classical thin-airfoil theory · approaches for modeling viscous effects in potential flows follow...

Classical

Thin-Airfoil Theory

Enrique [email protected]

Extracted from [1].

2 – Thin-airfoil theory

• As an extension of the point vortex, imagine an infinitestraight vortex filament, of strength , which in anyperpendicular plane induces the same field as a pointvortex (i.e. the latter is simply a section of the vortexfilament).

• Then, a vortex sheet may be thought as an infinitenumber of vortex filaments, each of which hasinfinitesimal strength and extends to infinite or to aboundary of the flow. For this arrangement, the localvortex sheet strength (or circulation density) isdefined by

where s is the width of the sheet enclosed by theevaluation contour.

Preliminary: vortex filaments and vortex sheets (I)

0

1lims

ds

V s (1)

Taking an elemental section of the vortex sheet, the velocity induced at a point Pcan be written as

2dsdVr

(normal to r)

Similarly to a point vortex, the resulting velocity field satisfies continuity and it isirrotational (except at the sheet itself). In addition, the velocity field is finite andcontinuous at any point of the flow. Near the vortex sheet we can state

1 2 2 1

1 2 1 2( ) ( )

s d u s v n u s v n

u u s v v n

V s

where is an averaged strength. When s0, v1 andv2 become identical, so that (v1-v2)0 (normalcomponent is continuous). Furthermore, if n0 then

1 2( )s u u i.e. the local strength of the sheet is equal to the jump in the tangential velocity.

Extracted from [2].

(2)

(3)


As the velocity field due to a vortex sheet satisfies continuity and is irrotational(and decays away from the airfoil satisfying the uniform flow condition), thesolution of the airfoil problem can be reduced to find the circulation (s) whichmakes the airfoil contour to be a streamline of the flow (+Kutta condition!).Then,

ds

l V

However, for arbitrary shape airfoils no analytical solution exists for (s) andthis approach could not be solved until numerical computation allowed theintroduction of panel methods. For the time being, some simplifications will beintroduced for its solution.

It is interesting to note that vortex sheets placed along the airfoil contour can mimic to some extent the real flow behavior,because the resulting vorticity can be assimilable to that generated by viscosity inside the boundary layer. Someapproaches for modeling viscous effects in potential flows follow this idea.

Extracted from [1].


The thin-airfoil approximation• If the thickness is small (usually t/c<15%),

its effects can be neglected (when focusingon lift!) and the vortex sheet can be placedalong the camber line.

• In addition, if the camber is also small(usually f/c<4%), the vorticity distribution (s)can be placed along the chord line (planarwing approximation). Hence, = (x).

• For such arrangement, the boundaryconditions are similar as before: the meancamber line is a streamline of the flow (butwith some simplification) and the Kuttacondition is enforced at the trailing edge; i.e.(TE)=0. The latter avoids having infinitevelocity at the edge of the vortex sheet.

Extracted from [1].5 – Thin-airfoil theory

Problem statement

If the mean camber line is a streamlineof the flow, at any point P along z(x)we have

0n niV V

where Vin is the normal component of

the total velocity induced at P by theelemental circulations on the sheet, i.e.

2 330 0

0 1

cos cos1 ( ) 1 ( )cos2 2 cos

c cni

x ds x dxVr x x

3

( ) cos2

ni

x dsdVr

and V∞n is the normal component of the freestream velocity at P, i.e.

1sin tann dzV Vdx

(4)

(5)

(6)

Extracted from [2].


Using Eqs. (5) and (6), the streamline boundary condition (4) results

12 30

0 1

cos cos1 ( ) sin tan2 cos

c x dx dzVx x dx

which must be complemented with the Kutta condition

( ) 0TE

As the camber is small, we can assume that cos1cos2cos31. In addition, ifthe angle of attack is also small, sintan (in rad). Then, Eq. (7) results

1

00

1 ( ) tan2

c x dx dzVx x dx

which is the fundamental equation of the thin-airfoil theory. The solution ofthe problem consists in solving Eq. (9) for (x) subject to Kutta condition (Eq. (8)).

