Theodorsen theory for deformable airfoils full derivation

7/29/2019 Theodorsen theory for deformable airfoils full derivation

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0.1 Derivation of Theodorsen Aerodynamic Model

Theodorsen assumed the magnitude of oscillation to be small and the oncom-ing velocity to be constant. In the present work, Theodorsen model is slightlymodified to include time varying oncoming velocity and for large magnitude ofoscillation. It is assumed that the trailing wake is harmonic in this case.

Theodorsen model is developed for thin airfoils essentially a flat plate. Con-formal mapping is used to model the flow around a flat plate from the flowaround a 2D circle. The non-circulatory lift is generated with a source-sinksheet distributed along the chord. The strength of the source-sink sheet is de-termined from the boundary condition that there is no flow penetration throughthe surface of the airfoil. The circulatory lift is generated with a vortex sheeton the airfoil and in the wake that extends upto infinity. The strength of thevortex sheet is determined from the kutta condition that the flow leaves the

trailing edge smoothly.

The Joukowski conformal mapping that is used to map a circle of radius bcentered at the origin to a flat plate of semichord 2b is given by

x + iy = X+ iY +b2

X+ iY(0.1)

where(x, y)- mapped domain(flat plate)(X, Y)-unmapped domain(circle)

This transformation maps all points outside the circle to a location outside

the flat plate, points inside the circle to a location outside the flat plate but ondifferent Rieman surface, points on the circle to locations on the flat plate.For an airfoil, the deformation is given by

y =h

b+ x (0.2)

wherex and y are non-dimensionalized with respect to semi-chord(b).

y and h are positive downwards and is positive clockwose.

Non-Circulatory Flow

The Non-circulatory flow is represented by source-sink pair sheet.The net velocity potential due to source-sink sheet distributed along the chordof the flat plate is given by

=

1c

sourcesinkbdx1 (0.3)


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wherec is the location of the leading edge of flat plate.

The velocity potential at a point on the unit circle (X, Y) due to a sourceor sink placed at origin is given by

=

4log

X2 + Y2

(0.4)

where- Strength of source or sink(X, Y)- Spatial location in the unmapped domain(circle)

The velocity potential of a source or sink placed at (X1, Y1) is given by

=

4log(XX1)2 + (Y Y1)2 (0.5)

Putting the double source of strength of 2 at (X1, Y1) and a double sink ofstrength of2 at (X1,Y1) as shown in [1], we get the flow around a circle[Figure 0.1]. The combined velocity potential is obtained by the summation ofindividual potentials. The combined velocity potential in the unmapped domainis given by

=

2log

(XX1)2 + (Y Y1)2

(XX1)2 + (Y + Y1)2(0.6)

where is defined on a unit circle and it implies Y =

1X2 and Y1 =1X12.

Using Joukowski mapping, we map the points on the circle directly to locationson the flat plate.

x + iy = X+ i

1X2 + b2

X+ i

1X2(0.7)

x + iy = X+ i

1X2 + X i

1X2X2 + (1X2) (0.8)

x + iy = X+ i

1X2 + X i

1X2 (0.9)x = 2X (0.10)

By Joukowski transform of a unit circle, the flat plate domain is from 2 to 2.Applying change of variables to change the domain from

1 to 1, x becomes

2x.This implies

x = X (0.11)

Y =

1 x2 (0.12)


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Fig. 0.1: Conformal mapping of flat plate by a circle

Applying the transformation to [Eqn 0.6], we get the velocity potential due tosource-sink pair as

sourcesink =

2

log(x x1)2 + (y y1)2

(x x1)2

+ (y + y1)2 (0.13)

Put [Eqn 0.13] in [Eqn 0.3], we get

=

1c

2log

(x x1)2 +

1 x2

1 x122

(x x1)2 +

1 x2 +

1 x122 bdx1 (0.14)

The strength of source or sink as determined by the boundary condition thatthere is no flow penetrating the airfoil [Figure 0.2]. At the airfoil surface, thevelocity must be perpendicular to the surface and the velocity of the fluid flowingout of the source is the strength of the source. To satisfy this, the velocitygenerated by the source or sink must be equal and opposite to the componentof freestream minus the transverse motion of the airfoil.

v x

y = ty (0.15)

Changing to the non-dimensionalised quantities,

= v

xy + b

ty (0.16)


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wherev-free stream velocity

y-motion of the airfoil which is positive upwards non-dimensionalised with re-spect to semi-chord(b).

