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Transcript of Chapter8 Latest Combined Complejo
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058:0160 Chapter 8Professor Fred Stern Fall 2009 1
Chapter 8: Inviscid Incompressible Flow: a Useful Fantasy
8.1 Introduction
For high Re external flow about streamlined bodies viscous effects are confined toboundary layer and wake region. For regions where the B.L is thin i.e. favorable pressuregradient regions, Viscous/Inviscid interaction is weak and traditional B.L theory can beused. For regions where B.L is thick and/or the flow is separated i.e. adverse pressuregradient regions more advanced boundary layer theory must be used includingviscous/Inviscid interactions.
For internal flows at high Re viscous effects are always important except near the
entrance. Recall that vorticity is generated in regions with large shear. Therefore, outsidethe B.L and wake and if there is no upstream vorticity then =0 is a good approximation.Note that for compressible flow this is not the case in regions of large entropy gradient.Also, we are neglecting noninertial effects and other mechanisms of vorticity generation.
Potential flow theory
1) Determine from solution to Laplace equation02 =
B.C:
at BS : . 0 0V nn
= =
at S : V =
Note:F: Surface Function
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10 . 0 .
DF F FV F V n
Dt t F t
= + = =
for steady flow . 0V n =
2) Determine V from V = and p(x) from Bernoulli equation
Therefore, primarily for external flow application we now consider inviscid flow theory (0= ) and incompressible flow ( const= )
Euler equation:
. 0
. ( )
V
DVp g
Dt
VV V p z
t
=
= +
+ = +
2
.2
: 2
VV V V
Where V vorticity fluid angular velocity
=
= = =
21( )2
0 0 :
Vp V z V
t
If ie V then V
+ + + =
= = =
1( )
2p z B t
t
+ + + =
Bernoullis Equation for unsteady incompressible flow, notf(x)
Continuity equation shows that GDE for is the Laplace equation which is 2nd orderlinear PDE ie superposition principle is valid. (Linear combination of solution is also asolution)
2
1 2
2
2 2 2 2 1
1 2 1 2 2
2
0
00 ( ) 0 0
0
V
= = == +
= = + = + =
=Techniques for solving Laplace equation:
1) superposition of elementary solution (simple geometries)2) surface singularity method (integral equation)3) FD or FE4) electrical or mechanical analogs5) Conformal mapping ( for 2D flow)6) Analytical for simple geometries (separation of variable etc)
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8.2 Elementary plane-flow solutions:
Recall that for 2D we can define a stream function such that:
x
y
v
u
=
=
0)()( 2 ==
== yxyxz
yxuv
i.e. 02 =Also recall that andare orthogonal.
yx
xy
v
u
==
==
udyvdxdydxd
vdyudxdydxd
yx
yx
+=+=
+=+=
i.e.const
const
dx
dyv
u
dx
dy
=
=
==
1
8.2 Elementary plane flow solutions
Uniform stream
yx
xy
v
constUu
===
==== 0
i.e.yU
xU
=
=
Note: 022 == is satisfied.V U i = =
Say a uniform stream is at an angle tothe x-axis:
cosu Uy x
= = =
sinv Ux y
= = =
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After integration, we obtain the following expressions for the stream function andvelocity potential:
( )cos sinU y x =
( )cos sinU x y = +
2D Source or Sink:
Imagine that fluid comes out radially at origin with uniform rate in all directions.(singularity in origin where velocity is infinite)Consider a circle of radius r enclosing this source. Let v r be the radial component ofvelocity associated with this source (or sink). Then, form conservation of mass, for acylinder of radius r, and unit width perpendicular to the paper,
3
A
LQ V d A
S
=
( ) ( )2
,
2
r
r
Q r b v
Or
Qv
br
=
=
0, == vr
mvr
Where:2
Qm
b= is the convenient constant with unit velocity length
(m>0 for source and m
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. 0
1 1( ) ( ) 0r
V
rv vr r r
=
+ =
i.e.:
rv
rrvr
=
==
===
r
1velocityTangential
1velocityRadial
Such that 0V = by definition.
Therefore,
rv
rrvr
=
==
=
==
r
1
0
1
r
m
i.e.x
ymm
yxmrm
1
22
tan
lnln
==
+==
Doublets:
The doublet is defined as:
==source
2sink
1source
2sink
1 )( m m
2tan
1tan1
2tan
1tan
tan21
tan
)
m()(
+
==
1 2
2 2
sin sintan tancos cos
sin sin
2 sincos costan( )sin sin
1cos cos
;r rr a r a
r rarr a r a
r rm r a
r a r a
= = +
+ = =
+ +
For small value of a
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1
2 2 2 2 20
2 sin 2lim tan ( )a
ar amy ym
r a r x y
= = = + cos
2 ( )am Doublet Strengthr
= =
By rearranging:222
22)
2()
2(
=+++
= yxyx
y
It means that streamlines are circles with radius
2=R and center at (0, -R) i.e. circles
are tangent to the origin with center on y axis. The flow direction is from the source tothe sink.
