Nsf Ideal Fluids

8/10/2019 Nsf Ideal Fluids

http://slidepdf.com/reader/full/nsf-ideal-fluids 1/98

II. Ideal – fluid flow

● Ideal fluids are

● Inviscid

● Incompressible

● The only ones decently understood mathematically

● Governing equations

∇⋅u=0

∂u∂ t +(u⋅ )u=−1

ρ p+ f

Continuity

Euler



Boundary conditions

u⋅n=U ⋅n

ormal to surface

!elocity of surface

Free-slip "velocity is parallel

to surface#

$otential flow "special case#

u = (u = / x, v = / y, w = / z )

$otential flow is irrotational

Continuity equation for potential flow

2 = 0

Continuity equation "with boundary conditions#can be solved alone for velocity



Then plug into momentum equation "Bernoulliform# to solve for pressure



%. &' potential flows%.(. )tream function ● &' ideal continuity equation

∂u∂ x

+∂ v∂ y

=0

u=∂φ

∂ x , v=

∂φ

∂ y

● !elocity potential

● Introduce streamfunction "counterpart ofpotential# so that

u=

∂ψ

∂ y , v=−

∂ψ

∂ x



)treamfunction satisfies continuity equation byconstruction

∂2 ψ∂ x∂ y

− ∂2 ψ∂ y∂ x

=0

)treamfunction e*ists for any ideal &' flow

Before going further+ consider vorticity in &' flow

=∇×u=det[ i j k

∂∂ x ∂∂ y ∂∂ z u v w

]





!orticity in &' flow

=k

∂v

∂ x−

∂u

∂ y =k ω

,or &'+effectivelya scalar

ow consider an irrotational &' flow

ω=∂v∂ x− ∂

u∂ y=0

E*press velocity in terms of streamfunction

ω= ∂∂ x −∂ψ∂ x − ∂∂ y −∂ψ∂ y =0

∇ 2

ψ=0



$roperties of streamfunction

● )treamlines are lines of const

● 'ifference in the value of between twostreamlines equals the volume of fluid flowing

between them

● )treamlines const and potential lines

const are orthogonal at every point in the

flow



)treamlineequation/

0hy const is a streamline

d s

dx d y

d ψ=d ψds

ds= ∂ψ∂ x ∂ x∂ s +∂ψ∂ y

∂ y∂ sds=−vdx+udy

d ψ=0 means v dx=udy ;dy

v

=dx

u



,low rate between two streamlines

1

2

1

B

u

u

v

!olume flow rate

Q=∫ A

B

u⋅nds=∫ A

B

u⋅d n=∫ A

B

u dy−∫ A

B

v dx

ds

'irection along 1B2

ds "dx +dy #

'irection normal to 1B2dn "dy +3dx #

dn

Q=∫ A

B

d ψ=ψ1−ψ2



4rthogonality between streamlines and potentiallines

d ψ=−vdx+udy=0 1long a streamline

1long an isopotential line " const #...

d φ=∂φ∂ x dx+

∂φ∂ y dy=u dx+v dy=0

ormal to streamline2 "3v + u#

ormal to isopotential line2 "u+ v #

They are orthogonal2 "3v + u#"u+ v# 5





Comple* potential constructed from velocitypotential and streamfunction

F( z ) = ( x, y) + i ( x, y)

Cauchy36iemann condition satisfied byconstruction

Advantages of using complex potential● If and are the real and imaginary parts of

any holomorphic function+ 2

= 0 and 2

= 0

automatically● Comple* velocity w = dF /dz = u - iv – directly

related to flow velocity



8agnitude of comple* velocity

w*w = (u + iv)(u – iv) = u2+v2 = uu =

$olar coordinates in comple* plane

x

y

r

x + iy = r (cos + i sin ) = rei

q

e r

e

u = ur cos - u sin

v = ur sin + u

cos

w = (ur - iu) e-i



%.9. :niform flow

F ( z )=Ce−iα

z

w ( z )=dF

dz =Ce

−iα=C cosα−iC sin α

u=C cosα , v=C sinα

x

y

a



%.%. )ource+ sin;+ and vorte*

F ( z )=C log z =C log (r e

iθ

)=C (log r +i θ)

w ( z )=dF

dz =

C

z =

C

r e−iθ

ur =C

r , uθ=0 x

y

,irst+ let C be real and positive

φ=C log r , ψ=C θ

)ource at z = 0



)ource strength "discharge rate#

m=∫0

2π

ur r d θ=∫0

2π

Cd θ=2πC

Comple* potential of a source of strength m at z = z

0

F ( z )= m2π

log( z − z 0)

