer thickness is not to scale, turbulent outer layer accounts for 80% – 90% of the region). 7 1 STEADY UNIFORM FLOW IN OPEN CHANNELS Engineering classi? ation In the turbulent logarithmic layer the measurements show that the turbulent shear stress is constant and equal to the bottom shear stress. By assuming that the mixing length is proportional to the distance to the bottom ( = ? z), Prandtl obtained the logarithmic velocity pro? le. Various expressions have been proposed for the velocity distribution in the transitional layer and the turbulent outer layer. None of them are widely accepted. However, By the modi? cation of the mixing length assumption, cf. next section, the logarithmic velocity pro? le applies also to the transitional layer and the turbulent outer layer.Measurement and computed velocities show reasonable agreement. Therefore in engineering point of view, a turbulent layer with the logarithmic velocity pro? le covers the transitional layer, the turbulent logarithmic layer and the turbulent outer layer, cf. Fig. 8. As to the viscous sublayer. The e? ect of the bottom (or wall) roughness on the velocity distribution was ? rst investigated for pipe ? ow by Nikurase. He introduced the concept of equivalent grain roughness ks (Nikurase roughness, bed roughness). Based on experimental data, it was found 1) Hydraulically smooth ? ow for u ? ks ? ?5Bed roughness is much smaller than the thickness of viscous sublayer. Therefore, the bed roughness will not a? ect the velocity distribution. 2) Hydraulically rough ? ow for u ? ks ? ? 70 Bed roughness is so large that it produces eddies close to the bottom. A viscous sublayer does not exist and the ? ow velocity is not dependent on viscosity. 3) Hydraulically transitional ? ow for 5 ? u ? ks ? ? 70 The velocity distribution is a? ected by bed roughness and viscosity. Fig. 8. Engineering classi? cation of ? ow region (Layer thickness is not to scale). 8 1 STEADY UNIFORM FLOW IN OPEN CHANNELS 1. 5 Velocity distributionTurbulent layer In the turbulent layer the total shear stress contains only the turbulent shear stress. The total shear stress increases linearly with depth (eq (1-10) or Fig. 6), i. e. ?t (z) = ? b 1 ? z h By prandtl’s mixing length theory ? t = ? 2 du dz 2 and assuming the mixing length = ? z 1 ? z h 0. 5 where Von Karman constant ? = 0. 4 1 , we get du = dz ? b /? ?z = u? ?z Integration of the equation gives the famous logarithmic velocity pro? le u (z) = u? z ln ? z0 (1-15) where the integration constant z0 is the elevation corresponding to zero velocity (uz=z0 = 0), given by Nikurase by the study of the pipe ? ws. ? ? ? 0. 11 u ? ? ? ? ? ? ? ? ? ? ? Hydraulically smooth ? ow Hydraulically rough ? ow u ? ks ? u ? ks ? ?5 ? 70 u ? ks ? Z0 = ? 0. 033 ks ? ? 0. 11 ? u? (1-16) + 0. 033 ks Hydraulically transition ? ow 5< < 70 It is interesting to note that the friction velocity u? , which, by de? nition, has nothing to do with velocity, is the ? ow velocity at the elevation z = z0 e? , i. e. uz=z0 e? = u? In the study of sediment transport, it is important to know that the friction velocity is the ? uid velocity very close to the bottom, cf. Fig. 9. 1 ? = 0. 4 is obtained experimentally in pipe ? ow 9 1STEADY UNIFORM FLOW IN OPEN CHANNELS Viscous sublayer In the case of hydraulically smooth ? ow there is a viscous sublayer. Viscous shear stress is constant in this layer and equal to the bottom shear stress, i. e. ?? = ? ? du = ? b dz Integrating and applying u|z=0 = 0 gives u (z) = ?b ? ? z = u2 ? z ? (1-17) Thus, there is a linear velocity distribution in the viscous sublayer. The linear velocity distribution intersect with the logarithmic velocity distribution at the elevation z = 11. 6? /u? , yielding a theoretical viscous sublayer thickness ?? = 11. 6 ? u? The velocity pro? le is illustrated in Fig. , with the detailed description of the ? uid velocity near the bottom. Fig. 9. Illustration of the velocity pro? le in hydraulically smooth and rough ?ows. 10 1 STEADY UNIFORM FLOW IN OPEN CHANNELS Bed roughness The bed roughness ks is also called the equivalent Nikurase grain roughness, because it was originally introduced by Nikurase in his pipe ? ow experiments, where grains are glued to the smooth wall of the pipes. The only situation where we can directly obtain the bed roughness is a ? at bed consisting of uniform spheres, where ks = diameter of sphere. But in nature the bed is composed of grains with di? erent size.Moreover, the bed is not ? at, various bed forms, e. g. sand ripples or dunes, will appear depending on grain size and current. In that case the bed roughness can be obtained indirectly by the velocity measurement, as demonstrated by the following example. Example 2 Given Wanted Solution Find the bed roughness from velocity measurement. Flume tests with water depth h = 1m, the measured velocities at the elevation of 0. 