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Chapter 5 Frictional Boundary Layers

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Chapter 5 Frictional Boundary Layers Chapter 5 Frictional boundary layers 5.1 The Ekman layer problem over a solid surface In this chapter we will take up the important question of the role of friction, especially in the case when the friction is relatively small (and we will have to find an objective measure of what we mean by small). As we noted in the last chapter, the no-slip boundary condition has to be satisfied no matter how small friction is but ignoring friction lowers the spatial order of the Navier Stokes equations and makes the satisfaction of the boundary condition impossible. What is the resolution of this fundamental perplexity? At the same time, the examination of this basic fluid mechanical question allows us to investigate a physical phenomenon of great importance to both meteorology and oceanography, the frictional boundary layer in a rotating fluid, called the Ekman Layer. The historical background of this development is very interesting, partly because of the fascinating people involved. Ekman (1874-1954) was a student of the great Norwegian meteorologist, Vilhelm Bjerknes, (himself the father of Jacques Bjerknes who did so much to understand the nature of the Southern Oscillation). Vilhelm Bjerknes, who was the first to seriously attempt to formulate meteorology as a problem in fluid mechanics, was a student of his own father Christian Bjerknes, the physicist who in turn worked with Hertz who was the first to demonstrate the correctness of Maxwell’s formulation of electrodynamics. So, we are part of a joined sequence of scientists going back to the great days of classical physics. Another intriguing figure in the story is the great Norwegian explorer/scientist Fridtjof Nansen. Trained as a botanist, a heroic arctic explorer, Nansen was a larger than life figure who played a leading role in Norway’s independence movement (from Sweden) and later, working under the auspices of the League of Nations was involved in many of the early 20th centuries most difficult human dramas. His biography is well worth reading. His relation to our study is more homely. He came to see V. Bjerknes to describe to him some unusual observations of ice-drift on one of Nansen’s expeditions on the Norwegian vessel , the Fram. Nansen noted that the ice drifted at an angle to the wind rather than directly downwind. He realized that it must be the earth’s rotation that was the cause and he asked Bjerknes if one of his students could look into the problem. Ekman was chosen, solved the problem for his doctoral thesis ( and indeed his solution was more general and elaborate that the one presented here and usually associated with his name.) This work was done in 1902, more than 100 years ago and is one of the very first boundary layer problems worked out in fluid mechanics and is essentially contemporary with the work of Prandtl who confronted a similar issue for nonrotating fluids. Now, although we have not completed our formulation of the equations of fluid mechanics in the general case, we do have a complete set of equations for the case where the density can be considered a constant. This will give us a pretty good picture of the phenomenon of interest and we will return to the problem of the general formulation afterwards and be able to check how sensible our assumption of constant density is. The problem to be discussed is the shown in Figure 5.1.1 z U g Ω y x Figure 5.1.1 The fluid flows in the x direction with speed U far from a solid wall at z=0. 2 Chapter 5 Far from a solid, horizontal plane which we idealize as having infinite extent, a uniform flow in the x direction has the magnitude U. What can we say about the structure of the resulting flow? We will use a Cartesian coordinate frame and use the notation (x,y,z) for (x1,x2, x3) and the associated components of velocity will be (u, v, w). ! Since the density is constant the continuity equation is simply !iu = 0 or !u !v !w + + = 0 (5.1.1) !x !y !z while the momentum equation for this incompressible fluid, ! du ! ! 