Category Theoretical and Applied Aerodynamics

Fig. 3.22 Location of center of pressure versus a for parabolic cambered plate

(d)

xa. c. 1

c 4

The aerodynamic center is located at quarter-chord for all thin airfoils. The moment at the aerodynamic center is

independent of a.

Application:

• flat plate Cm, a.c. — 0

• parabolic plate Cm, a.c. — – nd.

Let A and B be two points located at xA and xB along the x-axis. M’A and M’B are the moments and L’ the resulting force. The following formula will be used to transfer the moment from one point to the other

M[b = M ‘a + (xb — xa ) L’ (3.79)

The elementary moment of a small aerodynamic force located at x about the leading edge reads (Fig.3.19)

dM o = – dF’ x = —pU Г ‘(x )xdx (3.74)

the minus sign is needed for the “nose up” moment to be positive. Replacing Г’ by its expansion and integrating in t from zero to n gives

Rearranging and using orthogonality of sine modes one gets

1 П П 1 П

M’o = — pU 2c2 A0 sin2 tdt + Ai sin2 tdt — A2 sin22tdt

’ 2 0 0 2 0

The moment coefficient is

Note that the moment depends on one more mode, mode 2, than lift. Application:

• moment coefficient for the flat plate: Cmo = – f a

• moment coefficient for the parabolic plate: Cm, o = – f (a + 4 f )

The lift force is obtained by application of the Kutta-Joukowski lift theorem, where one accounts for the net resulting circulation

With the elimination of sin t in the first term and using orthogonality of the sines in the second, there remains only two terms

L’ = npU2c (A0 + Ai^ (3.67)

The lift coefficient reads

L’ 2Г / A1

° = Трит = Uc = n A0+t) (368)

The important result is that the lift coefficient only depends on the first two modes, A0 which depends on a and aadapt, and A1 which reflects the contribution to the camber of a parabolic plate, mode 1.

Fig. 3.19 Drag calculation

Application:

• the lift coefficient of a flat plate is C; = 2na

• the lift coefficient of a parabolic plate is Ci = 2n(a + 2 d)

Note, that at incidence of adaptation, the lift of the flat plate is zero, but it is 4n d for the parabolic plate.

The drag is zero (d’Alembert paradox). This seems quite surprising when con­sidering the case of a flat plate at incidence a, since the integration of pressure will produce a force F’ normal to the plate, hence a drag D’ = F’ sin a ~ F’a.

The proof of zero drag can be carried out in the general case as follows. A small element of a cambered plate is subject to a pressure difference, hence a small force dF’ perpendicular to the surface element

dF’ = 1 pU2 (C~(x) – C + (x}) ndl (3.69)

where ndl = (—dz, dx). See Fig. 3.19.

The projection of the force onto the x-axis gives the drag contribution

dD = dF’x = —1 pU2 (c— (x) – C + (x)^ dz

= -2pU2 (c— (x) — C+ (x)) dZdx (3.70)

Now using Cp = —2U, < u > = Г’ and = ™(Ц/0), one can write

dD’ = —pr'(x)w(x, 0)dx (3.71)

Upon integration and replacing w(x, 0) by its expression in terms of Г’ one gets

The last expression corresponds to the integration on a square domain [0, c] x [0, c] of an antisymmetric function, i. e. replacing x by £ and vice-versa changes the integrand in its opposite, as shown in Fig. 3.20 for M and M’. The result is D’ = 0 or Cd = 0 for all thin cambered plates.

Now the question remains concerning the result of the pressure integration along the flat plate and the apparent drag calculated to be

(3.73)

where w(x, 0) has been replaced by —aU. This apparent mystery is explained by the presence of a suction force acting at the leading edge of thin cambered plates in the direction of the tangent to the camberline, whenever the solution is singular at the leading edge. This suction force is not accounted for in the integration of pressure along the plate, but is found in a momentum balance on any contour that excludes the leading edge singularity (for example by using a small circle around it). The suction force FS balances exactly the “drag” from the pressure integration, i. e. F’s = —L’a, see Fig. 3.21.

