4.9 The Joukowski mapping: circles to ellipses

A particularly useful application of the mapping idea concerns the flow around bodies. Wehave solved the problem of the flow around a cylinder. Thus if we can find a conformalmapping between the unit circle and any given shape, we have solve the flow problemaround this shape.

z

ζ

R

2c

2d

z = ζ + b2

ζ

Consider the (inverse) mapping

z = ζ +b2

ζ, (30)

called the Joukowski mapping. Consider a circle of radius R, whose surface in the ζ-planeis described by the polar representation ζ = Reiθ. Under the Joukowski mapping,

z = Reiθ +b2

Re−iθ =

(

R +b2

R

)

cos θ + i

(

R− b2

R

)

sin θ = c cos θ + id sin θ

is an ellipse with axes length 2c and 2d.The semiaxes come out to be

c = R +b2

R, d = R− b2

R.

Now consider the uniform flow past an ellipse. To model an arbitrary angle α betweenthe direction of the flow and the semi-major axis, we consider the flow around a cylinderthat approaches the x-axis under an angle α:

w1(ζ) = U

(

ζe−iα +R2

ζeiα)

.

In principle, one can find w(z) using the inverse of the Joukowski mapping

ζ = f(z) =1

2(z +

√z2 − 4b2),

so that w(z) = w1(f(z)).However, the resulting expressions are often not so useful. For example, to find the

streamlines, it is much easier to find the streamlines of the w1(ζ) in the ζ-plane, and thento transform them using (30). This is how the pictures were produced. If one wants tocalculate the velocity, one uses

u− iv =dw

dz=

dw1

dζ

1

dz/ dζ=U(e−iα − R2eiα/ζ2)

(1− b2/ζ2).

On the cylinder, ζ = Reiθ, so

u− iv =U(e−iα − eiαe−2iθ)

(1− (b2/R2)e−2iθ)=

2iU sin(θ − α)

(eiθ − (b2/R2)e−iθ).

This means there are stagnation points at θ = α and θ = −π + α. This point is wherea streamline leaves the surface. In other words, this streamline (plotted in red) has thesame value of the streamfunction ψ then the surface of the ellipse.

4.10 Lift

Now we want to fly! In principle, we know how to construct the flow around a wing ofarbitrary shape, we only have to find the transformation, starting from a circle. We haveseen already that the key ingredient is to have circulation around the wing. We havecalculated the lift in the case of a cylindrical cross-section, but what is it for an arbitraryshape?

4.10.1 Blasius’ theorem

Ct = (cosχ, sinχ)

n = (sinχ,− cosχ)

χ

x

y

We now calculate the force on a fixed body in a stream, using complex notation. Thisis a close relative of the calculation done in section (3.6), leading to (25). Suppose a fixedrigid body, boundary C is in a steady flow, generating a potential w(z). We know

dw

dz= u− iv = qe−iχ

so |u| = q (reminder: χ is the angle the flow direction makes to the horizontal). So,according to Bernoulli (no gravity):

p = p0 − 12ρq2,

where p0 = patm + ρU2/2 is a constant.Let s be the arclength along C, which we use to integrate over the surface of the body.

Now the line element along C can be written in complex notation as

dz ≡ dx+ idy =

(

dx

ds+ i

dy

ds

)

ds.

Since the surface of the body is a streamline, the velcocity vector (u, v) is parallel tothe tangent on the surface. Thus using the angle χ we have dz = dx + idy = eiχ ds =(cosχ + i sinχ) ds. Multiplying by −i we achieve a rotation by ninety degrees in theclockwise direction, which gives the direction of the outward normal: n = (sinχ,− cosχ).Now we are in a position to write the total force F = (Fx, Fy)

F = −∮

C

pn ds

in complex form, defining a complex force F = Fx − iFy. Then

F = Fx − iFy = −∮

C

p(sinχ+ i cosχ) ds = −i∮

C

pe−iχ ds

= −i∮

C

p0(dx− idy) +iρ

2

∮

C

q2e−2iχeiχ ds

=iρ

2

∮

C

(qe−iχ)2 dz =iρ

2ρ

∮

C

(

dw

dz

)2

dz

(the term proportional to p0 vanishes because the integral of a total differential over aclosed loop is zero). This is Blasius’ theorem.

Example A cylinder in a stream with circulation:

w(z) = Uz + UR2

z− iΓ

2πlog z

sodw

dz= U − U

R2

z2− iΓ

2πzSo Blasius around the circle C says

Fx − iFy =iρ

2

∮

C

(

U − UR2

z2− iΓ

2πz

)2

dz

=iρ

2

∮

C

(

U − iUΓ

πz+ terms {z−2, z−3, z−4}

)

dz

Now use Cauchy’s residue theorem

∮

C

dz

z= 2πi,

∮

C

dz

zn= 0, n 6= 1 and

Fx − iFy = iρUΓ,

which is the same as in 4.3.

