06 Theodorsen Aerodynamics - · PDF fileAERO0032-1, Aeroelasticity and Experimental Aerodynamics, Lecture 6 3 2D wing oscillations FConsider a 2D airfoil oscillating sinusoidally in

AERO0032-1, Aeroelasticity and Experimental Aerodynamics, Lecture 6

Lecture 6: Unsteady Aerodynamics –

Theodorsen

G. Dimitriadis

Aeroelasticity

1

AERO0032-1, Aeroelasticity and Experimental Aerodynamics, Lecture 6 2

Unsteady Aerodynamics

  As mentioned earlier in the course, quasi-steady aerodynamics ignores the effect of the wake on the flow around the airfoil   The effect of the wake can be quite significant   It effectively reduces the magnitude of the

aerodynamic forces acting on the airfoil   This reduction can have a significant effect on

the values of the flutter


2D wing oscillations   Consider a 2D airfoil oscillating sinusoidally in an

airflow.   The oscillations will result in changes in the circulation

around the airfoil   Kelvin’s theorem states that the change in circulation

over the entire flowfield must always be zero.   Therefore, any increase in the circulation around the

airfoil must result in a decrease in the circulation of the wake.

  In other words, the wake contains a significant amount of circulation, which balances the changes in circulation over the airfoil.

  It follows that the wake cannot be ignored in the calculation of the forces acting on the airfoil.


Kelvin’s Theorem

  The theorem states that:

  For the oscillating airfoil problem, this means that:

  Where Γ0 is the total circulation at time t=0.

∂Γ∂t

= 0

Γairfoil t( ) + Γwake t( ) = Γ0


Pitching and Heaving

Wake shape of a sinusoidally pitching and heaving airfoil.

Positive vorticity is denoted by red and negative by blue


Experimental results

Wake vorticity is a real-world phenomenon. Here is a comparison between numerical simulation results (top) and flow visualization in a water tunnel (bottom) by Jones and Platzer.


How to model this?   The simulation results are useful but

–  Not always accurate (there can be problems concerning starting vortices for example)

–  Not practical. If the motion (or any of the parameters) is changed, a new simulation must be performed.

  Analytical mathematical models of the problem exist. They were developed in the 1920s and 1930s.

  Most popular models: –  Theodorsen –  Wagner


Simplifications   In Theodorsen’s approach, only three major

simplifications are assumed: –  The flow is always attached, i.e. the motion’s

amplitude is small –  The wing is a flat plate –  The wake is flat

  The flat plate assumption is not problematic. In fact Theodorsen worked on a flat plate with a control surface (3 d.o.f.s), so asymmetric wings can also be handled.

  If the motion is small (first assumption) then the flat wake assumption has little influence on the results.


Basis of the model   The model is based on elementary solutions of the Laplace equation:

  Such solutions are: – The free stream: – The source and the sink: – The vortex: – The doublet:

∇2φ = 0

φ =U cosαx +U sinαy

φ =σ2πln r =

σ2πln x − x0( )2 + y − y0( )2

φ =µ2πcosθr

=µ2π

xx 2 + y 2

φ = −Γ2π

θ = −Γ2πtan−1 y − y0

x − x0

'

( )

*

+ ,


Circle   Theodorsen chose to model the wing as a circle that can be mapped onto a flat plate through a conformal transformation:


Joukowski’s conformal transformation

R

x

y

xa

ya

-2R 2R

  Define the complex variable z as z=x+iy.   Then consider the new complex variable za, given by:

za = xa + iya = z +R2

z


Singularities   Theodorsen chose to use the following singularities: – A free stream of speed U and zero angle of

attack – A pattern of sources of strength +2σ on the top

and surface of the flat plate, balanced by sources of strength -2σ on the bottom surface

– A pattern of vortices +ΔΓ on the flat plate balanced by identical but opposite -ΔΓ vortices in the wake


Complete flowfield

b

x

y

2σ 2σ

2σ

x1,y1

x1,-y1 -2σ -2σ -2σ

X0,0 b2/X0,0

-ΔΓ +ΔΓ

The circle radius, b, is equal to the wing’s half-chord, b=c/2.

