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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 Kelvinsβs therem of starting vortex
Big Goal: Create method to lift and moment at airfoil to develop further design of
complete wing
Definition
Nomenclature βNACA WXYZβ
W = (Nilai Maks Camber/100) x Panjang ChordX = (Titik Maks Camber/10) x Panjang ChordYZ = (Ketebalan Maks/100) x Panjang Chord
βNACA VWXYZβV = [1.5(Koefisien Lift)/10]WX = (Nilai Maks Camber/200) x Panjang ChordYZ = (Ketebalan Maks/100) x Panjang Chord
βNACA UV-XYZβU = 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
Ξ=β«πΎ ππ πΏβ²=πβπ β Ξ
Kutta Condition
This wonβt be achieved in nature
πΎ (ππΈ )=0
Kelvinβs Circulation Theorem
Circulation
Kelvinβs 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 Lβ
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
π β ,π+πβ² (π )=0[ 4.12]
The Symmetric Airfoil
π β,π=π β sin [πΌ+π‘ππβ1( ππ§ππ₯ )] [4.13]
π β,π=π β(πΌβ ππ§ππ₯ )[ 4.14]
For small angle, sin ΞΈ = tan ΞΈ = ΞΈ
The Symmetric Airfoil
- Since camber line close to chord line, wβ(s) = w(x) [4.15]
- Elemental vortex strength , locate at distance , ()
- Velocity dw induced by elemental vortex at
ππ=β πΎ ππ 2ππ
using [4.16]
[4.17]
The Symmetric Airfoil
- Recall [4.12]
π β(πΌβ ππ§ππ₯ )ββ«
0
π πΎ (π ) ππ2π (π₯βπ )
=0
or
12β«0
π πΎ ( π ) πππ (π₯βπ )
=π β(πΌβ ππ§ππ₯ )[ 4.18]
Since airfoil is thin, taken to be no cambered, dz/dx = 0
12β«0
π πΎ (π ) πππ (π₯βπ )
=π βπΌ[4.19 ]
π=π2 (1βπππ π)
π₯=π2 (1βπππ ππ)
d
Transportation equation
The Symmetric Airfoil
12β«0
π πΎ (π )π ππππππππ πβπππ π0
=π βπΌ [4.23 ]
πΎ (π )=2πΌπ β(1+πππ π)
π πππ [4.24 ]
Calculation of lift
Ξ=β«0
π
πΎ ( π ) ππ [4.28 ]
Ξ=π2β«0
π
πΎ (π )π πππππ [4.29]
Using transportation equation
Using equation [4.24]
Ξ=πΌππ ββ«0
π
(1+πππ π ) ππ=ππΌππ β [4.30]
The Symmetric Airfoil
=Lift
Lift Coefficient
π π=πΏ β²
πβπ=
ππΌπ πβπ β2
12πβπ β2 π
π π=2ππΌ πππ
ππΌ=2π
βlift coefficient is linearly proportional to angle of attackβ
The Symmetric Airfoil
π β² πΏπΈ=ββ«0
π
π (ππΏ )=βπβπ ββ«0
π
ππΎ (π ) ππ
Using transforming equation and some mathematical relation, obtain
ππ , ππ=βπ π
4
ππ ,π/4=ππ ,πΏπΈ+π π
4And since
ππ , π/4=0 βcenter of moment and aerodynamic center is at quarter-length of chord
Momentum coefficient
The Cambered Airfoil
π π=2π [πΌ+ 1πβ«0
π ππ§ππ₯ (πππ π 0β1 )ππ0]
Lift coefficient
Equation for determining angle of attack that produce zero lift (Ξ±L=0)βlift coefficient is linearly proportional to angle of attackβ
Momentum coefficient
ππ ,π/4=π4 (π΄2βπ΄1)
βquarter-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
π βπππ π½πββπ=1
π πΎ π
2π π½ π , π=0
πΎπ=βπΎπβ1
Ξ=βπ=1
π
πΎ ππ π πΏβ²=πβπ ββπ=1
π
πΎ ππ π
Classic β Modern Ways of
Classic Shape Aerodynamic Characteristics
Modern Characteristics wanted Shape