Lecture 20 Notes: Classical Aerofoil Theory

18.354J Nonlinear Dynamics II: Continuum Systems 20 Lecture 20 Classical aerofoil theory We now know that through conformal mapping it is possible to transform a circular wing into a more realistic shape, with the bonus of also getting the corresponding inviscid, irrotational flow field. Let’s consider some more realistic shapes and see what we get. 20.1 An elliptical wing First let’s rotate our cylinder by an angle α. The complex potential becomes   R2 iα iΓ −iα w(z) = u0 z e + e ln z. − z 2π (486) Now, using the Joukowski transformation we want to turn our circular wing into an elliptical wing. The transformation stipulates that Z = z + c2 /z, so that r Z Z2 z± (Z) = ± − c2 . (487) 2 4 Considering z+ (Z)  w(z) = u0 z+ (Z)e −iα  R2 iα iΓ + − ln z+ (Z). e 2π z+ (Z) (488) If we choose c = R, then the ellipse collapses to a flat plate. The velocity components in the Z plane are dW i(Γ + 2πu0 Z sin α) √ U − iV = = u0 cos α − , (489) dZ 2π Z 2 − 4R2 which can also be written as dw/dz U − iV = = dZ/dz  u0 e−iα − 1−  eiα R2 z2 R2 /z 2 − iΓ 2πz On the surface of the body we have z = Reiθ , so the velocities become
−iθ u0 e−iα − e−2iθ eiα − iΓe 2πR U − iV = . 1 − e−2iθ (490) (491) At θ = 0 and θ = π we are in trouble because the velocities are infinite. Notably, however, this problem can be removed at θ = 0 if the circulation is chosen so that the numerator vanishes   iΓ e−iθ u0 (ei(θ−α) − e−i(θ−α) ) − 2πR U − iV = . (492) 1 − e−2iθ Thus for a finite velocity at θ = 0 we require u0 (e−iα − eiα ) − iΓ = 0, 2πR (493) giving Γ = −4πu0 R sin α. (494) In this case flow leaves the trailing edge smoothly and parallel to the plate. Note that it is not possible to cancel out singularities at both ends simultaneously, as we have to rotate in the opposite direction to cancel out the singularity at θ = π. 20.2 Flow past an aerofoil What if we could now construct a mapping with a singularity just on one side ? This we can do by considering a shifted circle, that passes through z = R but encloses z = −R. In this case we obtain an aerofoil with a rounded nose but a sharp trailing edge. The boundary of the appropriate circle is prescribed by z = −λ + (a + λ)eiθ , (495) where θ is a parameter. First we must modify the complex potential for flow past a cylinder to take account of this new geometry. We have that   iΓ (R + λ)2 iα e − ln(z + λ). (496) w(z) = u0 (z + λ)e−iα + (z + λ)2 2π T po find the complex potential for the aerofoil one must then substitute in z = Z/2 + Z 2 /4 − R2 . Determining the velocities as before we find that dw/dz dW = = u0 u − iv = dZ dZ/dz e−iα −  R+λ z+λ 1− 2 − iΓ 2π(z+λ) R2 z2 . (497) The value of Γ that makes the numerator zero at the trailing edge is Γ = −4πu0 (R + λ) sin α. (498) The flow is then smooth and free of singularities everywhere (because we have successfully trapped the rogue singularity inside the wing), and this is an example of the Kutta-Joukowski condition at work. Our goal in the reminder of this part is to show that our earlier results FL = ρ`u0 Γ and FD = 0 are unaffected by the wing shape. To this end, we first derive Blasius’ lemma and then the Kutta-Joukowski theorem. 20.3 Blasius’ lemma To derive Blasius’ lemma, we consider the force acting per unit length on the wing, f = F /` = (fx , fy ), which is obtained by integrating the pressure over the (now arbitrary) surface contour ∂S I f =− p n ds (499) ∂S where n is the outward surface normal vector and ds the arc length. Denote by dz = dx+idy a small change along the curve ∂S. In complex notation, the normal element n ds can then be expressed as −idz = dy − idx, (500) and Eq. (499) can be rewritten as I f := fx + ify = i p dz. (501) From Bernoulli we have that p = p0 − ρ|v|2 /2, where v = vx − ivy = dw , dz |v|2 = vv̄ = vx2 + vy2 = dw dz 2 . (502) Thus, we find f =i I   ρ ρ p0 − |v|2 dz = −i 2 2 I |v|2 dz (503) Taking the complex conjugate, we have ρ f¯ = fx − ify = i 2 I |v|2 dz̄. (504) Furthermore, since v is parallel to z on boundary (which is a stream line), we may use 0 = vx dy − vy dx (505) to rewrite v 2 dz = (vx − ivy )2 (dx + idy) = (vx2 − vy2 − 2ivx vy )(dx + idy) = vx2 dx − vy2 dx − 2ivx vy dx + vx2 idy − vy2 idy + 2vx vy dy = vx2 dx − vy2 dx − 2ivx2 dy + vx2 idy − vy2 idy + 2vy2 dx = vx2 dx + vy2 dx − vx2 idy − vy2 idy = (vx2 + vy2 )dx − (vx2 + vy2 )idy = (vx2 + vy2 )(dx − idy) = |v|2 dz̄ (506) Hence, ρ f¯ = i 2 I ρ v dz = i 2 2 I  dw dz 2 dz (507) which is the statement of the Blasius lemma. 20.4 Kutta-Joukowski theorem We now use Blasius’ lemma to prove the Kutta-Joukowski lift theorem. For flow around a plane wing we can expand the complex potential in a Laurent series, and it must be of the form dw a−1 a−2 (508) = a0 + + 2 + ... z dz z Higher powers of z cannot appear if the flow remains finite at |z| → ∞ and, in this case, we can identify a0 = vx (∞) − ivy (∞). (509) In particular, if the wing moves along the x-axis and surrounding gas is at rest, then simply a0 = vx (∞). To obtain the physical meaning of a−1 , we note that by virtue of the residues theorem26 I 1 dw a−1 = dz (510) 2πi dz 26 Let’s assume some otherwise analytic function f (z) has a pole at z = 0. The residue is the coefficient P k a−1 of the Laurent series f (z) = ∞ k=−∞ ak z . The residue theorem states that for a positively oriented simple closed curve γ around z = 0 I f (z)dz = 2πi a−1 . γ Computing the integral on the rhs. gives I I dw dz = (vx − ivy )(dx + idy) dz I I = (vx dx + vy dy) + i (vx dy − vy dx) The last integral vanishes as the boundary is stream line, see Eq. (505), so that I 1 Γ a−1 = v · dx = . 2πi 2πi (511) (512) where Γ is the circulation defined above. To evaluate the rhs. in Eq. (507), we note that to leading order  dw dz 2 ' a0 + 2 a0 Γ a0 a−1 = a0 + z πiz Thus, using the residue theorem, we find   ρ a Γ 0 f¯ = fx − ify = i 2πi = iρΓa0 = ρΓvy (∞) + iρΓvx (∞). πi 2 (513) (514) Recall that FD = `fx and FL = `fy , this is indeed the generalization of our earlier results for drag and lift on a cylinder, if we identify vy (∞) = 0 and vx (∞) = −u0 . Note that the results FD = 0 is again a manifestation of d’Alembert’s paradox (now for arbitrarily shaped wings), which can be traced back to the fact that we neglected the viscosity terms in the Navier-Stokes equations.