Black dots: sources and sinks Red dots: vortices


Complete flowfield – flat plate

b x

y

-b

Wing Wake

Points inside the circle are transformed outside the flat plate. Therefore, the vortices inside the circle are mapped on the wake. The wing’s chord should have been 4b but this has been divided by 2 because we are using sources of strength 2σ.


About the wing and wake   The wing is a flat plate with a source distribution that changes in time.   The +2σ and -2σ source contributions do not cancel each other out.   The wake of the wing is a flat line with vorticity that changes both in space and in time.   The +ΔΓ and -ΔΓ vorticity contributions do not cancel each other out.

AERO0032-1, Aeroelasticity and Experimental Aerodynamics, Lecture 6

Wing and wake are slits

  Different parts of the circle map to different parts of the wing

b x

y

-b

Outside circle

Inside circle

Circle upper surface

Circle lower surface


Boundary conditions   As with all attached flow aerodynamic problems there are two boundary conditions: –  Impermeability: the flow cannot cross the solid

boundary – Kutta condition: the flow must separate at the

trailing edge   Kelvin’s theorem must also be observed.


Boundary conditions 2   The impermeability condition is fulfilled by the source and sink distribution   The Kutta condition is fulfilled by the vortex distribution   Kelvins’ theorem is automatically fulfilled because for every vortex +ΔΓ there is a countervortex -ΔΓ . Therefore, the total change in vorticity is always zero.


Impermeability   Impermeability states that the flow normal to a solid surface is equal to zero.   For a moving wing, the velocity induced by the source distribution normal to the wing’s surface must be equal to the velocity due to the wing’s motion and the free stream, i.e.

  Where n is a unit vector normal to the surface and w is the external upwash.

∂φ∂n

= −w


Impermeablity (2)   Across the solid boundary of a closed object the source strength is given by

  (assuming that the potential of the internal flow is constant)   Therefore, σ=w   This means that the strength of the source distribution is defined by the wing’s motion.

σ = −∂φ∂n


Wing motion   Assume that the wing has pitch and plunge degrees of freedom.   The total upwash due to its motion is equal to

  where xf is the position of the flexural axis and goes from -1 to +1.

w = − Uα + h + b x 1 +1( ) − x f( )α ( )

x =xb

x 1and x is measured from the half-chord


Potential induced by sources   The potential induced by a source at x1 , y1 is given

by

  By noting that we are using sources of strength 2σ, the potential induced by a source at x1 , y1 and a sink at x1 , -y1 is given by

  The value of this potential does not change if we use non-dimensional coordinates

dφ x1,y1( ) =σ2πln x − x1( )2 + y − y1( )2 =

σ4πln x − x1( )2 + y − y1( )2[ ]

dφ x1,±y1( ) =σ2πln

x − x1( )2 + y − y1( )2

x − x1( )2 + y + y1( )2&

' ( (

)

* + +

dφ x 1,±y 1( ) =σ2πln

x − x 1( )2 + y − y 1( )2

x − x 1( )2 + y + y 1( )2&

' ( (

)

* + +

where x =xb

, y = 1− x 2


Total source potential

  The total potential induced by the sources and sinks is given by

  Substituting for σ from the upwash equation we get

φ x ,y ( ) =b2π

σ lnx − x 1( )2 + y − y 1( )2

x − x 1( )2 + y + y 1( )2&

' ( (

)

* + + −1

1

∫ dx 1

φ x, y( ) = b2π

Uα + h+ b x1 +1( )− x f( ) α( ) ln x − x1( )2 + y − y1( )2

x − x1( )2 + y + y1( )2"

#$$

%

&''−1

1

∫ dx1


After the integrations   A long sequence of hardcore integration sessions has been censored. Such scenes are unsuitable for 2nd year Master students and middle-aged engineering professors.   The result on the upper surface is:

  The result on the lower surface is

φ x , y ( ) = b Uα + h − x fα ( ) 1− x 2 +b2α

2x + 2( ) 1− x 2

φ x ,y ( )lower = −φ x ,y ( )

(1)


Pressure on the surface   From the unsteady Bernoulli equation:

  Where p is the static pressure, ρ the air density and q the local air velocity

  For calculating the forces on the wing we need to apply this equation to the wing’s surface.

