BOUNDARY LAYER THEORY HIGH RENOLDS NUMBER FLOW BOUNDARY LAYERS (Re ∞) BOUNDARY LAYER Thin region adjacent to surface of a body where viscous forces dominate over inertia forces ⎛ inertia forces ⎞ Re = ⎜ ⎟ Re >> 1 ⎝viscous forces ⎠ Boundary layer separation Wake: viscous effects not Outer flow Flow fieldimportant Viscouseffect aroundvorticityan not s negligible Inner flowarbitraryzero Vorticity zero Strongshape (Inviscid viscous potential flow) effects Steady ,incompressible 2-D flow with no body ddθ * 1 Uτ 0 ∂u n ++(2δθ) = forces. Valid for laminar flow τ 0 ∼ () dx θρdx U 2 dy O.D.E for θ () x To solve eq. we first ”assume” an approximate velocity profile inside the B.L Relate the wall shear stress to the velocity field Typically the velocity profile is taken to be a polynomial in y, and the degree of fluid this polynominal determines the number of boundary conditions which may be satisfied u EXAMPLE: =+abη +cηη2 =f () LAMINAR FLOW OVER A FLAT PLATE: U U∞ U ≈ 0,99U∞ • Laminar boundary layer predictable • Turbulent boundary layer poor predictability UL • Controlling parameter Re = ν • To get two boundary layer flows identical match Re (dynamic similarity) • Although boundary layer’s and prediction are complicated,simplify the N-S equations to make job easier High Reynolds Number Flow 2-D , planar flow uv, xy, u* = v * = , x*, y*= U∞ L → Dimensionless gov. eqs. ∇.0V = **** 2*2* ∂∂uu**∂u∂P1 ∂u∂u X ; ++u ν =−+()22+ ∂∂tx∂y ∂xxRe∂∂y viscous terms * P P = 2 22 ∂∂ν νν∂ ∂P 1 ∂ν∂ν ρU∞ Y; ++u ν =−+()+ ∂∂tx∂y∂yRe ∂x22∂y “Naïve” way of solving problem for 1 Re →∞ → 0 Re If you drop the viscous term Euler’s eqs. (inviscid fluid) • We can not satisfy all the boundary B.C.s because order of eqs. Reduces by 1 Inside B-L can not get rid of viscous terms U∞ U∞ U (x,y) y δ δ 1 δ * =〈 L 100 L Derivation of B-L eqs. From the N-S eqs • Physically based argument :determine the order of terms in N-S • Limiting procedure as Re ∞ eqs. and throw out small terms Assumption 1 δ δ * =〈〈1 L Term Order * (1) ∂u =1 (1) ∂x* δ * ∂ν * =1 δ * ∂y* v* δ * * * ∂ν δ * = δ ∂x* 1 ∂2*u 1 2 ∂y*2 δ * * du ∂u* u* =1 dt* ∂x* ∂∂uu**∂u*∂P*1 ∂2u*∂2u* ++u**ν =−+()+ ∂∂tx∂y∂xRe ∂x22∂y (1) (1) * 1 (1) (1) δ =1 (1) *2 * =1 δ 2 *2 δ (1) (1) ()δ Neglect since of order (1) *2 >>>1 Also for y –direction ()δ * * * * 2* 2* ∂∂ν **νν∂∂P 1 ∂ν∂ν ++u ν =−+()22+ ∂∂tx**∂y*∂y*Re ∂∂xy** * * ()δ * ()δ ** (1) ()δ *2 δδ * (1) * ()δ { + } ()δ (1) 2*(δ ) 2 * ()δ * * * ()δ ()δ ()δ ∂P* ∂P* ()δ * small relative to (1) ∂y* ∂x* To good approximation