Category Theoretical and Applied Aerodynamics

For this case, Joukowski transformation has e = 8 = 0. The flat plate is mapped into a circle with its center at the origin in the (X, Z)-plane, Fig. 2.22. The mapping can be interpreted as a conformal transformation from the cylinder plane to the physical plane such that x = Ф/U, z = Ф/U, where Ф and Ф are the velocity potential and stream function of the flow past a circular cylinder without circulation. Indeed, Eqs. (2.30) and (2.31) written in terms of X and Z read

In this transformation, the cylinder of radius a centered at the origin, which cor­responds to the stagnation streamline & = 0, is mapped onto the x-axis (z = 0), in the interval [-2a, 2a].

This case corresponds to b = a = | in Sects.2.8.1 and 2.8.2, and the results derived previously can be used.

In the mapping, the flat plate parametric representation is given by

c

x = 2acos0 = cosQ, z = 0
2

and the velocity components in the physical plane are

дф_ (since-a)+r )

dx = 2U sin в

Forcing the circulation to be zero yields an anti-symmetric flow that does not satisfy the Kutta-Joukowski condition, Fig. 2.23. The lift is zero, but the moment coefficient about the mid-plate reduces to

Making use of the mapping, the velocity on the flat plate now reads

where the minus (plus) sign corresponds to the upper (lower) surface. The stagnation points are at x = ±2 cos a, z = 0.

When the Kutta-Joukowski condition is applied, the velocity on the flat plate becomes

(2.85)

The streamline in this case leaves the trailing edge “smoothly”, Fig.2.24.

Note that the velocity is infinite at в = n (leading edge) in all cases except for a = 0. The stagnation point corresponds to в = n + 2a, i.e. x = -§cos2a. Itisalso worthy to note that the exact solution is consistent with the thin airfoil theory result that will be derived in the next chapter, since the flat plate is the ultimate thin airfoil, having zero thickness and zero camber. Another assumption that will be made is that of small a. If we map the flat plate to the segment [0, c] with x’ = | (1 + X) the horizontal velocity component becomes

cosa ± sina — ~ U – (2.86)

The plus (minus) sign corresponds to the upper (lower) surface. This last result is the thin airfoil result with a perturbation u above and below the plate that is symmetric about the x’-axis.

Fig. 2.24 Flat plate at 25c incidence (Г = 4nasina)

Fig. 2.25 Pressure coefficient along the plate at a = 10°

The pressure coefficient is given by

The result for-Cp is displayed in Fig. 2.25. It is not symmetric w. r.t. the Ox’-axis. If we anticipate again on thin airfoil theory, the linearized Cp will be symmetric, i. e. C – = —C+

p ^ p ■

The lift coefficient is obtained from the Kutta-Joukowski lift theorem

Ci = = 2n sina

2 pU 24a

It is interesting, however, to consider the integration of the pressure along the flat plate directly. The result is expected to be normal to the plate, i. e. in the z-direction. Indeed, integrating in в the elementary forces due to pressure around the plate yields the pressure contribution Fp = (0, F’p), where

F’p = 4npU2asina cosa (2.89)

On the other hand, the lift per unit span, as given by the Kutta-Joukowski theorem is

L’ = 4npU 2asina (2.90)

acting in a direction perpendicular to the incoming velocity vector as L’ = (-L’sina, L’cosa). The difference between these results is due to a horizontal suction force FS = (F’s, 0) acting at the leading edge, in the direction of the negative x-axis, that is

due to the leading edge singularity. This missing component, parallel to the camber line and of magnitude

F’s = —4npU 2asin2a (2.91)

points to the fact that integration of pressure does not capture the localized force that exists at a sharp leading edge.

The moment w. r.t. the leading edge is the sum of elementary moments

12

dM, o = —PU Cp(e)(x(в) + 2a)dx(e) (2.92)

Here, O stands for the leading edge. The moment is given by

(1 + cose)sine dв (2.93)

Some simplifications occur when the half-angle | is used. The result is

22

M, o = —2npU a sin2a and the moment coefficient reads

n

Cm, o = 4sin2a

For small values of a, the center of pressure for the flat plate is located at the quarter-chord.

