PanelMethods

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1、Click to edit Master title style,Click to edit Master text styles,Second level,Third level,Fourth level,Fifth level,*,*,*,Panel Methods,What are panel methods?,Panel methods are techniques for solving incompressible potential flow over thick 2-D and 3-D geometries.,In 2-D,the airfoil surface is divi

2、ded into piecewise straight line segments or panels or“boundary elements”and vortex sheets of strength,g,are placed on each panel.,We use vortex sheets(miniature vortices of strength,g,ds,where,ds,is the length of a panel)since vortices give rise to circulation,and hence lift.,Vortex sheets mimic th

3、e boundary layer around airfoils.,Analogy between boundary layer and,vortices,Upper surface boundary layer contains,in general,clockwise rotating,vorticity,Lower surface boundary layer contains,in general,counter clockwise,vorticity,.,Because there is more clockwise,vorticity,than counter clockwise,

4、Vorticity,there is net clockwise circulation around the airfoil.,In panel methods,we replace this boundary layer,which has a small but finite thickness with a thin sheet of,vorticity,placed just outside the airfoil.,Panel method treats the airfoil asa series of line segments,On each panel,there is v

5、ortex sheet of strength,DG,=,g,0,ds,0,Where ds,0,is the panel length.,Each panel is defined by its two end points(panel joints),and by the control point,located at the panel center,where we will,Apply the boundary condition,y,=Constant=C.,The more the number of panels,the more accurate the solution,

6、since we are representing a continuous curve by a series,of broken straight lines,Boundary Condition,We treat the airfoil surface as a streamline.,This ensures that the velocity is tangential to the airfoil surface,and no fluid can penetrate the surface.,We require that at all control points(middle

7、points of each panel),y,=C,The stream function is due to superposition of the effects of the free stream and the effects of the vortices,g,0,ds,0,on each of the panel.,Stream Function due to,freestream,The free stream is given by,Recall,This solution satisfies conservation of mass,And,irrotationalit

8、y,It also satisfies the Laplaces equation.Check!,Stream function due to a,Counterclockwise,Vortex of,Strengh,G,Stream function Vortex,continued.,Pay attention to the signs.,A counter-clockwise vortex is considered“positive”,In our case,the vortex of strength,g,0,ds,0,had been placed on a panel with

9、location(x,0,and y,0,).,Then the stream function at a point(,x,y,)will be,Panel whose center,point is(x,0,y,0,),Control Point whose center,point is(,x,y,),Superposition of All Vortices on all Panels,In the panel method we use here,ds,0,is the length of a small segment of the airfoil,and,g,0,is the v

10、ortex strength per unit length.,Then,the stream function due to all such infinitesimal vortices at the control point(located in the middle of each panel)may be written as the interval below,where the integral is done over all the vortex elements on the airfoil surface.,Adding the,freestream,and vort

11、ex effects.,The unknowns are the vortex strength,g,0,on each panel,and the value of the,Stream function C.,Before we go to the trouble of solving for,g,0,we ask what is the purpose.,Physical meaning of,g,0,Panel of length ds,0,on the airfoil,Its circulation=,DG=g,0,ds,0,V=Velocity of the flow just o

12、utside the boundary layer,If we know,g,0,on each panel,then we know the velocity of the flow,outside the boundary layer for that panel,and hence pressure over that panel.,Sides of our contour,have zero height,Bottom side has zero,Tangential velocity,Because of viscosity,Pressure distribution and Loa

13、ds,Since V=-,g,0,Kutta,Condition,Kutta,condition states that the pressure above and below the airfoil trailing edge must be equal,and that the flow must smoothly leave the trailing edge in the same direction at the upper and lower edge.,g,2,upper,=V,2,upper,g,2,lower,=V,2,lower,F,From this sketch ab

14、ove,we see that pressure will be equal,and the flow,will leave the trailing edge smoothly,only if the,voritcity,on each panel,is equal in magnitude above and below,but spinning in opposite,Directions relative to each other.,Summing up.,We need to solve the integral equation derived earlier,And,satis

15、fy,Kutta,condition.,Numerical Procedure,We divide the airfoil into N panels.A typical panel is given the number j,where J varies from 1 to N.,On each panel,we assume that,g,0,is a piecewise constant.Thus,on a panel numbered j,the unknown strength is,g,0,j,We number the control points at the centers

16、of each panel as well.Each control point is given the symbol“i”,where i varies from 1 to N.,The integral equation becomes,Numerical procedure,continued,Notice that we use two indices i and j.The index I refers to the control point where equation is applied.,The index j refers to the panel over which the line integral is evaluated.,The integrals over the individual panels depends only on the panel shape(straight line segment),its end points and the control point,.,Therefore this integral may be c