See discussions, stats, and author profiles for this publication at: https://www.researchgate.net/publication/335620750 HYDRODYNAMICS Book · September 2019 CITATIONS READS 0 1,880 1 author: Jorgen Fredsoe Technical University of Denmark 234 PUBLICATIONS 12,769 CITATIONS SEE PROFILE Some of the authors of this publication are also working on these related projects: HYDRALAB plus View project Generating waves for OpenFoam: waves2Foam View project All content following this page was uploaded by Jorgen Fredsoe on 05 September 2019. The user has requested enhancement of the downloaded file. HYDRODYNAMICS Jørgen Fredsøe MEK-DTU 2008 jf@mek.dtu.dk 2 3 List of Content 1. The kinematics and forces of flows 7 1.1 Velocities and accelerations 8 1.2 The stresses in a fluid 15 2. Laminar flows 25 2.1 The constitutive equations 26 2.2 The deformation of fluid elements in laminar flows 31 2.3 The equation of continuity 34 2.4 The flow equations (the Navier-Stokes equations) 36 2.5 Other examples of laminar flows 38 3. Potential flow 44 3.1 Definition of the potential 44 3.2 The Bernoulli equation 45 3.3 The Laplace equation 47 3.4 An example of a potential flow: flow over a wavy bed 47 4 4.0 Progressive sinusoidal waves 60 4.1 Shallow water theory 60 4.2 Sinusoidal wave at an arbitrary water depth 65 4.3 Wave energy 75 4.4 Waves over variable water depth 80 4.5 The wave reaction force 86 4.6 Wave induced currents 96 5. Laminar Boundary Layers 102 5.1 The boundary layer equations 103 5.2 The general momentum equation 104 5.3 The momentum equation for boundary layers 105 5.4 Boundary layers with outer pressure gradients 119 6. Turbulent flows 129 6.1 The stability of a laminar flow 131 6.2 Turbulent fluctuations 138 6.3 Reynolds stresses 139 6.4 The mixing length theory 143 6.5 Velocity profiles in uniform channel flows 146 6.6 The structure of turbulence in a pipe 161 5 7. Turbulent boundary layers 165 7.1 The turbulent boundary layer equation 166 7.2 Stationary boundary layer along a plate 168 7.3 Examples of non-stationary boundary layers 175 7.4 Free turbulence: the turbulent jet 179 7.5 Free turbulence: the plane wake 186 8. Turbulence modelling 193 8.1 Mixing length 193 8.2 The 1-equation model (k-model) 198 8.3 2-equation models (k-epsilon model) 207 Appendix References Index 6 1. The kinematics and forces of flows The solution of a flow problem does in principle imply solving Newtons second law G G du ∑ F = ∫ ρ dt dX (1.1) G G where ΣF is the sum of all forces acting on the volume X with the density ρ . u is the velocity vector for the small elements dX of this volume. In generalthe velocity and the force vectors have three components. In hydrodynamics the starting point will often be to consider the forces acting G on a small element with the velocity u , and thereafter by integration obtain (1.1). For a small element dX Newtons second law reads G G du dF = ρ dX dt (1.2) In the following the velocity term and force term for the forces acting on the small volume. To simplify we will normally consider the plane situation only. This is defined G as a flow where all the velocity vectors of the fluid particles u are parallel to a fixed plane, and do not change by a shift in the direction normal to this plane. 7 1.1 Velocities and accelerations G In equation (1.1) we see the velocity vector u which is the velocity of a fluid particle, and the total derivative G G du a= dt (1.3) which describes the acceleration of a fluid particle. In the following we consider a flow, where a fluid particle at the time t0 is in the point A, cf. figure 1.1. The particle moves along a curve called the particle path. If the flow is stationary, the flow pattern is the same at all times, which means that at any point (x,y) in the flow there is no change in the velocity with time, so that G ∂u =0 ∂t (1.4) In this case a new particle that passes point A at a new time t2 also pass through point B at a later time: t = t2 + (t1 – t0). Fig. 1.1 Definition of particle path and velocity vector. 8 Example 1.1: Example of non-stationary flow. Figure 1.2 shows a flow through a narrow duct where a block moves up and down between the walls of the duct. The discharge (coming from left towards right) is constant, but it is easy to see that a particle passing point A while the block is at the top will have a particle path different from that of the particle passing A while the block is at the low position. Fig. 1.2 Example of a non-stationary flow: the block is moving up and down in a narrow duct. Streamlines A streamline is a curve which has the velocity vector as tangent. In a nonstationary, or unsteady, flow the streamlines are in general different from the particle paths, cf. example 1.1, while the streamlines and the particle paths are identical in a stationary flow, cf. figure 1.1. Consider figure 1.1; at time t the particle has the position G x = ( x, y ) (1.5) G G and at time t + dt it will have the position x + dx = ( x + dx, y + dy ) . In the flow situation the streamline is therefore defined by G G dx = const × u or 9 dx dy = u v (1.6) where u is the velocity component in the x-direction and v is the velocity component in the y-direction, G u = (u , v) (1.7) Let us then calculate the velocity of a fluid particle. Consider once more figure 1.1, where the particle starting from A at t=t0 later passes through B. In the general non-stationary case we have seen, that a particle passing A at a later time t2 will not necessarily pass through B, cf. figure 1.3. This means that the particle paths given by (1.5) in general will be functions of time G G x = x (t ) = ( x(t ), y (t )) (1.8) The velocity is now found as the change in the position of a fluid particle per unit time, or Fig. 1.3 Non-stationary flow: particle paths through A at different times. 10 G G dx ⎛ d x(t ) d y (t ) ⎞ =⎜ u= , ⎟ dt ⎝ dt dt ⎠ (1.9) In all of the x-y-plane a velocity vector is therefore defined for every point: G u = (u ( x, y, t ), v( x, y, t )) (1.10) Example 1.2: The velocity field in a plane drain. Fig. 1.4 A plane drain. Figure 1.4 shows the flow which is formed when water is sucked out in point 0 with a discharge Q = Q(t), where the discharge may vary with time. All velocity vectors are directed towards 0. This particular flow is characterised as being axisymmetric, because the velocity field is not changed by an arbitrary rotation about an axis through O and normal to the plane of the paper. The velocity at a distance r from 0 can be calculated by using the continuity equation, which in this case expresses that the amount of water flowing in through the circle with radius r, shown in figure 1.4, must also be sucked out in point 0. 11 Because of the axi-symmetry it is much simpler to to work with polar coordinates rather than a cartesian coordinate system. Fig. 1.5 The velocity components in a polar coordinate system. As shown in figure 1.5 a point A can be characterised by either (x,y) or (r , θ) where x = r cos(θ) and (1.11) y = sin(θ) The velocity in the direction of the r-axis is called v r and the velocity normal to it is called v θ (see figure 1.5). In the flow of the plane drain the component v θ is seen to be zero due to symmetry, and v r is independent of θ due to symmetry. The continuity equation reads − v r 2πr = Q (1.12) or vr = − Q 2πr (1.13) which describes the velocity field. The velocity field can also be specified in a cartesian coordinate system with point 0 as origo: 12 u = v r cos(θ) − vθ sin(θ) = − Qx Q x =− 2π r 2π( x 2 + y 2 ) v = v r sin(θ) + vθ cos(θ) = − Qy Q y =− 2π r 2π( x 2 + y 2 ) and From the velocity field, equation (1.9), the acceleration field, defined by (1.3) can be found. As described a fluid particle follows the particle path given by (1.8). The acceleration of the particle is found as G G du ⎛ d (u ( x, y, t )) d (v( x, y, t )) ⎞ a= , =⎜ ⎟ dt ⎝ dt dt ⎠ (1.14) where the coordinates for the particle are given by (1.8), so that the acceleration in the direction of the x-axis is found as d d (u ( x, y, t )) = (u ( x(t ), y (t ), t )) dt dt ∂u dx(t ) ∂u dy (t ) ∂u = + + ∂x dt ∂y dt ∂t ax = (1.15) where the last part of the equation has been obtained by making a total differentiation. The term ∂u / ∂x × dx(t ) / dt has the following physical interpretation: dx(t ) / dt (= u ) describes how far the fluid particle has moved along its path in the xdirection per unit time. The quantity ∂u / ∂x describes how much the u-velocity varies in the x-direction at the location of the particle per unit length. The product of the two terms gives the change in u for the particle, because it is moved from the position x where the velocity is u to another position x + dx / dt with a new velocity u + du / dx . 13 Example 1.5: Flow through a constriction. Figure 1.6 shows a plane flow in a narrow conduit with a constriction. For simplicity it is assumed that the velocity u in the direction of the x-axis is constant across the conduit. Due to continuity (mass/volume conservation) it is easy to see that the velocity u2 is larger at the constriction than the velocity u1 at the original crosssection of the conduit. This means that, even if the fluid discharge Q is constant in time, a fluid particle must be accelerated from u1 to u2 as it moves into the constriction. This acceleration represents the term (∂u / ∂x)(dx / dt ) = (∂u / ∂t ) u , which is also know as an advective- or convective term, because it represents the acceleration due to changes in the flow direction. Fig. 1.6 Flow in a narrow, non-uniform conduit. Equation (1.15) can thus be written ax = u ∂u ∂u ∂u +v + ∂x ∂y ∂t (1.16) Similarly the acceleration in the y-direction is written ay = u ∂v ∂v ∂v +v + ∂x ∂y ∂t (1.17) The two first terms on the right hand side represents the convective terms. The last term ∂u / ∂t expresses that the flow in general is not stationary, so that in a fixed point 14 (x,y) there may also be changes with time so that fluid particle in a fixed point are also changing their velocity. Fig. 1.7 Flow forced by a moving piston. Example 1.6: Unsteady flow. Figure 1.7 shows the flow in a narrow conduit between two parallel walls. The flow is forced from left to right by a moving piston. The velocity profile shown to the right of the piston is taken to be constant over the depth. Due to the continuity (the fluid is incompressible) the fluid velocity must everywhere be identical to the velocity of the piston. This means that the convective terms are identical to zero. This always the case in a uniform flow, which is defined as a flow where the conditions are the same from one cross section to the next, i.e. there is no variation in the flow direction. The acceleration of the fluid in this case is du ∂u ∂ ds d 2 s = = = dt ∂t ∂t dt dt 2 (1.18) where s it the position of the piston at the time t. 1.2 The stresses in a fluid Let us consider a small unit cube of the fluid, cf. figure 1.8. At the left part of the figure the forces acting on the four sides of the cube are shown. These forces, F1 to F4, can be resolved into two components: normal forces acting in the direction normal area and tangential forces acting in the direction of the plane of the area. 15 The tangential force per unit area is known as shear stress. The definition of the direction for a positive shear stress is shown in figure 1.9. Fig. 1.8 Forces on a unit cube, resolved into normal and shear forces at the cube to the right. Fig. 1.9 Definition of shear stresses. Fig. 1.10 Normal stresses acting on a fluid element. 16 The notation in figure 1.9 is as follows: τ xy is the shear stress in the ydirection acting on a surface with the area vector (cf. example 1.7) in the direction of the x-axis. This can easily be extended to the more general case of a three-dimensional stress condition. Analogously the normal stress σ xx and σ yy are defined as the normal force per unit area, cf. figure 1.10. Example 1.7: Area vector Let us consider a volume X as shown in figure 1.11. This volume is limited by a surface with the area A. On this surface we can cut out an infinitesimal part of the area: dA. G This part area is characterised by the area vector dA , which has a magnitude equal to the part area (dA) and a direction normal to the surface at the part area and taken to be positive in the direction out of the volume X. Fig. 1.11 Definition of the area vector. 17 If consider the specific case of a plane infinitesimal cube with sides parallel to the coordinate axes, the area vectors per unit width (the width is in the G direction G normal to the plane of the drawing) are as shown in figure 1.12, where i and j are the unit vectors in the x- and y-direction respective Fig. 1.12 The area vectors of an infinitesimal cube. 18 Example 1.9: Pressure in a fluid at rest. In a fluid at rest the shear stresses are zero. This is because if τ ≠ 0 , it will in one way of the other set the fluid in motion, which intuitively can be seen from figure 1.9. In a fluid at rest we therefore have only normal stresses. Let us consider a triangular body, figure 1.13, where dx and dy are of equal length, forming an isosceles triangle. Figure 1.13 Forces acting on a triangle in a fluid at rest. The normal stress at the hypotenuse is simply called σ . The y-axis is pointing directly up, and the force of gravity has the magnitude ρ g × volume , where ρ is the density of the fluid and g is the acceleration of gravity. The vertical force balance reads − σ yy dx − 12 ρ g dx dy + σ 2dx × 2 =0 2 If we make the triangle shrink to become infinitely small (dx → 0, and dx = dy ) we get σ yy = σ (1.19) The horizontal force balance reads 19 − σ xx dy + σ 2dx × 2 =0 2 or, using dx = dy, we find σ xx = σ (1.20) It is thus seen that σ xx = σ yy . Following this principle it is easy to demonstrate that the normal stress is the same in all directions. In a fluid at motion σ xx is in general not equal to σ yy (but in practise often very close to being equal). A measure for the strength of the normal stresses is called pressure, which is defined as p = − 13 (σ xx + σ yy + σ zz ) (1.21) where the z-axis is normal to the plane of the drawing. The pressure is thus the mean value of the normal stresses in the three directions of the coordinate axes, but with opposite sign, so that the p is positive when the normal forces are directed into the fluid volume considered. Example 1.10: Proof that τ xy = τ yx (1.22) Determine the moment of the forces (torque) acting on the cube shown in figure 1.9 and neglect higher order terms: Moment = surface area × stress × arm = ⎡ ⎡ ∂τ yx ⎞ dx ⎤ ∂τ xy ⎞ dy ⎤ dy ⎛ dx ⎛ + ⎜⎜ τ yx + + ⎜⎜ τ xy + dx ⎟⎟ ⎥ dx ⎢τ xy dy ⎟⎟ ⎥ − dy ⎢τ yx ∂ 2 y 2 2 x ∂ ⎠ 2 ⎥⎦ ⎝ ⎠ ⎥⎦ ⎝ ⎣⎢ ⎣⎢ 20 After division by the factor dydx we have terms with and with out the factors dx or dy. The terms with the factors dx or dy are infinitely small and can be neglected and we end with 1.22. The moment must be zero because the moment of inertia of an infinitesimal cube around its axis normal to the x-y plane is smaller (of a higher order in dx and dy) than the moment of the forces. Example 1.11: The shear stress distribution in infinitely long narrow plane gap. Let us consider two horizontal plates of infinite extension. The lower is at rest while the upper is moving with the constant velocity v0 in its own plane. There is a fluid between the two plates. A condition for the fluid is that it adheres to the solid surfaces, so that the fluid particles adjacent to the fixed plate at the bottom has the velocity 0, while the fluid particle adjacent to the upper plate has the velocity vo. This means that there will be a flow in the gap between the two plates: the fluid will move with a certain velocity u parallel to direction of motion of the upper plate, which is taken to be the x-axis, cf. figure 1.14. Fig. 1.14 The velocity profile between two plates. The upper plate moves with the constant velocity vo for a very long time, and the induced flow is therefore steady or stationary, which means that 21 ∂u = 0 , u = u ( x, y ) ∂t (1.23) Furthermore the flow is uniform because the conditions do not vary from one cross section to the next. Mathematically this is written ∂u =0 ∂x (1.24) which means that u is only a function of y. Due to symmetry the vertical velocities are identical to zero everywhere. By applying eqs. (1.16) and (1.17) we find that the acceleration of any fluid particle in the flow is identical to zero. If we consider a unit cube in the flow the resulting force acting on it must be equal to zero, cf (1.1) and (1.2). The plates are placed horizontally, and the force of gravity acts in the negative y-direction. In figure 1.15 the force balance in the direction of the x-axis is considered. The conditions are fully uniform and the flow is driven only by the moving lid. The horizontal forces in the fluid are therefore not varying in the flow direction, and we can take ∂p =0 ∂x Fig. 1.15. The forces acting in the direction of the x-axis (horizontal). 22 The horizontal force balance the reads ⎡ ⎤ ∂ ⎢τ xy + (τ xy ) dy ⎥ dx − τ xy dx = 0 ∂y ⎣ ⎦ (1.25) or ∂τ xy ∂y =0 τ xy is often simply written as τ τ xy ≡ τ (1.26) Like the pressure p, τ cannot vary with x, and we get: ∂τ dτ = ∂y dy (1.27) τ = const (1.28) or The variation in the pressure in the vertical can be found by a projection of the forces on the direction of the y-axis. The forces are shown in figure 1.16. The two shear stresses are of the same magnitude and cancel each other out. The can be sen directly from the conditions of symmetry or mathematically from (1.28) together with (1.22). The vertical force balance therefore reads: − ∂p dy dx − ρg dx dy = 0 ∂y or p = −ρ g y + const (1.29) which is the hydrostatic pressure distribution. 23 Fig. 1.16 The forces acting in the direction of the y-axis (vertical). 24 2. Laminar flows In general a distinction is made between laminar and turbulent flows. If colour is injected into a flow by a thin tube it can be seen, that at small flow velocities the colour forms a thin thread, which keeps its identity and can be followed over a long distance in the flow and is only dispersed very slowly. This type of flow is known as a laminar flow and is found to occur for small values of the Reynolds number (cf. chapter 6) or at small flow velocities. The very slow dispersion of the thread is because the molecules in the fluid - in addition to the motion of the flow – are moving among each other, the so-called Brownian motions. Fig. 2.1 Laminar (a) and turbulent (b) flow. At larger flow velocities the nature of the flow changes. A large number of irregular eddies are superposed and added to the mean flow. The eddies are of very different magnitude and character. The thread of colour will be spread much faster and cannot be observed as an entity over a long stretch of the flow. This type of flow is known as turbulent flow and will be treated in chapter 6. 2.1 The constitutive equations By a constitutive equation we mean an equation giving a relation between the forces and deformations of the fluid. Let us continue from example 1.11, where we 25 considered the flow in a closed channel between two plates and the upper plate moved forward in its own plane while the lower plate was stationary. As described in the example there is a variation of the velocity u across the channel because of the no-slip boundary conditions due to the adhesion u=0 u = v0 for for y=0 y=a (2.1) where a is the width of the channel. This is approximately the you find (disregarding the small effects of the curvature) between two concentric cylinders with slightly different cylinders, cf. figure 2.2. Fig. 2.2 Experimental set-up for determination of the viscosity of fluids. The void between the two cylinders is filled with fluid and the outer cylinder is made to rotate with the angular velocity ω , so that the velocity along the outer cylinder becomes v0 = ω (r + a) (2.2) The inner cylinder may rotate as well, but is restricted by a spring. The tension in the spring increases as the outer cylinder rotates because the shear stress τ is transferred through the fluid to the inner cylinder. For given values of a and r ordinary fluids will show a linear relation between the speed of rotation of the outer cylinder ω (or vo) and the deformation of the spring, which is a measure of the force τ 2πr acting on the inner cylinder, cf. figure 2.3. for different fluids, for example air and water, the deformation will be quite different for the same vo (air will give much less deformation of the spring). 26 Fig. 2.3 The relation between the velocity of the outer cylinder and the shear stress acting on the inner cylinder for given a and r. If in the other hand a is changed for the same fluid while maintaining r, the deformation of the spring will turn out to be inversely proportional to the width of the gap for the same velocity vo. From these two experiments we can conclude that τ= v0 μ a (2.3) where μ is found to be different for different fluids. The ratio vo/a is a measure of the velocity gradient of the flow du v0 ~ dy a (2.4) where the y-axis is in the radial direction (normal to the gap). Normally (2.2) and (2.4) are written as du τ=μ (2.5) dy which is known as Newton's formula. The quantity μ is the dynamic viscosity of the fluid and is larger the more thick and viscous the fluid is. In addition to μ it is useful to introduce the kinematic viscosity ν defined by μ = ρν (2.6) where ρ is the density of the fluid. 27 The relation between vo and τ for different widths of the gap. Fig. 2.4 In a general two-dimensional flow (2.5) reads ⎛ ∂u ∂v ⎞ τ xy = μ⎜⎜ + ⎟⎟ ⎝ ∂y ∂x ⎠ (2.7) Example 2.1 The velocity distribution in a closed channel between two plane parallel walls. In (2.4) it was assumed that the velocity gradient was constant over the gap. This is in fact the case as described in example 1.11 du τ = τ0 = μ (2.8) dy which implies u= τ0 y+k μ (2.9) By use of the boundary conditions (2.1) we get τ 0 v0 = a μ (2.10) and k = 0, i.e. u = v0 y a (2.11) 28 Example 2.2 The power exerted by the outer cylinder. We consider again the experiment illustrated in figure 2.2 where the outer cylinder is rotated, while the inner one I kept stationary by the tension of the spring. The outer cylinder may be driven by a motor, and the question addressed is how much power the motor must supply to rotate it. There are 2 different types of energy loss, a) loss due friction in bearings and transmission and b) loss due to friction I the fluid. Only b) is considered here. The force acting on the outer cylinder from the fluid is F = τ 0 2π ( r + a ) (2.12) or, from (2.10) F= v0 μ 2π ( r + a ) a (2.13) The power is defined as the work performed per time unit, where the work is force time path. The total power exerted E is therefore force times velocity (r + a) (2.14) E = Fv0 = v02 μ 2π a Calculation example: Consider a flow with vo = 0.10 m/s, a = 2 mm and water as the fluid. For water the viscosity is ν = 10-6 m2/s at the temperature 10O C. The power is found to be ⎛ 0.102 m ⎞ ⎟ E = (0.1 m / s ) 2 ⋅ 10 3 kg / m 3 ⋅ 10 −6 m 2 / s ⋅ 2π ⋅ ⎜⎜ ⎟ 0 . 002 m ⎠ ⎝ or E = 0.0032 kg m s −3 = 0.0032Watt / m width The 'm width' is included in the unit for power because the power exerted is proportional to the width of the gap (or height of the cylinder) taken normal to the plane of the drawing in figure 2.2. This transfer of energy to the fluid between the two cylinders must be spent in the fluid. Since the flow is stationary it does not gain any kinetic or potential energy and the power supplied must be dissipated to heat. 29 Fig. 2.5 The distribution of τ , u and the energy dissipation in the gap between the two cylinders. From figure 2.5 it is seen that the term ∂ ∂u (τ u ) = τ ∂y ∂y (2.15) is constant across the gap. This must represent the power per unit volume, which therefore is equal to the loss of energy to heat (the dissipation ε ). In this example the dissipation is constant over the depth and is given as ⎛ ∂u ⎞ ∂u ε=τ = μ⎜⎜ ⎟⎟ ∂y ⎝ ∂y ⎠ 2 (2.16) 2.2 The deformation of fluid elements in laminar flows In section 1.1 we considered the paths of fluid particle in a flow. I this section we shall, investigate the deformation of a small volume of fluid. The starting point is the flow between two parallel walls introduced in example 1.1. The velocity profile is linear (cf. figure 2.5) and is given by u = v0 y a (2.17) where vo is the velocity of the upper wall and a is the width of the gap. The vertical velocities are identical equal to zero. It is thus seen that the streamlines are straight lines parallel to the walls. In figure 2.6 an element ABCD is shown. At time t = 0 the element is quadratic with sides parallel to the x- and y-axis. 30 Fig. 2.6 The deformation of a quadrate in a flow with a linear velocity profile. First of all it is noted that the centre of gravity of the element has moved in the direction of the flow. This part of the motion is called translation. Next it can be seen that point A has moved closer to point C while the distance between points B and D is increased. This is called the deformation, cf. figure 2.7a. Finally the deformed quadrate as a whole has been given a small rotation in the clockwise direction, cf. figure 2.7b Fig. 2.7 a: The deformation of a quadrate. b: The rotation of the deformed quadrate. In a more general flow situation the deformation can be described as follows: In figure 2.8 it is shown how a quadrate ABCD is deformed to A'B'C'D' during the short time interval dt. The latter is also shown in figure 2.8 with point D' placed over point D, so that the translation (u dt , v dt ) has been compensated for. The rotation, which is taken to be positive in the counterclockwise direction, is now determined as a sort of average of the change in angle per unit time G (α + α 2 ) rot v = 1 dt (2.18) 31 The angle α1 is determined as ∂v dt dx ∂ x α1 = ∂u dt dx dx + ∂x or, using that (1 + α) −1 ≈ 1 − α α1 = Fig. 2.8 (2.19) for ∂v ⎛ ∂u ⎞ dt ⎜1 − dt ⎟ ∂t ⎝ ∂x ⎠ α << 1 (2.19) The general deformation of a unite quadrate in a plane flow. Since the term dt is very small, the term with dt2 is one order smaller and can be neglected. Equation (2.19) then becomes α1 = ∂v dt ∂t (2.20) 32 The angle α 2 is determined similarly, so the result of (2.18) is G ⎛ ∂v ∂u ⎞ rot v = ⎜⎜ − ⎟⎟ = 2 ω3 ⎝ ∂x ∂y ⎠ (2.21) The rotation or vorticity is commonly written as 2ω and in a three-dimensional flow it will become a vector, where (2.21) designates the component of this vector 2ω3 in the direction normal to the x-y-plane. The deformation or strain is determined in an analogous way as G ∂v ∂u def v = (α1 − α 2 ) / dt = + ∂x ∂y (2.22) The deformation of a fluid element is primarily the result of the shear stresses. Newton's formula (2.7) indicates the relation between the shear stresses and the deformation. 