Boundary Layer Theory. v = 0, ( v)v = p + 1 Re 2 v. Consider a cylinder of length L kept in a inviscid irrotational flow.

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1 1 Boundary Layer Theory 1 Introduction Consider the steady flow of a viscous fluid. The governing equations based on length scale L and velocity scale U is given by v =, ( v)v = p + 1 Re 2 v For small Re we have Stokes flow. We consider boundary layer theory when Re is large. 2 D Alembert s Paradox Consider a cylinder of length L kept in a inviscid irrotational flow. We consider a U x cylinder of radius R with an imposed velocity Ue 1 far from the cylinder. Since fluid can not penetrate the surface, we have v r (r, θ) =

2 Since the flow is irrotational, we have v = and thus we can write v = φ. Using continuity equation we get which in cylindrical coordinate gives 1 r r ( r φ r 2 φ = ) + 1 r 2 2 φ θ 2 = The boundary condition is φ Ux as r or 2 φ Ur cos θ as r At the surface of the cylinder we have φ r = r=r Let us assume the solution as φ = f(r) cos θ which gives ( 1 d r df ) 1 r dr dr r f = 2 Taking f(r) = r n we obtain f(r) = ( Ar + B ) cos θ r Applying the boundary conditions at and on the surface we get A = U B = UR 2 and hence The velocity components are given by ) φ = U (r + R2 cos θ r v r (r, θ) = φ r = U v θ (r, θ) = 1 r φ θ = U (1 R2 r 2 (1 + R2 ) cos θ r 2 ) sin θ To find the pressure on the cylinder we square and the above two relations at r = R to get v 2 (R, θ) = 4U 2 sin 2 θ Now we use the Bernoulli s equation. Since the flow is irrotational, the constant is same everywhere in the fluid and we have p ρv2 (R, θ) = p ρu 2

3 3 and hence the pressure on the surface is p = p ρu 2 (1 4 sin 2 θ) Due to symmetric nature of the pressure, the total drag on the cylinder is zero. 3 Boundary layer concept Consider the following model problem for f(y): ɛf + f = a f() =, f(1) = 1 where a > and ɛ is small positive parameter. The exact solution satisfying the boundary conditions are f = ay + Let f = O(1) i.e. y is not small then 1 a 1 e 1/ɛ (1 e y/ɛ ) f ay + 1 a = f o (y) This solution (outer solution) is valid away from zero. This could have been obtained by setting ɛ = and using the boundary conditions at y = 1. However it does not satisfy the boundary condition at y =. For y near we introduce y = ɛy so Y = O(1) when y is small. Introducing into the governing equations we get f + f = ɛa where f = f(y ). Again ignoring the ɛa term we get f = A(1 e Y ) = A(1 e y/ɛ ) = f i This is called inner solution which satisfy the boundary conditions at y =. Thus we have a solution which satisfy the boundary condition at y = and another solution which satisfy the boundary condition at y = 1. We use matching condition to find the constant A. The matching condition is lim Y f i = lim y f o

4 4 This gives A = (1 a) and thus f i = (1 a)(1 e Y ) If we plot the f i, f o and the exact solution, then we find that f i performs well near y = and f o performes well away from zero. The key feature of the above problem is the development of a thin region near y = where there is a rapid change in the feature of the solution. In this thin region it is not appropriate to drop the ɛf term. This thin region near y = is called boundary layer for the above model problem. Very much similar characteristics developed for high speed Re 1. We may neglect the 2 v/re term in most part of the flow but must kept near the surface of the body where there is a boundary layer and this term is no longer negligible. 4 The boundary layer equations First consider the classical problem of fluid streaming at high speed over a plate. The flow is steady and the external stream consists of fluid velocity U ˆx. Let L be y U δ L x the characteristic length scale and the Reynolds number Re = UL/ν is very large. We thus anticipate that the term 2 v/re will be important in a thin boundary layer δ(x) above the plate. We aim to derive the equations valid in the boundary layer based on the estimate of the order of the quantities. Since u varies from at the flat plate to U in the external stream we assume that u U. Changes in the boundary layer is expected at a length scale L parallel to axis of x and δ(x) perpendicular to the plate. Thus x L and y δ. Thus x 1 L, y 1 δ

