part iii: aerodynamics of airfoils and wingsaeweb.tamu.edu/aero301/lecture_notes/lecture pack 5,...

AERO 301: Spring 2011 III.1: Kutta Condition

Part III: Aerodynamics of Airfoils and Wings

III.1: The Kutta Condition

Previously we established the link between lift and circulation using the Kutta–Joukowski

Theorem: L′ = ρU∞Γ and also claimed that this theorem applies for objects of any shape,not just cylinders.

However, not many planes (or birds!) use cylinders to

create lift. How do airfoils do the job?

A: They have sharp trailing edges (and we introduce

a tiny bit of viscosity). This makes the stagnation

streamline (body streamline) leave the trailing edge of

the airfoil smoothly.

True Inviscid Flow: Γ = 0

Reality, Almost Inviscid Flow: Γ > 0

The smooth stagnation streamline that leaves from the trailing edge is an experimental

observation called the Kutta Condition.

AERO 301: Spring 2011 III.1: Kutta Condition

In the case of flow over a cylinder, sources and sinks define the body shape. Any amount

of circulation can be added without changing the shape. This is much the same case with

an airfoil. We can define a shape and add any amount of circulation and, in principle, get

any amount of lift we want.

However, the Kutta condition tells us that because the stagnation streamline must leave the

trailing edge smoothly, there is only one value of the circulation that is actually observed in

practice.

Why does the existence of viscosity (and the fact that we can’t have negative absolute

pressures) give rise to the Kutta condition?

AERO 301: Spring 2011 III.2: Thin Airfoils

III.2: Thin Airfoil TheoryNow, the real problem. How do we distribute basic flows (uniform flow, sources, sinks,

doublets and vortices) to generate a flow over an airfoil shape while satisfying the Kutta

condition?

First, let us revisit what an airfoil looks like:

Chord Line

Camber Line

chord, c

We will restrict ourselves to talking

about thin airfoils. This means that the

thickness (top to bottom) is much less than cand, effectively, the top and bottom surfaces

collapse onto the camber line.

Typically, saying it is a thin airfoil also implies that the maximum amount of camber (distance

between the camber line and chord line) is much smaller than the chord length.


Aside: The NACA 4-Series airfoil designations are codes that give airfoil shapes. The two

most common are the NACA 0012 and NACA 2412.

• The first digit is the maximum camber in units of c/100.The 0012 is symmetric, that is, it has no camber.

The 2412 only has 2% camber.

• The last two digits give the maximum thickness in units of c/100.Both these airfoils have a maximum thickness of 12%.

• The second digit is the location of maximum camber in units of c/10.


In the context of a thin airfoil we do not need to use any sources or sinks to make a body

shape because the top and bottom surfaces touch each other.

Instead, we will distribute many vortices along the camber line

and set their strengths such that no-penetration is satisfied along the camber line.

Each vortex will be fairly weak so we call their strengths dΓi where i is the index of eachvortex.

Then, we add up all the circulation associated with all of the vortices and get then net

circulation: Γ = ∑i dΓi and the lift: L′ = ρU∞Γ .

Instead of a discrete distribution of vortices, imagine that we have an infinite number of

infinitesimally weak vortices distributed along the chord line. Then

dΓ = γ(s)ds

where ds is an infinitesimally short piece of the camber line andγ is the vortex strength per unit length.


With this approach, the total lift on the airfoil is given by

L′ = ρU∞

∫ TE

LEγ(s)ds

Our job becomes determining the function γ(s) and its integral because these give the lift.The function for γ(s) is chosen so that we satisfy no-penetration along the camber line.(We’ll return to this point.)

Physically, what does γ represent? Consider a small piece of the camber line and computethe circulation around a path that encloses this piece.

utop

ubottom And, γ =


Actually determining the function γ(s) makes use of the fact that the distributed vorticesalong the camber line must satisfy the no-penetration condition along this line. How do we

do this?

The function γ(s) must be such that thevelocity normal to the camber line is

zero at all s.

Consider the diagram to the right

Camber Line

δ1 δ

2

δ3

z

xx x0

Chord Line

The small amount of velocity normal to

the camber dvn line at point x0 due to

the vortex at x is

dvn =γ(s)ds

2π rcosδ3 where r =

x0− xcosδ2

and ds =dx

cosδ1

so

dvn(x0,x) =1

2πγ(x)dxx0 − x

cosδ2 cosδ3cosδ1


What are the angles? Remember, we are considering thin airfoil theory and this means

that both the thickness and maximum camber are small. Therefore, all of the δ ’s are smalland cosδ1,2,3 ≈ 1.