(7)

(8)

(9)

Note that Eq. (9) is an integral equation whichhas a singularity at x = x0. Thus, the principalvalue (finite part) of the integral is taken (seeAppendix C in [3]).7 – Thin-airfoil theory

Case of analysis I: symmetric airfoils

0 0

1 cos2sin2

1 cos2

cx

cdx d

cx

Then, the transformed thin-airfoil equations (8) and (9) can be written for thesymmetric airfoil (dz/dx=0) as

( ) 0

00

1 ( )sin2 cos cos

d V

(10)

(11)

(12)

Extracted from [4].For convenience, the following transformationof coordinates is firstly introduced


It can be shown that a solution satisfying Eqs. (11) and (12) is

1 cos( ) 2sin

V

Hint: introduce Eq. (13) into Eq. (11) and use the result obtained for the Glauert’sintegral (see a demonstration in Appendix E of reference [4]), given by

(13)

00

0 0

sincos 0,1,2,...cos cos sin

nn d n

For the The Kutta condition, introducing Eq.(13) into Eq. (12) we have

1 cos 0( ) 2 2sin 0

V V

indeterminate form

Thus, applying L’Hopital’s rule

sinlim ( ) 2 2 (0) 0cos

V V

(14)


Once () is determined, the total circulation can be computed by

0 0( ) ( )sin

cx dx d V c

and using the Kutta-Joukowsky theorem (l=V), the lift coefficient (per unitspan) results

2lC

where 2 is the lift slope. The moment coefficient can be obtained as follows

0 0

0( )

c c

LE

c

M x dL V x d

V x x dx

Then, using the transformation (10) it is possible to obtain

2 4LE

lm

CC

(15)

(16)

(17)

(18)

Extracted from [1].


From the definition of the center of pressure, we can state

( 4) 14

LEmcp l

l l

Cx Cc C C

Therefore, the moment about c/4 = constant = 0. Hence, xac = xcp = ¼ c.

Remarks:

• The lift is proportionally linear to the angle of attack. Lift slope = const. = 2(1/rad) 0.11 (1/deg).

• As expected, the angle of zero lift is L0 = 0.

• The moment (e.g. about the LE) is also linear with the angle of attack (or lift).

• The center of pressure (M=0) is located at a quarter chord point (c/4).

• The aerodynamic center (M does not depend on alpha) is coincident with thecenter of pressure. The Cmac (or Cm0) is zero for symmetric airfoils.

(19)


Although the Cp distribution cannot be computed because thickness wasneglected, the difference of pressure across the mean line can be obtained, e.g.taking the lift per unit area generated. Alternatively, from Eq. (3) we can state

VU

VL

1( ) ( )21 ( )2

U L U

L

x V V V V x

V V x

Then, using Bernoulli’s equation for two points above and below the vortexsheet we can state

2 21 12 2U U L L Up V p V V

and the jump in Cp results

(*) Hint: use Eq. (13) with thetransformation (10) and thefollowing half-angle formulae

sin 1 costan2 1 cos 1 cos

(20)

( ) 22 2

4

L Up p x c xCp Vq V V x

c xx

(*)


Note that Eq. (20) is singular at the LE.This causes an error when evaluating theforces through integration of the pressuredistribution. The contribution of theleading edge suction force can becomputed by alternative methods, see forinstance [3].

= L/x


Example: NACA 0006 airfoil

Experimental values (Re = 3-9E6)

alpha Cl

-4.0 -0.45

4.0 0.45

0.112lC

Note that for this airfoil the lift curveloses it linearity about 8 deg. Ingeneral, the scope of the idealtheory must be restricted to lowalpha. Typically, for > 8-10 degviscous effects are not negligible,and the ideal flow model losesaccuracy ( limits depend on Re,Mach and the airfoil geometry).

2lC

1/40mC 0 0l


Case of analysis II: cambered airfoilThe thin-airfoil equations to be solved are (see Eqs. (8) and (9))

( ) 0

00

1 ( )sin2 cos cos

d dzVdx

(21)

(22)

where x = c/2(1-cos) and 0. A vorticity distribution satisfying theseequations can be expressed as

01

1 cos( ) 2 A A sinsin n

n

V n

(23)

In Eq. (23), the leading term is very similar to Eq. (13) and accounts for alphaand camber effects. The second term is a Fourier series summation, thataccounts for camber effects, where the coefficients A0, A1,... must be determinedto fulfill Eq. (21). Note that Eq. (22) is satisfied by the definition of Eq. (23).


