4. INTEGRAL ANALYSIS OF THE BOUNDARY LAYER SPRING 2009 4.1 Characteristic integral lengths 4.2 The momentum-integral relation 4.3 Application to laminar flow 4.4 Application to turbulent flow Examples The integral analysis of boundary layers was originally proposed by Theodore von Kármán and K. Pohlhausen in separate papers in 1921. boundary-layer equivalent profile displaced profile 4.1 Characteristic Integral Lengths streamline Displacement Thickness δ h h h+ * The flow near the surface is retarded, so that the streamlines must be displaced δ* outwards to satisfy continuity. To reduce boundary the total mass flow rate of a frictionless fluid by the same amount, the surface y y would have to be displaced outward by a distance *, called the displacement thickness : ∞ ⌠ U * = ( U − U ) dy = mass flux deficit e e e e Ue Ue ⌡0 δ* Momentum Thickness equal areas The loss of momentum flux for the mass flux U dy between adjacent streamlines is − × (U e U ) U dy . Hence the total loss of momentum flux is equivalent to the removal of momentum through a distance , called the momentum thickness : ∞ 2 = ⌠ − = eU e (U e U ) U dy momentum f lux defici t ⌡0 For incompressible flow (with uniform free-stream density) these take particularly simple forms: ∞ ⌠ U Displacement thickness : * = 1( − ) dy (1) ⌡ 0 U e ∞ ⌠ U U Momentum thickness : = 1( − ) dy (2) ⌡ 0 U e U e Turbulent Boundary Layers 4 - 1 David Apsley Shape Factor The ratio of displacement thickness to momentum thickness is called the shape factor H: * Shape factor : H ≡ (3) U U U Since 1− > 1( − ) the shape factor is always greater than 1. U e U e U e A large shape factor is an indicator of a boundary layer near separation. small large y H y H U 4.2 The Momentum-Integral Relation 4.2.1 Zero-Pressure-Gradient Boundary Layer streamline CONTROL VOLUME τw 0 x For a zero-pressure-gradient boundary layer ( e and Ue constant) the rate of loss of momentum equals the total drag on the surface up to that point; i.e., per unit width: ⌠ x U 2 = D(x) = dx 1e2e3 123 w ⌡0 momentum flux deficit total drag force Differentiating: d ( U 2 ) = dx e e w 2 Dividing by eU e we can also express it in a simpler non-dimensional form: d c = f dx 2 Integrating, gives the total drag coefficient (= average skin-friction coefficient) over length L: 2 (L) c = D L The drag coefficient can then be deduced entirely from the downstream velocity profile. Turbulent Boundary Layers 4 - 2 David Apsley 4.2.2 General Case The following derivation applies to a steady, compressible or incompressible, 2-dimensional boundary layer with arbitrary free-stream pressure gradient. For complete generality we will allow for a wall transpiration velocity Vw as well as a wall shear stress w. Start with the integral momentum flux deficit: ∞ 2 = ⌠ − eU e (U e U) U dy ⌡0 Differentiate with respect to x, taking the differential operator through the integral sign and using the product rule, ∞ ∂ d 2 = ⌠ {}− ( eU e ) U(U e U ) dy dx ⌡0 ∂x ∞ ⌠ ∂( U ) dU ∂U = (U −U ) + U e − U dy ∂ e ∂ ⌡0 x dx x In the first underlined term substitute from the continuity equation: ∂( U ) ∂( V ) = − ∂x ∂y In the second underlined term substitute from the boundary-layer momentum equation: ∂U ∂U dU ∂ U = − V + U e + ∂x ∂y e e dx ∂y Then ∞ d ⌠ ∂( V ) dU ∂U dU ∂ ( U 2 ) = − (U −U ) + U e + V − U e − dy e e ∂ e ∂ e e ∂ dx ⌡0 y dx y dx y ∞ ⌠ ∂( V ) ∂U dU ∂ = − (U −U ) + V − ( U − U ) e − dy ∂ e ∂ e e ∂ ⌡0 y y dx y Collecting y derivatives together, and simplifying using the product rule and the fact that Ue is independent of y, ∞ ∞ d ⌠ ∂ dU ⌠ ( U 2 ) = − {}V (U −U ) + dy − e ( U − U ) dy e e ∂ e e e dx ⌡0 y dx ⌡0 ∞ d dU ⇒ 2 = − − + − e ( eU e ) V (U e U ) ( eU e *) dx 0 dx Since Ue – U and both vanish at y = ∞, whilst U = 0, V = Vw and = w at y = 0, d dU ( U 2 ) = V U + − e ( U *) dx e e w w e w dx e e i.e. d dU ( U 2 ) + U e * = + V U dx e e e e dx w w w e Turbulent Boundary Layers 4 - 3 David Apsley Kármán’s integral relation d dU ( U 2 ) + U e * = + V U (4) dx e e e e dx 12w3 142w w43e 14243 1424 434 wall wall rate of loss effect of free -stream stress transpirat ion of momentum pressure gradient For constant-density flow the most compact form comes from expanding the first differential 2 and dividing by U e : d dU e c f Vw + 2( + H ) = + (5) dx U e dx 2 U e where c = w skin-friction coefficient f 1 2 2 U e * H = shape factor