5 5 Let V be the typical size for v. Hence from the continuity equations we get Now consider the x-momentum equation. Thus using the above arguments u x + v y = U L V δ V δ L U uu x + vu y = p x /ρ + ν(u xx + u yy ) U U L + δ L U U δ p x + ν Now according to Prandtl s hyposthesis and thus U 2 L νu δ 2 δ L uu x ν 2 u y 2 ( U L + U ) 2 δ 2 ( ) 1/2 νu δ L(Re) 1/2 L So the thickness of the boundary layer is O(Re 1/2 ) which is very small since Re 1. Also it is clear that at the most p/(ρl) U 2 /L p ρu 2 So retaining the dominant term in the x momentum equation we have uu x + vu y = p x /ρ + νu yy Next we consider the y-momentum equations. We have uv x UV/L = U 2 δ/l 2 and vv y V 2 /δ = U 2 δ/l 2. Also p y = ρu 2 /δ. Thus Also and uv x vv y U 2 δ/l 2 ρu 2 /δ νv xx νv/l 2 = νδu/l 3 = 1 δ Re L U 2 L ρu 2 /δ νv yy νv/δ 2 = 1 Reδ 2 δu 2 ρu 2 /δ where due to Prandtl s hypothesis Reδ 2 = O(1). Hence the y-momentum equation reduces to just p y ρ =

6 This states that the pressure does not vary across the boundary layer. Thus the Prandtl s boundary layer equations are u x + v y = uu x + vu y = p x ρ + νu yy p y = The boundary condition include no-slip at the plate and we must match the free stream velocity. Thus u = v = at y =, u U as y The same derivation will hold even if the free strem is U(x)ˆx i.e. non-uniform. Now can determine p x term as follows. Taking y in the x-momentum equation we get p x = ρuu x where the external stream U is determined from the invscid theory. 6 5 Blasius solution In this case external stream U is constant and hence p x =. The governing equations become u x + v y = uu x + vu y = νu yy The boundary condition include no-slip at the plate and we must match the free stream velocity. Thus u = v = at y =, u U as y The boundary conditions hold for x > i.e. at the plate. We seek a similarity solution for the above problem. To motivate this note the the x-momentum equation in the boundary layer is parabolic i.e. there is no upstream influence. First we show that the length scale L does not appears in the solution of the problem. For the moment suppose that the length of the plate is of finite length L. For a point with coordinate x, the solution does not affected by

7 what is happening further downstream (due to parabolic nature of the flow). Thus the oncoming flow is ignorant about the length L. We have shown that δ L Re But since δ does not depend on L and [x] = [L], we replace L by x ( Ux δ ν ) 1/2 Now since inside the boundary layer y δ we get ( ) 1/2 Ux y δ i.e. y x ν Thus we look for solution in terms of a single variable η as η = ( ) 1/2 U y 2νx 7 Let us assume that u = Uf (η) Introducing stream function ψ we get Hence ψ(x, η) = 2νxUf(η) v = ψ ( ψ x = x + ψ ) η νu = η x 2x (ηf f) Putting into the x-momentum equation we finally obtain The boundary conditions are f + ff = f = f = at η = f 1 as η Thus we have a nonlinear third order differential equations with three boundary conditions. Thus the solution is completely determined. The above equation can be solved numerically for example using shooting method. For this let us apply Runge-Kutta method. Instead of solving the boundary value problem, an initial value problem with values f() = f () = and an estimated value f () = f e is solved. The estimated value f e is changed until the bondary