The motion of aileron is represented by

y = (x c) (0.17)

wherec-location of aileron hinge from the midpoint.

Fig. 0.2: Airfoil with no penetration boundary condition Ref.[2]

To account for the effect of flap deflection by an angle (positive clockwise), is replaced by

= v (0.18)Put [Eqn 0.18] in [Eqn 0.14], we get

=

1c

log(x x1)2 +

1 x2

1 x12

2(x x1)2 +

1 x2 +

1 x12

2 bdx1 (0.19)

From [1], we have1c

v2

log(x x1)2 +

1 x2

1 x12

2(x x1)2 +

1 x2 +

1 x12

2 dx1 =2 (x c) (log) N 2

1 x2 cos1 (c)

(0.20)


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where

N= 1 cx1 x21 c2x c (0.21)

The net velocity potential due to is

=v b

1 x2 cos1 (c) (x c)log N

(0.22)

where

N=1 cx

1 x2

1 c2

x c (0.23)

To account for the effect of flap deflecting at an angular velocity ddt is repre-

sented by = (x1 c) ddtb which gives

ddt =

1

c

(x1 c) ddtb2

log(x x1)2 +

1 x2 1 x122(x x1)2 +

1 x2 +

1 x12

2 bdx1 (0.24)From [1], we have

1c

(x1 c)log(x x1)2 +

1 x2

1 x12

2(x x1)2 +

1 x2 +

1 x12

2 dx1 =

1 x2

1 c2 cos1 (c) (x 2 c)

1 x2 + (x c)2 logN (0.25)

The net velocity potential due to ddt is

ddt =ddtb

2

2

1 x2

1 c2 + cos1 (c) (x 2 c)

1 x2 (x c)2 logN

(0.26)

To account for the effect of angle of attack of the airfoil , put c = 1 in [Eqn0.22] which gives

= v b

1 x2 (0.27)To take into account, the effect of airfoil in downward motion with a velocityddth, replace in [Eqn 0.27] by tan

1

d

dth

v

which gives

ddth(t) = tan1

ddthv

vb

1 x2 (0.28)

To account for the rotation of airfoil around a at an angular velocity ddt, weshould consider two cases.


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1. Effect of rotation of the entire airfoil around the leading edge obtained bysubstituting c =

1 in [Eqn 0.26].

2. Subtract the effect of the vertical velocity of the actual point of rotation

which is obtained by substitutingd

dt(1+a)b

v for in [Eqn 0.27] .

ddt =

1

2

d

dt

b2 (x + 2)

1 x2

d

dt

(1 + a) b2

1 x2 (0.29)

which gives

ddt =

d

dtb2

x2 a

1 x2 (0.30)

The various integrals evaluated are listed here.

1

c

dx =1

2bvT4 (0.31)

11dx =

1

2bv (0.32)

d

dt

11dx =

b

2

vd

dt +

d

dtv

(0.33)

1c

ddthdx =

12

bd

dthT4 (0.34)

11 ddthdx =

b

2 v tan1

ddth

v

(0.35)

d

dt

11 ddthdx =

b

2

d

dtv

tan1

ddth

v

+

b

2 v

d2

dt2h

v (ddth) ddtv(v)2

1 +( ddth)

2

(v)2

(0.36)1c

ddtdx =

d

dtb2T9 (0.37)

11 ddtdx =

12

d

dtb2 a (0.38)

d

dt

11 ddtdx =

12

d2

dt2b2 a (0.39)

1

c

dx =12

bv T5

(0.40)

11dx =

12

bv T4 (0.41)


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d

dt 1

1

dx = T4b

2 vd

dt+

d

dtv (0.42)1

c

ddtdx =

12

b2d

dtT2 (0.43)

11 ddtdx =

12

b2d

dtT1 (0.44)

d

dt

11 ddtdx =

12

b2d2

dt2T1 (0.45)

where T are Theodorsen constants which are given below.

