AERO3630-Lec_Lifting Line_Finite Wing Theory

7/31/2019 AERO3630-Lec_Lifting Line_Finite Wing Theory

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AERO3630 Lecture on : Lifting Line/Finite Wing Theory

•You are now starting to learn how to

determine lift on a wing (3D)

Start from where you left off in Thin AirfoilTheory

Imagine the wing is made up of infinitesimal

airfoil sections spanning a distance ‘b’




Make the same Assumptions:

Ideal Flow (Potential Flow)

Incompressible flow

Further assumptions: •Low angle of incidence

•Negligible airfoil thickness

Using thin airfoil theory obtain the total Circulation for each

section so that the circulation varies in the y direction ,)( y




In the thin airfoil theory, i.e., for infinite wing

span, the camber line was replaced by a string

of line vortices of infinitesimal strengths

Now in a similar manner we are looking for a

vortex strength distribution produced by the

flow field around a 3D lifting body




Now a Wing can be replaced by a system of

vortex systems:

•Starting Vortex

•Free Trailing Vortex

•Bound Vortex




The starting vortex is created when

the wing starts its motion or when air starts to flowover a wing.

The free trailing vortex is also created during themotion

Hence, starting and trailing vortices are physical

entities and can even be seen if conditions are right

The difficulty lies with bound vortex.






All we will consider is that:

The bound vortex system is a hypotheticalarrangement of vortices which replaces the

real wing in every way except that of

thickness.

It is largely concerned with developing the

bound vortex system which simulatesaccurately or at least a little distance away, allthe properties, effects, disturbances, force

systems etc., due to the real airfoil.




WARNING!

Do not interpret the representation of the

wing by a bound vortex system to be a

rigorous model. What it does is to allow a

relationship to be established between the:

•physical load distribution for the wing

•trailing vortex system




Prandtl hypothesised that:

Each section of the wing acts as though it is an

isolated 2D section provided the span wise

flow is not great.

Let us now see how the vortex system that

replaces the wing looks like in the next slide







Now further assumptions:

We can always argue that after a long time thestarting vortex system will be pushed away at such a

great distance from the wing that its effects would

have been mitigated by viscosity and other factors

and become negligible

Let us see how this looks like in the next slide

Note: Pay attention to the lifting line and the natureof the Circulation distribution on this line







We are now going to apply some logical argument

based on practical observations to justify the natureof circulation in the span wise direction

The flow is going to behave :•more like a 2D flow at the centre of the wing , i.e.,

Circulation is maximum at the middle or mid-span.

•and less 2D or become more 3D as it moves

towards the wing tip

•Since the flow has to break away at wing tip, the

Circulation will be zero at wing tip

)( y




Then:

The Maximum is at mid point, i.e., at y=0

And at the two ends , i.e., at the tips:

and

)(0 y

0)( ss

0)( ss




Now since this is finite wing, according to Newton’s Laws

every action has an equal and equal and opposite reaction.

This reaction is essentially the measure of the induced drag or

the downwash created.

Using Biot-Savart law, the flow velocity induced or associated

with a 3D vortex filament element can be expressed as:

Or

r

y ywi

4

)()(

)(2

1)(

2

1)(

0 y ydy

yd ydwi




If the origin is taken at the centre of the bound

vortex, then the velocity at any point y along thebound vortex induced by the trailing semi infinite

vortices is:

Or

)2 / (4)(

)2 / (4)()(

yb y

yb y ywi

22)2 / (4

)()(

yb

y ywi




Let us now see how the effects of the trailing vortices

can be superimposed on the lifting line graphically.

We will see that this looks like a horse shoe vortex

system in the next slide







Now we are in a position to work out the lift andinduced drag on a finite wing

Obtain the circulation for each wing section (airfoil).

Place the total circulation on the quarter chord

length of each airfoil section on the wing.

The locus of these points is the ‘Lifting Line’

Let us see how this looks like in the next slide







Once the circulation distribution and the induced velocity

distribution are known or defined, By putting the co-ordinate system at the centre of the wing:

the total lift and the drag for the wing can be obtained in

the following manner

Where s and –s are the semi span or b/2 and –b/2

s

s

dy yU y L )()(

s

s

ii dy y yw y D )()()(




CONSIDR AN EXAMPLE:

The lift and induced drag for:Elliptic Lift Distribution

So that:

And:

2

1)(

s

y y o

s

s

dys

yU y L

2

0 1)(

s

s

i dys

y yw y D

2

01 1)()(




Similar to thin airfoil theory, it is better to work in angles:

Where , the location of any point on the lifting line is given by

coss y

dysU y L

s

s

sin)cos1()(2

0

0

2

0

04

sin)( U bd sU y L




The lift co-efficient then becomes:

Where S is the wing area

02

S U

bC L




Before working out the induced drag let us first work out

the total downwash velocity:

So that:

dy y y ys

y

sw

s

s

i

)(40

22

0

dy y y ys

y

sdy

d s

s

)( 0

222

0

dy y y ys

y

y y ys

y y

sw

s

s

i

)

)()

)((

40

22

0

0

22

00




Integrating between -s and s with respect to y:

Note: The value is constant across the span

ss

wi4

)0(4

00

s

wi

4

0




Now:

Where, for elliptic distribution:

giving:

Or:

s

s

ii dys

y

yw D

2

0 1)(

s

wi

4

0

2

0

8

i D

0

2

00 sincos1

4d s

s Di




Now:

Where S is the area of the wing

Recalling, aspect ratio is given by: b2/S

S U

C i D

20

4




Recalling, aspect ratio to be given by: b2/S

This is the expression for minimum induced drag

2

22

2

0 2

44

b

S U C

S U S U C L Di

)(

2

2

2

AR

C

b

S C C L L Di




Some other points of note:

The resultant induced velocity at a point is, in general in thedownward direction, and is called ‘downwash’, where

Or

i

i

U

w

U

w y y

i )(tan1







The downwash has the effect of ‘tilting’ the undisturbed air,

so the effective angle of attack of the aerodynamic centre

(i.e., the quarter chord) is

Where is the effective angle of attack (3D)

is the downwash angle or induced angle

is the angle of attack (2D)

ie

e

i




The downwash has the effect of ‘tilting’ the undisturbed air,

so the effective angle of attack of the aerodynamic centre

(i.e., the quarter chord) is

ie

e

i




The lift curve slope relations are:

)(1 2

2

3

ARa

a

d

dC a

D

D L

D



AERO3630 Lecture on : Techniques for General Circulation Distribution

Consider a span wise circulation distribution than can berepresented by a Fourier sine series consisting of n terms

Then:

Note, here too applies:

and

coss

y

]sin[4)(

1

n BsU n

n

]sin[4)()(1

2 n BU sU Ln

n




But:

And

Replacing by

)(2

cU

C L

)( 03 e D L aC

LC )(2 cU

)()]()([)(

)(20

U aU ac e




But

Then:

dy y y

dy

d

wU

s

s

)(4

1

0

sin

sin1

n

n nnBU w

)]()([

)(

)(20

aU

ac e

sin

sin1

n

n nnBU




Since:

Substituting and rearranging:

Let

And rearranging:

This is the Mono-plane equation

]sin[4)(1

n BsU n

n

n Bcas

n

n

e

sin81 )]()( 0 a

sin

sin1

n

n nn B

)]()([sin 0 a

s

cae

8

)sin(sin1

nnnBn

n




224 U s

d n Bn sinsin0 L

If we consider the symmetrical loading distributions, only odd terms of the

series need considering:

......]5sin3sin]sin[4)( 531 B B BsU

L22

4

U s

0

1

4

2sin

2

B

224 U s

03 1

)1sin(

1

)1sin(

n

n

n

n Bn

n




Now:

And:

Giving:

But

Giving

24

2sin

2 01

B

1,01

)1sin(

1

)1sin(

03

n

n

n

n

n Bn

n

1

22)

2

1)(4( BU s L

S U C L L )21(

2

)(1 AR BC L




Similarly for Induced drag:

But

Thus:

s

s

ii dy y yw D )()(

0

22

4 U s Di d n B

n

n sin1

0

nnBn

n sin1

d n Bn

n sin1

n

nnB1

2

2

)( ARC i D

n

nnB1

2




Now recall:

Giving

But

Hence:

Or

)(1 AR BC L

)(1

AR

C B L

)( ARC i

D n

nnB1

2

)(

2

AR

C C L

Di

nn

B

Bn

1

2

1

)(

)(

2

AR

C C L

Di .......)]

753(1[

2

1

2

7

2

1

2

5

2

1

2

3 B

B

B

B

B

B




Remember: If we consider the symmetrical loading distributions, only odd terms of

the series need considering:

So that

Or

where:

And [can you guess why??]

Then:

)(

2

AR

C C L

Di .......)]

753(1[

2

1

2

7

2

1

2

5

2

1

2

3 B

B

B

B

B

B

)(

2

ARC C L

Di )1(

0

.......)753

(2

1

2

7

2

1

2

5

2

1

2

3

B

B

B

B

B

B

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AERO3630-Lec_Lifting Line_Finite Wing Theory