4 ds C- 2: Unit C - 1: List of Subjects

AE301 Aerodynamics I

UNIT C: 2-D Airfoils

ROAD MAP . . .

C-1: Aerodynamics of Airfoils 1


C-3: Panel Methods

C-4: Thin Airfoil Theory


Unit C-1: List of Subjects

Wings and Airfoils

Lift Curves

Airfoils

NACA Conventional Airfoils

NACA Airfoil Data

NOMENCLATURE FOR WINGS (3-D)

3-D Wing Geometry Nomenclature:

• Leading edge (LE)

• Trailing Edge (TE)

• Airfoil (a cross section of wing)

Recall that the lift, drag, and moment coefficient for 3-D wing can be defined as:

L

LC

q S

= D

DC

q S

= M

MC

q Sc

=

NOMENCLATURE FOR AIRFOILS (2-D)

2-D Airfoil Geometry Nomenclature:

• Chord Line

• Mean Camber Line

• Chord (c)

• Thickness (t)

• Camber: (difference between chord line and mean camber line)

For 2-D airfoil, the aerodynamic coefficients are “per unit span” basis:

( )

( )'

1l

L b Lc

q c q c

= = ( )

( )'

1d

D b Dc

q c q c

= = ( )

( ) 2

'

1m

M b Mc

q cq c c

= =

Note: b = wing span

Unit C-1Page 1 of 10

Wings and Airfoils

LIFT OF AIRFOILS

Lift on an airfoil depends on the following properties:

• V (freestream velocity)

• (freestream density)

• S (wing area)

Hence, lift coefficients are normalized by these properties: L

LC

q S

= and '

l

Lc

q c

= .

LIFT CURVE OF AIRFOILS

The behavior of lift (“lift curve” characteristics) depends on the following properties:

• (angle of attack)

Lift-curve (or often called, “cl - ” curve) provides important relationship between angle of attack and

lift coefficient, under a certain condition of Reynolds number. Interestingly, lift-curve is fairly close to a

linear line, as long as it is not under the “stall” condition (hence, we often assume it is a simple “linear

function” in our aerodynamic analysis for simplification).

• (viscosity)

Lift curve depends on the Reynolds number.

• a (freestream speed of sound, or “compressibility”)

Lift curve will also depend on the compressibility of the flow field (Mach number).


Lift Curves

AIRFOILS

• The “shape” of airfoil: the design of 2-D airfoil will have a significant impact on aircraft

performance.

• Airfoils represent performance of a given cross-section of a wing. The shape of an airfoil has

tremendous effects on the overall performance of wing (thus, airplane).

• Airfoils can be considered as a model for a “unit span” of an infinite wing of constant cross-

section. The performance of an airfoil can be determined by a “quasi-2-D” wind tunnel tests.

• A Quasi-2-D is actually a 3-D, but constant cross section. Thus, the cross-sectional properties (i.e.,

lift, drag, and moment “per unit length”) can be determined.

GEOMETRIC AND AERODYNAMIC TWISTS OF WINGS

• Geometric twist of wing is varying angle of attack along the span, but retains the same airfoil.

• Aerodynamic twist of wing is varying airfoil (cross section of the wing) along the span, but retains

the angle of attack.


Airfoils (1)

http://www.ae.uiuc.edu/m-selig/ads.html

CAMBERED V.S. SYMMETRICAL AIRFOIL

The camber in airfoil is the asymmetry between the top and the bottom curves of an airfoil. Cambered

airfoils generate lift at positive, zero, or even small negative angle of attack, whereas a symmetric airfoil

only has lift at positive angles of attack.

0ldc

ad

= : lift curve slope – the slope of the cl – curve (straight line)

The lift curve slope of a “thin” airfoil (either symmetric or cambered) is:

0a = 2 (1/rad) = 0.10966 (1/deg)

0L = : zero lift angle of attack (“alpha zero-lift”) – the angle of attack (negative value), where the

cambered airfoil generates no lift.

LIFT EQUATIONS

Assuming the “linear” relationship between angle of attack () and lift coefficient (cl), one can

“estimate” the lift coefficient at a given angle of attack:

• For symmetrical airfoil ( 0 0L = = ):

0lc a =

• For cambered airfoil ( 0 0L = ):

( )0 0l Lc a == −

• NOTE: these lift equations are based on assumptions that angle of attack () and lift coefficient (cl)

are perfectly in linear relationship (these are simple linear algebraic equations, such as: y ax b= + ).

Is it always true???