Taking the derivative w. r.t. a of the same general formula for the moment at point D and now setting the result to zero, gives the aerodynamics center

d Cm D П X тл

— =——– cos2a +-D 2n cosa = 0 (2.97)

da 2 c

For small values of a one obtains the small perturbation result derived in the next chapter

To close this section, the limiting case of the flow coming perpendicular to the plate (a = 90°) is considered, with the plate having zero circulation or satisfying the Kutta-Joukowski condition. The two results are shown in Fig.2.26.

In Fig.2.26a, the flow is perfectly symmetric about the mid-plate and the net resulting force and moments are zero. Again, this is an idealized flow representation and viscosity will radically change the picture by having separation at the sharp edges of the plate. The same can be said of Fig. 2.26b in terms of realism of the picture, but it is intriguing to observe that there is circulation, hence lift, and the force is not perpendicular to the flat plate, but aligned with the plate and to the left. The occurrence of such a force, again a suction force, will be discussed in more details in the next chapter.

We proceed in the same way as before. The potential, with the flow coming at an angle a from the X-axis, reads

( b2 Г

Ф = U r + cos(0 – a) – (в – a) (2.70)

r 2n

In this case, on the cylinder, the velocity components are

Г

V = 0, Ve = —2U sin (в – a) – (2.71)

2n b

The circulation corresponding to a stagnation point at в = 0 is given by Г = 4Ub sin a. On the surface of the ellipse, after some algebra, one finds

Since the ellipse has a rounded trailing edge, there is no mechanism to control the circulation in inviscid flow and an ellipse at incidence will in general produce a flow with zero circulation. For Г = 0 the two stagnation points are located at в = a and в = ж + a, see Fig. 2.18.

The pressure coefficient is given by

This is represented in Fig. 2.19 as – C + and-Cp for the upper and lower surfaces, respectively.

Fig. 2.19 Pressure coefficients for the ellipse at incidence (Г — 0)

From anti-symmetry, the lift is zero. The moment however is not. It is obtained from

where O stands for the center of the ellipse, where the moment is calculated. After lengthy algebra, the result reads

The two stagnation points are now located at в — 0 and в — n + 2a. The corresponding flow is sketched in Fig. 2.20.

The surface pressure distribution for the ellipse with a small flap is shown in Fig. 2.21.

Fig. 2.20 Flow past an ellipse with a small flap at the trailing edge

Fig. 2.21 Pressure

distribution on the ellipse with small flap at the trailing edge

The lift coefficient is given by the Kutta-Joukowski lift theorem

where c = 2(b2 + a2)/b. Substituting the value of the circulation gives

b2 ґ e

Ci = 4n sin a = 2n 1 + sin a (2.78)

b2 + a2 c’

The results for the cylinder and the flat plate are obtained for e/c = 1 and e/c = 0, respectively. In small perturbation theory, the term e/c can be neglected as a second – order term when multiplied by sina ~ a.

1.8.1 The Ellipse at Zero Incidence

Consider the circle centered at the origin and of radius b > a in the cylinder plane. Its image in the physical plane, through the Joukowski transformation, Eq. (2.39), is an ellipse centered at the origin, with main axis along Ox, from -2a to 2a and with parametric representation

x = b~+a2 cos в = 2 cos в z = bl-a2 sin в = 2 sin в

where в represents the polar angle in the cylinder plane. The ellipse has thickness e (small axis) and chord c (large axis). The potential function in the cylinder plane is

and the velocity components are

дФ b2 1 дФ b2

Vr = 17 = U 1 – 72 cosв, ув = r — = – U 1 + ^ sinв (2.63)

On the surface of the cylinder, this reduces to

Vr = 0, Ув = -2U sin в (2.64)

Using the chain rule

дФ ___ дф dx і дф ді_

dr дx дг + дz дг

дФ _ дф д£ , дф 3z

дв = дx дв + дz дв

(2.66)
the partial derivatives Ц – and дфф can be solved for, on the ellipse as
(2.67)
V U
(2.69)
N

The center of pressure (c. p.) is the point about which the moment of the aerodynamic forces is zero.