  The local velocity on the surface is tangential to the surface. As the wing lies on the x-axis:

p = −ρq2

2+∂φ∂t

&

' (

)

* + + Constant

q =U cosα + u =U cosα +∂φ∂x

≈U +∂φ∂x


Pressure difference

  The pressure on the upper surface is then

  And on the lower surface:

  The pressure difference is simply

pu = −ρ12U +

∂φ∂x

& '

( )

2

+∂φ∂t

&

' *

(

) + + Constant

pl = −ρ12U −

∂φ∂x

& '

( )

2

−∂φ∂t

&

' *

(

) + + Constant

Δp = pu − pl = −2ρ U∂φ∂x

+∂φ∂t

' (

) *

= −2ρ Ub∂φ∂x

+∂φ∂t

' (

) *


Non-circulatory lift

  The non-circulatory lift is given by

  Substituting for Δp we get

  Because we set that φ(1)=φ(-1)=0. Carrying out the integrations we get

lnc = Δpdx0

c

∫ = b Δpdx −1

1

∫

lnc = −2ρbUb∂φ∂x

+∂φ∂t

& '

( ) dx

−1

1

∫ = −2ρbφ−11− 2ρb

∂φ∂tdx

−1

1

∫ = −2ρb∂φ∂tdx

−1

1

∫

lnc = ρπb2 h − x f −c2

% &

' ( α + Uα

% & *

' ( +


Non-circulatory moment   The non-circulatory moment around the flexural axis is given

by:

  Substituting for Δp we get

  where the first integral was evaluated by parts. Carrying out the integrations we get:

mnc = Δp x − x f( )dx0

c

∫ = b Δp b x +1( ) − x f( )dx −1

1

∫

mnc = −2ρbU x ∂φ∂x dx

−1

1

∫ − 2ρb∂φ∂t

x b + b − x f( )dx −1

1

∫

= 2ρbU φdx −1

1

∫ − 2ρb∂φ∂t

x b + b − x f( )dx −1

1

∫

mnc = ρπb2 x f −c2

% &

' (

h − x f −c2

% &

' ( α

% & *

' ( + −

ρπb4

8α + ρπb2Uh + ρπb2U 2α


Circulatory forces   Up to now we’ve only satisfied the impermeability condition.   Now we need to satisfy the Kutta condition using the vortex distribution.

x

y

X0,0 b2/X0,0

-ΔΓ +ΔΓ


Potential induced by vortices

  The potential induced by the vortex pair at (X0,0) and (b2/X0,0) is

  As before, non-dimensional coordinates can be used

  Define   And remember that on the circle   Then:

φΔΓ =ΔΓ2π

tan−1 yx − X0

− tan−1 yx − b2 /X0

'

( )

*

+ ,

X 0 +1/ X 0 = 2x 0 or X 0 = x 0 + x 02 −1

y = 1− x 2

φΔΓ = −ΔΓ2πtan−1

1− x 2 x 02 −1

1− x x 0

φΔΓ =ΔΓ2π

tan−1 y x − X 0

− tan−1 y x −1/X 0

'

( )

*

+ ,

(2)


Pressure difference

  The pressure difference caused by this potential is, as before

  Theodorsen assumes that the vortices propagate downstream at the free stream velocity. Then

  So that

Δp = pu − pl = −2ρ U∂φ∂x

+∂φ∂t

' (

) *

∂φ∂t

=∂φ∂x0

U

Δp = −2ρU∂φ∂x

+∂φ∂x0

'

( )

*

+ , = −2ρ

Ub

∂φ∂x

+∂φ∂x 0

'

( )

*

+ ,

where x0 = bx 0


Pressure difference (2)   After obscene amounts of algebra and calculus we obtain

  But this is the contribution to the pressure difference at one point on the flat plate by only one vortex. In order to obtain the full circulatory aerodynamic loads we need to integrate for all vortices over all the wing.

