8.3 Method of Images058:0160 Chapter 8Professor Fred Stern Fall 2009 1Chapter 8: Inviscid Incompressible Flow: a Useful Fantasy8.1 IntroductionFor 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 theentrance. Recall that vorticity is generated in regions with large shear. Therefore, outsidethe B.L and wake and if there is no upstream vorticity then ω=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 theory1) Determine from solution to Laplace equation02B.C:at BS: . 0 0V nnf�= � =� at S: V f=�Note: F: Surface Function1058:0160 Chapter 8Professor Fred Stern Fall 2009 210 . 0 .DF F FV F V nDt t F t� �= � + � = � =-�Ѷ for steady flow . 0V n =2) Determine V from V f=� and p(x) from Bernoulli equationTherefore, primarily for external flow application we now consider inviscid flow theory (0) and incompressible flow (const)Euler equation:. 0. ( )VDVp gDtVV V p ztr rr r g� ==- � +�+ � =- � +�2.2: 2VV V VWhere V vorticity fluid angular velocityww� =� - �=Ѵ==�21( )20 0 :Vp V z VtIf ie V then Vr r g r ww f�� +� + + = ��=Ѵ==�1( )2p z B ttfr r ff g�+ + � �� + =� Bernoulli’s Equation for unsteady incompressible flow, not f(x)Continuity equation shows that GDE for is the Laplace equation which is 2nd order linear PDE ie superposition principle is valid. (Linear combination of solution is also a solution)21 222 2 2 211 2 1 222000 ( ) 0 00V ffff ffff ff ff�� =��� =� == +�� =� = � � + = � � +� = ��� =�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 analogs 5) Conformal mapping ( for 2D flow)6) Analytical for simple geometries (separation of variable etc)2058:0160 Chapter 8Professor Fred Stern Fall 2009 38.2 Elementary plane-flow solutions:Recall that for 2D we can define a stream function such that:xyvu0)()(2yxyxzyxuvi.e. 02Also recall that and are orthogonal.yxxyvuudyvdxdydxdvdyudxdydxdyxyxi.e. constconstdxdyvudxdy18.2 Elementary plane flow solutionsUniform streamyxxyvconstUu0i.e. yUxUNote: 022 is satisfied.ˆV U if�=� =Say a uniform stream is at an angle tothe x-axis:cosu Uy xy fa�� �= = =� �sinv Ux yy fa�� �= =- =� �3058:0160 Chapter 8Professor Fred Stern Fall 2009 4After integration, we obtain the following expressions for the stream function andvelocity potential: ( )cos sinU y xy a a�= -( )cos sinU x yf a a�= +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 width perpendicular to the paper,3ALQ V d AS� �= �� �� ��( ) ( )2,2rrQ r b vOrQvbrpp= � �=0, vrmvrWhere: 2Qmbp= is the convenient constant with unit velocity × length (m>0 for source and m<0 for sink). Note that V is singular at (0,0) since rv � �In a polar coordinate system, for 2-D flows we will use:1Vr rfffq� �=� = +� �And:4058:0160 Chapter 8Professor Fred Stern Fall 2009 5. 01 1( ) ( ) 0rVrv vr r rqq� =� �+ =� �i.e.: rvrrvrr1 velocityTangential1 velocityRadialSuch that 0V�� = by definition.Therefore, rvrrvrr101rmi.e. xymmyxmrm122tanlnlnDoublets:The doublet is defined as:source2sink1source2sink1)(mψ m2tan1tan12tan1tantan21tanθθθθ)m()θ(θ1 22 2sin sintan tancos cossin sin2 sincos costan( )sin sin1cos cos ; r rr a r ar rarr a r ar rm r ar a r aq qq qq qq qy qq qq qq q= =- +-- +- = =-+- +For small value of a5058:0160 Chapter 8Professor Fred Stern Fall 2009 612 2 2 2 202 sin 2lim tan ( )aar amy ymr a r x yq ly-�� �= - =- =-� �- +� �cos2 ( ) am Doublet Strengthrl ql f= =By rearranging:22222)2()2( yxyxyψIt means that streamlines are circles with radius 2R and center at (0, -R) i.e. circlesare tangent to the origin with center on y axis. The flow direction is from the source to thesink.2D vortex: 6058:0160 Chapter 8Professor Fred Stern Fall 2009 7Suppose that value of the and for the source are reversal.01rvKvr r rqf yq=� �= =- =� �Purely recirculating steady motion, i.e. ( )v f rq=. integration results in:ln K=constantKθψ K rf ==-2D vortex is irrotational everywhere except at the origin where V and V ��are infinity.CirculationCirculation is defined by:cC closed contourΓ V d s == ��� For irrotational flowOr by using Stokes theorem: ( if no singularity ofthe flow in A). 0c A A Γ V d s V d A ndAw=�=Ѵ�==� � �� � �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 rK{{2 20 0ˆ ˆ( ) 22 and V d sKv e rd e rd K Krp pq q qq q ppGG= � = = =� �Note: for point vortex, flow still irrotational everywhere except at origin itself where V, i.e., for a path not including
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