Airfoils (2)

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(a) Assuming the linear relationship between cl – (valid only in the certain range of ):

0 0( ) 0.11[10 ( 3)]l Lc a == − = − − = 1.43

(b) Upside-down means that the airfoil is now “negatively” cambered. The zero lift AOA is now + 3

degrees. Thus, 10 degrees AOA is essentially equivalent to only 7 degrees AOA, so:

0 0( ) 0.11[10 ( 3)]l Lc a == − = − + = 0.77

(c) In order to maintain the same lift coefficient (1.43 at 10 degrees AOA), the upside-down airfoil must

be pitched to a higher AOA.

0 0( )l Lc a == −

=> 0

0

1.43(3)

0.11

lL

c

a == + = + = 16 (degrees)


Class Example Problem C-1-1

Related Subjects . . . “Airfoils”

Can airplane fly upside-down?

To answer this question, make the following simple calculation. Consider a positively

cambered airfoil with a zero-lift angle of attack of −3 degrees. The lift slope of this

airfoil is 0.11 per degree.

(a) Calculate the lift coefficient at an angle of attack of 10 degrees.

(b) Now imagine the same airfoil turned upside-down, but at the same 10 degrees angle

of attack as part (a). Calculate its lift coefficient.

(c) At what angle of attack must the upside-down airfoil be set to generate the same lift

as that when it is right-side-up at a 10 degrees angle of attack?

NACA AIRFOIL DATA

NACA airfoils are airfoil shapes developed by the National Advisory Committee for Aeronautics

(NACA). The lift, drag, and moment coefficients for these airfoils were obtained through wind tunnel

tests conducted in 1950s (NACA Report 824).

NACA FOUR/FIVE DIGIT SERIES AIRFOILS

The normalized coordinates (x/c, y/c) of NACA 4 digit or 5 digit series conventional airfoils can be

computer-generated (NASA-96-TM4741).

• NACA 4/5 digit series airfoils are usually called as: NACA Conventional Airfoils.

NACA conventional airfoils have maximum thickness (usually) at quarter chord. For a conventional

airfoil, the maximum thickness location is designed to be at the location of aerodynamic center. If the

maximum thickness location is moved (usually “aft” not “forward”), the airfoil design is usually

considered “non-conventional” (i.e., NACA 6-series Natural Laminar Flow or “NLF” airfoils).


NACA Conventional Airfoils

NACA 4-Digit Series: NACA X X XX

One digit describing maximum camber (in % of

chord).

One digit describing the distance to the maximum

camber location measured from the leading edge (in 10% of chord).

Two digits describing maximum thickness of the

airfoil (in % of chord).

NACA 5-Digit Series: NACA X XX XX

One digit, when multiplied by 1.5, gives the lift

coefficient in 1/10.

Two digits, when divided by 2, describe the

distance to the maximum camber location measured from the leading edge in 1/10 of chord.

Two digits describing the maximum thickness of

the airfoil in % of chord.

http://www.ppart.de/aerodynamics/profiles/NACA4.htmlhttp://www.ppart.de/aerodynamics/profiles/NACA5.html

OTHER NACA AIRFOILS

NACA 1-Series:

Mathematically derived airfoil shape from the desired lift characteristics. Prior to this, airfoil shapes

were only determined using a wind tunnel.

NACA 6-series (Natural Laminar Flow, or “NLF” airfoil):

An improvement over 1-series airfoils with emphasis on maximizing natural laminar flow (NLF).

Maximum thickness is moved (close) to the half chord location.

NACA 7-series:

Further advancement in maximizing laminar flow achieved by separately identifying the low pressure

zones on upper and lower surfaces.

NACA 8-series (NASA-Super Critical, or “NASA-SC” airfoil):

Supercritical (SC) airfoils designed to optimize transonic flow characteristics.


NACA Airfoil Data

THIS FIGURE IS FOR EXPLANATION PURPOSES ONLY = NOT VERY ACCURATE !!!

−1.2 −.8 −.4 0 .4 .8 1.2 1.6

cl, section lift coefficient

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(a) At standard sea-level condition with 44 m/s airspeed:

= 1.225 kg/m3

= 17.8910−6 kg/ms

c = 1.0 m

The test section Reynolds number: 6

(1.225)(44)(1)Re

17.89 10

Vc

−= =

= 3106

Let us look at the NACA 4412 airfoil data (Reynolds number 3106):

At 2 degrees of AOA:

cl = 0.6 cd = 0.0067 cm,c/4 = −0.08



Related Subjects . . . “NACA Airfoil Data”

A model wing of constant chord length is placed in a low

speed subsonic wind tunnel, spanning across the test

section (this is called, a quasi-2-D test). The wing has a

NACA 4412 airfoil and a chord length of 1.0 m. The test

section airspeed is 44 m/s at standard sea-level condition.