2D vortex:
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Suppose that value of theand for the source are reversal.0
1
rv
Kv
r r r
=
= = = Purely recirculating steady motion, i.e. ( )v f r = . integration results in:
ln K=constant
K
K r
==
2D vortex is irrotational everywhere except at the origin where V and V are infinity.
Circulation
Circulation is defined by:
cC closed contour
V d s
=
=
For irrotational flow
Or by using Stokes theorem: ( if no singularity ofthe flow in A)
. 0c A A
V d s V d A ndA= = = =
Therefore, for potential flow 0= in general.However, this is not true for the point vortex due to the singular point at vortex corewhere Vand V are infinity.
If singularity exists: Free vortexr
K=
{ {
2 2
0 0 ( ) 2
2and
V d s
Kv e rd e rd K K
r
= = = =
Note: for point vortex, flow still irrotational everywhere except at origin itself whereV, i.e., for a path not including (0,0) 0 =
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Also, we can use Stokes theorem to show the existence of :
'
C
ABC AB C
V d s V d s = = Since'
. 0ABCB A
V d s =Therefore in general for irrotational motion:
.V d x =
Where: se =unit tangent vector along curve xSince se is not zero we have shown:V = i.e. velocity vector is gradient of a scalar function if the
motion is irrotational. ( 0V d s = )The point vortex singularity is important in aerodynamics, since,
extra source and sink can be used to represent airfoils and wings and we shall discussshortly. To see this, consider as an example:
an infinite row of vortices:
==
=
)2
cos2
(cosh2
1ln
2
1ln
1 a
x
a
yKrK
i
i
Where ir is radius from origin of ith vortex.
Equally speed and equal strength (Fig 8.11 of Text book)
For y a? the flow approach uniform flow with
a
K
yu
=
=
+: below x axis-: above x axisNote: this flow is just due to infinite row of vortices and there isnt any pure uniformflow
8
.
( ). 0
s
s
V d x d
d x d d xV
ds ds ds
d xe
ds
V e
=
= =
=
=
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Vortex sheet:
From afar (i.e. y a? ) looks that thin sheet occurs which is a velocity discontinuity.
Define ==a
K
2strength of vortex sheet
dxa
Kdxuudxudxud
ulul
2)( ===
i.e.dx
d= =Circulation per unit span
Note: There is no flow normal to the sheet so that vortex sheet can be used to simulate abody surface. This is beginning of airfoil theory where we let )(x= to represent bodygeometry.
Vortex theorem of Helmholtz: (important role in the study of the flow about wings)
1) The circulation around a given vortex line is constant along its length2) A vortex line cannot end in the fluid. It must form a closed path, end at a
boundary or go to infinity.3) No fluid particle can have rotation, if it did not originally rotate
Circular cylinder (without rotation):
In the previous we derived the followingequation for the doublet:
2 2
sinDoublet
y
x y r
= = = +
When this doublet is superposed over a
uniform flow parallel to the x- axis, we get:
2
sin 1sin 1 sinU r U r
r U r
= =
Where: = doublet strength which is determined from the kinematic body boundarycondition that the body surface must be a stream surface. Recall that for inviscid flow it is
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no longer possible to satisfy the no slip condition as aresult of the neglect of viscousterms in PDEs.
The inviscid flow boundary condition is:F: Surface Function
10 . 0 0DF F FV F V nDt t F t
= + = = = (for steady flow)
Therefore at r=R, V.n=0 i.e. Rrrv ==0 .
r rV v e v e = + ,
2 2
r
r
r
F Fe e
F rn eF F F
+ = = =
+
2
11 cosrV n v U
r U r
= = =
=0
2U R =
If we replace the constantU
by a new constant R2, the above equation becomes:
2
21 sin
RU r
r
=
This radial velocity is zero on all points on the circle r=R. That is, there can be novelocity normal to the circle r=R. Thus this circle itself is a streamline.