Introducing Eq. (23) into the streamline condition given by Eq. (21)

00 0

10 0

A A1 cos sin sincos cos cos cos

n

n

n dzd ddx

(24)

Recalling Glauert’s integral (see Appendix E in [4])

00

0 0

000

sincos 0,1,2,...cos cos sinsin sin coscos cos

nn d n

n d n

Using the integrals above, Eq. (24) results

0 01

A A cosnn

dzndx

(25)

(26) 1sin sin cos( 1) cos( 1)2

n d n n

(27)


Eq. (27) can be also expressed as

0 01

A A cosnn

dzndx

which is in the form of a Fourier cosine series expansion for dz/dx. In generalform, the latter is

01

( ) B B cos 0nn

f n

with0 0

0

1 ( )

2 ( )cosn

B f d

B f n d

Then, assuming f()=dz/dx, by analogy of Eqs. (28) and (29), the An coefficients(satisfying the streamline condition) result

(28)

(29)

0 0 0 00 0

1 2; cosndz dzA d A n ddx dx

Hence, Eq. (23) can be solved for a given airfoil and angle of attack by using thecoefficients (30). Note that () reduces to the symmetric case (Eq.(13)) if dz/dx=0(A0=). The leading terms A1,2,... depend only of the shape of the mean line.

(30)


The contribution of A0 and the An terms to the circulation are sketched below

• The terms A0, A1, A2,… satisfy the Kutta condition at the trailing edge.

• The terms An form a trigonometric expansion with enough capability toreproduce an arbitrary distribution of circulation.

• The term A0 provides more weight to the leading edge zone (where a highsuction area is expected). Note that A0 is higher in that zone and goes to zerotowards the trailing edge.

• The purpose of the constant 2V introduced in () is simply to cancel the 2V

term in the right-hand-side of Eq. (21).

Extracted from [3].


Remarks:

The total circulation can be obtained by

0 0( ) ( )sin

2c cx dx d

and using the Kutta-Joukowsky theorem (l=V), the lift coefficient (per unitspan) results

0 12lC A A

(31)

(32)

Introducing Eq. (23) into Eq. (31)

0 0 01

(1 cos ) A sin sinnn

cV A d n d

/2 for n=1 ; 0 for n1

the circulation results

0 1 2cV A A

(33)


Introducing Eqs. (30) into Eq. (33), the lift coefficient can be written as

0 00

12 (cos 1)ldzC ddx

It can be observed that, similarly to the symmetric case, the Cl varies linearlywith , with slope 2 (1/rad). Eq. (34) also shows that for cambered airfoils Cl0at =0. Making Eq. (34) equal to zero, the angle of zero lift L0 results

(34)

Hence, the lift coefficient can be also expressed as

where the term (-l0) is the absolute angle of attack. Note that in general l0 isnegative.

(35)

(36)


0 0 0 00

1 (cos 1)l ldz ddx

02 ( )l lC

The moment characteristics for a cambered airfoil can be obtained similarly tothe symmetrical case by using Eq. (17). This procedure yields

It is possible to note that the terms A1 and A2 in Eqs. (37) and (38) do not dependon or Cl; therefore, they only contribute to the airfoil free moment Cm0.

Introducing Eq. (33) into Eq. (37), the moment coefficient can be written as

(37)

(38)

A line parallel to V∞, which passes through the TE making anangle LO with respect to the chord is called the Zero-Lift-Line.The airfoil generates zero lift moving in that direction. Forsymmetrical airfoils the ZLL is coincident with the chord line. Extracted from [2].


20 1 2 2LEm

AC A A

2 14 4LE

lm

CC A A

From the definition of center of pressure, for the cambered airfoil we have

1 214 4

LEmCP

l l

Cx A Ac C C

where it is possible to observe that xcp varies with lift and tends to infinite whenCl0 (these characteristics make xcp useless for practical analyses). Takingmoment with respect to the quarter chord point, it is possible to obtain

which, as seen before, does not depend on alpha (A1 and A2 are only a functionof the shape of the camber line). Hence, it can be concluded that theaerodynamic center for the cambered airfoil is located at xac= xc/4.

(39)

(40)


1/4 2 1 4mC A A

• Similar to the symmetrical case, the lift is proportionally linear to the angle ofattack. The lift slope is constant = 2 (1/rad) 0.11 (1/deg).

• The angle of zero lift is different from zero (L0 < 0) and its value dependsonly on the shape of the mean line.

• The moment (e.g. about the LE) has two contributions: one due to Cm0,which depends on the camber line and is constant, and another which isproportional to the angle of attack (or lift).