Comple* potential of a sin; of strength m at z = z 0

F ( z )=− m2π

log( z − z 0)

x

y



ow consider a purely imaginary constant in thelogarithmic potential2

F ( z )=−iC log z =−iC log(reiθ)=−iC log r +C θφ=C θ , ψ=−C log r

w ( z )=

dF

dz =−i

C

z =−i

C

r e−iθ

ur =0, uθ=C

r

x

y

$oint vorte*



!orte* strength "circulation#

Γ=∮ L u⋅d l =∫0

2π

uθ r d θ=2πC

Comple* potential of a vorte* with circulation G at z = z

0

F ( z )=−i Γ2π log ( z − z 0)

ote (. z = z 0 is a singularity "u

q #

ote &. This flow field is called a free vortex2

Γ L' =∮ L '

u⋅d l ≡0 1ny contour not

including z 0



%.<. ,low in a sector

F ( z )=U z

n

, n⩾1

1braham de 8oivre=s formula

1braham de 8oivre(>>?3(?<%

1uthor of The Doctrine of Chances



%.<. ,low in a sector

F ( z )=U z

n

, n⩾1

1braham de 8oivre=s formula

e

i nθ

=(cosθ+i sinθ)

n

=cos (nθ)+i sin(nθ):se polar coordinates

z =r eiθ

F ( z )=U r n

cos(nθ)+i U r n

sin (nθ)

$otential and stream function

φ=U r n

cos (nθ) , ψ=U r n

sin(nθ)



q=p/n

Comple* velocity

w ( z )=nUz n−1=nU r

n−1e

i(n−1)θ=

= nU r n−1 cosnθ+i nU r n−1 sin nθ e−iθ

!elocity components

ur =nU r

n−1

cos nθuθ=−nU r

n−1sin nθ

n (2 uniform flow

n &2 flow in aright3angle corner

n 92 shown



%.>. ,low around a sharp edge

F ( z )=C z

1 /2

=C r

1 /2

e

iθ/2

w ( z )=dF

dz =

1

2C z

−1 /2=C

2 r 1 /2 e

−iθ/2=

ur =

C

2 r 1/2 cos

θ2 , uθ=−

C

2 r 1 /2 sin

θ2

$otential and streamfunction

φ=C r 1 /2

cos θ2 , ψ=C r

1/2sin θ

2

= C

2 r 1 /2 e

−iθeiθ/2= C

2 r 1 /2 (cos θ2+i sin θ

2)e−iθ

Comple* velocity



x

y

ur =

C

2 r 1/2 cos

θ2 , uθ=−

C

2 r 1 /2 sin

θ2

= 0, q = 0

= 0, q = 2p

)ingularity



%.?. 'oublet

)ource at x = -e )in; at x = +e

y

x

ow lete

0



Comple* potential of source and sin;

F ( z )=m

2π log( z +ε)−

m

2π log( z −ε)

F ( z )=m

2π log

z +ε z −ε

=m

2π log

1+ε/ z 1−ε/ z

,or small /z + e*pand denominator into series2

(1−ε/ z )−1=1+ε/ z +…$lug that into F ( z )

F ( z )=m

2π log((1+ε/ z )(1+ε/ z +…))

F ( z )=

m

2π log 1+2

ε z +…



:se series e*pansion for logarithm near (

F ( z )=m

2π log

(1+2 ε

z +…

)=

m

2π (2 ε

z +…

)If we ta;e the limit of this as 0+ the result willbe trivial2 F ( z ) = 0

,or a non3trivial result+ let limε →0

mε=πμ

Then

limε →0

F ( z )=μ z = μ

x+iy=μ x−iy( x+iy)( x−iy)=μ

x−iy x

2+ y2

φ=μ x

x2

+ y2 , ψ=−μ

y

x2

+ y2



Consider a streamline = !nst

ψ=−μ

y

x2+ y2

ψ( x2+ y2)=−μ y

x2

+ y2

+μ ψ y=0

x2+ y

2+μ ψ y+

(

μ2 ψ

)

2

=

(

μ2 ψ

)

2

x2+( y+ μ2 ψ )