Fig. 3.21 Suction force and z

resulting lift force ‘

With d(x) = 4df(1 – c) and d'(x) = 4d(1 – 2c) = 4d cos t, as seen earlier, the Fourier coefficients are easily identified to be

d

A0 = a, A1 = 4 , An = 0, n > 2

c

hence, the vorticity distribution reads

Upon elimination of t in terms of x, the result becomes

(3.64)

Note that this result can be interpreted as the superposition of the flow past a flat plate at incidence (first term proportional to a) plus a parabolic cambered plate at zero incidence (second term proportional to d). It is interesting also to realize that the

Fig. 3.16 Cp distribution for the parabolic plate at zero incidence

Fig. 3.17 Streamlines of the flow past a parabolic plate at zero incidence

parabolic plate at zero incidence satisfies two K – J conditions, one at the trailing edge but also one at the leading edge. Indeed, the angle of adaptation is aadapt = 0. At zero incidence, the parabolic plate has lift and the Cp distributions are half ellipses, as shown in Fig. 3.16.

The flow in this case can be depicted as in Fig.3.17.

Due to the flow symmetry w. r.t the axes x = 2, the lift force will be located at x = xc. p. = §, the center of pressure. When the incidence is positive, the flow will go around the leading edge clockwise and the stagnation point will be located on the lower surface. The opposite is true if the incidence is negative.

The angle of adaptation has a practical interest for thin cambered plates: at the ideal angle of attack, the viscous effects are minimized because the flow does not have to go around the leading edge with very high velocities and then slow down abruptly, which often provokes separation. However, since it is not possible to control perfectly the incidence of the profile of a wing, thickness is added in the leading edge region that allows the stagnation point to move about the ideal position, without major penalties in viscous effects. See the sketch in Fig.3.18.

(a) (b)

For the flat plate, d'(x) = 0. The above equations for the Fourier coefficients reduce to A0 = a, Ai = … = An = 0, n > 1. The vorticity distribution reduces to the singular term

The leading edge of the flat plate is singular for a = 0. The angle of adaptation is aadapt = 0, which corresponds to the uniform flow (zero perturbation). The result for C± (x) at a = 5° is shown in Fig. 3.15.

The solution to the fundamental integral equation of thin airfoil theory has been obtained by using a change of variable that plays an important role in this theory, as well as in the Prandtl lifting line theory, as will be seen later. It is due to Glauert [1] and reads

x = §(1 – cos t), 0 < t < п (3 5

£ = §(1 – cos в), 0 < в < п (3.51)

This transformation maps a uniform distribution of points on a half circle (with steps At or Ав) by projection onto its diameter (x or £). The new distribution clusters the points near the leading and trailing edges. See Fig. 3.13.

The vorticity distribution is assumed to be of the form

1 + cos в ^

Г'[£(в)] = 2U Ao + An sin пв (3.52)

I sin в n=1 J

Fig. 3.13 Geometric interpretation of the change of variable

The first term in the bracket is singular at в = 0 (leading edge) when A0 = 0. It represents the leading edge singularity. The first term by itself represents the flow past a flat plate, as will be seen shortly. Notice that this term is regular at the trailing edge as Г ‘(в) ^ 0ase ^ n. The infinite summation, the Fourier series, is the superposition of regular solutions that satisfy two K — J conditions, at the leading and trailing edges, since all the terms vanish there. In other words, the solution proposed for Г’ enforces, by construction, the K — J condition. The solution procedure consists in calculating the Fourier coefficients A0, Ai,…,An,…using the fundamental integral equation.

Substitution of the expression for Г’ into the fundamental integral equation, and using the change of variable results in w(x, 0)

or, after some rearranging

TO

d'[x(t)] — a = —A0 + An cos nt, 0 < t < n (3.55)

n=1

The left-hand-side is well defined for 0 < t < n. The right-hand-side is a Fourier series of period 2n. It is a series of cosines, hence even in t. Knowing its value in [0, n] is sufficient to define it for all values of its argument —to < t < to. Consider, for example, a parabolic plate at incidence a. The left-hand-side is

d x d

d [x(t)]— a = 4 1 — 2 — a = 4 cos t — a (3.56)

c V c> c

This is shown in Fig. 3.14 for d = 0.03 and a = 5° = 0.087rd.

