Blasius Theorem

AOE 5104Advanced Aero- and Hydrodynamics

Dr. William Devenport andLeifur Thor Leifsson

2

Blasius Theorem - BasicsFinding the force and moment on any region in ideal flow

x

iyp

diydxds

γ

dsypxpM

dsipiFF

dspF

dspF

O

yx

y

x

)sincos(

)sincos(

cos

sin

γγ

γγ

γ

γ

+−=

−=−

−=

−=

∫∫∫∫

Substituting for pressure using Bernoulli and introducing W(z) and z:

Force and Moment on a Body:

{ }∫

∫−=

=−

zdzzWM

dzzWiiFF

O

yx

2

2

)(Re2

)(2ρ

ρTHE BLASIUS RELATIONSCan also be shown to apply to any region of the flow (not just a body)

Fx

Fy

MO

3

Blasius Theorem - Explanation

x

iy

{ }∫

∫−=

=−

zdzzWM

dzzWiiFF

O

yx

2

2

)(Re2

)(2ρ

ρ

Fx

Fy

MO

• These integrals will be zero unless the indefinite integrals ∫W(z)2dz and ∫W(z)2zdz are functions that can have different values at the start and the end of the loop

• Only one type of function has this property logez = loger + iθ since may increase by 2π in traveling around the loop. In general loge(z-zi) increases by 2πi when passing around any loop enclosing zi

• Only functions of the form (z-zi)-1 integrate to loge(z-zi)

zi

4

Blasius Theorem - Procedure

x

iy

{ }∫

∫−=

=−

zdzzWM

dzzWiiFF

O

yx

2

2

)(Re2

)(2ρ

ρ

Fx

Fy

MO

1. Express W(z)2 or W(z)2z as a sum of terms of the form Bi(z-zi)n

e.g.

2. Extract n=-1 terms. ∫ will be 2πi × Σ of coefficients Bi for which zi lies inside the body or region

i.e. For W(z)2 :

For W(z)2z :

( ) ∑∑ −=−= −==− 1122

n

loopinzin

loopinziyxii

BBiiiFF ρππρ

{ } { }∑∑ −=−= =−= 11 Im2Re2

n

loopinzin

loopinziOii

BBiM ρππρ

zmz

iziz

Az

zzzz

z /,)2(

43,sin,)3(

,1

,)5(,)2(

1,7 3223

++

−−+

−−

5

Example 1 – Force and Moment on a Circular Cylinder with Circulation

zi

zaVVzW

π2)( 2

2 Γ−−= ∞

∞

a

∑ −=−=− 1n

loopinziyxi

BiFF ρπ

{ }∑ −== 1Im n

loopinziOi

BM ρπ

6

Example 2 – Force on a Vortex with Free Stream in Ground Effect

∑ −=−=− 1n

loopinziyxi

BiFF ρπ

V∞

Γ

h

7

Example 3 – Moment generated by a Joukowski airfoil

dzdWzW ζζ )(~)( =For foil

Write in terms of z

Take W(z)2z

Write as Laurent series

{ } )2sin(2)cos(Im 221 απραδρπ CVlmBM n

loopinziOi

∞−= −−== ∑ Positive counterclockwise

)cos()2sin(2 22 αδαπρ −−= ∞ lmCVM O Positive clockwiseor, by convention,

or, )cos()2sin(42

2

αδαπ−−=

cmC

cCC lM O

Pure moment

Moment due to lift acting on lever arm )cos( αδ −−m

ζ-plane z-plane

αV∞

z=ζ+C2/ ζ

C

δm

δζ ime=1

αV∞

l=-ρV∞Γ

MO c

Positive clockwise

8

αV∞

Lift l

rz=reiθ

θ -αα

)cos( αδ −−m

f, g

)cos()2sin(2 22 αδαπρ −−= ∞ lmCVM O

)cos()2sin(42

2

αδαπ−−=

cmC

cCC lM O

MOTransferring the moment:

For a general point z, lever arm has a length so

Example 3 – Interpretation

)cos()cos( αδαθ −−− mr

)cos()cos()2sin(2 22 αδαθαπρ −−−+= ∞ lmlrCVM Z

)cos()cos()2sin(42

2

αδαθαπ−−−+=

cmC

crC

cCC llM Z

Center of pressure: Point about which moment is zero (located by distance f).

Setting Mz=0 gives)sin(2

)2sin(2

βαα+

−=aCf

Aerodynamic center: Point about which moment is independent of α (located by g).

Setting gives andaCg /)cos(2 αβ −−=0=∂∂α

ZM )2sin(2| 22 βπρ CVMcenterAeroZ ∞−=

9

Results for Momentζ-plane

CαV∞

z=ζ+C2/ ζ

2

2

1ζζC

ddz

−=ζ1

aβ

1. The lift acts almost at mid chord (but don’t say this in public)

2. For a thin uncambered airfoil, the center of pressure f is close to the ¼ chord. The lift ‘appears’ to act at this point.

3. For any thin airfoil the aerodynamic center will be close to the quarter chord, and very gradually moves upstream with α

4. The moment about the aerodynamic center varies with camber (almost linearly with the zero lift angle of attack

αV∞

Lift l

r

z=reiθ

θ -α

α

)cos( αδ −−m

f, g

MO

4/4

cCaCc

Ca

≈≈≥≥

Whereand a and c increase slowly with camber and thickness

for a thin airfoilCcam ,,<<

and

)cos()2sin(42

2

αδαπ−−=

cmC

cCC lM O

)sin(2)2sin(2

βαα+

−=aCf aCg /)cos(2 αβ −−=

)2sin(2| 22 βπρ CVM centAeroZ ∞−=