T1 = 1/31 c2

2 + c2

+ c cos1 (c) (0.46)

T2 = c

1 c2

1 c2

1 + c2

cos1 (c) + c

cos1 (c)2 (0.47)

T3 =

1/8 + c2

cos1 (c)2

+ 1/4 c

1 c2 cos1 (c)

7 + 2 c2

1/8

1 c2

5 c2 + 4 (0.48)

T4 = cos1 (c) + c

1 c2 (0.49)

T5 = 1 + c2

cos1 (c)2

+ 2 c

1 c2 cos1 (c) (0.50)T6 = T2 (0.51)

T7 =

1/8 + c2

cos1 (c) + 1/8 c

1 c2

7 + 2 c2

(0.52)

T8 =

1/31 c

2 2 c2 + 1+ c cos1 (c) (0.53)T9 = 1/2p + 1/2 aT4 (0.54)

wherep = 1/3

1 c2

3/2T10 =

1 c2 + cos1 (c) (0.55)

T11 = cos1 (c) (1 2 c) +

1 c2 (2 c) (0.56)

T12 =

1 c2 (2 + c) cos1 (c) (2 c + 1) (0.57)T13 = 1/2T7 1/2 (c a)T1 (0.58)

T14 =1

16+

1

2ac (0.59)

Non-Circulatory Forces

The forces can be calculated by finding out the pressure difference and integrat-ing over the surface. Bernoulli equation for unsteady flow is given by

p0 = p + 1/2 V2 +

t (0.60)


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Applying for free stream condition, we get

p0 = p + 1/2 v2 (0.61)

where v is the free stream velocity.

Bernoulli equation is linearized for small disturbances along x and y namelyv

and w

respectively.

V =v + v

i + w

j (0.62)

V2 = v2 + 2 vv

(0.63)

The linearised Bernoulli equation is given by

pp = v

x +

t (0.64)

Non-dimensionalising x with respect to semi-chord b, we get

pp = v

b

x +

t

(0.65)

The pressure difference between the upper and lower surface is given by

p = 2v

b

x +

t

(0.66)

The lift is obtained by integrating the pressure difference over the entire airfoilas follows

LNC = b11pdx (0.67)

which gives

LNC = b2 v

d

dt +

d

dtv +

d

dtv tan1

ddth

v

b2 v

d2

dt2h

v d

dth ddtv

v2

1 +( ddth)

2

v2

ba d2

dt2

+ b2T4

vd

dt+

d

dtv

bT1

d2

dt2

(0.68)

which is positive downwards.The non-circulatory flow gives forces due to accelerations only.

Circulatory Flow

The non-circulatory lift force is derived by considering the boundary conditionthat there is no flow penetrating the airfoil. The non-circulatory flow model does


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not satisfy kutta condition which states that the flow should leave the trailingedge smoothly. The circulatory flow is developed using a vortex sheet on the

airfoil as well as in the wake that extends upto infinity as show in [Figure 0.3].These pair of vortices does not violate the no penetration boundary conditionas the normal velocities on the unit circle due to this vortex pair is zero.

Fig. 0.3: Vortex locations for the airfoil-aileron combination

The velocity potential of a point vortex is given by

=

2 tan1 Y Y1XX1 (0.69)

where is the Vortex strength(X, Y) is the spatial location in the unmapped domain(unit circle)(X1, Y1) is the location of point vortex in the unmapped domain

The combined velocity potential for a pair of point vortices, one outside cir-cle at a distance X0 and the other inside the circle at a distance

1X0

is givenby

=2

tan1

Y Y0XX0

tan1

Y Y0X 1X0

(0.70)

whereX and Y are spatial variables in the unmapped plane is the vortex strength at X0


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Using properties of tangents, the above equation gets simplified to

=2

tan1X0 1X0

Y

X2 X0 +

1X0

X+ Y2 + 1

(0.71)Applying Joukowski transformation,For X0,

x + iy = X0 +1

X0(0.72)

x0 = X0 +1

X0(0.73)

For 1X0 ,

x + iy =1

X0 +11

X0

(0.74)

x0 =1

X0+ X0 (0.75)

Both X0 and1X0

in the unmapped domain get transformed to the same locationx0 in the mapped domain.We have,

x0 =1

X0+ X0 (0.76)

2X= x (0.77)

The mapped domain is from -2 to 2 for x and from 2 to for x0. Apply changeof variables to change the domain for x from -1 to 1 and for x0 from 2 to , weget

x0 2x0 (0.78)x 2x (0.79)

This implies,

X0 +1

X0= 2x0 (0.80)

X= x (0.81)

Y =

1 x2

(0.82)

X0 = x0 +x02 1 (0.83)

1

X0= x0

x02 1 (0.84)


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Substitute [Eqns 0.81-0.84] in [Eqn 0.71], we get

