058:0160 Professor Fred Stern Fall 2009 Chapter 8 1 Chapter 8: Inviscid Incompressible flow: a useful fantasy Introduction For high Rn external flow about streamlined bodies viscous effects are confined to boundary layer and wake region. For regions where the B.L is thin i.e. forward pressure gradient regions, Viscous/Inviscid interaction is weak and traditional B.L theory can be used. For regions where B.L is thick and/or the flow is separated i.e. adverse pressure gradient regions more advanced boundary layer theory must be used including viscous/Inviscid interactions. For internal flows at high Rn viscous effects are always important except near the entrance. Recall that vorticity is generated in regions with large shear. Therefore, outside the B.L and wake and if there is no upstream vorticity then w=0 is a good approximation. Note that for compressible flow this is not the case in regions of large entropy gradient. Now, we are neglecting noninertial effects and the mechanism of vorticity generation. Potential flow theory 1) Determine from solution to Laplace equation 2 0 B.C: 0 n at S : V 0 at S B : V .n 0 Note: F: Surface Function 1 058:0160 Professor Fred Stern Fall 2009 Chapter 8 2 DF F 1 F 0 V .F 0 V .n Dt t F t For steady flow: V .n 0 2) Determine V from V and p(x) from Bernoulli equation Therefore, primary for external flow application we now consider inviscid flow theory ( 0 ) and incompressible flow ( const ) Euler equation: .V 0 DV p g Dt V V .V ( p z ) t V2 V .V V 2 Where : w V vorticity V 1 ( p V 2 z ) V t 2 If 0 ie V 0 then V : 1 p z B(t ) : Bernoulli’s Equation for unsteady incompressible flow t 2 Continuity equation shows that PDE for is the Laplace equation which is 2nd order linear PDE ie superposition principle is valid. (Linear combination of solution is also a solution) .V . 2 0 1 2 21 0 2 2 2 2 1 2 0 (1 2 ) 0 0 2 2 0 Techniques for solving Laplace equation: 1) superposition of elementary solution 2) surface singularity method (integral equation) 3) Fr or FE 4) electrical or mechanical analogs 5) Conformal mapping ( for 2D flow) 6) Analytic for simple geometry ( separation of variable etc) 2 058:0160 Professor Fred Stern Fall 2009 Chapter 8 3 Elementary plane-flow solutions: Recall that for 2D we can define a stream function such that: u y v x z vx u y ( x ) ( y ) 2 0 x y i.e. 2 0 Also recall that and are orthogonal. u y x v x y d x dx y dy udx vdy d x dx y dy vdx udy i.e. dy u 1 dx const v dy dx const Elementary plane flow solutions Uniform stream u U y x const v 0 x y Ux U y Note: 2 2 0 is satisfied. V U iˆ i.e. For a uniform stream moving at an angle, , relative to the x-axis, we can write u Ucos y x 3 058:0160 Professor Fred Stern v Usin Chapter 8 4 Fall 2009 x y After integration, we obtain the following expressions for the stream function and velocity potential: U y cos xsin U x cos ysin 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 vr be the radial component of velocity associated with this source (or sink). Then, form conservation of mass, for a cylinder of radius r, and unit height perpendicular to the paper, Q 2r 1 v r Or , vr Q 2r m , v 0 r Q Where: m m= convenient constant (m>0 for source and m<0 for sink) 2 vr In a polar coordinate system, for 2-D flows we will use: 1 V r r And: .V 0 1 1 (rv r ) (v ) 0 r r r 4 058:0160 Professor Fred Stern Chapter 8 5 Fall 2009 i.e.: 1 r r 1 v Tangential velocity r r Such that V . 0 by definition. v r Radial velocity Therefore, m 1 r r r 1 v 0 r r vr m ln r m ln x 2 y 2 i.e. m m tan 1 y x Doublets: The doublet is defined as: ψ m( 1 2 ) 1 2 sink source sink source m tan θ tan θ 1 2 tan (θ θ ) tan ( ) 1 2 m 1 tan θ tan θ 1 2 r sin r sin tan 1 ; tan 2 r cos a r cos a r sin r sin 2ar sin tan( ) r cos a r cos a 2 r sin r sin m r a2 1 r cos a r cos a 2 ar sin m tan 1 ( 2 ) r a2 5 058:0160 Professor Fred Stern Chapter 8 6 Fall 2009 For small value of a : ψ 2amr sin θ 2amy 2 2 2 r a x y2 a2 ma constant (m ; a 0 ) ψ 2am ( Doublet Strength) sin θ r y x y2 2 cos r By rearranging: y 2 ψ 2 x2 (y ) ( )2 2 2 2 x y It means that