2

=( μ2 ψ)2

Circle of radius m/(2y) and center at x = 0, y = 3m/(2y)



y

x



y

x

w ( z )=−μ

z 2=−

μ

r 2e−2iθ=−

μ

r 2e−iθ (cosθ−i sinθ)

ur =−

μ

r 2 cos

θuθ=−

μ

r 2 sinθ



'oublet of strength m at z = z 0

F ( z )=

μ

z − z 0



%.@. Circular cylinder flow

Aet uniform flow go past a doublet

F ( z )=Uz +μ z

$otential and stream function

F ( z )=Ureiθ+ μ

r eiθ= Ur +

μr

cosθ+i Ur −μr

sinθ

$otential )tream function

Consider streamline y = 0

Ur = m/r means that this streamline is a circle of

radius " = (m/U )1/2



Can rewrite comple* potential as

F ( z )=U z +"

2

z

z →∞ , F ( z )→U z

:niform flow dominates the far field

z →0, F ( z )→U "

2

z

'oublet dominates the flow near the origin



,low symmetry2 F (- z ) = - F ( z )

!elocity5"rearstagnationpoint#

!elocity5"forwardstagnationpoint#

)ingularity at origin



%.. Cylinder with circulation

Ta;e cylinder flow+ add rotation around the origin

F ( z )=U z +"

2

z +

iΓ2π

log z +C

!orte* at origin

Constant to

;eep y = 0 at r = "

$retty easy to find C + tuc; it into the logarithm

F ( z )=U z +"

2

z

+iΓ

2π

log z

"Comple* velocity

w=dF

dz

=U 1−"

2

z 2 +

iΓ

2π

1

z



w=U 1−"

2

z 2 +

iΓ2π

1

z =U 1−

"2

r 2e−2 iθ +

iΓ2π

1

r e−iθ

w=[U (1−"2

r 2 )cosθ+i(U (1+

"2

r 2 )sinθ+ Γ

2π r )]e−iθ

w=[U (e iθ−"

2

r 2e−iθ)+ iΓ

2π1

r ]e−iθ

6emember that w = (ur -iuq)e-iq

ur =U 1−"

2

r 2

cosθ , uθ=−U 1+"

2

r 2

sinθ− Γ2π r



4n the surface "r = "#+

ur =0, uθ=−2U sin θ− Γ

2π"Boundary/

,ind stagnation points "velocity 5+ r = "#

sinθ s=−

Γ4πU "

$ossibilities2

& stagnation points on the cylinder

( stagnation point on the cylinder

5 stagnation points on the cylinder "but maybesomewhere else in the flow#



Two stagnation points

0< Γ4πU"<1

4ne stagnation point

Γ

4πU"=1







r s= Γ4πU

!"( Γ4πU )

2

−"2

Two stagnation points

3 inside the cylinder "so who cares#

D outside the cylinder "good stuff#



%.(5. Blasius integral laws

● ,ind potential

● ,ind velocity components

● $lug velocity into Bernoulli equation to find

pressure on body surface

● Integrate to find

● ydrodynamic force on the body

● ydrodynamic moment on the body

● MUCH simpler with complex potential!



c.g.

Body of an arbitrary shape

)urface2 streamline y = 0

1ny contour fully enclosing the body

C 5

, o r c e , o

r c e

x

y X

Y

M

Comple* force2 $ % i&

Blasius first law

$ −i& =iρ2∮C 0

w2dz

Blasius second law

=ρ2# ∮

C 0

z w2dz



valuating complex integrals

Taylor series "real variable#

( ( x− x0)=$n=0

∞"n( x− x0)

n , "n= ( (n)( x0)

n)

This e*pansion is valid in an interval | x - x0| < d x



valuating complex integrals

Aaurent series "comple* variable#

( ( z − z 0)=$n=−∞

∞ "n( z − z 0)n ,

"n= 1

2π i∮C

( (% )(%− z 0)−n−1

d %

This e*pansion is valid in an annulus where ( isholomorphic2

1 < | z - z

0| <

2

If 1 = 0+ z 0 – isolated singularity

C

z 0

Coefficient "-1 of Aaurent series2

residue of ( at z 0



Cauchy theorem

If comple* function ( ( z ) is holomorphic

everywhere inside contour C +

∮C

( ( z )dz =0

Cauchy residue theorem

If comple* function ( ( z ) is holomorphiceverywhere inside contour C + e*cept isolated

singularities+∮C

( ( z )dz =2iπ$+

"−1, +



xamplee z – holomorphic everywhere in a dis; of radius r with center at z = 0

e z =1+ z +

z 2

2 +

z 3

6 +…

e z

/z – holomorphic everywhere in a dis; of radius r with center at z = 0+ e*cept at it center

e z

z

=1

z

+1+ z

2

+ z

2

6

+…

"-1=1

Note. "%m

0, "%m%

0, = 1,2, ... at z = z 0 –

z 0

is a pole of order m



%.((. ,orce and moment on a circularcylinder

Comple* potential

F ( z )=U z +"