The aerodynamics center (a. c.), also called neutral point, is the point about which the moment of the aerodynamic forces is independent of incidence.

2.7.1 Results for the Circular Cylinder

The reference length for the circular cylinder is the diameter c = 2a.

The pressure coefficient on the cylinder can be obtained from the velocity derived earlier

where Г (a) = 4n Ua sin a. See Fig. 2.16 for the two cases a = 0 and a = §.

The lift coefficient is

Ci (a) = 4п sin a (2.58)

This is a purely theoretical result, as real flow effects will dramatically change the flow field, compared to that shown in Fig. 2.9, even for small incidence angles. From the inviscid flow point of view, however, the cylinder at low incidence (near a = 0) has a lift slope which is twice the result we will find in thin airfoil theory. The maximum lift that could be achieved theoretically with the small flap device is Clmax = 4п, a very large number, when compared with high lift profiles, as we will see. This result corresponds to a = п.. See Fig.2.17.

Fig. 2.17 Ideal cylinder-with-flap lift curve

The drag coefficient for the cylinder, as for all airfoils in inviscid, incompressible flow is (d’Alembert paradox)

Cd = 0 (2.60)

In frictionless flow, the contact force between the fluid and the obstacle is normal to the surface. Hence, the center O of the cylinder is the center of pressure, i. e. Cm, o = 0 for all a’s. This result indicates that O is also the aerodynamic center.

With the Joukowski transformation, the mapping of the cylinder to the profile is singular at B, X = a, Z = 0 (the other singular point is at X = – a, Z = 0). This point corresponds to the cusped trailing edge B of the Joukowski profile, see Fig.2.12. It has been found experimentally that the flow will leave the profile at a sharp trailing edge (cusp or small angle). This is due to viscosity. As the boundary layer grows from the leading edge to the trailing edge, the fluid particles do not have enough momentum to come around the trailing edge, and hence separate from the profile there. The sharp trailing edge of the profile plays the same role as the flap for the cylinder. This condition makes the inviscid flow solution past a profile unique by fixing the circulation.

The Kutta-Joukowski (K-J) condition states the the flow must leave the profile at the sharp trailing edge “smoothly”. See Fig.2.15.

Revisiting the Joukowski transformation of Sect. 2.5.1, one can see that with an incoming flow making an angle a with the X-axis and with an arbitrary circulation around the cylinder, the rear stagnation point will be located at an arbitrary point such as P in Fig.2.12, therefore the stagnation streamline will leave the profile at point p and the K-J condition will not be satisfied. The velocity will be infinite at the trailing edge. In order to enforce the K-J condition, the circulation must be adjusted such that the rear stagnation point is located at point B. This uniquely determines the value of the circulation Г as a function of incidence a.

1.5 Definitions

The chord c of a profile is the radius of the largest circle centered at the trailing edge and touching the leading edge.

The incidence angle or angle of attack is defined as the angle between the profile chord and the incoming flow velocity vector.

It is convenient to use dimensionless coefficients to represent pressure, forces and moments. The pressure coefficient is defined as

where the denominator, 2 p y° is called the dynamic pressure (has dimension of a pressure, unit Pa) and use has been made of the Bernoulli equation.

The lift, drag and moment coefficients per unit span are made dimensionless, in 2-D flow, with the dynamic pressure and a reference length or length squared as

where the prime indicates that the force or moment is per unit span and the chord is the reference length in all cases in 2-D. Note that lower case subscripts will be used for two-dimensional coefficients to leave upper-case subscripts notation for wing, tail and complete configurations. Here, the moment coefficient is taken about point O, which in general will represent the leading edge of a profile or the nose of the airplane.

Other important quantities are:

• the lift slope d1

• the zero incidence lift coefficient Ci, o

• the moment slope d Jf0

• the zero incidence moment coefficient Cm,0,0.