streamlines are circles with radius R and center at (0, -R) i.e. circles 2 are tangent to the origin with center on y axis. 6 058:0160 Professor Fred Stern Chapter 8 7 Fall 2009 2D vortex: Suppose that value of the and for the source are reversal. Kθ ψ Kθ r 0 K constant 1 K r r r 2d vortex is irrotational anywhere except at the origin where v and V are infinity. Circulation Circulation is defined by: For irrotational flow Γ V d s c C closed contour Or by using Stokes theorem: ( if no singularity of the flow in A) Γ V d s V dA .ndA 0 c A A Therefore, for potential flow 0 in general. However, this is not true for the point vortex due to the singular point at vortex core where v and V are infinity. If singularity exists: Free vortex K r K (rd ) 2K and K r 2 Note: for point vortex, flow still irrotational everywhere except at origin itself where v=infinity. 2 0 7 058:0160 Professor Fred Stern Fall 2009 Chapter 8 8 Now, we can use Stokes theorem to show the existence of : V .ds V .ds ABC ' AB C C Since V .ds 0 ABCB' A Therefore in general for irrotational motion: V .dx V .dx d dx d dx . ds ds ds dx es ds (V ).e s 0 V Where: e s =unit tangent vector along curve x Since e s is not zero we have shown: V i.e. velocity vector is gradient of a scalar function if the motion is irrotational. ( V .ds 0 ) The point vortex singularity is important in aerodynamics,since, extra source and sink can be used to represent airfoils and wings and we shall discuss shortly. To see this, consider as an example: an infinite row of vortices: 1 2y 2x 1 K ln ri K ln (cosh cos ) 2 a a 2 i 1 Where ri is radius from origin of ith vortex. Equally speed and equal strength (Fig 8.7 of Text book) 8 058:0160 Professor Fred Stern Fall 2009 Chapter 8 9 For y a the flow approach uniform flow with K y a +: below x axis -: above x axis Note: this flow is just due to infinite row of vortices and there isn’t any pure uniform flow u Vortex sheet: From afar looks that thin sheet occurs which is a velocity discontinuity. 2K strength of vortex sheet a 2K d u l dx u u dx (u l u u )dx dx a d i.e. =Circulation per unit span dx Note: There is no flow normal to the sheet so that vortex sheet can be used to simulate a body surface. This is beginning of airfoil theory where we let (x ) to represent body geometry. Define 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 the length 2) 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 following equation for the doublet: Doublet sin r 9 058:0160 Professor Fred Stern Chapter 8 10 Fall 2009 When this doublet is superposed over a uniform flow parallel to the x- axis, we get: 1 r sin u 1 2 r u r Where: doublet strength which is determined from the kinematic body boundary condition that the body surface must be a stream surface. Recall that for inviscid flow it is no longer possible to satisfy the no slip condition as aresult of the neglect of viscous terms in PDEs. sin u r sin The inviscid flow boundary condition is: F: Surface Function DF F 1 F 0 V .F 0 V .n 0 (for steady flow) Dt t F t Therefore at r=R, V.n=0 i.e. vr 0 r R . vr 1 cos =0 u 1 2 r u R u R 2 If we replace the constant u by a new constant R2, the above equation becomes: R2 r sin 2 r This radial velocity is zero on all points on the circle r=R. That is, there can be no velocity normal to the circle r=R. Thus this circle itself is a streamline. u 1 We can also compute the tangential component of velocity for flow over the circular cylinder. From equation, R2 v u 1 2 sin r r On the surface of the cylinder r=R, we get the following expression for the tangential and radial components of velocity: v 2u sin vr 0 How about pressure p? look at the Bernoulli's equation: p p 1 2 1 2 v r v u 2 2 2 10 058:0160 Professor Fred Stern Chapter 8 11 Fall 2009 After some rearrangement we get the following non-dimensional form: 2 2 v v p p Cp 1 r 2 1 2 u u 2 For the circular cylinder being studied here, at the surface, the only velocity component that is non-zero is the tangential component of velocity. Using v 2u sin , we get at the cylinder surface the following