2.3 The equation of continuity The continuity equation for a fluid expresses the conservation of mass. For a non-compressible fluid the density ρ is constant and the conservation of mass will in this case be the same as conservation of volume. For a plane flow the continuity equation will in this way express that the areas of ABCD and A'B'C'D' in figure 2.8 are identical. The simplest way to derive the continuity equation mathematically is by considering the flow of water in and out of a small unit cube, which is placed in a fixed position. Figure 2.9 shows the plane case. The volume flux per time through the side AD is u dy, because the flow component v is parallel with AD and does not contribute to the transport across AD. The remaining contributions for the three other sides are given in the figure. The continuity equation now expresses that the total flow into the area through AD and DC must correspond to the flux out through AB and BC. Hereby we find ∂u ∂v + =0 ∂x ∂y (2.23) 33 Fig. 2.9 The flux of fluid through a unit element in a fixed position. In the general three-dimensional case the continuity equation is written ∂u ∂v ∂w + + =0 ∂x ∂y ∂z (2.24) In mathematical terms the continuity equation expresses that the divergence of the velocity field is 0. Example 2.3 The continuity equation for a plane flow in polar coordinates In many cases flow is described most easily by use of polar coordinates, cf. figure 1.4 illustrating a plane drain. The polar velocity components are shown in figure 1.5. To derive the continuity equation in polar coordinates it is convenient to consider the element shown in figure 2.10. AB and DC are here sections of a circle with O as its centre, while AD and BC are parts of straight lines passing through O. The continuity equation now reads, cf. figure 2.10 ∂v ∂ (v r r ) + θ = 0 (2.25) ∂r ∂θ or vr + r ∂v r ∂vθ + =0 ∂r ∂θ (2.26) 34 Fig. 2.10 The flux of fluid in an element in polar coordinates. The first term, which makes the equation differ from the Cartesian version in (2.23), is included because the length of the circular sections increases in the r-direction. 2.4 The flow equations (the Navier-Stokes equations) We can now formulate Newton's second law (1.2) for a fluid particle by use of the forces due to the pressure and the shear stresses. Figure shows a unit volume with sides parallel to the x- and y- axes. The x-axis forms the angle β with horizontal. Newton's second law applied in the x-direction now reads ρ ax = ∂τ ∂p − + ρ g sin β ∂y ∂x (2.27) Similarly we find for the y-direction ρay = ∂τ ∂p − + ρ g cos β ∂x ∂y (2.28) 35 Fig. 2.11 Forces acting in the x-direction on a unit quadrate. The two components of the force of gravity in the x- and y-directions can be written g x = g sin β and g y = g cos β (2.29) By use of (2.29), Newton's formula (2.7) and the kinematic equations (1.16) and (1.17) equations (2.23) and (2.24) can formulated as the general flow equations, know as the Navier-Stokes equations. The derivation of the completely correct equations requires a more accurate description of the normal stresses in the fluid than the simple application of the averaged pressure term p, cf. (1.21). The formal derivation is given in Appendix 1. The complete Navier-Stokes equations read ρ ⎛ ∂ 2 u ∂ 2 u ⎞ ∂p ⎛ ∂u du ∂u ∂u ⎞ + ρ gx = ρ⎜⎜ + u + v ⎟⎟ = μ⎜⎜ 2 + 2 ⎟⎟ − dt ∂x ∂y ⎠ ∂y ⎠ ∂x ⎝ ∂t ⎝ ∂x (2.30) ρ ⎛ ∂ 2 v ∂ 2 v ⎞ ∂p ⎛ ∂v dv ∂v ∂v ⎞ +ρgy = ρ⎜⎜ + u + v ⎟⎟ = μ⎜⎜ 2 + 2 ⎟⎟ − dt ∂ y ∂x ∂y ⎠ x y ∂ ∂ ⎝ ∂t ⎠ ⎝ (2.31) 2.5 Other examples of laminar flows Poiseuille flow: The plane Poiseuille flow is the flow between two parallel plates, where the plates are fixed and the flow is driven by a pressure gradient (for example when a fluid flows from one reservoir to another). Due to symmetry it is seen that the flow in the y-direction is 0. As the flow is uniform and stationary the velocity u is only a function of y 36 u = u( y) v=0 (2.32) The Navier-Stokes equations (2.30) and (2.31) now are reduced to μ d 2 u ∂p − + ρ g sin β dy 2 ∂x (2.33) and ∂p + ρ g cos β = 0 ∂y Fig. 2.12 (2.34) The flow between two parallel stationary walls. It is often found to be convenient to introduce the excess pressure p+, defined as p + = p + ρg z (2.35) where z is a vertical coordinate relative to a horizontal base level, cf. figure 2.11. We have therefore ∂p + ∂p = + ρg cos β ∂y ∂y and (2.36) + ∂p ∂p = − ρg sin β ∂x ∂x Using this (2.33) and (2.34) become ∂ 2 u ∂p μ 2 = ∂x ∂y (2.37) 37 and ∂p + =0 ∂y (2.38) Physically the pressure p+ is seen from (2.35) to be the pressure in excess of the hydrostatic pressure. The hydrostatic pressure is not of interest for the description of the flow because it does not contribute to the driving forces (the hydrostatic pressure corresponds to the conditions in stagnant water). It is thus the pressure in excess of the static, which can put a fluid in motion. Equation (2.38) expresses that the pressure p+ is constant across the gap p + = p x (x) (2.39) ∂p + = const ∂x (2.40) so that since ∂p + / ∂x must be the same for all values of x (uniform flow conditions). The constant in (2.40) may be written − γI , where I is called the hydraulic gradient. Equation (2.37) then can be written μ ∂ 2u = − γI ∂y 2 or ∂ 2u Ig =− 2 ν ∂y (2.41) which can be integrated twice to give u=− Ig y 2 + C1 y + C 2 ν 2 (2.42) Where the constants C1 and C2 can be found from the two boundary conditions u = 0 for y= 0 (2.43) u = 0 for y = a (the width of the gap) from which we get 38 u= Ig (a − y ) y 2ν (2.44) giving a parabolic velocity profile as shown in figure 2.12. Unsteady flow, the wave boundary layer: In this example we consider the flow which occurs when a plate is given a harmonic oscillation in its own plane. The fluid above the plate is taken to be of infinite extent. The velocity of the plate is given as t⎞ ⎛ u = u 0 cos⎜ 2π ⎟ ⎝ T⎠ (2.45) u = u 0 cos(ωt ) , ω = 2π / T (2.46) or Here T is known as the wave period since the motion of the plate is seen (cf. (2.45)) to be periodic, repeating itself after the time T, 2T, 3T etc. The quantity ω is known as the cyclic frequency or the angular velocity. The motion of the plate in its own plane cannot cause any pressure gradients so ∂p / ∂x = 0 . Further it is seen that ∂u / ∂x = 0 and v = 0. The flow equation (2.30) is therefore reduced to + ρ ∂ 2u ∂u =μ 2 ∂t ∂y (2.47) where the hydrostatic pressure has been neglected as in the previous example. Fig. 2.13 The forces acting on a unit quadrate above an oscillating bed and the resulting velocity profile. 39 Equation (2.47) expresses that the fluid is accelerated by the shear stresses, as indicated in figure 2.13. When the plate oscillates with the periodically (with the period T) a natural solution would be that the velocity u at a given distance y from the bed also varies periodically in time. The period must be the same everywhere, but the fluid can be out of phase with the motion of the plate. The solution can therefore be written u = f ( y ) cos(ωt ) + g ( y ) sin(ωt ) (2.48) A simpler way of treating the problem is by the use of complex numbers in which case (2.48) can be written u = h( y ) exp(i ωt ) (2.49) which written in full reads u = [hr cos(ωt ) − hi sin(ωt )] + i[hr sin(ωt ) + hi cos(ωt )] (2.50) where hr is the real part of h and hi is the imaginary part of h. When using complex number two equations emerge, one from the real and one from the imaginary. Normally only the real part of the solution is taken to have physical significance. Insertion of (2.49) into (2.47) gives d 2h ρ iω h = μ 2 dy (2.51) having the solution ⎛ iω ⎞ ⎛ iω ⎞ h = C1 exp⎜⎜ y ⎟⎟ + C 2 exp⎜⎜ − y ⎟⎟ ν ν ⎝ ⎠ ⎝ ⎠ (2.52) Using that i = (1 + i ) / 2 it is seen that the first term (with C1) will grow to infinity away from the wall. As the boundary condition must be that the motion is limited infinitely far away from the bed, we must have C1 = 0. Equation (2.49) then becomes ⎛⎛ ⎛ ω ⎞ ω ⎞ ⎞⎟ u = C 2 exp⎜⎜ − y ⎟⎟ exp⎜ i⎜⎜ ωt − y ⎟⎟ (2.53) ⎜ ⎟ ν ν 2 2 ⎝ ⎠ ⎠⎠ ⎝⎝ C2 is found from the condition expressing that the fluid must follow the plate, given in (2.46) for y = 0. This gives C2 = uo. The term δ= 2ν ω (2.54) 40 is known as the Stokes length and is a measure of how far from the plate its motion can be felt. For ν = 10-6 m2/s (i.e. water at 200) and T = 5 s we find δ = 1.26 mm, a quite small quantity. Equation (2.53) describes how u decreases exponentially from the wall (while it oscillates) with the length scale δ , cf. figure 2.13. The shear stress can be found by application of Newton's formula ⎛⎛ ∂u νu 0 ⎧ τ y ⎞⎞ ⎛ y⎞ = =ν ⎨− exp⎜ − ⎟ exp⎜⎜ i⎜ ωt − ⎟ ⎟⎟ δ ⎠⎠ δ ⎩ ∂y ρ ⎝ δ⎠ ⎝⎝ (2.55) ⎛⎛ y ⎞ ⎞⎫ ⎛ y⎞ − i exp⎜ − ⎟ exp⎜⎜ i⎜ ωt − ⎟ ⎟⎟⎬ δ ⎠ ⎠⎭ ⎝ δ⎠ ⎝⎝ or τ0 νu = − 0 [cos(ωt ) − sin (ωt )] ρ δ (2.56) at the bed, where y = 0. It is seen that τ 0 has a phase shift of 450 in the dimensionless time ωt relative to the velocity of the plate. The wall shear stress does not have its maximum at the same time as the velocity because we have an unsteady flow and fluid has to be accelerated to follow the oscillatory motion. 41 3. Potential flow Potential flow is a particular type of flow, which can be described in mathematical terms by application of the so-called potential φ . Physically this means that the shear stresses can be neglected, i.e. the flow can be described as an equilibrium between inertial forces (mass times acceleration), pressure forces and gravity. This is also known as an ideal flow or an ideal fluid. One consequence of neglecting the shear stresses is that the no-slip boundary condition at a solid surface cannot be fulfilled. If we consider the wave boundary layer, cf. chapter 2, equation 2.45 will degenerate to ρ du =0 dt (3.1) which with the boundary condition u = 0 for y → ∞ has the solution u = 0. This describes actually the flow over the entire area except for the very thin layer near the plate, where the flow occurs because the water adheres to the oscillating plate. This illustrates the general strength of the theory for ideal fluids: in a relatively simple way it can describe the flow in the main part of the domain, so that you only needs to take the effect of the shear stresses into account in thin layers along the solid boundaries. These thin layers are known as boundary layers and are treated in chapter 5 and 7. 3.1 Definition of the potential The potential is a scalar, called φ . It is related to the velocity field through the relation ∂ϕ (3.2) u=− ∂x and v=− ∂φ ∂y (3.3) There are a number of mathematical condition associated to the existence of a potential: it can be shown that it is a necessary and sufficiently condition that the rotation of the velocity field is zero in all points of the flow domain, i.e. the flow must be irrotational. 44 Example 3.1 It is seen from the definition of rotation, equation (2.21), that this is a necessary condition G ∂v ∂u ∂ ⎛ ∂φ ⎞ ∂ ⎛ ∂φ ⎞ − = − ⎜⎜ ⎟⎟ + ⎜ ⎟ = 0 2 rot v = ∂x ∂y ∂x ⎝ ∂y ⎠ ∂y ⎝ ∂x ⎠ (3.4) G That rot v = 0 is also a sufficient condition, is for example shown in [1]. 3.2 The Bernoulli equation As mentioned shear streses are neglected in ideal fluids, which means that the viscosity is taken to be zero. I this way the Navier-Stokes equation (2.30) and (2.31) turns into ⎛ ∂u ∂u ∂p ∂u ⎞ ρ⎜⎜ + u + v ⎟⎟ = − + ρg x ∂x ∂x ∂y ⎠ ⎝ ∂t (3.5) ⎛ ∂v ∂v ∂v ⎞ ∂p ρ⎜⎜ + u + v ⎟⎟ = − + ρg y ∂x ∂y ⎠ ∂y ⎝ ∂t (3.6) and These two equations are known as the Euler equations. For a potential flow the rotation is also known to be zero, so that ∂v ∂u = ∂x ∂y (3.7) cf. equation (3.4). Inserting this into (3.5) and (3.6) gives ∂p ⎛ ∂u ⎞ ∂ ⎛ 1 ⎞ ρ⎜ ⎟ + ⎜ V 2 ⎟ = − + ρg x ∂x ⎝ ∂t ⎠ ∂x ⎝ 2 ⎠ (3.8) ∂p ⎛ ∂v ⎞ ∂ ⎛ 1 ⎞ ρ⎜ ⎟ + ⎜ V 2 ⎟ = − + ρg y ∂y ⎝ ∂t ⎠ ∂y ⎝ 2 ⎠ (3.9) and where V 2 = u2 + v2 (3.10) 45 If we apply (3.2) and (3.3) in (3.8) and (3.9) and integrates with respect to x and y respectively, both equation give z+ p V 2 1 ∂φ = C (t ) + − γ 2 g g ∂t (3.11) where z is a vertical coordinate (cf. figure 2.11 and equation (2.27)). Equation (3.11) is the general version of Bernoulli’s equation. For a stationary flow ( ∂ / ∂t = 0 ) it turns into z+ p V2 + = const γ 2g (3.12) 3.3 The Laplace equation While laminar (and turbulent) flows are solved by use of the dynamic flow equation and the boundary conditions, a potential flow is determined by solving the continuity equation with the corresponding boundary conditions. The continuity equation (2.21) can by applying (3.2) and (3.3) be written ∂ ⎛ ∂φ ⎞ ∂ ⎛ ∂ϕ ⎞ ⎜− ⎟ + ⎜− ⎟ = 0 ∂x ⎝ ∂x ⎠ ∂y ⎜⎝ ∂y ⎟⎠ (3.13) Δφ = 0 (3.14) or Where Δ is the Laplace operator ∂2 ∂2 Δ= 2 + 2 ∂x ∂y (3.15) and (3.14) is know as the Laplace equation, which is seen simply to be the continuity equation. 3.4 An example of a potential flow: flow over a wavy bed In this (and the following) chapter we consider potential flows, where the flow is composed of a simple basic flow, on which we superpose a small additional flow field, also known as a perturbation. In this example the flow over a sinusoidal bed (or a ripple bed) is considered. The technique applied is as follows: a: First the basic flow is determined. In this case it is the flow over a horizontal bed. The Laplace equation with boundary conditions are applied. 46 b: The disturbance to the basic flow due to the bed undulations is then determined. This is also done by application of the Laplace equation and boundary conditions. The boundary conditions expanded in Taylor series and it is used that the perturbation is small, so that only the most significant terms are maintained. Fig. 3.1 The basic flow and the perturbed flow. Figure 3.1a shows the basic flow, which is simply a uniform flow over a horizontal bed. The Laplace equation reads ∂ 2φ ∂ 2φ + =0 ∂x 2 ∂y 2 (3.16) Since the flow is uniform we have that ∂u / ∂x = 0 , or by use of (3.2) ∂ 2φ =0 ∂x 2 (3.17) then (3.16) turns into ∂ 2φ =0 ∂y 2 (3.18) Integration of (3.17) gives φ = f1 ( y ) x + f 2 ( y ) (3.19) while the integration of (3.18) similarly gives 47 φ = g1 ( x ) y + g 2 ( x ) (3.20) Combination of (3.19) and (3.20) gives φ = k1 xy + k 2 x + k 3 y + k 4 (3.21) The boundary condition is: the vertical velocity is zero at y = 0 (no flux through the bed), i.e. ∂φ =0 ∂y for y=0 (3.22) k1 x + k 3 = 0 for y=0 (3.23) v=− or which implies that both k1 and k3 must be 0, since (3.23) must hold for all values of x. Adding a constant to potential gives no change in the velocity field, cf. (3.2) and (3.3). The constant k4 can therefore have any value and is arbitrarily set to zero. The potential is then φ = k2 x (3.24) From (3.24) we find u=− ∂φ = −k 2 ∂x (3.25) v=− ∂φ =0 ∂y (3.26) and it is seen that –k2 is equal to the undisturbed velocity V in the flow direction, and the potential is therefore given as φ 0 = −Vx (3.27) where the index 0 shows that this is the basic (plane bed) flow. The velocity is constant and equal to V in the x-direction over the entire water depth D, which isbecause the shear stresses have been neglected so that the velocity cannot become 0 at the bed. 48 We are now at point b in the procedure, and the bed is no longer plane, but has now superposed a small sinusoidal perturbation. The bed level is given by h = h0 sin( kx) (3.28) where ho is the amplitude of the bed undulation and k is the so-called wave number. It is seen from (3.28) that h jhas the same value for kx and for kx + 2π , we therefore find that kL = 2π or k= 2π L (3.29) where L is the wave length of the bed undulation. This perturbation of the initially plane bed introduces a change in the potential ~ from φ 0 to φ . The change is called φ so that ~ φ = φ0 + φ (3.30) ~ ~ For h0 → 0 we see that φ → 0 , and therefore the order of magnitude of φ is ~ φ = O (h0 ) Normally the amplitude is described by a dimensional term like ho/D or ho/L. In the following we choose ~ φ = O(h0 / D) (3.31) where the quantity ε= h0 D (3.32) is a small quantity, ε << 1 . This will be used in the following when determining the ~ boundary conditions. First φ is found by solving the Laplace equation, which gives ~ ~ ~ Δφ = 0 = Δ ( φ 0 + φ ) = Δφ 0 + Δ φ = Δ φ (3.33) or ~ Δφ = 0 Since the bed oscillation oscillates periodically with the wave number k in the ~ x-direction it is reasonable to assume that φ behaves similarly. Therefore a qualified guess for the solution to (3.33) could be 49 ~ φ = f ( y )[cos(kx) + a sin(kx)] (3.34) which, when inserted into (3.33) gives f ′′ − k 2 f = 0 (3.35) f = c1 exp( ky ) + c 2 exp( −ky ) (3.36) or The two constants c1 and c2 are found from the two boundary conditions, one at the water surface and one at the bed. 1) Boundary condition at the bed: As we use potential theory we cannot require that u = 0 (the water adheres to the solid boundary) at the bed. But no water can pass the bed, which is impermeable. This can be expressed by saying that the velocity Fig. 3.2 Kinematic boundary condition at the bed. vector must be parallel to the bed. As seen from figure 3.2 this means that v ∂h = u ∂x for y=h (3.37) The vertical velocity v is found as ~ ⎛ ∂φ ⎞ ⎛ ∂φ ⎞ v = ⎜⎜ − ⎟⎟ = ⎜⎜ − ⎟⎟ ⎝ ∂y ⎠ y =h ⎝ ∂y ⎠ y = h ~ ⎛ ∂ ⎛ ∂~ ⎛ ∂φ ⎞ φ ⎞⎞ ⎜ ⎟ ≈ ⎜− ⎟ + h⎜ ⎜⎜ − ⎟⎟ ⎟ + O (ε 3 ) ⎜ ⎟ ⎝ ∂y ⎠ y =0 ⎝ ∂y ⎝ ∂y ⎠ ⎠ y =0 (3.38) In (3.38) expansion in a Taylor series has been used to express the boundary condition at y = h by quantities given at y = 0. The two first terms at the right hand side of (3.38) has the following orders of magnitude 50 ~ ⎛ ∂φ ⎞ ⎜− ⎟ = O(h0 / D) ⎜ ∂y ⎟ ⎠ y =0 ⎝ (3.39) cf. (3.31) and ~ h ∂ ⎛ ∂φ ⎞ ⎜− ⎟ = O((h0 / D) 2 ) ⎜ ⎟ D ∂y ⎝ ∂y ⎠ y =0 (3.40) which shows that the last term in (3.38) consists of terms with ho/D in the power 3 or higher. The horizontal velocity can in a similar way be written ~ ~ ⎛ ∂φ ⎞ ⎛ ∂φ ⎞ ⎛ ∂φ ⎞ = V − ⎜⎜ ⎟⎟ = V − ⎜⎜ ⎟⎟ u = ⎜⎜ − ⎟⎟ + O((h0 / D) 2 ) ⎝ ∂y ⎠ y =h ⎝ ∂y ⎠ y =h ⎝ ∂y ⎠ y =0 (3.41) Equation (3.37) can now be written ~ ⎛ ∂φ ⎞ ∂h ⎜− ⎟ + O((h0 / D) 2 ) = [V + O(h0 / D)] ⎜ ∂y ⎟ ∂x ⎝ ⎠ y =0 where ∂h = O(h0 / D) ∂x (3.42) (3.43) When it is assumed that ho/D << 1 we are allowed only to keep terms which are proportional to ho/D and neglect all terms of higher order, which means terms that contain ho/D raised to a power of 2 or higher. In principle (3.37) can be written as 2 3 h ⎛h ⎞ ⎛h ⎞ A 0 + B⎜ 0 ⎟ + C ⎜ 0 ⎟ + ..... = 0 D ⎝D⎠ ⎝D⎠ (3.44) In a linear theory it is only required that A = 0, which in our case gives ~ ⎡ ∂φ ⎤ ∂h =V ⎢− ⎥ ∂x ⎣ ∂y ⎦ y =0 (4.45) [In a higher order theory additional requirements are to be met: 2. order theory: B = 0, 3. order theory: C = 0 etc.]. By application of (3.28) and (3.36) equation (3.45) can now be written (−c1 + c 2 )k [cos(kx) + a sin(kx)] = V h0 k cos(kx) (3.46) from which we get 51 a=0 and c 2 − c1 = h0 V (3.47) 2) Boundary condition at the surface Flow with a free surface are complicated to describe because the shape of the surface (and therefore the geometry of the flow) is not know in advance (in contrast to for example flow in a pipeline). In general the deviation of the surface from the mean water surface (cf. figure 3.1) can be written as a Fourier expansion ∞ η = ∑ [a n cos(nkx) + bn sin(nkx)] (3.48) n =0 The kinematic boundary condition at the surface can be expressed as, cf. (3.37) v ∂η = u ∂x for y = D+η (3.49) This can (similar to (3.37)) be linearised to give v y=D = V ∂η ∂x or − k (c1 exp(kD) − c 2 exp(−kD)) cos(kx) = ∞ V ∑ [a n cos(nkx) + bn sin( nkx)] (3.50) n =0 where it has been used that a = 0 in (3.34). It follows therefore that only b1 ≠ 0 , while all the other coefficients (an and bn) are identically equal to 0. Equation (3.50) therefore is reduced to − V b1 = c1 exp(kD) − c 2 exp(− kD) (3.51) One additional boundary condition is needed for the surface, because we have only 2 equations to determine the three unknowns c1, c2 and b1. The last boundary condition expresses that the pressure at the surface is constant and equal to the atmospheric pressure. p = const for y = D+η (3.52) This is know as a dynamic boundary condition, because it is using information on the forces – in contrast to a kinematic boundary condition, which is related to the direction of motion only. 52 The relevant equation for the dynamics of the potential flow is the Bernoulli equation (3.12), which in combination with ((3.52) gives 1 2 (u + v 2 ) + gη = const for y = D + η (3.53) 2 As before the boundary condition are to be linearised. We shall identify and maintain only the most significant terms of the square velocities u2 and v2. First we consider u2 ~ 2 ~ ⎡⎛ h ⎞ 2 ⎤ ⎛ ∂φ ⎞ ⎛ ∂φ ⎞ 2 ⎟ ⎟ ⎜ ⎜ u = ⎜V − ⎟ = V − 2V ⎜ ⎟ + O ⎢⎜ 0 ⎟ ⎥ ∂x ⎠ ⎢⎣⎝ D ⎠ ⎦⎥ ⎝ ∂x ⎠ y = D ⎝ 2 (3.54) where the term V2is constant and may be included in the constant in (3.53). Similarly we have ~ 2 ⎡⎛ h ⎞ 2 ⎤ ⎛ ∂φ ⎞ v = ⎜⎜ − ⎟⎟ = O ⎢⎜ 0 ⎟ ⎥ ⎢⎣⎝ D ⎠ ⎥⎦ ⎝ ∂x ⎠ y = D + η 2 (3.55) Hereby (3.53) turns into ~ ⎛ ∂φ ⎞ + gη = 0 − V ⎜⎜ ⎟⎟ ⎝ ∂x ⎠ y = D (3.56) or V k [c1 exp(kD) + c 2 exp(−kD)]sin( kx) + gb1 sin( kx) = 0 from which we get c1 exp(kD) + c 2 exp(−kD) = − g b1 Vk (3.57) From the equations (3.47), (3.51) and (3.57) the coefficients c1, c2 and b1 can be determined. The coefficient b1 is thus found to be described by b1 ⎡ 1 ⎤ sinh(kD)⎥ = ⎢cosh(kD) − 2 h0 ⎣ kDF ⎦ −1 (3.58) where F= V gD (3.59) is the Froude number. 53 The variation in the surface elevation with the flow velocity can now be found from the expression η= b1 h h0 (3.60) where b1 is given by (3.59). It is seen that the surface is either in phase or 180 degree out of phase with the bed oscillation. For small values of F (or for small flow velocities at a given water depth) we find b1 kD → −F 2 sinh(kD) h0 (3.61) which means that the water surface is plane (almost as for a non-moving fluid). For F2 < tanh(kD) = Fc2 kD (3.62) b1/ho will be negative as shown in figure 3.3b, which means that the surface is in counter-phase (180 degree out of phase) with the bed oscillation until the critical Froude number Fc is exceeded. For larger values of the Froude number the surface will be in phase with with the bed as shown in figure 3.3c. Theoretically b1 → ∞ for F → Fc , which is because the present theory is linear and terms like b12 and higher powers have been neglected. These terms become important as F approaches Fc. Shallow water theory For small values of kD (that is for very long oscillations of the bed) the previous calculations can be simplified considerably. In this case the pressure can locally be taken to vary hydrostatically over the vertical, increasing from zero as you move down from the water surface p = ρg ( D − y + η) (3.63) 54 Fig. 3.3 The variation of the water surface with the flow velocity. This different from the general case considered until now, where the pressure cannot be taken to be hydrostatic because of the curvature of the streamlines, cf. figure 3.4. For very weak curvature the pressure variations indicated in figure 3.4 can be neglected. Equation (3.63) implies that the horizontal velocity u is constant over a given vertical, because the horizontal pressure gradient, which accelerates the flow, is the same at all levels for a given x u = U = U (x) (3.64) Fig. 3.4 The pressure variation associated with curvature of the streamlines. 55 The flow equation reads ρ dU ∂p =− dt ∂x (3.65) cf. (2.28). Since the flow is stationary and the pressure is given by (3.63) equation (3.65) can be written U dU dη +g dx dx (3.66) The continuity equation can be formulated for the entire discharge across a vertical U ( D + η − h) = Q (3.67) which simply expresses that the discharge Q is constant in the flow direction. As done for the general case it is assumed that the bed undulation is small and causes only a small perturbation to flow. The velocity can then be written U ( x) = U 0 + u~ ( x) (3.68) where u~ is the perturbation caused by the bed oscillation. When linearised (3.66) and (3.67) now reads du~ dη (3.69) U0 +g =0 dx dx and U 0 (η − h) + u~D = const (3.70) By differentiating (3.70) with respect to x the quantity u~ can be eliminated from (3.69) which gives U 02 d (η − h) dη − =0 gD dx dx (3.71) ⎛ F2 ⎞ ⎟ η = h⎜⎜ 2 ⎟ ⎝ F − 1⎠ (3.72) or equation (3.72) is the shallow water version of (3.58). As seen the change in the phase of the water surface now occurs for a critical Froude number of Fc = 1 instead of the general expression (3.62), where (cf. figure 3.5) Fc is always less than 1. 56 Fig. 3.5 The variation in Fc with kD. 57 4. Progessive sinusoidal waves Small progressive water waves is an example, where potential flow theory can give a very good representation of nature. As it is also an important subject in coastal and offshore engineering a whole chapter has been devoted to these waves. 4.1 Shallow water theory Contrary to chapter 3.4 we start with the simple case where the pressure distribution can be taken to be hydrostatic and proceeds thewn to the derivation of the general expressions for an arbitrary water depth. Consider a wave with the height H propagating with the celerity c towards right, cf. figure 4.1. The water surface elevation η can then be written as η= H sin(ωt − kx) 2 (4.1) ω= 2π T (4.2) where Is called the cyclic or angular frequency and T is the wave period. If you are at a fixed place, for example x = 0, (4.1) is reduced to: H t⎞ ⎛ sin ⎜ 2π ⎟ (4.3) 2 ⎝ T⎠ From this it is seen that two successive wave crests will arrive with a time interval of T (= the wave period). If you consider the wave train at a given time, for example t = 0, (4.1) gives η= η= H H ⎛ 2π ⎞ x⎟ sin(kx) = sin ⎜ 2 2 ⎝ L ⎠ (4.4) cf. (3.39), it is seen that the distance between two successive wave crests is L (= the wave length). 