Incompresible Flow over Airfoils

Ranggi SahmuraRevan Difitro

Road Map

Incompressible flow over airfoils

Singularity distribution over airfoil

surface

Singularity distribution

over camber line or

chord lineFundamental

s of airfoil (characteristics, the vortex sheet, Kutta

condition and Kelvins’s therem of starting vortex

Big Goal: Create method to lift and moment at airfoil to develop further design of

complete wing

Definition

Nomenclature “NACA WXYZ”

W = (Nilai Maks Camber/100) x Panjang ChordX = (Titik Maks Camber/10) x Panjang ChordYZ = (Ketebalan Maks/100) x Panjang Chord

“NACA VWXYZ”V = [1.5(Koefisien Lift)/10]WX = (Nilai Maks Camber/200) x Panjang ChordYZ = (Ketebalan Maks/100) x Panjang Chord

“NACA UV-XYZ”U = Nomor SeriV = (Tekanan Min/10) x Panjang ChordX = Koefisien Lift/10YZ = (Ketebalan Maks/100) x Panjang Chord

Nomenclature

Characteristics

Angle of AttackVsLift Coefficient

Angle of AttackVsDrag Coefficient

Vortex Sheet Theorem

Γ=∫𝛾 𝑑𝑠 𝐿′=𝜌∞𝑉 ∞ Γ

Kutta Condition

This won’t be achieved in nature

𝛾 (𝑇𝐸 )=0

Kelvin’s Circulation Theorem

Circulation

Kelvin’s Theorem of Circulation

The vortex sheet at instant of times becomes vortex sheet of all times

Starting Vortex

Starting vortex explain how the Kutta condition achieved in nature

Classical Thin Airfoil Theory

Classical thin airfoil

theory

Symmetric Airfoil

Cambered Airfoil

Chord lineCamber line

Singularity distribution

over camber line or

chord line

Grand Step

1. Find that satisfies 2 condition

2. from , we get

3. from , by Kutta-Jokowski theorem, we get L’

Kutta condition satisfied (TE = 0)Body surface acts as streamline of the flow, Vn = 0

The Symmetric Airfoil

The Symmetric Airfoil

- For thin airfoil, vortex sheet over airfoil surface will look almost the same as vortex sheet on camber line

- Since airfoil is thin, camber line will be closed to chord line vortex will fall approximately over chord line

(x); to satisfy Kutta condition, (c) = 0

- In order to achieve this, camber line has to be streamline of the flow

- If camber line is streamline, velocity normal to camber line has to be zero

𝑉  ∞ ,𝑛+𝜔′ (𝑠 )=0[ 4.12]

The Symmetric Airfoil

𝑉 ∞,𝑛=𝑉 ∞ sin [𝛼+𝑡𝑎𝑛−1( 𝑑𝑧𝑑𝑥 )] [4.13]

𝑉 ∞,𝑛=𝑉 ∞(𝛼− 𝑑𝑧𝑑𝑥 )[ 4.14]

For small angle, sin θ = tan θ = θ

The Symmetric Airfoil

- Since camber line close to chord line, w’(s) = w(x) [4.15]

- Elemental vortex strength , locate at distance , ()

- Velocity dw induced by elemental vortex at

𝑑𝑉=− 𝛾 𝑑𝑠2𝜋𝑟

using [4.16]

[4.17]

The Symmetric Airfoil

- Recall [4.12]

𝑉 ∞(𝛼− 𝑑𝑧𝑑𝑥 )−∫

0

𝑐 𝛾 (𝜉 ) 𝑑𝜉2𝜋 (𝑥−𝜉 )

=0

or

12∫0

𝑐 𝛾 ( 𝜉 ) 𝑑𝜉𝜋 (𝑥−𝜉 )

=𝑉 ∞(𝛼− 𝑑𝑧𝑑𝑥 )[ 4.18]

Since airfoil is thin, taken to be no cambered, dz/dx = 0

12∫0

𝑐 𝛾 (𝜉 ) 𝑑𝜉𝜋 (𝑥−𝜉 )