If the wing is at a 2 degrees angle of attack, determine:

(a) cl, cd, and cm,c/4

(b) the lift, the drag, and the moment about the quarter

chord (per unit span)

NACA 4412

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(b) The dynamic pressure is: 2 21 1(1.225)(44)

2 2q V= = = 1,185.8 N/m2

Therefore,

' (1,185.8)(1)(0.6)lL qcc= = = 711.48 N (per unit span)

' (1,185.8)(1)(0.0067)dD qcc= = = 7.945 N (per unit span)

2 2

/4 , /4' (1,185.8)(1) ( 0.08)c m cM qc c= = − = −94.86 N (per unit span)


Class Example Problem C-1-2 (cont.)


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Upside-down flight is equivalent to the NACA data in the range of negative angle of attack. The only

difference is that the negative angle of attack produces “downforce” (means negative lift coefficient),

while upside-down flight produces “lift” (positive lift coefficient). Thus, one can read-out the lift

coefficient from the NACA data (in the range of negative angle of attack).

(a) From NACA 4412 data (assuming the Reynolds number 3106):

At 10 = degrees => 1.34lc =

(b) If the data is read-off from “negative” angle of attack:

At 10 = − degrees => 0.64lc =

(c) Impossible.

The airfoil stalls out at about “negative” angle of attack of −12 degrees (maximum 0.8lc = )




Once again, can airplane fly upside-down?

To answer this question, we will use the same wind tunnel test as in Class Example

Problem C-1-2. Consider NACA 4412 airfoil.

(a) Determine the lift coefficient at an angle of attack of 10 degrees.

(b) Now imagine the same airfoil turned upside-down, but at the same 10 degrees angle

of attack as part (a). Determine its lift coefficient.

(c) At what angle of attack must the upside-down airfoil be set to generate the same lift

as that when it is right-side-up at a 10 degrees angle of attack?

NACA 4412



ROAD MAP . . .



C-3: Panel Methods




Pressure Coefficient

Obtaining Lift from CP

Compressibility Effects

Critical Mach Number

Drag-Divergence Mach Number

Supercritical Airfoil

Wave Drag

PRESSURE COEFFICIENT

Pressure coefficient (Cp) is a dimensionless quantity

−

=q

ppC

p

Convention of Cp plot: opposed vertical axis (negative up positive down)

• Cp = 1: means that the location where the velocity (V) is equal to zero (stagnation point)

Note that the “highest” positive value of pressure coefficient is: Cp = 1 (Cp cannot become more than

1: it is impossible, by definition).

• Cp = 0: means that the location where the static pressure at that point (p) becomes equal to the

freestream static pressure (p∞), commonly called: the static port location.

• Cp = negative: means that the surface pressure is lower than the freestream static pressure. The

surface of negative pressure coefficient is called “suction surface.”

• Cp = positive (but less than 1): means that the surface pressure is higher than the freestream static

pressure. The surface of positive pressure coefficient is called “pressure surface.”

CP DISTRIBUTION OVER AN AIRFOIL (WITH A SMALL POSITIVE AOA)

Upper surface: Cp at the leading edge starts from 1 (stagnation point). Cp starts to decrease (favorable

pressure gradient) very rapidly (p < p∞) and reaches the minimum pressure point. After this minimum

pressure point, Cp increases (adverse pressure gradient) toward the trailing edge.

Lower surface: Cp at the leading edge starts from 1 (stagnation point). Cp starts to decrease (favorable

pressure gradient) and then slightly increase (adverse pressure gradient) toward the trailing edge.


Pressure Coefficient

Cp = 0: pressure is the same as p

Cp = 1: stagnation point

V

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At standard sea-level condition,

= 0.0023769 slug/ft3

p = 2,116.2 lb/ft2

The test section airspeed is: 70 mph = 88 ft/s

70 mph60 mph

= 102.667 ft/s

Thus, the dynamic pressure is:

2 21 1(0.0023769)(102.667)

2 2q V = = = 12.527 lb/ft2

2,100 2,116.2

12.527p

p pC

q

− −= = = 1.293−



Related Subjects . . . “Pressure Coefficient”

Consider an airfoil model mounted in a subsonic wind tunnel. The test section

airspeed is 70 mph, and the condition is the standard sea-level. If the pressure

measured at a point on the airfoil (using a static pressure tap connected to a U-tube

manometer) is 2,100 lb/ft2, what is the corresponding pressure coefficient?