expression for the pressure coefficient: C p 1 4 sin 2 Where is the angle measured from the rear stagnation point (at the intersection of the back end of the cylinder with the x- axis). Cp Cp over a Circular Cylinder 1.5 1 0.5 0 -0.5 0 -1 -1.5 -2 -2.5 -3 -3.5 30 60 90 120 150 180 210 240 270 300 330 360 Theta, Degrees From pressure coefficient we can calculate the fluid force on the cylinder: 1 F ( p p ).nds u 2 C p ( R, ).nds 2 A A ds ( 2Rd )b b=span length 2 1 F u 2 b(2R) (1 4 sin 2 )(cos iˆ sin ˆj )d 2 0 CL Lift 1 2 u b(2R) 2 2 = (1 4 sin 2 ) sin d 0 (due to symmetry of flow around x 0 axis) CF Drag 2 = (1 4 sin 2 ) cos d 0 (dÁlembert paradox) 1 2 0 u b(2R) 2 We see that C L C D 0 is due to symmetry of Cp. But how realistic is this potential flow solution. We started explain that viscous effects were confined to thin boundary layer and it has negligible effect on the inviscid flow for high Rn flow about streamlined 11 058:0160 Professor Fred Stern Fall 2009 Chapter 8 12 bodies. A circular cylinder is not a streamlined body (like as airfoil) but rather a bluff body. Thus, as we shall determine now the potential flow solution is not realistic for the back portion of the circular cylinder flows ( 90) . For general: CD CD C DSKIN FRICTION FORM DRAG CD : drag due to viscous modification of pressure destruction FORM DRAG CDSKIN FRICTION : Drag due to viscous shear stress Streamlined Body : CDSK CDFO Bluff Body :CDFO CDSK Circular cylinder with circulation: The stream function associated with the flow over a circular cylinder, with a point vortex of strength placed at the cylinder center is: sin u r sin ln r r 2 From V.n=0 at r=R: u R 2 Therefore, u r sin u R 2 sin ln r r 2 The radial and tangential velocity is given by: R2 1 vr u 1 2 cos r r R2 u 1 2 sin r 2r r On the surface of the cylinder (r=R): R2 1 vr u 1 2 cos 0 r R v 2u sin r 2R v Next, consider the flow pattern as a function of . To start lets calculate the stagnation points on the cylinder i.e. : v 2u sin 0 2R K sin /2 4u R 2u R K Note: K 2 u R 12 058:0160 Professor Fred Stern Chapter 8 13 Fall 2009 So, the location of stagnation point is function of . s (stagnation point) K u R 0 ( sin 0 ) 1( sin 0.5 ) 2 ( sin 1 ) >2( sin 1 ) 0,180 30,150 90 Is not on the circle For flow patterns like above (except a), we should expect to have lift force in –y direction. Summary of stream and potential function of elementary 2-D flows: In Cartesian coordinates: u y x v x y In polar coordinates: 1 vr r r 1 v r r Flow Uniform Flow Source,Sink U x K ln r 2 U y K 2 13 058:0160 Professor Fred Stern Dipole Vortex 90 Corner flow Solid-Body rotation Chapter 8 14 Fall 2009 cos r 2 1 / 2 A( x 2 y 2 ) Doesn’t exist sin r ln r 2 Axy 1 2 r 2 These elementary solutions can be combined in such a way that the resulting solution can be interpreted to have physical signification;that is. Represent the potential flow solution for various geometers. Also, methods for arbitrary geometries combine uniform stream with distribution of the elementary solution on the body surface. Some combination of elementary solutions to produce body geometries of practical importance Body name Elemental combination Flow Patterns Rankine Half Body Uniform stream+source Rankine Oral Uniform stream+source+sink Kelvin Oral Uniform stream+vortex point Circular Cylinder Uniform stream+dipole Circular Cylinder with circulation Uniform stream+dipole+vortex 14 058:0160 Professor Fred Stern Chapter 8 15 Fall 2009 Keep in mind that this is the potential flow solution and may not well represent the real flow 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 the radial component of velocity is zero), we can find the pressure field over the surface of the cylinder from Bernoulli’s equation: p vr2 v2 p u2 2 2 2 Therefore: 2 1 2 1 2 2 p p u 2u sin p u 2 2R 2 2 2 R 2 u 2 u 2 sin 2 sin R A B sin C sin 2 where, 1 2 2 A p u 2 2 2 2 R u B R C 2 u 2 Calculation of Lift: Let us first consider lift. Lift per unit span, L (i.e. per unit distance normal to the plane of the paper) is given by: L pdx pdx Lower upper