1, 0. 2, 0. 4 and 0. 6 m are 0. 53, 0. 58, 0. 64 and 0. 67 m/s, respectively. bed roughness ks The ? tting of the measured velocities to the logarithmic velocity pro? le by the least square method gives u? 0. 03 m/s and z0 = 0. 0000825 m. Hence, ks = z0 /0. 033 = 0. 0025m ? b = ? u2 = 0. 9 N/m2 ? we can con? rm that it is hydraulically rough ? ow by u? ks /? = 75 > 70 Fig. 10. Fitting of the measured velocities to logarithmic velocity pro? le. Comment The logarithmic velocity pro? le suggests that the maximum velocity occurs at the ? ow surface. However, the measurements reveal that the maximum velocity occurs some distance under the ? ow surface due to the surface shear from the air. Moreover, the logarithmic velocity is basically developed for the logarithmic turbulent layer which is close to the bottom.Therefore, the velocity measurement in connection with the determination of ? b and ks is preferred to take place at the elevation ks < z < 0. 2 h The following ks values have been Concrete bottom Flat sand bed Bed with sand ripples suggested based on ? ume tests ks = 0. 001 ? 0. 01 m ks = (1 ? 10) ? d50 ks = (0. 5 ? 1) ? (height of sand ripple) 11 1 STEADY UNIFORM FLOW IN OPEN CHANNELS 1. 6 Ch? zy coe? cient e Ch? zy proposed an empirical formula for the average velocity of steady uniform channel e ? ow v U = C RS (1-18) where R S C Hydraulic radius, i. e. area of cross section divided by wetted parameter Bed slope Empirical coe? ient called Ch? zy coe? cient. C was originally e thought to be constant. Various formulas for C have been proposed Here we will see that C can be theoretically determined by averaging the logarithmic velocity pro? le. Recalling that the friction velocity is (eq (1-13)) u? = gRS and applying it into eq (1-18), we get the expression of C C = U v g u? (1-19) Averaging the logarithmic velocity pro? le gives U = 1 h u? ? h u(z) dz = z0 u? ?h h ln z0 z z0 ? dz = ln h z0 ? 1 + z0 h u? h ln ? z0 e (1-20) Inserting the above equation into eq (1-19) gives v C = g h ln ? z0 e Hydraulically smooth ? ow u ? ks ? (1-21) ? ? ? ? 12 h ? 18 log ? ? 3. 3 ? /u? ? ? 18 log 12 h ks ?5 (1-22) ? 70 Hydraulically rough ? ow u ? ks ? where the expression for z0 has been used and Ln has been converted to Log. Moreover v the inclusion of g = 9. 8 m/s2 means that C has the unit m/s. 12 1 STEADY UNIFORM FLOW IN OPEN CHANNELS Example 3 Given Ch? zy coe? cient and bottom shear stress. e A project is to be located at the water depth h=5 m in Kattegat strait. The measured tidal current velocity is U=1 m/s (Havnecon a/s). The sediment size is d90 =0. 15 mm (Danish Geotechnic Institute). It is estimated that the height of sand ripples is app. 0 cm. 1) Bottom shear stress ? b , when there are sand ripples with height of app. 10 cm on the bed 2) Bottom shear stress ? b , if the ? ow is hydraulically smooth. Wanted Solution 1) When there are sand ripples on the bed Bed roughness Ch? zy coe? cient e Friction velocity Bed shear stress ks ? 0. 75 ? (height of sand ripple) = 0. 075 m C = 18 log u? = U C 12 h ks 1 52. 3 = 18 log v 12? 5 0. 075 = 52. 3 v m/s v g = 9. 8 = 0. 06 m/s ?b = ? u2 = 1000 ? 0. 062 = 3. 6 N/m2 ? The ? ow is hydraulically rough as u? ks 0. 06 ? 0. 075 = = 4500 > 70 ? 10? 6 2) If we assume that the ? w is hydraulically smooth (not the case in reality), we have C = 18 log By inserting u? 12 h 3. 3 ? /u? v = U g/C into the above equation, we get 11. 4 U h ? C = 18 log v 11. 4 ? 1 ? 5 10? 6 C v m/s C = 18 log The solution of the equation is C = 103 Friction velocity Bed shear stress Comment u? = U C m/s v 9. 8 = 0. 03 m/s v g = 1 103 ?b = ? u2 = 1000 ? 0. 032 = 0. 9 N/m2 ? The example shows that if we know only the average velocity, which is often the case, it is easier working on C. For turbulent ? ow over ripple bed, The bottom shear stress obtained in the above example is consisted of skin friction shear stress ? and form pressure of ripples ? b . It is ? b which drives grains as bed-load transport. More details will be given in the next chapter. 13 1 STEADY UNIFORM FLOW IN OPEN CHANNELS 1. 7 Drag coe? cient, lift coe? cient and friction coe? cient Drag and lift coe? cients A real ? uid moving past a body will exert a drag force on the body, cf. Fig. 11. Fig. 11. Drag force and lift force. Drag force is consisted of friction drag and form drag, the former comes from the projection of skin friction force in the ? ow direction, and the latter from the projection of the form pressure force in the ? w direction. The total drag is written as FD = 1 ? CD A U 2 2 (1-23) The lift force is written in the same way FL = where A CD , CL 1 ? CL A U 2 2 Projected area of the body to the plane perpendicular to the ? ow direction. Drag and lift coe? cients, depend on the shape and surface roughness of the body and the Reynolds number. They are usually determined by experiments (1-24) 14 1 STEADY UNIFORM FLOW IN OPEN CHANNELS Friction coe? cient Fig. 12 illustrates ? uid forces acting on a grain resting on the bed. The drag force FD = 1 ? CD A (? U )2 2 where ? is included because we do not know the ? id velocity past the grain, but we can reasonably assume that it is the function of the average velocity and other parameters. Fig. 12. Fluid forces acting on a grain resting on the bed. We can also say that the grain exerts a resistant force FD on the ? ow. If A is the projected area of the grain to the horizontal plane, the bottom shear stress is ? b = FD 1 = ? A 2 CD ? 