2

z +

iΓ2π

log z

"

Comple* velocity

w=dF

dz =U 1−

"2

z

2 +

iΓ2π z

Blasius first law

$ −i& =iρ2∮C

0

w2dz



w2=U

2−2

2"

2

z 2 +

U 2"

4

z 4 +

i U Γπ z

−iU Γ"2

π z 3 − Γ2

4π2 z 2

0 -2 -4 -1 -3 -2Term order in z

"−1=iU Γπ

$ −i& =2iπ$+

"−1,+ =2iπiU Γπ =−iρU Γ

z = 0 – sole isolated singularity of w2

+ thus

$ = 0 "'=1lembert=s parado*#& = U G "Fhu;ovs;y3utta law#

)imilar analysis for zw

2

produces ' = 0



%.(&. Conformal transformations

elps deal with boundaries

x

y

"

# = ( ( z )

z = x+iy # = ! +i"

It=s only good if the Aaplace equation is alsotransformed into something nice...



● Consider ( – holomorphic function mapping " x, y#

into "!,"#

● In " x, y# plane+ let 2( x, y)

= 0● Then in "!,"# plane+ &"!,"# = 0

"proof2 p. 9#

● Aaplace equation is preserved by conformalmapping

● 0hat happens with comple* velocity

w ( z )=dF

dz =

dF (%)d %

d %dz =

d %dz

w(%)

!elocity scales during conformal mapping



Aet=s prove that conformal mapping preservessources+ sin;s+ etc.

C

d l

dx dy

Γ=∮C u⋅dl =∮C (u dx+v dy)Circulation of all

point vortices inside

C

m=∮C

u⋅dn=∮C

(udy−v dx)

)trength of allsourcesHsin;s inside

C



∮C

w( z )dz =

=∮C (u−iv )(dx+idy )=∮C (u dx+v dy)+i∮C (udy−vdx )==Γ+i m

Could have proven the same with residuetheorem...

ow consider a conformal mapping " x,y# "!,"#

(Γ+i m

)& z =

∮C & z

w( z )dz =

=∮C &%

w(%)d %=(Γ+i m )&%

=∮C & z

w(%) d %dz

dz =



Conformal mapping preserves strength ofsources+ sin;s+ and vortices



%.(9. Fhu;ovs;y transformation

i;olai EgorovichFhu;ovs;y

"(@%?3(&(#“Man will fly using the

power of his intellectrather than the strengthof his arms.”

z =%+

2

%

&%&→∞ , z →%

dz d %

=1− 2

%2

# = 02 singularity "let=s contain it

inside the body#

%=! ,dz

d %=0

# = $2 critical points "angle not preserved#



Critical points of Fhu;ovs;y transform

"# = ! +i"

x

y z = x+iy

3&c D&c

z 0

3c Dc

#0

%1%

2q

1q

2

%=! z =!+

2

!=!2cCan prove2 q

1- q

2= 2 (%

1- %

2)

1 smooth curve passing through # = will

correspond to a curve with a cusp in z 3plane



E*ample2 # = ei%

"# = ! +i"

x

y z = x+iy

3&c D&c 3c Dc

z =e

i'

+

2

e i'= (ei '

+e

−i '

)=2ccos '

hu"ovs"y transform recipe# )tart with flow

around a cylinder in #3plane+ map to something



8aor semia*is 8inor semia*is

%.(%. ,low around ellipses

Circle in #3plane+ radius " ! + center at origin%="e

i '

z ="ei '+

2

"e−i'= "+

2

" cos'+i "−

2

" sin '

$arametric equation of an ellipse



"# = ! +i"

3c Dc

x

y z = x+iy

3&c D&c



,low past a cylinder

F ( z )=U z +"