1.4.1 Kutta-Joukowski Lift Theorem

The theory of conformal mapping tells us that it is possible to map any simply connected profile onto a circular cylinder. In the transformation, the flow past the profile maps into the flow past a circular cylinder derived earlier. The far field flow is preserved. The inverse transformation will therefore describe the exact solu­tion of the flow past the original profile. One such transformation is the Joukowski transformation

z	z
Fig. 2.12 Cylinder plane (X, Z) and physical plane (x, z)
(2.39)

where (x, z) represent the physical plane of the profile and (X, Z) the plane of the cylinder. Notice that the transformation preserves the far field (i. e. x ^ X and z ^ Z far from the origin). A cylinder of radius r0, where r2 = (a + e)2 + S2 centered at X = —e, Z = S and passing through point B located at X = a, Z = 0 maps onto a family of Joukowski airfoils depending on e and S. The profile has a cusp at the trailing edge, point b, image of point B. See Fig. 2.12. In the figure, p is the image of P through the transformation.

The polar coordinate representation of the cylinder is

X = r (в) cos в, Z = r (в) sin в (2.40)

where r (в) = —(e cos в — S sin в) + л/a2 + 2ea + (e cos в — S sin в)2.

Through the mapping, the parametric representation of the Joukowski profile is obtained

(2.41)

In the mapping, the doublet and potential vortex contribute to a distribution of singularities (sources, sinks and vorticity distributions) inside the profile. In the far field, however, the leading terms in the perturbation velocity field correspond again to a doublet (closed obstacle) and a potential vortex with the same net circulation Г. To calculate the force on an arbitrary profile, one can apply the momentum theorem to the domain inside a large circle C of radius R centered at the origin

Fig. 2.13 Control surface for application of the momentum theorem

where F’ represents the force per unit span applied by the fluid to the profile. See Fig. 2.13.

We will assume that any solution near C can be expanded asymptotically as

Let n = (cos в, sin в) then

The force will be evaluated in the Cartesian coordinate system. The Cartesian form of V is

where i and k are the unit vectors of the x and z-axis respectively. Hence, the first term in the integrand Eq. (2.42) reads

When integrated on the circle, the cos в and sin в cos в terms vanish and there remains only

A uniform flow produces on a cylinder of arbitrary cross-section a lift force per unit span L’ = pU Г that is proportional to the circulation Г around the profile. The orientation of the force is obtained by rotating the incoming flow velocity vector 90° opposite to the circulation.

Another way to derive this theorem is to consider a control volume as in Fig. 2.14.

If the top and bottom boundaries of the control volume are far from the obstacle (distance H), the pressure will have the atmospheric value and the velocity will be the undisturbed velocity. The circulation, however, will not vanish as it remains constant over any closed curve enclosing the obstacle. The momentum theorem, Eq.(2.42), in projection on the z-axis reduces to

where pressure does not contribute because it is atmospheric on the upper and lower surfaces and its action is perpendicular to the z-axis along the side boundaries. Since u and w are of order 0 ^-^===^ or higher, where D (H > D) is the distance of the side boundaries to the z-axis, the result can be expressed as

L ‘ = pU Г + o(DT+H5) <2-54)

The final result is obtained by letting D ^ro. A similar approach, with a control volume where now D > H can be used to prove that the drag D’ = 0. This result is consistent with the calculation of the lift over the cylinder, since with the Joukowski transformation or any transformation of the obstacle to a cylinder that preserves the far field, the lift over the obstacle is the same as the lift over the cylinder at the same angle of incidence. Furthermore, as a consequence of the conservation of momentum, the lift is proportional to the circulation as L’ = pU Г and the drag D’ = 0.

1.4.2 The d’Alembert Paradox

The d’Alembert paradox results from the fact that when a profile is moving at uniform velocity in a fluid, the resulting force is perpendicular to the velocity. The drag is zero. This result is consistent with the inviscid, frictionless fluid assumption. But in fact, with all real fluids, there is friction along the surface of the profile and there will be <a small) drag. The analysis of viscous forces in the boundary layer will resolve this paradox.