2 A A U2 = 1 ? f U2 2 (1-25) where f is the friction coe? cient of the bed, which is a dimensionless parameter. By applying the Ch? zy coe? cient we get e ? 0. 06 ? ? ? ( log( 12 h ) )2 ? ? 3. 3 ? /u? ? ? ? ? ? 2 ( log( 12 sh ) ) k Hydraulically smooth ? w f = 2g = C2 u ? ks ? ?5 (1-26) ? 70 0. 06 Hydraulically rough ? ow u ? ks ? Example 4 Solution Calculate the bed friction coe? cient in Example 3. The bed friction coe? cient is f = log 0. 06 12 h ks 2 = log 0. 06 12? 5 0. 075 2 = 0. 0071 Therefore, the bed shear stress is ? b = Comment 1 1 ? f U2 = 1000 ? 0. 0071 ? 12 = 3. 6 N/m2 2 2 8 g c2 f is preferred over C because f is non-dimensional. Darcy-Weisbach friction coe? cient obtained in pipe ? ow is fDW = 15 1 STEADY UNIFORM FLOW IN OPEN CHANNELS 1. 8 Exercise 1) A bridge across a river is supported by piers with a square cross section (length=width=B=1 m).The water depth is h=10 m. The velocity distribution in the river can be expressed by (DS449 del 2, s 15) u(z) = u0 h+z h 1/7 ?h? z ? 0 where the velocity on the surface is u0 = 0. 8 m/s. The square piers can be placed in two ways with di? erent drag coe? cients, see the ? gure. 1) Which placement of the pier gives minimum ? uid force ? 2) Calculate this minimum force. 2) You want to measure the average velocity of the ? ow in a river with a water depth of 10 m by use of only one velocity probe. How would you locate the velocity probe ? 3) A project is to be located at the water depth h=5 m in Kattegat strait.The measured tidal current velocity is U=1. 5 m/s (Havnecon a/s). The sediment size is d90 =0. 15 mm (Danish Geotechnic Institute). It is estimated that the height of sand ripples is app. 10 cm. 1) Ch? zy coe? cient C. e 2) friction coe? cient f . 3) bottom shear stress ? b . 16 2 SEDIMENT TRANSPORT IN OPEN CHANNELS 2 2. 1 Sediment transport in open channels Sediment properties Density The density of natural sediments is ? s = 2650 kg/m3 . Therefore, the relative density is s = ? s /? = 2. 65. Size and shape of a grain Generally grains are triaxial ellipsoids, having a long diameter da , intermediate diameter db and short diameter dc .Corel shape factor gives the most useful description of the shape of a grain SCorel = v dc da db (2-1) For natural grains typically SCorel = 0. 7. The diameter of a grain can be presented as ds Sieve diameter, obtained by sieve analysis dn Nominal diameter, which is the diameter of the sphere having the same volume and weight as the grain. dn ? db and dn is slightly larger than ds . df Fall diameter, which is the diameter of the smooth sphere having the same fall velocity in still water at 240 C as the grain. Fall diameter is the best description of the grain size, because it takes into account the grain shape.Grain size distribution The most useful and convenient method for the analysis of the grain size distribution is the sieve analysis, cf Fig. 1. The median diameter of the sample is d50 , i. e. 50% of the grains by weight pass through. Fig. 1. Grain size distribution. 17 2 SEDIMENT TRANSPORT IN OPEN CHANNELS Settling velocity When a grain falls down in still water, it obtains a constant velocity when the upward ? uid drag force on the grain is equal to the downward submerged weight of the grain. This constant velocity is de? ned as the settling velocity (fall velocity) of the grain.Considering now the settling of a sphere with diameter d, cf. Fig. 2 Fig. 2. Settling of a sphere in still water. The force balance between the drag force and the submerged weight gives ? d3 ? d2 2 1 ? CD ? s = (? s ? ?) g 2 4 6 Therefore the settling velocity of the sphere is ? s = 4(s ? 1)gd 3 CD (2-2) The drag coe? cient of a sphere depends on the Reynolds number (Re = ? s d/? ) Laminar (Re < 0. 5) Turbulent (Re > 103 ) CD = 24 Re (by theory) ?s = ? s = 1 18 (s ? 1) g d2 ? CD ? 0. 4 (by experiment) 3 (s ? 1) g d As to natural grains, Fredse et al. (1992) gives the empirical expression for the drag coef? ient CD = 1. 4 + 36 Re (2-3) Inserting the equation into eq (2-2) and solving for ? s give 36 ? dn 2 ?s = + 7. 5 (s ? 1) g dn ? 2. 8 36 ? dn (2-4) 18 2 SEDIMENT TRANSPORT IN OPEN CHANNELS 2. 