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Elements of the Boundary Layer Theory – Part 1

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Elements of the Boundary Layer Theory – Part 1 AERODYNAMICS I LECTURE 4 ELEMENTS OF THE BOUNDARY LAYER THEORY – PART 1 AERODYNAMICS I Concept of a boundary layer. Prandtl’s Equation Boundary layer (BL) – thin layer of a fluid adjacent to the surface of an immersed body. The BL features huge transversal gradient of the flow velocity. Within the BL, the dynamical effect due to fluid viscosity are comparable in magnitude to inertial effects. General assumption – the BL is thin in comparison to a characteristic longitudinal extension of the flow. Hence, we have L or 1 L AERODYNAMICS I We will present rudiments of the BL theory as formulated by Ludwig Prandtl in the beginning of the XX century. The idea is to introduce a simplification of the Navier-Stokes Equations by careful estimation of magnitude of different terms. The departure point is the set of equations governing steady 2D motion of a viscous incompressible fluid 1 ux u y u x p () xx u yy u 1 upx y y () xx yy xyu 0 One can assume that the flow inside the (wall-attached) BL is nearly parallel. We expect that the velocity component parallel to the wall is much larger than the component normal to the wall. Let U U() x denote the velocity distribution (along the wall) at the external “edge” of the BL. Roughly speaking, the velocity U can be identified with the velocity of an external (potential) flow computed at the boundary of the immersed body. For slender bodies, we typically have UU , where U is the free-stream velocity. AERODYNAMICS I Hence, the following scaling holds true UUU L uU , u, u u xLL y x Next, from the continuity equation we have U u U U x y y LL Thus, the terms in the equation of motion in x-direction can be scaled as follows 1 ux u y u x p () xx u yy u 2 U U UU U 2 U 2 L L L L 2 Note that, in the viscous term 2 xxu()L yy u yy u AERODYNAMICS I Inside the BL, the dominating viscous term must be of the same order as the convective terms. Hence 222 UUU yyu 22~ LLLUL It follows that the (average) relative thickness of the BL is related to the Reynolds number by the formula 11 L UL ReL mm5 2 Example: Estimate average BL’s thickness for L0.5 m , U 50ss , 10 . 50 0.5 6 1 We have: Re5 2.5 10 L 0.63 mm (!) 10 2.510 6 We see that the thickness of the BL is really small !!! AERODYNAMICS I Consider the equation of motion in the direction y (normal to the wall). It is clear that all terms are smaller than analogical ones in the equation for the direction x by the factor L 1. We conclude that the general order of magnitude of the equations of motion is 2 1 U ux u y u x p yy u , O()L 2 1 U upx y y yy , O()L L Automatically, one can conclude that y p 0, i.e., the pressure across the BL is nearly constant and equal to the pressure at the external border of the BL. The distribution of pressure along this border can be computed from the Bernoulli Equation (which is applicable as the external flow is virtually inviscid), namely 1122 p22 U p()() x U x After differentiation with respect to x , one obtains 1 p()()() x U x U x AERODYNAMICS I After insertion to the equation of motion in x direction, we obtain the Prandtl Equation (for a laminar BL) ux u y u U()() x U x yy u In order to obtain closed system, we add the continuity equation xyu 0 We also need to specify the boundary (or wall) and far-field conditions u, u , ( x ,0) 0 , limu(,)() x y U x wall y From the mathematical point of view the Prandtl Equation is the partial differential equation (PDE) of the parabolic type (like, for instance, the equation of heat conduction). The spatial variable x plays here a role of a time-like variable. Thus, in order to solve this equation we need to specify also the “initial” condition, i.e., the velocity profile at some location xx 0 u(,)() x00 y u y AERODYNAMICS I In general, the solution of the PE is obtained numerically and the result of such calculations is the velocity and pressure inside the BL for xx 0. Ideally, we would choose x0 at the front stagnation point, where the BL starts to develop. There is, however, some dose of trickiness in such choice, as the initial thickness of the BL is zero and uy0 () is not properly defined. The way to overcome this “singularity” is to use approximate analytical solution of the BL flow near