Static Pressure Taps

LIFT COEFFICIENT CALCULATION FROM CP DISTRIBUTION

Lift force (per unit span) can be found by integrating Cp over the surface: TE TE

LE LE

' cos cosl uL p ds p ds = − ,

where, lp = Pressure on the lower surface

up = Pressure on the lower surface

also, cosds dx =

0 0

'

c c

l uL p dx p dx= −

Adding and subtracting p , we have: ( ) ( )0 0

'

c c

l uL p p dx p p dx = − − −

From the definition of lift coefficient:

0 0

' 1 1c c

l ul

p p p pLc dx dx

q c c q c q

− −= = − = , ,

0

1( )

c

p l p uC C dxc

−

NORMALIZED COORDINATE LOCATION

Often, it is convenient to specify the coordinate “along the chord”: x

xc

=

1

, , , ,

0 0

1( ) ( )

x cc

p l p u p l p u

x c

xC C dx C C d

c c

=

=

− = − =

1

, ,

0

( )p l p uC C dx−


Obtaining Lift from CP

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Related Subjects . . . “Obtaining Lift from Cp”

Consider an airfoil with chord length c and the running distance x measured along the

chord line. The leading edge is located at x/c = 0 and the trailing edge at x/c = 1.

Calculate the lift coefficient for this airfoil. The pressure coefficient variations over the

upper and lower surfaces are given respectively as:

COMPRESSIBILITY CORRECTION FOR PRESSURE COEFFICIENT

As the freestream Mach number is increased (M > 0.3), the compressibility effects can no longer be

ignored. A correction for compressibility is called, the Prandtl-Glauert rule:

,0

21

p

p

CC

M

=−

• This is simply a “correction factor” and will only provide “estimated” pressure coefficient at a given

Mach number

• In order to more accurately determine pressure coefficient, one needs to apply energy equation (if it

is reasonable to assume that the flow is still isentropic and ideal gas of air).

CRITICAL MACH NUMBER

Critical Mach number (Mcr): the freestream Mach number at which sonic flow is first obtained

somewhere on the airfoil surface.

Critical pressure coefficient (Cp,cr): the specific value of Cp that corresponds to the presence of sonic

flow (M = 1).


Compressibility Effects

Prandtl-Glauert Rule:

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EFFECTS OF AIRFOIL THCKNESS

If the airfoil is thick, the pressure at the minimum pressure point on the surface of the airfoil becomes

lower. As a result, the sonic flow begins to appear at much lower Mach number. Therefore, the critical

Mach number is lower.

EQUATION OF CRITICAL MACH NUMBER (1)

Starting from the pressure coefficient: 1p

p p p pC

q q p

−= = −

eqn. 1

Recall, the definition of dynamic pressure:

( )( )

( )2

2 21 1 1

2 2 2

Vq V p V p

p p

= = = eqn. 2

Also, recall, speed of sound is ( )2a RT p = = , and eqn. 2 becomes:

( )( ) ( )

2 22 2

2

1 1 1

2 2 2 2

V Vq V p p p M

p a

= = = = eqn. 3


Critical Mach Number (1)

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EQUATION OF CRITICAL MACH NUMBER (2)

Recall, for isentropic flow (between freestream and stagnation point):

( )1

20 11

2

pM

p

−−

= +

eqn. 4

Applying eqn. 4 between freestream and stagnation point:

( )1

20 11

2

pM

p

−

− = +

eqn. 5

Combining eqns. 4 & 5 yields: ( )

( )

( )1

2

0

20

11 1

21

1 12

Mpp p

p p pM

−

+ −

= = + −

eqn. 6

Now, substituting eqn. 6 into eqn. 1:

( )

( )

( )

( )

( )

( )1 1

2 2

22 2 2

1 11 1 1 1

22 21 1 11 1 1

1 1 1 12 2 2

p

M Mp pp

Cq p M

p M M M

− −

+ − + −

= − = − = − + − + −

Cp = Cp,cr, where M = 1, therefore the Cp at critical Mach number is:

( )( )1

2

, 2

2 121

1p cr

MC

M

−

+ − = −

+


Critical Mach Number (2)

DRAG DIVERGENCE MACH NUMBER

Drag divergence Mach number: the freestream Mach number at which drag coefficient begins to

increase rapidly.

Mcr < Mdrag divergence < 1.0

Drag divergence is based mainly on the formation of strong normal shock wave at the minimum

pressure point on the upper surface of the airfoil and associated flow separation behind the shock wave,

called the “shock-induced” flow separation.

• Due to the drag divergence, the early attempts of supersonic flight (NASA / Bell X-1) faced

difficulties of achieving the successful supersonic flight: this is commonly known as a “SONIC

BARRIER.”


Drag-Divergence Mach Number

SUPERCRITICAL AIRFOIL

A supercritical (SC) airfoil (NASA-SC or NACA 8-series) is an airfoil designed, primarily, to delay the

onset of wave drag in the transonic speed range. The supercritical airfoil was created in the 1960s, by

NASA scientist Richard Whitcomb.

While the design was initially developed as part of the supersonic transport (SST) project at NASA, it

has since been mainly applied to increase the fuel efficiency of many high subsonic aircraft.