2

z



ow consider freestream flow at an angle

Can get this by conformal

mapping too "in plane z "2 z = ei" z " 3 rotation# x =

y =

a

Correspondingly+ z " = e%i" z



In plane z "

F ( z ' )=U z' +"

2

z '

F =U z e−iα+

"2

z e−iα =U z e

−iα+"

2

z e

iα

Aet=s have this flow in #3plane2

F (%)=U % e−iα+

"2

%e

iα

ow recall that

z =%+

2

%



E*press # in terms of z 2

%2+2−% z =0

%= z

2!"( z 2)

2

−2

6ecall that for z + z #. Thus select

%= z

2

+

"( z

2

)

2

−2

$lug this into F (#) to get F ( z )... "s;ip derivation#

[ ]



F ( z )=U [ ze−iα+

(

"2

2e

iα−e−iα

)( z

2

−

"( z

2

)

2

−2

)]

:niform flow at angle a approaching an ellipsewith maor semia*is " + 2/" and minor semia*is

" - 2/"



a

)tagnation points2 # =$"eia

)t ti i t i l



a

)tagnation points in z 3plane...

z =!"eiα!

2

"e−iα

2 2



z =! "+2

" cosα!i "−

2

" sinα

x=! "+

2

"cosα

y=! "− 2

"sinα

3 forward stagnation point

D downstream stagnation pointa 52 horiJontal flow approaching horiJontalellipse

a = p/22 vertical flow+ horiJontal ellipse "or

horiJontal flow+ vertical ellipse#



%.(<. utta condition and the flat3plate airfoil

% (< Fhu;ovs;y Chaplygin postulate and



%.(<. Fhu;ovs;y3Chaplygin postulate andthe flat3plate airfoil

,low around a sharp edge "section %.>#...

F ( z )=C z 1 /2

w ( z )=dF

dz =C

2 z 1 /2

z 52 singularity

● 1t a sharp edge+ velocity goes to infinity

● This is not the case in e*periment+ luc;ily

● eed a fi* for theory near sharp edges

●

That=s not the only problem though...

z %+2



a

# = ! +i" z = x+iy

a

r =

z =%+%

erein liesthe problem/



)mo;e visualiJation of wind tunnel flow past a lifting surface 1le*ander Aippisch+ (<9

)tagnation point is 1A01K)at the trailing sharp edge/

h " " Ch l i t l t



hu"ovs"y-Chaplygin postulate$ ,or bodies with sharp trailing edges at modestangles of attac; to the freestream+ the rearstagnation point will stay at the trailing edge

%ealing with trailing-edge singularity In modeling real lifting surfaces+ trailing edge hassharp but finite curvature

ow to Lfi*M the flat plate flow



a

# = ! +i"

ow to fi* the flat3plate flow

1ngle of attac;



z = x+iy

a

# = ! +i"

1dd circulation... z = x+iy

a

...to move the

stagnation point to thetrailing edge/

0e want to move the rear stagnation point to



0e want to move the rear stagnation point to z = 2

That would correspond to # = in the J3planeeed to move it there from # = eia

,or cylinder flow with circulation...

sinθ s=− Γ4πU "

If sin q s = - sin a+

Γ=4πU " sinα

6ecipe for constructing a comple* potential for



6ecipe for constructing a comple* potential forcorrected flat3plate flow "Eq. %.&&b#

●

Cylinder flow● 1dd circulation G = &p " U sin a

● 6otate the plane a degrees countercloc;wise

● Fhu;ovs;y transform

●

● $rofit/

Aift on a flat plate airfoil e*tending from 2" to 2"



Aift on a flat3plate airfoil e*tending from -2" to 2"

& =ρU Γ

Blasius law for cylinder flow2

In our case

& =4πρU

2

" sinα

Introduce dimensionless lift coefficient



Introduce dimensionless lift coefficient

C L=&

12ρU 2

Characteristic length scale"for wings – chord length#

wing

chord

,or our flat plate+ = 4" and

C L=2π sinα

1t small angles of attac;+ lift coefficient on a flatplate increases with angle of attac;/





"# = ! +i"

3c Dc

%.(>. )ymmetrical Fhu;ovs;y airfoil

Goal2 airfoil with sharp trailing edge and bluntleading edge

Center2

%m = %e

small-( + 2m)

r = " = (1 + e)

x

y z = x+iy

3&c D&c

t

Aeading edge in # plane2 ( + 2m)