• The center of pressure (M=0) varies with lift. It is located at the quarter chordpoint (c/4) when Cl = 0, and tends to infinite when Cl0.

• Similar to symmetric airfoils, the aerodynamic center (M does not depend onalpha) is located at c/4. The Cmac (or Cm0) depends on the shape of themean line and it is generally a negative number.

Summarizing, for cambered airfoils we have seen that


Example: NACA 23012 airfoil

Comparison between theoretical andexperimental data (Re = 3-9E6). See[1], example 4.6, pp. 313. 2lC

1/40mC

0 0l Theoretical Experimental

Cl @ = 4º 0.559 0.55

l0 -1.09 -1.1

Cm1/4 -0.0127 -0.01


Effects of camber on some aerodynamic variables

Extracted from [2].

Extracted from [6].

Extracted from [6].


(a)+(b)+(c)

• As seen before, for a general cambered airfoilthe distribution of circulation can bedecomposed into a component due to a flatplate (c) with angle of attack, and a componentdepending on the shape of the camber line (a).

• Similarly, profiting from the linearity of theproblem, the effect of trailing edge devices canbe studied by an additional camber of themean line caused by deflection of the flap (b).Thus, its circulation can be added to theprevious ones in order to obtain the resultsought (a+b+c).

Case of analysis III: extension to trailing edge flaps


The next analysis focuses particularly on plain-type trailing edge flaps.

Profiting from the linearity of the problem, we can write

0

0

l ll l

l ll

C CC

C C

where all the angles are small and the viscous effects are assumed to be negligible.Then, since the angle of attack is not affected by flap deflection, Eq. (41) results

(41)

(42)

The underlined term in Eq. (42) can be seen as a l0, caused by flap deflection,which generates a lift increment Cl. This quantity can be computed and added tothe basic camber line solution (without flap) to obtain the solution of the airfoil withflap deflected.


00

l ll l

CC

Focusing on the effects of flap deflection on the zero-lift angle, from Eq. (35) wehave that, for a general cambered airfoil

0 0 0 00

1 (cos 1) with 0ldz ddx

(43)

As it is possible to observe, the term (cos0-1) is zero at the LE and achieves itmaximum when =, i.e. at the TE. This indicates that the part of the mean linenear the TE has the highest impact on l0. The design of devices like trailing-edgeflaps or ailerons is based on this fact.

Example: a downward deflection of the mean linenear the TE makes l0 more negative, henceincreases the lift at the same geometrical .Although stall is reduced with flap deflection, thisreduction is lower than the gain achieved in lift,and Clmax increases. The lift slope is not affectedby flap deflection.

0

2l

lC


0 0 00 0

0 0 0

1 1(cos 1) ... ...

1 ( )(cos 1) sin

h

h

h

l

l h h

dz ddx

d

dz/dx=0

In order to compute the effect of the flap deflection, the following geometry isproposed

The l0 term can be evaluated from Eq. (43) as follows

(44)

1

1

0 x c E

x c E

dzdx

with E = 1-xh/c (flap-chord ratio) and tan

E1-E

h

c/2 xh (hinge)

Then, the incremental variation of the zero-lift angle due to flap deflection is

00

sin1l h hl

(45)

V

l0/ is usually calledthe flap efficiency factor.29 – Thin-airfoil theory

From Eq. (42), using Eq. (45) the incremental lift results

02 2 sinl l h hC

The change in the moment coefficient can be evaluated in a similar manner. Forexample, using Eq. (40) we can write

1/4 2 1 0 0 0 00 0

0 0 0 0

1 cos 2 cos4 2

1 ( )cos 2 ( )cos2 h h

mdz dzC A A d ddx dx

d d

and after integration Eq. (47) yields(*),

1/4

1 sin (1 cos )2m h hC

(*) Use the identity sin2A = 2·sinA·cosA

(46)

(47)

As it can be observed in Eqs. (45), (46) and (48), the incremental values for l0, Cl

and Cm vary linearly with the flap deflection .