Supercritical airfoils have three main benefits:

• Higher (or “delayed”) critical Mach number,

• Develop shock waves further aft than traditional airfoils, and thus,

• Greatly reduce shock-induced boundary layer separation.

NASA-SC airfoil is relatively “thick” airfoil and not suitable for supersonic flight. The design purpose

is to delay formation of “shock-induced flow separation” to avoid drag divergence at the high-end of

transonic flight.

Boeing (MD) C-17 Globemaster III (US Patent

No: 4,858,853, McDonnell Douglas Corporation,

1987)


Supercritical Airfoil

NASA-SC

Characteristics of NASA-SC Airfoil:• Flattened upper surface,• Highly cambered (curved) aft section, and• Greater leading edge radius as compared to traditional airfoils

WAVE DRAG

Wave drag is the pressure drag due to the formation of shock waves. At supersonic flight, the entire

vehicle is placed “behind” the shockwave. Under this condition, the flow behind the shockwave is not

the same condition of freestream: it is much higher pressure, density, and temperature.

Wave drag is usually the order of magnitude higher than other drag components (such as skin friction &

pressure drags). Super-cruising (cruising at supersonic) is technically very challenging, due to the

massive increase of drag (due to the formation of shock waves: wave drag), usually requires higher

thrust to compensate the higher drag for supersonic cruising.

AERODYNAMIC COEFFICIENTS AT SUPERSONIC FLIGHT

Lift and drag at supersonic flight is mainly dependent upon Mach number (as well as the angle of

attack). For a thin supersonic airfoil (close to a flat plate), the lift and wave drag coefficients can be

estimated as:

2

4 or

1l Lc C

M

=−

2

, ,2

4 or

1d w D wc C

M

=−


Wave Drag

2

, ,2

4 or

1d w D wc C

M

=−

2

4 or

1l Lc C

M

=−

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Lift and wave drag coefficients can be calculated as:

= 5 degrees = 5(/180) = 0.087266 radians

2 2

4 4(0.087266)

1 (3) 1lc

M

= =− −

= 0.1234

2 2

,2 2

4 4(0.087266)

1 (3) 1d wc

M

= =− −

= 0.01077

At 22,000 ft:

= 0.0011836 slug/ft3

a = 1,028.6 ft/s => M = 3 means that V = (1,028.6)(3) = 3,085.8 ft/s

2 21 1(0.0011836)(3,085.8)

2 2q V = = = 5,635.215 lb/ft2

Hence, lift and wave drag (per unit span) are:

' (5,635.215)(10)(0.1234)lL q cc= = = 6,953.85 lb

,' (5,635.215)(10)(0.01077)w d wD q cc= = = 606.91 lb



Related Subjects . . . “Wave Drag”

Consider a thin supersonic airfoil with chord length c = 10 ft in a Mach 3 freestream at

an altitude of 22,000 ft. The airfoil is at an angle of attack of 5 degrees. Calculate the

lift & wave drag coefficients and the lift & wave drag per unit span.

Figure: X-43 Hyper X Prototype(NASA Ames Research Center)



ROAD MAP . . .



C-3: Panel Methods




Problem Solutions? How?

Source Panel Method

Vortex Panel Method

This is a tutorial (self-study) material for the COURSE PROJECT

• Not covered in class as a standard lecture of AE301

• Online lecture (YouTube) Available

PANEL METHODS / THIN AIRFOIL THEORY: PURE THEORETICAL SCHEMES

Panel methods and thin airfoil theory are built upon potential flow analysis. In other words, these are

pure theoretical schemes. Similar to the unit B-3, as we deal with the pure theory, one must be very

carefully recognize (and keep track of) the followings:

(1) What are the assumptions made to simplify the equation?

• Steady-state?

• Inviscid?

• No body forces?

• Incompressible?

• Irrotational?

(2) Because of the assumptions made, how your theoretical solution different (or apart) from the actual

(or real) flow field phenomena?

Unit CUnit C--33Page Page 11 of 12of 12

Problem Solutions? How?Problem Solutions? How?

SOURCE PANEL METHOD

A source sheet can be defined as infinite number of line sources (side by side) where the strength of

each line source is infinitesimally small.

Source strength per unit length can be defined as: ( )s = .

Consider an arbitrary point ( , )P x y . . . a small section of the source sheet of strength ds induces an

infinitesimally small velocity potential d :

ln2

dsd r

=

Thus, the complete velocity potential at point P induced by the entire source sheet is:

( , ) ln2

b

a

dsx y r

=


Source Panel Method (1)Source Panel Method (1)

Let us approximate the source sheet by a series of straight panels, moreover, let the source strength per

unit length be constant over a given panel.