Fig. 2.15 a Profile that does not satisfy the K-J condition and b Profile that does satisfy the K-J condition

Finite bodies require that the sum of all sources and sinks be zero. Doublets, as limit of a source and a sink of equal and opposite intensity, have zero net volume flow. The superposition of a uniform flow and a doublet (D > 0) produces a flow field given by

cos в sin в

Vr = U cos в — D Ve = —U sin в — D — (2.29)

The velocity potential and streamfunction are

(2.30)

(2.31)

By inspection of the streamfunction, one can see that Ф = 0for r = a = UD. This solution represents the flow past a circular cylinder of radius a. At the surface of the cylinder, the tangency condition, Vr = 0, is satisfied. We have also seen earlier that the potential vortex admits streamlines which are circles centered at the origin. In fact, the most general solution of the flow past a circular cylinder of radius a is given by the superposition of the uniform flow, a doublet of intensity D = Ua2 and a potential vortex. The corresponding velocity field is

Note that the circulation Г is arbitrary. The tangency condition is satisfied by Vr(a, в) = 0. The asymptotic condition is also satisfied as V ^ Vro when r ^ro. Stagnation points are found on the cylinder by satisfying the condition

Г1

Ve(a, es) = -2U sines – = 0 (2.33)

2n a

This admits two solutions provided |Г| < 4nUa, namely es = arcsin (-4ra) and n + arcsin (4]fUa), where – П < es < П^. The stagnation points are points A and B in Fig. 2.9. The streamlines are given by

Fig. 2.9 Flow past a circular cylinder with circulation

(2.34)

A remarkable streamline is the stagnation streamline, V = Гл. It is made of the cylinder and the two branches attached to the stagnation points.

The question of how to determine the value for Г is important. Within the inviscid flow framework, a possible mechanism for controlling the circulation is the use of a small flap, a thin flat plate placed perpendicular to the cylinder at es forcing the rear stagnation point to locate at point B, as shown in Fig. 2.9. Once the rear stagnation point is established, the front stagnation point A will settle at n – es. In a sense, es relates to the “incidence” a = —es of the cylinder. The previous equation for the stagnation points can be written as

Г (a) = 4n Ua sin a

which reflects the dependency of the circulation on the flap location. The pressure field is obtained from the Bernoulli equation

p(r, в) = p^ + 2p(U2 – Vr2 – V2)

On the cylinder surface this reduces to

Knowing the surface pressure distribution, one can calculate the lift and drag over the cylinder

Substituting for р(в) in the above formulae yields L’ = pUГ and D = 0, the latter being an obvious consequence of symmetry.

Another, more practical means of controlling the circulation is by giving the cylinder a rotational speed Q. Any real fluid having viscosity, the particles in contact with the cylinder will rotate at speed Q, thus entraining the nearby par­ticles through viscous shear forces to rotate also. As a result, a circulatory flow is created. The relationship between the rotation speed Q and the circulation Г is not straightforward, but can be studied numerically or experimentally. See for example the discussion, in the book by White [1], of the Magnus effect and the application to the Flettner rotor ship where the sails are replaced by rotating cylin­ders. According to Wikipedia (http://en. wikipedia. org): “Assisted by Albert Betz,

Fig. 2.10 Buckau Flettner rotor ship (from https://en. wikipedia. org/wiki/File: Buckau_Flettner_Rotor_ Ship_LOC_37764u. jpg)

Fig. 2.11 JJC demonstrating the Magnus effect

Jacob Ackeret and Ludwig Prandtl, Flettner constructed an experimental rotor ves­sel, and in October 1924 the Germaniawerft finished construction of a large two-rotor ship named Buckau. The vessel was a refitted schooner which carried two cylinders (or rotors) about 15 m (50 ft) high, and 3 m (10 ft) in diameter, driven by an electric propulsion system of 50hp (37kW) power,” Fig.2.10.

A simple experiment with a small cardboard cylinder and the flat surface of a table demonstrates the Magnus effect. A small string is wound around the cylinder and its extremity is pulled briskly to accelerate the cylinder both in translation and rotation. Depending on the sense of rotation, the cylinder will fall quickly to the ground, or fly for a short time well above the table, see Fig. 2.11.