60 Fig. 4.1 Definition of progressive shallow water waves and the pressure distribution. The wave described by (4.1) is propagating with the velocity c= ω L = k T (4.5) Which can be seen by considering (4.1) at two closely spaced points in time, t and t + Δt . During this time you have to move the distance Δx forward to maintain the same value of η and stay at the same place in the wave, so that H H sin(ωt − kx) = sin(ωt − kx + ωΔt − kΔx) (4.6) 2 2 where ωΔt − kΔx = 0 (4.7) Δx ω =c= Δt k (4.8) giving As an alternative (4.5) can be obtained directly by realizing tha a wave will move one wave length forward during one wave period. The propagation velocity c is also know as the phase velocity. The deformation of the water surface as the wave moves causes velocities in the fluid below. These velocities are periodic in time and space and are normally much smaller than the propagation velocity c. Consider the point A at location between the wave crest and the wave trough, figure 4.1. As mentioned we can for shallow water waves take the pressure to be hydrostatic – as indicated under the crest and trough in figure 4.1. This means that there is a negative pressure gradient from left to right in point A, because the pressure at a given distance from the bottom is large at the left side than at the right side of A due to the difference in the water surface elevation. The hydrostatic pressure is described as 61 p = ρg ( D − y + η) (cf. (3.63)). The pressure gradient in the x-direction at all water depths is therefore found as ∂p ∂η = ρg ∂x ∂x (4.9) This pressure gradient is giving an acceleration of the fluid ⎡ ∂u ∂p du ∂u ∂u ⎤ = −ρ = −ρ ⎢ + u +v ⎥ ∂x dt ∂x ∂y ⎦ ⎣ ∂t (4.10) Here the term u is small (being of the order H/L, as u → 0 for H → 0). The term v is also small, and the two convective terms (the two last at the right hand side) in 4.10 are therefore Small of second order (~(H/L)2), while the term ∂u / ∂t is proportional to H/L and will therefore be the dominant term on the right hand side of (4.10). When linearized (4.10) becomes ρg ∂u ∂η = −ρ ∂t ∂x (4.11) since ∂η / ∂x is not a function of y we have the same pressure gradient the entire water depth. (4.11) therefore shows that we can take u to be constant over the depth. The continuity equation now reads ∂ [( D + η)u ]dx dt = − ∂η dx dt ∂t ∂x or u ∂u ∂η ∂η ∂u +η +D = ∂x ∂x ∂x ∂t The first two terms on the right hand side are small of 2. order (~(H/D)2) and can be neglected reducing the equation to D ∂u ∂η =− ∂x ∂t (4.12) 62 Fig. 4.2 The continuity equation for a water column From (4.11) and (4.12) second order partial differential equations can be formed for either u or η . For η the equation reads: 1 ∂ 2η ∂ 2η = ∂x 2 gD ∂t 2 (4.13) By inserting (4.1) we get −k2 = − 1 2 ω gD or c= ω = gD k (4.14) which expresses that (small) shallow water waves propagates with a velocity depending only on the water depth and not on for example the wave length or the wave height. The fluid velocity u may for example be derived from (4.12) ∂u 1 ∂η 1 H =− =− ω cos(ωt − kx) D ∂t D 2 ∂x (4.15) giving u= 1 H ω sin(ωt − kx) D 2 k or simply 63 u= c η D (4.16) It is thus seen that the water particles under the wave crest have velocities in the direction of the wave propagation and the particle under the troughs have velocities in the opposite direction, see figure 4.3. Fig. 4.3 The wave induced velocities in shallow water waves. 4.2 Sinusoidal wave at an arbitrary water depth The expression (4.16) is of course not general: at large water depths surface water waves cannot be felt at the bottom, so the velocity field and the propagation velocity must depend on other parameters than the water depth (in this case the wave length). At large water depth the pressure variation under the wave is spread out and the assumption of hydrostatic pressure is no longer valid. In this case vertical accelerations are of importance and a more detailed description of the flow field is necessary. This can be obtained by use of the potential flow theory. We consider once more a wave given by η= H sin(ωt − kx) 2 (4.17) propagating with the velocity c over a fluid with the water depth D. The bed is horizontal at y=0. By expressing the balance between pressure forces and the acceleration, i.e. neglecting the effect of viscosity, potential flow theory can be applied. The Laplace equation reads Δφ = 0 (4.18) At the bottom the kinematic condition is v=0 for y=0 or − ∂φ =0 ∂y for y=0 (4.19) 64 At the surface y = D + η the pressure (in excess of the atmospheric pressure) must be zero, in which case the generakised Bernoulli equation (3.11) reads D +η+ u 2 + v 2 1 ∂φ = C (t ) − 2g g ∂t for y = D+η (4.20) Finally there is the kinematic condition expressing that a water particle at the water surface stays at the water follows the water surface. Mathematically this can be expressed as v= dη dt for y = d +η (4.21) In (4.21) the total derivative related to a water particle has been applied, where ∂η ∂η dη ∂η = +u ≈ dt ∂t ∂x ∂t (4.22) using that the convective terms can be neglected in a linearized analysis. (4.21) now reads v= ∂η ∂t for y = D+η (4.23) The vertical velocity v at the surface can be written as an expansion ⎛ ∂v ⎞ v(d + η) = v( D) + η⎜⎜ ⎟⎟ + ...... y ∂ ⎝ ⎠ y =D (4.24) In a linear theory the last terms on the right hand side can be neglected as thy are of the order (H/D)2 or smaller, so (4.23) becomes: ⎛ ∂φ ⎞ ∂η = v( D) = ⎜⎜ − ⎟⎟ ⎝ ∂y ⎠ y = D ∂t Analogous (4.20) can be linearized to give η− 1 ⎛ ∂φ ⎞ =0 ⎜ ⎟ g ⎝ ∂t ⎠ y =D (4.25) (4.26) where C(t) has been included in φ . Changing φ with a constant does not affect the velocity field. The water surface has been taken to vary periodically in time and x (4.17). It is therefore natural to assume the induced velocities – and therefore also φ - to vary periodically in the same way, possibly with a phase shift, introducing also a sine term 65 φ = f ( y )[a sin(ωt − kx) + cos(ωt − kx)] (4.27) where f(y) is an unknown function and a is a constant. By inserting (4.27) into the Laplace equation (4.18) we get ∂2 f −k2 f = 0 2 ∂y (4.28) f = c1 exp(ky ) + c 2 exp(−ky ) (4.29) having the solution The three constant, c1, c2 and a are found from the three boundary condition (4.19), (4.25) and (4.26). Equation (4.19) gives c1k − c 2 k = 0 or c1 = c 2 (4.30) so that f can be written f = c1 [exp( ky ) + exp( −ky )] = 2c1 cosh( ky ) (4.31) (4.25) now gives − 2c1k sinh( kD)[a sin(ωt − kx) + cos(ωt − kx)] = (4.32) H ω cos(ωt − kx) 2 from which we find a = 0 (4.33) and 2c1k sinh( kD) = − H ω 2 (4.34) Finally the boundary condition (4.26) gives H 1 sin(ωt − kx) + ω sin(ωt − kx)2c1 cosh(kD) = 0 2 g (4.35) or 66 2c1 ω H cosh(kD) = − 2 g (4.36) From (4.34) and (4.36) we find the relation kg tanh(kD) = ω 2 (4.37) Noting that the equation (4.8) for the wave celerity is general, (4.37) can be written c= ω = k g tanh(kD) k (4.38) which is the general equation describing the celerity of small sinusoidal waves at an arbitrary water depth. For shallow water kD will tend to zero, so (4.38) approaches c = gD tanh(kD) → gD kD for kD → 0 (4.39) which is the expression for shallow water waves, (4.14). For deep water we have kD → ∞ so that tanh(kD) → 1 which gives c→ g k (4.40) In deep water the wave celerity for a linear theory is thus only a function of the wave length. The particle velocities can be found from the potential φ , from (4.27), (4.31), (4.33) and (4.36) is found as φ=− H g cosh(ky) cos(ωt − kx) 2 ω cosh(kD) (4.41) From this the velocities can be found as u=− ∂φ kH g cosh(ky) = sin(ωt − kx) ∂x 2 ω cosh(kD) (4.42) v=− ∂φ kH g sinh(ky ) = cos(ωt − kx) ∂y 2 ω cosh(kD) (4.43) and It can be noticed that in a linear theory the velocities u and v are proportional to the wave height and for moderate values of H they will be much smaller than the propagation velocity c. The velocity profiles under the wave crest and trough are shown in figure 4.4. 67 Fig. 4.4 Wave induced velocities under a sinusoidal wave. Compare with figure 4.3. Example 4.1 Refraction As waves propagates towards a coastline with an angle different from 0o, the wave fronts will be deflected (a wave front is a curve indicating the position of the wave crest of a wave at a given time). This is because the wave celerity c decreases with decreasing water depth. In figure 4.5 the wave fronts at time t and t + dt are shown for a wave approaching a straight coast with depth contour lines parallel with the coastline. During this time interval the wave has moved from AC to BD, where AB and CD are known as wave orthogonals, defined as curves that are orthogonal to the wave fronts. The distance between the two wave orthogonals is called Δb . Fig. 4.5 Deflection of wave fronts approaching a coast. 68 The distances A-B and C-D are given as A − B = c dt (4.44) ∂c ⎛ ⎞ C − D = ⎜ c − Δb sin β ⎟ dt ∂x ⎝ ⎠ (4.45) Where c is the wave celerity at x = 0, cf. figure 4.5. From this it is seen that CD is slightly longer than AB, namely by ds = − ∂c Δb sin β dt ∂x (4.46) ds dc = sin β dt Δb dx (4.47) By defining dβ as dβ = − it is found that the deflexion dβ that takes place as the wave moves towards the coastline by the distance dx = (c dt cos β ) is given as ∂β 1 ∂c = tan β ∂x c ∂x (4.48) This equation can be integrated to give c = const sin β (4.49) which is know as Snell’s law Example 4.2 Particle paths in waves. From the particle velocities given by (4.42) and (4.43) it is possible to calculate they position at different times. Consider a particle with a mean position in (xo,yo). The particles position in the wave motion can then be found from dx dy (4.49) =u , =v dt dt By nitration of (4.42) and (4.43) the particle coordinates are found as 69 x = x0 − H cosh(ky ) cos(ωt − kx) 2 sinh(kD) (4.50) y = y0 + H sinh(ky ) sin(ωt − kx) 2 sinh(kD) applying also (4.37). The particle paths are shown in figure 4.6 for deep, intermediate and shallow water. In deep water the particle paths are circular (in the linear theory), while the particles in shallow water move back and forth in an almost horizontal oscillation. It may be noted that if higher order terms in H/D are included in the analysis the particle paths are no longer closed. Fig. 4.6 Particle paths in deep, intermediate and shallow water waves. Example 4.3 Wave groups In nature waves are generated by the blowing over the water surface. The water surface can be shown to be unstable under the action from the wind develops infinitely small undulations which grow to become waves. These waves are not all identical but vary in length and height. Wave with different periods and heights form a quite complicated and time-varying ocean surface. Here we consider a simple example, the pattern formed by superposition of two sinusoidal waves with the same wave height but with slightly different wave lengths, having the wave numbers k and k + Δk respectively. The first wave is described as η1 = H H sin(ωt − kx) = sin( k (ct − x)) 2 2 (4.51) while the second wave is given by 70 η2 = H sin[(k + Δk )((c + Δc)t − x)] 2 (4.52) By superposition and application of formulas for addition of sines we get η = η1 + η 2 = (4.53) ⎡ Δk ⎧⎛ ⎫⎤ Δc ⎞ H cos ⎢ ⎨⎜ c + k ⎟t − x ⎬⎥ sin[k (ct − x)] Δk ⎠ ⎭⎦ ⎣ 2 ⎩⎝ The surface described by (4.53) is illustrated in figure 4.7 Fig. 4.7 Wave group. The last tewrm on the right hand side of (4.53) represents the normal propagating wave, while the cosine factor makes the amplitude of the wave varies in the x direction so that the wave amplitude is modulated by the broken curve shown in figure 4.7. This curve has the wave length Lg, given as Lg = 4π Δk (4.54) A difference in the wave number by Δk corresponds to a difference in the wave length of ΔL , which is seen from Δk = ( k + Δk ) − k = 2π 2π 2π ⎛ ΔL ⎞ ΔL − ≈ − 1⎟ = −k ⎜1 − L + ΔL L L ⎝ L L ⎠ (4.55) so that Lg L = −2 L ΔL (4.56) where ΔL < 0 for Δk > 0. It is seen from (4.56) that Lg >> L for small relative differences in the wave length of the two waves. The two points A and B, where the wave length is 0, moves forward with the velocity cg (the group velocity), determined as 71 cg = c + k Δc ∂c ≈c+k ∂k Δk (4.57) cf. (4.53). From (4.38) we have c2 = g tanh( kD) k giving 2c ⎡ 2kD ⎤ dc g g D g = − 2 tanh(kd ) + = − 2 tanh(kD) ⎢1 − ⎥ 2 dk k cosh (kD) k k ⎣ sinh( 2kD) ⎦ (4.58) From which (4.57) gives cg = c − c⎡ 2kD ⎤ ⎢1 − ⎥ 2 ⎣ sinh( 2kD) ⎦ cg = c⎡ 2kD ⎤ ⎢1 + ⎥ 2 ⎣ sinh( 2kD) ⎦ or (4.59) The wave group is seen to propagate with a velocity varying between c/2 in deep water (kD → ∞) and c in shallow water (kD → 0) . The individual waves (drawn with a full cure in figure 4.7) propagate of course with the celerity c. It may be noted that large natural waves are often observed to form groups of 3-8 waves, and that the groups are important for the design of offshore as well as coastal structures. The explanation for cg being smaller than c in deep water and equal to c in shallow water is that all waves regardless of the period or wave length have the same celerity in shallow water (no frequency dispersion), while short wave propagate at a slower rate than long waves in deep water. 4.3 Wave energy When compared to an undisturbed water surface a wave possesses 2 types of energy, namely potential energy and kinetic energy. The potential energy is present because water so to speak has been moved up from the wave trough to the wave crest, see figure 4.8. The total potential energy can thus be found by calculating the potential energy gained by a single strip with the length dx, the height ηand the centre of gravity η /2 beneath the still water level when moved to a position with the centre of gravity placed η/2 above the still water level, giving 72 E pot 1 = L L/2 ∫ ρgη(η dx) = 0 1 ρgH 2 16 (4.60) The kinetic energy is found by integrating over the entire volume of water (using that the upper limit of integration D + η can be approximated by D) E kin L D ⎤ 1 ⎡ 1 2 = ρ ∫ ⎢ ∫ (u + v 2 )dy ⎥ dx L 0 ⎣⎢ 0 2 ⎦⎥ (4.61) Considering first the depth integrated kinetic energy we get D 2 1 2 1 ⎛ kHg ⎞ 1 2 ∫ 2 (u + v )dy = 2 ⎜⎝ 2ω ⎟⎠ cosh 2 (kD) × 0 (4.62) D D ⎤ ⎡ 2 2 2 sin ( t kx ) cosh ( ky ) dy cos ( t kx ) sinh 2 (ky ) dy ⎥ ω − + ω − ⎢ ∫ ∫ ⎥⎦ ⎢⎣ 0 0 by applying (4.42) and (4.43). Using the rule cosh2(ky) = 1 + sinh2(ky) (4.62) is reduced to D ⎞ ⎛ 1 2 1 ⎛ kHg ⎞ 1 2 2 2 ⎟ (4.63) ⎜ ( u v ) dy D sin ( t kx ) sinh ( ky ) dy + = ω − + ⎜ ⎟ ∫2 ∫ 2 ⎟ ⎜ 2 2 ω cosh ( kD ) ⎝ ⎠ 0 0 ⎠ ⎝ D Using that sinh2(ky) = ½(cosh(2ky)-1) (4.63) can be further modified to Fig. 4.8 Determining the potential energy 2 ( ) ⎞ ⎡ 1 2 kHg 1⎛ D ⎤ 2 2 ∫ 2 (u + v )dy = 2 ⎜⎜⎝ 2ω cosh(kD) ⎟⎟⎠ ⎢⎣ D sin (ωt − kx) − 12 + 4kD sinh(2kD)⎥⎦ (4.64) 0 D Averaging (in time or space) make the term ( sin 2 (ωt − kx) − 1 / 2 ) = 0, and the average kinetic energy per unit bed area is found as (cf. (4.61)) 73 2 E kin ⎞ D kHg 1 ⎛ ⎟ sinh( 2kD) = ρ⎜⎜ 2 ⎝ 2ω cosh(kD) ⎟⎠ 4kD (4.65) The relation sinh(2kD) = 2sinh(kD)cosh(kD) inserted in (4.65) reduced this to E kin = ρ 1 gH 2 16 (4.66) Exactly identical to the result for the potential energy. Example 4.4 Transport of wave energy, wave energy flux As the wave moves forward with the celerity c it also transports energy in the same direction, but the transport velocity ce of the energy is generally smaller than the velocity of propagation. The energy transport velocity ce can be determined in two ways, by considering the kinematics or the dynamics. The kinematic analysis has in fact already been made in example 4.3 considering the wave pattern resulting from the superposition of two waves of the same amplitude, but with slightly different wave numbers. There it was found that the modulation curve shown as broken lines in figure 4.7 moved forward with the velocity cg. In the points A and B the wave energy is zero, as the amplitude here is zero. A ‘parcel’ of energy is contained between A and B. This ‘parcel’ is transported forward with the same velocity as point A and B; that is cg. The amount of energy contained between A and B is constant as long as energy dissipation is not considered. The dynamic analysis of the transport and transport velocity of energy in waves is more fundamental and can be made as follows: The significant part of the energy transport can be expressed through the work per time unit (power) made by the wave motion at the vertical line A-A, i.e. the work made by the fluid at the left hand side of A-A, cf. figure 4.9. This work can be expressed as W = D+η D 0 0 ∫ pu dy ≈ ∫ pu dy (4.67) using that η<< D. This work results in a transfer of energy to the fluid at the right hand side of A-A, i.e. a flux of energy through A-A from left to right. The energy flux Ef through A-A can thus be written Ef =W (4.68) It may be noted that there is also a flux of kinetic energy through A-A. This contribution is of higher order (larger than (H/D)2) than the work, and can therefore be neglected in this analysis. 74 Fig. 4.9 The rate of work (power) made on A-A by the fluid at the left hand side of A-A. The pressure in (4.69) can (as described for (2.35)) be decomposed as p + = p − ρg ( D − y ) (4.69) so that p+ describes the excess pressure above the mean hydrostatic pressur. This pressure can be found from the generilized Bernoulli equation (3.11), where the potential is known from (4.41). After some calculations the pressure is found as p + = ρg H cosh(ky ) sin(ωt − kx) 2 cosh(kD) (4.70) (4.67) now gives D W = D + ∫ p u dy − ∫ ρgyu dy 0 (4.71) 0 where u can be obtained from (4.42). Only the first term on the right hand side of (4.71) gives a contribution with a time average different from zero, expressing the transport of energy through the line A-A. Physically the energy transport can be explained by the relation between the excess pressure, the instantaneous water surface elevation and the horizontal water velocity. When the surface elevation is high (wave crest) the pressure is high and the water moves forward, and correspondingly when the elevation is low the pressure is also low (p+ is negative) and the water moves backwards. The product of p+ and u is therefore always positive. The term can be written D + ∫ p u dy = 0 2kD ⎞ 2 ρgH 2 c ⎛ ⎟ sin (ωt ) ⎜⎜1 + 8 ⎝ sinh(2kD) ⎟⎠ (4.72) Which gives the time averaged rate of work 75 T D ⎤ ⎛ 1 ⎡ + 1 2kD ⎞ ⎟⎟ W = ∫ ⎢ ∫ p u dy ⎥ dt = ρgH 2 c⎜⎜1 + T 0 ⎣⎢ 0 16 sinh( 2 kD ) ⎝ ⎠ ⎦⎥ (4.73) Introducing the group velocity from (4.59) and that 1 E = E kin + E pot = ρgH 2 8 into (4.73) we get W = E f = cg E (4.74) (4.75) Which expresses directly that the energy flux corresponds to the wave energy being moved with the group velocity cg. 4.4 Waves over variable water depth When waves approach a coast their dimensions (height and length will change due to the decreasing water depth. For example will the wave height at sufficiently shallow water increase with a decreasing water depth until the wave breaks. The breaking begins when the wave height is comparable to the water depth (often the criterion H/D = 0.8) is used). Simultaneously the wave length will be smaller as the water depth decreases. Change in the wave length Equation (4.39) gave the relation for the variation in the wave celerity at an arbitrary water depth c = gD tanh(kD) kD (4.76) 76 Fig. 4.10 Change in the propagation velocity, the wave length and wave height with the water depth. Subscript o indicates infinitely deep water. As shown in figure 4.10 the wave celerity decreases as the wave propagates into more shallow water. At the same time we have from (4.5) T= L c (4.77) where T is the wave period. This we assume to be constant, which just means that the same number of waves pass a given cross-section. From (4.77) together with (4.76) we get tanh(kD) L = gD T kD (4.78) For deep water we specifically get Lo = T g = k0 gLo 2π (4.79) or Lo = gT 2 2π (4.80) which inserted into (4.78) gives L = Lo tanh(kD) (4.81) 77 Fig. 4.11 Variation in wave height over variable water depth. Analogously (4.80) and (4.77) gives c = co tanh(kD) (4.81) This relation is shown in figure 4.10. Equation (4.80) the general expression for the variation in the wave length with the water depth, provided that the slope of the bed is not too large. If the bed slope is large other phenomena, such as reflection, may be of importance. Change in the wave height The variation in the wave height over a gently sloping bed is determined by requiring that there is no energy loss in the wave motion. Normally there is a small energy loss in the near-bed wave boundary layer, which is negligible over a length of some wave lengths. breaking waves there is a considerable energy loss, and the considerations made in the following cannot be applied in the surf zone. Let us consider a two-dimensional plane situation where the waves propagate over a bed with depth contours which are normal to the plane of the paper, see figure 4.11. When there is is no energy loss the energy flux in A and B must be the same and (4.75) then gives ( c g E ) A = (c g E ) B (4.82) E is found from (4.60) and (4.66) 1 E = E pot + E kin = ρgH 2 8 (4.83) The group velocity cg may be found from (4.59) and (4.81) cg = ⎡ 1 2kD ⎤ co tanh(kD) ⎢1 + ⎥ 2 ⎣ sinh( kD) ⎦ (4.84) 78 Using (4.84) and (4.83) in (4.82) and that in deep water cgo = ½co we get ⎡ 2kD ⎤ 2 H 2 tanh(kD) ⎢1 + ⎥ = const = H o sinh( 2 ) kD ⎣ ⎦ (4.85) where Ho is the wave height in (infinitely) deep water. The variation in the wave height can therefore be written ⎡ H 2kD ⎤ = coth(kD) ⎢1 + ⎥ Ho ⎣ sinh(2kD) ⎦ (4.86) which expresses the relative change in wave height as function of the dimensionless parameter kD. This parameter can for a given wave period be found from (4.80). Example 4.5: Wave table. The variation in wave height and wave length as a wave propagates from deep to more shallow water can be found from (4.79), (4.80) and (4.86). These equations may be solved use of a hand calculator or by use of a wave table as given in figure 4.12. Consider for example a 2 m high wave in 40 m of water and with a period of 11 s. We wish to determine the height and length at 10 m water depth. The table can be used as follows. 1) The wave length at infinitely deep water is calculated: Lo = 1.56T2 = 189 m 2) D/Lo is calculated: 40m/189m = 0.212 Now the value for column 1 is known, and H/Ho can be read from column 8 H = 0.921 Ho giving a wave height Ho in infinitely deep water of Ho = 2m/0.921 = 2.17 m The deep water wave height is seen to be larger than at 40 m water depth. This is because cg and thus the energy flux has a maximum for D/L = 0.19, cf. (4.84) and figure 4.10. 79 Wave table: Formulae and relations General: ω = 2π / T Abbreviation: G = 2kD/sinh(2kD), θ = ωt − kx Wave surface: η = H / 2 cos θ = h / 2 cos(ωt − kx) , k = 2π / L , c = L / T , Lo = 1.56T 2 Formula No. Wave propagation velocity, celerity: c= g tanh(kD) k (4.38) Group velocity: cg = c (1 + G ) 2 (4.59) Horizontal orbital velocity: u= Vertical orbital velocity: v=− Horizontal particle motion: x = xo + H cosh(ky ) sin(θ) 2 sinh(kD) (4.50) & (4.80) Vertical particle motion: y = yo + H sinh(ky ) sin(θ) 2 sinh(kD) (4.50) & (4.80) Excess pressure: p + = ρg H cosh(ky) cos(θ) 2 cosh(kD) (4.70) Reaction force, pressure: ρgH 2 G Fp = 16 Reaction force, momentum: Fm = ρgH 2 (1 + G ) 16 (4.97) Energy flux: Ef = ρgH 2 (1 + G )c 16 (4.59) & (4.75) πH cosh( ky ) cos(θ) T sinh( kD ) (4.42) & (4.80) πH sinh( ky ) cos(θ) T sinh(kD) (4.43) & (4.80) (4.92) Fig. 4.12 Wave table (from [3]. Note that the water surface in these formulae is shifted by 900 relative to the formulae in the text, cos(θ) instead of sin θ ). 80 81 At 10 m water depth the parameter for column 1 is found as 10 m D = = 0.0529 Lo 189 m And column 3 and 8 then gives D = 0.0971 L ⇒ L = 10 m / 0.0971 = 103 m H = 1.015 Ho ⇒ H = 2.17 ⋅ 1.015 m = 2.20 m and Finally it should be noted that the limit for using shallow water wave theory is D/L < 1/20, while the limit for using deep water wave theory is D/L > ½. 4.5 The wave reaction force In the last part of this chapter more time-averaged quantities will be considered. Time averaged terms are of higher order than one (H/L)2, (H/L)3 etc. In the following only terms of second order will be retained. In section 4.3 the energy in the wave motion was considered, now we will analyse the momentum in water waves. Consider a shallow water wave. The pressure distribution is hydrostatic as shown in figure 4.13. Now we will show that the presence of the waves makes the fluid at the left hand side of the vertical section A-A exert a force F on the fluid to the right, which is larger than the hydrostatic pressure force po, we would find if there were no wave motion po = 1 ρgD 2 2 (4.87) The force F consists of two contributions, a pressure force Fp and a momentum force Fm. 82 Fig. 4.13 The pressure distribution under a shallow water wave. A: The pressure force Fp The instantaneous pressure force acting on A-A is D +η Fp = ∫ p dy (4.88) 0 For a shallow water wave the pressure p is given as p = ρg (d + η − y ) (4.89) by which (4.88) is found to give D +η 1 ⎤ ⎡ F p = ρg ⎢( D + η) y − y 2 ⎥ 2 ⎦0 ⎣ (4.90) = ρg ( D + η) 1 = po + ρgDη + ρgη 2 2 2 2 The water surface elevation is H η = sin(ωt − kx) 2 and we find therefore the following time-averaged value of Fp (averaging over a wave period) T 1 1 F p = ∫ F p dt = po + ρgH 2 T0 16 (4.91) 83 As seen in (4.91) the pressure force is larger than po. Physically the force is centred around the level of the mean water surface. At all levels below the wave trough there is fluid during the entire wave period and the time averaged pressure force corresponds to the hydrostatic pressure for quiescent water. This is seen from p = ρg ( D + η − y ) so that p = ρg ( D − y ) At the levels D – H/2 < y < D+H/2 there is only water during a part of the wave period (at the trough level practically all the time, at the crest level practically none of the time). If we consider the level above the mean water level there is no water under quiescent conditions (pressure = 0) and therefore no force. With the wave ther is water part of the time and therefore a positive pressure force. Similarly for the levels between the mean water surface and the wave trough level: Here we have a higher pressure when the wave crest passes, but not a pressure lower than zero when the trough passes, this gives also a positive pressure force. In case of intermediate or deep water waves the analysis of the pressure force is considerably more complicated because the mean pressure distribution deviates from the hydrostatic. The general first order (sinusoidal) wave theory then gives F p = po + 1 2kD ρgH 2 16 sinh(2kD) (4.92) Equation (4.92) is seen to approach (4.91) for kD → 0 , and for deep water waves: Fp → 0 for kD → ∞ , so there is no additional pressure force for deep water waves. B: The momentum force In the case of shallow water waves the horizontal particle velocity is u= c η D (4.93) (cf. (4.16)). This horizontal velocity gives rise to a flux of momentum which acts as a force. From basic hydraulics the momentum force is know, and can be written as G G G Fm = −ρ ∫ v (v ⋅ dA) (4.94) G where dA is an infinitesimal area vector on the control surface, for the vertical surface A-A the area vector is horizontal, pointing towards left. The horizontal component of (4.94) for A-A is 84 D +η Fm = ρ ∫u 2 dy (4.94a) 0 (4.93) and (4.95) gives Fm = ρ c2 2 η ( D + η) D (4.95) Time-averaging of (4.96) then gives Fm = ρ c2 H 2 1 = ρgH 2 D 8 8 (4.96) using that c = gD . In contrast to F p Fm is evenly distributed over the vertical from the bed to the mean water surface. In the general case of arbitrary water depth the horizontal wave orbital velocity (4.42) inserted into (4.94a) gives (after time-averaging) Fm = ⎛ 1 2kD ⎞ ⎟⎟ ρgH 2 ⎜⎜1 + 16 ⎝ sinh( 2kD) ⎠ (4.97)In total the resulting time-averaged force F acting on section A-A in figure 4.13 is F = F p + Fm = po + 1 ⎛ 2kD ⎞ ⎟ E ⎜⎜1 + 2 ⎝ sinh(2kD) ⎟⎠ (4.98) = po + Fr where E is the wave energy given by (4.74). By deducting po (the hydrostatic pressure under quiescent conditions) from F we get Fr which is normally known as the reaction force of the wave. Example 4.6: The change in the mean water surface for waves approaching a coast. 