For the total number of n panels, the source strengths per unit length for each panel can be represented

by: 1 2 3, , , , ,j n

Then, solve: , 1, 2,3 ,j j n = such that the body surface becomes a streamline of the flow: means that

the boundary condition is imposed numerically by j ’s such that the normal component of the flow

velocity is zero at the midpoint of each panel (called, the “control point”).

Again, consider a point P (x,y) in the flow field, and let pjr be the distance from any point on the jth

panel to P.

The velocity potential induced at P due to the jth panel is:

ln2

j

j pj j

j

r ds

=



The potential at P due to all panels is:

( )1 1

ln2

n nj

j pj j

j j j

P r ds

= =

= =

where, the distance 2 2( ) ( )pj j jr x x y y= − + −

Next, let us put P at the control point of the jth panel ( , )i iP x y :

( )1

, ln2

nj

i i ij j

j j

x y r ds

=

= 2 2( ) ( )ij i j i jr x x y y= − + −

This is the contribution of all panels to the potential at the control point of the ith panel.

The free stream component normal to the panel is:

,ˆ cosn i iV V n V = =

The normal component of velocity induced at ( , )i ix y by a panel is:

( , )n i i

i

V x yn

=



Combining this with the potential ( , )i ix y , which is contribution of all the panels to the potential at the

control point of the ith panel:

( )1

( 1)

ln2 2

nji

n ij j

j ijj

V r dsn

=

= +

Applying the boundary condition, , 0n nV V + = , yields:

( )1

( 1)

ln cos 02

ni

ij j i

j ijj

r ds Vn

=

+ + =

Let: , ,(ln )i j i j j

ij

I r dsn

=

=> ( )

,

1 1

cos 02 2

nji

i j i

j j

I V

=

+ + =

This is a linear algebraic equation with n unknowns ( 1 2 3, , , , ,j n ) => these unknowns can

be solved (n equations with n unknowns)

Once ’s are all solved ( , 1,2,3 ,i i n = ) . . .

The component of free stream velocity tangent to the surface is:

, sins iV V =

The tangential velocity sV at the control point of the ith panel induced by all the panels is:

( )1

ln2

ni

s ij j

j j

V r dss s

=

= =

The total surface velocity at the ith control point iV is the sum of the contribution from the free stream

and from the source panels:

( ),

1

sin ln2

ni

i s s i ij j

j j

V V V V r dss

=

= + = +

Finally, the pressure coefficient at the ith panel can be given as: 2

, 1 ( )p i iC V V= −



Source panel code has two fundamental problems:

• Source panel does not simulate circulation. Therefore, the source panel code’s application is

inherently limited to a nonlifting flow around a body only.

• Source panel cannot enforce the Kutta condition at the trailing edge. Therefore, the stagnation

point is formed at an arbitrary location as the angle of attack is increased, and this causes the

nonphysical solutions.


Class Example Problem CClass Example Problem C--33--11

Related Subjects . . . Related Subjects . . . ““Source Panel MethodSource Panel Method””

There is a “theoretical” airfoil, called a “Joukowski” airfoil. The shape of this airfoil

can be obtained from a circle by a mathematical transformation between two domains.

Since this airfoil is a theoretically derived shape, the surface pressure coefficient

distribution (Cp) can also be obtained theoretically. Using source panel method,

calculate the pressure coefficient distribution (Cp) of symmetic Joukowski airfoil at

AOA = 0. Compare the theoretical distribution of Cp (Joukowski airfoil: known

values) against Cp computed by the source panel method.

Symmetric Joukowski Airfoil

VORTEX SHEET

A vortex filament can be defined as a straight line perpendicular to the page, going through point O

(extend to infinity both sides).

A vortex sheet can be defined as infinite number of vortex filaments side by side, with the strength of

each filament is infinitesimally small.

Consider an arbitrary point P (x, z). The small section of vortex sheet of strength ds induces a small

velocity potential and velocity dV :

2

dsd

= − and

2

dsdV

r

= −

The velocity potential at P due to entire vortex sheet from a to b is, therefore:

1( , )

2

b

a

x z ds

= − where, the circulation is:

b

a

ds =


Vortex Panel Method (1)Vortex Panel Method (1)

Consider a dashed path enclosing a vortex sheet of strength ds :

2 1 1 2 1 2 1 2( ) ( ) ( )C

v dn u ds v dn u ds u u ds v v dn = − = − − − + = − + − V ds

Let the top and bottom of the dashed line approach the vortex sheet:

0dn → , and 1 2u u = −

Important conclusion: the local jump in tangential velocity across the vortex sheet is equal to the local

sheet strength. This is the fundamental concept of thin airfoil theory.

The vortex panel method is built upon the idea that, if we replace the 2-D airfoil shape by a vortex sheet,

it is possible to obtain the lift using the Kutta-Joukowski theorem.