In this particular case, the Fourier coefficients are easily found by inspection:

d

A0 = a, A1 = 4-, An = 0, n > 2 (3.57)

c

This approach can be used whenever the profile camber is a polynomial of low degree, because d'(x) can be expressed as a sum of cosines by elimination of x with t through x = 2 (1 — cos t) and use of some identities to replace cos21 by cos 2t,

Fig. 3.14 Fourier series representation for a thin parabolic cambered plate at incidence

etc.. .If the profile camber is not a polynomial, or is defined by piecewise continuous functions, the Fourier coefficients have to be calculated using the general method of integration, using orthogonality of the modes as

(3.58)

(3.59)

Note that A0 is the only coefficient which depends on a, and it is of the form

A0 = a – aadapt

where aadapt = П JC d'[x(t)]dt is a geometry related constant. When a = aadapt, the singular term vanishes (A0 = 0) and the leading edge is “adapted”, in other words satisfies the K – J condition. aadapt is called the angle of adaptation or ideal angle of attack.

We consider a small element along the x-axis of length d£ with a distribution of vorticity dГ = Г'(£)d£. The contribution of this element to the perturbation stream – function at a point M (x, z) is given in Cartesian coordinates by

Upon integration, this becomes

From this we obtain the perturbation velocity field as

._дф_ 1 Г ГQz

U x’ z dz 2n J0 (x – £)2 + z2

Note that и is odd and w is even in z, hence < и > = 0 and < w >= 0 across the cut [0, c]. This is consistent with the vortex singularity, see Fig. 3.11.

For the application of the tangency condition on the thin cambered plate at inci­dence, we let z ^ 0, while 0 < x < c. The condition becomes

Fig. 3.11 Lifting flow z

induced by a vortex element

MX x,-z)

1 c Гf (£)

w(x, 0) = — —Щd£ = U(d'(x) – a (3.49)

2W0 x – £

The equation above is the fundamental integral equation of thin airfoil theory. In general the airfoil shape d(x) is given as well as the flow conditions U, a. The unknown vorticity Г'(x) is the solution of an integral equation. Once Г’ is found, which satisfies the Kutta-Joukowski condition, Г’ (c) = 0, one can obtain the velocity field everywhere and in particular u(x, 0±) and hence the pressure coefficients on the thin cambered plate, C± (x). Using the result derived earlier with the same singular integrand, the limiting process yields

, , 1 fc Г'(£)г ^ Г ‘(x) f r+

lim u(x, z) = 2 2d£ = = u(x, 0+) (3.50)

z^0+ 2n 0 (x – £)2 + z2 2

It is easy to show by symmetry that u(x, 0-) = – . Across the thin cambered

plate (or cut along the x-axis) the u-component has a jump, < u (x) >= u (x, 0+) – u(x, 0-) = Г'(x). There is also a pressure (or Cp) jump, except at the trailing edge, where the flow leaves the profile “smoothly”. In contrast, the leading edge sustains in general an infinite velocity (Г'(0) = ±to), the only exception being when the leading edge is “adapted”, in which case the flow attaches “smoothly” and satisfies a K – J condition there (Г'(0) = 0), see Fig. 3.12 for typical flow fields at the leading and trailing edges.

The mathematical form of the Kutta-Joukowski condition can be explained by the fact that, if the flow leaves the trailing edge smoothly, the pressures above and below the trailing edge must be equal, hence u+(c) = u -(c) which implies that Г'(c) = 0.

In the following, the streamfunction will be chosen as an intermediate function to solve for the flow velocity field. Similar results would be obtained with the velocity potential.

We consider a small element along the x-axis of length d£ with a distribution of source/sink dQ = Q'(£)d£. The contribution of this element to the perturbation streamfunction at a point M (x, z) is given in Cartesian coordinates by

, , Q(£)d£ t

d ф = arctan l

2n x – £,

The perturbation velocity components are obtained by taking derivatives with respect to x and z. Note that the derivatives will be carried out under the integral sign, which requires some regularity of the integrand and convergence of the integral. We shall assume that these conditions are fulfilled.

Note that u is even and w is odd in z. This is consistent with the symmetry of the flow about the x-axis. For that reason, u will be continuous across the cut [0, c], whereas w will be allowed to jump. Let < a >= a+ – a~ represent the jump of a across the cut, then < u >= 0 and < w >= 0. Since the pressure is related to u, the pressure will be continuous across the cut, and w will have a jump corresponding to the change of sign of the profile slope (± ) between the upper and lower surfaces.

See Fig. 3.8.