=𝑉 ∞𝛼[4.19 ]

𝜉=𝑐2 (1−𝑐𝑜𝑠𝜃)

𝑥=𝑐2 (1−𝑐𝑜𝑠 𝜃𝑜)

d

Transportation equation

The Symmetric Airfoil

12∫0

𝑐 𝛾 (𝜃 )𝑠𝑖𝑛𝜃𝑑𝜃𝑐𝑜𝑠𝜃−𝑐𝑜𝑠 𝜃0

=𝑉 ∞𝛼 [4.23 ]

𝛾 (𝜃 )=2𝛼𝑉 ∞(1+𝑐𝑜𝑠𝜃)

𝑠𝑖𝑛𝜃 [4.24 ]

Calculation of lift

Γ=∫0

𝑐

𝛾 ( 𝜉 ) 𝑑𝜉 [4.28 ]

Γ=𝑐2∫0

𝜋

𝛾 (𝜃 )𝑠𝑖𝑛𝜃𝑑𝜃 [4.29]

Using transportation equation

Using equation [4.24]

Γ=𝛼𝑐𝑉 ∞∫0

𝜋

(1+𝑐𝑜𝑠𝜃 ) 𝑑𝜃=𝜋𝛼𝑐𝑉 ∞ [4.30]

The Symmetric Airfoil

=Lift

Lift Coefficient

𝑐 𝑙=𝐿 ′

𝑞∞𝑆=

𝜋𝛼𝑐 𝜌∞𝑉 ∞2

12𝜌∞𝑉 ∞2 𝑐

𝑐 𝑙=2𝜋𝛼 𝑑𝑐𝑙

𝑑𝛼=2𝜋

“lift coefficient is linearly proportional to angle of attack”

The Symmetric Airfoil

𝑀 ′ 𝐿𝐸=−∫0

𝑐

𝜉 (𝑑𝐿 )=−𝜌∞𝑉 ∞∫0

𝑐

𝜉𝛾 (𝜉 ) 𝑑𝜉

Using transforming equation and some mathematical relation, obtain

𝑐𝑚 , 𝑙𝑒=−𝑐 𝑙

4

𝑐𝑚 ,𝑐/4=𝑐𝑚 ,𝐿𝐸+𝑐 𝑙

4And since

𝑐𝑚 , 𝑐/4=0 “center of moment and aerodynamic center is at quarter-length of chord

Momentum coefficient

The Cambered Airfoil

𝑐 𝑙=2𝜋 [𝛼+ 1𝜋∫0

𝜋 𝑑𝑧𝑑𝑥 (𝑐𝑜𝑠𝜃 0−1 )𝑑𝜃0]

Lift coefficient

Equation for determining angle of attack that produce zero lift (αL=0)“lift coefficient is linearly proportional to angle of attack”

Momentum coefficient

𝑐𝑚 ,𝑐/4=𝜋4 (𝐴2−𝐴1)

“quarter-length of chord is aerodynamic centre but not centre of pressure

)

Limitation of the classic theory

Thin airfoil theory Only of thin airfoil ( <12% ) Small angle of attack

Area of interest Low speed plane wings using airfoil

those are thicker than 12% High angle of attack for take off and

landing Generation of lift over other body

shape

Vortex Panel Numerical Method

Vortex Panel Numerical Method

Singularity distribution over airfoil

surface

Vortex Panel Numerical Method

- Determining using numerical method, using equation below

𝑉 ∞𝑐𝑜𝑠 𝛽𝑖−∑𝑗=1

𝑛 𝛾 𝑗

2𝜋 𝐽 𝑖 , 𝑗=0

𝛾𝑖=−𝛾𝑖−1

Γ=∑𝑗=1

𝑛

𝛾 𝑗𝑠 𝑗 𝐿′=𝜌∞𝑉 ∞∑𝑗=1

𝑛

𝛾 𝑗𝑠 𝑗

Classic – Modern Ways of

Classic Shape Aerodynamic Characteristics

Modern Characteristics wanted Shape