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H534 chapter 19SurfaceGv x = du/dy = V 0 w=0v y = -du/dx = 0FInletyv x = du/dy = V 0w=0v y = -du/dx = 0w=u=0x Eu=0CHalfBeamDBcenter lineu=0Adu/dx = 0dw/dx = 0w=u=0OutletFigure 19.6 Boundary conditions for flow around the beam in Figure 19.4. The flow issymmetric about the centerline, and the beam has length L in the x direction (along flow).19.9.2 Boundary Conditions for a BeamA well-defined solution of these elliptic PDEs requires a combination of (less thanobvious) boundary conditions on u and w. Consider Figure 19.6, based on theanalysis of [Koon 86]. We assume that the inlet, outlet, and surface are far from thebeam, which may not be evident from the not-to-scale figure.Freeflow: If there were no beam in the stream, then we would have free flowwith the entire fluid possessing the inlet velocity:v x ≡ V 0 , v y =0, ⇒ u = V 0 y, w =0. (19.69)(Recollect that we can think of w =0 as indicating no fluid rotation.) Thecenterline divides the system along a symmetry plane with identical flowabove and below it. If the velocity is symmetric about the centerline, then itsy component must vanish there:v y =0, ⇒ ∂u =0 (centerline AE). (19.70)∂xCenterline: The centerline is a streamline with u = constant because there isno velocity component perpendicular to it. We set u =0according to (19.69).Because there cannot be any fluid flowing into or out of the beam, the normalcomponent of velocity must vanish along the beam surfaces. Consequently,the streamline u =0is the entire lower part of Figure 19.6, that is, the centerline−101COPYRIGHT 2008, PRINCET O N UNIVE R S I T Y P R E S SEVALUATION COPY ONLY. NOT FOR USE IN COURSES.ALLpup_06.04 — 2008/2/15 — Page 534