Let Q = <fC (V. n)dl and Г = <fC (V. dl), where dl represents a small oriented element of the contour and dl = |dl| is the length of the element. The first integral represents the net volume flow rate out of contour C and the second, the circulation along the same contour, taken in the clockwise direction. It is easy to show that if the contour C contains sources and sinks of intensity Q1, Q2,… the result will be

Q = Q1 + Q2 +—— the algebraic sum of the sources and sinks inside the contour.

If the contour C contains potential vortices of circulation Г1, Г2,… the result will be Г = Г1 + Г2 + ■■■ the algebraic sum of vortices inside the contour.

2.3.1 Example of Superposition: Semi-infinite Obstacle

The superposition of the uniform flow and a source (Q > 0) at the origin produces a semi-infinite body, called Rankine body. Stagnation points and stagnation streamlines play an important role in the description of the flow topology. Here, the velocity field is given by

Q 1

V = U cos в + , Ve = – U sin в (2.26)

2n r

Fig. 2.8 Half-body and flow vector field

The main flow features can be studied with the streamfunction

Q

& = Ur sin в + в (2.28)

2п

A remarkable streamline is that which corresponds to the half-body. It corresponds to & = ■§. It is also the stagnation streamline. It is made of several pieces, the negative x-axis and the half-body itself. The flow is sketched in Fig. 2.8.

Note that any streamline can be materialized as a solid surface. The flow inside the half body is also represented, however it is singular at the origin and cannot represent a realistic flow near that point. The maximum thickness of the half-body is Q. One can calculate the force on a solid surface coinciding with the separation streamline using the momentum theorem and Bernoulli’s equation applied to a contour composed of the half-body and a large circle of radius R to close the contour. The lift vanishes due to symmetry, but the axial force on the half-body is given by D = —pU Q, a thrust.

From the property of linear governing equations, if ф1 and ф2 are each solution of the Laplace equation, ф1 + ф2 or any linear combination will be a solution of the Laplace equation. The same is true for the stream functions. This is the basis of the principle of superposition. Note that the elementary solutions for ф and f are singular at their centers (infinite velocity), hence they are also called singular solutions and the superposition of singular solutions is called the method of singularities. These singular solutions are regular everywhere in the flow field when the centers are excluded, for example by enclosing them within the obstacle.

Mathematically one can prove that it is possible to solve for the flow past an arbitrary obstacle by superposing elementary solutions, but, in general, the resulting pressure field is not the sum of the individual pressure fields, due to the nonlinear­ity of Bernoulli’s equation. For small disturbances, the Bernoulli equation can be linearized, allowing for superposition of pressure fields as well.

Consider a cylinder rotating clockwise in a fluid at rest at infinity. The particles will be entrained, through viscous shear forces, to move around the cylinder in concentric circles. It can be shown that, at steady-state, such a flow is irrotational and the velocity is given by

vr = 0, 2nrvg = const. = —Г (2.24)

Г is called the vortex strength or circulation and has unit (m2/s). Such a potential flow around the cylinder is called a potential vortex (this terminology may be con­fusing since the vorticity in such a flow is zero outside the cylinder otherwise it is not a potential flow). Now, the cylinder radius can be vanishingly small. The potential flow solution however is singular at r = 0, a point inside the cylinder, no matter how small the radius is.

Notice that a potential vortex corresponds to the flow obtained when the potential and stream lines of a source/sink flow are exchanged. The solution is given by

ГГ

ф = — в, f = ln r (2.25)

2n 2n

The potential lines are now the rays and the streamlines the circles centered at the origin. See Fig. 2.7. The vortex solution is a building block for lifting airfoils to represent incidence and camber as will be seen later.

Note that the velocity components (u, w) (obtained from ф and f by taking the partial derivatives with respect to x and z when using the Cartesian coordinates, or d – and 1 dg when using polar coordinates), decay in the far field as 1, which is consistent with the asymptotic condition.

Fig. 2.7 Potential vortex: potential lines and streamlines