On the surface of the cylinder, x = Rcos. Thus, dx = -Rsind, and the above integrals may 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 L R A B sin C sin sin d R A B sin C sin 2 sin d 2 Reversing the upper and lower limits of the second integral, we get: 2 2 L R A B sin C sin 2 sin d R A sin B sin 2 C sin 3 d BR 0 0 Substituting for B ,we get: L u This is an important result. It says that clockwise vortices (negative numerical values of ) will produce positive lift that is proportional to and the freestream speed. Kutta and Joukowski generalized this result to lifting flow over airfoils. Equation L u is known as the Kutta-Joukowski theorem. 15 058:0160 Professor Fred Stern Chapter 8 16 Fall 2009 Drag: We can likewise integrate drag forces. The drag per unit span, D, is given by: D pdy pdy Front rear Since y=Rsin on the cylinder, dy=Rcosd. Thus, as in the case of lift, we can convert these 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 can show that D0 This result is in contrast to reality, where drag is high due to viscous separation. This contrast between potential flow theory and drag is the dÁlembert Paradox. The explanation of this paradox are provided by Pranttl (1904) with his boundary layer theory i.e. viscous effects are always important very close to the body where the no slip boundary condition must be satisfied and large shear stress exists which contributes the drag. Lift for rotating Cylinder: We know that L u therefore: L CL 1 2 u (2R) u R 2 2 1 1 Define: v Average v d 2 0 2 R CL Note: Γ V d s V dA V Rd c A A 2 v u a vera g fe 16 058:0160 Professor Fred Stern Fall 2009 Chapter 8 17 Theoretical and experimental lift and drag of a rotating cylinder Experiments have been performed that simulate the previous flow by rotating a circular cylinder in a uniform stream. In this case v R which is due to no slip boundary condition. - Lift is quite high but not as large as theory (due to viscous effect ie flow separation) - Note drag force is also fairy high Fletttern (1924) used rotating cylinder to producee forward motion. 17 058:0160 Professor Fred Stern Fall 2009 Chapter 8 18 18 058:0160 Professor Fred Stern Fall 2009 Chapter 8 19 19 058:0160 Professor Fred Stern Chapter 8 20 Fall 2009 Complex variable and conformal mapping This method provides a very powerful method for solving 2-D flow problems. Although the 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 provides closed form analytic solution which provides basic solutions and can be used to validate numerical methods. 1. Function of a complex variable Conformal mapping relies entirely on complex mathematics. Therefore, a brief review is undertaken at this point. A complex number z is a sum of a real and imaginary part; z = real + iimaginary i 1 The term i, refers to the complex number i 1, i 1, i i, i 1 so that; Complex numbers can be presented in a graphical format. If the real portion of a complex number is taken as the abscissa, and the imaginary portion as the ordinate, a twodimensional plane is formed. 2 3 z = real + iimaginary = x + iy 4 y, imaginary x, real - A complex number can be written in polar form 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) r1e i1 r2 e i 2 r1r2 e i (1 2 ) - Conjugate: z = x + iy z x i. y z.z x 2 y 2 -Complex function: w(z) = f(z)= (x,y) + i (x,y) 20 058:0160 Professor Fred Stern Fall 2009 Chapter 8 21 If function which is differentiable for all values of z in a region of z plane is said to be regular and analytic in that region. Since a complex function relates two planes, a point can be approached always an infinite number of paths, and thus, in order to define a unique derivative f(z) must independent of path. pp1(y 0) : w w1 w 1 i 1 ( i ) z z1 z x1 x dw x i x dz 1 Note : w1 1 ( x x, y ) i 1 ( x x, y ) w w2 w 2 i 2 ( i ) pp 2(x 0) : z z2 z i ( x 2 x) dw i y y dz 2 dw to be unique and independent of path: dz x y y x Cauchy Riemann Eq. For Recall that the velocity potential and stream function were shown to satisfy this relationship as a result of their definitions. xx yx yy xy i.e. xx yy 0 Thus, complex function w i represents 2-D flows. 