2 Threshold of sediment Let us consider the steady ? ow over the bed composed of cohesionless grains. The forces acting on the grain is shown in Fig. 3. Fig. 3. Forces acting on a grain resting on the bed. The driving force is the ? ow drag force on the grain FD = 1 ? d2 ? CD (? u? )2 2 4 where the friction velocity u? is the ? ow velocity close to the bed. ? is a coe? cient, used to modify u? so that ? ? forms the characteristic ? ow velocity past the grain. The stabilizing force can be modelled as the friction force acting on the grain. If u? ,c , critical friction velocity, denotes the situation where the grain is about to move, then the drag force is equal to the friction force, i. e. ? d2 1 ? CD (? u? ,c )2 = f 2 4 which can be re-arranged into u2 f 4 ? ,c = 2 (s ? 1) g d ? CD + f ? 2 CL 3 ? 2 Shields parameter is de? ned as ? = u2 ? (s ? 1) g d (2-5) (? s ? ?) g ? d2 ? d3 1 ? ? CL (? u? ,c )2 6 2 4 We say that sediment starts to move if u? > u? ,c or or ? b > ? b,c ? > ? critical friction velocity u? ,c critical bottom shear stress ? b,c = ? u? ,c ? ,c critical Shields parameter ? c = (s? 1) g d u2 19 2 SEDIMENT TRANSPORT IN OPEN CHANNELS Fig. 4 shows Shields experimental results which relate ? c to the grain Reynolds number de? ned as Re = u? dn ? (2-6) The ? gure has 3 distinct zones corresponding to 3 ? ow situations 1) Hydraulically smooth ? ow for Re = u? dn ? ?2 dn is much smaller than the thickness of viscous sublayer. Grains are embedded in the viscous sublayer and hence, ? c is independent of the grain diameter. By experiments it is found ? c = 0. /Re 2) Hydraulically rough ? ow for Re ? 500 The viscous sublayer does not exist and hence, ? c is independent of the ? uid viscosity. ?c has a constant value of 0. 06. 3) Hydraulically transitional ? ow for 2 ? Re ? 500 Grain size is the same order as the thickness of the viscous sublayer. There is a minimum value of ? c of 0. 032 corresponding to Re = 10. Note that the ? ow classi? cation is similar to that of the Nikurase pipe ? ow (Fig. 8 of Chapter 1), where the bed roughness ks is applied in stead of dn . Fig. 4. The Shields diagram giving ? c as a function of Re (uniform and cohesionless grain). 2 2The critical Shields parameter of sand in air is 0. 01 ? ?c ? 0. 02. 20 2 SEDIMENT TRANSPORT IN OPEN CHANNELS It is not convenient to apply the Shields diagram because the friction velocity u? appears in both axes. Madsen et al. (1976) converted the Shields diagram into the diagram showing the relation between the critical Shields parameter ? c and the so-called sediment-? uid parameter S? S? = d (s ? 1) g d 4? (2-7) Fig. 5. The Shields diagram giving ? c as a function of S? . Example 1 Given Wanted Solution Threshold of sediment Sediment is quartz sand with ? s = 2650 kg/m3 and d = 0. 2mm. Fluid is sea water with ? 1025 kg/m3 and ? = 10? 6 m2 /s. Critical shear stress ? b,c The relative density is s = ? s /? = 2. 59 The sediment-? uid parameter is S? = d 0. 0002 (s ? 1) g d = 4? (2. 59 ? 1) ? 9. 8 ? 0. 0002 = 2. 79 4 ? 10? 6 From Fig. 5 it is found ? c = 0. 052, therefore, u? ,c ? b,c = ? c (s ? 1) g d = 0. 0127 m/s = ? u2 = 0. 165 N/m2 ? ,c 21 2 SEDIMENT TRANSPORT IN OPEN CHANNELS 2. 3 Bedforms, bed roughness and e? ective shear stress Bedforms Once sediment starts to move, various bedforms occur. In laboratory ? umes the sequence of bedforms with increasing ? ow intensity is Flat bed ? Ripples ? Dunes ? High stage ? t bed ? Antidunes Ripples Ripples are formed at relatively weak ? ow intensity and are linked with ? ne materials, with d50 less than 0. 7 mm. The size of ripples is mainly controlled by grain size. By observations the typical height and length of ripples are Hr ? 100d50 Lr ? 1000d50 At low ? ow intensity the ripples have a fairly regular form with an upstream slope 60 and downstream slope 320 . With the increase of ? ow intensity, ripples become three dimensional. Dunes The shape of dunes is very similar to that of ripples, but it is much larger. The size of dunes is mainly controlled by ? ow depth.Dunes are linked with coarse grains, with d50 bigger than 0. 6 mm. With the increase of ? ow intensity, dunes grow up, and the water depth at the crest of dunes becomes smaller. It means a fairly high velocity at the crest, dunes will be washed-out and the high stage ? at bed is formed. Antidunes When Froude number exceeds unity antidunes occur. The wave height on the water surface is the same order as the antidune height. The surface wave is unstable and can grow and break in an upstream direction, which moves the antidunes upstream. Fig. 6. Illustration of ? ow over ripples, dunes and antidunes, and their movement. 