Kinematics of Vorticity

 Aldous Gallagher
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1 Kinematics of Vorticity
2 Vorticity Ω Ω= V 2 circumferentially averaged angular velocity of the fluid particles Sum of rotation rates of perpendicular fluid lines Nonzero vorticity doesn t imply spin.ω=0. Incompressible? Direction of Ω? y U
3 irculation Γ Γ = V.ds Macroscopic rotation of the fluid around loop Nonzero circulation doesn t imply spin onnected to vorticity flux through Stokes theorem Stokes for a closed surface? y S U Ω. nds = V. ds = nds Γ c Open Surface S with Perimeter
4 Kinematic oncepts  Vorticity Boundary layer growing on flat plate ds n ylinder projecting from plate Ω Vortex Line: A line everywhere tangent to the vorticity vector, so ds Ω=0. Vortex lines may not cross. Rarely are they streamlines. Thread together axes of spin of fluid particles. Given by ds Ω=0. ould be a fluid line? Vortex sheet: Surface formed by all the vortex lines passing through the same curve in space. No vorticity flux through a vortex sheet, i.e. Ω.ndS=0 Vortex tube: Vortex sheet rolled so as to form a tube.
5 Vortex Tube. Ω = 0 Ω. nds = V. ds S Vorticity as though it were the velocity field of an incompressible fluid The flux of vorticity thru an surface is equal to the circulation around the edge of that surface = Γ c ds n Section 2 Section 1
6 Implications (Helmholtz Vortex Theorems, Part 1) The strength of a vortex tube (defined as the circulation around it) is constant along the tube. The tube, and the vortex lines from which it is composed, can therefore never end. They must extend to infinity or form loops. The average vorticity magnitude inside a vortex tube is inversely proportional to the crosssectional area of the tube
7 But, does the vortex tube travel along with the fluid, or does it have a life of it s own? If it moves with the fluid, then the circulation around the fluid loop shown should stay the same. Fluid loop at time t (V+dV)dt Vdt Same fluid loop at time t+dt
8 So the rate of change of Γ around the fluid loop is DΓ Dt = DV. ds Dt Now, the N.S. equation tell us that DV Dt = f p 1 + ρ ρ [(. τ ) i + (. τ ) j+ (. τ ) k] x y z Body force per unit mass Pressure force per unit mass Viscous force per unit mass, say f v So, in general DΓ Dt = p f fv. ds ρ. ds +. ds + Fluid loop at time t (V+dV)dt Vdt Same fluid loop at time t+dt
9 Body Force Torque Stokes Theorem. s = f d f. nds For gravity f = g k f = 0 S So, body force torque is zero for gravity and for any irrotational body force field Therefore, body force torque is zero for most practical situations
10 Pressure Force Torque If density is constant 1 ρ p. ds = 1 ρ S p. nds = 0 So, pressure force torque is zero. Also true as long as ρ = ρ(p). Pressure torques generated by urved shocks Free surface / stratification
11 Viscous Force Torque Viscous force torques are nonzero where viscous forces are present ( e.g. Boundary layer, wakes) an be really small, even in viscous regions at high Reynolds numbers since viscous force is small in that case The viscous force torques can then often be ignored over short time periods or distances
12 Implications In the absence of bodyforce torques, pressure torques and viscous torques the circulation around a fluid loop stays constant. Kelvin s irculation Theorem a vortex tube travels with the fluid material (as though it were part of it), or a vortex line will remain coincident with the same fluid line the vorticity convects with the fluid material, and doesn t diffuse fluid with vorticity will always have it fluid that has no vorticity will never get it Helmholtz Vortex Theorems, Part 2 Body force torque DΓ Dt = p f fv. ds ρ. ds +. ds + Pressure force torque Viscous force torque
13 Vorticity Transport Equation The kinematic condition for convection of vortex lines with fluid lines is found as follows D Dt ( Ω ds) = 0 After a lot of math we get... DΩ Dt = Ω. V
14 Example: Irrotational Flow? onsider a vehicle moving at constant speed in homogeneous medium (i.e. no free surfaces) under the action of gravity, moving into a stationary fluid. Apparent uniform flow (V = const.) Far ahead of the sub we have that (since V = const.) V = 0 and the flow here is irrotational. Now, the flow generates no body force or pressure torques and, except in the vehicle boundary layer and wake. So...
15 Example: The Starting Vortex onsider a stationary arifoil in a stationary medium. Fluid loop V = 0 Γ = V.ds = 0 Now suppose the airfoil starts moving to left. (Using the fact that that a lifting airfoil in motion has a circulation about it). L Γ V = const. Γ = 0 by Kelvin s Theorem
16 Starting Vortex
17 Example: Flow over a depression in a river bed h 2h A river flows over a depression locally doubling the depth. The river contains turbulence that is too weak to change the overall flow pattern. An turbulent eddy convects from upstream over the depression. Estimate its strength in the depression if the eddy is initially (a) vertical, and (b) horizontal. Solution: Need to assume that the viscous torques are not significant for the eddy so that the fluid tube it occupies remains coincident with the same fluid tube. (a) The vertical fluid tube occupied by the eddy will double in length in the depression and so (by continuity) it will halve its cross sectional area. By Helmholtz theorems, halving the cross sectional area of a vortex tube will double the vorticity. Hence. (b) Assuming the depression and flow is 2D, the flow speed will halve as it goes over the depression. The horizontal fluid tube containing the eddy will thus halve its length and (by continuity) double its crosssectional area. By Helmhotz theorems the vorticity will...
18 Example: Evolution of turbulence in a shear flow y U=ky, k=const Turbulence is convected and distorted in a shear flow, as shown. onsider an eddy that at time t=0 is vertically aligned and has a vorticity Ω o. Estimate the vorticity magnitude and angle of an eddy, as functions of time. Solution: Need to assume that the viscous torques are not significant for the eddy so that the fluid tube it occupies remains coincident with the same fluid tube. Also need too assume that the eddy is to weak to influence the flow that is convecting it. onsider a segment of the eddy of length l. In time t the top of the eddy will convect further than the bottom by y l an amount equal to the difference in the velocity (kl) times the time. Time 0 The angle of the fluid tube containing the eddy will thus be Time t The fluid tube also grows longer by the factor The cross sectional area of the tube thus reduces by this factor, and therefore the vorticity increases by this factor, i.e. klt
19 Example: Flow around a corner in a channel Air flows through a duct with a 6 o corner. An initially vertical eddy is introduced into the otherwise uniform flow upstream and convects around the corner. Estimate its orientation downstream. Solution: Need to assume that the viscous torques are not significant for the eddy so that the fluid tube it occupies remains coincident with the same fluid tube, and so that no vorticity is generated in the turn. Also need to assume that the eddy is too weak to influence the flow that is convecting it. onsider two initially perpendicular fluid lines, one in the streamwise direction and one vertical, coincident with the eddy. Since there is no vorticity component perpendicular to the page, and no torques to generate any, the sum of the rotation rates of these two fluid lines must remain zero. 6 o The streamwise fluid line will follow the streamline and thus rotate counterclockwise by 6 o. The vertical fluid line and thus the eddy must therefore rotate clockwise by 6 o, as shown. 6 o
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