85 Let us consider waves approaching a long straight coast in the direction normal to the coastline, cf. figure 4.14. As described in section 4.4 the wave height will increase towards the shoreline until the waves start to break. The breaking criterion being expressed by the wave height and depth at the breaker point H b = Khb ≈ KDb (4.99) where K is approximately 0.8. Fig. 4.14 The variation in the wave height as the waves propagate towards the coast. Outside the breaker point the loss of wave energy is very small and may for practical purposes be neglected (the energy loss only takes place in the bed boundary layer). When the energy loss is neglected the Bernoulli equation (4.20) may be applied. It reads u2 + v2 1 ∂φ + ( D + η) − = c(t ) 2g g ∂t for y = d +η (4.100) In this equation we include terms of second order (rather than linearizing as done in (4.26)) and get u 2 ( D) + v 2 ( D) 1 ⎛ ∂φ ⎞ 1 ⎛ ∂ 2φ ⎞ ⎟ +η− ⎜ ⎟ − η⎜⎜ = c(t ) g ⎝ ∂t ⎠ y = D g ⎝ ∂t∂y ⎟⎠ y = D 2g (4.101) Time-averaging of (4.101) over a wave period gives η=− u 2 ( D) + v 2 ( D) η ∂ 2 φ + + c(t ) 2g g ∂t∂y (4.102) Inserting φ, η, u and v from the solution found earlier (4.41), (4.42) and (4.43) gives 86 η= ⎡⎛ kH g ⎞ 2 1 ⎛ kH g ⎞ 2 1⎤ 2 − ⎢⎜ +⎜ ⎟ ⎟ tanh (kD) ⎥ 2 ⎦⎥ ⎣⎢⎝ 2 ω ⎠ 2 ⎝ 2 ω ⎠ 2g (4.103) + H H g 1 k tanh(kD)ω 2 2 ω 2 g + c(t ) By use of (4.78) k 2 g / ω 2 can be written as k 2g T 2 k = 2 g= 2 tanh(kD) ω L (4.104) by which (4.103) may be reduced to 2 ⎞ k 1 ⎛H ⎞ ⎛1 η = −⎜ ⎟ ⎜⎜ − k tanh( kD) ⎟⎟ + c(t ) ⎝ 2 ⎠ ⎝ 4 tanh(kD) 4 ⎠ 2 2 ⎛ H ⎞ ⎛ cosh (kD) − sinh (kD) ⎞⎟ + c(t ) = −⎜ ⎟ k ⎜⎜ ⎝ 2 ⎠ ⎝ 4 cosh(kD) sinh( kD) ⎟⎠ 2 (4.105) 2 k ⎛H⎞ = −⎜ ⎟ + c(t ) ⎝ 2 ⎠ 2 sinh( 2kD) If we put η = 0 in deep water we get c(t ) = 0 giving 2 k ⎛H⎞ η = −⎜ ⎟ ⎝ 2 ⎠ 2 sinh( 2kD) (4.106) From this it is seen that there is a small lowering of the mean water surface as the wave propagate into shallow water, provided we are outside the surf zone with breaking waves. This phenomenon is called set-down. That it is in fact a small variation can be seen from (4.106) and (4.99). Just before the point of wave breaking we find 1 ⎛H = − ⎜⎜ b Db 16 ⎝ Db η 2 ⎞ 2kD ⎟⎟ ⎠ sinh( 2kD) (4.107) 1 ⎛H ≈ − ⎜⎜ b 16 ⎝ Db 2 ⎞ K2 ⎟⎟ = − ≈ −0.04 16 ⎠ that is about 4% of the water depth at the point of wave breaking. 87 Inside of the point of wave breaking there is a considerable loss of energy (associated with a large amount of turbulence in the surface rollers). Here ther is a a strong decrease in the wave height as the waves approach the shoreline. A some distance inshore of the point of wave breaking experiments show that the local wave height is approximately equal to the local water depth h times a constant. For the sake of simplicity we take this constant to be equal to K H = Kh , h = D + η (4.108) In reality the coefficient K decreases from the value 0.8 just after the point of wave breaking to approach 0.5 in zone of similarity. In the following we will use K = 0.8 to make the calculations as simple as possible. Because of the loss of energy the Bernoulli equation cannot be used in the surf zone. In stead a force balance may be used. The idea is that due to the decrease in the wave height towards the shoreline, the wave reaction force will also decrease towards the shore. To have a balance of the forces (averaged over time) for a water column there must therefore also be a force directed offshore. This force can be obtained from a slope in the mean water surface. First consider the forces on a column with height D (equal to the still water depth) and of width dx due to the pressure in quiescent water, figure 4.15a. Here a horizontal projections of the hydrostatic pressure forces acting on the column are in equilibrium: the total horizontal component of the hydrostatic pressure is 0. If we give a small perturbation η to this horizontal water surface, figure 4.15b, a resulting horizontal force on the water column emerges. This force can be found a the resulting force of the three hatched areas in figure 4.15b. The force can be found by assuming hydrostatic pressure and calculate the areas in figure 4.15b, or it may be found from the gradient theorem, which is derived from Gauss’s theorem, saying that the integral of a quantity of the area of a closed surface is identical to the integral of its gradient over the volume enclosed by the surface Fig. 4.15. Pressure acting on a water column with horizontal water surface (a), and sloping water surface (b). G ∫ φ dA = ∫ grad (φ) dX A (4.109) X If we apply (4.109) on the pressure (i.e. p = φ ) we get that the resulting pressure force on a surface (in our case the water column) is identical to the the pressure 88 gradient over the volume enclosed by the surface. If we consider the horizontal force component, the pressure gradient over the entire water column is given by ∂p ∂η = ρg ∂x ∂x (4.110) given a total horizontal pressure force of ∂p ∂η dX = −ρg h dx ∂x ∂x X P = −∫ (4.111) Let us consider a water column, figure 4.16, in a situation where sinusoidal waves propagate towards the shore over decreasing water depth. On the water column there is a resulting reaction force from the wave with varying wave height. Taken to be positive towards the right hand side, the reaction force is − ∂Fr dx ∂x Fig. 4.16 Forces acting on a water column. The magnitude of the set-up shown is exaggerated. Since the resulting reaction force is different from 0, a compensating force must be introduced to obtain a time-averaged total force on the water column of 0. The compensating force is obtained by having a pressure gradient from the sloping mean water surface. This gives a resulting mean pressure force (4.111) and we will have a force equilibrium if the slope of the mean water surface fulfils ρgh ∂F ∂η =− r ∂x ∂x (4.112) Inserting Fr from (4.98) and using the shallow water approximation 2kD ≈1 sinh(2kD) we now get 89 ρgh ∂η ∂ ⎛31 ∂ ⎛31 ⎞ ⎞ =− ⎜ ρg H 2 ⎟ = − ⎜ ρg (kh) 2 ⎟ ∂x ∂x ⎝ 8 8 ∂x ⎝ 8 8 ⎠ ⎠ (4.113) inserting (4.108) to describe the variation in the wave height in the surf zone (known as the wave set-up). Equation (4.113) can be reduced ∂η 3 1 ∂(h 2 ) 3 ∂h = − K2 = − K2 ∂x 16 h ∂x 8 ∂x (4.114) from which 3 η = − K 2 h + const 8 (4.115) The constant may be determined from the mean water level at the point of wave breaking (4.107), giving η=− K 2 Db 3 2 + K ( Db − h) 16 8 At the shoreline h = 0, and here we find 5 η(0) = K 2 Db ≈ 0.20 Db 16 (4.116) (4.117) So the maximum in the wave set-up is 20% of the water depth at the point of wave breaking, a quantity which often has to be taken into consideration. 4.6 Wave induced currents In the previous sections we have considered the forces acting on a section normal to the direction of wave propagation, considering a plane situation with depth contours normal to the direction of propagation. This corresponds to forces Fm and Fp on section A-A in figure 4.17. If we in stead consider the forces on section B-B which is normal to the wave fronts it is easy to realize that the momentum force now is 0. Because all velocities are parallel to B-B there cannot be any flux of momentum through this section. On the other hand pressure is acting in all directions, and the pressure force Fp is the same on section B-B as on A-A. Therefore the reaction force of the wave on a given section will depend on the angle α between the section and the wave fronts. For α = 0 0 and 90 0 90 Fig. 4.17. Variation in the reaction force with the angle between the section considered and the waves (waves seen from above) we have only a normal force, but since these two normal forces are not identical (being Fm+Fp and Fp respectively) ther will for other directions be normal forces Sxx as well as shear forces Ssz, these forces are defined as stresses, integrated over the water depth, having the units of force per length of the section considered. For an arbitrary angle α (see section C-C in figure 4.17) the variation in Sxx and Ssz can be found from a force balance for triangular body, cf. figure 4.18. By expressing force equilibrium in the x-direction we find S xx dz − S11 cos α dz cos α − S 22 sin α dz sin α = 0 or , using S11 = ( Fm + F p ) and S 22 = − F p 91 Fig. 4.18 Forces acting on a triangular body. we get S xx = −( Fm + F p ) cos 2 α − F p sin 2 α or S xx = − F p − Fm cos 2 α (4.118) Similarly a force balance for the triangular body in figure 4.18 in the z-direction gives S xz dz − S11 cos α dz sin α + S 22 sin α dz cos α = 0 or S xz = − Fm sin α cos α (4.119) We consider now wave approaching a coastline at an angle. The resulting force on the longshore direction (the z-direction, cf. figure 4.19) on a water column with a height equal to the water depth D and covering an area of dxdz is F= ∂S xz ∂S dx dz + zz dx dz ∂x ∂z (4.120) The coast is assumed to be infinitely long with constant wave conditions along the coast. For such uniform conditions the term ∂S zz / ∂z become 0, giving 92 F= ∂S xz dx dz ∂x (4.121) As long as there is no energy dissipation (i.e. when the potential flow theory is valid) it can be shown that Sxz is constant, even when the waves are shoaling and refracting as they propagate towards the coast. This can also be realised just by considering that we cannot have shear stresses in a potential flow. Outside the surf zone the energy dissipation is very small and can be disregarded. After the waves have broken there is a considerable energy loss. Here Sxz changes rapidly because the decreasing wave height (and wave energy flux) makes the magnitude of Sxz also decease towards the shoreline. Since Sxz < 0 (cf. (4.119)) this implies that ∂S xz >0 ∂x (4.122) Therefore a resulting force is acting in the alongshore direction. The force in the direction wave propagation, projected on the coastline, i.e. in the positive z-direction in figure 4.19. The force generates a longshore wave-driven current also know as the littoral current. The strength of the current is determined by a force balance: the bed friction τ b must of the same magnitude (and in the opposite direction) as the gradient in Sxz. If we use a relation as τ b = ρK1U c (4.123) where Uc is the longshore current velocity, figure 4.20. The force balance for the unit water column, see figure 4.21, can be written Fig. 4.19 The wave forces acting on a unit water column in the alongshore direction. 93 ∂S xz dx dz = ρK1U c dx dz ∂x (4.124) In general this equation have to be solved numerically, using (4.119). A simple assessment of the variation of Uc across the surf zone may be made by using the assumptions that the angle between the waves and the coastline is small: cos β ≈ 1 (4.125) By application of shallow water theory Sxz can be written S xz = − ρg 2 ρg 2 sin β H sin β = − H c 8 8 c (4.126) =− ρg 2 sin β H gD 8 c using (4.119) and (4.96). Due to the effect of refraction, described by Snell’s law (4.48a) the quantity sin β / c is constant through the surf zone. Fig. 4.20 Schematic illustration of the distribution of the wave-driven longshore current. If we use the approximation H =0.8D in the surf zone, where D is the local water depth (4.126) gives, for a coast with constant beach slope: ∂S xz = K2 D3/ 2 ∂x (4.127) where K2 is a positive constant. From the force balance (4.124) we now find 94 U c = K3D3/ 2 (4.128) where K3 is a positive constant. It is seen that the wave driven current is strongest just after the point of wave breaking, where the water depth is largest. The velocity Uc decreases towards the coast as indicated in figure 4.20. In field conditions Uc can easily be 1 or 2 m/s, and this longshore current is obviously very important for the sediment transport along the coast, also know as the littoral drift. It may be noted that the wave induced stresses Sxx and Sxz here have been defined as positive when expressing a tension (cf. 4.118 and figure 4.18), this is similar to the conventions for stresses in hydrodynamics or structural engineering. In many books on coastal engineering these stresses, known as the radiation stresses associated with the wave motion, are defined with an opposite sign, being positive when expressing a normal force equivalent to a positive pressure. Fig. 4.21 Forces acting on a unit water column in the alongshore direction. 95 5. Laminar Boundary Layers Let us consider a thin plate moving in its own plane in a quiescent fluid with a constant velocity uo. Due to the adhesion of the fluid to the plate some of the surrounding fluid will be carried along with the plate, see figure 5.1. The flow is only occurring in a thin layer along the boundary to the plate. This thin layer is known as a boundary layer. If we in stead consider the completely analogous case where the fluid flows along a fixed plate, figure 5.2, we see a reduction in the flow velocity near the plate due to the adhesion. The zone where the velocity decreases from the free outer velocity uo down to 0 at the wall is also known as a boundary layer. Fig. 5.1 The boundary layer formed along a plate moving through a quiescent fluid. A boundary layer can in this way be described as a domain where the flow velocity has large variations when you move across the main flow direction. Inside this layer the viscous forces must be included in the flow description while the shear stresses can be neglected in the flow outside. In the outer domain the flow may therefore be described by potential flow theory. This is of course a simplification of the real flow conditions, because there must be a gradual transition from the domain where the viscosity is of importance to the domain where it may be totally neglected. 102 Fig. 5.2 The boundary layer along a fixed thin plate. 5.1 The boundary layer equations The assumption that the thickness of the boundary layer is very small implies that some of the terms in the Navier-Stokes equations are small and may be neglected. If we disregard the gravity, the Navier-Stokes equation in the flow direction reads (cf. figure 5.2) ⎛ ∂ 2u ∂ 2u ⎞ 1 ∂p ∂u ∂u ∂u +u +v =− + ν⎜⎜ 2 + 2 ⎟⎟ ∂t ∂x ∂y ρ ∂x ∂y ⎠ ⎝ ∂x (5.1) Colse to the wall the v is skmall, because the stream lines in the boundary layer must be approximately parallel to the vall. In contrast the term ∂u / ∂y must be large because the velocity grows from 0 to the outer velocity over a very short distance, namely the thin boundary layer. The only term which a priori may be neglected is therefore ν ∂ 2 u / ∂x 2 , which must be very small compared to ν ∂ 2 u / ∂y 2 . In this way (5.1) is reduced to ∂u ∂u ∂u 1 ∂p ∂ 2u (5.2) +u +v =− +ν 2 ∂t ∂x ∂y ρ ∂x ∂y If we consider the Navier-Stokes equation in the direction normal to the wall, ⎛ ∂ 2v ∂ 2v ⎞ ∂v ∂v ∂v 1 ∂p +u +v =− + ν⎜⎜ 2 + 2 ⎟⎟ ∂t ∂x ∂y ρ ∂y ∂y ⎠ ⎝ ∂x (5.3) we can from the same argumentation conclude that all terms except ∂p / ∂y are small, so that (5.3) can be approximated by ∂p =0 ∂y (5.4) 103 This a result of great importance because it shows that the pressure inside the boundary layer is determined by the pressure outside it (i.e. in the potential flow). This means that the pressure gradient in the flow direction in (5.2) is determined from the outer potential flow. Equations (5.2) and (5.4) are known as the boundary layer equations for laminar boundary layers. 5.2 The general momentum equation The momentum equation is one of the most frequently used equations in hydraulics (cf. [2]). It is derived by expressing Newtons 2. law for a finite volume of fluid X, which is enclosed by the control surface A. Since Newtons 2. law expresses that mass times acceleration inside the control surface is identical to the resulting force acting on X , we find dvi ∫ ρ dt dX = − ∫ p dAi + ∫ ρg i dX + si X A (5.5) X Here the index i can have the values i = 1,2 and 3, so that v1 is the velocity in the direction of the x1-axis (similar to u and x), v2 is the velocity in the x2 direction (similar to v and y) etc. Equation (5.5) is in this way actually three equations, corresponding to a force balance in each of the three directions of the axes in the coordinate system. In (5.5) the resulting force on the right hand side is composed of 3 contributions: the pressure force (which acts as positive into the volume - that is in the direction opposite the area vector dAi), the force of gravity (which is a volume force, other volume forces such as the Coriolis force could have been included as well) and finaly si, which is the vector sum of the shear stresses acting on the control surface plus the stresses in any solid bodies, which are cut by the control surface. Equation (5.5) can be modified, using that the left hand side can be written dvi ∫ ρ dt X ⎛ ∂v ⎞ ∂v dX = ∫ ρ⎜⎜ i + i v k ⎟⎟ dX ∂t ∂x k ⎠ X ⎝ (5.6) cf. (1.16) and (1.17). It should be noted that when the same index is used twice in the same term (as does index k in (5.6)), a summation is to be made over this index. The last term in (5.6) can be modified by application of Gauss’s gradient theorem (cf. (4.109)), which applied on the term vivk gives ∂ ∫ vi vk dAk = ∫ ∂xk (vi vk ) dX A X ⎡ ∂v ⎤ ∂v = ∫ ⎢ i v k + k vi ⎥ dX ∂x k ∂x k ⎦ X⎣ (5.7) Here it follows from the continuity equation that the last term in the integral equal to zero. 104 Equation (5.5) can thus be written ∂vi ∫ ρ ∂t dX + ∫ ρvi v k dAk = − ∫ p dAi + ∫ g i dX + si X A A (5.8) X Equation (5.8) can be written in vector form G G G G G G G ∂v ρ dX + ρ v ( v ⋅ d A ) = − p d A ∫ ∂t ∫ ∫ + ∫ ρg dX + s X A A X (5.9) 5.3 The momentum equation for boundary layers We will now apply the general momentum equation (5.9) specifically on boundary layers. This is because the momentum equation has proven to be a very efficient tool for calculating the wall shear stress in boundary layers and the thickness of boundary layers. In the following the term with the acceleration of gravity will be neglected, so p represents the pressure in excess of the hydrostatic pressure. As an approximation the boundary layer is taken to have a well-defined finite thickness δ and the control surface is taken as the closed surface A-A’-B’-B-A, shown in figure 5.3. The two sides A-A’ and B-B’ are normal to the wall, A-B follows the wall, and A’-B’ follows the outer boundary of the boundary layer: y = δ(x ) . As mentioned (5.9) is a vector equation and for this application we use the projection along the x-axis. The contribution from each term in the momentum equation is found individually. Fig. 5.3 The control surface for the momentum equation, and the definition of the displacement thickness. 1. The first term on the left hand side of (5.9) represents the change in momentum due to the flow being non-steady. The contribution in the direction of the x-axis is δ dx ∫ ρ 0 ∂u dy ∂t (5.10) 105 2. The second term on the left hand side of (5.9) is known as the convective term, which is found in non-uniform flows where the fluid particles change G G their velocity as thay move along the stream lines. The term ρv ⋅ dA is the mass flux through a surface element. This can be found as follows: first the mass flux through the surface A-A’ is given as δ M 1 = − ∫ ρu dy (5.11) 0 G the minus sign is because dA is directed out of the control surface and therefore against the positive x-direction. Through B-B’ we have δ M 2 = ∫ ρu dy + 0 δ ∂ ρu dy dx ∂x ∫0 (5.12) From this the mass flux down through A’-B’ can be found from the requirement that the total flux mass flux must be zero. δ ∂ M 3 = −( M 2 + M 1 ) = − ∫ ρu dy ∂x 0 (5.13) The contributions to the term ∫A G G G ρv (v ⋅ dA) can now be found for the four sides of the control surface. Along A-B we have no flux and U = 0 so the contribution is zero. Along A’-B’ at the top of the boundary layer the velocity is equal to the outer velocity: u = uo. When making the projection on the x-axis the momentum flux is found to be δ G G ∂ ⋅ = − u ( v d A ) u ρu dy dx 0 ∫ 0 ∂x ∫0 (5.14) From A-A’ the contribution is δ ∂ 2 ∫ ρu (−u ) dy = −∫ ρu dy 0 (5.15) 0 And finally the contribution from B-B’ we get 106 δ 2 ∫ ρu dy + 0 ∂ ∂ ρu 2 dy dx ∫ ∂x 0 (5.16) giving a total momentum flux of ∂ δ ∂ ∂ ρu 2 dy dx − u 0 ρu dy dx ∫ ∂x 0 ∂x ∫0 3. (5.17) The pressure term in (5.9) can by use of the gradient theorem be written G − ∫ p dA = − ∫ grad p dX A X As the pressure does not vary with y in the boundary layer (cf. (5.4)) the f contribution in the x-direction is found to be − 4. ∂p δ dx ∂x (5.18) G The term s represents the shear stresses along the control surface and stresses in any solid bodies piercing the control surface. In the present case there are no shear stress at the top of the boundary layer, forming the boundary to the outer frictionless flow. Along the wall the shear stress contribution is found to be − τ 0 dx (5.19) From the contributions (5.10), (5.17), (5.18) and (5.19) the momentum equation (5.9) is found to give δ ∂ δ ∂u ∂ ∂ ∂p 2 ∫ ∂t dy + ∂x ∫ ρu dy − u 0 ∂x ∫ ρu dy = −δ ∂x − τ 0 0 0 0 (5.20) which is known as the momentum equation for boundary layers. _____________________________________________________________________ Example 5.1 Flow along a plate with a sudden movement. Let us consider a plate which is suddenly accelerated from the velocity 0 to the velocity uo. The plate moves in its own plane and is taken to be of infinite extension. 107 Fig. 5.4 The development of the velocity profile over a plate with a sudden movement. This plate will entrain more and more and fluid as time passes, as shown in figure 5.4. This flow will be solved in two basically different ways: by an exact solution and by a boundary layer approximation. A. The exact solution: This is done by use of the principles given in chapter 2 by formulating the flow equation with the appropriate boundary conditions. The flow equation in the flow direction reads (cf. (2.30)) ∂u ∂ 2u =ν 2 ∂t ∂x (5.21) with the boundary conditions u = u0 u=0 for for y=0 t≤0 and t>0 (5.22) (5.23) and u→0 for y→∞ (5.24) (5.21) is the so-called equation for heat conduction. It can be reformulated from being a partial differential equation to an ordinary differential equation by introduction of dimensionless parameters. The velocity u is made dimensionless by division by uo U= u u0 (5.25) Another dimensionless parameter can be found as a combination of y, t and ν 108 Y= y (5.26) νt ( ν has the dimension m2/s). Insertion of (5.25) and (5.26) into (5.21) gives d 2U 1 dU + Y =0 dY 2 2 dY (5.27) where the following derivations have been applied ∂U dU ∂Y dU ⎡ 1 Y ⎤ = = − dY ∂t dY ⎢⎣ 2 t ⎥⎦ ∂t ∂U dU ∂Y dU Y = = ∂y dY ∂y dY y ∂ 2U Y dU 1 d ⎛ dU ⎞ ∂Y Y 2 d 2U = − + = ⎜Y ⎟ ∂y 2 y 2 dy y dY ⎝ dY ⎠ ∂y y 2 dY 2 Equation (5.27) can be integrated once to give ⎡ Y2⎤ dU = c exp ⎢− ⎥ dY ⎣ 4 ⎦ (5.28) which, by a second integration yields Y ⎛ s2 ⎞ u = c ∫ exp⎜⎜ − ⎟⎟ ds + c1 u0 ⎝ 4⎠ 0 (5.29) In equation (5.29) we recognise the error function F (Y ) = 2 π Y −x ∫ e dx 2 (5.30) 0 which as shown in figure 5.5 asymptotically approaches ± 1 By insertion of (5.30) into (5.29) we obtain for Y → ±∞ . 109 Fig. 5.5 The error function F(Y): u π ⎛Y ⎞ π ⎛ y ⎞ F ⎜ ⎟ + c1 = 2c F⎜ ⎟ + c1 = 2c u0 2 ⎝ 2⎠ 2 ⎜⎝ 2 νt ⎟⎠ (5.31) where the constants c and c1 is determined from the boundary conditions. Equation (5.22) gives c1 = 1 and (5.24) gives 0 = c π +1 from which (5.31) is found to give ⎛ y ⎞ u = 1 − F ⎜⎜ ⎟⎟ u0 ⎝ 2 νt ⎠ (5.32) We see from figure 5.5 that this expression describes a velocity profile of the shape indicated in figure 5.4. The error function F(Y) approaches 1 asymptotically for Y → ∞ and the movement of the wall can therefore in principle be felt infinitely far from it at any positive time. The main part of the flow is, however, restricted to a zone near the wall (the boundary layer!). The definition of this boundary layer can be made somewhat arbitrarily. As an example the thickness may be defined as the distance where the velocity is reduced to one percent of the velocity of the wall, or u = 0.01 u0 ⇒ F (Y ) = 0.99 (5.33) From tables we find that (5.33) means Y = 1.82. (If the choice had been one per mille we had found Y = 2.31). 110 The boundary layer thickness is by this definition found to be δ = 1.82 × 2 νt = 3.64 νt (5.34) From which we see that the boundary layer thickness starts from zero at t = 0 when the plate is set into motion, then δ grows as the square root of the time. The shear stress at the wall τ 0 can be found as ⎛ dU ∂Y ⎞ ⎛ ∂u ⎞ ν u0c τ0 1 u0 ν ⎟⎟ = = ν u 0 ⎜⎜ =− = ν⎜⎜ ⎟⎟ ρ νt π νt ⎝ dY ∂y ⎠ y ,Y =0 ⎝ ∂y ⎠ y =0 (5.35) It is seen that τ 0 is infinitely large at t = 0 (where the plate is given an infinite acceleration) and thereafter decreases with time while δ increases. B. Approximate solution by use of the momentum equation: The advantage of using the momentum equation for boundary layer calculations is that the calculations are made considerably simpler. This is obtained at the expense of knowledge of the details in the velocity profile. The momentum equation (5.20) is integrated over the boundary layer, and the exact shape of the velocity profile is therefore not crucial for the assessment of the variation in the boundary layer thickness and the bed shear stress. In this example we can, for example, make the guess that the velocity profile is given by a polynomium, which fulfils the two boundary conditions 1: u = u0 2: u=0 for y=0 (5.36) and for y=δ (5.37) These conditions are met by the first order polynomium f1 (η) = u y = 1− = 1− η u0 δ (5.38) where the dimensionless coordinate η = y / δ has been introduced. This profile is of course a very crude approximation to the actual velocity distribution. It has not been used that the boundary layer flow has a smooth transition to the outer flow (where the velocity in this example is 0), so that the gradient in the velocity is zero at the outer boundary 3: ∂u =0 ∂y for y=δ (5.39) This boundary condition can in addition to the two first be satisfied by the polynomium of second order 111 f 2 (η) = 1 − 2η + η 2 (5.40) but also, for example, by the function ⎛π ⎞ f 3 (η) = 1 − sin ⎜ η ⎟ ⎝2 ⎠ (5.41) In the following all three assumed profiles (5.38), (5.40) and (5.41) are used with the momentum equation. Because of the uniformity in the x-direction the momentum equation (5.20) can be reduced to δ ∂u ∫ ρ ∂t dy = −τ 0 (5.42) 0 where the wall shear stress τ 0 in this case will be negative. As uo is constant the timederivative of u is found as df ⎛ y ⎞ dδ df ∂η ∂f ∂u = u0 = u0 = u0 ⎜− ⎟ dη ⎝ δ 2 ⎠ dt dη ∂t ∂t ∂t (5.43) In a similar way τ 0 is found as τ0 = ρ ν ∂u df ∂η = ρ ν u0 , η=0 ∂y dη ∂y (5.44) or τ0 = ρ ν u0 df 1 dη δ , η=0 (5.45) Insertion of (5.43) and (5.45) into (5.42) gives 1 dδ df f ′(0) η dη = ρνu 0 ρu 0 ∫ dt 0 dη δ (5.46) or d dt ( δ )= ν 1 2 f ′(0) 2 1 (5.47) df ∫ dη η dη 0 which by integration gives 112 δ 2 = 2ν t f ′(0) (5.48) 1 ∫ η f ′ dη 0 By inserting the three estimated velocity profiles f1, f2 and f3 into (5.48) the sensitivity of the momentum equation for different choices of the profile can be assessed. All the profiles give the correct variation of δ and τ 0 with respect to time and viscosity, but as seen in table 5.1 the constants involved differ. In the column next to δ the variation in so-called displacement thickness δ * is given. This is introduced to have a more well-defined measure of the boundary layer thickness, applicable for the exact solution as well as the approximation obtained from the momentum equation. In connection with the exact solution we found that the thickness of the boundary layer depended on the choice of definition, based on u = 0.01uo or u = 0.001uo. This way of defining the boundary layer thickness is also very difficult to apply in connection with analysis of measurements. To overcome this ambiguity δ * is defined as the distance from the wall, corresponding to the 'displacement' of the outer flow due to the deficit in the velocity inside the boundary layer, giving ∞ u 0 δ = ∫ u dy * (5.49) 0 cf. figure 5.6. δ/ νt δ* / ν t f1 (η) = 1 − η 2.00 1.000 0.500 f 2 (η) = 1 − 2η + η 2 3.46 1.155 0.578 f 3 (η) = 1 − sin(ηπ / 2) 2.94 1.068 0.534 exact (u δ / u 0 = 0.01) 3.64 1.129 0.564 exact (u δ / u 0 = 0.001) 4.62 1.129 0.564 Velocity distribution τ 0 ν t / ρν u 0 Table 5.1 It is seen from table 5.1 that the variation in δ * is quite moderate when changing from one velocity profile to the other. 