22 SEDIMENT TRANSPORT IN OPEN CHANNELS If we know the average velocity of the current, water depth and sediment size, the bed forms can be predicted by empirical diagrams, e. g. the one by Znamenskaya (1969), cf. Fig. 7, where the sediment size is represented by the fall velocity of the sediment (? s ). The ripples speed (c) is also given so that the ? gure can be used to estimate the bed-load transport. Fig. 7. Bed roughness Bed forms given by Znamenskaya (taken from Raudkivi, 1976). The bed roughness ks is also called the equivalent Nikurase grain roughness, because it was originally introduced by Nikurase in his pipe ? w experiments, where grains are glued to the smooth wall of the pipes. The only situation where we can directly obtain the bed roughness is a ? at bed consisting of uniform spheres, where ks = diameter of sphere. Generally the bed roughness can be obtained indirectly by the velocity measurement, as demonstrated by Example 2 in Section 1. 5. The large collection of bed roughness values, obtained by velocity measurement and ? tting, covering various ? ow regions with di? erent sediment size, shows ks ? ? 100d50 = Hr ? ? (1 ? 10)d50 ? at bed (2-8) rippled bed 23 2 SEDIMENT TRANSPORT IN OPEN CHANNELS E? ctive shear stress In the presence of ripples, the resistance to the ? ow consists of two parts, one originating from the skin friction, another due to the form pressure of the ripples, i. e. ?b = ? b + ? b (2-9) where ? b is also called e? ective shear stress, because it is ? b which is acting on single sediment. Fig. 8. The resistance to ? ow over a rippled bed. In the case of ? at bed, ? b = 0, and the bed roughness is usually taken as 2. 5d50 , the e? ective shear stress is3 ? ? 2? ?b = ? b = 1 1 ? 0. 06 ? f U2 = ? ? 12 2 2 log 2. 5 dh 50 ? U2 (2-10) where h is water depth and U current average velocity.In the case of a rippled bed, ? b is the same as above, but the total stress is larger due to form pressure. ? ? 2? ?b = 1 ? 0. 06 ? ? 2 log 12 rh H ? U2 (2-11) where the bed roughness is assumed equal to the height of ripples (Hr ). The distinction between ? b and ? b explains the phenomenon that with the appearance of rippled bed, and hence the increase of ? b , the bed-load transport does not increase. 3 Assume the ? ow is always hydraulically rough. 24 2 SEDIMENT TRANSPORT IN OPEN CHANNELS 2. 4 Transport modes There are three sediment transport modes Wash load very ? e particles which are transported by the water, but these particles do not exist on the bed. Therefore the knowledge of bed material composition does not permit any prediction of wash load transport. Hence, wash load will not be considered in this book. the part of the total load which has more or less continuous contact with the bed. Thus the bed load must be determined in relation to the e? ective shear stress which acts directly on the grain surface. Bed-load Suspended load the part of the total load which is moving without continuous contact with the bed as the result of the agitation of the ? uid turbulence.The appearance of ripples will increase the bed shear stress (? ow resistance). On the other hand, more grains will be suspended due to the ? ow separation on the lee side of the ripples, cf. Fig. 8. Thus the suspended load is related to the total bed shear stress, The basic idea of splitting the total sediment load into bed-load and suspended load is that, as described above, two di? erent mechanisms are e? ective during the transport. As to the boundary between the bed-load and the suspended load, argument is still going on. Einstein (1950) suggests the boundary to be some grain diameters, typically 2d50 , above the bed.But this is not realistic when the bed is rippled, which is almost always the case. Therefore Bijker (1971) proposed that the bed-load transport takes place inside a layer with a thickness being equal to the bed roughness (height of ripples). m The SI unit for sediment transport q is m? s , read cubic meter of sediment per meter width per second. Note that q is expressing volumn of sediment. Sand porosity should be taken into account in converting q into dredging volumn. 3 Moreover, only cohesionless sediment will be treated in this book. 25 2 SEDIMENT TRANSPORT IN OPEN CHANNELS 2. 5Bed-load transport formulae Bed-load transport qB is often expressed in the dimensionless form ? B = qB d (s ? 1) g d (2-12) Kalinske-Frijlink formula Kalinske-Frijlink (1952) formula is a curve ? tted to all data available at that time qB = 2 d50 ? b exp ? ?0. 27 (s ? 1) d50 ? g ? b (2-13) where ? b and ? b are bottom shear stress and e? ective shear stress, respectively. Meyer-Peter formula The ? tting of large amount of experimental data by Meyer-Peter (1948) gives ? B = 8 (? ? ? c ) where ? ?b ? c 1. 