(48)

Computation of the hinge moment coefficient

The aerodynamic moment at the flap hinge can be computed by

( ) ( )

( ) ( )

h h

h

c c

h h hx x

c

hx

M x x dL V x x d

V x x x dx

Then, using the transformation (10) and dividing by q∞c2(1), the dimensionlessform of Eq. (49) results

(49)

01

1 (cos cos ) ( )sin2

1 cos(cos cos ) A A sin sinsin

h

h

h h

h nn

C dV

n d

(50)


From Eqs. (30), the coefficients A0 and An are

02sin1 2(1 ) ; cos

h h

h hn

ndz dzA d A n ddx dx n

and replacing into Eq. (49), the hinge moment coefficient can be written as

1

(cos cos ) (1 ) (1 cos )

2sin(cos sin cos sin ) sin

h

h

hh h

hh

n

C d

n n dn

Collecting terms we can obtain

1 2

1 2 3 41

cos I I

2sin1 cos I I cos I I

h h

h hh h

n

C

nn

(52)


(51)

The integral terms I1-4 in Eq. (52) are:

1

2

3

4

I (1 cos ) sin

sin 2I (1 cos ) cos sin2 4

sin( 1) sin( 1)1I sin sin2 1 1

sin( 2) sin( 2)1I sin sin cos2 2 2

h

h

h

h

h h

h hh

h h

h h

d

d

n nn dn n

n nn dn n

Note that the hinge moment coefficient (52) is written in terms of the angle ofattack (which can be put in terms of the lift) and the flap deflection. Thus, it is acommon practice to express

0h h

h l hl

C CC C CC

where the theoretical derivatives usually include experimental corrections. Theterm Ch0 in Eq. (53) is added to account for the moment caused by the basic formof the airfoil, which is not accounted for in this development.


(53)

Other comparisons with experimental resultsI. Lift slope

0.107lC

• For t/c 24%, the Cl, predicted is within 15% of accuracy with respect to experimentalvalues (but note that such thick airfoils are not very frequent in practice).

• For typical airfoils with t/c about 12%, the accuracy is within 3%, which is a very goodapproximation. Comparable results can be obtained for other airfoil families, see [5].

Extracted from [5].


Extracted from [5].

II. Aerodynamic center

• For t/c 16%, the theoretical values forxac are within 4% error.

• The error increases with thickness andcamber. The maximum error observed inthe figures is about 12% (max. thickness).


Extracted from [5].

• In the NACA 6-series, particularly for the thickerairfoils, the aerodynamic center is located aft thequarter chord point.

• The delayed chordwise position of the minimumCp has an effect on xac.

A note on the vertical position of the ACIn real flows, the aerodynamic center isslightly above the chord line. The zcoordinate can be found by enforcing linearbehavior of the Cm and computing thedeviation. The vertical position of the AC iscomputed to compensate this deviation. Itcan be found [6] that z/c -7Cm/Cl2.


Extracted from [2].

• When E is low, the relation betweenthe boundary layer thickness and theflap size increases, thus the efficiencyachieved in practice is reduced.

• As expected, the agreement withtheory improves as E increases.

Extracted from [5].

III. Flap deflection


In order to account for real flow effects, some empirical corrections are usuallyapplied to the theoretical flap efficiency factor. For instance, the followingcorrection is proposed in [8]

results valid for t/c 12% and E 0.4

corre

ctio

n fa

ctor

0 0l l

C T

factor


(54)

where the subscripts C and T indicatecorrected and theoretical resultsrespectively, and factor is an experimentalvalue obtained for different types of trailingedge devices (see Figure on the left).

39 – References and complementary material

References

1. Anderson J. D. Jr. Fundamentals of aerodynamics, 3rd edition. McGraw-Hill BookCompany (1984).

2. Kuethe, A. M., Chow, C. Y. Foundations of aerodynamics. Bases of aerodynamicdesign. 5th edition. John Wiley and Sons (1998).

3. Katz J., Plotkin A. Low speed aerodynamics: from wing theory to panel methods.McGraw-Hill series in aeronautical and aerospace engineering (1991).

4. Karamcheti K. Principles of ideal-fluid aerodynamics. R. E. Krieger Pub. Co(1980).

5. Abbott, I. H., Doenhoff, A. E. V. Theory of wing sections. Including a summary ofairfoil data. New York: Dover (1959).

6. Hoerner, S. F., Borst, H. V. Fluid-dynamic lift: practical information on aerodynamicand hydrodynamic lift (1985).

7. Houghton, E. L., Carpenter, P. W. Aerodynamics for engineering students. 5th

edition. Butterworth-Heinemann (2003).

8. McCormick, B. W. Aerodynamics, aeronautics, and flight mechanics. 2nd edition.John Wiley and Sons (1995).

Enrique [email protected]

classical thin-airfoil theory · approaches for modeling viscous effects in potential flows follow...

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