Δp x ,x 0( ) = −ρUΔΓbπ

x 0 + x 1− x 2 x 0

2 −1

'

( )

*

+ ,


Lift - Integrate over wing   Integrating over the wing is easy

  Substituting:

  This is an uncharacteristically easy integral leading to

lc x 0( ) = Δp x ,x 0( )0

c

∫ dx = b Δp x ,x 0( )−1

1

∫ dx

lc x 0( ) = −ρUΔΓ

π x 02 −1

x 0 + x 1− x 2

'

( )

*

+ ,

−1

1

∫ dx

lc x 0( ) = −ρUΔΓx 0

x 02 −1


Lift - Integrate over the wake   Integration over the wake starts at the trailing edge and extends to infinity

  We can define

  So that the circulatory lift becomes

lc = −ρUx 0

x 02 −1

ΔΓ1

∞

∫

ΔΓ = bVdx 0

lc = −ρUbx 0

x 02 −1

Vdx 01

∞

∫

(3)


Circulatory Moment   The circulatory moment around the flexural axis becomes

  After substituting for Δp, the integrals become much more complicated.   Here’s the result:

mc x 0( ) = Δp x ,x 0( ) x − x f( )0

c

∫ dx = b Δp x ,x 0( ) b x +1( ) − x f( )−1

1

∫ dx

mc = −ρUbb2

x 0 +1x 0 −1

− ecx 0

x 02 −1

$

% & &

'

( ) ) Vdx 01

∞

∫


The nature of V   V is a non-dimensional measure of vortex strength at a point x0 in ‘flat plate space’.   As we’ve assumed that vortices don’t change strength as they travel downstream, V is a function of space.   In fact, it is stationary in value if we use a reference system that travels with the fluid.   If we use a fixed system, then V is a function of both time and space, i.e.

V = f Ut − x0( )


Kutta condition   Nevertheless, we still don’t know what value to give to

V.   This value can be obtained from the Kutta condition.   One of the forms of the Kutta condition is that ‘the

local velocity at the trailing edge must be finite’.   We can restrict this to the horizontal velocity

component since the wing lies on the x-axis.   In mathematical form:

  Where φtot stands for the total potential caused by both the sources and vortices.

∂φtot

∂x x =1

= finite


Kutta condition (2)   From equations (1), (2) and (3), the total potential is

  The horizontal airspeed is

φtot x ( ) = b Uα + h − x fα ( ) 1− x 2 +b2α

2x + 2( ) 1− x 2

−b

2πtan−1 1− x 2 x 0

2 −11− x x 0

Vdx 01

∞

∫

∂φtot∂x

= −b Uα + h− x f α( ) x1− x 2

−b2 α22x 2 + 2x −1

1− x 2

+b2π

x02 −1

1− x 2 x − x0( )V dx01

∞

∫


Kutta condition (3)   The total potential can be written as

  At the trailing edge . Therefore, the denominator becomes zero there.

  For the horizontal velocity to be finite, the numerator must also become zero at the trailing edge. Therefore:

∂φtot∂x

=11− x 2

−b Uα + h− x f α( ) x − b2 α2

2x 2 + 2x −1( )"

#$

+b2π

x02 −1

x − x0( )V dx01

∞

∫'

())

Uα + h − x fα ( ) +3bα

2=

12π

x 02 −1

1− x 0Vdx 01

∞

∫

x = 1


Kutta condition (4)   Cleaning up we get

  Now we have a relationship that gives us the necessary vortex strength for the Kutta condition to be satisfied.   This is the most important result of Theodorsen’s approach.

−1

2πx 0 +1x 0 −1

Vdx 01

∞

∫ = Uα + h +3c4− x f

' (

) * α (4)


Circulatory lift   Remember that the circulatory lift is given by

  Divide this lift by equation (4) to obtain:

  Where

lc = −ρUbx 0

x 02 −1

Vdx 01

∞

∫

lc = πρUcC Uα + h +3c4− x f

& '

( ) α

& ' *

( ) +

C =

x 0x 02 −1

Vdx 01

∞

∫x 0 +1x 0 −1

Vdx 01

∞

∫


Circulatory moment   Similarly:

  Divide by equation (4) to obtain

mc = −ρUbb2

x 0 +1x 0 −1

− ecx 0

x 02 −1

$

% & &

'