Notes . (1) The results also hold along a curved-wall boundary layer, provided that the radius of curvature is much greater than the thickness of the boundary layer. (2) The integral relations hold whether the boundary layer is laminar, turbulent or transitional. (3) The integral relation is exact only for the correct mean velocity profile. However, the power of the scheme is that it provides an excellent estimate of drag with only a half-decent approximation for U(y). (4) For a flat-plate boundary layer with zero pressure gradient and no wall transpiration, Kármán’s integral relation takes the particularly simple form: Zero pressure-gradient boundary layer : d c = f (6) dx 2 D(L) ⌠ L In this case the total-drag coefficient c (L) = where D(L) = dx is given by D 1 2 w 2 U e L ⌡0 (L) c (L) = 2 (7) D L The total drag is entirely determined by the downstream momentum thickness . (5) The shape factor H is only relevant when there is a free-stream pressure gradient (d Ue/d x ≠ 0). A large shape factor indicates a boundary layer approaching separation. (6) An alternative proof of Kármán’s integral relation – starting from the differential boundary-layer equations – is given in the Examples Section. Turbulent Boundary Layers 4 - 4 David Apsley 4.3 Application to Laminar Flow A fully-developed profile may be approximated by: y U f ( ), y ≤ U = e (8) > U e y where f )0( = ,0 f ′ )0( ≠ 0 or ∞, f )1( = ,1 f ′ )1( = 0 . (Why is the second condition important?) The integral boundary-layer depths are proportional to : * = A , = B (9) and the skin-friction coefficient dU dy y=0 2 f ′ )0( C c = = = (10) f 1 2 2 U e U e Re where A, B and C are constants determined by the particular profile adopted: ⌠1 ⌠1 A = 1( − f ) d , B = f 1( − f ) d , C = 2 f ′ )0( (11) ⌡0 ⌡0 Substituting the expressions for and cf into the integral relation: 2 d c f d C d C = ⇒ B = ⇒ = dx 2 dx 2U e dx B U e Hence: 2/1 2/1 2/1 = C x = C − 2/1 or Re x (12) B U e x B (x is measured from some virtual origin; in practice, this can be taken as the leading edge). d 1 Since ∝ x 2/1 we have = and hence dx 2 x d c = 2 = f dx x Summary of Results For a Zero-Pressure-Gradient Laminar Boundary Layer A * = A , = B , H = B c = , c (L) = 2c (L) f x D f where ⌠1 ⌠1 A = 1( − f ) d , B = f 1( − f ) d , C = 2 f ′ )0( ⌡0 ⌡0 C 2/1 x = B U e Turbulent Boundary Layers 4 - 5 David Apsley Thus, assuming only that the profile is self-similar , but without any knowledge about its precise form, we can confirm that ∝ x 2/1 (or ∝ Re − 2/1 ) and c (L) ∝ Re − 2/1 x x D L The simplest profile satisfying U = 0 on y = 0 dU U = U , = 0 on y = e dy is (the channel-flow profile) = y − y = − 2 U U e 2( ) or f ( ) 2 = 1 = 2 = This has A 3 , B 15 , C 4 and gives the following approximations. Approximation Blasius solution x x * 83.1 72.1 U e U e x x .0730 .0664 U e U e H 2.5 2.59 In both cases, c = f x = cD (L) 2c f (L) Thus, the drag is correct to within 10% and the shape factor to within 3.5%, even for a simple approximation. Much better approximations are given in the Examples section. Turbulent Boundary Layers 4 - 6 David Apsley 4.4 Application to Turbulent Flow The momentum integral relation for a zero-pressure-gradient boundary layer is d c = f dx 2 To make this tractable we adopt the power-law approximations that we have met before: U y Velocity profile: = ( ) 7/1 (13) U e − 6/1 U e = ¡ = Skin-friction coefficient: c f .0 0205 Re (where Re ) (14) From the first of these we can compute displacement and momentum thickness: U 7/1 = , = y / U e Displacement thickness: ¢ ⌠ U ⌠1 1 * ≡ 1( − ) dy = 1( − 7/1 ) d = (15) ⌡ 0 U e ⌡0 8 Momentum thickness: ¢ ⌠ U U ⌠1 7 ≡ 1( − ) dy = 7/1 1( − 7/1 ) d = (16) ⌡ 0 U e U e ⌡0 72 Shape factor: * 9 H ≡ = (17) 7 Substituting for skin-friction coefficient cf and momentum thickness in the momentum integral relation, 7 d 6/1 = .0 01025 ( ) 72 dx U e 7 6/1 6/1 ⇒ d = .0 01025 ( ) dx 72 U e 7 6 6/7 6/1 ⇒ × = .0 01025 ( ) x 72 7 U e 6/7 6/1 ⇒ ( ) = .0 123 ( ) x U e x Hence − 7/1 7/6 = .0 166 Re or Re £ = .0 166 Re (18) x x x Thus, the boundary layer grows as ∝ x 7/6 d 6 1 Since ∝ x 7/6 we have = = , and hence: dx 7 x 12 x Turbulent Boundary Layers 4 - 7 David Apsley d 1 − c = 2 = 2× = .0 0277 Re 7/1 (19) f dx 12 x x − 6/1 = £ (Alternatively, use c f .0 0205 Re and substitute for Re .) 2 7 − c (L) = = c (L) = .0 032 Re 7/1 (20) D L 6 f L i.e.