5 (2-14) ?b /? (s? 1) g d e? ective Shields parameter ? = e? ective shear stress critical Shields parameterEinstein-Brown formula The principle of Einstein’s analysis is as follows: the number of deposited grains in a unit area depends on the number of grains in motion and the probability that the hydrodynamic forces permit the grains to deposit. The number of eroded grains in the same unit area depends on the number of grains in that area and the probability that the hydrodynamic forces are strong enough to move them. For equilibrium conditions the number of grains deposited must be equal to the number of grains eroded, which, together with experimental data ? tting, gives ? B = 40 K (? )3 36? 2 2 + ? 3 (s ? 1) g d3 50 36? 2 (s ? ) g d3 50 (2-15) K = Bagnold formula Bagnold proposed a formula based on the work done by current. The formula has the same form as the modi? ed Meyer-Peter formula. 26 2 SEDIMENT TRANSPORT IN OPEN CHANNELS Example 2 Given Bed-load transport A river with sea water ? ow sediment ? = 1025 kg/m3 U = 1 m/s ? s = 2650 kg/m3 ? = 10? 6 m2 /s h=2 m d50 = 0. 2mm. Wanted Solution Bed-load transport qB 1) critical Shields parameter The relative density is s = ? s /? = 2. 59 The sediment-? uid parameter is S? = d50 0. 0002 (s ? 1) g d50 = 4? (2. 59 ? 1) ? 9. 8 ? 0. 0002 = 2. 79 4 ? 10? 6 From Fig. 5 it is found the critical Shields parameter is ? = 0. 052. 2) E? ective Shields parameter The e? ective shear stress is ? ?b = 1 ? ? ? 2 0. 06 log 12 h 2. 5 d50 ? 2? ? U 2 = 1. 40 N/m2 The e? ective Shields parameter is ? = ?b /? (s? 1) g d50 = 0. 44 As we do not have information on ripple height, we take Hr = 100 d50 = 0. 02 m, the bottom shear stress is ? ? ? b = 1 ? ? ? 2 0. 06 log 12 h Hr 2? ? U 2 = 3. 24 N/m2 The coe? cient in the Einstein-Brown formula is K = 36? 2 2 + ? 3 (s ? 1) g d3 50 36? 2 = 0. 44 (s ? 1) g d3 50 3) Calculate qB by formulae formula qB (m /(m ? s)) Comment 3 Kalinske-Frijlink 0. 0000121 Meyer-Peter 0. 0000215 Einstein-Brown 0. 000167 The total bed-load transport in the river depends on the width of the river. When the accuracy of sediment transport formulae is concerned, experts say that if a formula gives the correct order of magnitude, it is a good formula. It is not surprising that the formulae give more or less the same result, because all formulae include parameters to be determined by the ? tting of experimental results. 27 2 SEDIMENT TRANSPORT IN OPEN CHANNELS 2. 6 Suspended load Sediment concentration in a steady current Consider a steady ? ow in a open channel. The sediment is kept in suspension by turbulent ? uctuations.Sediment concentration c has the unit m3 /m3 , i. e. the volume of sediments in 1 cubic meter water. The classical approach to calculate the vertical distribution of suspended sediment is to apply Prandtl’s mixing length theory, cf. Fig. 9 Fig. 9. Suspended sediment in steady turbulent ? ow. Consider a uniform sand with a settling velocity ? s . In a unit time, through a unit area on the horizontal plan A-A, the volume of sediment travelling upward and downward are qu = (w ? ?s ) qd = (w + ? s ) c ? c + 1 dc 2 dz 1 dc 2 dz In a steady situation, qu and qd must be equal to each other, which gives c ?s + 1 w 2 dc = 0 dz (2-16)By assuming that 1 w 2 = ? u? z 1 ? z h where ? = 0. 4 and u? is the friction velocity, we get c ? s + ? u? z 1 ? z h dc = 0 dz (2-17) which is integrated with the integration constant given by c|z=a = ca c(z) = ca s ( ?? u? ) h? z a z h? a (2-18) 28 2 SEDIMENT TRANSPORT IN OPEN CHANNELS Reference elevation and reference sediment concentration a and ca in eq (2-18) are called reference elevation and reference sediment concentration, respectively. The reference elevation a is the boundary between the bed load and the suspended load. Bijker (1992) suggests that a is taken as the bed roughness ks and relates ca to the bed-load transport qB .It is assumed that bed-load transport takes place in the bed-load layer from z = 0 to z = a = ks , and in the bed-load layer there is a constant sediment concentration ca . Fig. 10 shows the velocity pro? le applied by Bijker. He argues that in hydraulically rough ? ow there is still a viscous sublayer, which starts from z = 0 to z = z0 e where the linear velocity distribution is tangent with the logarithmic velocity distribution. Note the thickness of the viscous sublayer is much smaller than that in hydraulically smooth ? ow (Fig. 9 in Chapter 1). Fig. 10. Viscous sublayer in hydraulically rough ? ow. By the logarithmic velocity pro? e we get u|z=z0 e = u? /? The averaged velocity in the bed-load layer is Ub = 1 ks 1 u? Z0 e + 2 ? ks z0 e u? z ln ? z0 dz ? 6. 34 u? therefore, the bed-load transport is qB = Ub ks ca hence we obtain the reference sediment concentration qB qB ca = = Ub ks 6. 34 u? ks 29 (2-19) 2 SEDIMENT TRANSPORT IN OPEN CHANNELS Suspended sediment transport Now we know the vertical distribution of both the suspended sediment concentration and the ? uid velocity, cf. Fig. 11. Fig. 11. Illustration of vertical distribution of c and u. the suspended sediment transport can be calculated as h qS = = u(z) c(z) dz a ? h a u z ? ? ln ? 