8 condition f ( ) = 1 is satisfied. In practice this condition is satisfied at position with finite but large enough η value (η = 5 gives a deviation from 1 of less than 1 4. Suppose we want to calculate the friction drag D in the x-direction on a portion of the plate of length L. To do this we first calculate σ 12 y= = µ u ( ) 1/2 U y = µ f () y= 2νx Thus D = L µ u y dx = U 2µρULf () y= If the width of the plate is b, then the above expression is multiplied by b. Boundary layer thickness: There is no unique boundary layer thickness since the effect of the viscosity in the boundary layer decreases asymptotically as we move outwards from the wall. If we defined the boundary layer thickness to be the position where u =.99U, the we find that η 99 = 3.6. Therefore the boundary layer thickness defined in this way is Displacement thickness: νx δ U By this we understand the thickness by which the inviscid outer flow is displaced outwards by the drop in the velocity in the boundary layer. The reduction in volume flux due to the action of viscosity is (U u)dy and thus the defining equation for displacement thickness δ 1 is Thus we have Uδ 1 = δ 1 = (U u)dy or δ 1 = 2νx U (1 u U )dy νx [1 f (η)] dη 1.72 U (Note: We can define the displament thickness from other argument as well. Let χ is at a great distance from the wall so that the velocity at y = χ is U. Let us calculate volume flux Q between y = and y = χ. Now Q is less than Uχ since near the wall u < U. Thus we defined δ 1 such that U(χ δ 1 ) = Q = χ udy or δ 1 = χ (1 u/u)dy 8

9 Taking limit of χ we obatin the same result.) Momentum thickness of boundary layer: The momentum thickness δ 2 is defined similarly. The momentum between y = and y = χ is M = χ Hence for a fluid of constant density δ = ρu 2 dy = ρu 2 (χ δ ) χ The momentum thickness δ 2 is defined by and we let χ. δ 2 = δ δ 1 = (1 u 2 /U 2 )dy χ u U (1 u U )dy (Note: We can define the momentum thickness from other argument as well. The mass flux for a fluid in a small area perpendicular to x-direction is ρudy. The loss of momentum fluix for this is ρu(u u)dy. And thus the total loss of momentum is The momentum thickness is defined by ρu 2 δ 2 = Thus we arrive at the same equation.) or ρu(u u) dy ρu(u u) dy or δ 2 = u U (1 u U )dy The momentum thickness for the Blasius solution is 2νx νx δ 2 = f (1 f ) dη.664 U U Energy thickness of boundary layer: Energy thickness δ 3 is defined by ρu 3 δ 3 = For Blasius solution we have 2νx δ 3 = U δ 3 = ρu(u 2 u 2 ) dy u u2 (1 U U )dy 2 f (1 f 2 ) dη 1.44 νx U 9

10 6 Falkner-Skan solutions 1 The Blasious solution is a special case of a broader class of solutions known as Falkner-Skan solutions. For this consider the outer stream of the general form U(x). We seek a solution of the form ψ = U(x)g(x)f(η), n = y/g(x) By an analogy with the Blasius solution, g(x) represents the thickness of the boundary layer. The boundary layer equations to be solved are with boundary conditions ψ y ψ xy ψ x ψ yy = UU x + νψ yyy ψ = ψ y = at y = ; ψ y U(x) as y Now u = ψ x = Uf, v = ψ x = (Ug) x f ηug x f Similarly ψ yy = Uf /g, ψ yyy = Uf /g 2, ψ xy = 1 g [(Ug) xf Ug x (ηf + f )] Substituting into the boundary layer x momentum equation we get f + αff + β(1 f 2 ) = where να = g(ug) x, νβ = g 2 U x Thus f will satisfy an ordinary differential equations provided α and β are constant. Otherwise the equations depend on x and η. Hence we assume that α and β. Now (g 2 U) x = 2g(gU) x g 2 U x = ν(2α β) If 2α = β then g 2 U = constant = νc. Hence g 2 U x g 2 U = β/c U = Beβx/C where B is a constant. These type of flows are of no physical interest. Hence we consider the cases in which 2α β. Then (g 2 U) x = ν(2α β) g 2 U = ν(2α β)x

11 where we can choose the constant of intergration equal to zero by shifting the origin. Now g 2 U g 2 U x = where U(x = L) = U. Now β (2α β)x = λ ( x ) λ x U = U L [ ] 1/2 ν(2α β)l ( x ) (1 λ)/2 g 2 U x = νβ g = L If α = 1 then β = 2λ λ + 1 and the Falkner-Skan solutions are given by U = U ( x L U ) ( ) λ 1/2 ( ) 1/2 λ + 1 U, η = y 2 νx 11 where f satisfies with ψ = ( ) 1/2 2 (Uνx) 1/2 f(η) λ + 1 f + ff + β(1 f 2 ) = f = f = at η = f 1 as η 7 Inviscid flow in a sector Using complex theory, consider the complex potential F (z) = φ + iψ = Uz n Substituting z = re iθ we get get φ = Ur n cos(nθ), ψ = Ur n sin(nθ) From this it is evident that ψ = when θ =, π/n. Now u iv = Unz n 1 = (nur n 1 cos(nθ) + inur n 1 sin(nθ))e iθ Decomposing u, v into u r and u θ componets we get u = u r cos θ u θ sin θ, v = u r sin θ + u θ cos θ