We can also compute the tangential component of velocity for flow over the circular
cylinder. From equation, 2
21 sin
Rv U
r r
= = +
On the surface of the cylinder r=R, we get the following expression for the tangential andradial components of velocity:
2 sinv U =
0=rv
How about pressure p? look at the Bernoulli's equation:
( )2 2 21 12 2r
ppv v U
+ + = +
After some rearrangement we get the following non-dimensional form:
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( )2 2
22
, 11
2
r
p
v vp pC r
UU
+= =
For the circular cylinder being studied here, at the surface, the only velocity componentthat is non-zero is the tangential component of velocity. Using 2 sinv U = , we get atthe cylinder surface the following expression for the pressure coefficient:
Cp = 1 42sin
Where is the angle measured from the rear stagnation point (at the intersection of theback end of the cylinder with the x- axis).
Cp over a Circular Cylinder
-4
-3
-2
-1
0
1
2
0 30 60 90 120 150 180 210 240 270 300 330 360
Theta, Degrees
Cp
From pressure coefficient we can calculate the fluid force on the cylinder:21( ) ( , )
2 pA A
F p p nds U C R nds = = ( )ds Rd b= b=span length
2
2 2
0
1 (1 4sin )(cos sin )2
F U bR i j d
= +
2 2
1 1
2 2
L
Lift F jC
U bR U bR
= =
= 0sin)sin41(2
0
2 =
d (due to symmetry of flow
around x axis)
2 2
1 1
2 2
F
Drag F iC
U bR U bR
= =
= 0cos)sin41(2
0
2 =
d (dlembert paradox)
We see that 0== DL CC is due to symmetry of Cp. But how realistic is this potentialflow solution. We started explain that viscous effects were confined to thin boundarylayer and it has negligible effect on the inviscid flow for high Rn flow about streamlined
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bodies. A circular cylinder is not a streamlined body (like as airfoil) but rather a bluffbody. Thus, as we shall determine now the potential flow solution is not realistic for theback portion of the circular cylinder flows )90( > .For general:
FRICTIONSKINDRAGFORMDDD
CCC +=
DRAGFORMDC : drag due to viscous modification of pressure destruction
FRICTIONSKINDC : Drag due to viscous shear stress
:
:DSK DFO
DFO DSK
Streamlined Body C C
Bluff Body C C
> >
Circular cylinder with circulation:The stream function associated with the flow over a circular cylinder, with a point vortexof strength placed at the cylinder center is:
sin
sin ln2U r rr
= From V.n=0 at r=R: 2U R =
Therefore,2 sin
sin ln2
U RU r r
r
=
The radial and tangential velocity is given by:2
2
11 cos
r
Rv U
r r
= =
2
21 sin
2Rv U
r r r
= = + + On the surface of the cylinder (r=R):
2
2
11 cos 0r
Rv U
r R
= = =
2 sin2
v Ur R
= = +
Next, consider the flow pattern as a function of . To start lets calculate the stagnationpoints on the cylinder i.e.:
2 sin 02
v UR
= + =
sin / 24 2
K
U R U R
= = =
Note:2
KK
U R
= =
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So, the location of stagnation point is function of.
2
K
U R U R
= = s
(stagnation point)
0 ( 0sin = ) 0,1801( 5.0sin = ) 30,1502 ( 1sin = ) 90>2( 1sin > ) Is not on the circle but where 0rv v= =
For flow patterns like above (except a), we should expect to have lift force in ydirection.
Summary of stream and potential function of elementary 2-D flows:In Cartesian coordinates:
yx
xy
v
u
==
==
In polar coordinates:
rv
rrvr
=
=
=
=
r
1
1
Flow Uniform Flow xU yUSource (m>0)Sink (m
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Doubletr
cos
r
sin
Vortex K - lnK r90 Corner flow )(2/1 22 yxA AxySolid-Body rotation Doesnt exist 2
2
1r
These elementary solutions can be combined in such a way that the resulting solution canbe interpreted to have physical signification;that is. Represent the potential flow solutionfor various geometers. Also, methods for arbitrary geometries combine uniform streamwith distribution of the elementary solution on the body surface.
Some combination of elementary solutions to produce body geometries of practicalimportanceBody name Elemental combination Flow Patterns
Rankine Half Body Uniform stream+source
Rankine Oral Uniform stream+source+sink
Kelvin Oral Uniform stream+vortex point
Circular Cylinderwithout circulation
Uniform stream+doublet
Circular Cylinder withcirculation
Uniformstream+doublet+vortex
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Keep in mind that this is the potential flow solution and may not well represent the realflow especially in region of adverse px.
