Kinematics of Vorticity

Kinematics of Vorticity The Equations of Motion Differential Form (for a fixed volume element)   The Continuity equation D   .V  Dt  The Navier Stokes’ equations    Du p   u    v u    u w  1      f x   2 (  3 .V)   (  )   (  ) Dt  x x  x  y  x y  z  z x   Dv p   v u    v    v w      1     f y    (  )  2 (  3 .V)   (  ) Dt  y x  x y  y  y  z  z y  Dw p   u w    v w    w     1  f z    (  )   (  )  2(  3 .V) Dt z x  z x  y  z y  z  z  The Viscous Flow Energy Equation   1 2  D(e  2 V )   u 1 v u u w   f.V  ( pV)  .(kT)  2 u(  3 .V)  v(  )  w(  ) Dt x  x x y z x 

  v u v 1 v w    u w v w w 1    u(  )  2 v(  3 .V)  w(  )   u(  )  v(  )  2w(  3 .V) y  x y y z y  z  z x z y z 2  Vorticity 

No spin, but a net rotation rate

3 Circulation 

Ω.ndS  V.ds     c S C Open Surface S ndS with Perimeter C

4 Flow Past a ‘Cookie-Tin’

Top view

Side view

Re ~ 4,000 ‘Horseshoe vortex’

5 Pictures are from “An Album of Fluid Motion” by Van Dyke Large Eddy Simulation Re=5000

6 George Constantinescu IIHR, U. Iowa Kinematic Concepts - Vorticity Boundary layer Cylinder projecting growing on flat plate Vortex line from plate dS n

Vortex sheet



Vortex tube

Vortex Line:

7 Kinematic Concepts - Vorticity Boundary layer Cylinder projecting growing on flat plate Vortex line from plate dS n

Vortex sheet



Vortex tube

Vortex sheet:

Vortex tube:

8 .Ω  0 Ω.ndS  V.ds   Vortex Tube   c S C • 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

Section 2

dS n

Section 1

9 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 cross- sectional area of the tube

10 But, does the vortex tube travel along with the fluid, or does it have a life of it’s own?

Same fluid loop Fluid (V+dV)dt at time loop C ds t+dt at time t Vdt Dds ds  dt Dt 11 D DV DV p 1 So  .ds and     f    (.τ x )i  (.τ y )j (.τ z )k Dt C Dt Dt 

Same fluid loop (V+dV)dt at time Fluid t+dt loop at time t Vdt

12 Body Force Torque f.ds C

13 1 Pressure Force Torque  p.ds  C

14 Viscous Force Torque

15 Implications

In the absence of body-force torques, pressure torques and viscous torques… D • the circulation around a fluid loop stays constant:  0 Kelvin’s Circulation Theorem Dt • 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 D p – fluid that has no vorticity will never get it  f.ds  .ds  f .ds    v Dt C C  C Helmholtz’ Vortex ‘Body force Theorems, Part 2 torque’ Viscous force ‘Pressure force ‘torque’ 16 torque’ Vorticity Transport Equation

• The kinematic condition for convection of vortex lines with fluid lines is found as follows

D Ωds  0 Dt

After a lot of math we get....

DΩ  Ω.V Dt

17 Example: Irrotational Flow?

Consider 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.)

http://www.lcp.nrl.navy.mil/~ravi/par3d.html

18 Example: The Starting Vortex Fluid loop C Consider a stationary arifoil in a stationary medium.   V.ds  0 C  C

V = 0

Now suppose the airfoil starts moving to left. (Using the fact that that a lifting airfoil in motion has a circulation about it).

V = const.

Fluid loop C 19 Starting Vortex

20 Pictures are from “An Album of Fluid Motion” by Van Dyke Example: Flow over a depression in a river bed

A river flows over a depression locally doubling the depth. The river contains turbulence that is too h weak to change the overall flow pattern. An 2h 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)

21 Example: Flow over a depression in a river bed

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.

22 The River Avon nearing flood stage, Salisbury, Wiltshire, UK

23 Example: Evolution of turbulence in a shear flow

U=ky, k=const Turbulence is convected and distorted in a shear flow, as shown. Consider an eddy that at time t=0 is y vertically aligned and has a vorticity 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. klt Consider a segment of the eddy of length l. In time t the top of the eddy y l will convect further than the bottom by

an amount equal to the difference in Time 0 Time t the velocity (kl) times the time. The angle of the fluid tube containing the eddy will thus be arctanklt l arctankt

The fluid tube also grows longer by the factor (klt)2  l 2 l  (kt)2 1 The cross sectional area of the tube thus reduces by this factor, and 24 therefore the vorticity increases by this factor, i.e. 2   o (kt) 1 Eddies in a turbulent channel flow

DNS Simulation Thomas Bewley, Edward Hammond & Parviz Moin Stanford University

25 Example: Flow around a corner in a channel

Air flows through a duct with a 6o corner. An initially vertical eddy is introduced into the otherwise uniform flow upstream and convects around the corner. Estimate its orientation downstream. 6o

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.

Consider 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. 6o The streamwise fluid line will follow the

streamline and thus rotate counterclockwise 6o by 6o. The vertical fluid line and thus the eddy must therefore rotate clockwise by 6o, as shown. 26 Turbulent flow in a pipe through a 180o bend. Hitoshi Sugiyama, Utsunomiya University