12 12 Thus u iv = (u r iu θ )e iθ Hence comparing we get u r = nur n 1 cos(nθ), u θ = nur n 1 sin(nθ) From ψ and u θ expressions, it is clear that θ =, π/n constitutes the sector. Also for < θ < π/2n, u r is positive and for π/2n < θπ/n, u r is positive. u θ remains negative throughout. 8 Falkner-Skan solutions contd. Note that for we have F (z) = cz γ+1 u r = (γ + 1)Ur γ cos((γ + 1)θ), u θ = (γ + 1)Ur γ sin((γ + 1)θ) Since u = u r cos θ u θ sin θ, v = u r sin θ + u θ cos θ we say that near y =, θ is near and hence we have u u r, v i.e. uniform flow along the x direction. For y = we get u = (γ + 1)Ux γ, v = [Note: The u-velocity is uniform and function of x and v = near y =. Away from x axis both u and v could be of comparable magnitude.] To find the angle of the wedge 2φ corresponding to the flow U = U ( x L ) λ we note that 2φ + 2π/(λ + 1) = 2π φ = λπ λ + 1 In the special case of λ = we recover the Blasius solution discussed before.

13 13 y x π/(λ+1) For λ = 1 we have φ = π/2 which is the stagnation point flow. Now β = 1 and the equations are f + ff + 1 f 2 = subject to f() = f () =, f ( ) = 1 In this case g(x) is constant and hence the boundary layer has constant thickness. Boundary layer flow in a convergent channel: If we take α = and β = 1 then we have (g 2 U) x = ν, g 2 U x = ν Thus we get U x /U = 1 x U = C x For C > we get g = ν/cx The solution for U corresponds to potential flow in a convergent channel. It corresponds to F (z) = C log z Thus the solution obtained here corresponds to a boundary layer on the wall of a convergent channel. For C <, solution for g is not possible. Thus boundary layer in a divergent channel separate (more later) due to the adverse pressure gradient.

14 14 Thus for α =, β = 1 and C > we get f + 1 f 2 = subject to Introduce f = F, then f() = f () =, f ( ) = 1 F = F 2 1 The boundary condition on F is that F () = and F ( ) = 1. Its solution is given by f = F = 3 tanh 2 { η/ 2 + tanh 1 ( 2/3 )) 2 9 Flow separation We have seen that in the boundary layer p/ y = thus variation of pressure with x in the boundary layer must be that of outer stream i.e. p(x) = p o (x). Thus pressure is impressed on the boundary layer from outside. Let us consider the boundary layer equation. Since u = v = at y = we have ( ) 1 dp o 2 µ dx = u y 2 Thus at the body surface 2 u/ y 2 has the same sign as that of dp o /dx. On the other hand, in the exterior edge of the boundary layer, we must have 2 u/ y 2 < since 2 u/ y 2 from below as the profile matches smoothly with the exterior flow. At an x station with favourable pressure gradient (dp o /dx < ) has 2 u/ y 2 < throghout the boundary layer. y= On the other hand, for x station with adverse pressure gradient (dp o /dx > ) it has an inflexion point between the crossover of 2 u/ y 2 < and 2 u/ y 2 >. Note the pressure on a cylinder is in a uniform stream is given by p = p ρu 2 (1 4 sin 2 θ) Thus pressure is maximum at θ =, π and minimum at θ = π/2, 3π/2. Thus betwen θ = π and θ = π/2 we have favourable pressure gradient and between π/2 < θ < unfavourable. Similar characteristic observed for a flow over a blunt body. As one

15 15 y dp /dx < o y dp o /dx > u o u o u u proceeds along the further along the boundary layer, the profile becomes so distorted that we have ( ) u = y y= at the surface at the point of separation (point S in the figure). Beyond the separation point we have recirculation and the boundary layer becomes thicker. Thus boundary layer approximation becomes invalid after the separation. 1 Approximate solution of boundary layer eq. In the previos discussion about the solutions of boundary layer, the solutions obtained are all exact in the sense that a similarity form reduce the problem to a nonlinear ordinary differential equations that could be solved to a high degree of accuracy. For the cases where exact solutions does not exist, we sought approximate solutions of the problem. One of the classical method widely used is Kármán-Pohlhausen method.

