問題一覧
1
The vortex generated behind the trailing edge of an aerofoil when it begins moving from a stationary position
2
For a sealed (no fluid exchanged) enclosed space with a certain amount of enclosed fluid, the rate of change of circulation is zero (DΓ / Dt = 0) if the potential flow conditions are met
3
For an enclosed space C¹ containing an aerofoil at rest with no circulation, accelerating the aerofoil will change the shape of the enclosure into C² but no mass will have been exchanged, This aerofoil will generate an anti-clockwise starting vortex, but total circulation must stay at 0, so there must be an equal clockwise vortex generated around the aerofoil (bound vortex), This means that Γ³ must equal – Γ⁴, in order to satisfy the condition Γ¹ = Γ² = 0
4
A simplification to the vortex sheet, where for a thin aerofoil, the individual vortices can all be placed along the chord line rather than the camber line, γ(x) along the sheet needs to be such that the camber line is a stream line (making the velocity normal to the camber line 0) and the Kutta condition is satisfied
5
The component of velocity normal to the camber line must be zero along the entire line, V∞_n + w'(s) = 0 Where V∞_n is the component of the freestream velocity normal to the camber line, and w'(s) is the component of the velocity induced by the vortex sheet normal to the camber line
6
V∞_n = V∞ • (α – dz / dx) Where V∞ is the freestream velocity, α is the angle of attack, and dz/dx is the gradient of the camber line, V∞_n = – w'(s), where w'(s) is the component of the velocity normal to the camber line induced by the vortex sheet along the chord line
7
As the curvature is small, the velocity normal to the camber line is approximately equal to the velocity normal to the chord line (w'(s) ≈ w(x) = –V∞_n)
8
w(x) = – (0 to c) ∫ γ(ξ) dξ / 2π(x – ξ)
9
(1 / 2π) • (0 to c) ∫ (γ(ξ) dξ / (x – ξ)) = V∞ • (α – dz / dx) Where c is the chord length, γ is the vortex strength per unit length, dξ is a small portion of the vortex sheet, x is the horizontal distance along the chord, V∞ is the freestream velocity, α is the angle of attack (radians), and dz/dx is the gradient of the camber line
10
The camber line is coincident with the chord line, so dz/dx = 0
11
γ(θ) = 2 • α • V∞ • ((1 + cos θ) / sin θ) Where γ is the vortex strength per unit length, θ is the anti-clockwise angle, α is the angle of attack, and V∞ is the freestream velocity
12
Γ = π • α • c • V∞ Where α is the angle of attack (radians), c is the chord length, and V∞ is the freestream velocity
13
ξ = (c / 2) • (1 – cos θ), dξ = (c / 2) • (sin θ dθ), As x is a Fixed Point: x = (c / 2) • (1 – cos θ⁰)
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11問 • 2年前問題一覧
1
The vortex generated behind the trailing edge of an aerofoil when it begins moving from a stationary position
2
For a sealed (no fluid exchanged) enclosed space with a certain amount of enclosed fluid, the rate of change of circulation is zero (DΓ / Dt = 0) if the potential flow conditions are met
3
For an enclosed space C¹ containing an aerofoil at rest with no circulation, accelerating the aerofoil will change the shape of the enclosure into C² but no mass will have been exchanged, This aerofoil will generate an anti-clockwise starting vortex, but total circulation must stay at 0, so there must be an equal clockwise vortex generated around the aerofoil (bound vortex), This means that Γ³ must equal – Γ⁴, in order to satisfy the condition Γ¹ = Γ² = 0
4
A simplification to the vortex sheet, where for a thin aerofoil, the individual vortices can all be placed along the chord line rather than the camber line, γ(x) along the sheet needs to be such that the camber line is a stream line (making the velocity normal to the camber line 0) and the Kutta condition is satisfied
5
The component of velocity normal to the camber line must be zero along the entire line, V∞_n + w'(s) = 0 Where V∞_n is the component of the freestream velocity normal to the camber line, and w'(s) is the component of the velocity induced by the vortex sheet normal to the camber line
6
V∞_n = V∞ • (α – dz / dx) Where V∞ is the freestream velocity, α is the angle of attack, and dz/dx is the gradient of the camber line, V∞_n = – w'(s), where w'(s) is the component of the velocity normal to the camber line induced by the vortex sheet along the chord line
7
As the curvature is small, the velocity normal to the camber line is approximately equal to the velocity normal to the chord line (w'(s) ≈ w(x) = –V∞_n)
8
w(x) = – (0 to c) ∫ γ(ξ) dξ / 2π(x – ξ)
9
(1 / 2π) • (0 to c) ∫ (γ(ξ) dξ / (x – ξ)) = V∞ • (α – dz / dx) Where c is the chord length, γ is the vortex strength per unit length, dξ is a small portion of the vortex sheet, x is the horizontal distance along the chord, V∞ is the freestream velocity, α is the angle of attack (radians), and dz/dx is the gradient of the camber line
10
The camber line is coincident with the chord line, so dz/dx = 0
11
γ(θ) = 2 • α • V∞ • ((1 + cos θ) / sin θ) Where γ is the vortex strength per unit length, θ is the anti-clockwise angle, α is the angle of attack, and V∞ is the freestream velocity
12
Γ = π • α • c • V∞ Where α is the angle of attack (radians), c is the chord length, and V∞ is the freestream velocity
13
ξ = (c / 2) • (1 – cos θ), dξ = (c / 2) • (sin θ dθ), As x is a Fixed Point: x = (c / 2) • (1 – cos θ⁰)