113 Definition of the displacement thickness δ * : the two hatched areas shall be of the same size. Fig. 5.6 Example 5.2 Stationary, non-uniform flow along a plate. In the previous example the flow was identical for all values of x because the plate was taken to be of infinite extension. In the following we will instead consider the boundary layer formed along a plate when it is met by a flow of constant velocity uo parallel to the plane of the plate, cf. figure 5.2. Since uo is constant the pressure p outside the boundary layer is also constant giving ∂p / ∂x = 0 outside the boundary layer. According to the boundary layer equation (5.4) it follows immediately that the pressure gradient in the boundary layer also is zero and the momentum equation (5.20) can be reduced to ∂ δ ∂ ∂ ρu 2 dy − u 0 ρu dy = −τ 0 ∫ ∂x 0 ∂x ∫0 (5.50) We try as in the previous example to make a guess for the velocity profile. We assumed a profile described by u ⎛ y⎞ = f⎜ ⎟ u0 ⎝δ⎠ (5.51) in which the time-variation is included only through the variation of δ with t. The profiles are in this way said to be similar or affine. As we shall see when considering flow separation this assumption is not valid in general. For the present situation we can, however, use the assumption (5.51) because the outer pressure gradient is 0. The procedure is exactly as in the previous example. Possible velocity profiles to use can be ⎛π ⎞ f1 (η) = sin ⎜ η ⎟ ⎝2 ⎠ (5.52) similar to (5.41), or 114 f 2 (η) = 3η − 3η 2 + η3 (5.53) Equations (5.52) and (5.53) both satisfy the boundary conditions u=0 for η=0 u = u0 for η =1 (5.54) and du =0 dx for η =1 Equation (5.53) is an extension of (5.40) from a polynomium of sendond to third order. In this way the additional boundary condition d 2u (5.55) = 0 for η = 1 dy 2 is also satisfied. Only the calculations for the first approximation, the sinus-shaped profile f1, will be carried out here. The two first terms in (5.50) gives δ π/2 ⎤ ∂ ⎡ 2 2δ ∂ u ( u u ) dy u sin ξ (1 − sin ξ) dξ⎥ ρ − = − ρ ⎢ 0 0 ∫ ∫ ∂x ⎢⎣ π 0 ∂x 0 ⎥⎦ =− ∂ ⎡ 2 2δ ⎛ π ⎞⎤ ⎜1 − ⎟ ⎥ ⎢ρu 0 ∂x ⎣ π ⎝ 4 ⎠⎦ = −ρu 02 (5.56) 2 ⎛ π ⎞ dδ ⎜1 − ⎟ π ⎝ 4 ⎠ dx while τ 0 is given as ⎛ ∂u ⎞ πu = ρν 0 τ 0 = ρ ν⎜⎜ ⎟⎟ 2δ ⎝ ∂y ⎠ y =0 (5.57) the momentum equation (5.50) is thus reduced to ρu 0 πu 2 ⎛ π ⎞ dδ = ρν 0 ⎜1 − ⎟ π ⎝ 4 ⎠ dx 2δ (5.58) which has the solution 115 δ= π 2 − π/ 2 νx νx = 4.8 u0 u0 (5.59) taking x = 0 at the leading edge of the plate, cf. figure 5.2, where the boundary layer thickness starts from 0. As done in the previous example the displacement thickness can be determined. The definition is slightly different when the wall is fixed, compare figures 5.6 and 5.7 ∞ u 0 δ * = ∫ (u 0 − u ) dy (5.60) 0 Fro the sinus profile we get νx π−2 δ* = δ = 1.74 π u0 (5.61) Table 5.2 gives, similarly to table 5.1, the variation in the displacement thickness and wall shear stress for the two assumed profiles and for the exact solution, which for this case can only be obtained numerically. Velocity distribution δ* / ν x / u 0 τ 0 u 0 x / ν / ρu 02 f1 (η) = sin(ηπ / 2) 1.741 0.328 f 2 (η) = 3η − 3η 2 + η3 1.740 0.323 1.721 0.332 exact Table 5.2 Fig. 5.7 Definition of the displacement thickness: the two hatched areas must be of equal size. 116 5.4 Boundary layers with outer pressure gradients As mentioned in example 5.2 the assumption of similarity is not valid in general. The calculational work increases dramatically when the pressure gradient is different from zero, and the problem will only be treated cursory in the following. Let us consider a plane flow condition as sketched in figure 5.8. The water flows from left to right, first through a channel with converging walls. Here the flow velocity increases in the flow direction, which means that the pressure decreases in the flow direction, cf. the Bernoulli equation. At the right side the water flows through diverging walls so that the pressure increases in the flow direction. In other words: in the converging part we have ∂p / ∂x < 0 and in the diverging part ∂p / ∂x > 0 . It is only in the converging section that the flow has the character of a boundary layer flow. The flow in the diverging section will tend to separate with a return flow close to the wall. The velocity profiles are thus completely different in the two sections and may not be approximated by similar profiles. A more general way of describing the velocities in stationary flows with pressure gradients is the Pohlhausen method. The main ideas of this method are outlines in the following. It is assumed that the velocity profiles can be described by a polynomium such as u = a η + b η 2 + c η3 + d η 4 u0 , η= y δ (5.62) which automatically fulfils the condition u = 0 for η = 0. To determine the coefficients in (5.62) we apply as before the following boundary conditions u = u0 for η =1 ∂u =0 ∂y for η =1 ∂ 2u =0 ∂y 2 Fig. 5.8 for (5.63) η =1 The velocity profiles in converging and diverging flow. 117 To introduce the pressure gradient in the description we now revert to the boundary layer equation (5.2), which at the wall degenerates to − 1 ∂p ∂ 2u +ν 2 =0 ρ ∂x ∂x (5.64) using that u and v are exactly 0 at the wall. Using (5.64) the last boundary condition may be formulated as ∂ 2 (u / u 0 ) ∂η 2 = δ 2 ∂p = −Λ , ρν u 0 ∂x η=0 (5.65) where Λ is a measure of the pressure gradient, which we know is related to the outer flow velocity uo by ∂u 1 ∂p = −u 0 0 ρ ∂x ∂x (5.66) Using the boundary conditions (5.63) and (5.65) the four coefficients in (5.62) are determined, giving u = F (η) + Λ G (η) u0 (5.67) where F = 2η − 2η3 + η 4 and (5.68) 1 G = (η − 3η 2 + 3η3 − η 4 ) 6 It is thus seen that the velocity profiles in plane stationary boundary layers can be described as a one-parameter family with Λ as the shape-parameter. Examples of possible shapes of profiles are shown in figure 5.9. In general the velocity profile will vary along the wall corresponding to the variation in Λ . The parameter Λ is constant only for affine boundary layers, as for example in the flow along a plane plate parallel to the flow direction (example 5.2) where Λ = 0, By use of (5.67) and the momentum equation for the boundary layer any plane stationary boundary layer flow may in principle be calculated. Figure 5.9 illustrates the very important difference between flow with a positive (opposing) and a negative (favourable) pressure gradient: 118 Fig. 5.9 Velocity profiles in boundary layers with positive and negative pressure gradients. In a boundary layer flow with a flow in the direction of a pressure drop Λ will be positive and the velocity profiles will show very little variation from one crosssection to the other. Conversely, if the flow is in the direction of a pressure increase Λ will be negative, which drastically changes the shape of the velocity profile. For Λ = -12 the velocity gradient is zero at the wall corresponding to the wall shear stress being zero. For even larger negative values of Λ , there is a return flow near the wall, a situation described as the boundary layer separating from the wall. The point where Λ = -12 is known as the separation point. In a separating flow the assumptions behind the boundary layer theory normally breaks down: the thickness of the boundary layer increases drastically and the assumption that velocity component normal to the wall is always small compared to the longitudinal velocity is no longer valid. 119 Example 5.3 Flow around a circular cylinder Fig. 5.10 Flow around a circular cylinder. Top: potential flow. Bottom: laminar flow, photo: Re (= 2u 0 R / ν) = 26 (from [4]). The potential flow around a circular cylinder may be calculated based on the principles given in chapter 3. The solution for the potential becomes ⎡ R2 ⎤ φ = −u 0 x ⎢1 + 2 2⎥ ⎣ x +y ⎦ (5.69) The stream lines are shown in figure 5.10 (top). This flow pattern is, however, only a good approximation to the real flow, figure 5.10 (bottom), at the upstream part of the cylinder – and then outside the boundary layer. At the downstream part of the cylinder the flow is diverging, and separation will take place at the two points S, figure 5.10 (bottom). The flow separation is crucial for the force from the current – the drag force – acting on a cylinder, which is fixed in flowing water. If there are no separation, there is a thin boundary layer along the entire perimeter of the cylinder, and the only force acting on the cylinder is due to the wall friction. The pressure in the boundary layer flow is symmetrical around the y-axis as shown in figure 5.11, because the pressure outside the boundary layer can be determined from Bernoulli's equation. The flow separation will change the pressure distribution drastically. The velocities in the separation zone are much smaller than 120 outside it. The separation zone thus forms a wake with a rather small variation in the pressure. The transition between the outer flow and the wake is very sharp as Fig. 5.11 Sketch of the pressure distribution around a circular cylinder. indicated in the photo of figure 5.12. As an approximation the pressure in the wake may be taken to be equal to the pressure at the separation points. The actual pressure distribution is therefore as sketched with the dashed line in figure 5.11, which is seen to result in pressure force acting on the cylinder in the direction of the flow. Figure 5.13 shows an example of the measured pressure distribution around the perimeter of a cylinder submerged in flowing water. The measurement indicates the so-called pressure coefficient, which is defined as ( p − p0 ) Cp = (5.70) 2 1 U ρ 0 2 where po is the undisturbed pressure far upstream. In this case the measurements are made with a high value of the Reynolds number (The Reynolds number is defined in chapter 6, cf. (6.19)), and the boundary layer at the upstream side is turbulent (cf. Fig. 5.12 Photo of laminar and turbulent separation around a cylinder, from [4]. 121 chapter 6 and 7), but qualitatively we find the same pressure distribution at high and low Re numbers, with the exception of very small Re numbers (<5), where separation does not occur. This is because the velocities in that case are so small that the square of the velocity (non-linear terms) can be neglected when compared to the shear stresses. This means that the inertia force (the convective term u ∂u / ∂x ) is neglected. The flow in this regime is known as creeping, cf. figures 5.14 and 5.15. When the Reynolds number is larger than approximately 40 the wake downstream of the cylinder becomes unstable. Small disturbances will be amplified and one of the two downstream circulation vortices will be shed from the cylinder and carried with the flow. While a new vortex is being formed, the remaining vortex at the other side of the cylinder will be shed, and the process continues. This is known as rhythmically vortex shedding, cf. figure 5.14 and the photography in figure 5.16. Fig. 5.13 Measured pressure distribution around a circular cylinder, Re = 3.6 106. The vortex shedding occurs with a frequency f determined by the relation fd =S U0 where d is the diameter, Uo the undisturbed velocity and S is a dimensionless number known as the Strouhals number, which for a wide range of Re has the value 0.2. 122 Fig. 5.14 The flow pattern around a cylinder at different values of Reynolds number. Fig. 5.15 Photo of creeping flow. The undisturbed velocity is 1 mm/s. The flow takes place between two parallel glass plates with a mutual distance of 1 mm. (Hele-Shaw apparatus). 123 Fig. 5.16 Rhythmic vortex shedding forming a vortex street. Fig. 5.17 The variation in Strouhals number with Reynolds number, based on physical experiments. 124 125 6. Turbulent flows If we consider the flow through a pipe, two different flow situations can occur for the same fluid depending on the flow velocity. For low flow velocities the flow is laminar (cf. chapter 2), but when the velocity exceeds a certain critical value, Fig. 6.1 Reynolds' experiment, showing the transition from laminar to turbulent flow in a pipe (from [4]). The flow changes to become turbulent. This is illustrated in figure 6 by O. Reynolds' famous experiment from the 1860-ties, where the transition from the regular organized laminar flow to the more irregular and stochastic turbulent flow is clearly visible. If the pipe diameter or the viscosity of the fluid is changed it will be found that the critical flow velocity is also changed. It was shown that the point of transition from laminar to turbulent flow can be characterized by the dimensionless number 129 Re = UD ν (6.1) where U is a characteristic velocity (for example the mean velocity in the pipe), D a typical dimension (for example the diameter of the pipe) and ν is the kinematic viscosity of the fluid. For flow in a pipe the critical Reynolds number is approx. 2000. If we consider a pipe with a diameter of 0.1 m we find for water ( ν = 10-6 m2/s) the flow will be turbulent for flow velocities larger than about 0.02 m/s. In the field of hydraulic engineering the flows to be dealt with are therefore in most cases turbulent. An important example of laminar pipe flow is found in the human blood vessels where the Reynolds number is moderate due to the small dimensions (the flow in aorta can, however, be turbulent). Fig. 6.2 A turbulent spot (from[4]). The transition from laminar to turbulent flow is gradual. If the velocity is is only slightly above the critical the turbulence may emerge locally as shown in figure 6.2, which is a picture of a 'turbulent spot'. The Reynolds number has to be considerably larger (by a factor 10-50) for the flow to evolve to become fully turbulent. 6.1 The stability of a laminar flow The emergence of turbulence can be explained as the result of an instability in the laminar flow: basically any flow could be laminar since the laminar flow fulfils the flow equations and the proper boundary conditions. It can, however, be shown that under some circumstances (for large values of the Reynolds number) any small perturbation to the laminar flow will grow in time so that the nice regular laminar flow breaks up to become turbulent. This section treats the mathematical formulation of such a stability analysis. Normally the starting point of a hydrodynamic stability analysis is the socalled vorticity transport equation, which can be derived from the Navier-Stokes equation (2.30) and (2.31), cf. also (A8), appendix I: 130 ⎛ ∂ 2u ∂ 2u ⎞ du 1 ∂p =− + g x + ν⎜ 2 + 2 ⎟ ⎜ ∂x ρ ∂x dt ∂y ⎟⎠ ⎝ and (6.2) ⎛∂ v ∂ v⎞ dv 1 ∂p =− + g y + ν⎜ 2 + 2 ⎟ ⎜ ∂x ρ ∂y dt ∂y ⎟⎠ ⎝ 2 2 By making a differentiation of the first equation with respect to y and similarly a differentiation of the second with respect to x the pressure p can be eliminated to give ⎛ ∂u ⎞ ∂ ⎛ ∂u ⎞ ∂ ⎛ ∂u ∂u ⎞ ⎜ ⎟ + ⎜u + v ⎟⎟ − νΔ⎜⎜ ⎟⎟ = ⎜ ⎟ ⎜ ∂t ⎝ ∂y ⎠ ∂y ⎝ ∂x ∂y ⎠ ⎝ ∂y ⎠ (6.3) ∂v ⎞ ∂ ⎛ ∂v ⎞ ∂ ⎛ ∂v ⎛ ∂v ⎞ + v ⎟⎟ − νΔ⎜ ⎟ ⎜ ⎟ + ⎜⎜ u ∂y ⎠ ∂t ⎝ ∂x ⎠ ∂x ⎝ ∂x ⎝ ∂x ⎠ By application of the equation of continuity ∂u ∂v =0 + ∂x ∂y equation (6.3) can be modified to ∂ω ∂ω ∂ω = νΔω +v +u ∂y ∂x ∂t (6.4) where 2ω = ∂v ∂u − ∂x ∂y (6.5) cf. equation (2.21). Equation (6.4) can be written dω = νΔω dt (6.6) which is known as the vorticity transport equation. In a general three-dimensional flow we find similarly 131 dω1 ∂u = νΔω1 + ω k dt ∂x k dω 2 ∂v = νΔω 2 + ω k dt ∂x k (6.7) dω3 ∂w = νΔω3 + ω k dt ∂x k which is quite similar to (6.6) except for the last terms, expressing the effect of stretching of vortices, which can not occur in a purely two-dimensional flow. We now consider a laminar flow, for example the laminar boundary layer forming along a plane plate as described in section 5.2. The streamlines in this laminar flow are parallel to the plate. These streamlines are given a small perturbation, see figure 6.3, so that the original velocity field u = U ( y) v=0 (6.8) u = U ( y ) + u~ v = v~ (6.9) is changed to where u~ and v~ are the small velocity perturbations introduced to the flow. Fig. 6.3 A perturbation of a boundary layer flow. The basic flow has a vorticity Ω , determined from (6.8) to be Ω=− 1 ∂U 2 ∂y (6.10) while the vorticity of the perturbation is called ω . For the combined flow the vorticity transport equation now reads, cf. (6.4) 132 ∂ ∂ω ∂ + (U + u~ ) (Ω + ω) + v~ (Ω + ω) = ν(ΔΩ + Δω) ∂t ∂x ∂y (6.11) For the unperturbed basic flow the vorticity transport equation is all valid and it reads U ∂Ω = νΔΩ ∂x (6.12) By subtracting (6.12) from (6.11) and using that the basic flow is a boundary layer where ∂Ω ∂Ω << ∂x ∂y we finally get dω ∂Ω ~ + v = νΔω dt ∂y (6.13) By multiplying (6.13) by ω and using (6.10) equation (6.13) can be reformulated as d ⎛1 2⎞ ⎜ ω ⎟= dt ⎝ 2 ⎠ I 1 U ′′ωv~ + νωΔω 2 II III (6.14) The physical significance of the three terms is as follows I: Represents the time variation of the rotational energy of a fluid particle, given as 1/2 ω 2 (equivalent to the kinetic energy). II: Is the production of rotational energy. This can be seen from figure 6.4: We consider a fluid particle at the distance a from the bed, which at the time t = 0 has the rotation Ω + ω . After the time interval dt the particle is now at the level (a+vdt) above the bed, but the vorticity will be practically the same because the effect of the viscosity almost negligible for so short a time. This means that because of the introduced velocity perturbation v the fluid particle will now have a perturbed vorticity of 2 dΩ ~ ⎞ 1⎛ 1 dΩ ~ ⎜⎜ ω − v dt ⎟⎟ = ω 2 − ωv dt + .... dy 2⎝ 2 dy ⎠ (6.15) From this it is seen that the term − dΩ ω ν = U ′′ω ν dy (6.16) 133 which is representing II, signifies the increase in rotational energy ω 2 / 2 due to transfer from the basic flow per unit time. III: is the work made by the viscous forces, whereby kinetic energy is transformed to heat, cf. (2.16). This term (dissipation) is always negative because any motion will dye out with time if no energy is transferred from outside. Fig. 6.4 Production of rotation Equation (6.14) is now integrated of the thickness of the boundary layer, and all quantities are made dimensionless as follows ω Ω , Ω′ = ω′ = Ω0 Ω0 v U x x′ = δ v′ = , , Ut δ y y′ = δ t′ = (6.17) where Ω 0 is a typical value of the undisturbed vorticity in the boundary layer and U is a characteristic velocity, for example the velocity immediately outside the boundary layer. In this way (6.14) turns into 1 1 ⎛ ⎞ ′ d 1 2 ′ ′ (ω ) dy = ∫ ⎜⎜ − dΩ ω′v ′ ⎟⎟dy ′ ∫ dt 0 2 dy ′ ⎠ 0⎝ (6.18) ν ω′Δω′ dy ′ Uδ ∫0 1 + 134 It is seen from (6.18) that for identical boundary conditions all types of flows will dynamically similar if the number Re = Uδ ν (6.19) is the same. This is Reynolds number, cf. (6.1). If two flows are dynamically similar (that is have the same Reynolds number) then both flows will be either stable (that is they will remain laminar) or unstable (they will break up and become turbulent). The Re number can therefore be used as the criterion for a flow being laminar or turbulent. As mentions term III (the heat loss) will always be negative. From (6.18) it is seen that for small values of the Re number I will also be negative because term II (the production of energy) is negligible. When I is negative the vorticity of the perturbation will decrease with time meaning that the flow is stable and will remain laminar. For larger Re numbers the term I will become positive, provided that the term 1 dΩ ′ ∫ − dy ′ ω′v′ dy ′ 0 is positive. Normally ther will always exist a perturbation with a combination of ω′ and v ′ for which this integral becomes positive. As I in that case will be positive for large Re numbers the perturbation will grow with time and the flow will become turbulent. Example 6.1: Boundary layer along a flat plate If we consider a flow without any pressure gradients along a plate (cf. example 5.2) the Reynolds number will grow along the plate because the boundary layer thickness δ increases in the flow direction. When the Reynolds number reaches a critical value the flow will break up and become turbulent as show at section A-A in figure 6.5. Refined stability analyses has shown the critical Reynolds number for this flow to be about 400. Inserting δ = 4.8 νx U from (5.55) the position of the transition can be estimated because 135 Fig. 6.5 Transition from laminar to turbulent boundary layer flow along a plate. Re = 400 = Uδ 4.8U = ν ν νx U or x= 160000 ν 4.8 2 U For ν = 10-6 the transition to turbulent flow will take place 0.7 cm from the leading edge for U = 1 m/s and 7 cm from the edge for U = 0.1 m/s. 6.2 Turbulent fluctuations A fully developed turbulent flow may be interpreted as a regular mean motion superposed by irregular turbulent fluctuations associated with eddies or vortices. The flow defined as stationary when the mean motion does not change with time. If we as an example consider a stationary pipe flow, the flow in the direction of the axis will vary with time as sketched in figure 6.6A. The instantaneous velocity u can be written as u = U + u′ (6.20) wher U is the mean value and u' is the turbulent fluctuation. The mean flow normal to the axis of the pipe is zero, and the other two velocity components are given as (cf. figure 6.6B) v = v′ (6.21) w = w′ (6.22) and In a turbulent flow the turbulent fluctuations will always be three-dimensional. 136 Fig. 6.6 A registration of the velocity components as function of time in a pipe flow. A deterministic description of the turbulent fluctuations u', v' and w' is not possible, but by use of turbulence models it is today possible to describe a many of the mean values characterising the turbulence, such as mean flow velocities and mean shear stresses. 6.3 Reynolds stresses Consider a uniform, stationary open channel flow over a plane bed, cf. figure 6.7. In this flow the shear stress τ is determined by a static force balance and given by τ = ρg ( D − y ) sin β (6.23) cf. [2], p. 40. At small velocities the flow is laminar, and vi have τ du =ν ρ dy (6.24) From (6.23) and (6.24) we find 137 U= Fig. 6.7 g sin β ⎛ y2 ⎞ ⎜ Dy − ⎟ ν ⎜⎝ 2 ⎟⎠ (6.25) The distribution of the shear stress and velocity profiles in uniform channel flow. When the flow becomes turbulent equation (6.23) is still valid because τ as mentioned is determined by a static force balance. Equation (6.24) is, however, changed drastically due to the turbulent velocity fluctuations. The eddies cause an exchange of momentum in the direction normal to the flow direction, quite similar to the momentum exchange due to the heat motion of the molecules which is the mechanism behind the viscosity. In a turbulent flow the exchange due to the eddies is normally several orders of magnitude larger than the molecular exchange, and the viscosity can often be neglected. To analyse the effect of turbulent fluctuations on the equation of motion for the mean flow in a turbulent flow situation we take the Navier-Stokes eqautions as starting point. The Navier-Stokes equations are valid for the instantaneous flow conditions in turbulent as well as laminar flows. The turbulent velocity fluctuations are three-dimensional and it is therefore convenient to derive the flow equations for the mean flow by application of the index (or tensor) notation, cf. [2] p. 29. In index notation the Navier-Stokes equation reads dvi ∂ 2 vi ∂p (6.26) +μ ρ =− dt ∂x j ∂x j ∂xi Consider the left hand side, which can be written ∂v j ∂v ∂v dvi ∂vi ∂ = +vj i = i + (v i v j ) − vi ∂t ∂x j ∂t ∂x j dt ∂x j (6.27) Here the last term on the right hand side is zero, because the continuity equation must be satisfied. By introduction of (6.27) into (2.26) we get vi = U i + u i′ (6.28) where U i represents the mean motion and u i′ the turbulent velocity components we get 138 ρ ∂ (U i + u i′ ) ∂ ((U i + u ′)(U i + u ′) = +ρ ∂t ∂x j ∂ ( P + p ′) ∂2 (U i + u ′) − +μ ∂xi ∂x j ∂xi (6.29) Each term in (6.29) is time averaged by taking T φ= 1 φ dt T ∫0 (6.30) where T is a sufficiently large time interval and φ can be taken as any of the variables in (6.29). The second term in (6.29) can now be written ∂ (U iU j + U i u ′j + u i′U j + u i′u ′j ) = ∂x j ∂ (U iU j + U i u ′j + u i′U j + u i′u ′j ) = ∂x j ∂ ∂ ∂ ∂ (U iU j ) + (U i u ′j ) + (u i′U j ) + (u i′u ′j ) = ∂x j ∂x j ∂x j ∂x j (6.31) ∂ ∂ (U iU j ) + (u i′u ′j ) ∂x j ∂x j because Ui = Ui and u i′ = 0 In this way equation (6.29) can - when time averaged - be written ρ ∂U i ∂ +ρ (U i U j ) ∂t ∂x j =− ∂ 2U i ∂ ∂P +μ −ρ ( u i′u ′j ) ∂x i ∂x j ∂x j ∂x j (6.32) or dU i ∂ 2U i ∂P ∂ ρ =− +μ −ρ (u i′u ′j ) dt ∂xi ∂x j ∂x j ∂x j (6.33) where the continuity equation once more has been applied on the left hand side of (6.33). Equation (6.33) may also be written 139 ρ dU i ∂ ∂P σ ij + =− dt ∂xi ∂x j (6.34) where σ ij = ρν ∂U i ∂x j − ρu i′u ′j (6.35) When we compare (6.34) to (6.26) we see that the equations for the mean motion are identical to the Navier-Stokes equations except for some additional stresses in the form of the term u i′u ′j . These additional stresses are know as the Reynolds stresses, which emerge because the turbulent velocity fluctuations causes an exchange of momentum between the different parts of the flow. This is easily seen when only two dimensions are considered: If we consider a section parallel to the x-axis in the direction of the mean flow, the flow through this section per time will be ρv ′ . This corresponds to a transport of momentum of ρv ′(U + u ′) or in average ρv ′u ′ . This exchange of momentum must in the equation for the mean motion be represented as an apparent stress, which on average is given as − ρv ′u ′ . Fig. 6.8 Transverse exchange of momentum. We return to the flow shown in figure 6.7 where the shear stress is determined from a static force balance. In a laminar flow the shear stress is due to the molecular heat motion, but in the case of a turbulent flow it can also be carried by the exchange of momentum of the eddies, equation (6.24) is then changed to τ dU =ν − u ′v ′ ρ dy (6.36) Since τ is unchanged, the mean velocity profile must be modified when the flow becomes turbulent. This change can be calculated if the term u ′v ′ is known. At present only semi-empirical theories are available for making simple estimates of this term. Some of the most simple theories will be described in the following. 