The Kutta Joukowski lift theorem:
Since we know the tangential component of velocity at any point on the cylinder (and theradial component of velocity is zero), we can find the pressure field over the surface ofthe cylinder from Bernoullis equation:
22 2
2 2 2
r vv p Up
+ + = +
Therefore:
2 22 2 2 2
2 2
2
1 12 sin 2 sin sin
2 2 2 8
sin sin
Up p U U p U U
R R R
A B C
= + = + +
= + +
where
22
2 2
1
2 8A p U
R
= +
U
BR
=
22C U =
Calculation of Lift: Let us first consider lift. Lift per unit span, L (i.e. per unit distancenormal to the plane of the paper) is given by:
=Lower upper
pdxpdxL
On the surface of the cylinder, x = Rcos. Thus, dx = -Rsind, and the above integralsmay be thought of as integrals with respect to . For the lower surface, varies between and 2. For the upper surface, varies between and 0. Thus,
( ) ( ) +++++=
2 0
22 sinsinsinsinsinsin dCBARdCBARL
Reversing the upper and lower limits of the second integral, we get:
( ) ( ) =++=++=
2
0
322
0
2 sinsinsinsinsinsin BRdCBARdCBARL
Substituting for B ,we get:= uL
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This is an important result. It says that clockwise vortices (negative numerical values of) will produce positive lift that is proportional to and the free stream speed withdirection 90 degrees from the stream direction rotating opposite to the circulation. Kuttaand Joukowski generalized this result to lifting flow over airfoils. Equation = uL is known as the Kutta-Joukowski theorem.
Drag: We can likewise integrate drag forces. The drag per unit span, D, is given by:
=Front rear
pdypdyD
Since y=Rsin on the cylinder, dy=Rcosd. Thus, as in the case of lift, we can convertthese two integrals over y into integrals over. On the front side, varies from 3/2 to/2. On the rear side, varies between 3/2 and /2. Performing the integration, we canshow that
0=D
This result is in contrast to reality, where drag is high due to viscous separation. Thiscontrast between potential flow theory and drag is the dlembert Paradox.
The explanation of this paradox are provided by Prandtl (1904) with his boundary layertheory i.e. viscous effects are always important very close to the body where the no slipboundary condition must be satisfied and large shear stress exists which contributes thedrag.
Lift for rotating Cylinder:
We know that L U = therefore:
21 (2 )2
L
LC
U RU R
= =
Define:
== Rdvv Average
2
1
2
12
0
Note:c c c
V d s V d A V Rd = = =
2averagfeL
C vU
=
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Velocity ratio:a
U
Theoretical and experimental lift and drag of a rotating cylinder
Experiments have been performed that simulate the previous flow by rotating a circularcylinder in a uniform stream. In this case Rv = which is due to no slip boundarycondition.
- Lift is quite high but not as large as theory (due to viscous effect ie flowseparation)
- Note drag force is also fairy high
Fletttern (1924) used rotating cylinder to produce forward motion.
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8.3 Method of Images
The method of image is used to model slip wall effects by constructing appropriateimage singularity distributions.
Plane Boundaries:
2-D: ( ) ( )2 22 2ln 1 1
2
mx y x y = + + +
3-D: ( ) ( )1 1
2 22 22 2 2 21 1M x y z x y z = + + + + + +
Similar results can be obtained for dipoles and vortices:
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Spherical and Curvilinear Boundaries:
The results for plane boundaries are obtained from consideration of symmetry. Forspherical and circular boundaries, image systems can be determined from the Sphere &
Circle Theorems, respectively. For example:
Flow field Image SystemSource of strength M at c outside sphere ofradius a, c>a Sources of strength c
ma atc
a 2 and line
sink of strength am extending from center
of sphere toc
a 2
Dipole of strength at l outside sphere ofradius a, l>a dipole of strength la
3
at la2
Source of strength m at b outside circle ofradius a, b>a equal source at b
a 2 and sink of same
strength at the center of the circle
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Multiple Boundaries:
The method can be extended for multiple boundaries by using successive images.
(1) For example, the solution for a source equally spaced between two parallel planes
( )[ ] ( )[ ][ ]
( ) ( ) ( ) ( ) ( ) ( )[ ]
++++++++++++=
+++= =
azazazazazazm
anzanzmzwm
2ln4ln6ln4ln2lnln
24ln4ln)(,2,1,0
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(2) As a second example of the method of successive images for multiple boundariesconsider two spheres A and B moving along a line through their centers atvelocities U1 and U2, respectively:
Consider the kinematic BC for A:( ) ( ) 2222, azyytxtxF ++=
1 1 0 or cosR R R
DFV e U k e U
Dt = = =
where 2 cosR = ,
3
32 cos
2RUa
R = =
Similarly for B 2 cos 'R U =
This suggests the potential in the form
1 1 2 2U U = +
where 1 and 2 both satisfy the Laplace equation and the boundary condition:
1 1
'
cos , 0'R a R bR R
= =
= = (*)
2 2
'
0, cos ''R a R bR R
= =
= = (**)
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1 = potential when sphere A moves with unit velocity towards B, with B at rest2 = potential when sphere B moves with unit velocity towards A, with A at rest
If B were absent.3
01 2 2
cos cos2
a
R R
= = ,
2
3
0
a=
but this does not satisfy the second condition in (*). To satisfy this, we introduce theimage of 0 in B, which is a doublet 1 directed along BA at A1, the inverse point of Awith respect to B. This image requires an image 2 at A2, the inverse of A1 with respectto A, and so on. Thus we have an infinite series of images A 1, A2, of strengths 1 ,
2 , 3 etc. where the odd suffixes refer to points within B and the even to points withinA.