21 058:0160 Professor Fred Stern Fall 2009 Chapter 8 22 3. Application to potential flow W ( z ) i Complex potential where : velocity potential, : stream function dW z i z u iv u r iu e i Complex velocity dz r ' e i ' re i ( ) where r ' r (magnification) and ' (rotation) xxx about z0 is transformed into a similar xxx in the ζ-plane which is magnified and rotational. Implication: -Angles are presented between the intersections of any two lines in the physical domain xxx in the mapped domain. -The mapping is one-to-one, so that to each point in the physical domain, there is one and only one corresponding point in the mapped domain. For these reasons, such transformations are xxxx As mentioned above, usually the flow-field solution in the ζ-plane is known: W ( ) ( , ) i ( , ) Then 22 058:0160 Professor Fred Stern Fall 2009 Chapter 8 23 w( z) W f z ( x, y) i ( x, y) or & 4. Conformal mapping The real power of the use of complex variables for flow analysis is throughout the application of conformal mapping: techniques whereby a complicated geometry in the physical z-domain is mapped onto a simple geometry in the ζ-plane (circular cylinder) for which the flow-field solution is known. The flow-field solution in the z-plane is obtained by xxx. The ζ-plane solution to the z-plane through the conformal transformation ζ=f(z) (or inverse mapping z=g(ζ)). Before xxx the application of the technique, we shall review some of the more important properties and theorems associated with it. Consider the transformation, ζ=f(z) where f(z) is analytic at a regular point z0 where f’(z0)≠0 δζ= f’(z0) δz r ' e i ' , t re i , f ' z 0 e i The streamlines and equipotential lines of the ζ-plane (Φ, Ψ) become the streamlines of equipotential lines of the z-plane (φ, ψ). 2z 2 0 2z 2 0 i.e. Laplace equation ins the z-plane transforms into Laplace equation is the ζ-plane. The complex velocities in each plane are also simply xxx 23 058:0160 Professor Fred Stern Chapter 8 24 Fall 2009 dw dw d dz d dz dw ( z ) u iv (U iV ) f ' ( z ) dz i.e. velocities in two planes are xxx. Two independent theorems xxx conformal transformations are: (1) Closed curves map to closed curves (2) Xxxx mapping theorem: an arbitrary xxx xxx can be mapped onto the unit circle. More theorems are given and discussed in AMF Section 43. Note that there are for the interior problems, but we equally xxx for the exterior problems through the inversion mapping. Many transformations have been investigated and are xxx in xxx. The xxx (particularly AMF) contains many examples: 1) Elementary transformations: az b , ad bc 0 a) linear: w cz d b) xxx flow: w Az n 2 c) Jowkowsky: w c d) exponential: w e e) w z s , s xxx n 2) Flow field for specific geometries a) circle theorem b) flat plate c) circular area d) ellipsoid e) Jowkowski foils f) xxx (two circular areas) g) Thin foil theory [ solutions by mapping flat plate with thin foil b.c. onto xxx circle (xxx)) h) multiple xxx 3) Schwarx-Cristoffel mapping 4) xxx-streamline theory The techniques of conformal mapping are best xxx through their applications. Hence we shall consider two transformations (xxx flow or Joukowski) and briefly the SchwarzCristoffel mapping of free-streamline theory. For further details, consult the course book. 24 058:0160 Professor Fred Stern Fall 2009 Chapter 8 25 A simple example: Corner flow 1. In ζ-plane, let W ( ) i.e. uniform stream 2. Say f ( z ) z 3. w( z ) W ( f ( z )) z i.e. corner flow Note that 1-3 are unit uniform stream w( z ) Uz n UR n cos iUR n sin n , where z Re i i.e. UR n cos n , UR n sin n UR n sin n =const.=streamlines UR n cos n =const.