140 6.4 The mixing length theory Let us again consider the uniform open channel flow in figure 6.7. The flow is assumed to be turbulent, so that the mean flow is superposed by eddies of different size and rotational speed. The section A-A parallel to the bed (cf. figure 6.9) is considered. Across this section there is an exchange of fluid due to the eddies. These eddies are assumed to have Fig. 6.9 The exchange of momentum normal to the mean flow direction. a characteristic size, described by the length scale A . An eddy with this length scale will be able to transport fluid from one level I to another higher level II with a mutual distance of the order A . At the same time there will due to continuity be transported the same amount of water down from II to I. Through section A-A there will be a mass flux per unit area of the order of magnitude ρq / 2 - going up as well as down. Her q may be interpreted as a characteristic value of the turbulent fluctuations normal to the mean flow direction. The fluid particles transported from level I are assumed to maintain their longitudinal flow velocity unchanged until they reach level II, where they are mixed with the surrounding fluid and obtain the mean flow velocity at this level (this is the reason for the name 'mixing length'). As seen from figure 6.9 these particles will increase their velocity by U II − U I ≈ A dU dy (6.37) Since there is a flux of q up per unit time (and similarly a flux q down) the total exchange of momentum per unit area through A-A is found to be dU ΔM = ρqA dy When formulating an equation of motion for the mean flow, in which the turbulent fluctuations are averaged out, you must as seen in (6.36) include the effect of the turbulent fluctuations as a shear stress given by 141 τ = ρ qA dU dy (6.38) q is as mentioned of the same order of magnitude as the vertical velocity fluctuations. For reasons of continuity these must be of the same order of magnitude as the horizontal velocity fluctuations, which are assumed to be of the same order of magnitude as the difference in the mean flow velocity over the distance of the mixing length, giving q≈A dU dy whereby (6.38) becomes τ = ρA 2 dU dU dy dy (6.39) To get further it is, however, necessary with knowledge of the mixing length A . This may be expected to be a function of the distance from the wall A = A (y) and must decrease towards 0 at the wall where the turbulence ceases. Since A is a length scale the most simple assumption is to take it to vary linearly with the distance y from the wall, A = κy (6.40) where κ is a constant (know as von Kármáns constant; κ is close to 0.40. The transfer of momentum due to the eddies in a turbulent flow is as described analogous to the transfer by the motions of the molecules. The latter gave rise to Newton's formula (2.5), and because of the analogy you operate with technical turbulence theories where an expression similar to (2.5) is introduced to describe the shear stress, namely dU τ = ρν t (6.41) dy where ν t is called the eddy viscosity. When comparing (6.41) with (6.39) we find νt = A2 dU dy (6.42) In the derivation of (6.42) the moleculous viscosity has been neglected. It is easily seen that if the kinematic viscosity is to be included (6.41) is modified to τ = ρ(ν t + ν) dU dy (6.43) 142 where ν t is still found from (6.42). normally ν t >> ν , so that ν can be neglected, except in situations, as will be described in the following. 6.5 Velocity profiles in uniform channel flows The velocity distribution in an infinitely wide channel can easily be calculated by use of the mixing length theory, i.e. from (6.40), (6.42) and (6.43). The shear stress is given by τ τ0 ⎛ Y ⎞ = ⎜1 − ⎟ ρ ρ ⎝ D⎠ (6.44) where τ 0 is the shear stress at the bed. This must balance the longitudinal component of the gravity force of the water and is therefore given by τ 0 = ρgDI (6.45) where I is the slope of the water surface. The friction velocity (or shear velocity) Uf is by definition given as τ0 Uf = (6.46) ρ From (6.44), (6.45) in combination with (6.40), (6.41) and (6.42) we get' y⎞ dU ⎛ U 2f ⎜1 − ⎟ = (ν t + ν) dy ⎝ D⎠ 2 ⎞ dU ⎟⎟ + ν = A ⎜⎜ dy ⎝ dy ⎠ 2 ⎛ dU (6.47) 2 ⎞ dU ⎟⎟ + ν = κ y ⎜⎜ dy ⎝ dy ⎠ 2 A. 2 ⎛ dU A smooth bed Let us consider the flow over a completely smooth bed. It is now seen that no matter how small ν is, the viscous term (2. term on the right hand side of (6.47)) will be larger than the turbulent contribution (1. term on the right hand side of (6.47)) when we are sufficiently close to the bed and y tends towards zero. If the turbulent contribution is neglected, equation (6.47) is reduced to (valid for small values of y) 143 U 2f = ν dU dy (6.48) which with the boundary condition U = 0 for y = 0 gives Uf U = y ν Uf (6.49) However, we only have to move a small distance from the wall to have ν t >> ν . Away from the wall (6.47) reads dU ⎞ ⎟⎟ U = κ y ⎜⎜ ⎝ dy ⎠ 2 f 2 2⎛ 2 (6.50) where the weak variation in τ with y has been neglected on the left hand side of (6.50). Equation (6.50) can then be written dU 1 dy (6.51) = Uf κ y which can be integrated to give 1 U = ln( y ) + const Uf κ (6.52) We have thus a layer very close to the wall where the velocity varies linearly (6.49), while the velocity a little further from the wall varies logarithmicaly as illustrated in figure 6.10, where the dimensionless velocity U/Uf is show as function of the dimensionless vertical coordinate y+, given by y+ = y Fig. 6.10 Uf ν (6.53) The velocity profile close to the bed for a smooth bed. 144 Figure 6.11 shows similar measurements made near the wall in a pipe flow. Away from the wall the measurements follow a straight line with a logarithmically scaled y+-axis, cf. figure 6.11B. Fig. 6.11 Measurements of the mean velocity profile over a smooth wall. If equation (6.52) is rewritten as 1 U = ln y + + K1 Uf κ (6.54) we see from figure 6.11 that all measurements are following the same straight line, and that K1 in (6.54) shall be chosen to be K1 = 5.7 (6.55) This value is therefore determined experimentally, but today it may also be derived theoretically by application of more sophisticated models. Very close to the wall the points deviate from the straight line given by (6.54), which is because the viscous forces become dominant. The distance from the wall to the point of intersection between the two velocity profiles (6.54) and (6.53) is know as the thickness of the viscous sub-layer δ v . The thickness of the viscous sub-layer is thus found from U f δv ν ⎛ U f δv ⎞ ⎟ = 5.7 + 2.5 ln⎜⎜ ⎟ ν ⎠ ⎝ (6.56) 145 Fig. 6.12 Photography of the viscous syb-layer. a: seen fromt the side, b: seen from above, c: detail from b. From [4]. or U f δv ν = 11.7 (6.57) From this the thickness of the viscous sub-layer is found to be δ v = 11.7 ν Uf (6.58) This description of the near-wall conditions is naturally based on simplifications. In reality the nature of viscous sub-layer is not like a laminar flow, as indicated in the photograph in figure 6.12, where a a complex pattern with a large number of 'streaks' in this layer are clearly visible. _____________________________________________________________________ Example 6.2 The flow resistance and the thickness of the viscous sub-layer in a smooth channel flow. Let us consider a channel flow with a discharge of Q = 1 m3/m/s and a slope I of 10 . We want then to determine the thickness of the viscous sub-layer. For this we need to know the shear velocity Uf, cf. (6.58). If we assume the thickness of the viscous sub-layer is very small compared to the total water depth, the total discharge can be related to Uf and D by integration of (6.54) over the depth -4 146 D Q = ∫ Udy ≈ 0 D Uf ⎛ yU f ⎞ ⎟ dy + 5.7 DU f ν ⎟⎠ ∫ ln⎜⎜⎝ κ 0 (6.59) or DU f Q= Uf ν κ Uf ν ∫ ln ς dς + 5.7 DU f 0 DU f ν = [ς ln ς − ς]0 ν κ DU f ⎡ ⎛ DU f = ⎢ln⎜ κ ⎣⎢ ⎜⎝ ν = 3.2 DU f + DU f κ + 5.7 DU f (6.60) ⎞ ⎤ ⎟ − 1⎥ + 5.7 DU f ⎟ ⎠ ⎦⎥ ⎛ DU f ln⎜⎜ ⎝ ν ⎞ ⎟ ⎟ ⎠ This equation corresponds to the formula for flow resistance know from hydraulics V 1 ⎛ DU f = 3.2 + ln⎜⎜ κ ⎝ ν Uf ⎞ ⎟ ⎟ ⎠ (6.61) where V is the mean flow velocity. Equation (6.60) has two unknown, Uf and D. These are, however, correlated by the relation U 2f = gDI (6.62) 2 cf. (6.45) and (6.46). By using the given discharge Q = 1 m /s and a viscosity of ν = 10 −6 m 2 / s iteration of equations (6.62) and (6.61) gives the following quantity U 3f gI = 0.034 m 2 / s From this we find Uf = 0.032 m/s, while the depth is found to be 1.06 m. From (6.58) we find δ v = 11.7 × 10 −6 m = 0.37 mm 0.032 (6.63) which I very small compared to the water depth. For a larger discharge the viscous sub-layer becomes even thinner. B. A rough bed 147 If the bed is not completely smooth the irregularities in the bed may cause an important change in the velocity distribution over the entire water depth. As an example let us consider the flow over a sand bed where the sand is assumed not to move. If the sand grains lie evenly distributed along the bed as sketched in figure 6.13, the bed has a roughness k, which is the same as the grain diameter. When δ v >> k the roughness elements of the wall are completely embedded in the viscous sub-layer and the turbulent flow is not affected by the presence of the roughness of the wall. This situation therefore corresponds to smooth bed conditions. When δ v << k (as indicated by the dotted curve in figure 6.13) the viscous sub-layer loses its significance because the shear stress is no longer transferred to the bed through viscous forces. In stead the shear stress is transferred as forces acting on the roughness elements though a higher pressure on the front of the elements and a lower pressure on the back quite similar to the drag force on a cylinder, cf. example 5.3. In a region just above the rough bed the mean flow in non-uniform and can in this way transfer a part of the stress, while the Reynolds' stresses transfer the rest. As an approximation it can be assumed that the turbulent Reynolds' stress carries the entire shear stress at a distance of the order k from the top of the roughness elements. Fig. 6.13 Flow over a rough bed. When the bed shear stress is transferred as drag forces the viscosity is of very small significance, cf. example 5.3, and over a rough bed the flow resistance and velocity distribution is practically independent of the viscosity of the fluid. As mentioned the shear stress is transferred as Reynolds' stresses just a small distance above the roughness elements and by similar arguments as for a smooth bed (outside the sub-layer) the velocity profile can be described by (6.52), that is 1 U = ln y + const Uf κ (6.64) In the case of a rough bed it is of course more difficult to make a definition of the position where y is 0. It is close to the top if the sand grains and the exact position is of minor practical importance. The vertical dimensionless coordinate may be taken as y/k, in which case experiments have shown that the constant in (6.64) shall be 8.6. Giving 148 1 ⎛ y⎞ 1 ⎛ y ⎞ U = ln⎜ ⎟ + 8.6 ≈ ln⎜ ⎟ Uf κ ⎝ k / 30 ⎠ κ ⎝k⎠ (6.65) This expression may as an approximation be used all the way from the bed to the surface and gives rise to formula for the flow resistance quite analogous to (6.61) V ⎛ D⎞ = 6.1 + 2.5 ln⎜ ⎟ Uf ⎝k⎠ (6.66) which as described is valid for δ v << k . There is a complex area of transition between the conditions of smooth and rough bed, but the expression ⎛k 3.2ν ⎞⎟ V = 6.1 − 2.5 ln⎜ + ⎜ D DU f ⎟ Uf ⎝ ⎠ (6.67) gives an adequate description of the flow resistance and includes (6.66) and (6.61) as asymptotic cases. Example 6.3 Suspended transport of sand in water Let us consider sand of uniform grain size, which in quiescent water has the settling velocity (i.e. fall velocity of the individual sand grains) w. Even though each gains settles due to the effect of gravity experience shows that if the current (or wind velocity) is strong enough considerable amounts of sand may be carried in suspension. This can be explained by the presence of turbulent eddies. Following the basic idea of the mixing length theory there is an upward flux of fluid through section A-A per unit area of q., cf. figure 6.14. This fluid will carry an amount of sand taken on average from the level ½ A below the section, where the average concentration is c − ½ A dc / dy , where c(y) is the mean concentration of sand by volume. The upward flux of sand is then ⎛ dc ⎞ (v − w)⎜⎜ c − 12 A ⎟⎟ dy ⎠ ⎝ (6.68) where v is the vertical velocity of the fluid. In (6.69) it is expressed that the sand is settling with the velocity w relative to the motion of the water. The upward flux of sand is countered by a downward flux due to the downward fluid velocity, which is also q. The downward flux is analogous to (6.68) and given as 149 ⎛ dc ⎞ (v + w)⎜⎜ c + 12 A ⎟⎟ dy ⎠ ⎝ Fig. 6.14 (6.69) Suspension of sediment in a turbulent flow. In the case of a stationary distribution of sediment the flux up and down through section must be equal, so that (6.68) and (6.69) are of the same magnitude, giving cw + 12 vA dc =0 dy (6.70) From (6.38) we have (using that ½v = q) that qA = 12 vA = τ/ρ ⎛ dU ⎞ ⎟⎟ ⎜⎜ ⎝ dy ⎠ (6.71) The velocity gradient is here found as dU U f = dy κy (6.72) cf. (6.65), and τ / ρ is taken from (6.44) by which (6.70) can be written y ⎞ dc ⎛ cw + κU f ⎜1 − ⎟ y =0 ⎝ D ⎠ dy The variables can be separated to give dc w D dy =− c κU f D − y y (6.73) (6.74) which can be integrated to give 150 ⎛D− y⎞ ⎟⎟ c = const × ⎜⎜ ⎝ y ⎠ Z (6.75) where the coefficient Z has been introduced as Z= w κU f (6.76) It is seen from (6.75) that the heavier the sand (i.e. larger w) the closer the sand will be to the bed. Very fine sand (small w) will be more unifromly distributed over the vertical. Similarly strong turbulence (large Uf) will tend to carry more sediment away from the bed. The eddy viscosity, defined by (6.41) is as seen from (6.71) given as y⎞ ⎛ ν t = qA = κU f y⎜1 − ⎟ ⎝ D⎠ (6.77) Insertion of ν t in (6.70) makes this read cw + ν t dc =0 dy (6.78) A simple physical interpretation can be made of each term in (6.78). The first term cw is the flux of sand through a unit area due to the settling of the grains. The second term expresses the concept known as 'turbulent diffusion' with the eddy viscosity as the diffusion coefficient. Typically diffusion is a process where a scalar (concentration, temperature etc.) is spread when its magnitude varies from one place to another, and the flux of the scalar is proportional to its gradient ( in the present case the gradient is dc/dy). In the case of quiescent water or a laminar flow we have molecular diffusion, because the spreading is cause by the heat motion (Brownian motion) of the molecules (in addition to any advection). In general the molecular diffusivity is of the same order as the kinematic viscosity of the fluid. Turbulence is a much stronger mechanism for mixing than the molecular motion and ν t is therefore most often many orders of magnitude larger than ν . The quantity ν t can be interpreted as the diffusivity for the momentum. The diffusivity for suspended or dissolved matter, temperature etc. is normally assumed to be of the same order of magnitude as ν t . 151 Example 6.4 The energy balance in a turbulent flow. If we consider the flow in a pipe or a uniform channel flow the turbulence in a given point will be maintained, which means that the intensity and the statistical properties of the turbulent fluctuations are not changing with time in a given point, or in the downstream direction. This is in contrast to turbulence generated by a grid, which is moved swiftly through a fluid at rest or alternatively by letting a fluid flow through a fixed grid. In these cases the turbulence will decay with time or space respectively. The reason for the turbulence being maintained at the same level is that energy is constantly being transferred from the mean motion to the turbulence due to the gradient in the mean flow velocity profile. In section 6.1 it was explained how a small perturbation for sufficiently large values of the Reynolds' number received energy from the basic flow and therefore grew in strength to form turbulence. It is an analogous mechanism, which causes a continuous production of turbulent energy and maintains the turbulence. Let us consider a fluid particle with the components u' and v' of the turbulent velocity fluctuations. During a small time increment Δt this particle has moved the distance v ′Δt away from the wall. If we disregard the effect of outer forces on the particle it will keep the same absolute velocity after the time Δt . But since it is now at a level with a different mean velocity the relative velocity fluctuation will be changed to dU v ′Δt (6.79) u′ − dy as indicated in figure 6.15. Fig. 6.15 The production of turbulent kinetic energy. The other two velocity fluctuations v' and w' are not affected. The turbulent kinetic energy k per unit volume is now defined as k = 12 (u ′ 2 + v ′ 2 + w′ 2 ) (6.80) At the end position k is found as 152 k= ⎛ 2 1⎜ ′ u − 2⎜ ⎝ ⎞ dU v ′Δt ⎟⎟ + 12 (v ′ 2 + w′ 2 ) dy ⎠ (6.81) dU = k 0 − u ′v ′ Δt dy where ko is the initial kinetic energy. When time averaged it is seen that the term − u ′v ′ dU dy (6.82) describes the increase in the turbulent kinetic energy due to the gradient in the mean flow velocity. The factor − ρu ′v ′ is equal to the Reynolds stress which for fully developed turbulent flow is equal to the shear stress. The term (6.82) may therefore be written − ρu ′v ′ dU dU =τ dy dy (6.83) which thus described the rate of production of turbulent kinetic energy. The term at the right hand side expresses the the work made by the shear stress on a unit cube. The dissipation (heat loss9 is according to (2.16) given as ⎛ ∂u ⎞ ε = ρν⎜⎜ ⎟⎟ ⎝ ∂y ⎠ 2 (6.84) (a more general proof can be found in appendix I). For a laminar flow (6.84) can also be written ε=τ ∂u ∂y (6.85) so that the energy dissipation in a laminar is directly found as the work by the shear stress. In a turbulent flow the picture is more complicated because the work by the shear stress (6.83) first is transformed to turbulent energy, that is stronger motion of the eddies. Experiments show that the eddies containing most of the turbulent energy are the larger ones corresponding to a smaller wave number. The energy is, however, dissipated to heat in connection with the small eddies because the velocity gradients are largest for these (cf. the expression for the dissipation (6.84)). From theoretical considerations as well as by experiments it has been found that the the energy dissipated per time and per volume can be approximated by k 3/ 2 ε=A Ad (6.86) 153 where A is a dimensionless factor of the order 0.08 and A d is the dissipation length – also know as the length scale of the turbulence. This is a measure of the size of the large energy containing eddies and is of the same order of magnitude as the mixing length A , cf. section 8.2. It may seem to be a paradox that the expression (6.86) for the total dissipation does not depend on the viscosity, since we know that in the end it is the viscous forces which cause the dissipation. The explanation can found in a model proposed by Kolmogorov: It must be expected that eddies of widely different orders of magnitude cannot interact directly. A very large eddy can transport very small eddies by advection without any other direct influence. Eddies can only exchange substantial amounts of energy between them if they are of comparable size. In broad terms the pattern is so that the large eddies, which contain most energy, transfer some of this to eddies which is one order of magnitude smaller. These the pass the energy to new eddies yet one order smaller. This process continues to smaller and smaller eddies until the energy has been passed to eddies so small that the energy can be dissipated through the action of viscous forces. If the amount of energy passing through this chain of processes ('the energy cascade') is determined by the transfer between the larger eddies, then only the last part of the process is affected by the viscosity. The model can be summarized: The energy containing large eddies is a reservoir of energy, which is drained gradually by transfer of energy from the larger to the smaller eddies (or from the smaller to the larger wave numbers) through a sort of cascade process. The largest wave numbers (the smallest eddies) represent an energy drain, where the viscous forces transform the turbulent energy to heat. It is thus quite natural to call the expression (6.83) the 'production of turbulent energy'. From these considerations it is seen that contribution to the turbulent energy immediately is given so that only the longitudinal fluctuations u' are increased. Thereafter the 'surplus' is gradually distributed to the other components through a process involving the pressure fluctuations. Figure 6.16 shows a series of measurements of the energy balance in smooth pipe flow. The balance of energy is quite similar in open channel flow, showing very small deviations from figure 6.16. The energy balance can be expressed through an equation, which mainly consists of the terms production + resulting diffusion = dissipation Fig. 6.16 Measurements of the energy balance in pipe flow. 154 The dissipation is largest near the pipe wall and smallest – but not zero – at the axis of the pipe. Production and dissipation are dominant in the main part of the crosssection of the pipe, but there exist a contribution of some significance which is due to the diffusion of turbulent energy toward the axis across the direction of the mean flow. The diffusion is of relative importance in the central area away from the wall. As seen from the figure the production is very large close to the wall and decreases to zero at the axis of the pipe. 6.6 The structure of turbulence in a pipe In the following the most important characteristics of a turbulent pipe flow will be briefly described. As mentioned the characteristics of turbulence in a pipe and in an open channel flow are almost identical. Figure 6.17 shows five distribution functions for turbulent quantities in pipe flow. These distributions are almost universal and resemble also open channel flow conditions. Figure 6.17a shows the distribution of the mean flow velocity, which as described in section 6.5 is logarithmic. In the form given the logarithmic distribution is, however, not universal, but it is easy to make a universal formulation in the form of the so-called deficit law U0 −U Uf = 2.45 ln r y′ (r = the radiusof the pipe) which is valid for smooth as well as for rough pipes, when Uo is taken as the velocity at the pipe axis. Fig. 6.17 Measurements of the characteristics of turbulence in a pipe. 155 Figure 6.17b shows the distribution of the shear stress τ . τ is zero at the axis and attains the value τ 0 = ρU 2f at the wall. In a turbulent flow most of τ is transferred through the Reynolds stresses, that is τ = −ρu ′v ′ . However, for a smooth pipe this is not the case in the viscous wall layer, where the viscosity is dominant, especially very close to the wall. Figure 6.17c shows the distribution of the RMS (root mean square) of the turbulent fluctuations in the three principal directions. The RMS value (or standard fluctuations) is a measure of the variation around the mean value and is defined as u = u′2 (6.87) In formula 6.17c u is in the direction of the axis and v is in the direction normal to the wall. Close to the wall u is largest and v smallest, but close to the axis the three standard fluctuations are practically of the same magnitude which means that the turbulence is almost isotropic (i.e. has the same characteristics in all three directions). Figure 6.17d shows the distribution of the turbulent kinetic energy k. The variation is approximately linear except in a region close to the axis. This has been used as basis for some turbulence models especially for boundary layers where proportionality between τ and k has been assumed, cf. chapter 8. Finally the correlation coefficient between u' and v' is shown in figure 6.17e. This is defined as R= u ′v ′ uv (6.88) R lies between –1 and +1. In turbulent pipe flows R is constant (= -0.4) over most of the cross-section, increases towards the axis to zero. By full correlation we mean in practice that there is a linear relation between the two variables, for example u' = -v', which is seen to give R = -1. If all the measured values are plotted together in a diagram as in figure 6.18a, the points will on a straight line passing through origo. Korrelation zero is found when the two velocity flutuations are independent variables u ′v ′ = 0 Fig. 6.18 and R=0 Examples of different degrees of correlation. 156 If simultaneous measurements are plotted, the point will be positioned at random locations around origo, and no orientation can be distinguished. The concept of correlation is of most interest when two fluctuations are partly correlated as is the case is a point in the pipe flow. Simultaneous values of u' and v' will then mainly fall within an ellipse-shaped area as indicated in figure 6.18c. A simple illustration of this relationship in turbulent pipe flow can be made by plotting the end points of the velocity vectors at different times as shown in figure 6.19. Statistically the points will then lie in areas of variation, which close to the wall are ellipse-shaped (as in figure 6.18c) and near the axisis is circular (cf. figure 6.18b). It is the (partial) correlation between u' and v' which makes the transfer of shear stresses as Reynolds stresses in a turbulent flow possible. 