Let &n nf AA AB c= =
c
bcf
2
1 = ,1
2
2f
af = ,
2
2
3fc
bcf
= ,
=
3
3
01c
b,
=
3
1
3
12f
a,
( )
=
3
2
3
23fc
b,
where 1 = image dipole strength, 0 = dipole strength 33
sistance
radius
0 1 1 2 21 2 2 2
1 2
cos cos cos
R R R
= K with a similar development procedure for 2.
Although exact, this solution is of unwieldy form. Lets investigate the possibility of anapproximate solution which is valid for large c (i.e. large separation distance)
22 2 2 2
2
1 12 22 2
2 2
2' 2 cos 1 cos
1 1 11 2 cos 1 2 cos
'
c cR R c cr R
R R
c c R R
R R R R c c c
= + = +
= + = +
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Considering the former representation first definingc
R= and cosu =
[ ] 21
221
1
'
1 += u
RR
By the binomial theorem valid for 1 = + + +
L
L
Going back to the two sphere problem. If B were absent
3
1 2cos
2
a
R =
using the above expression for the origin at B and near B'
1R
c
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( )33 3 12 2 3
3
3
' cos1cos
2 2
cosR
a R Pa a
R c c
a
c
= = + +
= +
L
L
which can be cancelled by adding a term to the first approximation, i.e.3 3 3
1 2 3 2
1 cos 1 cos '
2 2 '
a a b
R c R
=
to confirm this3 3 3
1 2 3 3
1 3 'cos ' 1 cos
2 2
a a R a b
c c c R
=
3
3 3 3 ''1
' 3 3 '
cos ' cos( ) 0
a a bR b hot
R c c R
= = + = +
Similarly, the solution for f2 is
2
3 ' 3 3
2 3 2'
1 cos 1 cos
2 2
b a b
c RR
=
These approximate solutions are converted to ( )3c .
To find the kinetic energy of the fluid, we have
1
2A B
n nS SK dS dS
= + 2 2
11 1 12 1 2 22 2
1
2 2A B
n
S S
K A U A U U A U dS
+
= + + = 1
11 1 A
A
A dSn
=
,2
22 2 B
B
A dSn
=
,1 1
12 2 1A B
A B
A dS dSn n
= =
where 22 sindS R d =
311
3
2aA = ,
3 3
12 3
2 a bA
c
= , 322
3
2bA = ,
3 32 2
1 1 1 2 2 23
1 2 1' '4 4
a bK M U U U M Uc
= + + : using the approximate form of the potentials
where 2 21 1 2 21 1
' , '4 4M U M U : masses of liquid displaced by sphere.
8.4 Complex variable and conformal mapping
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This method provides a very powerful method for solving 2-D flow problems. Althoughthe method can be extended for arbitrary geometries, other techniques are equally useful.Thus, the greatest application is for getting simple flow geometries for which it providesclosed form analytic solution which provides basic solutions and can be used to validatenumerical methods.
Function of a complex variable
Conformal mapping relies entirely on complex mathematics. Therefore, a brief review isundertaken at this point.
A complex numberzis a sum of a real and imaginary part; z= real + i imaginary
The term i, refers to the complex number
so that;
Complex numbers can be presented in a graphical format. If the real portion of a complexnumber is taken as the abscissa, and the imaginary portion as the ordinate, a two-dimensional plane is formed.
z = real +i imaginary = x + iy
-A complex number can be written in polarform using Euler's equation;
z = x + iy = rei = r(cos + isin)
Where: r2 = x2 + y2
- Complex multiplication:z1z2 = (x1+iy1)(x2+iy2) = (x1x2 - y1y2) + i(x1y2 + y1x2)
- Conjugate: z = x + iy z x iy= 22. yxzz +=-Complex function:
w(z) = f(z)= (x,y) + i (x,y)
27
1=i
1,,1,1
432====
iiiii
y, imaginary
x, real
)(
21212121 +== iii errerer
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If function w(z) is differentiable for all values of z in a region of z plane is said to beregular and analytic in that region. Since a complex function relates two planes, a pointcan be approached along an infinite number of paths, and thus, in order to define a uniquederivative f(z) must be independent of path.
