calulatind drag on airfoil

8/2/2019 Calulatind Drag on Airfoil

1/15

Calculating Drag on Airfoil

Aerodynamic forces and

moments on a solid object

due to the flow come from

only two sourcesPressure distributionp(s)

Shear stress distribution

t(s)

where s is a curvilinearcoordinate measured along

the surface of the body from

some fixed reference point

16/01/12 Drag 1

p(s) is normal to the surface, and

t(s) is tangential to the surface


2/15

16/01/12 Drag 2

The total aerodynamic forceR and momentM

acting on the body, are the resultant force and

moment arising fromp(s) andt(s). See Fig 1.9


3/15

Now,

N= normal force, perpendicular

to c

A = axial force, parallel to c

Where the chord c is the linear

distance from the LE to the TE

a= angle between v and c Hence ais also the angle between

L andNand betweenD andA

The relationship between the two

sets of force components is

L =NcosaA sina(1.1)

D =Nsina+A cosa(1.2)

16/01/12 Drag 3


4/15

16/01/12 Drag 4


5/15

Integrating the Pressure and Shear Stress

Let us examine in detail the

integra-tion of the pressure and

shear stress distributions to obtain

the aerodyna-mic forces and

moments Start with Fig 1.11 and note:

The chord line is drawn horizontal

Hence v is at an angle a

below the chord line

xy coordinate system oriented

parallel and perpendicular to the

chord

su= distance from LE measured

along upper surface to any pointA

pu, tu= pressureand shear stress

distributions on upper surface atA.

Both are functions ofsu

16/01/12 Drag 5


6/15

sl,pl, tl are similarly defined for an arbitrary pointB on the lower

16/01/12 Drag 6


7/15

Deriving the Force and Moment Equations

The pressure is inclined at an

angle qw.r.t. the vertical.

Therefore, the shear stress is

inclined at an angle q w.r.t. the

horizontal (Ref next slide) q is positive when

measured clock-wise from the

vertical line to the direction ofp,

and from the horizontal line to the

direction oft We are now ready to derive the

aero-dynamic force and moment.

16/01/12 Drag 7


8/15

16/01/12 Drag 8

q is positive when measured clock-wise from thevertical line to the direction ofp, and from the

horizontal line to the direction oft


9/15

Deriving the Force and Moment Equations

Consider an elemental strip of

width ds and unit span as shown in

Fig 1.12

(Ref 2nd next slide) The area of

this elemental strip is dS = ds(1) =ds

This is shown shaded in Fig. 1.12

16/01/12 Drag 9


10/15

16/01/12 Drag 10


11/15

NandA represent the total normal and axial forces on the wing

N and A represent the total nor-mal and axial forces per unit

span

For the upper surface we have

16/01/12 Drag 11

uuuuu dsdspNd qtq sincos

uuuuu dsdspAd qtq cossin


12/15

And for the lower surface

16/01/12 Drag 12

lllll dsdspNd qtq sincos

lllll dsdspAd qtq cossin


13/15

The total normal and axial forces per unit span are obtained by

integrating these equations from LE to TE for both surfaces.

Then

And

Therefore

16/01/12

Drag

13

lu NdNdNd

)( TE

LE

luNdNdN

TE

LE

uuu dspN )sincos( qtq TE

LE

lll dsp )sincos( qtq


14/15

Axial Force per Unit Span

The axial force perunit span is

from which is

and therefore

The lift and drag per unit span can

then be calculated by substituting

(1.7) and (1.8) into (1.1) and (1.2)

Now lets move on to obtain the

moment of these aerodynamic

forces

Clearly, the moment depends on

the point about which the momentis taken. We shall take moments

about the LE

16/01/12 Drag 14

lu AdAdAd

TE

LE

uuu dspA )cossin( qtq TE

LE

lll dsp )cossin( qtq


15/15

Finding Axial Component

16/01/12 Drag 15

The relationship between the two sets of force components is

L =NcosaA sina (1.1)

D =Nsina+A cosa (1.2)

As N and A are known

L and D can be determined

TE

LE

uuu dspN )sincos( qtq TE

LE

lll dsp )sincos( qtq

TE

LE

uuu dspA )cossin(qtq

TE

LE

lll dsp )cossin( qtq

calulatind drag on airfoil

Documents