4.10.2 The Kutta-Koukowski lift theorem

From the previous example one appreciates that Blasius’ theorem yields a much moregeneral result, since only the residue of the integral comes into play. Consider a body ofarbitrary cross-section C, in a uniform stream, U , which generates circulation of strengthΓ. Far away from the body,

w(z) → Uz − iΓ

2πlog z

(origin inside C) sodw

dz≈ U − iΓ

2πz, z → ∞

and it must be analytic outside C and so

∮

C

(

dw

dz

)2

dz = limR→∞

∮

|z|=R

(

dw

dz

)2

dz

since there are no singularities in between C and |z| = R, a circle of large radius (basictheorem from C.F.T). So

∮

C

(

dw

dz

)2

dz =

∮

|z|=R

(

U − iΓ

2πz

)2

dz = 2UΓ,

using the same reasoning as for the cylinder. Hence we have shown that

Fx − iFy = iUρΓ.

So the drag on an arbitrary body is zero and the lift force is Flift = −UρΓ. This is theKutta-Joukowski lift theorem. Note that the lift force is by definition the force in thedirection normal to the flow.

4.11 Oblique flow past plates

.

We have developed a wonderful theory for lift. The only problem is that we don’tknow how to determine the value of Γ which determines the lift. Indeed, the wings of anairplane do not have circular cross section: if cylinder is placed in a uniform stream (leftside of the Figure), there is no reason why the flow should turn either way, and thus thereis no circulation. The secret is to fashion the shape so as to induce lift. Namely, the tailof wing is shaped to have a sharp corner, so as to force the flow to separate at that point.

As the simplest possible model for such a situation consider the flow around a flatplate, which is placed in a uniform stream at an angle α. A flat plate is obtained from anellipse by letting d → 0, that is putting b = R in the Joukowski transformation (30):

z = ζ +R2

ζ.

The the width of the plate is 2c = 4R. As is seen from the Figure, the flow separates fromthe plate beyond the trailing edge, so that fluid particles very close to the plate have togo around the sharp corner before leaving the plate.

This situation is reflected by the velocity on the surface of the plate, which is

u− iv =U sin(θ − α)

sin θ,

and so u→ ∞ as θ → 0, π (at the sharp edges of the plate). This is a physically untenablesituation; fluid particles are not likely to make such a sharp turn without leaving the platealtogether, especially not at infinite speed! The same conclusion can be drawn from theconcept of “adverse pressure gradient”, introduced earlier. The speed is very great at thetrailing edge and decreases as one moves up the plate. According to Bernoulli, this meansthat the pressure increases, in other words fluid particles experience an adverse pressuregradient. Instead of following a path of adverse pressure, fluid particles prefer to leavethe body, producing a point of separation at the trailing edge. This requirement is knownas the Kutta condition.

As we have seen in our calculation of the flow around a cylinder with circulation, thepoint of separation can be made to move by adding circulation:

w1(ζ) = U

(

ζe−iα +R2

ζeiα)

− iΓ

2πlog ζ.

Then the velocity is

u− iv =U sin(θ − α)

sin θ− Γ

4πR sin θLet us focus on the trailing edge, corresponding to θ → 0. (assume for example that thefront of the plate is slightly ‘rounded’, so that separation is not as important there). Thuswe want u− iv to be finite for θ → 0, a requirement which is another incarnation of theKutta condition. This can be achieved by choosing

Γ = −4πRU sinα,

which is the case shown in the above Figure. Now the singularity cancels and the velocitybecomes

u− iv =U

sin θ(sin(θ − α) + sinα) → U cosα, as θ → 0.

This means that the flow leaves the plate smoothly in the direction of the orientation ofthe plate, as seen in the above Figure.

The wonderful thing is that we have now determined the circulation uniquely, so wecan calculate the lift using Blasius’ theorem:

Flift = −UρΓ = 4πRρU2 sinα,

where 4R is the width of the plate. Once more, note that this is the force acting normalto the flow, so it is at an angle α relative to the orientation of the plate. It is customaryto define a lift coefficient cL by

Flift = cLAw

ρU2

2,

where Aw = 4R is the area of the wing (per unit length), also called the chord. Thus wehave found that for the plate

C(plate)L = 2π sinα ≈ 2πα

In the Figure, this result is compared to experiment, and it works really well, as long asthe angle of attack α is small. If however α becomes too large the theory fails abruptly.The reason is clear from the two photographs below. As long as α is small, the flowremains laminar and attached to the wing. As α is too great, the flow separates and acompletely different type of description must be sought.