Aeading edge in #3plane2 -( + 2m)

In z 3plane+ the leading edge is...

z =− (1+2ε)−

1+2ε=−2c+-(ε2)−2c

Chord length = 4

)imilarly "more series e*pansions+ lineariJation#thic;ness

t =3" 3ε ,t

=3" 3

4

ε

8a*imum thic;ness occurs at x = %

Thic;ness ratio

E*tra ,lugJeugbau E1955+ (@?+ 0alterE t d i Fh ; ; i fil



E*tra design+ Fhu;ovs;y wing profile

Can find e in # plane from desired and t in z



1t Jero angle of attac;+ stagnation point is attrailing edge+ lift 5

1dd angle of attac; a...

Can find e in #3plane from desired and t in z 3plane2

ε=

4

3" 3t

0.##

t

Equation for symmetric Fhu;ovs;y profile in z 3plane

y

=!

2

3" 3(1−2 x

)"1−(2 x

)2

To satisfy the Fhu;ovs;yHuttaHwhatever



To satisfy the Fhu;ovs;yHuttaHwhatevercondition...

a

# = ! +i"

r = "

eed to move thisstagnation point...

!

...here/

,or a cylinder of radius "+ the needed amount ofcirculation is "same as for flat plate...#

G = &p " U sin a

E*press radius " in terms of and t



,or an angle of attac; a+ circulation we need toadd is...

Γ=4πU " sinα=πU 1+ 43" 3

t

sinα

E*press radius " in terms of and t ...

"=+m= (1+ε)=

4

1+ 4

3" 3

t

Aift coefficient for symmetrical Fhu;ovs;y airfoil

C L2π 1+0.## t

sinα

t 0+ this reduces to lift coefficient of flat plate

Fhu;ovs;y symmetrical profile has better lift/

% (? 1 i f il



x y

"# = ! +i"

3c Dc

%.(?. 1rc airfoil

1irfoil of Jero thic;ness but finite curvature

m

a

r

%

:se cosine theorem toget r

"

2

=r

2

+m

2

−2 rm cos

π2−'

In z 3plane+

z =r e

i'

+

2

r e

−i '

=

= r +2

r cos '+i r −

2

r sin '

4

4



x2= r

2+2c2+

r 2

cos2' , y

2= r 2−2c

2+

r 2

sin2'

' sin

2

% ' cos

2

%

r 2

cos2'sin

2'= x2

sin2'− 2

2+

4

r 2

cos2'sin

2 '

r 2

cos2'sin

2'= y2

cos2 '+ 2

2−4

r 2

cos2'sin

2'

x2

sin2

'− y2

cos2

'=4c2

cos2

'sin2

'

.

:se cosine theorem2

sin '=r 2−

2

2 rm = r −

2

r

1

2m=

y

2msin '



[ ( )]2

[ ( )2

]



x2+[ y+(

m−

m

)] =2[4+(

m−

m

) ] y⩾0

Equation of an arc in the z 3plane

"# = ! +i"

3c Dc

m

a

r

% x

y

z = x+iy

3&c D&c

h – will find

4tto Ailienthal and his glider+ (@<



4tto Ailienthal and his glider+ (@<



)tagnation point needs to rotate by a + t$n-1(m/)



)tagnation point needs to rotate by a + t$n (m/)

1ngle of attac; !ertical shift

AineariJe2

t$n-1(m/) m/ = e, "

1mount of circulation to be added2

Γ=4πU " sin α+m

4πU sin α+

m

Aift coefficient2

C L=2πU sin α+m

=2πU sin α+2

.

1gain+ more lift than flat plate/

% (@ Fhu;ovs;y airfoil



%. (@. Fhu;ovs;y airfoil

● now how to create lifting surfaces with2● )traight chord+ finite thic;ness

● Fero thic;ness+ small finite curvature "camber#

● Both improve lift+ compared with flat plate● Create a lifting surface with both thickness and

camer "Fhu;ovs;y profile#

# = ! +i"



"# ! i"

3c Dc

a

r

. / 2

0.## t/

– chordt – ma*. thic;ness. – ma*. camber

y z = x+iy



x

y

3&c D&c

t .



Circulation



Γ=πU 1+0.##t

sin α+2.

thic;ness

camber

Aift coefficient

C L=2π 1+0.##t

sin α+

2 .



Date post:	02-Jun-2018
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