1 1 1
1 1
1
1 1 1
( )1( 0) :
: ( , ) ( , )
x x
w w i iwPP y
z z z x x
dwi
dz
Note w x x y i x x y
+ += = =
= +
= + + +
2 2 2
2 2
2
( )2( 0) :
( )
y y
w w i iwPP x
z z z i y y
dwi
dz
+ += = =
= +
Fordz
dwto be unique and independent of path:
x y y xand = = Cauchy Riemann Eq.
Recall that the velocity potential and stream function were shown to satisfy this
relationship as a result of their othogonality. Thus, complex function iw += represents 2-D flows.
xx yx yy xy = = i.e. 0=+ yyxx and similarly for . Therefore if analytic andregular also harmonic, i.e., satisfy Laplace equation.
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Application to potential flow
( )w z i = + Complex potential where : velocity potential,: stream function
( ) ix x rdw
i u iv u iu e
dz
= + = = Complex velocity
)('' += ii reer where rr =' (magnification) and +=' (rotation)Triangle about z0 is transformed into a similar triangle in the -plane which ismagnified and rotated.
Implication:
-Angles are preserved between the intersections of any two lines in the physical domainand in the mapped domain.
-The mapping is one-to-one, so that to each point in the physical domain, there is one andonly one corresponding point in the mapped domain.
For these reasons, such transformations are called conformal.
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Usually the flow-field solution in the -plane is known:
),(),()( += iWThen
( )( ) ),(),()( yxiyxzfWzw +== or == &
Conformal mapping
The real power of the use of complex variables for flow analysis is through theapplication of conformal mapping: techniques whereby a complicated geometry in thephysical z-domain is mapped onto a simple geometry in the -plane (circular cylinder) forwhich the flow-field solution is known. The flow-field solution in the z-plane is obtainedby relating the -plane solution to the z-plane through the conformal transformation=f(z) (or inverse mapping z=g()).
Before considering the application of the technique, we shall review some of the moreimportant properties and theorems associated with it.
Consider the transformation,=f(z) where f(z) is analytic at a regular point Z0 where f(z0)0= f(z0) z
'' ier= , iz re = , ( ) iezf =0
'
The streamlines and equipotential lines of the -plane (, ) become the streamlines ofequipotential lines of the z-plane (, ).
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2 2
2 2
0
0
z
z
= =
= =i.e. Laplace equation in the z-plane transforms into Laplace equation
is the -plane.
The complex velocities in each plane are also simply related
' ( )dw dw d dw
f zdz d dz d
= =
( ) ( ) '( ) ( ( ))dw dW
z u iv U iV f z f zdz d
= = = =
i.e. velocities in two planes are proportional.
Two independent theorems concerning conformal transformations are:(1) Closed curves map to closed curves(2) Rieman mapping theorem: an arbitrary closed profile can be mapped onto the unit
circle.
More theorems are given and discussed in AMF Section 43. Note that these are for theinterior problems, but we equally valid for the exterior problems through the inversionmapping.
Many transformations have been investigated and are compiled in handbooks. The AMFcontains many examples:
1) Elementary transformations:
a) linear: 0, ++= bcad
dcz
bazw
b) corner flow: nAzw =
c) Jowkowsky:
2
cw +=
d) exponential: new =e) szw = , s irational
2) Flow field for specific geometries
a) circle theoremb) flat platec) circular arcd) ellipsee) Jowkowski foilsf) ogive (two circular areas)g) Thin foil theory [solutions by mapping flat plate with thin foil BC onto unit
circle]
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h) multiple bodies
3) Schwarz-Cristoffel mapping4) Free-streamline theory
The techniques of conformal mapping are best learned through their applications. Herewe shall consider corner flow.
A simple example: Corner flow
1. In -plane, let =)(W i.e. uniform stream
2. Say
zzf == )(
3.
zzfWzw == ))(()( i.e. corner flow
Note that 1-3 are unit uniform stream.
niURURUzzw nnn sincos)( +== , where iz Re=i.e. nUR n cos= , nUR n sin=
nUR n sin= =const.=streamlines nUR n cos= =const.=equipotentials
1 1 ( 1) 1 1( cos sin )
( )
n n i n n n i
i
r
dw dW d nUz nUR e nUR n inUR n e
dz d dz
u iu e
= = = = +
=
nnURu
nnURu
n
n
r
sin
cos
1
1
=
=
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(n2
0 ru , 0
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)( 02 xxG = in V+V (i.e. entire domain) where is the Dirac delta function.