0 I1 ln ca s ( ?? u? ) h? z a ? dz z h? a ? = 11. 6 u? ca a h 0. 033 ks + I2 where I1 and I2 are Einstein integrals given by I1 A(z? ?1) = 0. 216 (1 ? A)z? A(z? ?1) = 0. 216 (1 ? A)z? ks h 1 A 1 A 1? B B 1? B B ? s ? u? z? dB z? I2 ln B dB where A = B= z h z? = By applying Bijker’s recommendation on a and ca , we get qS = 1. 83 qB I1 ln h 0. 033 ks + I2 (2-20) 30 2 SEDIMENT TRANSPORT IN OPEN CHANNELS Example 3 Given Suspended sediment transport As in Example 2, i. e. ) sea water ? ow sediment ? = 1025 kg/m3 U = 1 m/s ? s = 2650 kg/m3 ? = 10? 6 m2 /s h=2 m d50 = 0. 2mm. 3 Wanted Solution Suspended sediment transport qS In Example 2 by Meyer-Peter formula we get qB = 0. 0000215 m? s , and by assuming Hr = ks = 100 d50 = 0. 02 m, we obtain ? b = 3. 24 N/m2 The fall velocity 36 ? d50 2 + 7. 5 (s ? 1) g d50 ? 2. 8 ?s = 36 ? d50 = 0. 02 m/s The friction velocity Therefore A= Hr h u? = = 0. 01 ?b ? = 0. 056 m/s z? = ?s ? u? = 0. 89 and we get the Einstein integrals by numerical integration I1 = 0. 216 A(z? ?1) (1 ? A)z? A(z? ?1) (1 ? A)z? 1 A 1 A 1? B B 1? B B z? dB = 1. 00 z? I2 = 0. 216 ln B dB = ? 2. 50 The suspended sediment transport is qS = 1. 83 qB I1 ln h 0. 033 ks 1. 00 ? ln + I2 2 0. 033 ? 0. 02 ? 2. 50 = 1. 83 ? 0. 0000215 m3 m? = 0. 000217 The ratio between the bed-load and suspended transport is Q = qS = 1. 83 qB I1 ln h 0. 033 ks + I2 = 10 31 2 SEDIMENT TRANSPORT IN OPEN CHANNELS 2. 7 Total sediment transport There are numerous formulae, cf. Raudkivi (1976). Two of them are Bijker qT = qB + qS = qB qT = 0. 05 U 2 d50 (s? 1) g 1 + 1. 83 I1 ln 1. 5 ? b (? s ?? ) g d50 h 0. 033 ks + I2 Engelund Example 4 Given Total sediment transport As in Examples 2 & 3, i. e. ) sea water ? ow sediment ? = 1025 kg/m3 U = 1 m/s ? s = 2650 kg/m3 ? = 10? 6 m2 /s h=2 m d50 = 0. 2mm. Wanted Solution Total sediment transport qT In Examples 2 & 3 we obtained ? qB qS = = = 3. 24 N/m2 m3 m? s m3 m? s 0. 0000215 0. 000217 The total sediment transport Bijker qT = qB + qS = 0. 0000215 + 0. 000217 = 0. 000239 m3 m? s Engelund qT = 0. 05 U 2 d50 (s? 1) g ?b (? s ?? ) g d50 1. 5 = 0. 05 ? 12 0. 0002 (2. 59? 1)? 9. 81 3. 24 (2650? 1025)? 9. 81? 0. 0002 1. 5 = 0. 000179 m3 m? s 32 2 SEDIMENT TRANSPORT IN OPEN CHANNELS 2. 8 Exercise 1) A river with sea water ? ow sediment ? = 1025 kg/m3 U = 1. 5 m/s ? s = 2650 kg/m3 ? = 10? 6 m2 /s h=3 m d50 = 0. 2mm. 1) threshold of sediment 2) bed roughness 3) bed-load transport 4) suspended-load transport 5) total transport 33 3WAVE BOUNDARY LAYER 3 Wave boundary layer This chapter is written with a view to coastal sediment transport. The main outcome is the bottom shear stress of sea bed. 3. 1 Concept of wave boundary layer Linear wave theory gives the amplitude of water particle oscillation on the bottom A = H 1 2 sinh 2 (3-1) ? h L and the horizontal velocity of water particle ub = Um sin(? t) where Um is the maximum horizontal velocity Um = ? H 1 T sinh 2 = A? (3-3) (3-2) ? h L Fig. 1. Horizontal velocity pro? le and water particle orbit by linear wave theory. Linear wave theory assumes that the ? uid is ideal (no viscosity), so here is water particle movement on the bottom, which is not the case in reality. Unfortunately, for the study of sediment transport, the ? ow pattern close to the bottom is of great interest. To overcome this contradiction, the concept of wave boundary layer is introduced. Prandtl developed the concept of ? uid boundary layer in general: For ? uids having small viscosity, the e? ect of internal friction in the ? ow is appreciable only in a thin layer surrounding the ? ow boundaries. Under wave action, this thin layer is called wave boundary layer. 34 3 WAVE BOUNDARY LAYER 3. 2 Laminar wave boundary layer on smooth bedFirst we will get some impression of wave boundary layer by looking at the laminar wave boundary layer on smooth bed, which can be described theoretically. Oscillating water tunnel Lundgren and Srensen (1956) invented the oscillating tunnel to model the wave boundary layer, cf. Fig. 2. Note that the piston movement is the same as the water particle on the sea bed given by linear wave theory. The ? ow in the test section is horizontal and uniform. The thickness of the boundary layer changes with time, but remains very thin due to oscillation. Outside the boundary layer, the ? ow is undisturbed.Fig. 2. Oscillating water tunnel. Formulation of equation of motion We start from the Navier-Stokes equation in the horizontal direction ? ?u ? u ? u + u + w ? t ? x ? z = ? ?p ?? + ? z ? z (3-4) The ? ow in the test section is horizontal (w = 0) and uniform ( ? u = 0) ? x ? ?u ? p ?? = ? + ? t ? z ? z (3-5) Outside the boundary layer we have u = u0 and ? = 0, therefore ? ?u0 ? p = ? ?t ? z (3-6) which, minus eq (3-5), gives ? ?(u ? u0 ) ?? = ? t ? z 35 (3-7) 3 WAVE BOUNDARY LAYER Because the ? ow is laminar, shear stress can be expressed by Newton’s law of viscosity ? = ?? ?u ? z (3-8) e get the equation of motion ? (u ? u0 ) ? 