=equipotentials dw dw d nUz n 1 nUR n 1e i ( n 1) (nUR n 1 cos n inUR n 1 sin n )e i dz d dz (u r iu )e i u r nUR n1 cos n u nUR n1 sin n 2n u 0 , u 0 2n n u 0 , u 0 0 r r i.e. w( z ) Uz n represents corner flow: n=1uniform stream, n=2xxx corner 25 058:0160 Professor Fred Stern Fall 2009 Chapter 8 26 Introduction to Surface Singularity methods (also known as Boundary Integral and Panes Methods) Next, we consider the solution of the potential flow problem for an xxx geometry. Consider the B.V.P for a body of arbitrary geometry fixed in a uniform stream of an xxx, incompressible, and irrotational fluid. The surface singularity method is formulated on the xxx form of Greens theorem and what is known as Greens function. G G dV 2 2 V G G dS (1) n n S S S S b S where Φ and G are any two xxx xxx in V (control volume) of xxx xxx application. Φ= velocity potential G= Green’s function Say, 2 G ( x x0 ) in V+V’ (i.e. entire domain) where δ is the xxx xxx function. G0 on S∞ (xxx xxx Former Transforms) Solution for G is: G ln r , r x x0 , i.e. elementary 2-D source at x x0 of unit strength, and (1) becomes G G dS n n S SB By externally the definition of Φ into V’ xxx can be shown that Φ can be xxx represented by distributions of sources, defined as 26 058:0160 Professor Fred Stern GdS : source distribution, : source strength S SB or Fall 2009 Chapter 8 27 S SB G dS : dipole distribution, : dipole strength n Also, it can be shown that a source distribution representation can only be used to represent the flow for a xxx body; it is , for xxx xxx dipole or vortex distributions must be used. As this stage, let’s consider the solution of the flow about a xxx body of xxx gemometry fixed in a uniform stream. Note that since G0 on S∞ Φ xxx solution the condition S∞. The xxx condition is the condition that the xxx xxx is a stream xxx need to determine the k source distribution. Strength is the xxx notation. 2 First, consider a source distribution meeting for representing xxx flow around a body of arbitrary geometry. V U Q : total velocity U : uniform stream, Q : perturbation potential due to xxx of body U U (cos iˆ sin ˆj ) : note that for xxx flow must be given (i.e. for a symmetric field 0 or for xxx filed o ) K Q ln rds : source distribution on body surface 2 SB Now, K is determined from the body boundary condition. Q U n V n 0 i.e. U n Q n 0 or n 27 058:0160 Professor Fred Stern Chapter 8 28 Fall 2009 i.e. normal velocity xxx by xxx must cancel xxx K ln rds U n n SB 2 This singular integral equation for K is xxx by xxx the into a number of panelsover which K is assumed constant, i.e. we write ni M no. of panels j 1 Ki 2 x where rij Si ln rij dS i U ni , i=1,M, j=1,M x j z i z j =distance from ith panel control xxx to ri position 2 i 2 vector along jth panel. Note that the integral equation is singular since 1 rij ln rij ni rij ni at for rij 0 then integration blows up j that is when j=i at we trying to determine the contribution of the panel to its own source strength. Speed xxx must be taken that can be shown that the xxx does exist at the integral xxx can be written ki 2 ni S j Si Ki M K j 2 i 1 2 j i where ln rij dS i ki 2 1 rij r nij dS j U ni Si ij rij 1 rij 1 1 rij 1 i rij ni n xi n zi 2 xi x j n xi z i z j n zi rij ni rij rij xi z i rij where rij2 xi x j z i z j Consider the ith panel 2 2 S i cos i iˆ sin i ˆj , ni S i ˆj sin i iˆ cos i ˆj n xi sin i , n zi cos i K ln rds U n n SB 2 28 058:0160 Professor Fred Stern Fall 2009 Chapter 8 29 r j r j 0 S S i , r j 0 : origin of jth panel xxx xxx, S Si : distance along jth panel V U Q K Q ln rds 2 SB V n 0 Ki M K j 2 i 1 2 j i Let K ln rds U n n SB 2 1 rij S rij nij dS j U ni U sin i , ni sin i iˆ cos i ˆj i 1 rij r nij dS j I j and RHS i U sin i , then Si ij xi x j n xi zi z j n zi Ki M K j dS j I j RHS i , I j 2 2 2 i 1 2 S j xi x j z i z j j i 29 058:0160 Professor Fred Stern Chapter 8 30 Fall 2009 x x S n n z z S n n dS x x S n z z S n x x n z z n S n n n n x x z z 2S n x x n z z S Ii li i j0 j zj xi 0 i j0 li i i 0 j j0 j xj j0 zi j 2 zj i xi j0 i 2 0 j j0 zi xj j xj zi zj xi 2 j0 i CS j D li i 2 S 2 AS j B 2 j j0 j zi i j0 zj i j0 2 j dS j dS j where A cos j xi x j 0 sin j z i z j 0 B xi x j 0 z i z j 0 2 2 C sin i j D xi x j 0 sin i z i z j 0 cos i li S 1 dS j C i dS j DI i1 CI i 2 where S 2j 2 AS j B 0 0 I j1 depends on if q<0 or >0 where q 4B 4 A2 I j D I j2 li 1 ln AI i1 2 Ki M K j 2 i 1 2 j i x x n z x x z i j xi i 2 Sj i j z