AMS 212B Perturbation Methods Lecture 20 Part 1 Copyright by Hongyun Wang, UCSC. is the kinematic viscosity and ˆp = p ρ 0

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1 Lecture Part 1 Copyright by Hongyun Wang, UCSC Prandtl boundary layer Navier-Stoke equation: Conervation of ma: ρ t + ( ρ u) = Balance of momentum: u ρ t + u = p+ µδ u + ( λ + µ ) u where µ i the firt coefficient of vicoity and λ i the econd coefficient of vicoity (dynamic vicoity) Incompreible flow (ρ = cont = ρ) Conervation of ma: u = Balance of momentum: t + ( u ) u = ˆp+ νδ u where ν = µ ρ i the kinematic vicoity and ˆp = p ρ. Non-dimenionalization We ue a characteritic length and a characteritic velocity to do the nondimenionalization. L: Characteritic length U: Characteritic velocity (uually the velocity at infinity) A characteritic time i T = L/U We conider the non-dimenionalization t = U L t, t = L U t - 1 -

2 x = 1 L x, x = L x y = 1 L y, y = L y z = 1 L z, z = L z 1 u = u, U u = U u = L, = 1 L Δ = L Δ, Δ = 1 Δ L Conervation of ma: u = Balance of momentum: U L t + U L ( u ) u = 1 L ˆp+ U L ν Δ u ==> t + u Reynold number: Re = U L ν = ρu L µ Normalized preure: p = ˆp U = p ρ U u = ˆp U + 1 Re Δ u, Re U L ν The momentum equation become t + u u = p + 1 Re Δ u Non-dimenional Navier-Stoke equation (For implicity, we drop the prime in notation) - -

3 u = t + ( u ) u = p+ 1 Re Δ u Prandtl boundary layer Conider the two-dimenional teady tate flow over a emi-infinite plate. The emiinfinite plate i parallel to the x-axi, tarting at x = and extending to x =. The boundary condition are At infinity: u, y On the plate: = u ( x, ) = ( 1,) = (,) for x u x, Below we conider the teady tate flow in the limit of Re. Outer expanion: The (leading term) outer expanion atifie u ( out ) = u ( out ) ( ) u ( out ) out = p u ( out )(, y) = 1,, u ( out )( x, ) = 1, It i traightforward to verify that the expreion below i a olution. u ( out )( x, y) = ( 1,), p ( out )( x, y) = p ( ) = cont. Suppoe the ytem ha a unique olution. Then the olution given above i the unique olution. Inner expanion: For mathematical convenience, we with the component of velocity. u( x, y) = u( x, y), v( x, y) The equation become + v y = (E1) u + v y = p + 1 Re u + u y (E) - 3 -

4 u v + v v y = p y + 1 Re v + v y (E3) Now we ue caling to tudy the inner expanion. Let δ = O(Re) -α be the thickne of the boundary layer. We conider the caling = y( Re) α, y = ( Re) α. After the caling, equation (E1) become + ( Re ) α v = Inide the boundary layer, we aume u = O 1 = O ( 1 ), u = O ( 1 ) = O ( 1 ) and u = O ( 1 ) Remark: (E1B) The aumption above i reaonable, but i not completely trivial. For example, ince vertical velocity v = both in the outer expanion and on the boundary in the inner expanion (on the plate in the boundary layer), inide the boundary layer it i unreaonable to expect OR OR v = O(1) v = O ( 1 ) v = O ( 1 ) v inide the boundary layer i caued by the incompreibility. Thi can be een from the continuity equation (conervation of ma) u varie with x at fixed inide the boundary layer ==> / ==> v/ = (Re) -α (/) Integrating (E1B) from to and uing v(x, ) = =, we get - 4 -

5 v( x,) = ( Re) α x, d For mathematical convenience, we write v(x, ) a v( x,) = ( Re) α v * ( x,) where v * (x, ) = O(1). v * x, ( x,) = d = O 1 Next, we look at equation (E) after the caling. u + v * = p + 1 Re u + Re α u The two term on the left ide are of the order O(1). The principle of leat degeneracy tell u 1 ( Re Re ) α u ~O ( 1 ) ==> (Re) α 1 = O(1) ==> α 1 = ==> α = 1/ To the leading order, (E) become u + v * = p + u (EB) Next, we look at equation (E3) after the caling. With α = 1/, we have = (Re) 1/ y, / y = (Re) 1/ ( / ), and v = (Re) 1/ v * After caling, equation (E3) become 1 Re u v * + 1 Re v v * * The leading order of the equation: Re p = O 1 Re = Re p + 1 Re Therefore, to the leading order, (E3) become 1 Re v * + Re v * - 5 -

6 p = (E3B) Matching to the outer expanion yield = cont = p p x, (true to the leading order) Subtituting into (EB), we obtain that u atifie u x, + d = u (EC) with boundary condition = 1 = 1 and u( x,) = for x > u, u x, Self-imilar olution (imilarity olution): To how the elf-imilarity, we conider the caling u ( x,) = u αx, α It i traightforward to verify that u ( x,) atifie equation (EC) and the aociated boundary condition. Under the aumption that olution i unique, we conclude u ( x,) = u( x,) ==> u( x,) = u αx, α Thi i true for all value of α. Selecting α = 1/x, we arrive at = u 1, u x, x u 1, η, η x That i, the olution depend only on the combined variable η x. Let u introduce function We have η u 1, η f η u 1, η d η = f ( η) - 6 -

7 = u 1, u x, ( x,) = f x = f x x, d = f ( x,) u( x,) = = f = f = f x 3/ x x x d f d = f x 3/ x d d x x f x 1 x d x x f 1 x x x x 1 x 1 x Subtituting into equation (EC) and etting x = 1, we obtain that ƒ(η) atifie f ( )+ 1 f f ( ) = with boundary condition =, f f which i baed on definition of f η =, f ( ) = 1, which i baed on u x, = f η u 1, η x d η Numerical imulation how that the correponding initial value are - 7 -

8 =, f ( ) =, f ( ) =.331 f Drag on the plate: The drag on the plate are expreed in phyical quantitie before caling. Let (x (p), y (p) ) and (u (p), v (p) ) be the phyical coordinate and the phyical velocitie before the caling. The phyical quantitie are related to the non-dimenional quantitie a x = x ( p ) L = Re y ( p ) L = U vl y ( p ) u ( p ) x ( p) (, y p ) = U u x, = U f x The hear tre at x (p), denoted by τ, i given by τ = µ ( p ) y ( p ) y ( p) = = ρ νu = y ( p ) = ρ νu f ( ) 1 x U vl =.331ρ νu U v x p Drag on the portion of plate from x (p) = to x (p) = X i U Drag( X ) =.331ρ νu v X 1 dx ( p ) x p = 4.331ρ νu U X v Thi i a good etimate for the total drag on a plate of finite length X