So, the velocity normal to the camber line at any location x0 due to the vortices at all xlocations is given by

vn(x0) =1

2π

∫ c

0

γ(x)x0 − x

dx

The vn calculated for each x0 must exactly cancel the

contribution to the normal velocity due to the uniform

freestream flow. The freestream flow is oriented at an angle of

attack α from the chord line and the camber line is oriented atan angle θ to the chord line as shown.

Camber Line

x0

U∞

α

θ

Chord Line

where

θ = tan−1(

dzdx

)

x0

≈

(

dzdx

)

x0

(Note that the θ in the diagram is negative.)

So, we add the contribution of the vortices and the freestream flow and, because these must


equal zero, we obtain the condition for γ(x)

12π

∫ c

0

γ(x)x0− x

dx =U∞

[

α −

(

dzdx

)

x0

]

This is called the Fundamental Equation of Thin Airfoil Theory.

Solving this equation gives the function γ(x). We integrate γ(x) to find the total circulationand, using the Kutta–Joukowski theorem, the lift.

Note that this equation guarantees that the no-penetration condition will be satisfied but

what about the Kutta condition?

Well, consider what happens to the top and bottom surface streamlines at the trailing

edge. . .


Vtop

Vbottom

γ(c) =

AERO 301: Spring 2011 III.3: Symmetric Airfoils

III.3: Symmetric Airfoils

Now that we have derived the fundamental equation of thin airfoil theory, let us use it to

find the γ distribution on a thin airfoil such as a NACA 0012 and its friends with NACA00xx designations.

This is the simplest case we can consider because we do not need to consider the camber

distribution z(x) and the fundamental equation reduces to

12π

∫ c

0

γ(x)x0− x

dx =U∞α

Note that this formula is only valid for small angles of attack!

In general, it’s very difficult to solve integral equations. In this case there is a series of tricks

that help.

First, change coordinates: x=c2(1−cosθ)

so x = 0 → θ = 0, x = c → θ = π and dx=c2

sinθdθ


The equation becomes

Next, we guess that the solution is

γ(θ) = 2αU∞1+ cosθ

sinθ

To verify that this guess is correct (would I give you an incorrect guess?) substitute into

the integral equation.

To accomplish this, a useful integral to know is

∫ π

0

cos(nθ)cosθ − cosθ0

dθ = πsin(nθ0)

sinθ0

It works! And, better still, it satisfies the Kutta Condition.


Now that we know our γ(θ) distribution we can calculation lift using Kutta-Joukowski:

L′ = ρ U∞

∫ c

0γ(x)dx

L′ = παρU2∞c

or

cl = 2παThis is always the lift coefficient

for a thin, symmetric airfoil at a

small angle of attack.


We can compute streamlines (in a coordinate system aligned along the airfoil) as

ψ =U∞(cosα y− sinα x)+U∞ c4π

∫ π

0(1+ cosθ) ln

{

[

x−c2(1− cosθ)

]2+ y2

}

dθ

and this gives the following pattern for α = 10◦

What is γ at the leading edge?

What does this imply for the velocities at the leading edge?

Is this physical? Do we care?


In addition to the lift generated by our airfoil we would also like to know the pitching

moment per unit span, M ′

We can calculate the moment about the leading edge using dM′ = −xdL′ to account for

the moment arm associated with the vortices’ contributions to lift.

So we have

M′

LE =−

∫ c

0xdL′ =−

∫ c

0ρ U∞ xγ(x)dx

M′

LE =−π4

ρ U2∞ c2 α

or

cmLE =−π α2


Instead of finding pitching moment about the leading edge, we could find the center of

pressure, the location about which there is zero pitching moment.

We could find this by solving for xc.p. in 0=−

∫ c

0(x−xc.p.)γ(x)dx

Actually, because we know the moment about the leading edge we can accomplish the same

objective by finding the equivalent moment M′

ξ at another location x = ξ and solving for theξ that gives M′

ξ = 0.