157 7. Turbulent boundary layers As mentioned in example 6.1 a laminar boundary layer will be unstable at sufficiently large Reynolds numbers and go through a transition to become turbulent. Such a turbulent boundary formes along a plane plate is shown in the photograph in figure 7.1. Fig. 7.1 Photograph of a turbulent boundary layer, Re ≈ 3000. The transition between the turbulence in the boundary layer and the outer potential flow is very irregular, as seen in figure 7.1. In a point places near the edge of the boundary layer there will at some instances be a fully turbulent flow and at other instances practically no turbulence at all. This is called intermittent turbulence. At the same time the transition between areas with turbulence and areas with potential flow is remarkable sharp. In these transitions the viscosity is very significant, while the outer flow (potential flow) and the fully developed turbulent flow are independent of the viscosity (except for the small turbulent structures). The outer potential flow in the vicinity of a turbulent boundary layer is more irregular because of these fluctuations of boundary between the turbulence and the potential flow. 7.1 The turbulent boundary layer equation The boundary layer equation for a laminar boundary layer was described in section 5.1. The main idea was that the variation in the different terms (velocity, 165 pressure) is much larger across the flow direction (the y-direction) than in the flow direction (the x-direction). In the turbulent case the x-component of the Navier-Stokes equation (6.33) reads dU 1 ∂ 2U 1 ∂ ∂ ∂P − ρ (u ′) 2 − ρ (u ′v ′) + ρν ρ =− 2 dt ∂x ∂y ∂x ∂y (7.1) where the term ρν ∂ 2U 1 / ∂x 2 has been neglected (just as in the laminar case). The ycomponent of the flow equation reads similarly 0=− ∂ ∂P − ρ (v ′) 2 ∂y ∂y (7.2) Here the term ρ ∂ (u ′v ′) / ∂x has been neglected, as is involves a differentiation in the x-direction. Integration of (7.2) gives P = −ρ(v ′) 2 + const (7.3) Outside the boundary layer in the potential flow (v ′) 2 = 0 , and here the pressure is called po. Equation (7.3) can therefore be written (7.4) P = p0 − ρ(v ′) 2 In most cases the last term in (7.4) can be neglected, so that we in turbulents as well as in laminar boundary layers can assume that the pressure inside the boundary layer is determined by the pressure outside in the potential flow, i.e. P = p0 (7.5) Insertion of (7.4) into (7.1) gives ∂p0 dU 1 ∂ 2U 1 ∂ ρ =− + ρν − ρ (u ′v ′) 2 dt ∂x ∂y ∂y [ ∂ −ρ (u ′) 2 − (v ′) 2 ∂x ] (7.6) In practice the last term in the right hand side can also be neglected, so that (7.6) can be reduced to ρ ⎞ ∂p dU 1 ∂ ⎛ ∂U = − 0 + ρ ⎜⎜ ν 1 − (u ′v ′) ⎟⎟ ∂x ∂y ⎝ ∂y dt ⎠ (7.7) Using (6.35) this gives 166 ρ ∂p dU 1 ∂τ =− 0 + dt ∂x ∂y (7.8) which can be seen as the equation of motion for a unit cube. On the left hand side we have mass times acceleration and on the right hand side we have the resulting forces from the pressure and the shear stress in the flow direction. In addition to the boundary layer equations (7.5) and (7.8) the integrated momentum equation (5.20) can also be applied. The derivation of (5.20) is not based on the assumption of the flow being laminar. But it should be kept in mind that Newton's formula (2.5) for τ is normally not applicable for turbulent flows. In stead one should apply the more general expression for the shear stress at the wall τ 0 = ρU 2f (7.9) The actual application of (7.9) illustrated by examples in the following. 7.2 Stationary boundary layer along a plate In the following we shall consider the variation in the boundary layer thickness for a turbulent flow along a plate. The plate is rough, and there is no outer pressure gradient so the case is analogous to the laminar flow treated in example 5.2. Corresponding to (5.50) the integrated momentum equation now reads δ δ ∂ ∂ ρU 2 dy − u 0 ∫ ρUdy = − ρU 2f ∫ ∂x 0 ∂x 0 (7.10) Just as in the laminar case we can now assume a velocity profile for the mean flow velocity U. In order to make a reasonable guess for the velocity profile in the boundary layer we can compare the conditions in the boundary layer to the conditions in pipe and channel flow as described in chapter 6. The conditions in the boundary layer are illustrated in figure 7.2. Fig. 7.2 Typical velocity profile and shear stress distribution in a boundary layer. 167 Apart from the fluctuating boundary, the main difference from the pipe flow is that the shear stress is not determined from a static balance. Further the boundary layer flow is non-uniform, because the boundary layer is growing in the flow direction. Close to the wall experiments do, however, show that an equilibrium layer exists similar to the channel flow, where the production and the dissipation of turbulent energy practically balance each other. Mathematically this can be expressed as τ dU k 3/ 2 =A ρ dy A (7.11) where the production is given by (6.83) and the dissipation of turbulent energy is given by (6.86). Measurements has further confirmed that the turbulent fluctuations shown in figure 6.17 and the distinction between smooth and rough walls in all essentials can be used in boundary layer flows as well as channel- and pipe flows. This is, however, only valid for the affine or self-similar boundary layers (cf. example 5.2 for the laminar case), where the velocity profiles can be formulated as ⎛ y⎞ u0 − U = U f f ⎜ ⎟ ⎝δ ⎠ (7.12) uo is the velocity outside the boundary layer, see figure 7.2. For such an affine boundary layer measurements as shown in figure 6.17 can be used to estimated the near-bed turbulent kinetic energy to be k = 12 ρ (u ′ 2 + v ′ 2 + w′ 2 ) ≈ 3.5U 2f (7.13) Equation (7.11) can now be used to assess the velocity profile in the equilibrium layer, using that τ τ0 ≈ = U 2f ρ ρ (7.14) where τ 0 is the bed shear stress, cf. figure 7.2. By inserting (7.13) and (7.14) together with A = κy (7.15) Into equation (7.11) we get Uf dU = const1 × y dy (7.16) or 168 U = cons1 × ln( y ) + const 2 Uf (7.17) In other words the velocity profile in an affine boundary layer is logarithmic, as it is in channel- and pipe flows. Because the conditions in a boundary layer are quite similar to a channel flow, we can in the case of a hydraulically rough bed write (7.17) as 1 ⎛ y ⎞ U = ln⎜ ⎟ U f κ ⎝ k / 30 ⎠ (7.18) (7.18) is completely analogous to (6.65). This expression is now used in the momentum equation (7.10). First we determine the term δ ∫ U dy = ⎡ ⎛ δ ⎞ ⎤ ⎟ − 1⎥ ⎣ ⎝ k / 30 ⎠ ⎦ Uf δ ⎢ln⎜ κ 0 (7.19) Using that U = uo for y = δ , we find u0 1 ⎛ δ ⎞ = ln⎜ ⎟ U f κ ⎝ k / 30 ⎠ (7.20) which inserted into (7.19) gives δ ∫ U dy = δ u 0 − δUf 0 (7.21) κ Similarly we find δ ⎛U f ∫ U dy = ⎜⎜⎝ κ 0 2 ⎛U f = ⎜⎜ ⎝ κ = u 02δ ⎞ k 30δ / k 2 ⎟ ⎟ 30 ∫ ln (ς ) dς ⎠ 0 2 2 [ ⎞ k 2 ⎟ ⎟ 30 ς (ln ς ) − 2ς ln ς + 2ς ⎠ − 2u 0 δUf κ + 2δ ] 30δ / k 0 (7.22) U 2f κ2 Insertion of (7.21) and (7.22) into (7.10) gives 169 Uf δU f ∂ ⎛⎜ 2 + 2δ 2 u 0 δ − 2u 0 ∂x ⎜⎝ κ κ 2 ⎞ ⎟ ⎟ ⎠ (7.23) − u0 U ∂ ⎛ ⎜δ u0 − δ f ∂x ⎜⎝ κ ⎞ ⎟ = −U 2f ⎟ ⎠ In the present case uo is constant and (7.23) can be reduced to ∂ (2Z 2δ − Zδ ) = −κ 2 Z 2 ∂x (7.24) Where Z is a new dimensionless variable, defined as Z= Uf κ u0 (7.25) equation (7.20) can be written ⎛ δ ⎞ Z ln⎜ ⎟ =1 ⎝ k / 30 ⎠ (7.26) together equation (7.24) and (7.26) form a system of equations in Z, from which δ can be determined. Differentiation of equation (7.26) gives ⎛ δ ⎞ Z ′ Z ′ ln⎜ ⎟+ δ =0 ⎝ k / 30 ⎠ δ or Z′ δ ′ + =0 Z2 δ (7.27) Differentiation of (7.24) gives (4Z − 1) δ Z ′ + (2 Z 2 − Z )δ ′ = −κ 2 Z 2 (7.28) Which together with (7.27) gives κ 2Z 2 dδ = dx 4 Z 2 − 3Z 2 + Z (7.29) 170 Fig. 7.3 The variation in the boundary layer thickness along a rough plate. this equation must (together with (7.26)) be solved numerically. Figure 7.3 shows such a solution, where the boundary condition has been taken as: δ = δ 0 for x = 0 , where δ 0 is the thickness of the boundary layer at its transition from being laminar to being turbulent, cf. figure 6.5. It is seen from figure 7.3 that δ grows approximately linearly with x (in contrast to the laminar case, where δ ∝ x ). This can be realized by the following considerations: the quantity Z defined in (7.25) is normally small because the friction velocity is only a few percent of the outer flow velocity. Further the variation in Z along the x-axis is weak as can be seen from (7.26) where the ln-function varies very slowly. In this way (7.29) can as an approximation be written as dδ ≈ κ 2 Z ≈ const dx (7.30) which gives δ ≈ const × x Example 7.2: Wind-driven flow. As an even more simple example of the use of the momentum equation we can calculate how the wind creates a flow in a body of water with land (in the form of a vertical wall) to the left and extending infinitely to the right. On the surface (along the x-axis) a wind is blowing, which creates a constant shear stress τ 0 , figure 7.4. This shear stress creates a boundary layer flow near the surface. For simplicity the flow is taken to be turbulent everywhere, why it is natural to assume the velocity profile to be logarithmic. Since the velocity must be zero for y = δ , this can be written 171 U 1 ⎛δ⎞ = ln⎜⎜ ⎟⎟ Uf κ ⎝ y⎠ Fig. 7.4 (7.31) Stationary wind-driven flow behind an edge. The flow is stationary, and the outer flow velocity is in this case equal zero. The integrated momentum equation (5.20) now reads: δ ∂ ρU 2 dy = S ∫ ∂x 0 (7.32) where S is the resulting outer force in the x-direction. Along a plate we have S = −τ 0 , where τ 0 is the bed shear stress acting against the flow direction, cf. (7.10). In the present case S is an external force, which is actually acting in the x-direction, see figure 7.4. Therefore S will now be given as S = τ 0 = ρU 2f (7.33) By inserting (7.33) into (7.32) an integration and a few calculations gives ∂ ⎛ ρ 2 ⎞ 2 ⎜ 2 U f 2 δ ⎟ = ρU f ∂x ⎝ κ ⎠ (7.34) Since the τ 0 and therefore U 2f is constant in this example, (7.34) is esily integrated to give δ= κ2 x = 0.08 x 2 (7.35) using that δ = 0 for x = 0. We thus see that once again δ grows linearly with x, which is a characteristic for turbulent flows. Until now we have assumed that there is no flow outside the boundary layer. Because of the mass flux in the boundary layer, which grows in the wind direction, this does not hold exactly. Due to continuity there must therefore be a flow of water 172 from the stagnant water and up into the boundary layer. Figure 7.5 shows this balance of continuity. As seen in the figure the continuity equation reads dq = v0 dx (7.36) Fig. 7.5 The continuity for the area surrounding the boundary layer. Here the discharge q is given as δ q = ∫ u dy = 0 1 U f δ = 0.2U f x κ (7.37) from which we get v0 = 0.2U f (7.38) It is thus seen that more and more water is flowing up to be part of the boundary layer flow as illustrated in figure 7.6. This process is called entrainment. Fig. 7.6 Entrainment. 7.3 Examples of non-stationary boundary layers In this section we shall continue to work on the two cases considered in section 7.2. In that section the flow was non-uniform in the x-direction, but stationary. Now the flow is taken to be uniform but unsteady. 173 Example 7.3: Wind-driven flow (continued from example 7.2) We now consider a body of water with a free surface of infinite extent. From time t = 0 a wind is blowing over the surface, causing a constant surface shear stress. This shear stress generates a boundary layer flow which is assumed to be turbulent with a velocity profile described by U 1 ⎛δ⎞ = ln⎜⎜ ⎟⎟ Uf κ ⎝ y⎠ (7.39) Due to the uniformity the boundary layer thickness δ is constant for all x, but varies with time, as more and more of the fluid takes part in the boundary layer flow. The momentum equation (5.20) now reads (taking uo = 0) δ ∫ρ 0 Fig. 7.7 dU = S = ρ U 2f dt (7.40) Wind-driven current. Where the sign of the outer force is discussed in example 7.2. From (7.39) we find dU U f 1 dδ = κ δ dt dt (7.41) this is inserted into (7.40) which after integration gives U f dδ = U 2f κ dt (7.42) From (7.42) we get (with initial condition δ = 0 for t = 0) δ = κU f t (7.43) 174 The boundary layer thickness δ is seen to grow linearly in time. Since there is no variation in the x-direction there is no vertical velocity, but more and water is still entrained to take part in the boundary layer flow. Example 7.4: The boundary layer along a plate suddenly put in motion This example is the turbulent version of example 5.1 where the laminar boundary layer along a plate suddenly put into motion was treated. So we consider a plate which at time t = 0 is given a sudden acceleration from velocity 0 to velocity uo. The plate moves in its own plane and of infinite extent. The boundary layer flow over the plate is assumed to be turbulent for all times t > 0, and the velocity profile is assumed to be described by U − u0 1 ⎛ y ⎞ = − ln⎜ ⎟ κ ⎝ k / 30 ⎠ Uf (7.44) which fulfils the boundary condition U = uo at the bed. At the top of the boundary layer we have u0 Uf = 1 ⎛ δ ⎞ ln⎜ ⎟ κ ⎝ k / 30 ⎠ (7.45) in accordance with the boundary condition U = 0 for y = δ . Fig. 7.8 The boundary layer along a plate suddenly given the velocity uo. By combining (7.449 and (7.45) U can as an alternative be written as 1 ⎛δ⎞ U = ln⎜⎜ ⎟⎟ Uf κ ⎝ y⎠ (7.46) The expression for U is now inserted into the momentum equation (5.20) (see also (5.42)), which in this case reads 175 δ ∂U ∫ ρ ∂t dy = τ 0 = ρU 2f (7.47) 0 the shear stress from the plate is in the positive direction and τ 0 is positive in (7.47). By differentiation of (7.46) we get ∂U U f dδ dU f U + = ∂t κ δ dt dt U f (7.48) This can be inserted into (7.47), which by integration gives U f dδ 1 dU f + δ = U 2f κ dt κ dt (7.49) An additional relation between Uf and δ can be obtained by differentiation of (7.45) with respect to time, which gives ⎛ δ ⎞ U f dδ ln⎜ =0 ⎟+ dt ⎝ k / 30 ⎠ δ dt dU f (7.50) If we introduce Z= Uf κ u0 similarly to (7.25) in section 7.1, we can write (7.50) as dZ Z 2 dδ =0 + dt δ dt Similarly equation (7.49) can be written dδ dZ Z +δ = u0 κ 2 Z 2 dt dt (7.51) (7.52) 176 Fig. 7.9 Variation in δ as function of time, rough bed case. From (7.51) and (7.52) we may eliminate the term dZ/dt and get dδ [1 − Z ] = u 0 κ 2 Z dt (7.53) which together with (7.26) yields a first order differential equation in δ that has to be solved numerically to describe the growth in δ with time. Figure 7.9 shows results from such a numerical solution. It is seen that δ grows faster with time the larger the roughness of the bed. 7.4 Free turbulence: the turbulent jet In this and the next section we shall look at flows characterized by being far away from any solid boundaries. Yet the flow is still assumed to have boundary layer characteristics, which means that the variation across the mean flow direction is much stronger than the longitudinal variation. Let us consider a fluid flowing through a circular hole into a space filled with a stagnant fluid with similar properties (density, viscosity) as the fluid coming from the hole. For sufficiently large discharges the fluid coming through the hole will form a submerged turbulent jet. In the following we shall consider the spreading of this jet at some distance from the orifice, where we use some assumptions regarding the turbulence. An example of a turbulent jet can be seen in the photograph in figure 7.11. Just as for the boundary layer formed along a wall there is also an irregular fluctuating boundary between the turbulence n the jet and the quiescent fluid outside the jet, which will cause intermittency in the measured turbulence. 177 Fig. 7.10 The spreading of a jet. When we are at a distance from the orifice (typically 50-100 times its diameter) the flow in the jet has reached a stage where the velocity profiles can be taken to be similar, which means the velocity can be described as ⎛r⎞ u = U 0 ( x) f ⎜ ⎟ ⎝δ⎠ (7.54) where Uo is the velocity at the centreline and δ is half the width of the jet. Because of rotational symmetry it is convenient to use polar coordinates normal to the x-axis (cylindrical coordinates) and r is the coordinate normal to the x-axis. To calculate the behaviour of the jet we shall first apply the momentum equation for boundary layers (5.20), which is written δ ∂ ρ2πru 2 dr = 0 ∂x ∫0 (7.55) where it has been used that there are no outer forces (shear stresses), the flow is steady, the outer velocity is zero and there are no pressure gradients. Fig. 7.11 Photograph of the spreading of a jet. 178 By inserting (7.54) into (7.55) we get 1 ∂ ⎡ 2 2 r 2 ⎛ r ⎞ ⎛ r ⎞⎤ δ U ⎢ 0 ∫ f ⎜ ⎟d ⎜ ⎟⎥ ∂x ⎢⎣ δ ⎝ δ ⎠ ⎝ δ ⎠⎥⎦ 0 (7.56) or U 0 δ = const (7.57) Additional assumptions have to be introduced to proceed further in the analysis of the free jet. This is in contrast to a boundary layer along a wall because there is no outer force in the form of the wall shear stress to control the development in the boundary layer. Let us consider the flow equation for the boundary layer. In equation (7.8) it was interpreted as the equation of motion for a unit cube, where mass times acceleration was given at the resulting pressure- and shear stresses in the flow direction. In the present case the pressure gradient is zero, and the flow equation can in polar coordinates – be written ρ2πr dr du ∂ = (τ2πr ) dr dt ∂r (7.58) which is the force balance for a ring-shaped element (torus) with the cross-section of a unit cube, cf. figure 7.10. Equation (7.58) can be written u ∂u ∂u 1 ∂ ⎛ r τ ⎞ ⎜ ⎟ = +v ∂x ∂r r ∂r ⎜⎝ ρ ⎟⎠ (7.59) where v is the velocity on the r-direction. In the boundary layer flow τ is not known, and we have to introduce assumption no. 2 to utilize equation (8.59). This assumption is that the distribution of τ across the boundary layer has a given shape which is similar in the x-direction. As τ / ρ has the dimension (m/s)2 we can now write τ as τ ⎛r⎞ = U 02 g ⎜ ⎟ ρ ⎝δ⎠ (7.60) Since τ in a turbulent flow is due to the Reynolds stresses, which are formed by the eddies, equation (7.60) says that the pattern of the energy containing eddies is self-similar along the jet axis. For large Reynolds numbers detailed theoretical studies and experimental investigations has shown that this is in fact the case. Equation (7.60) can now be inserted into (7.59) expressing v through the continuity equation for polar coordinates ∂u 1 ∂ (rv) + ∂x r ∂r (7.61) cf. example 2.3. From (7.61) and (7.54) we find that the velocity v at the distance r from the x-axis is given as 179 r ⎛r⎞ ⎛ r ⎞ δ′( x) r ′ r v = − ∫ rU 0 ( x) f ⎜ ⎟ dr + ∫ rU 0 ( x) f ′⎜ ⎟ 2 dr ⎝δ⎠ ⎝ δ ⎠ δ ( x) 0 0 r ζ ζ = −U 0′ ( x) δ 2 ( x) ∫ s f ( s ) ds + U 0 ( x) δ( x) ∫ s 2 f ′( s ) δ( x) ds 0 (7.62) 0 = −U 0′ ( x) δ 2 ( x) F (ζ ) + U 0 ( x) δ( x) G (ζ ) where ζ = r / δ has been introduced as a new dimensionless coordinate and F and G are unknown functions in ζ . Now (7.59) reads r ⎤ ⎡U 0′ δ 2 F δ′U 0 δ G ⎤ f′ ⎡ ′ ′ − U 0 f ⎢U 0 f − U 0 f 2 δ′⎥ − ⎢ ⎥U 0 δ r δ ⎣ ⎦ ⎢⎣ r ⎥⎦ g′ 1 = U 02 g + U 02 r δ (7.63) Another relation between U 0′ and δ′ is obtained by differentiation of (7.57), which gives U 0′ δ + U 0 δ′ = 0 (7.64) We now have two rather complicated expressions with U 0′ and δ′ , namely (7.63) and (7.64). Equation can (by multiplication with δ /U 02 ) be written U 0′ U0 δ f 2 − δ′f f ′ζ − U 0′ δ 2 δ g F f ′ + δ′G f ′ = + g ′ U0 r r ζ or by use of (7.64) ⎡ F f ′ G f ′⎤ g δ′⎢− f 2 − f f ′ζ + + ⎥ = + g′ ζ ζ ⎦ ζ ⎣ (7.65) This relation is in the form dδ G1 (ζ ) = G2 (ζ ) dx (7.66) from which it is easily seen that 180 dδ = const dx (7.67) because G1 and G2 are functions of ζ only. The width of the jet is thus proportional to x δ∝ x (7.68) The assumption of self similarity is not valid immediately after the orifice, and a suitable choice of the origin for the x-axis has to be made for (7.68) to be valid. From (7.57) and (7.68) we find the corresponding variation of the velocity at the centreline to be U 0 ( x) ∝ 1 x (7.69) More detailed information on the characteristics of the jet, such as velocity profiles and the distribution of the shear stress, can be obtained by use of the technical turbulence theory if we introduce the eddy viscosity ν t . For the flow along a wall (section 6.4) we found according to (6.41) that 2 τ/ρ U f y⎞ ⎛ νt = = = κU f y⎜1 − ⎟ du Uf ⎝ D⎠ dy κy (7.70) It is thus seen that the eddy viscosity close to the wall grows linearly with the distance y from it. In a free jet there is no solid boundary damping the free turbulent motions and measurements does show that it is a good approximation to take ν t to be constant across the entire width of the jet. Generally ν t is proportional to a typical velocity scale (here: Uo) and to a typical length scale (here: δ ), so ν t may be written νt ∝ U 0δ (7.71) According to (7.57) this product is constant, and we can therefore also expect ν t to be constant and independent of x. The most convenient way to proceed is by application of (7.68) and (7.69) so that u can be written ⎛r⎞ U δ ⎛r⎞ k ν ⎛ r⎞ u = U 0 f ⎜ ⎟ = 0 f ⎜ ⎟ = 1 t f1 ⎜ k ⎟ δ ⎝δ⎠ x ⎝δ⎠ ⎝ x⎠ (7.72) = ν t k1r ⎛ kr ⎞ ν t f1 ⎜ ⎟ = H ′(ζ )k r x ⎝ x⎠ r 181 where the new function H ′(ζ ) = k1 k 2 ζ f 1 (ζ ) where ζ= kr x has been introduced. By inserting into the continuity equation (7.61) it is seen that v= νt (ζH ′ − H ) r (7.73) is the solution for the radial velocity v. By inserting (7.72) and (7.73) together with ν t2 ∂u ν t2 k 2 τ H ′′ − 2 H ′k =νt = ρ r x ∂r r (7.74) into the flow equation (7.59) we get the following ordinary differential equation in H HH ′ ( H ′) 2 HH ′′ d ⎡ H ′⎤ − − = ⎢ H ′′ − ⎥ 2 ζ ζ dζ ⎣ ζ ⎦ ζ (7.75) This can be integrated to give HH ′ = H ′ − ζH ′′ (7.76) which has the solution H= ζ2 1+ ζ2 / 4 (7.77) The velocity profile can the be found by differentiation of (7.77), which gives 1 ζ u∝ r (1 + ζ 2 / 4) 2 or u∝ 1 1 x (1 + ζ 2 / 4) 2 (7.78) Since Uo is the velocity at the centreline (7.78) can be written u= U0 (1 + ζ 2 / 4) 2 (7.79) Which give a bell-shaped profile as shown in figure 7.12. 182 Fig. 7.12 Measurements of the velocity profile, compared to equation (7.79). As seen from figure 7.12, the measurements are in good agreement with (7.79) except at the outer edges of the jet where there is intermittence and the assumption of ν t being constant is a crude assumption. 7.5 Free turbulence: the plane wake In this section we consider the flow field downstream of a cylinder which is kept in a fixed position. Far upstream of the cylinder the flow is a parallel flow with the constant velocity uo, cf. figure 7.13. Close to the cylinder the flow is strongly disturbed as described in example 5.3. For example in most cases there will be vortex shedding downstream of the cylinder. These vortices will die out when we are sufficiently far downstream from the cylinder (approximately 100 timed the diameter or more), but around the x-axis the flow will still be affected by the cylinder, and the velocity at the centreline will be smaller than uo. The cylinder thus creates a lee zone for the current, and this is the flow field we shall consider in the following. Fig. 7.13 The flow around a circular cylinder: definition of control surfaces for the momentum equation. As we have done before, we can start with formulating the momentum equation. This is used for the control surface A'BCD', which is quite analogous to the control surface shown in figure 5.3 used for the derivation o fthe integrated momentum equation for boundary layers, equation (5.20). In our case there are no shear stresses, and (5.20) can therefore be written 183 δ δ ∂ ∂ ρu 2 dy − u 0 ρu dy = 0 ∫ ∂x −δ ∂x −∫δ (7.80) or, using that uo is constant δ ∫ u (u − u 0 ) dy = const (7.81) −δ At this stage we must (as in the previous example) introduce additional assumptions to get further with the analysis. The first assumption is that the velocity profiles are affine or self similar, which can be written as ⎛ y⎞ u0 − u = U 0 f ⎜ ⎟ ⎝δ⎠ (8.82) where Uo is the maximum velocity deficit at the give position at the x-axis, cf. figure 7.13. Inserting (7.82) into (7.81) gives δ ∫U 0 −δ ⎛ y⎞ f ⎜ ⎟ dy = const ⎝δ⎠ (8.83) where it has been assumed that Uo/uo is small (i.e. the velocity reduction in the wake this far downstream is small), so that the term (Uo/uo)2 may be neglected. Equation (7.83) can be written 1 U 0 δ ∫ f (ζ ) dζ = const −1 which gives that U 0 δ = const (7.84) To determine the variation both in Uo and in δ with x we must (as also done in section 7.4) introduce the assumption of affinity (self similarity) in the distribution of the shear stress distribution τ ⎛ y⎞ = −u ′v ′ = U 02 g ⎜ ⎟ ρ ⎝δ⎠ (7.85) This expression is now inserted into the flow equation for boundary layers, which in this case can be written 184 u ∂u ∂u ∂ ⎛ τ ⎞ +v = ⎜ ⎟ ∂x ∂y ∂y ⎜⎝ ρ ⎟⎠ (7.86) Her v is found from the equation of continuity y ∂u dy ∂ y 0 v = −∫ (7.87) From (7.82) it is seen that v must be proportional to Uo, and similarly the term ∂u / ∂y . By neglecting higher order terms in Uo/uo equation (7.87) may be approximated as u ∂u ∂ ⎛ τ ⎞ = ⎜ ⎟ ∂x ∂y ⎜⎝ ρ ⎟⎠ (7.88) which by inserting (7.82) and (7.85) can be written U0 y U 02 ⎛ ⎞ ′ ′ ′ U 0 ⎜U 0 f − f δ ⎟=− g′ δ δ δ ⎝ ⎠ (7.89) From (7.84) we find by differentiation U 0′ δ + U 0 δ′ = 0 (7.90) From this δ′ can be expressed and then eliminated from (7.89), which now can be written U 0′ [ f + ζ f ′] = − U 03 U 0δ g′ (7.91) or U 0′ U 03 =− g′ 1 U 0δ f + ζ f ′ (7.92) Since U 0 δ is constant the right hand side of (7.92) is only a function of the dimensionless parameter ζ = y / δ , while the left hand side is only a function of x. This means that both must be constant, which gives U0 ∝ 1 x (7.93) and δ∝ x (7.94) 185 If we finally want to know the shape of the velocity profile, this can be obtained by introducing the eddy viscosity ν t , which may be taken to be constant. Now the flow equation (7.88) reads ∂u ∂ ⎛ ∂u ⎞ ∂ 2u ⎜ ⎟ u = νt = νt 2 ∂y ∂y ⎜⎝ ∂y ⎟⎠ ∂y A solution to (7.95) fulfilling the boundary conditions u → u 0 (7.95) for ⎛ u0 y 2 ⎞ ⎟ exp⎜⎜ − u0 − u = K ⎟ 4 νt x ν x t ⎠ ⎝ u0 x, y → ∞ is (7.96) where K is a constant. This velocity profile is shown in figure 7.14. The eddy viscosity has the dimension m2/s, and experiments have shown that the velocity profile (7.96) shows the best agreement if ν t is put to ν t = 0.016 u 0 d (7.97) where d is the diameter of the cylinder. The integrated velocity deficit δ ∫ (u 0 − u ) dy (7.98) −δ has also the dimension m2/s and is constant. It can be shown that a more general expression for ν t may be based on the velocity deficit, writing ν t as δ ν t = β ∫ (u 0 − u ) dy (7.99) −δ where β is a dimensionless constant. 