Rayleigh's Problem - Solid Wall as a Source of Vorticity

Lecture Notes on Fluid Dynamics (1.63J/2.21J) 1.8 Rayleigh’s Problem - solid wall as a source of vorticity Owing to terms representing convective inertia, the Navier-Stokes equations are highly nonlinear. Explicit solutions are usually limited to a class of problems where inertia is identically zero. This happens when the flow is unidirectional and uniform. Flow quantities depend only on a transverse coordinate. We discuss one such example with a view to examining the role of viscosity. Consider a two-dimensional flow in the upper half plane of (x, y) bounded below by a rigid plate coinciding with the x axis. At t = 0 the plate suddenly moves in the tangential direction at constant velocity U . Find the development of the fluid motion in the region y > 0. ∂ = 0. It Because the plate is infinite in extent, the flow must be uniform in x, i.e. ∂x follows from continuity that ∂v = 0, y > 0 ∂y implying that v = constant in y. Since v(0, t) = 0 , v ≡ 0 for all y. Therefore, the only unknown is u(y, t) which must satisfy the momentum equation, ∂u ∂2u =ν 2 ∂t ∂y (1.8.1) where µ ρ denotes the kinematic viscosity. The boundary conditions are : ν= u = U, y = 0, t > 0; The initial condition is no slip (1.8.2) (1.8.3) u = 0, y ∼ ∞, t > 0 (1.8.4) u = 0, t = 0, ∀y (1.8.5) Mathematically this is the heat conduction problem for a semi-infinite rod. The solution is well-known (Carlaw & Jeager, Conduction of Heat in Solids or Mei, Mathematical Analysis in Engineering) , ! y u = U 1 − erf √ (1.8.6) 2 νt 5 4 y  2 √ νt = 5 3 4 3 2 2 1 1 0 0 0.25 0.5 0.75 1 u/U Figure 1.8.1: Velocity profile due to impulsive motion of x-plane where 2 erf ζ = √ π Z ζ 2 e−λ dλ. (1.8.7) 0 is the error function. As shown in Figure 1.8.1, fluid momentum is diffused away from the √ plane y = 0. The region affected by viscosity (the boundary layer) grows in time as δ ∼ νt. This observation can be confirmed, indeed anticipated, merely by a scaling argument based on the momentum equation (1.8.1) without solving it. Let U, t, δ denote the scales of velocity, time and region of viscosity respectively. For viscosity to be important, the two terms in (1.8.1) must be comparable in order of magnitude, i.e., U U ∼ν 2 t δ It follows that √ δ ∼ νt Let us use this simple example to study the role of vorticity ζ~ = ! ∂v ∂u ~ − k ∂x ∂y (1.8.8) In this problem there is only one vorticity component, ζ3 = − ∂ y 2U ∂u 2 =U erf √ = √ e−y /4νt . ∂y ∂y 2 νt 4πνt (1.8.9) which is just the velocity shear. Mathematically (1.8.9) is the solution to the diffusion equation ∂ζ ∂2ζ = ν 2. (1.8.10) ∂t ∂y which follows from (1.8.1), and the initial condition that there is a plane source of at y = 0: ζ3 (y, 0) = 2U δ(y). (1.8.11) Thus vorticity is diffused away from the solid wall which acts as a voriticity source. Note that the shear stress at the wall is ∂u τxy (0, t) = µ ∂y y=0 = −ρU r ν . πt (1.8.12) which is initially infinite but decays with time. Why is the wall a source of vorticity? Just after the plane started to move there is a velocity discontinuity at y = 0+. The associated velocity gradient is ∂u/∂y = −U δ(y) hence the vorticity is a highly concentrated function of y: −∂u/∂y = U δ(y). Furthermore the half space (0 < y < ∞) problem can be thought of as one half of the whole plane problem for −∞ < y < ∞ if the top of the fluid in the lower half plane suddenly moves to the left at the speed U . This would give an initial vorticity U δ(y) at y = 0−. Thus for the a whole space problem there is a vorticity source of total strength 2U δ(y) at the initial instant. As time proceeds, half of the released vorticity is diffused to the region of y > 0 and half to y < 0. Thus, the solid wall is the source of vorticity. The reader can verify the solution (1.8.9) by assuming a similarity form, C ζ3 (y, t) = √ f t y √ t ! (1.8.13)