The tangency condition on the upper surface is applied by letting z ^ 0+ while

0 < x < c and

w(x, 0+) = Uf’+(x) = 1 Ue'(x) (3.29)

which will also, by symmetry, enforce the tangency condition on the lower surface. In the process, the integral goes to zero, almost everywhere, except at £ = x, where the integrand is of the form 0. One can show that the limit is finite and equal to

lim w(x, z) = Q'(x)

z^0+ 2

where e > 0 is a small number. The first and last integrands are regular and

z2 2 ^ 0 as z ^ 0, in other words

(x — £)2+z2

In the middle integral, the dummy variable £ is replaced by t s. t. £ = x + zt. The integral becomes

We(x, z)

Now, if Q ’ is a smooth function, it can be expanded in Taylor series as Q (x+zt) = Q(x) + ztQ"(x) + O(e2). Note that — e < zt < e. Substitution of this last result into the integral yields

Q ‘(x) z dt Q'(x) –

We(x, z) = 2 + O(e) = [arctan t] 1 e + O(e) (3.34)

2n e 1 +12 2n z

z

Letting z ^ 0+, followed by e ^ 0, gives the anticipated result.

By identification with the previous equation, the source/sink strength is found to be Q'(x) = Ue'(x). When the thickness increases, sources are used and when it decreases, sinks are used. For a closed profile the net sum of sources and sinks is zero, a consequence of e(0) = e(c) = 0.

Fig. 3.9 Limiting process e ‘ (x)

for principal value integral x – f

Now that the singularity distribution is known, the perturbation velocity field can be calculated using the above equations for u(x, z) and w(x, z). We are especially interested by the velocity on the obstacle. We know w already and u is given by

U Ґ e’ (i)

u(x, 0) = di (3.35)

2n J0 x – І

Note that the integrand is singular at І = x. The singularity is a “simple pole”. A finite value can be obtained according to the limiting process

Г-e e'(£) c e'(0 ^

——d e + – – d Є

0 x e x+e x e

In this process, the two diverging contributions cancel out because the areas, as depicted in Fig.3.9, are equal and opposite. Such integrals are called “principal value” integrals.

The proof is easily obtained by studying the following limit for the remainder

x – e e£) *+e e’(£)

r = lim d e +

e^° x-e x – e x+e2 x – І

and making use of e2 ^ e and of the Taylor expansion

(i x )2

e£) = £(x) + (І – x)e"(x) + 2 e"x) + O(e3) (3.38)

where e has been written as e = x + (e – x).

The first term in R integrates as

X — Є2 P (£)

—-d£ = e'(x) ln є — ln є2 — e"(x) є — є2 + O(є3)

which vanishes in the limit є ^ 0.

Consider as example, the flow past the symmetric wedge of semi-angle в at zero incidence. The thickness distribution is e(x) = 2вх for small values of в. This results in Q'(x) = 2Uв = const. The velocity along the wedge is

, n, Uв rc d£ u(x, 0) =

n 0 x — £

Going through the limiting process in detail, let

Uв ( Г—є d£ + fc d£

= (ln x — ln є + ln є — ln(c — x))

Note that the terms in ln є cancel out and the limit for є ^ 0 is simply

U в x

и (x, 0) = — ln———-

The pressure coefficient, Cp = — 2u (x, 0)/U is shown in Fig.3.10. Note that the areas under the curve are equal and of opposite signs. For this reason, the resulting pressure force in the x-direction (drag) is zero.

To close this section, it is worth mentioning that, although thickness does not contribute to forces and moments, it will be beneficial in mitigating adverse viscous effects due to sharp leading edges.

Fig. 3.10 Pressure coefficient along the wedge

The small disturbance assumption has allowed for the complete linearization of the problem of the flow past a thin airfoil. It is therefore possible to use linear superposition of solutions. The general problem of the flow past a thin cambered airfoil at incidence can be viewed as the sum of a symmetric problem corresponding to the flow past a symmetric profile having the same thickness distribution at zero incidence, and a lifting problem corresponding to the flow past a thin cambered plate having the same camberline and incidence. This is depicted in Fig. 3.7.

The symmetric problem will be solved by distributing sources and sinks on a cut along the x-axis, from zero to c. Due to the symmetry of the problem, there will be no net force in the z-direction and no net moment. The drag will also be zero (d’Alembert paradox).

The lifting problem will be handled by a distribution of vortices on a cut along the x-axis, from zero to c. In general, the flow will produce lift and moment, but again zero drag.