1 ! + 2! " u = # %p # gkˆ + &%2u (5.1.2) dt $ where k is a unit vector in the vertical direction, along the z axis and parallel to the rotation axis. The sum of the gravitational and centrifugal accelerations is written as g , the effective acceleration due to gravity and yields a force anti-parallel to the rotation. Recall that ν is the kinematic viscosity, µ/ρ. The three components of (5.1.2) are !u 1 + uu + vu + wu " fv = " p + $(u + u + u ), !t x y z # x xx yy zz !v 1 + uv + vv + wv + fu = " p + $(v + v + v ), (5.1.3 a b, c) !t x y z # y xx yy zz !w 1 + uw + vw + ww = " p " g + $(w + w + w ), !t x y z # z xx yy zz Here I have used the notation that subscripts denote partial differentiation and f = 2! is the Coriolis parameter. Let’s look for steady solutions so that the local time derivative is zero. Furthermore, since the velocity at infinity is independent of x and y, let’s look for solutions that are independent of x and y. That would imply, from (5.1.1) that !w / !z = 0 . But since w 3 Chapter 5 must vanish on the lower solid surface, w must be zero everywhere. So, with u and v independent of x and y and with w zero, (5.1.3) reduces to: 1 ! fv = ! p + #u , " x zz 1 fu = ! p + #v , (5.1.4 a, b, c) " y zz 1 0= ! p ! g, " z Note that these equations are linear, a very great simplification. Now we have to see whether we can find solutions satisfying the boundary conditions. One solution to the equations is a rather simple one, u = U, v = 0, (5.1.5 a, b, c) p = !"gz ! " fUy. You can check that this satisfies the equations exactly. The flow is, everywhere, the constant velocity in the x direction, as at large z and the pressure field balances the hydrostatic force in the vertical and the Coriolis acceleration in the horizontal. y low pressure Ω U x high pressure pressure force Figure 5.1.2 The balance between the pressure gradient and the Coriolis acceleration in the solution ((5.1.5). The principal and inescapable problem with this solution is that it does not satisfy the no- slip boundary condition at z=0 where both u and v should be zero. The solution (5.1.5) 4 Chapter 5 satisfies the equation, even with the viscous term, but the solution is wrong because it does not satisfy the boundary condition and it is wrong no matter how small µ might be. So, what is the relation between (5.1.5) , that we would naively imagine to be a good solution for small viscosity, and the correct solution that satisfies the boundary condition? Let us return to (5.1.4 a, b, c) and find solutions for the velocity components u and v that are functions of z. Note that the presence of rotation then implies that although the flow is in the x direction far from the wall, the frictional term in (5.1.4a) will force a flow in the y direction under the influence of rotation. An x or y derivative of (5.1.4 c) shows immediately that ! " !p% ! " !p% $ ' = = 0 (5.1.6) !z # !x & !z #$ !y &' so that the horizontal pressure gradients are independent of z. As z ! the horizontal pressure gradient balances the Coriolis acceleration of the uniform flow U, so, !p !p = "# fU, = 0 (5.1.7 a, b) !y !x at infinity, and by (5.1.6) this is true everywhere in the flow, even down to the lower surface. Thus, (5.1.4 a, b) are just ! fv = "uzz , (5.1.8 a,b) fu = fU + "vzz So that finally, we have two simple, ordinary differential equations for the velocity components u and v. If we eliminate u between the two equations by taking 2 z derivatives of the second equation, we obtain, 5 Chapter 5 d 4v 4 + " = 0, dz4 ! 4 (5.1.9 a,b) 1/2 # 2" & ! = $% f '( The quantity δ is the Ekman layer thickness and it depends on the kinematic viscosity and the rotation. For very small viscosity, or rapid rotation, the thickness becomes smaller and smaller. The four independent solutions o f(5.1.9 a, b ) are: !z/" !z/" z/" z/" v = C1e sin z / " + C2e cos z / " + C3e sin z / " + C4e cos z / " (5.1.10 a) while from (5.1.8b) !z/" !z/" z/" z/" u = U ! C1e cos z / " + C2e sin z / " + C3e cos z / " ! C4e sin z / " (5.1.10b) Our boundary condition as z∞ is that (u,v)(U,0). To satisfy this we clearly have to take C3 = C4 = 0 . At z =0, both u and v are zero. This determines the remaining constants, C1 = U, C2 = 0 so that our final solution is, u = U(1! e!z/" cos z / "), (5.1.11 a,b) v = Ue!z/" sin z / " When z >> ! u approaches U exponentially rapidly and v goes to zero. That is, outside a region of O(! ) the solution approaches (5.1.5 a, b, c) which is the solution we would obtain completely ignoring friction. The effect of friction is limited to a region of O(! ) near the solid boundary; this is the Ekman boundary layer.
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