Download - 08_Class_KinematicsVorticity.pdf

7/27/2019 08_Class_KinematicsVorticity.pdf

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Kinematics of Vorticity


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Vorticity = V

2 circumferentially averagedangular velocity of the fluid particles

Sum of rotation rates of perpendicular fluid lines

Non-zero vorticity doesnt implyspin

. =0. Incompressible? Direction of ?

U

y


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Circulation

Macroscopic rotation of the fluidaround loop C

Non-zero circulation doesnt implyspin

Connected to vorticity flux throughStokes theorem

Stokes for a closed surface?

=

C

d sV.U

y

cC S

d dS == sVn ..Open Surface S

with Perimeter Cn dS


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Kinematic Concepts - VorticityBoundary layer growing on flat plate

Cylinder projectingfrom plate

Vortex Line: A line everywhere tangent to the vorticity vector, so ds =0.Vortex lines may not cross. Rarely are they streamlines. Thread together axesof spin of fluid particles. Given by d s =0. Could be a fluid line?

Vortex sheet: Surface formed by all the vortex lines passing through the samecurve in space. No vorticity flux through a vortex sheet, i.e. .n dS =0

Vortex tube: Vortex sheet rolled so as to form a tube.

ndS


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Vortex Tube

Section 1

Section 2ndS

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

0. =

cC S

d dS == sVn ..


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Implications(Helmholtz Vortex Theorems, Part 1)

The strength of a vortex tube (defined as thecirculation around it) is constant along the tube.

The tube, and the vortex lines from which it is

composed, can therefore never end. They mustextend to infinity or form loops. The average vorticity magnitude inside a vortex

tube is inversely proportional to the cross-sectional area of the tube


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But, does the vortex tube travel along with thefluid, or does it have a life of its own?

Fluidloop attime t

Same

fluid loopat timet+dt

V dt

(V +dV )dt

If it moves with the fluid, then thecirculation around the fluid loopshown should stay the same.


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Fluidloop attime t

Same

fluid loopat timet+dt

V dt

(V +dV )dt

[ ]k jif V ).().().(1 z y x p Dt D

+++=

=C

d Dt D

Dt D

sV

.So the rate of change of around the fluid loop is

Now, the N.S. equation tell us that

Body forceper unit mass

Pressure force

per unit mass

Viscous force per

unit mass, say f v

++=C C C

d d p

d Dt D

sf ssf v ...

So, in general


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Body Force Torque

k f g= = S C dS d nf sf ..Stokes Theorem

For gravity 0= f

So, body force torque is zero for gravity and for any irrotationalbody force field

Therefore, body force torque is zero for most practicalsituations


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Pressure Force Torque

0.1

.1

== S C dS pd p ns If density is constantSo, pressure force torque is zero. Also true as long as = (p).

Pressure torques generated by

Curved shocks

Free surface / stratification


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Viscous Force Torque Viscous force torques are non-zero where

viscous forces are present ( e.g. Boundary layer,wakes)

Can be really small, even in viscous regions at

high Reynolds numbers since viscous force issmall in that case

The viscous force torques can then often be

ignored over short time periods or distances


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ImplicationsIn the absence of body-force torques, pressure torques and

viscous torques the circulation around a fluid loop stays constant.

Kelvins Circulation 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 doesnt diffuse fluid with vorticity will always have it fluid that has no vorticity will never get it

++=C C C

d d p

d Dt D

sf ssf v ...

Body forcetorque

Pressure forcetorque

Viscous force

torque

Helmholtz VortexTheorems, Part 2


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Vorticity Transport Equation The kinematic condition for convection of

vortex lines with fluid lines is found as follows

( ) 0=ds Dt

D

V

= . Dt

D

After a lot of math we get....


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Example: Irrotational Flow?

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

Consider a vehicle moving at constant speed in homogeneous medium (i.e. nofree surfaces) under the action of gravity, moving into a stationary fluid.

Apparentuniform flow(V = const.)

Far ahead of the sub we have

that (since V = const.)0= V

and the flow here is

irrotational.Now, the flow generates nobody force or pressuretorques and, except in thevehicle boundary layer andwake.

So...


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Example: The Starting VortexConsider a stationary arifoil in a stationary medium.

L

V = 0

V = const.

Fluid loop C

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

== C C 0.dsV

C = 0 by Kelvins Theorem


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Starting Vortex


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Example: Flow over a depression in a river bed

A river flows over a depression locally doubling thedepth. The river contains turbulence that is tooweak to change the overall flow pattern. Anturbulent eddy convects from upstream over the

depression. Estimate its strength in the depressionif the eddy is initially (a) vertical, and (b) horizontal.

h2h

Solution: Need to assume that the viscous torques are not significant for the eddyso 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 thedepression and so (by continuity) it will halve its cross sectional area. ByHelmholtz theorems, halving the cross sectional area of a vortex tube willdouble the vorticity. Hence .

(b) Assuming the depression and flow is 2D, the flow speed will halve as it goesover the depression. The horizontal fluid tube containing the eddy will thushalve its length and (by continuity) double its cross-sectional area. By Helmhotztheorems the vorticity will ...


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Example: Evolution of turbulence in a shear flow

Turbulence is convected and distorted in a shear flow, as shown. Consider an eddy that at time t=0is vertically aligned and has a vorticity o. Estimatethe vorticity magnitude and angle of an eddy, as

functions of time.

Solution: Need to assume that the viscous torques are not significant for the eddyso 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 isconvecting it.

U=ky, k=const

y

y l

klt

Time 0 Time t

Consider a segment of the eddy of length l. In time t the top of the eddywill convect further than the bottom byan amount equal to the difference inthe velocity ( kl) times the time.The angle of the fluid tube containing the eddy will thus be

The fluid tube also grows longer by the factorThe cross sectional area of the tube thus reduces by this factor, andtherefore the vorticity increases by this factor, i.e.


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Example: Flow around a corner in a channel

Air flows through a duct with a 6 o corner. An initiallyvertical eddy is introduced into the otherwiseuniform flow upstream and convects around thecorner. Estimate its orientation downstream.

Solution: Need to assume that the viscous torques are not significant for the eddyso 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 andone vertical, coincident with the eddy. Since there is no vorticity componentperpendicular to the page, and no torques to generate any, the sum of the

rotation rates of these two fluid lines must remain zero.

The streamwise fluid line will follow thestreamline and thus rotate counterclockwiseby 6 o. The vertical fluid line and thus the

eddy must therefore rotate clockwise by 6 o, as shown.

6o

6o

Download - 08_Class_KinematicsVorticity.pdf

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