=2

tan1

1 x2

x02 1

1 xx0

(0.85)

The Pressure difference between the upper surface and lower surface is given by

p = 2 v

x +

t

(0.86)

Since the disturbance is assumed to move with free stream velocity, we have

t = v

x0 (0.87)

The pressure difference is given by

p = 2 v

x +

x0

(0.88)

Non-dimensionalising x and x0 with respect to semi-chord b, we get the pressuredifference as

p = 2 vb

x +

x0

(0.89)

Using [Eqn 0.85], we get from [1],

2 x

=

x02 1

1 x2 (x0 x)(0.90)

2 x0

=

1 x2

x02 1 (x0 x)(0.91)

The pressure difference is given by

p = vb

1 x2

x02 1 (x0 x)+

x02 1

1 x2 (x0 x)

(0.92)

which gets simplified to

p = vb

(x0 + x)1 x2

x02 1

(0.93)

Now the lift force on the airfoil is obtained by integrating over the domain

LC = v

11

x0 + x1 x2

x02 1

dx (0.94)

LC = vx0

x02 1(0.95)


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Replacing with Ubdx0 where U is the strength of the vortex per unit lengthand integrating [Eqn 0.95] from trailing edge to infinity gives the circulatory

lift.LC = vb

1

x0Ux02 1

dx0 (0.96)

Similarly, the velocity potential of the vortex sheet is given by

0 = b

2

1

tan1

1 x2

x02 1

1 xx0

Udx0 (0.97)

Circulatory Forces

The Magnitude of vortex strength is determined by kutta condition which statesthat the flow should leave the trailing edge smoothly. In other words, velocitiesat the trailing edge should not be infinite.

For finite trailing edge angle(t.e > 0) as in [Figure 0.4], the velocities on theupper and lower surfaces must be tangential to their respective surfaces. Thisimplies the existence of two different velocities at the trailing edge. So, the onlyrealistic option is,

Vupper = Vlower = 0 (0.98)

Fig. 0.4: Flow around trailing edge- Kutta condition

Thus at x = 1,

x +

x ddth +

x +

x ddt +

x +

x ddt = finite = 0 (0.99)

At x = 1, 1 x2b

x =

1

2

1

Ux0 + 1x0 1

dx0 (0.100)

1 x2b

x = v (0.101)


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1 x2b

x ddth = v tan1

ddth

v(0.102)

1 x2b

x ddt = b (1/2 a)

d

dt (0.103)

1 x2b

x =

vT10

(0.104)

1 x2b

x ddt =

1

2

bddt

T11

(0.105)

Substituting the terms, the [Eqn 0.99] gives an important expression.

1

2

1

x0 + 1Ux0 1

dx0 =

v + v tan1ddth

v + b1

2 a

d

dt + v

T10

+

1

2

bT11ddt

= Q

(0.106)

Multiply [Eqn 0.96] by Q and divide it by 121

x0+1x01

Udx0 we get

LC = 2vbQ1

x0x021

Udx01

x0+1x01Udx0

(0.107)

which is positive downwards.

LC = 2vbQC(k) (0.108)where C(k) is the Theodorsen Lift deficiency function given by

C(k) =

1

x0x021

Udx0

1x0+1

x01Udx0

(0.109)

and its solutions are bessels function.

The complete expression for Lift is given by

L = b2 v

d

dt +

d

dtv +

d

dtv tan1

ddth

v

b2 v

d2

dt2h

v d

dth ddtv

v2

1 +( ddth)

2

v2

ba d2

dt2

+ b2T4 vd

dt+

d

dtv bT1

d2

dt2

2vbC(k) v + tan1

ddth

v

+b12 a

ddt

v

2vbC(k) vT10

+

1

2

bT11ddt

v

(0.110)


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which is positive downwards.

The above equation is the Theodorsen equation modified for varying oncom-ing air stream and for large magnitude of oscillations.


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BIBLIOGRAPHY

[1] Theodorsen, Theodore General theory of aerodynamic instability and themechanism of flutter, Technical report, National Advisory Committee forAeronautics, Technical report No. 496, 1934.

[2] Walker, William.Paul. Unsteady aerodynamics of deformable thin air-foils, Masters thesis, Virginia Polytechnic Institute, 2009.

Theodorsen theory for deformable airfoils full derivation

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Transcript of Theodorsen theory for deformable airfoils full derivation