P ≅ Px() pressure at the edge of B- L. is equal to pressure on boundary layer. • Time – dependant P ≅ Px (, t ) known from the other flow • Pressure at all points is the same • Only need to consider x-direction B-L. eqs. Prandtl (1904) Outer flow (inviscid) y 2-D planar x Governing ∂∂uv +=0 1) ∂∂xy eqs.for B.L B-L.eqs. ∂∂uu∂u1 ∂P∂2u 2) ++=uv− +ν still non-linear ∂∂tx∂yρ ∂x∂y2 but parabolic type unknows u,v (x,y,t) PP≅ (,xt) known from the potential flow Need B.C.s & I.C.(time dependant) • 2-D, steady BCs • u=ν =0 at y=0 • u=u(y) at x=0 • u= U ∞ (x) y ∞ (y δ ) marching condition • B-L. eqs. can be solved exactly for several cases • Can approximate solution for other cases Limitation of B.L egs.: where they fail? (1) Abrupt chances (2) Eqs. are not applicable near the leading edge δ L is small δ * = 〈〈 1 invalid L (3) Where the flow separates not valid beyond the separation point Separation point Bernouilli eqs. ρ =constant PV2 11dP dV +=constant + 20U = ρ 2 ρ dx 2 dx Valid along the streamlines 1 dP dU −=U substitute the B.L eqs u,v can be found ρ dx dx known SIMILARITY SOLUTION TO B.L. EQS Example 1 Flow over a semi-infinite flat plate dp Zero pressure gradient = 0 p = constant dx dp Steady ,laminar & U=constant ( = 0 ) dx U y x 1 • Bernouilli eqs. outsideB.L pU+=ρ 2 cons. dp 2 U=constant , = 0 dx Governing (B.L. eqs.) become ∂∂u ν +=0 (1) ∂∂xy ∂∂uu∂2u uv+=ν (2) ∂∂x yy∂2 B.C. • y=0 u= v =0 (no-slip) & y ∞ , u U • x=0 u=U Blasuis(1908) : 1.Introduce the stream function ψ (x,y) • Recall ; ∂ψ ∂ψ u = ν =− ∂y ∂x note that ψ satisfies cont. eqs. substitute intoB.L. mom. Eqs ∂∂ψ 22ψψ∂∂ψ ∂3ψ ..−=ν (2’) ∂∂y x∂y ∂x ∂yy23∂ • Now, assume that we have a similarity “stretching” variable, which has all velocity profiles on plate scaling on . δ i.e uy = f () y U∞ δ δ x dimensional analysis δ = gU(,∞ x,ν ) δ Ux 1 ==gg()∞ (Re) ∼()δ 2 δ ∼ ν x ν Re ν x 2 δ ∼ mm δ 1 U ..sm= [] ∼ both ()δ ∞ sm x Rex Viscous dif. Depth Ux ν x Re = ∞ δ ≈ 5 ν U∞ y Let η = [-] similarity variable δ U u η = y ∞ = f ()η ν x U Use similarity profile assumption to turn 2 P.D.E 1 O.D.E ∂ ψ yy η ν x u = ψ ==udy Uf ()ηηdy =Uf () dη ∂ y x = fixed ∫∫ ∫ 00 0 U η ψ ==Uxνη∫ f()dηUxνF()η 0 ψ = UxνηF() F()η U ψ = UxνηF() η = y ∞ ν x y ∂ψ ∂ψ ψ −=ψ udy ddψ =+y dx 0 ∫ ∂∂yx 0 •Now, substitute ψ into P.D.E for ψ (x,y) to get O.D.E for F(η ) ∂∂ψ 1 U ν η =+∞ FUν xF' ' dF dF2 ∂∂xx2 ∞ x F = F '' = dη dη 2 ∂η 11U∞ 1 ∂ψ 1 U ν ' ∂ψ U =− y =− η =−∞ ()FFη ==Uxν F''∞ UF ∂xx22ν xx ∂xx2 ∂yx∞∞ν ∂2ψ U 2 3 2 =− ∞ ηF '' ∂ ψ U∞ '' ∂ ψ U∞ ''' = UF∞ = F ∂∂xy 2x ∂y2 ν x ∂yx3 ν Substituting into eq. (2’) 2 11 UU∞∞⎡⎤1 ν 22⎡U∞⎤U∞ UF∞∞'(−−ηηF''') ⎢⎥( ) (F−F') ⎢U( ) F''⎥=νF''' 22xx⎣⎦⎣ννx⎦x or U 2 1 U 2 1 U 2 U 2 − ∞ η F '' F '− ∞ FF'' + ∞ ηFF'' ' = ∞ F ''' 2x 2 x 2 x x 1 FF'''+=F'' 0 blasius eq. 3rd order , non linear ODE 2 U Note: FF ''' += F '' 0 for η = y ∞ BVP 2ν x BC’s are At y=0 u=v=0 η = 0 ∂ψ F’(0)=0 BC 1) u ==0 UF∞ '0= y=0 η =0 ∂y y=0 1 U ν F(0)=0 BC 2) ν = 0 − ∞ ('FF−=η )0 y=0 2 x BC 3) (x,y ∞ ) U ∞ ∂ψ →U UF' = U F '(η →∞) 1 F '(∞) = 1 ∞ ∞∞η→∞ ∂y y→∞ F( η ) dimensionless function Or At x=0 uU= UF∞∞' x=0 = U ∞ η→∞ F’( ∞ )=1 same with BC 3) Matching B.C • Solution to blasius eg a)power series b)runge-kutta • results tabulated form for F,F’,F’’,etc p.g 121 U u η = y ∞ F F ' = F '' ν x U∞ 0 0 0 0.33206 5.0 3.28329 0.99155 0.01591 F’’= 0.33206 From the solution • Velocity profile U = y ∞ γ x η 5 5 F’= u ν U∞ Re U x 0.8 ∞ 1 U ν ∂ψ 1 U ν ην→∞ = ∞ (5x1−3.28) ν =− = ∞ ('ηFF− ) ∞ 2 x ∂xx2 ν ∞ 1 ν 1 − 1 = 0.86 =−Re 2 ηFF' x [] U ∞ Rex U∞ 2 Shear stress distribution along the flat plate ∂∂u ν τµ=+( ) τ(x, y) 4 ν ∞ 1 ∂∂yx For Rex =⇒10 =0.00865 ≈ U∞ 100 ∂∂uu**ν ∂ τµ≅ 6 ν ∞ 1 ** For Re =⇒10 =0.000865 ≈ ∂∂yx ∂y x U∞ 1000 ∂u At the wall (y=0) τµ0 ()x = ∂y y=0 τ w ()x 2 3 ∂ ψ U∞ U τµ()xU==µ F'' ∞ 0 2 ∞ η=0 τµ0 (xF) = ''(0) ∂yxy=0 ν ν x Distribution along the wall 0.332 Non dimensionalize : τ 2F ''(0) 0.664 Ux. C ==0 = Re = ν fxC f = 0.664 1 2 ν ρU RexxRe Ux 2 ∞ Friction coef. Note : x →⇒0 τ 0 →∞ y ν →∞ B.L eqs.are not valid near the leading edge x x Up to the point we are considering Drag force acting on the flat plate We have to integrate shear stress x = τ ()ζζd τ FD ∫ 0 0 ↓ per unit width x 3 21FbD = .328()U∞µρx x dimensionless drag coef.( CD ) we have 2 wetted sides 2F C = D A=2bx D 1 ρUA2 2 ∞ Width normal to the blackboard 1.328 56 CDx=< valid for laminar flow i.e for Re 5.10 to 10 Rex 6 for Rex >10 → turbulent drag becomes considerably greater Boundary Layer Thickness : δ U u ηη==yy∞ at 5 ⇒ =0.99 → =δ (Table) ν xU UU∞∞5x x 5 ≅≅δδ Rex = ννx Rex δ:defined as the distance from the wall for which u=0.99U∞ Boundary Layer Parameter (thicknesses) Most widely used is δ but is rather arbitrary y=δ when u=0.99 U∞ hard to establish more physical parameters are needed Displacement thickness: δ U ∞ U∞ * δ δ δ * an imaginary displacement of fluid from the surface to account for “lost” mass flow in boundary layer * . ∞∞ ∞ δ mutot ==ρρdyUdy=ρUdy−ρUdy or ∫∫∞∞∫ ∫∞ 00y=δ * 0 * −ρδU∞ ∞∞u ρδUU**=−(ρ ρu)dy δ=(1−)dy ∞∞∫∫ 00U∞ if ρδ=>cons.