Laminar Boundary Layers. Answers to problem sheet 1: Navier-Stokes equations

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Isaac Preston
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1 Laminar Boundary Layers Answers to problem sheet 1: Navier-Stokes equations The Navier Stokes equations for d, incompressible flow are + v ρ t + u + v v ρ t + u v + v v = 1 = p + µ u + u = p ρg + µ v + v 3 1. Couette-Poiseuille flow a For steady flow t = v t =, i.e., we can ignore the time derivatives in equations and 3. The pressure gradient is applied along and the upper plate is moving in the direction, so the vertical velocity v =. The continuity equation 1 therefore yields, =, which tells us that the horizontal velocity u = uy only. We can therefore ignore all derivatives of u in Eqns. and 3, which are now reduced to p + µ u = 4 p ρg =. 5 These must be solved subject to the following boundary conditions: Integrating equation 5 gives, u = U, v = for y = a u = v = for y =. p = ρgy + A, where A is an integration constant, i.e. a function of only. We identify ρgy as the hydrostatic pressure and note that it does not interact with the flow: it does not appear in the equation for u since its derivative is zero. To find the velocity profile we integrate equation 4, keeping in mind that the applied pressure gradient p/ is constant: u = 1 p µ y + By + C.

2 To determine the constants B and C we apply the boundary conditions: uy = = C =, Hence, the solution is uy = a = U B = U a 1 p µ a. uy = 1 p µ yy a + U a y. b The velocity profile is a superposition of two components, shown as dotted lines in the figures. The straight line Uy/a is the shear flow resulting from the movement of the upper plate y = a at speed U. The parabola is due to the pressure gradient. It either points forward favourable gradient p/ < or backwards adverse gradient p/ >. In the former case the maimum velocity occurs above the centre line a/; in the latter case below. This can be seen from = y = a U µ a p/ If p/ < U y If p/ > U y

3 c Total rate of throughout: a Q = uydy = 1 [ p y 3 µ 3 ay = 1 p µ a Ua ] a + U a [ ] y a Mean velocity: < u >= Q a = 1 p µ a 6 + U Per unit area, the drag force on the plane y = is Π y = µ = p y a + µu a = µu a 1 p a.. Oscillatory plane a The oscillating plane is infinite, so there is a translational invariance in the direction: the velocity field does not depend on. There is no applied pressure gradient p/. So we can set all derivatives equal to zero in Eqns. 1 to 3. From the continuity equation, we then get v =, which tells us that the vertical velocity v = everywhere, since v = on the boundary. We therefore only need worry about the horizontal velocity field u = uy, t. To calculate this, we use the -momentum equation t = u ν, 6 where ν = µ/ρ is the kinematic viscosity. You might identify this as a diffusion equation. This must be solved subject to the boundary condition We seek a solution of the form ut, y = = U cosωt ut, y = fy cosωt = Refye iωt, where fy is a comple function. Substituting this into equation 6, we get for which the general solution is iωfy = ν f, fy = Ae 1+iky + Be 1+iky,

4 with k = ω/ν. Hence, uy, t = Re{Ae ky e iωt ky + Be ky e iωt+ky }. Physically, the velocity must remain finite for all positive y no matter how far we are from the oscillating plane, so B =. Also, on the solid boundary, ut, = U cosωt, so A = U, and we finally get ut, y = Ue ky cosωt ky. This indicates that the velocity oscillation imposed at the surface of the plane propagates towards the interior of the viscous fluid as an eponentially damped, transverse wave with velocity ω/k. b The penetration depth δ of the velocity field is the distance from the plane for which the amplitude of the disturbance is decreased by a factor e 1. Thus δ = 1 k = ν ω. For water, ν = 1 6 m /s and with ω = 1 6 Hz, we get δ = 1.4 µm. c The drag force on the plate per unit area is ω Π y = µ = µuksin ωt cosωt = µu ν coswt 3π because t ρe + ρ ρu i E = E i t + ρu i E + ρ i t + u E i i by the continuity equation. ρ t + ρu i i = = ρ DE Dt, 4. Consider a two-dimensional section of a cubic element of fluid i.e. a rectangle of fluid of sides δ, δy. Taking moments about the centre of the rectangle, gravity forces and normal stresses Π, Π yy contribute nothing since their line of action passes through. The net moment about is due to shear stresses only, and in a clockwise sense this moment is, Π y + Π y δy + Π yδ δy Π y + Π y δ + Π yδy δ = Π y Π y + 1 { Πy δy Π } y δ δδy. The moment of inertia of the cube about is, per unit depth in the paper, 1 1 ρδδy[δ + δy ].

5 y Π yy + Πyy δy Π y δy Π Π y + Πy δy O. δ Π yy Π y Π + Π δ Π y + Πy δ Denoting the angular acceleration about by a, Newton s second law for a system in rotation gives 1 1 ρδδy[δ + δy ]a = Π y Π y + 1 { Πy δy Π } y δ δδy. Taking δ and δy to both have the same order of magnitude δ, we notice that the LHS scales as aδ 4 and the RHS as Π y Π y δ +Oδ 3. To avoid a becoming infinite in the limit as δ, the prefactor to the δ term on the RHS must be zero, so we have Π y = Π y. This reasoning can be etended to a three-dimensional elementary cube of fluid to show that the stress tensor [Π] is symmetric.

REE Internal Fluid Flow Sheet 2 - Solution Fundamentals of Fluid Mechanics

REE Internal Fluid Flow Sheet 2 - Solution Fundamentals of Fluid Mechanics REE 307 - Internal Fluid Flow Sheet 2 - Solution Fundamentals of Fluid Mechanics 1. Is the following flows physically possible, that is, satisfy the continuity equation? Substitute the expressions for