G0 on S
Solution for G (obtained Fourier Transforms) is: rG ln= , 0xxr = , i.e. elementary
2-D source at0xx =
of unit strength, and (1) becomes
== BSS
dSn
G
nG
First term in integrand represents source distribution and second term dipole distribution,which can be transformed to vortex distribution using integration by parts. By extendingthe definition of into V it can be shown that can be represented by distributions ofsources, dipoles or vortices, i.e.
=
= BSS
GdS
: source distribution, : source strengthor
=
= BSS
dSn
G : dipole distribution, : dipole strength
Also, it can be shown that a source distribution representation can only be used torepresent the flow for a non-lifting body; that is, for lifting flow dipole or vortexdistributions must be used.
As this stage, lets consider the solution of the flow about a non-lifting body of arbitrary
geometry fixed in a uniform stream. Note that since G0 on S already satisfy thecondition S. The remaining condition, i.e. the condition is a stream surface is used todetermine the source distribution strength.
Consider a source distribution method for representing non-lifting flow around a body ofarbitrary geometry.
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V U = + : total velocity
U : uniform stream, : perturbation potential due to presence of body
)sin(cos jiUU +=
: note that for non-lifting flow must be zero (i.e. for asymmetric foil 0= or for cambered filed oLift = )
ln2
BS
Krds
= : source distribution on body surface
Now, K is determined from the body boundary condition.
0=nV i.e. 0U n n + = or U nn
=
i.e. normal velocity induced by sources must cancel uniform stream
nUrdsK
n BS=
ln
2
This singular integral equation for K is solved by descretizing the surface into a numberof panels over which K is assumed constant, i.e. we write
no. of panels
1
ln2
M
ij i iSi
ji
Kjr dS U n
n
=
=
=
, i=1,M, j=1,M
where ( ) ( ) 22 jijiij zzxxr += =distance from ith panel control point to jr =position
vector along jth panel.
Note that the integral equation is singular since
i
ij
ij
ij
i n
r
rr
n
=
1
ln
at for 0=ijr this integral blows up; that is, when i=j and we trying to determine thecontribution of the panel to its own source strength. Special care must be taken. It can beshown that the limit does exist at the integral equation can be written
2ln
2i
SSiij
i
i kdSrn
k
ij
=
= =
+
M
iji S
ijij
ij
ij
ji
i
nUdSn
r
r
KK
1
1
22
where ( ) ( )[ ]zijixijiij
zi
i
ij
xi
i
ij
ij
iiji
iji
ij
ij
nzznxxr
nz
rn
x
r
rnr
rn
r
r+=
+
==
2
1111
where ( ) ( )222 jijiij zzxxr +=
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Consider the ith panel
jiS iiisincos += , jijSn iiii cossin +==
ixin sin= , izin cos=
nUrdsK
nBS
=
ln2
ijj SSrr += 0
,:0jr
origin of j
th
panel coordinate system,i
SS
: distance along j
th
panel
V U = +
ln2
BS
Krds
=
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0=nV nUrdsK
nBS
=
ln2( )
=
==
+M
iji S
iij
ij
ij
ij
ji
i
UnUdSn
r
r
KK
1
sin1
22
, jin iii cossin +=
Let jS
j
ij
ij
ij
IdSn
r
ri
=1 and ( )ii URHS = sin , then
i
M
iji
j
ji RHSIKK
=+ =1 22 , ( ) ( ) jS jiji
zijixiji
j dSzzxx
nzznxxI
j
++
=22
( ) ( )( ) ( ) ( )
( ) ( ) ( ) ( )[ ]j
l
jj
j
j
l
jjizjjizijjiji
xizjzixjjzijixiji
j
l
xjjjizjjji
zixjjjixizjjji
i
dSBASS
DCS
dS
SzznxxnSzzxx
nnnnSnzznxx
dSnSzznSxx
nnSzznnSxxI
i
i
i
++
+=
++++
++=
+
+=
0
2
0 2
00
2
0
2
0
00
0 2
0
2
0
00
2
2
where
( ) ( )( )
( ) ( )ijiiji
ji
jiji
jijjij
zzxxD
C
zzxxB
zzxxA
cossin
sin
sincos
00
2
0
2
0
00
+=
=
+=
=
2100
1iij
li
j
l
j CIDIdSS
CdSDIii +=
+
= where BASS jj ++= 22
1jI depends on if q0 where 244 ABq =
12 ln2
1ij AII =
( ) ( )( ) ( ) ii
M
iji
j
S jiji
zijixijiji RHSnUdSzzxx
nzznxxKK
j
==+
++
= 1
2222
[ ] ( )iiii UURHS =+= sincossinsincos
i
M
iji
i
ji RHSIKK
=+ =1 22 : Matrix equation forKi and can be solved using Standard methods such
as Gauss-Siedel Iteration.