16 16 The boundary layer equations are u x + v y = uu x + vu y = UU x + νu yy Usually we find a functional form of solution which satisfy these equations at each point in space. If such a solution can not be found, it might be possible to satisfy these equations on the average rather than at each point of the space. Thus if the momentum boundary layer equations is integrated with respect to y across the boundary layer, then the resulting equation represnt balance between average intertia and average viscous forces for each x station. Then a velocity distribution may be found which that satisfies this average balance of forces but that does not satisfy the balance at each point across the boundary layer. 1.1 General momentum integral Performaing manipulation of the uu x term the boundary layer equations can be written as u x + v y = (u 2 ) x + (uv) y = UU x + νu yy Integrating the momentum equation across the boundary layer and using the BCs u(x, ) =, u(x, δ) = U, µ u y (x, ) = τ, u (x, δ) = y (where τ is the surface shear stress) gives δ (u 2 ) x dy + Uv(x, δ) = U x δ U dy τ ρ Note that since U is a function of x, it can be taken outside the integral. Integrating the continuity equation we get and hence we get δ (u 2 ) x dy U δ u v(x, δ) = x dy δ u x dy = U x δ U dy τ ρ

17 17 Using Leibnitz s rule, we get d dx δ (u 2 ) dy U 2 dδ dx U d δ u dy + +U 2 dδ δ dx dx = U x U dy τ ρ or d δ (u 2 ) dy U d δ δ u dy = U x U dy τ dx dx ρ Rewriting the second integral the momentum integral equation becomes d dx δ or combining terms we get (u 2 ) dy d δ Uu dy + du δ u dy = du δ U dy τ dx dx dx ρ d δ u(u u) dy + du dx dx δ (U u) dy = τ ρ or replacing δ by (since the integrand is zero for y > δ) d [U 2 u dx U (1 u ] U ) dy + du dx U (1 u U ) dy = τ ρ Introducing displacement and momentum thickness we get d dx (U 2 δ 2 ) + du dx Uδ 1 = τ ρ For any assumed form of the velocity profile across the boundary we can calculate δ 1, δ 2 and τ from their definitions. Then the above equation will provide ODE for the boundary layer thickness. 1.2 Kármán-Pohlhausen approximation The velocity profile is represented by a fourth order polynomial. Let η = y/δ(x) and u/u = a + bη + cη 2 + dη 3 + eη 4 The coefficient are in general function of x. The boundary conditions are u(x, ) =, u(x, δ) = U(x), u (x, δ) = y 2 u y = U(x) du 2 ν dx, 2 u (x, δ) = y2

18 18 In terms of u/u and η these BCs become u/u =, 2 (u/u) η 2 = δ2 ν u/u = 1, (u/u) η Imposing the boundary conditions we get The solution is given by = a Λ = 2c du dx = Λ(x), on η = = 2 (u/u) η 2 = on η = 1 1 = a + b + c + d + e = b + 2c + 3d + 4e = 2c + 6d + 12d a =, b = 2 + Λ/6, c = Λ/2, d = 2 + Λ/2, e = 1 Λ/6 Thus we can write the velocity profile in the form u/u = (2η 2η 3 + η 4 ) + Λ 6 (η 3η2 + 3η 3 η 4 ) or u/u = 1 (1 + η)(1 η) 3 + Λ 6 η(1 η)3 = F (η) + ΛG(η) The function F and G are shown in the figure. F is a monotonic increasing function and ranges from zero at η = to one at η = 1. On the other hand, G increases from zero at η = to a maximum of.17 approximately at η =.25 and then decreases to zero at η = 1. For Λ =, the plot of u/u provides the velocity profile representation by a fourth order polynomial over a flat surfrace. For values of Λ > 12 we observe that u/u > 1 which is not physically possible. Thus Λ < 12. Also for Λ < 12 we observe u/u < i.e. reverse flow. Although reverse flow occurs in reality, the assumption of boundary layer theory is not valid. Hence we require that 12 < Λ(x) < 12 Next we calculate the boundary layer thicknesses and the shear stress on the surface. Thus δ 1 = δ 1 ( 3 (1 u/u) dy = δ 1 Λ ) 2