( ) ) Vdx 01

∞

∫

mc = −πρUcb2− ecC%

& ' (

Uα + h +3c4− x f

% &

' ( α

% & *

' ( +


Total lift   The total lift, both circulatory and non-circulatory is easily obtained by adding the two contributions:

  Notice that the non-circulatory terms are the added mass terms.

l = lnc + lc = ρπb2 h − x f −c2

% &

' ( α + Uα

% & *

' ( +

+πρUcC Uα + h +3c4− x f

% &

' ( α

% & *

' ( + (5)


Total moment   The total moment is again obtained from the sum of the two

contributions:

  Happily, some terms drop out:

m =mnc +mc = ρπb2 x f −

c2

"

#$

%

&' h − x f −

c2

"

#$

%

&' α

"

#$

%

&'−

ρπb4

8α

+ρπb2U h+ ρπb2U 2α −πρUc b2− ecC

"

#$

%

&' Uα + h+

3c4− x f

"

#$

%

&' α

"

#$

%

&'

m = ρπb2 x f −c2

% &

' (

h − x f −c2

% &

' ( α

% & *

' ( + −

ρπb4

8α − ρπb2 3c

4− x f

% &

' ( α

+πρUec 2C Uα + h +3c4− x f

% &

' ( α

% & *

' ( + (6)


Discussion   Theodorsen’s approach has led to equations (5) and (6) for the full lift and moment acting on the airfoil.   The main assumptions of the approach are:

– Attached flow everywhere – The wake is flat – The wake vorticity travels at the free stream

airspeed   In all other aspects it’s an exact solution   However, it’s not complete yet. What is the value of C?


Prescribed motion   In order to carry out the integrals and define C we need to know V.   The only way to know V is the prescribe it.   However, prescribing V directly is not useful.   It’s better to prescribe the wing’s motion and then determine what the resulting value of V will be.


Sinusoidal motion   The most logical choice for prescribed motion is sinusoidal motion.

Slowly pitching and plunging airfoil

Vorticity variation with x/c in the wake. It is sinusoidal near the airfoil


More about sinusoidal motion

  For small amplitude and frequency oscillations, the form of V(x0,t) is sinusoidal near the airfoil.   Notice that V(x0,t) is periodic in both time and space: – V(x0,t)=V(x0,t+2π/ω) = – V(x0,t)=V(x0+U2π/ω,t) =

  This means that phase angle of V(x0,t) is given by ωt+ωx0/U.


Vortex strength

x0

V(x0,t)

α(t)

h(t)

The vortex strength of the wake behind a pitching and plunging airfoil can have any spatial and temporal distribution, V(x0,t).

There are two special motions for which Theodorsen’s function can be evaluated: steady motion and sinusoidal motion

For sinusoidal motion:

α = α0ejωt

h = h0ejωt

V = V0ej ωt +

ωU

x0$ % &

' ( )

= V0ej ωt +

bωU

x 0$ % &

' ( )


Theodorsen Function For sinusoidal motion Theodorsen’s function can be evaluated in terms of Bessel functions of the first and second kind.

A much more practical, approximate, estimation is:

With k=ωc/U

or k=ωb/U


Usage of Theodorsen   Theodorsen’s lift force is now given by

  Theodorsen’s function can be seen as an analog filter. It attenuates the lift force by an amount that depends on the frequency of oscillation   Theoretically, Theodorsen’s function can only

be applied in the case where the response of the system is exactly sinusoidal

lc = πρUcC k( ) Uα + h +3c4− x f

& '

( ) α

& ' *

( ) +


Example

lc = πρUcC k( )h0 jω exp jωt

lc = πρUch0 jω exp jωtQuasi-steady lift:

Theodorsen lift:

Consider the circulatory lift of a purely plunging flat plate, h=h0exp jωt.


Lift and moment   The full equations for the lift and moment around the flexural axis using Theodorsen are:


Aeroelastic equations

  The full aeroelastic equations are:

  For sinusoidal motion they become:

  Substituting for l(t) and m(t) yields …

m SS Iα#

$ %

&

' (

h α

) * +

, - .

+Kh 00 Kα

#

$ %

&

' (

hα) * +

, - .