j n zi zj 2 i dS j U ni RHS i RHS i U cos sin i sin cos i U sin i Ki M K j I i RHS i : Matrix equation for Ki xxx can be solved using Standard 2 i 1 2 j i methods such as Gauss-Sidel Iteration. In order to evaluate Ij, we make the substitution x j x j 0 S j S xj rj rj0 S j S j z j z j 0 S j S xj where Sj= distance along the jth panel 0 S j l i 30 058:0160 Professor Fred Stern Chapter 8 31 Fall 2009 S xj n zi cos j S zj n xi sin j After substitution, Ij becomes B A 2 xi x j 0 z i z j 0 2 2 cos 2 j xi x j 0 sin 2 j z i z j 0 2 cos j sin j xi x j 0 z i z j 0 2 2 xi x j 0 1 cos 2 j z i z j 0 1 sin 2 j 2 2 2 q 4 B A 2 4 xi x j 0 sin i z i z j 0 cos i i.e. q>0 and as a result, 2S 2 A 1 S A 2 I j1 tan 1 i tan 1 i where q 2 B A 2 2 E E E q q 2 C 1 I j DI j1 C ln AI j1 D CAI j1 ln l0i 2 2 D CA tan 1 li A tan 1 A C ln l j z 2 Al j B E E E 2 B where Si2 2 AS j B Therefore, we can write the integral equation in the form Ki M K j I j U sin i 2 i 1 2 j i 1 2 k i Ij 2 ki Ij 2 1 2 K = RHS i i Which can be solved by standard xxx for linear xxx of wquations with xxx xxx coefficient methods (Gauss-Sidel Iteration). Once Ki is known, V U Q And p is obtained from Bernoulli equation, i.e. V V p 1 2 U Xxx potential flow solution only xxx xxx in independent of flow condition, i.e. U∞, i.e. only xxx 31 058:0160 Professor Fred Stern Chapter 8 32 Fall 2009 On the xxx of the xxx Vn=0 so that V2 C p 1 S2 where VS V S =tangential surface velocity U VS S U S S U cos i QS S where U S i U cos iˆ sin ˆj cos iiˆ sin i ˆj VSi U S i QS Q S S M QSi j 1 j i K 2 ln rds SB ln r dS 2 S Kj ji j : j=i term is xxx since source panel induces no tangential i lj flow on its xxx. ( lj ln r ji dS j J j ) S i ln rij 1 rij 1 i rij S i S i rij S rij rij rij 1 S xi S zi 2 xi x j S xi z i z j S zi z i xi rij where S xi cos i , S zi sin i 1 rij VS C pi 1 i U li li x 0 2 K , V U cos i i J j j 1 2 i j xi x j n xi z i z j n zi 1 rij dS j dS j 2 2 rij S S j x i x j z i z j J ij i x j 0 S j S xj S xi z i z j 0 S j S zxj S zi x x j 0 S j S xj z i z j 0 S j S zj 2 i 2 dS j where S xi cos i , S zi sin i , xi x j 0 S j S xj z i z j 0 S j S zj S 2j 2 AS j B 2 x i i x j cos i z i z i 0 sin i S j S xj S xi S zj S zi cos j cos i sin j sin i S 0 2 CS j D 2 j AS j C dS i D xi x j 0 cos i z i z j 0 sin i 1 C cos i j DI j1 CI j 2 DI j1 C ln AI j1 2 32 058:0160 Professor Fred Stern Chapter 8 33 Fall 2009 2 C D AC 1 l j A C l j 2 Al j B 1 A ln tan tan ln 2 E E E 2 B D AC xi x j 0 cos i z i z j 0 sin i J j D AC I j1 xi x j 0 cos i z i z j 0 sin i cos j i x x cos z z sin sin sin cos cos x x cos sin sin cos cos cos z z sin sin sin cos cos sin x x cos 1 cos sin sin cos z z sin 1 sin cos cos sin x x sin cos sin sin cos z z cos sin i j0 i i j0 i i j i j 2 i j0 i i j j i i 2 i j0 i i j i j j 2 i j0 i j i j j 2 i j0 i where i j0 j j i i j j i j j i j0 j i cos j cos i sin j D AC sin i j E 33 058:0160 Professor Fred Stern Fall 2009 Chapter 8 34 34 058:0160 Professor Fred Stern Fall 2009 Chapter 8 35 35 058:0160 Professor Fred Stern Fall 2009 Chapter 8 36 36 058:0160 Professor Fred Stern Fall 2009 Chapter 8 37 37 058:0160 Professor Fred Stern Fall 2009 Chapter 8 38 38 058:0160 Professor Fred Stern Fall 2009 Chapter 8 39 39 058:0160 Professor Fred Stern Fall 2009 Chapter 8 40 40 058:0160 Professor Fred Stern Fall 2009 Chapter 8 41 41 058:0160 Professor Fred Stern Fall 2009 Chapter 8 42 42 058:0160 Professor Fred Stern Chapter 8 43 Fall 2009 Method of Image The method of image is useful for plane, spherical, and circular flow domain. Since we have already discussed this method through a variety of applications, here we shall provide a summary. Plane Boundaries: m 2 2 ln x 1 y 2 x 1 y 2 2 1 2 2 2 2 2 3-D: Q M x 1 y 2 z 2 x 1 y z Similar results can be obtained for dipoles and vortices. 