L = παρU2∞c

MLE=−π

4ρU2

∞c2α

L = παρU2∞c

Mξ

note that the moment arrows are drawn in thepositive direction (nose up) but the LE momentis negative so it’s a nose-down pitching moment


Using the diagram on the previous page,

−L′ξ +M′

ξ = M′

LE

For a symmetric airfoil xc.p. = c/, the quarter-chord point.

Note that the center of pressure is independent of α. This means that there is no momentabout c/4 at any angle of attack. Does this strike you as a good point about which to starta free-body diagram for flight dynamics?

The point at which the moment does not depend on α is called the aerodynamiccenter.

For a symmetric airfoil the aerodynamic center (where the moment does not depend on α)is the same as the center of pressure (where there is no moment). That is, the constant

moment at the aero center is zero. This is not always the case. In general, the center of

pressure can depend on α and the aero center can have a non-zero moment.

AERO 301: Spring 2011 III.4: Cambered Airfoils

III.4: Cambered Airfoils

What do we do if we want an airfoil with properties besides cl = 2π α and cm,c/4 = 0? Webuild an airfoil with camber. (Note that this is our only option within thin-airfoil theory.)

In this case, we are back to the full form of the fundamental equation:

12π

∫ c

0

γ(x)x0− x

dx =U∞

[

α −

(

dzdx

)

x0

]

How do we analyze this? Let us keep the same framework with x/c = (1− cosθ)/2.

We also keep the same basic approach to γ(θ) except we modify it in a way that permitslots of flexibility:

γ(θ) = 2U∞

[

A01+ cosθ

sinθ+

∞

∑n=1

An sin(nθ)

]


• The coefficient A0 is similar to what we had for a symmetric airfoil.

• If there is no camber we expect A0 = α and An>0 = 0.

• The sum is a Fourier sine series correction that permits us to solve for any realistic

camber distribution. The sines can represent any smooth function that is equal to zero

at θ = 0 and π.

• Note that any combination of An’s will satisfy the Kutta condition.

Put the preceding form for γ(θ) into the fundamental equation and using a trig substitutionwe get:

sin(θ)sin(nθ) =cos[(n−1)θ ]− cos[(n+1)θ ]

2and the handy integral from last time

∫ π

0

cos(nθ)cosθ − cosθ0

dθ = πsin(nθ0)

sinθ0

to find that

(

dzdx

)

x0

= α −A0+∞

∑n=1

An cosnθ0


How do we deal with the preceding expression?

Use our orthogonality tricks!

A0 = α −1π

∫ π

0

dzdx

dθ0

An = α −2π

∫ π

0

dzdx

cos(nθ0)dθ0

Note that α only appears in the A0 term. All the other terms only depend on the camber

distribution.

Also note (this is tricky) that dz/dx is a function of x while the integral is written as an

integral over θ. It’s up to you to be consistent and use only x OR θ when performingthe integral.


Before we compute An’s for any specific camber distribution, let us see how the coefficients

affect cl and cm.

To find cl we integrate cl =2

U∞ c

∫ c

0γ(x)dx

cl = 2π A0+π A1

Note that the absence of An terms with n > 1 is not an approximation.


To find cm,LE we integrate cm,LE =−2

U∞ c2

∫ c

0xγ(x)dx

cm,LE =−π2

(

A0 +A1−12

A2

)

Again, the absence of An terms with n > 2 is not an approximation.


Where is the center of pressure (the point about which there is zero pitching moment)?

xcp

c=

A0+A1−A2/24(A0+A1/2)

The center of pressure changes as α changes so the center of pressure of a cambered airfoilis not the aerodynamic center.

What is cm,c/4? We switch our reference point by c/4 and obtain

cm,c/4 =π4(A2 −A1)

cm,c/4 is not a function of α and is, therefore, the aerodynamic center (with a non-zeroconstant moment).


Example #1: A Piecewise-Linear Approximation to a NACA 24xx

A NACA 24xx can be approximated by two straight sections of camber line as indicated:

Approx. Camber Line

Chord Linex=0

(0.4c,0.02c)

x=c

What is cl and what is cm,c/4?


The piecewise-linear distribution has a constant dz/dx in two sections so its not too difficultto integrate. The results of this simplified approach are:

A0 = α −0.00940

π

A1 =0.163

π

A2 =0.0327

π

cl = 2πα +0.145

cm,c/4 = −0.0327

part iii: aerodynamics of airfoils and wingsaeweb.tamu.edu/aero301/lecture_notes/lecture pack 5,...

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