186 Fig. 7.14 The distribution of velocity and shear stress across a plane wake. Example 7.3: The force on a cylinder in a turbulent flow. The force acting on the cylinder from the flow can be determined by applying the general momentum equation (5.9) on the control surface ABCD shown in figure 7.13. Here AD is placed so far upstream of the cylinder that the velocity over the entire section can be taken to be the undisturbed velocity uo. The momentum equation (5.9) reads G G G G ρ ( ⋅ ) = − v v d A F ∫ (7.100) A G where F is the force acting on the cylinder from the fluid. We now look at the xcomponent of the momentum equation and calculate each term on the left hand side of (7.100). From section AD we get δ − ∫ ρ u 02 dy (7.101) −δ From BC we get δ ∫ρu 2 dy (7.102) −δ From the two sections AB and DC the sum of the contributions is ρ Q u0 (7.103) because the velocity here is undisturbed uo because we are outside the boundary layer. Q is the total volume flux through the two sides, which is there because of weak 187 vertical velocities in the outer flow due to entrainment, cf. section 7.4. Q is found as the difference between the volume flux through AD and BC δ δ −δ −δ Q = ∫ u 0 dy − ∫ u dy (7.104) By inserting (7.101)-(7.104) into (7.100) we get δ −F = ∫ [− ρ u0 + ρ u 2 2 ] + ρ u 0 (u 0 − u ) dy (7.105) −δ or δ f = ρ ∫ u (u − u 0 ) dy (7.106) −δ This expression is according to (7.81) a constant, which it should be because the force F cannot be dependent on the position of the sections of the control surface. Equation (7.106) can be linearized to give δ f = ρ u 0 ∫ (u − u 0 ) dy (7.107) −δ which by using (7.97) and (7.99) can be written F ∝ ρ u 02 d (7.108) It is thus seen that the force is proportional to the square of the undisturbed velocity and proportional to the diameter of the cylinder. 188 8. Turbulence modelling In the two previous chapters we have seen how turbulent flows may be described in more or less detail, depending on how much information we require about the flow. Since it is not feasible to solve the complete Navier-Stokes equations numerically in the turbulent case because of the presence of the very small eddies (of the scale where dissipation to heat takes place), it is necessary in stead to introduce physically plausible assumptions to construct an approximate model of the flow. This is actually what has been done in the last two chapters, where the concepts of mixing length and eddy viscosity have been introduced. 8.1 Mixing length The concept of the mixing length is explained in section 6.4. The basic idea is that eddies transfer momentum across the plane parallel to the direction of the mean flow. This exchange of momentum is equivalent to a shear force, which as shown in section 6.4 can be written as τ = ρ A2 ∂U ∂U ∂y ∂y (8.1) The eddy viscosity may be defined through the expression τ = ρν t ∂U ∂y (8.2) From which we seen that in a turbulent flow the eddy viscosity and the mixing length is related as ∂U ν t = A2 (8.3) ∂y By combining (8.1) and (8.3) we get alternatively νt = A τ ρ (8.4) This expression will later be used in section 8.2. Turbulence modelling is now aiming at making as good a representation of the length scale A as possible. The models are to a large degree based on dimensional 193 considerations and interpretation of experimental data. In the following some simple ways of representing A in different flow situations are described. Boundary layer along a wall: Disregarding a possible viscous sublayer at the wall, A will grow linearly with the distance from the wall: A =κ y for y <δ (8.5) which was also applied in chapter 6. For some cases rthe description can be improved by restricting the linear growth of A to a certain fraction λ of the boundary layer thickness, outside which A is taken to be constant. This can be formulated A =κ y A = κλδ for for y < λδ y > λδ (8.6) where we may take λ ≈ 0.25 . Turbulent pipe flow: Here A can be take to be parabolically distributed, described as 2 A y⎞ y⎞ ⎛ ⎛ = 0.14 − 0.08⎜1 − ⎟ − 0.06⎜1 − ⎟ D ⎝ D⎠ ⎝ D⎠ 4 (8.7) Where D is the pipe diameter. In equation (8.7) A varies as A = 0.4 y (= κy ) for small values of y. Plane jet in a waterbody at rest: In this flow the mixing length varies only weakly across the jet. Measurements has given the value A = 0.2δ where δ is half the width of the jet. The width and therefore also the mixing length increases in the flow direction. Example 8.1: The turbulent wave boundary layer. The turbulent wave boundary layer is the flow found immediately above the sea bed, because the wave-induced oscillatory flow has to obey the no-slip boundary condition at the bed. 194 Fig. 8.1 The turbulent wave boundary layer. We treated the laminar case in example 2.5, where we considered the flow generated by oscillating a plate in its own plane. In that case a small part of the fluid (with a thickness of 2ν / ω , cf equation 2.54) was taking part in the oscillatory motion. Under waves it is the fluid that oscillates back and forth over the immobile plate (the sea bed), cf. figure 8.1. For a laminar flow the solution to this case completely similar to flow treated in section 2.5. We now consider the turbulent wave boundary layer. Outside the boundary layer, away from the bed the shear stress can be neglected, so the flow equation can be written ρ dU 0 ∂U 0 ∂U 0 ⎞ ∂U 0 ∂p ⎛ =− = ρ ⎜U 0 + ⎟≈ρ dt ∂x ∂t ⎠ ∂t ∂x ⎝ (8.8) Here U0 is the horizontal velocity outside the boundary layer. The boundary layer is thin compared to the depth, and we are so close to the bed that vertical velocities can be neglected even outside the boundary layer. Inside the boundary layer the horizontal velocity is U, and here shear stresses are present so that the flow equation reads ρ ∂U ∂p ∂τ =− + ∂t ∂x ∂y (8.9) The pressure p is transferred unchanged from the outer flow to the boundary layer, and the pressure gradient in (8.9) can therefore be taken from (8.8) giving ρ ∂ (U 0 − U ) ∂τ =− ∂t ∂y (8.10) Now it is convenient to introduce the concept of velocity deficit, defined as Ud = U −U0 (8.11) 195 As shown in figure 8.1. By use of Ud (8.10) can be written ρ ∂U d ∂τ = ∂t ∂y (8.12) The distribution of the shear stress τ over the boundary layer is not known, and (8.12) cannot be solved directly. By introducing the mixing length concept from (8.1) a relation between τ and U can be established, so that τ can be eliminated. Equation (8.1) can be written ∂U d ∂U d τ = ρA 2 (8.13) ∂y ∂y because there are no shear stresses in the outer potential flow. Now the combination of (8.12) and (8.13) gives ∂U d ∂ ⎛ ∂U d ∂U d ⎞ ⎟ = ⎜⎜ A 2 ∂t ∂y ⎝ ∂y ∂y ⎟⎠ (8.14) This partial equation in Ud can be solved numerically if A is given. The simplest formulation to use is (8.5). The boundary conditions are as follows: at the top of the boundary layer the velocity deficit Ud must tend to zero Ud → 0 for y→∞ (8.15) If the bed is hydraulically rough, the velocity must be zero for y = kN/30, or U d = −U 0 for y = k N / 30 (8.16) where kN is the equivalent bed roughness. 196 Fig. 8.2 The instantaneous velocity profiles in a turbulent boundary layer. Finally the flow must be periodic with the period T, i.e. U d (t ) = U d (t + T ) (8.17) Figure 8.2 shows the variation of U over half a wave period. The variation of the eddy viscosity in time and space may be found from (8.3), which together with (8.5) gives νt = κ 2 y2 ∂U d ∂y (8.18) The time variation in ν t at two different levels is shown in figure 8.3. It is seen that ν t increases with the distance from the bed, but that ν t twice every period becomes zero. This is because the velocity gradient due to the oscillatory nature of the flow has to cross zero twice every wave period. 8.2 The 1-equation model (k-model) The mixing length model described ion section 8.1 has two obvious shortcomings. a) Equation (8.18) means that ν t is zero whenever the velocity gradient is zero. This for example not correct at the surface in an open channel flow, whereν t at the surface is about 80% of the maximum value. In the oscillatory boundary layer we also saw that ν t was zero twice during a wave cycle. Since 197 new turbulent eddies are produced all the time and it takes some time for the eddies to dissipate, it should be expected that ν t in reality always is positive in this boundary layer. b) No advection or diffusion processes of the turbulent energy are included in the description: the local level of the turbulence is normally not determined only by what is happening at the place in question, but depends also on what has happened some distance upsteam. In a recirculating flow processes occurring downstream may also be of significance c) Fig. 8.3 The variation in ν t for a / k N = U 1 / ω k N = 10 2 and 10 3 . Broken line: 1-equation model. Fully drawn curve: mixing length model. A description of these processes requires the representation of memory effects in the turbulence. The strength of the turbulence can be characterized by the turbulent kinetic energy, defined as k = 12 (u ′ 2 + v ′ 2 + w′ 2 ) (8.19) 198 cf. equation (6.80). The turbulence may now be described in more detail by adding an equation for the variation in the turbulent kinetic energy in the turbulent flow. The variation in k in the flow is described by the so-called k-equation, which is a transport equation for k, keeping track of the production of k (i.e. how many eddies are formed), the dissipation of k, and finally the transport of k to or away from the actual location. Such a transport equation may be formulated by taking the Navier-Stokes equation (A8) for each coordinate direction, multiply each equation with the corresponding turbulent velocity component ui and finally adding all three equations. This will give d (U i + u i ) ∂ 2 (U i + u i ) ∂p ρ ui = −u i + ui ρ g i + ui μ dt ∂xi ∂x j ∂x j (8.20) Equation (8.20) is time-averaged. The term on the left hand side is time-averaged as follows ui d (U i + u i ) ∂ (U i + u i ) ∂ (U i + u i ) = ui + u i (U j + u j ) dt ∂t ∂x j = ui ∂U i ∂u ∂U i + ui i + ui U j ∂t ∂t ∂x j ∂U i ∂u ∂u + ui u j + U j ui i + ui u j i ∂x j ∂x j ∂x j = 0+ (8.21) ∂U i ∂k ∂k ∂k + 0 + ui u j +U j +uj ∂t ∂x j ∂x j ∂x j When calculating the last term it has been used that the continuity equation must also hold for the turbulent velocity fluctuations: ui, j = 0 (8.22) The whole time-averaged equation (8.20) now becomes ρ ∂U i ∂k ∂k ∂ + ρU j +ρ (u j k ) + ρu i u j ∂t ∂x j ∂x j ∂x j ∂ 2ui ∂p = −u i + μ ui ∂xi ∂x j ∂x j or 199 ∂U i ∂ 2 ui ∂ dk = − + − + u p u k u u u ρ ( j ρ j ) ρ i j μ i ∂x j ∂x j dt ∂x j ∂x j (8.23) For boundary layer flow equation (8.23) can be reduced to ρ ∂ 2ui ∂U dk ∂ + μ ui = − (v ′p ′ + ρ v ′k ′) − ρ u ′v ′ ∂x j ∂x j ∂y dt ∂y I II III IV (8.24) V Where u’ and v’ are the longitudinal and transverse velocity fluctuations respectively. The parameter p’ is the fluctuating part of the pressure and k’ is the fluctuating part of the turbulent kinetic energy. The five terms in (8.24) can be interpreted as follows: I is the total rate of change of the turbulent energy for a fluid particle. II & III can be interpreted as net diffusion of energy. II is expressing the change in energy due to the work done by the fluctuating pressure, while III represents the net transport of energy to the particle due to the turbulent fluctuating velocity field. The diffusive character of these terms can be illustrated by integration of (8.24) over the boundary layer thickness. If we consider a flow along a wall with a free stream flow with no turbulence above it the integrals of II and III will be zero, so these terms do not contribute to a change in the total energy level, but redistribute the energy across the boundary layer. IV describes the production of turbulent energy due to the transfer of energy from the mean flow, cf. example 6.4, page 142. Finally V represents a drain of the turbulent energy due to dissipation to heat. Application of (8.24) requires that the terms on the right hand side are approximated by use of parameters which are known or can be determined. The diffusive terms II-III are often assumed to be modelled analogously to a normal gradient diffusion process by − (v' p' + ρ v' k ') = ρ ν t ∂k σ k ∂y (8.25) whereν t is the eddy viscosity. This expression is analogous to the second term in (6.78) which describe the distribution of suspended sediment. There the term represented the redistribution of sand caused by the turbulent eddies. The term IV in (8.24) is well known because − ρ u 'v' is the shear Reynolds stress, giving − ρ u ′v ′ ∂U ∂U =τ ∂y ∂y (8.26) 200 here τ may be modelled in several ways. We can use the concept of eddy viscosity which gives τ = ρν t ∂U ∂y (8.27a) But there are also other possibilities. As mentioned in chapter 6 it is for open channel flow found that k is proportional toτ over a large part of the cross-section. This is used in some turbulence models, because it has turned out that the relation τ = 0.3ρ k (8.27b) fits well with experiments. Equation (8.27b) has the advantage relative to (8.27a) that the local shear stress is not necessarily zero where the velocity gradient is zero. Finally the dissipation of energy (term V in (8.24)) is found from (6.86) as μ ui ∂ 2 ui k 3/ 2 = − Aρ Ad ∂x j ∂x j (8.28) where A from measurements are found to be about 0.08. If the eddy viscosity is used to model τ , equation (8.24) now reads 2 ⎛ ∂U ⎞ ∂ ⎛ ν ∂k ⎞ k 3/ 2 dk ⎟⎟ + ρν t ⎜⎜ ⎟⎟ − Aρ = ρ ⎜⎜ t ρ ∂y ⎝ σ k ∂y ⎠ dt Ad ⎝ ∂y ⎠ (8.29) In this equation there are three unknows, namely k, ν t , and U, and we therefore need two additional equations to describe the flow. One of these is the flow equation ⎛ ∂U ⎞ ⎜ν t ⎟ (8.30) ⎜ ∂x j ⎟ ⎝ ⎠ Here the pressure in the boundary layer is as usual determined from the potential flow outside the boundary layer. The other equation correlates the eddy viscosity to the turbulent kinetic energy k. The eddy viscosity has the units of m2/s and can be formulated as the product of a length scale and a velocity scale. A typical velocity scale in a turbulent flow is k , which is the square root of the mean square of the velocity fluctuations. k thus represents a typical velocity in the most energy containing eddies. A measure for the size of these eddies is the length scale A d , cf. equation (6.86). The eddy viscosity can now be written ∂ 1 ∂p dU + =− dt ρ ∂x ∂x j νt = Ad k (8.31) 201 Combined, equations (8.29), (8.30) and (8.31) are sufficient to determine k and U. Example 8.2: The k-equation in the equilibrium layer. In example 6.4 we considered the energy balance in a channel- or pipe flow. In general this balance expressed that the formation of eddies (the production of turbulent kinetic energy) corresponds to the decay of eddies due to the heat loss (dissipation) and dispersion of eddies (diffusion). Finally eddies can be transport by advection. In the equilibrium layer, which is found in the lower 10-20% of the flow (cf. figure 6.17), the dominating terms are the production and the dissipation, which approximately balance each other out. From this we get that the two terms IV (given by (8.26)) and V (given by (8.28)) must balance each other, giving ⎛ ∂U ν t ⎜⎜ ⎝ ∂y 2 ⎞ k 3/ 2 ⎟⎟ = A Ad ⎠ (8.32) Application of the eddy viscosity (8.27a) in (8.32) gives 2 ⎛τ ⎞ k 3/ 2 ⎜⎜ ⎟⎟ = ν t A = Ak 2 Ad ⎝ρ⎠ (8.33) τ = ρ Ak (8.34) or which is seen to be identical to (8.27b). In the equilibrium layer there is therefore no contradiction between the two different ways of modelling τ (by (8.27a) and (8.27b)). The distribution of τ can be related to A d and the velocity gradient as follows: From the eddy viscosity we have 2 2 ⎛ ∂U ⎞ ⎛ ∂U ⎞ ⎛τ ⎞ ⎟⎟ = A 2d k ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ = ν t2 ⎜⎜ ⎝ρ⎠ ⎝ ∂y ⎠ ⎝ ∂y ⎠ 2 (8.35) where (8.31) has been applied. By then inserting k= τ 1 ρ A (8.36) from (8.34) into (8.35) we get 202 τ 1 2 ⎛ ∂U ⎞ ⎟⎟ = A d ⎜⎜ ρ A ⎝ ∂y ⎠ 2 (8.37) As seen this expression is equivalent to the mixing length model (6.39) τ ∂U ∂U = A2 ρ ∂y ∂y (8.38) The mixing length theory can therefore be taken as a turbulence model for domains with local equilibrium. Further it is seen that the mixing length A and the length scale A d are closely related, since (8.37) and (8.38) gives A d = A1 / 4 A ≈ 0.53 A Example 8.3: The turbulent wave boundary layer modelled by a one-equation model. This example is a continuation of example 8.1, where the turbulent wave boundary layer was modelled by use of the mixing length theory. We thus once more consider the flow shown in figure 8.1, now with the following system of equations at our disposal A: The flow equation, cf. (8.12) ρ ∂U d ∂τ ∂ ⎛ ∂U d ⎞ ⎟ = ρ ⎜⎜ν t = ∂y ⎝ ∂y ⎟⎠ ∂y ∂t (8.39) where Ud is the deficit velocity, cf. figure 8.1. B: The transport equation for turbulent kinetic energy ⎛ ∂U d ∂ ⎛ ∂k ⎞ ∂k ⎟⎟ + ν t ⎜⎜ = ⎜⎜ν t ∂t ∂y ⎝ ∂y ⎠ ⎝ ∂y 2 ⎞ k 3/ 2 ⎟⎟ − A Ad ⎠ (8.40) where A d in the boundary layer is given by A d = 0.53A = 0.53κy = 0.21y C: (8.41) The eddy viscosity is given by 203 νt = Ad k (8.42) These three equations form a complete system for the three unknown: Ud,ν t , and k. The boundary conditions are that U close to the bed is described by the familiar logarithmic distribution U→ ⎛ y ⎞ ⎟ ln⎜⎜ κ ⎝ k n / 30 ⎟⎠ Uf for y→0 (8.43) where kn is the roughness of the bed. At the top of the boundary layer the velocity shall approach the outer velocity U (δ ) = U 0 (8.44) which also imply that the gradient must be zero ∂U =0 ∂y for y =δ (8.45) In addition two conditions for k are required. Very close to the bed we may assume local equilibrium between the production of energy and the dissipation. Then (8.33) can be applied to give the boundary condition k= 1 k νt ∂U ∂y for y= kn 30 (8.46) Finally there is no transport of turbulence out to the outer potential flow. Since the transport is expressed through gradient diffusion this can be written ∂k =0 ∂y for y =δ (8.47) This system of equations must be solved numerically, for example by use of a finite difference scheme. In practice the thickness of the boundary layer is not know in advance, but since it only enters via the boundary conditions these are just applied at the top of the model domain. The results of this calculation are then as follows: The velocity profiles are very close to those we obtained in example 8.1, cf. figure 8.2. On the other hand the oneequation model gives a more realistic description of the eddy viscosity and the turbulent kinetic energy. In figure 8.3 the dotted curves show how the one equation model predicts the variation in the eddy viscosity during a wave period. It is seen that ν t now is always 204 positive, which describes how the eddies do not have time to die out completely before new turbulent kinetic energy is produced in the wave motion. It is further seen that for very rapid oscillations (small values of a/kN) ν t is not far from being constant in time when we are just at a certain distance from the bed. Figure 8.4 shows the distribution of the turbulent kinetic energy in the boundary layer under a wave. Since the advective terms are not included in the analysis we can actually only calculate how k varies over the vertical at a given location. The picture in figure 8.4 is then made from the temporal variation by using the relation for progressive waves ∂ 1 ∂ =− ∂x c ∂t Fig. 8.4 (8.48) The distribution of kinetic energy in the wave boundary layer (a/kN = 103). It is seen from figure 8.4 that while close to the bed k is large when the free stream velocity is large, there is a certain delay in the maximum in k further away from the bed. This is because there is a transport of k away from the bed, and this transport implies a delay. 205 8.3 2-equation models (k-epsilon model) The advantage by the one-equation model is as described in section 8.2 that k , which is a good measure of the turbulent velocity scale, is calculated by use of a transport equation so that memory effects in k are included. However, it must be noted that the turbulent Length scale is not determined from a transport equation. Memory effects in A d can be just as important as the effects in k, why a one-equation model often is found not to give a better description of a flow situation than the simple mixing length theory. In 2-equation models the kinetic energy k as well as the length scale A d are determined from a transport equation. The eddy viscosity is then found from νt = Ad k (8.49) cf. (8.31). Often it is not A d which is calculated directly, but some quantity called z, which is the product of k and A d raised to different powers z = k m (A d ) n (8.50) The length scale A d is then determined from k and z. It is not convenient to calculate A d directly because the diffusion of the quantity A d cannot be taken to be proportional to ∂(A d ) / ∂y . The most frequently used combination of m and n in (8.50) is m = 1.5 and n = -1, which gives z= k 3/ 2 Ad (8.51) so that z becomes proportional to the dissipation ε , cf. equation (6.86). An equation for ε can be developed quite analogously to the k-equation (8.23) on the basis of the NavierStokes equation. Generally a z-equation will read dz ∂ ⎛ ν t ∂z ⎞ ⎡ ν t ⎟ + z ⎢c1 = ⎜ dt ∂y ⎜⎝ σ z ∂y ⎟⎠ ⎢ k ⎣ 2 ⎛ ∂U ⎞ k⎤ ⎟⎟ − c 2 ⎥ + S z ⎜⎜ νt ⎥ ⎝ ∂y ⎠ ⎦ (8.52) this equation may be compared to the k-equation (8.29), which can be written 206 ⎡ν dk ∂ ⎛ ν ∂k ⎞ ⎟⎟ + k ⎢ t = ⎜⎜ t dt ∂y ⎝ σ k ∂y ⎠ ⎢⎣ k 2 ⎛ ∂U ⎞ k⎤ ⎟⎟ − A ⎥ ⎜⎜ νt ⎥ ⎝ ∂y ⎠ ⎦ (8.53) In (8.52) the constants σ z , c1 and c2 and the term Sz must be modelled. Reference is made to the specialised literature, e.g [5]. Example 8.4: The turbulent wave boundary layer determined by the 2-equation model. Finally we calculate the turbulent wave boundary layer (cf. examples 8.1 and 8.3) by use of a k − ε model, wher the equation for k is combined with an equation for the dissipation ε . The ε -equation is obtained from (8.52), which for z= k 3/ 2 Ad (8.54) and using νt = Ad k (8.55) can be written (after multiplication by A (=0.08)) εν ∂ ⎛ ν ∂ε ⎞ dε ⎟⎟ + c1 t = ⎜⎜ t dt ∂y ⎝ σ ε ∂y ⎠ k 2 ⎛ ∂U ⎞ ε2 ⎜⎜ ⎟⎟ − c1c 2 k ⎝ ∂y ⎠ (8.56) The constants in this equation can according to [5] be chosen as c1 = 1.41, c2 = 1.92 and σ ε = 1.3. The k-equation can, cf. (8.40), be written ⎛ ∂U ∂ ⎛ ∂k ⎞ ∂k ⎟⎟ + ν t ⎜⎜ = ⎜⎜ν t ∂t ∂y ⎝ ∂y ⎠ ⎝ ∂y 2 ⎞ ⎟⎟ − ε ⎠ (8.57) Finally we have the flow equation (8.39) ∂U d ∂ ⎛ ∂U d = ⎜⎜ν t ∂y ⎝ ∂y ∂t ⎞ ⎟⎟ ⎠ (8.58) 207 where Ud = U - Uo (8.59) Equations (8.56), (8.57) and (8.58) is a complete set of equations for the three unknowns ε , k and U. The equations have to be solved numerically with appropriate boundary conditions. For the velocity and the turbulent kinetic energy k the boundary conditions are the same as in example 8.3: equations (8.43), 8.44) and (8.45) for U and (8.46) and (8.47) for k. For ε expressions analogous to those for k can be derived. The condition for local equilibrium near the bed will together with the logarithmic velocity distribution thus imply that the boundary condition for ε becomes ε = ( A) 3 / 4 k 3/ 2 κy for y = k N / 30 (8.60) At the top of the boundary layer the gradient in ε is taken to be zero, that is ∂ε =0 ∂y for y =δ (8.61) A numerical result from the analysis is shown in figure 8.5, which shows the variation in the length scale in time for a specific value of the parameter a/kN. In the mixing length Fig. 8.5 Time-variation in the turbulent length scale (a/kN = 53) from [6]. length theory and the one-equation model the length scale A increased with the distance from the bed, but was constant in time. From figure 8.5 it is seen that the k- ε model predicts a distinct time-variation of A . At present there are no measurements available to confirm the theoretical predictions shown in figure 8.5. 208 The variation in A implies that the theoretical mean level for k becomes smaller when the k- ε is applied in stead of the k-model, cf. figure 8.6. This is because the kequation (where k is taken to be constant) predicts less dissipation in the outer parts of the boundary layer because the dissipation is inversely proportional to A . On the other hand the k- ε model predicts practically the same velocity distribution in the boundary layer as the k-model and the mixing length model do. Fig. 8.6 The mean level in the turbulent kinetic energy, calculated from the 1- and the 2-equation model. a/kN = 1000. From [6]. 209 Index Acceleration Advective terms Affine boundary layer Angular velocity Area vector Axi-symmetric flow 12 14 117 41 17 11 Bernoulli equation Boundary layer Laminar Turbulent Boundary layer equation Laminar Turbulent 45 102 165 103 166 Cascade process Circular cylinder Continuity equation Convective terms Converging flow Correlation coefficient Constitutive equation Creeping flow Current induced pressure Current, wave-induced Cyclic frequency 160 123, 187 34,213 14 120 163 214 127 124 96 41 Deformation Diffusion, turbulent Displacement thickness Dissipation length Dissipation Divergence Diverging flow 31 156 115 159 31, 215 35 120 Eddy viscosity Energy equation, turbulent Energy flux Entrainment Equilibrium layer Error function 146 201 82 175 162, 203 111 Flow equations Fluctuations Free turbulence Froude number 186, 212 138 179 55 Group velocity 74 Hydraulic gradient Hydrostatic pressure 40 23 Ideal fluid Intermittence 44 166 Jet, turbulent 179 k- ε -model k-model 207 198 Laplace equation Length scale for turbulence 47 159 Maintained turbulence Mixing length Momentum equation for boundary layers Momentum force 157 143, 193 Navier-Stokes equations Newton's formula Normal stress 37, 166, 214 28 17 One-equation model 198 p+ Particle path Particle paths in waves Perturbation Plane flow 39 9 72 47 7 105 86 Pohlhausen's method Poiseulle flow Polar coordinates Potential Power Pressure force Pressure Pressure, current induced Production of energy 120 38 35 44 29 86 19 124 158 Reaction force Refraction Reynolds experiment Reynolds number Reynolds stresses RMS-value Root mean square Rough bed 86 70 129 136 139 162 162 152 Self similar boundary layers Separation Set-down Set-up Shallow water theory Shallow water wave Shear stress Sinusoidal waves Smooth bed Standard fluctuation Stationary flow Stokes length Stream line Strouhal number Sudden movement of plate Laminar flow Turbulent flow Suspended sand 117 124 90, 92 90, 95 60 61 16 60 147 162 8 43 65 127 Technical turbulence theory Translation Translatory motion Turbulence length scale Two-equation model 145 32 32 159 207 Uniform flow 15 109 177 155 Viscosity, dynamic Viscosity, kinematic Viscous sublayer Vorticity transport equation 28 28 149 132 Wake, plane Wave boundary layer Laminar Turbulent Wave energy flux Wave energy Wave groups Wave length Wave number Wave period Wave table Wave-induced current Wind –driven flow 186 40 195 82 75 72 50 50 60 84 96 173 References View publication stats [1] Engelund, F.: Hydrodynamik. Den prvate Ingeniørfond, Danmarks Tekniske Højskole, 1968. [2] Pedersen, Fl. Bo: Hydraulik for bygningsingeniører. Den private Ingeniørfond, Danmarks Tekniske Højskole, 1988. [3] Svendsen, I. A. and Jonsson, I. G.: Hydrodynamics of Coastal Regions. Den private Ingeniørfond, Danmarks Tekniske Højskole, 1976. [4] Van Dyke, M.: An Album of Fluid Motion. The Parabolic Press, Stanford, CA, 1982. [5] Launder, B. E. and Spalding, D. B.: Mathematical Models of Turbulence. Academic Press, London, 1972. [6] Justesen, P.: Turbulent wave boundary layers. Series Paper No. 43, ISVA, Danmarks Tekniske Højskole, 1988.