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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
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