Download - 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
-
8/2/2019 Calulatind Drag on Airfoil
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
-
8/2/2019 Calulatind Drag on Airfoil
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
-
8/2/2019 Calulatind Drag on Airfoil
4/15
16/01/12 Drag 4
-
8/2/2019 Calulatind Drag on Airfoil
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
-
8/2/2019 Calulatind Drag on Airfoil
6/15
sl,pl, tl are similarly defined for an arbitrary pointB on the lower
16/01/12 Drag 6
-
8/2/2019 Calulatind Drag on Airfoil
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
-
8/2/2019 Calulatind Drag on Airfoil
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
-
8/2/2019 Calulatind Drag on Airfoil
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
-
8/2/2019 Calulatind Drag on Airfoil
10/15
16/01/12 Drag 10
-
8/2/2019 Calulatind Drag on Airfoil
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
-
8/2/2019 Calulatind Drag on Airfoil
12/15
And for the lower surface
16/01/12 Drag 12
lllll dsdspNd qtq sincos
lllll dsdspAd qtq cossin
-
8/2/2019 Calulatind Drag on Airfoil
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
-
8/2/2019 Calulatind Drag on Airfoil
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
-
8/2/2019 Calulatind Drag on Airfoil
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