2u = ? 2 ? t ? z The boundary condition is u|z=0 = 0 u|z=? = Um sin(? t) (3-10) (3-9) Velocity pro? le and bottom shear stress The solution of eq (3-9) is ? ? ? ? u = Um sin(? t) ? Um exp ?? z 2? /? ? sin ?? t ? z 2? /? ? (3-11) The second term is a dampened wave, which decays quickly away from the bed, cf. Fig. 3. Fig. 3. The local velocity amplitude oscillating around Um . The bottom shear stress is given by ? b = ? ? ? u ? z z=0 v = ?? Um 2? /? [sin(? t) + cos(? t)] = ? b,max sin(? t + 450 ) where ? b,max = ? ? Um v ? /? As the free stream velocity is u0 = Um sin(? ), we can see that the phase shift between ? b and u0 is 450 , i. e. ?b,max will appear ahead of Um by 450 . 36 3 WAVE BOUNDARY LAYER 3. 3 Wave boundary layer thickness With the inclusion of the wave boundary layer, the velocity pro? le corresponding to wave crest is shown in Fig. 4. Fig. 4. wave boundary layer (laminar ? ow on smooth bed). The boundary layer thickness depends on how we de? ne the top of the boundary layer. Jonsson (1966) de? ned the top of the boundary layer as the minimum elevation where the velocity amplitude is equal to Um , which, by eq (3-11), gives ? j = ? 2 2? ? (3-12)Sleath (1987) de? ned the top of the boundary layer as the elevation where the velocity amplitude is 95% of Um , i. e. 5% relative di? erence, which gives the boundary layer thickness ? 0. 05 = 3 2? ? (3-13) In reality, the ? ow type is turbulent ? ow on rippled bed, the boundary layer thickness is a? ected by bed roughness, cf. Fig. 5. Sleath (1987) gives the empirical formula ? 0. 05 = 0. 26 ks A ks 0. 70 (3-14) where ks is the bed roughness and A the amplitude of water particle oscillation on the bottom. Fig. 5. wave boundary layer thickness over rippled bed. 37 3 WAVE BOUNDARY LAYERExample 1 Given Wanted Solution Wave boundary layer thickness Wave height H=2 m, wave length L= 80 m Water depth h=5 m, sea bed ripple height Hr =15 cm Wave boundary layer thickness ? 0. 05 By linear wave theory the amplitude of the water particle on the bottom A = 1 H 2 sinh 2 ? h L = 2. 48 m The bed roughness is taken as the ripple height ks = Hr =0. 15 m, then ? 0. 05 = 0. 26 Comment A ks 0. 70 ks = 0. 28 m Boundary layer thickness at one location varies with time. ?0. 05 = 0. 28 m is the one when wave crest passes the location (maximum boundary layer thickness). In Section 1. we have shown that the boundary layer ful? ls the whole ? ow depth in channel ? ow. However, the wave boundary layer will remains thin due to the oscillation of water particles. Let us imagine a progressive wave with a period of 8 seconds. First water particles close to the bottom move forward, wave boundary layer is developing, but the development is stopped after 4 seconds, because the water particles stop and start to move backward, and a new boundary layer starts to develop. 38 3 WAVE BOUNDARY LAYER 3. 4 Wave friction coe? cient De? nition of wave friction coe? cient The current friction coe? ient fc is de? ned as ? b = 1 ? fc U 2 2 (3-15) As to sea bed, the bottom shear stress varies with time. Jonsson (1966) de? ned the wave friction coe? cient fw as ? b,max = 1 2 ? fw Um 2 (3-16) where Um is the maximum horizontal velocity of water particle on sea bed, given by the Airy wave theory, fw is a ? ctional coe? cient because ? b,max and Um do not occur at the same time. Wave friction velocity is de? ned as u? ,w = ? b,max = ? fw Um 2 (3-17) fw : Laminar boundary layer and smooth bed The theoretical expression of ? b,max for laminar boundary layer on smooth bed is ? b,max = ? ? Um ? /?By comparison with eq (3-16), we get fw = 2 Um ? 2 ? ? = 2 A? ? 2 ? ? = 2 ? A2 ? 0. 5 By observation it is found that laminar boundary layer on smooth bed corresponds to A2 ? /? < 3 ? 105 fw : Turbulent boundary layer and smooth bed Justesen (1988) suggests fw = 0. 024 ? A2 ? ?0. 123 for 106 < A2 ? < 108 ? (3-18) 39 3 WAVE BOUNDARY LAYER fw : Turbulent boundary layer and rough bed In reality the ? ow is always hydraulically turbulent over rough bed. Jonsson (1966) gives an implicit empirical formula for fw , which is approximated by Swart (1974) in an explicit form ? fw ks = exp ? 5. 213 A 0. 194 ? 5. 977? (3-19) Nielsen (1992) means that Swart formula tends to overpredict fw for small ks /A, cf. Fig. 6. The new ? tting gives ? fw = exp ? 5. 5 ks A 0. 2 ? ? 6. 3? (3-20) Fig. 6. Observed wave friction coe? cient. Example 2 Given Wanted Solution Wave friction coe? cient fw Wave height H=2 m, wave length L= 80 m Water depth h=5 m, sea bed ripple height Hr =15 cm Wave friction coe? cient fw By linear wave theory the amplitude of the water particle on the bottom A = 1 H 2 sinh 2 ? h L = 2. 48 m The bed roughness is taken as the ripple height ks = Hr =0. 15 m, then fw = exp 5. 5 Comment ks A 0. 2 ? 6. 3