14 Circular Cylinders14. Circular Cylinders

1 Flow past a circular cylinder without1. Flow past a circular cylinder without circulation.

2. Adding circulation.

3 Circular cylinder in non-uniform flow3. Circular cylinder in non uniform flow

Why do we care so much about circular cylinders?cylinders?

1. Acyclic Flow Past a Circular CylinderNo circulation

1. Acyclic Flow Past a Circular Cylinder= Uniform flow + Opposing Doublet

C l V l it d P t ti l?Complex Velocity and Potential?

Fl i di n

General case

Cylinder radius?

Flow in x dirn.Cylinder at origin

2

1. Acyclic Flow Past a Circular Cylinder1. Acyclic Flow Past a Circular CylinderProve it’s circular?

21

2

)(zzeaVzeVzF

i

ii

21

2

)()(

zzeaVeVzW

ii

r=a

Pressure distribution on its surface?0 5

1

1 5

- 1

- 0 .5

0

0 .5C

p

30 6 0 1 2 0 1 8 0 2 4 0 3 0 0 3 6 0

- 3

- 2 .5

- 2

- 1 .5

T h e ta ( d e g r e e s )

1. Acyclic Flow Past a Circular Cylinder1. Acyclic Flow Past a Circular Cylinder

Comparison with real life

AOE 5104, Ideal Flow, TodayAOE 5104, Ideal Flow, Today

1

Experiment (Re~2000)Ideal Flow Theory

Re=2000 experiment, ONERARe=2000 experiment, ONERA, Werlé & Gallon 1972, Werlé & Gallon 1972

- 1

- 0 .5

0

0 .5

Cp

R 3900 Si l tiR 3900 Si l ti U E l M B 2001U E l M B 20010 6 0 1 2 0 1 8 0 2 4 0 3 0 0 3 6 0

- 3

- 2 .5

- 2

- 1 .5

Re=3900 Simulation, Re=3900 Simulation, Un. Erlangan, M Brewer 2001Un. Erlangan, M Brewer 2001T h e ta ( d e g r e e s )

4

2. Circular Cylinder with Circulation2. Circular Cylinder with Circulation= Uniform flow + Opposing Doublet + Vortex

S f li d di 2 ieaV i So, for a cylinder radius a, centered at z1 in a free stream of velocity V at angle to the x axis with circulation : )(2)(

)(

)(log2

)(

12

1

2

11

zzi

zzeaVeVzW

zzizzeaVzeVzFi

i

ei

)()( 11

Velocity on cylinder surface(take = z =0 and z=aei)

Pressure and Stagnation Points(take = z1=0 and z=aei)

2)(

2

2

2

ii eieaVz

izaVVzW

aV

aVv

stag

stag

4arcsin

2sin20

2

2)(

2

2

2

i

i

i

ii

iir

aeei

eaeaVeV

zei

zeaVeVezWivv

VvCsurfp /1

2

22

5

02

sin2

rva

Vv

aVaV

4sin8

44sin41 2

2. Circular Cylinder with Circulation2. Circular Cylinder with CirculationaV4

aVstag

4arcsin

1

aV4 aV4180-180 -1

0 0 1800 0 , 180

0 5 30 210-0.5 -30, 210

-1 -90, 270

6-1.5 ?

2. Circular Cylinder ith Ci l ti

0

with Circulation

aVaVC

surfp

4sin8

44sin41

22

-5

04

aV10

5.10.15.0

-10

Cp

-15Anton Flettner (1885-1961)

-20

70 60 120 180 240 300 360

-25

Theta (degrees)

3. Circular Cylinder in Non-Uniform Flow: Th Mil Th Ci l ThThe Milne-Thompson Circle Theorem

F(z) F1(z)

a

8

3. Milne-Thompson Theorem: Proofpz-plane)()()( 2

1 zaFzFzF

r=a

F(z) is the conjugate function of F(z) –the same except with all constants replaced by their conjugates. E.g

)(log2

)( 1zzizF e

i

9Note that

)(log2

)( 1zzizF e

)()( zFzF

3. Milne-Thompson Theorem: Examplesp p)()()( 2

1 zaFzFzF

(a) Circle in a uniform flow in x direction:

iy

(b) Circle in a source flow, source at z1 : z1

x

10

(b) Circle in a source flow, source at z1 (contd.)iy

F(z)z1

x

F1(z)1( )

+q+q-q

a

11Convection? Force on the cylinder?

Why Circular Cylinders?

MappingMapping

12

Panels

i

The Vortex Panel (or Sheet)

Consider a point vortex)(2

)(1zz

izW

Imagine spreading the vortex along a line WeImagine spreading the vortex along a line. We would then end up with a certain strength per unit length (s) that could vary with distance salong the line. Each elemental length of the line

ld b h lik i i ds dsz

ds would behave like a miniature vortex and so would produce a velocity field:

s

)()( dssizdW

z1

Where gives the coordinate of the line at s.The total flowfield produced by the line is then:

)(1 sz

))((2)(

1 szz

13

panel szzdssizW

))(()(

2)(

1

Panels

dssq )(1

The Source Panel (or Sheet)

Likewise, for a source panel

panel szzdssqzW

))(()(

21)(

1(and a doublet…)

Panels can be curved

ss

Vortex panels are more useful than others (?)

14

Since they can model flows with circulation.

Panel MethodsComplex Shapesp p

Since we can choose the strength of the panel at every point along its length (and thus indirectly the velocity here) we can satisfy a boundary condition on a continuous surface, such as an airfoil, by wrapping the airfoil in a curved y pp gpanel. With the free-stream added in this gives, e.g.

panel

i

szzdssieVzW

))(()(

2)(

1

z (s)panel

s

z1(s)Choose (s) so that the component of W(z) normal to the airfoil surface is zero. This condition can be written as:

0)(Im 1

dsdzzW

Could use panels of linearly varying

Example numerical implementation – A simple vortex panel method1. Break up the curved panel into N straight vortex panels of constant strength (can write down

velocity field of each panel algebraically). The N strengths are unknown.

linearly varying strength

15

2. Write an expression for the normal component of velocity at each panel centerpoint from the sum of all the velocities produced by the panels and the free stream. Gives N expressions.

3. Given that each expression must be equal to zero, solve the N equations for the N strengths