Thin airfoil theory and vortex panel method are built upon the same fundamental models of vortex sheet.

• If airfoil is thin => thin airfoil theory

• If airfoil is not thin => vortex panel method



1 2u u = −ds =

VORTEX PANEL METHOD

Very similar approach (as discussed in source panel method).

A series of vortex sheets approximates the body of arbitrary shape (vortex panels):

Total number of n panels with the vortex strengths per unit length of: 1 2 3, , , , ,j n

At the control points, the normal component of the velocity is zero:

, 0n nV V + =

1

cos 02

nj ij

i j

j ij

V dsn

=

− =

or

1

cos 02

nj

i ij

j

V I

=

− =

Note that the Kutta condition can be applied precisely at the trailing edge and is given by (TE) 0 = ,

means at trailing edge: 1i i −= −

This Kutta condition forces the flow field to have a proper stagnation point at (or near) the trailing edge

of the flow field.



Once the unknown vortex strength, 1 2 3, , , , ,j n , are obtained, the total circulation for the given

flow field can be calculated as:

1

n

j j

j

s=

=

Therefore, the lift per unit span is (from the Kutta-Joukowski theorem):

1

'n

j j

j

L V s

=

=



VARIETY OF SOURCE / VORTEX PANEL METHOD SCHEMES

• 1st order vortex panel method: the strength of vortex panel is constant for a given panel.

• 2nd order vortex panel method: the strength of vortex panel varies linearly over a given panel.

• Combination of source + vortex panels: use source panel to simulate the airfoil thickness, use vortex

panel to introduce circulation (lift).



There are a few important restrictions of using Vortex Panel Java Applet

• Airfoil coordinate data must be arranged that:

Trailing Edge (1,0) => Lower Surface Data => Leading Edge (0,0) => Trailing Edge

• Too much panels clustered near the trailing edge will make it difficult to properly apply Kutta

condition: you may want to manually remove some panels.

• Total number of panels should be in the range of 30 to 100. Using too many panels will slow down

computations and will not necessarily improve the accuracy.


Class Example Problem CClass Example Problem C--33--22

Related Subjects . . . Related Subjects . . . ““Vortex Panel MethodVortex Panel Method””

Using Virginia Tech vortex panel method (Java Applet), determine the lift coefficient

of NACA 0009 airfoil over a range of angles of attack from 0 to 16 degrees. Plot a lift

curve (cl v.s. ) and compare it against NACA airfoil data.

http://www.engapplets.vt.edu/

Lift Coefficient v.s. AOA

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

0 2 4 6 8 10 12 14 16

AOA (degrees)

Se

cti

on

Lif

t C

oe

ffic

ien

t

Vortex Panel Method NACA 0009 Data

AOA

(degrees) Vortex Panel Method NACA 0009 Data

0 0 0

1 0.118183

2 0.23633 0.2

3 0.354404

4 0.472371 0.4

5 0.590194

6 0.707837 0.6

7 0.825265

8 0.942441 0.85

9 1.05933

10 1.175896 1.08

11 1.292105

12 1.407919 1.26

13 1.523305

14 1.638227 1.3

15 1.752649

16 1.866538 1.14

Lift Coefficient



ROAD MAP . . .



C-3: Panel Methods




Thin Airfoil Approximation

Thin Airfoil Theory

Symmetric Thin Airfoil

Cambered Thin Airfoil

Real Flow over an Airfoil

This is for your REFERENCE purposes (self-study) only

• Useful for your future needs (aero-track courses)

• Not covered as a standard material in AE301

• Not included in any exam

• Online lecture (YouTube) Available

•

THIN AIRFOIL APPROXIMATION

If the airfoil is thin, a single vortex sheet can be used to approximate a thin airfoil by replacing it along

the camber line: thin airfoil theory.

PRANDTL’S CLASSICAL THIN AIRFOIL THEORY

The classical Prandtl’s thin airfoil theory is covered here:

• Rather than modeling the mean camber line by vortex sheet, the chord line (the straight line

connecting between leading edge and trailing edge) is represented by a vortex sheet as a function of

straight horizontal coordinate system (x): the vortex sheet is placed on the chord line: ( )x =

• The camber line is forced to be a streamline, and is calculated to satisfy this condition as well as

Kutta condition: ( ) 0c =

• For the camber line to be a streamline, the normal component of freestream is going to be cancelled

by the downwash generated by the vortex sheet (everywhere along the mean camber line);

mathematically: , '( ) 0nV w s + =


Thin Airfoil Approximation

At any point P on the camber line, where the slope of the camber line is dz dx , 1