=−l t( )m t( )) * +

, - .

−ω 2m + Kh −ω 2S−ω 2S −ω 2Iα + Kα

%

& '

(

) * h0α0

+ , -

. / 0 e jωt =

−l t( )m t( )+ , -

. / 0


Equations of motion?   As the system is assumed to respond

sinusoidaly there is no sense in writing out complete equations of motion.   Combining the lift and moment with the

structural forces gives


Validity of this equation   This algebraic system of equations is only valid

when the wing is performing sinusoidal oscillations.

  For an aeroelastic system such oscillations are only possible when: –  The airspeed is zero and there is no structural

damping – free sinusoidal oscillations –  There is an external sinusoidal excitation force –

forced sinusoidal oscillations –  The wing is flying at the critical flutter condition – self-

excited sinusoidal oscillations   The last case is very useful for calculating the

critical flutter condition.


Flutter Determinant   For the equation to be satisfied non-trivially, the 2x2

matrix must be equal to zero, i.e. D=0, where

  D is called the flutter determinant and must be solved for the flutter frequency ωF and airspeed, UF.

  As the determinant is complex, Re(D)=0 and Im(D)=0. Two equations with two unknowns.


Solution   The flutter determinant is nonlinear in ω and U.   It can be solved using a Newton-Raphson scheme   Given an initial value ωi, Ui, a better value can be

obtained from

  Where F=[Re(D) Im(D)]T.   The initial value of ωi is usually one of the wind-off

natural frequencies


Effect of Flexural Axis Since sinusoidal motion is assumed the Theodorsen equations are only valid at the flutter point (or when there is no damping at all). Therefore, the variation of the eigenvalues with airspeed cannot be calculated.

The flutter speeds calculated from Theodorsen are less conservative than the quasi-steady results.


Single-DOF flutter   Consider a wing with only one degree of freedom, the pitch.   The Theodorsen equation becomes:

  Or, splitting into real and imaginary parts:

Kα −ω2Iα +

34c− x f

"

#$

%

&'ρπb2Ujω −πρUec2C k( ) U +

34c− x f

"

#$

%

&' jω

"

#$

%

&'

− x f −c2

"

#$

%

&'2

ρπb2ω 2 −ρπb4

8ω 2 = 0

Kα −ω2Iα − x f −

c2

"

#$

%

&'2

ρπb2ω 2 −ρπb4

8ω 2 −πρUec2 Re C k( )( )U − Im C k( )( ) 34 c− x f

"

#$

%

&'ω

"

#$

%

&'= 0

34c− x f

"

#$

%

&'ρπb2Uω −πρUec2 Im C k( )( )U +Re C k( )( ) 34 c− x f

"

#$

%

&'ω

"

#$

%

&'= 0


Single-DOF flutter   These are two equations with two unknowns, U and ω.   If they have a real solution, then the system can undergo flutter, even if it only has a single degree of freedom.   The equations are nonlinear but can be solved using the same Newton-Raphson procedure.   Remember that the equations always have a solution at U=0. We are looking for a solution with non-zero airspeed and frequency.


Single-DOF flutter   Try the following case:

  with xf ranging between 0 and c. Remember that:   For these values only the U=0 solution exists, for all values of xf .   Now change m to m=50 Kg and solve for all xf <0.2c.

Kα =130 Nm/rad, m =10 Kg, c = 0.3 m,ρ =1.225 Kg/m3

Iα =m3c2 −3cx f +3x f

2( )


Single-DOF flutter Single-DOF flutter is possible for xf<0.11c. Notice that the flutter speed goes to infinity as 0.11c is approached.


Single-DOF flutter   Therefore, single-DOF flutter can occur but: – The mass distribution must be very particular.

Either the wing must be very heavy or most of the mass must be placed near the trailing edge.

– The pitch axis must lie very near the leading edge.

  Most aeroelastic textbooks ignore single-DOF flutter as it is such a special case.

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06 Theodorsen Aerodynamics - · PDF fileAERO0032-1, Aeroelasticity and Experimental Aerodynamics, Lecture 6 3 2D wing oscillations FConsider a 2D airfoil oscillating sinusoidally in