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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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    058:0160 Chapter 8Professor Fred Stern Fall 2009 2

    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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    058:0160 Chapter 8Professor Fred Stern Fall 2009 3

    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

    = = =

    3

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    058:0160 Chapter 8Professor Fred Stern Fall 2009 4

    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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    058:0160 Chapter 8Professor Fred Stern Fall 2009 5

    . 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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    058:0160 Chapter 8Professor Fred Stern Fall 2009 6

    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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    058:0160 Chapter 8Professor Fred Stern Fall 2009 8

    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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    058:0160 Chapter 8Professor Fred Stern Fall 2009 9

    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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    058:0160 Chapter 8Professor Fred Stern Fall 2009 11

    ( )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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    058:0160 Chapter 8Professor Fred Stern Fall 2009 12

    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

    = + = +

    = + = +

    24

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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

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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

    40

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