the stagnation point. AERODYNAMICS I Self-similar solutions of the Prandtl Equation For certain cases of U U() x , one can find for the PE so-called self-similar solutions. We say that the solution of the PE is self-similar if it can be expressed by a function of a single y variable applied to a “self-similar” coordinate , where the function ()x is to ()x be determined in the solution procedure. Since we concern only 2D case, let’s use the streamfunction. Assume that it can be written as follows y (,)xyUx ()()[ xf ()x ] Ux ()()[(,)] xf xy The corresponding velocity components are uxy(,) yy (,) xyUxxf ()()[(,)] xy (,) xy 1 Uxxf()()[(,)] xy Uxf ()[(,)] xy ()x AERODYNAMICS I and (,)xy x (,) xy Uxxf ()()[(,)] xyUx ()()[(,)] xf xy yx () U( x ) ( x ) f [ ( x , y )] 2 ()x Let’s calculate all necessary derivatives … yx () uxy(,) Uxf ()[(,)] xy Uxf () [(,)] xy x 2 ()x U()() x U x uxy(,) fxy [(,)], uxy (,) fxy [(,)] y()()xx yy 2 After insertion to the left side of the PE we get 2 u xy u u U( x ) U ( x ) f [ ( x , y )] ()x UxUxf() ()[(,)] xyf [(,)] xy Ux2 () f [(,)] xyf [(,)] xy ()x AERODYNAMICS I The right-hand side of this equation assumes the following form Ux() U( x ) U ( x ) f [ ( x , y )] 2 ()x Finally, equating to , we arrive at the following equation UxUxf() () 2 [(,)] xy UxUxf () ()[(,)] xyf [(,)] xy ()()x U x Ux2 () fxyf [(,)][(,)] xyUxUx ()() f [(,)] xy ()()xx2 Ux() After division by and simple algebra, the above equation reduces to 2 ()x UU (f2 f f 1) f f f Note that the left side of this equation depends on both x and . A self-similar solution is possible only when we remove the dependence on x, which is possible only for some forms of the function U U() x . AERODYNAMICS I Falkner and Skan notices (in 1930), that self-similar solutions exist if x m U()() x U L where mR is a parameter. m1 Then U() x U m x Lm x Assume that the function is equal ()x C()L L where the constants C and have to be determined. Then 2 2 2 U CUL 21m U CUL m x , ()x 21m L L One can see that the x-dependence disappears if and only if 1m 2m 1 0 2 AERODYNAMICS I Thus, we obtain the following expressions 2U CUL2 Um CULCUL221 m , 2 After insertion, we get the 3rd-order nonlinear ordinary differential equation for the unknown function f CULCUL221 m m( f2 f f 1) f f f 2 Equivalently CULCUL221 m m( f2 1) f f f 2 The constant C can be chosen accordingly to the “normalization condition” … CUL2 12 m 1 C 21m U L AERODYNAMICS I We obtain also the formula for the function ()x , namely 1m 1mm ()()x Cx2 L C2 x L 2 x L L LLLUL m1 x m 2 x L L2 2 x m1 LUL x m1 Ux() The second coefficient in the left side of the equation for the function f is now equal CUL2 2m m 1 m which leads to the final form of the famous Falkner-Skan Equation 2 f f f (1 f ) 0 yy where ff(), ()x 2 x m1 Ux() AERODYNAMICS I The parameter can be expressed in terms of the number m, namely 2m 2 1m m 21 m 2 Hence, the self-similar coordinate can be expressed by the formula y x (2 ) Ux() The boundary and “far-field” conditions for the Falkner-Skan Equation are uf0 (0) 0 wall 0 f (0) 0 wall u U( x ) lim f ( ) 1 y AERODYNAMICS I Physical interpretation of the Falkner-Skan self-similar solutions In order to understand the physical meaning of the FS solutions, consider the Laplace Equation for the streamfunction of an external potential flow. In polar coordinates, we have 2 11(r ) 0 r rr r22 The particular solution to this equation is (r , ) Krm1 sin[( m 1) ]. Indeed, we have m r K( m 1) r sin[( m 1) ] m1 rr K( m 1) r sin[( m 1) ] 2 m rrr K( m 1) r sin[( m 1) ] 1 21m r rr r K( m 1) r sin[( m 1) ] 2 0 1 K( m 1)21 rm sin[( m 1) ] r22 AERODYNAMICS I The polar components of the corresponding velocity field are 1 m r r K( m 1) r cos[( m 1) ] m r K( m 1) r sin[( m 1) ] Which radial lines const are the streamlines of this flow? Clearly, along such line we must have (r , ) 0. Hence, such that sin[(m 1) ] 0 ,1 0 , ,2 m1 Assume that m 0. Then [0,2) and 1 1 1 m11 2 ,2 2 Radial component of the velocity along the line ,1 0 is equal mm r (r , 0) K ( m 1) r U ( x L ) , m xm 2 U Lm AERODYNAMICS I Thus, we have just obtained the external flow corresponding to self-similar Falkner-Skan BL, i.e., m U( x )r ( x ,0) U ( x L ) Such flow corresponds to the flow in the vicinity of the flat plate oriented at the negative angle of attack 2 with respect to the free-stream direction.
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