In order to evaluate Ij, we make the substitution
xjjjj
xjjjj
SSzz
SSxx
+=
+=
0
0
jjjj SSrr += 0
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where Sj= distance along the jth panel ij lS 0
jxizj
jzixj
nS
nS
sin
cos
==
==
After substitution, Ij becomes
( ) ( )
( ) ( ) ( )( )[ ]( ) ( ) ( ) ( ) 2sin1cos1
sincos2sincos
22
0
22
0
00
2
0
22
0
2
2
0
2
0
2
+=
++
+=
jjijji
jijijjjijjij
jiji
zzxx
zzxxzzxx
zzxxAB
( ) ( ) ( )[ ]2002 cossin44 ijiiji zzxxABq ==i.e. q>0 and as a result,
E
AS
Eq
AS
qI iij
+=
+= 111 tan
122tan
2where EABq 22 2 ==
( )
( )
++
+
+
=
+=
+=
B
BAlzlC
E
A
E
Al
E
CAD
CICADAICDII
jji
l
jjjji
2
ln2tantan
ln2
ln2
1
11
0111
where BASS ji ++= 22
Therefore, we can write the integral equation in the form
( )iM
iji
j
ji UIKK
=+ = sin
22 1
2
1
2
22
1
j
j
j
j
Ik
Ik
iK =
iRHS
which can be solved by standard techniques for linear systems of equations with Gauss-Siedel Iteration.
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Once Ki is known,V U = + And p is obtained from Bernoulli equation, i.e.
21
=
U
VVp
Mentioned potential flow solution only depend on is independent of flow condition, i.e.U, i.e. only is scaledOn the surface of the body Vn=0 so that
2
2
1
=U
VC Sp where SVVS = =tangential surface velocity
SSUSVS +==
( ) SiiS QUS
SUVi
+=
+=
cos
where ( ) ( )jijiUSU iii sincossincos ++=
ln2B
S
S
K rdsS S = =
( )1
ln2i
j
Mj
S ji j
j ilj i
Kr dS
S
=
=
: j=i term is zero since source panel induces no tangential
flow on itself. ( ( ) jl
jji
i
JdSrS
j
=
ln )
( )
( ) ( ){ }iiii zjixji
ijz
i
ij
xi
ij
ij
iiji
ij
ij
ij
ij
i
SzzSxxrSz
r
Sx
r
r
SrrS
r
rr
S
+=
+
=
=
=
2
11
11ln
where ixiS cos= , iziS sin=
2
1
=
U
VC i
i
S
p ,( )
j
jij
i
iJ
KUV
=
+=1 2
cos
( ) ( )
( ) ( )( ) ( ) j
l
zjjjixjjji
zizxjjjixixjjji
j
S jiji
zijixiji
j
l
ij
ij
ij
dSSSzzSSxx
SSSzzSSSxx
dSzzxx
nzznxxdS
S
r
rJ
i
ji
+
+=
+
+=
=
0 2
0
2
0
00
22
1
where ixiS cos= , iziS sin= , ( ) ( ) BASSSSzzSSxx jjzjjjixjjji ++=+ 222
0
2
0
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( )
+
=
++i
i
jj
j
ijijzizjxixjjiiiiji
dSCASS
DCS
SSSSSzzxx
0
2
0 sinsincoscossincos
( (
( )
+=+==
+=
1121
00
ln2
1cos
sincos
jjjjji
ijiiji
AICDICIDIC
zzxxD
( )B
BAllC
E
A
E
Al
E
ACDCIACDJ
jjj
jj
+++
+
=+= 2
ln2
tantanln2
2
11
1
( ) ( )[ ] ( )( )( ) ( )[ ]( )( )[ ]
( )[ ]( ) ( )[ ]
( ) ( )[ ]( ) [ ] ( ) [ ]jijijjijijijji
jjijiji
jjijiji
jjijiiji
iijjiiji
jijiijiiji
ijijiiji
ijiiji
zzxx
zz
xx
zz
xx
zzxx
zzxx
zzxxACD
sincoscossincoscossinsincossin
sincoscossin1sin
cossinsincos1cos
sincoscossinsinsin
coscoscossinsincos
coscossinsinsincos
cossincos
sincos
00
2
0
2
0
20
2
0
00
00
00
+=
=
=
+
+=
where ( )jiE
ACD = sin
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