The vorticity equation and its applications · 2017. 10. 9. · The vorticity equation and its...

The vorticity equation and its applications

Felix KAPLANSKI Tallinn University of Technology

[email protected] Tallinn University of Technology, Estonia

Chapter 10: Approximate Solutions ME33 : Fluid Flow 2

Examples of vortex flows



VORTEX BREAKDOWN IN THE LABORATORY The photo at the right is of a laboratory vortex breakdown provided by Professor Sarpkaya at the Naval Postgraduate School in Monterey, California. Under these highly controlled conditions the bubble-like or B-mode breakdown is nicely illustrated. It is seen in the enlarged version that it is followed by an S-mode breakdown.



VOLCANIC VORTEX RING The image at the right depicts a vortex ring generated in the crater of Mt. Etna. Apparently these rings are quite rare. The generation mechanism is bound to be the escape of high pressure gases through a vent in the crater. If the venting is sufficiently rapid and the edges of the vent are relatively sharp, a nice vortex ring ought to form.


Vortex ring flow

Chapter 10: Approximate SolutionsME33 : Fluid Flow 7


A VORTEX RING At the right is a vortex ring generated by Professor T.T. Lim and his former colleagues at the University of Melbourne. The visualization technique appears to be by smoke.


Virtual image of a vortex ring flow www.applied-scientific.com/ MAIN/PROJECTS/NSF00/FAT_RING/Fat_Ring.html -

Force acts impulsively


Overview:

!  Derivation of the equation of transport of vorticity

!  Describing of the 2D flow motion on the basis of vorticity ω and

streamfunction ψ instead of the more popular (u,v,p)-system

!  Well-known solutions of the system (ω, ψ )


NSE


Vorticity


Vorticity transport equation


Helmholtz equation

ςω =z


Vorticity equation on plane

⎥⎦

⎤⎢⎣

⎡

∂∂

+∂∂

+∂∂

−=∂∂

+∂∂

+∂∂

∂∂

2

2

2

21:yu

xu

xp

yuv

xuu

tu

yν

ρ

⎥⎦

⎤⎢⎣

⎡

∂∂

+∂∂

+∂∂

−=∂∂

+∂∂

+∂∂

∂∂

2

2

2

21:yv

xv

yp

yvv

xvu

tv

xν

ρ

⎥⎦

⎤⎢⎣

⎡

∂∂

−∂∂

∂∂

+∂∂

−∂∂

∂∂

+∂∂∂

+∂∂∂

−=∂∂

∂∂

−∂∂

−∂∂

∂∂

−∂∂∂

−

∂∂

∂∂

+∂∂

∂+

∂∂

+∂∂

∂∂

+∂∂

−∂∂

∂∂

)()(

11

)(

2

2

2

2

2

2

2

2

2

yu

xv

yyu

xv

x

yxp

yxp

yu

yv

yuv

xu

yu

yxuu

yv

xv

yxvv

xvu

xv

xu

yu

xv

t

ν

ρρ

1 3

4

1

3

4

2 2

2)

1)

2)-1)=


Taking into account:

xv

yu

∂Ψ∂

−=∂Ψ∂

= ,

0

0

=∂∂Ψ∂

−∂∂Ψ∂

=∂∂

+∂∂

=∂∂

+∂∂

yxyxyv

xu

yv

xu

Continuity equation


Vorticity equation on plane

⎥⎦

⎤⎢⎣

⎡

∂∂

−∂∂

∂∂

+∂∂

−∂∂

∂∂

=

∂∂

−∂∂

∂+

∂∂∂

−∂∂

+∂∂

−∂∂

∂∂

)()(

)(

2

2

2

2

2

22

2

2

yu

xv

yyu

xv

x

yuv

yxvv

yxuu

xvu

yu

xv

t

ν

⎥⎦

⎤⎢⎣

⎡

∂∂

+∂∂

=∂∂

+∂∂

+∂∂

2

2

2

2

yxyv

xu

tωω

νωω

ω

3 3 4 4


Cylindrical coordinate system

In cylindrical coordinates (r , θ ,z ) with 0=∂∂ θ/-axisymmetric case


Vorticity equation: axisymmetric case

⎥⎦

⎤⎢⎣

⎡

∂∂

+∂∂

+∂∂

+∂∂

−=∂∂

+∂∂

+∂∂

∂∂

ru

rru

zu

zp

ruv

zuu

tu

r11: 2

2

2

2

νρ

⎥⎦

⎤⎢⎣

⎡−

∂∂

+∂∂

+∂∂

+∂∂

−=∂∂

+∂∂

+∂∂

∂∂

22

2

2

2 11:rv

rv

rrv

zv

rp

rvv

zvu

tv

zν

ρ

⎥⎦

⎤⎢⎣

⎡

∂∂

−∂∂

+∂∂

−∂∂

+∂∂

−∂∂

∂∂

+∂∂

−∂∂

∂∂

∂∂∂

+∂∂∂

−=∂∂

∂∂

−∂∂

−∂∂

∂∂

−∂∂∂

−

−+∂∂

∂∂

+∂∂

∂+

∂∂

+∂∂

∂∂

+∂∂

−∂∂

∂∂

zv

rru

rru

zv

rru

zv

rru

zv

z

rzp

rzp

ru

rv

ruv

zu

ru

rzuu

rv

rv

rv

zv

rzvv

zvu

zv

zu

ru

zv

t

222

2

2

2

2

2

2

2

2

11)(1)()(

11

)(

ν

ρρ

1 3

4

1

3

4

2 2

2)

1)

2)-1)=

1 2

Proof with Mathematica


Taking into account:

zrv

rru

∂Ψ∂

−=∂Ψ∂

=1,1

01111

)()(

0)()(

=∂Ψ∂

+∂Ψ∂

−∂∂Ψ∂

−∂∂Ψ∂

=+∂∂

+∂∂

=∂

∂+

∂∂

=∂

∂+

∂∂

zrzrrzrr

rzrr

vrvr

zur

rrv

zru

rrv

zru

Continuity equation


Vorticity equation: axisymmetric case

⎥⎦

⎤⎢⎣

⎡

∂∂

−∂∂

−∂∂

−∂∂

∂∂

+∂∂

−∂∂

∂∂

+∂∂

−∂∂

∂∂

=

∂∂

−∂∂

∂+

∂∂∂

−∂∂

+∂∂

−∂∂

∂∂

)(1)(1)()(

)(

22

2

2

2

2

22

2

2

ru

zv

rru

zv

rrru

zv

rru

zv

z

ruv

rzvv

rzuu

zvu

ru

zv

t

ν

⎥⎦

⎤⎢⎣

⎡−

∂∂

+∂∂

+∂∂

=∂∂

+∂∂

+∂∂

22

2

2

2 1rrrrzr

vz

ut

ωωωων

ωωω

3 3 4 4


⎥⎦

⎤⎢⎣

⎡−++=++ 22

2

2

2

rq

rrq

rzzu

rv

tω

∂∂ω

∂ω∂

∂ω∂

ν∂∂ω

∂∂ω

∂∂ω

ω∂∂

∂∂

∂∂ qr

rrq

zr−=

Ψ−

Ψ+

Ψ2

2

2

2

The Stokes stream function can be introduced as follows

zrv

rru qq ∂

Ψ∂−=

∂Ψ∂

=1,1

and gives second equation

Vorticity transport equation for 2D : q=1- axisymmetric vortices,

q=0 – plane vortices


For 3D problem: generalized Helmholtz equation

xzyxxxxx

zu

yu

xu

zw

yu

xv

tων

∂∂

ω∂∂

ω∂∂

ω∂∂ω

∂∂ω

∂∂ω

∂∂ω

Δ+++=+++

yzyxyyyy

zv

yv

xv

zw

yu

xv

tων

∂∂

ω∂∂

ω∂∂

ω∂

∂ω

∂

∂ω

∂

∂ω

∂

∂ωΔ+++=+++

zzyxzzzz

zw

yw

xw

zw

yu

xv

tων

∂∂

ω∂∂

ω∂∂

ω∂∂ω

∂∂ω

∂∂ω

∂∂ω

Δ+++=+++

222222 /// zyx ∂∂+∂∂+∂∂=Δwhere

For 3D problem we can not introduce streamfuction Ψ like for 2D problem.