, sin[ tan ( )]nV V dz dx −

= + −

For a thin airfoil, both and 1tan ( )dz dx− − are small values, and for a small angle: sin tan ,

so , [ ]nV V dz dx = −

The velocity dw at location x induced by the small elemental vortex d at location is given by:

( )

2 ( )

ddw

x

= −

−

Thus, the velocity ( )w x induced at location x by all the elemental vortices along the chord line can be

obtained by integrating dw from the leading edge ( 0 = ) to the trailing edge ( c = ) as:

0

( )( )

2 ( )

cd

w xx

= −

−


Thin Airfoil Theory (1)

Recall, for the camber line to be streamline:

, '( ) 0nV w s + = , where '( ) ( )w s w x for thin airfoil, so

0

( )0

2 ( )

cdz d

Vdx x

− − =

−

0

1 ( )

2

cd dz

Vx dx

= −

−

This is the governing equation of thin airfoil theory.


Thin Airfoil Theory (2)

SYMMETRIC THIN AIRFOIL SOLUTIONS (1)

For symmetric airfoil: 0dz dx = , therefore:

0

1 ( )

2

cd

Vx

=

−

Transformation from to : (1 cos )2

c = − , sin

2

cd d =

Transformation from x to 0 : 0(1 cos )2

cx = −

00

1 ( )sin

2 cos cos

dV

=−

The mathematical solution to this equation is:

(1 cos )( ) 2

sinV

+=


Symmetric Thin Airfoil (1)

SYMMETRIC THIN AIRFOIL SOLUTIONS (2)

The total circulation around the airfoil is: 0 0

( ) ( )sin2

cc

d d

= =

Solution of the thin airfoil theory is: 0

(1 cos )cV d cV

= + =

LIFT AND MOMENT COEFFICIENTS

From the Kutta-Joukowski theorem: 2'L V c V = =

The lift coefficient: 2

2

'

1

2

l

c VLc

q cV c

= = = 2

Moment at leading edge: 0 0

' ( ) ( )

c c

LEM dL V d = − = −

=> through transformation and integration: 2'

2LEM q c

= −

The moment coefficient: LE, le 2

'

2m

Mc

q c

= = − =4

lc−

The moment coefficient at quarter chord point (recall, from Unit A-5):

, 4 , le4 4 4

l l lm c m

c c cc c= + = − + = 0

CENTER OF PRESSURE AND AERODYNAMICS CENTER LOCATIONS

The location of center of pressure is: , 4

cp

1 1 0

4 4 2

m c

l

cx

c = − = − =

1

4

The location of aerodynamic center is: 0ac

0

1 0 1

4 2 4

mx

a = − + = − + =

1

4

• For a symmetric thin airfoil, the quarter-chord point is both the center of pressure and the

aerodynamic center.


Symmetric Thin Airfoil (2)

Using the thin airfoil theory:

At 4 degrees AOA: 4 degrees = 4( /180) = 0.06981 rad

(a) 2 2 (0.06981)lc = = = 0.4386

(b) , /4m cc = 0



Related Subjects . . . “Thin Airfoil Theory”

Consider a symmetric thin airfoil (such as NACA 0006) at 4 degrees angle of attack.

Using the thin airfoil theory, calculate the followings and compare them against NACA

airfoil data:

(a) The lift coefficient

(b) The moment coefficient at the quarter chord

From NACA 0006 airfoil data, at 4 degrees AOA: 4 degrees:

(a) lc = 0.4

(b) , / 4m cc = 0


Class Example Problem C-4-1 (cont.)

Related Subjects . . . “Thin Airfoil Theory”

CAMBERED THIN AIRFOIL SOLUTIONS

The generalization of the methods for a thin symmetric airfoil ( 0dz dx ) yields:

0

1 ( )

2

cd dz

Vx dx

= −

−

The mathematical solution to this equation is:

0

1

1 cos( ) 2 sin

sinn

n

V A A n

=

+ = +

The coefficients:

0 0

0

1 dzA d

dx

= − 0 0

0

2cosn

dzA n d

dx

=


Cambered Thin Airfoil

LEADING EDGE STALL

Example: NACA 4412

Characteristics of relatively thin airfoils with thickness ratios between 10 and 16 percent of the chord

length.

• Post-stall characteristics: rapid loss of the lift


Real Flow over an Airfoil (1)

NACA 4412

EFFECTS OF AIRFOIL THICKNESS ON STALL

Thin airfoil: thin airfoil is desired for high-speed applications (from high transonic to supersonic), in

order to minimize profile and wave drags.

Thick airfoil: thick airfoil is desired for low-speed applications (low subsonic), due to the favorable stall

characteristics.


Real Flow over an Airfoil (2)

NACA 4421

4 ds C- 2: Unit C - 1: List of Subjects

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Transcript of 4 ds C- 2: Unit C - 1: List of Subjects