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MIT 16 01 - Fluids

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� � � � Fluids – Lecture 2 Notes 1. Airfoil Vortex Sheet Models 2. Thin-Airfoil Analysis Problem Reading: Anderson 4.4, 4.7 Airfoil Vortex Sheet Models Surface Vortex Sheet Model An accurate means of representing the flow about an airfoil in a uniform flow is to place a vortex sheet on the airfoil surface. The total velocity �V (x, z), which is the vector sum of the freestream velocity and the vortex-sheet velocity, can be forced parallel to the airfoil surface γ(sγ (s) V V V by suitably setting the sheet strength distribution ). A panel method is normally used to numerically compute γ(s). By using a sufficient number of panels, this result can be made as accurate as needed. The main drawback of such numerical calculations is that they give limited insight into how the flow is influenced by changes in the angle of attack or the airfoil geometry. Such insight, which is important for effective aerodynamic design and engineering, is much better provided by simple approximate analytic solutions. The panel method can still be used for accuracy when it’s needed. Single Vortex Sheet Model In or der to simplify the problem sufficiently to allow analytic solution, we make the following assumptions and approximations: 1) The airfoil is assumed to be thin, with small maximum camber and thickness relative to the chord, and is assumed to operate at a small angle of attack, α ≪ 1. 2) The upper and lower vortex sheets are sup erimposed together into a single vort ex sheet γ = γu + γℓ, which is placed on the x axis rather than on the curved mean camber line Z = (Zu + Zℓ)/2. zz z γ (x)γ (x) γ (x) γ (x)l x x x Z(x)Z (x)u u Z (x)l Z(x) n 3) The flow-tangency condition �ˆV · n = 0 is applied on the x-axis at z = 0, rather than on the camber line at z = Z. But the normal vector ˆn is normal to the actual camber line shape, as shown in the figure. 4) Small-angle approximations are assumed. The freestream velocity is then written as follows. �k kV∞ = V∞ (cos α)ˆı + (sin α)ˆ≃ V∞ ˆı + αˆ1� � � � � � � On the x-axis where the vor t ex sheet lies, t he sheet’s velocity w(x), which is strictly in the z-direction, is given by integrating all the contributions along the sheet. c γ(ξ) dξ w(x) = − 0 2π(x − ξ) Adding this to the freestream velocity then gives the total velocity. � �ˆc γ(ξ) dξ ˆV (x, 0) = V∞ + wk ≃ V∞ˆı + V∞α − k (1) 0 2π(x − ξ) The normal unit vector is obtained from the slope of the camberline shape Z(x). dZ k ˆ�n(x) = − ˆı +ˆ, n = �n (2) dx |�n| z γ x 0 c dξ ξ V x α VdZ/dx n Vw To force the total velocity to be parallel to the camberline, we now apply the flow tangency condition �ˆV · n = 0. Performing this dot product between (1) and (2), and removing the unnecessary factor 1/|�n| gives the fundamental equation of thin airfoil theory . dZ c γ(ξ) dξ V∞ α − − = 0 (for 0 < x < c) (3) dx 0 2π(x − ξ) Thin-Airfoil Analysis Problem Flow tangency imposition For a given camberline shape Z(x) and angle of attack α, we now seek to determine the vor-tex strength distribution γ(x) such that the fundamental equation (3) is satisfied at every x location. As shown in the figure, this will result in the total velocity �V at every x-location to be approximately parallel to the local camberline, producing a physically-correct flow about this camberline. The thinner the airfoil, the closer the camberline is to the x-axis where the flow tangency is actually imposed, and the more accurate the approximation becomes. Com-pared to typical airfoils, the height of the camberline in the figure is exaggerated severalfo ld for the sake of illustration. .V n = 0 n V 2� � Coordinate transformation To enable solution of equation (3), it is necessary to first perform a tr ig onometric substitution for the coordinate x, and the dummy variable of integration ξ. c θο x = (1 − cos θo)2 c ξ = ( 1 − cos θ)2 c x0 d ξ θθ cc/2dξ � o ) Equation (3) dZ (for 0 < θ < πθ = πdξ = sin θ dθ θ = 0 2 x As shown in the figure, θ runs from 0 at the leading edge, to π at the trailing edge. Since x and θ are interchangable, functions of x can now be t r eated as functions of θ. then becomes 1 � π γ(θ) sin θ dθ = V∞ α − (4) 2π 0 cos θ − cos θo dx where the known camberline slope dZ/dx is now considered a function of θo. This is an integral equation which must be solved for the unknown γ(θ) distribution, with the additional requirement t hat it satisfy the Kutta condition at the trailing edge point, γ(π) = 0 Symmetric airfoil case In practice, the camberline slope dZ/dx can have any arbitrary distribution along t he chord. For simplicity, we will first consider a symmetric airfoil . This has a flat camberline, with Z = 0 and dZ/dx = 0. Equation (4) then simplifies to 1 � π γ(θ) sin θ dθ = V∞α (5) 2π 0 cos θ − cos θo Solution of this equation is still formidable, and is beyond scope here. Let us simply state that the solution is 1 + cos θc − x γ(θ) = 2α V∞ or γ(x) = 2αV∞sin θ x The shape of these distributions is shown in the figure below. 4 3 γ 2 1 0 2αV 0 π/2 θ π 0.0 0.5 x/c 1.0 3� � � � � � Note that at the trailing edge, γ = 0 as required by the Kutta condition, and that at t he leading edge γ → ∞. The latter is of course not physical, although the singularity is weak (integrable), a nd the integrat ed results for cℓ and cm are in fact valid. The load distribution pℓ − pu is obtained using the Bernoulli equation, together with the tangential velocity jump properties across the vortex sheet. 1 1 12 2 u =pℓ −


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