Lecture 12 - Marine Hydrodynamics Lecture 12 3.14 Lifting Surfaces 3.14.1 2D Symmetric Streamlined Body No separation, even for large Reynolds numbers. U stream line • Viscous effects only in a thin boundary layer. • Small Drag (only skin friction). • No Lift. 1 2.20 - Marine Hydrodynamics, Spring 20052.203.14.2 Asymmetric Body (a) Angle of attack α, chord line U α (b) or camber η(x), chord line mean camber line U (c) or both amount of camber U α angle of attack mean camber line chord line Lift ⊥ to Uand Drag to U 23.15 Potential Flow and Kutta Condition From the P-Flow solution for flow past a body we obtain P-Flow solution, infinite velocity at trailing edge. Note that (a) the solution is not unique - we can always superimpose a circulatory flow without violating the boundary conditions, and (b) the velocity at the trailing edge → ∞. We must therefore, impose the Kutta condition, which states that the ‘flow leaves tangentially the trailing edge, i.e., the velocity at the trailing edge is finite’. To satisfy the Kutta condition we need to add circulation. Circulatory flow only. Superimposing the P-Flow solution plus circulatory flow, we obtain Figure 1: P-Flow solution plus circulatory flow. 33.15.1 Why Kutta condition? Consider a control volume as illustrated below. At t = 0, the foil is at rest (top control volume). It starts moving impulsively with speed U (middle control volume). At t =0+ , a starting vortex is created due to flow separation at the trailing edge. As the foil moves, viscous effects streamline the flow at the trailing edge (no separation for later t), and the starting vortex is left in the wake (bottom control volume). t = 0 +t = 0for later t Kelvin’s theorem: After a while the ΓS Γ Γ Γ Γ S S S S U U no Γ starting vortex left in wake Γ = 0 starting vortex due to separation (a real fluid effect, no infinite vel of potetial flow) dΓ =0 → Γ = 0 for t ≥ 0ifΓ(t =0)=0 dt in the wake is far behind and we recover Figure 1. 43.15.2 How much ΓS ? Just enough so that the Kutta condition is satisfied, so that no separation occurs. For example, consider a flat plate of chord  and angle of attack α, as shown in the figure below. chord length Simple P-Flow solution Γ= πlU sin α L = ρUΓ= ρU2πl sin α |L|CL = =2π sin α ≈ 2πα for small α1 ρU2l � �� � 2 only for small α However, notice that as α increases, separation occurs close to the leading edge. Excessive angle of attack leads to separation at the leading edge. When the angle of attack exceeds a certain value (depends on the wing geometry) stall occurs. The effects of stalling on the lift coefficient (CL = ρU2 L span) are shown in the 1 2 following figure. 5C L This region independent of R, ν used only to get Kutta condition stall location f(R) stall 2π α O(5o ) • In experiments, CL < 2πα for 3D foil - finite aspect ratio (finite span). • With sharp leading edge, separation/stall to early. round leading edge to forstall sharp trailing edge to develop circulation stalling 63.16 Thin Wing, Small Angle of Attack • Assumptions – Flow: Steady, P-Flow. – Wing: Let yU (x), yL(x) denote the upper and lower vertical camber coordinates, respectively. Also, let x = /2, x = −/2 denote the horizontal coordinates of the leading and trailing edge, respectively, as shown in the figure below. y=yU(x) For thin wing, at a small angle of attack it is yU , yL << 1   dyU dyL , << 1 dx dx The problem is then linear and superposition applies. Let η(x) denote the camber line 1 t(x) η(x)= (yU (x)+ yL(x)),2and t(x) denote the half-thickness η(x) Camber line t(x) 1 t(x)= (yU (x) − yL(x)). 2For linearized theory, i.e. thin wing at small AoA, the lift on the wing depends only on the camber line but not on the wing thickness. Therefore, for the following analysis we approximate the wing by the camber line only and ignore the wing thickness. 7� • Definitions In general, the lift on the wing is due to the total circulation Γ around the wing. This total circulation can be given in terms due to a distribution of circulation γ(x) (Units: [LT −1]) inside the wing, i.e., /2 Γ= γ(x)dx −/2 γ (x) Γ U Noting that superposition applies, let the total potential Φ for this flow be expressed as the sum of two potentials Φ= −Ux + φ � �� � ���� Free stream Disturbunce potential potential The flow velocity can by expressed as v = ∇Φ=(−U + u, v) where (u, v) are given by ∇φ =(u, v) and denote the velocity disturbance, due to the presence of the wing. For linearized wing we can assume u v u, v << U ⇒ , << 1 U U Consider a flow property q, such as velocity, pressure etc. Then let qU = q(x, 0+) and qL = q(x, 0−) denote the values of q at the upper and lower wing surfaces, respectively. 8�� � � � � � � � � � � • Lift due to circulation Applying Bernoulli equation for steady, inviscid, rotational flow, along a streamline from ∞ to a point on the wing, we obtain 1 � � p − p∞ = − ρ |v|2 − U2 ⇒ 2 p − p∞ = − 1 ρ (u − U)2 + v 2 − U2 = − 1 ρ(u 2 + v 2 − 2uU ) ⇒ 2 2 1 u v v p − p∞ = − ρuU( + −2)2 U U u ���� ���� ���� <<1 <<1 ∼1 Dropping terms of order Uu , Uv << 1 we obtained the linearized Bernoulli equation for thin wing at small AoA p − p∞ = ρuU Integrating the pressure along the wing surface, we obtain an expression for the total lift L on the wing l/2 �� � � �� L = (p − p∞)nydS = p(x, 0−) − p∞ − p(x, 0+) − p∞ dx −l/2 l/2 l/2 L = p(x, 0−) − p(x, 0+) dx = ρU u(x, 0−) − u(x, 0+) dx (1) −l/2 −l/2 9� � �� � � To obtain the total lift on the wing we will seek an expression for u(x, 0±). Consider a closed contour on the wing, of negligible thickness, as shown in the figure below. γ (x) )0,( +xu 0→t )0,( −xu xδ x In this case we have γ(x)δx = |u(x, 0+)|δx + u(x, 0−)δx ⇒ γ(x)= |u(x, 0+)| + u(x, 0−) For small u/U we can argue that u(x, 0+) ∼= −u(x, 0−), and obtain γ(x) u(x, 0±)= ∓ (2)2 From Equations (1), and (2) the total lift can be expressed as l/2 L = ρU γ(x)dx = ρUΓ −l/2 =Γ The same result can be obtained from the Kutta-Joukowski law (for nonlinear foil)