Finally, we comment on the presence of circulation around the wing, which is thecrucial ingredient needed for flying. In the first instance, there is nothing wrong with thatfrom the point of view of a potential flow description. However, where is this circulationcoming from when one imagines starting up the flow, with zero circulation. Since thenet circulation in a large circle around the wing must vanish initially, and the Kuttacondition requires a circulation Γ < 0 around the wing, this means that in the processof the point of separation moving to the trailing edge, vortices of positive circulation(rotating counterclockwise) are shed from the wing. Eventually the vortices are convecteddownstream, and no longer matter for the problem.

4.12 Joukowski wings

wing

ζ

R

R−R − 2λ

λ

Now we try to model a wing in a slightly more realistic fashion, as proposed byJoukowski. In particular, we account for the fact that a real wing will be rounded at theleading edge (for the oncoming flow to go around it smoothly) and to be very sharp atthe trailing edge (for the flow to separate).

Mapping an ellipse, we have seen that if b < R, the Joukowski transformation mapsa circle to an ellipse; If b = R, the ellipse degenerates to a flat plate. Now let us take theJoukowski transformation

z = ζ +R2

ζ,

but shift the circle by λ to the left, and with radius R + λ. The equation of the circle inthe ζ-plane is

ζ = −λ + (R + λ)eiθ, θ = 0..2π.

Thus it touches the circle of radius R at the right, but lies inside it everywhere else. Inother words, the aerofoil is described by the equation

z = −λ + (R + λ)eiθ +R2

−λ + (R + λ)eiθ.

It is clear that this shape is rounded everywhere, but sharp at the trailing edge to theright, where it has a cusp (it looks locally line the sharp end of a plate), as seen in theFigure. The length of the wing in the horizontal is called the chord. From the above

mapping, back and front ends are at ζ = R and ζ = −R − 2λ, respectively. Thus thechord is

A = 2R +R + 2λ+R2

R + 2λ=

4(R + λ)2

R + 2λ.

Now the complex potential around the shifted circle is evidently

w1(ζ) = U

(

(ζ + λ)e−iα +(R + λ)2

(ζ + λ)eiα)

− iΓ

2πlog(ζ + λ),

and the complex velocity around the aerofoil is

dw

dz=

{

U

(

e−iα −(

R + λ

(ζ + λ)

)2

eiα

)

− iΓ

2π(ζ + λ)

}

(

1− R2

ζ2

)−1

Once more, there are potential singularities at ζ = ±R. However, the singularityζ = −R lies inside the wing, and is therefore harmless. On the other hand, the pointζ = R lies on the surface, and will lead to a singularity, unless Γ is chosen appropriately.At ζ = R, the term in braces becomes

U(

e−iα − eiα)

− iΓ

2π(R + λ)= −2iU sinα− iΓ

2π(R+ λ),

which vanishes forΓ = −4πU(R + λ) sinα.

In other words, using Blasius’ theorem, the lift coefficient for the Joukowski wing is

F(wing)lift = 4πρU2(R + λ) sinα.

and the lift coefficient is

c(wing)L =

2π(R + 2λ) sinα

R + λ.

wingζ

R

Rλ

µ

β

It is seen from the above formula that there is no lift if the angle of attack vanishes.In reality, wings are cambered: the top is rounded, and the bottom is hollowed out. Thiscan be achieved by shifting the center of the circle to the position −λ+ Iµ in the ζ-plane.To produce a sharp trailing edge, the circle still has to touch the point ζ = R, and thusits equation is

ζ = −λ + iµ+√

(R + λ)2 + µ2eiθ.

For future convenience, we define the angle

β = arctan

(

µ

R + λ

)

(see Figure). The chord is the same as for the profile without camber.

The resulting cambered profile is shown as the green outline above. To compute theflow, we note the complex potential in the ζ-plane:

w1(ζ) = U

(

(ζ + λ− iµ)e−iα +(R + λ)2 + µ2

(ζ + λ− iµ)eiα)

− iΓ

2πlog(ζ + λ− iµ),

and the velocity is now

u− iv =

{

U

(

e−iα − (R + λ)2 + µ2

(ζ + λ− iµ)2eiα)

− iΓ

2π(ζ + λ− iµ)

}(

1− R2

ζ2

)−1

.

As before, the term in braces has to vanish for ζ = R for the velocity to remain finite, sothis condition gives

U

(

e−iα − (R + λ)2 + µ2

(R + λ− iµ)2eiα)

− iΓ

2π(R + λ− iµ)= 0.

Noting that R + λ+ iµ =√

(R + λ)2 + µ2eiβ , we find

Γ = −4π√

(R + λ)2 + µ2 sin(α + β)

as the Kutta condition. The lift coefficient becomes

c(wing)L =

2π√

(R + λ)2 + µ2(R + 2λ) sin(α + β)

(R + λ)2,

and so indeed there is lift for α = 0. As a result, the angle of attack can be kept small,

and stall is less likely.The theoretical result for such a cambered profile is compared to experiment in the

Figure. The pressure distribution over the wing as well as the total lift predicted byJoukowski theory is in good agreement with experiment.