,

,

,

wvuzyx

kji

∂∂

∂∂

∂∂

=→

ω


For 3D problem: generalized Helmholtz equation in cylindrical coordinates

),2( 22 ∂θ∂ωω

ων∂∂

ω∂θ∂

ω∂∂

ω∂∂ω

∂θ∂ω

∂∂ω

∂∂ω θ

θθ rrzu

ru

ru

zu

ru

ru

tr

rr

zrr

rr

zrr

rr −−Δ+++=+++

),2( 22 ∂θ∂ωω

ωνω

∂∂

ω∂θ∂

ω∂∂

ωω

∂∂ω

∂θ∂ω

∂∂ω

∂∂ω θ

θθθθθ

θθθθθ

θθθ r

zrr

zr rrru

zu

ru

ru

ru

zu

ru

ru

t+−Δ+−++=−+++

2

2

2

2

22

2 11zrrrr ∂∂

+∂∂

+∂∂

+∂∂

=Δθ

where

,

,

,

,zzz

zzr

zz

zzr

z

zu

ru

ru

zu

ru

ru

tων

∂∂

ω∂θ∂

ω∂∂

ω∂∂ω

∂θ∂ω

∂∂ω

∂∂ω

θθ Δ+++=+++

,)(,,θ

ωωθ

ω θθ

θ

∂∂

−∂

∂=

∂∂

−∂∂

=∂∂

−∂∂

=ru

rrru

ru

zu

zu

ru r

zzrz

r


For 2D problem:

(u,v,p) (ω, ψ)

Winning: two variables instead of three Losses: difficulties with boundary conditions

for streamfunction


Vortex flow 2-D plane


Streamfunction and (u, v) through vorticity

and u, v are given by


!  Solutions, which contain vorticity expressed through delta-functions


Vortex flow.


The Biot-Savart Law

ω∂∂

∂∂

∂∂ r

rrzr−=

Ψ−

Ψ+

Ψ 12

2

2

2


!  Solutions, which contain vorticity expressed through delta-functions


Vortex flow.


!  Other solutions


Vortex flows. Hill’s ring


⎟⎟⎠

⎞⎜⎜⎝

⎛

∂

∂+

∂

∂=

∂

∂

rrrtωω

νω 1

2

2

We use polar coordinates (r, θ) and assume symmetry 0=∂∂ θ/


Solution

tr

et

c ν

πνω 4

2

4

−

=

crdret

c

rdr)t,r(

trr

tt

=∫=

∫==

−

==

ν

πνπ

ωπΓΓ

4

2

0

00

42

2

tr

et

ν

πνΓ

ω 4

2

4

−

=

Further we define constant c

and find solution



Appropriate tangent velocity

)e(r

rdrr

)t,r(u t/rr

ν

πΓ

ω 42

01

21 −−∫ ==


1 2 3 4 5

0.5

1

1.5

2

2.5

3

u

r


Burgers vortex (a viscous vortex with swirl)


Vorticity


Irrotational Flow Approximation

!   Irrotational approximation: vorticity is negligibly small

!   In general, inviscid regions are also irrotational, but there are situations where inviscid flow are rotational, e.g., solid body rotation (Ex. 10-3)


Irrotational Flow Approximation 2D Flows !   For 2D flows, we can also use the streamfunction !   Recall the definition of streamfunction for planar (x-y)

flows

!   Since vorticity is zero,

!   This proves that the Laplace equation holds for the streamfunction and the velocity potential


Elementary Planar Irrotational Flows Uniform Stream

!   In Cartesian coordinates

!   Conversion to cylindrical coordinates can be achieved using the transformation



Elementary Planar Irrotational Flows Line Source/Sink

!   Potential and streamfunction are derived by observing that volume flow rate across any circle is

!   This gives velocity components



!   Using definition of (Ur, Uθ)

!   These can be integrated to give φ and ψ

Equations are for a source/sink at the origin Proof with Mathematica



!   If source/sink is moved to (x,y) = (a,b)


Elementary Planar Irrotational Flows Line Vortex

!  Vortex at the origin. First look at velocity components

!  These can be integrated

to give φ and ψ

Equations are for a source/sink at the origin


Elementary Planar Irrotational Flows Line Vortex

!   If vortex is moved to (x,y) = (a,b)


Elementary Planar Irrotational Flows Doublet

!  A doublet is a combination of a line sink and source of equal magnitude

!  Source !  Sink


Elementary Planar Irrotational Flows Doublet

!  Adding ψ1 and ψ2 together, performing some algebra, and taking a→0 gives

K is the doublet strength


Examples of Irrotational Flows Formed by Superposition

!  Superposition of sink and vortex : bathtub vortex

Sink Vortex



!  Flow over a circular cylinder: Free stream + doublet

!  Assume body is ψ = 0 (r = a) ⇒ K = Va2



!   Velocity field can be found by differentiating streamfunction

!   On the cylinder surface (r=a)

Normal velocity (Ur) is zero, Tangential velocity (Uθ) is non-zero ⇒slip condition.

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