2-D: Q 1 2 Spherical and Curvilinear Boundaries: The results for plane boundaries are obtained from consultations of symmetry. For spherical and circular boundaries, image systems can be determined from the previously discussed Sphere&Circle Theorems, respectively, e.g. 43 058:0160 Professor Fred Stern Chapter 8 44 Fall 2009 Flow field Sources of strength M at C outside sphere of radius a, c>a Dipole of strength at l outside sphere of radius a, l>a Source of strength m at b outside circle of radius a, b>a Image System 2 Sources of strength ma at a and line c c sink of strength m externally from center a 2 of sphere to a c 3 2 dipole pf strength a at a l l 2 equal source at a and sink of some b strength at the center of the circle The method can be extended for multiple boundaries by using successive images. Recall the xxx for a source equally spaced between two parallel palnes 44 058:0160 Professor Fred Stern w( z ) m Fall 2009 Chapter 8 45 ln z 4n a ln z 4n 2 a m 0 , 1, 2 , mln z a ln z 2 a ln z 4 a ln z 6 a ln z 4 a ln z 2 a As a xxx example of the method of successive images consider two sphere moving along a line through their centers: Consider the kinematic b.c. for A: 2 F x, t x yt y 2 z 2 a 2 DF 0 V eˆ R U 1 kˆ eˆk or R U 1QR ' Dt 2 Ua 3 where Q 2 cos , QR 3 cos R 2 R Similarly for B QR U 2QR ' This suggests the potential in the form Q U1Q1 U 2 Q2 where Q1 and Q2 both satisfy the xxx equation and the boundary xxx: Q1 Q cos , 1 0 ---(*) R R a R' R 'b Q2 Q 0, 2 cos ' ---(**) R R a R' R 'b Q1 = potential where sphere A moves with unit velocity towards B, with B at rest Q2 = potential when sphere B moves with unit velocity towards A, with A at rest B were xxx 0 a3 a3 Q1 2 cos 2 cos , 0 2 2R R but this does not satisfy the second condition in (*). To satisfy this, we introduce the image of 0 in B, which is a doublet 1 directed along BA at A1, the image requires at image 2 at A2, the image of A1 with respect to A, and so on. Thus we have an infinite 45 058:0160 Professor Fred Stern Chapter 8 46 Fall 2009 series of images A1, A2, … of strength 1 , 2 , 3 etc. where the odd suffixes refer to points within B and the even to points within A. Let f n AAn & AB C b2 a2 b2 , f2 , f3 c ,… f1 c c f2 c f1 a3 b3 b3 ,… 1 0 3 , 2 1 3 , 3 2 3 c f1 c f 2 radius 3 where 1 = image dipole strength, 0 = dipole strength sistance 3 cos cos cos Q1 0 2 1 2 1 2 2 2 with a similar development procedure for Q2. R R1 R2 Although exact, this solution is of xxx form. Let’s investigate the possibility of an approximate solution which is valid for large C (i.e. large separation distance), 2C C2 R' R C 2Cr cos R 1 cos 2 R R 2 2 2 2 1 1 C C2 1 2 cos 2 R' R R R 1 2 1 R R2 1 2 cos 2 C C C 1 2 Considering the former representation first, 1 1 1 1 2u 2 2 R' R By the binomial theorem valid for x 1 1 x 2 1 0 1 x 2 x 2 3 x 3 , 0 1 and n 1 3 2n 1 2 4 2n Hence if 2u 2 1 2 1 0 1 2u 2 2 2 P0 u P1 u P2 u 2 After xxx terms in powers of , where the Pn are as before Legendre functions of the first kind (i.e. Legendre polynomials which are Legendre functions of the first kind of xxx xxx). Thus, 46 058:0160 Professor Fred Stern Chapter 8 47 Fall 2009 1 1 R R2 R C : 2 P1 cos 3 P2 cos R' C C C 1 1 C C2 R C : 2 P1 cos 3 P2 cos R' R R R Next, consider a doublet of strength at A cos Q R' 2 R cos C R 2 C 2 2CR cos 3 2 1 C 2 2 R C 2CR cos 1 2 Thus, 1 2 RP1 cos 3R 2 P2 cos cos R C :Q 2 2 3 C R' C C 1 2C 3C 2 R C : Q 2 P1 cos 3 P2 cos 4 P3 cos R R R Going back to the two sphere problem. If B were absent a3 Q1 2 cos 2R R' using the above expression for the xxx at B and near B 1 , RR’, ' C 3 3 3 1 a a R' P1 cos a Q 2 cos 2 2R C3 2 C a 3 cos C3 which can be cancelled by adding a term to the first approximation, i.e. 1 a 3 cos 1 a 3b 3 cos ' Q1 2 R2 2 C 3 R' 2 to confirm this 1 a 3 a3R' cos ' 1 a 3b 3 cos Q1 2 C2 2 C3 R C3 These approximate solutions are convert to C 3 . QR To find the kinetic energy of the fluid, we have 1 K QQ n dS QQ n dS 2 SA SB 1 K A11U 12 A12U 1U 2 A22U 22 QQ n dS 2 2 S A S 47 058:0160 Professor Fred Stern Fall 2009 Chapter 8 48 Q1 Q2 dS A , A22 Q2 dS B , n n A B Q1 Q1 A12 Q2 dS A Q1 dS B where dS 2R 2 sin rdr n n A B A11 Q1 2 3 2 2a 3b 3 A11 a , A12 , A22 b 3 , 3 3 3 C 3 3 1 2a b 1 K m'1 U 12 U 1U 2 M ' 2 U 22 : using the approximate form of the potentials 3 4 4 C 1 1 where m'1 U 12 , M ' 2 U 22 : masses of liquid xxx by sphere 4 4 48