19 G(t) F(t) and 1 ( u 37 δ 2 = δ (1 u/u) dy = δ U 315 Λ ) 945 Λ2 972 The surface shear stress is given by τ = µu δ (u/u) η = µu η= δ ( 2 + Λ ) 6 In the above expression boundary layer thickness δ(x) is still unknown. Now the momentum integral is written as or Uδ 2 ν 1 2 U d dx dδ 2 dx + (2δ 2 + δ 1 ) δ 2 du ν dx = τ δ 2 µu ( ) ( δ δ ) 1 δ 2 2 du ν δ 2 ν dx = τ δ 2 µu

20 2 Note that the nondimensional pressure parameter is Λ(x) = δ2 ν Using the previous result we get From the expression of δ 1 and δ 2 we get δ 2 2 ν du dx δ2 2 du ν dx = δ2 2 δ Λ 2 ( du 37 dx = 315 Λ ) Λ2 Λ = K(x) 972 δ 1 δ 2 = ( 3 ) Λ 1 2 ( 37 Λ 315 Λ ) = f(k) Note that δ 1 /δ 2 is a function of Λ and hence function of x. Also K in the previous expression is function of x. Hence we can treat δ 1 /δ 2 as function of K. From the expression of τ and δ 2 we get ( τ δ 2 µu = 2 + Λ ) ( Λ ) 945 Λ2 972 = g(k) Substituting these results into the momentum integral equations we get 1 2 U d ( ) δ [2 + f(k)] K = g(k) dx ν Introduce Z = δ 2 /ν as new dependent variavle and noting that we get K = ZdU/dX. Thus we have U dz dx δ 2 /ν du dx = K = 2[g(K) (2 + f(k))k] = H(K) where H(K) and K are both functions of Λ. For a given value of Λ we can calculate K and H(K). Thus we can plot H(K) against K. The function H(K) is approximately linear over the range of interest and we approximate H(K) by Hence we get U dz dx H(K) =.47 6K =.47 6K =.47 6Z du dx d dx (ZU 6 ) =.47U 5

21 21 Thus momentum integral reduces to This gives Z(x) =.47 U 6 δ 2 2 =.47ν U 6 x x U(ζ) 5 dζ U(ζ) 5 dζ The above method is applied to the boundary layer flow as follows. First we solved the potential flow problem to yield the outer velocity U(x). This U is used in the equation just derived to find δ 2. Next the pressure parameter Λ is obtained from δ 2 2 ν Having obtained Λ, we use to find δ and ( du 37 dx = 315 Λ ) Λ2 Λ 972 ( 37 δ 2 = δ 315 Λ ) 945 Λ2 972 ( 3 δ 1 = δ 1 Λ ) 2 to find δ 1. Next we compute u/u and surface shear stress. In practice it is difficult to find Λ for a given U(x) unless Λ is a constant. It is therefore much simpler to choose specific function Λ(x) and use the above equation to find U(x). The shape of the boundary will be fixed by U(x). As an application consider the flow over a flat surface in which U is a constant. The integral for δ 2 gives δ2 2 =.47 νx U Since U is constant Λ =. We also have δ 2 = δ Thus combining the last two equations we get δ. We also obatined δ 1 and τ. These values compare favourably with the exact solutions.

5.8 Laminar Boundary Layers

5.8 Laminar Boundary Layers 2.2 Marine Hydrodynamics, Fall 218 ecture 19 Copyright c 218 MIT - Department of Mechanical Engineering, All rights reserved. 2.2 - Marine Hydrodynamics ecture 19 5.8 aminar Boundary ayers δ y U potential