Solutions To Tutorial 2 PDF

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Aerodynamics Lectures and Tutorials...

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[SESA2022 Aerodynamics] Tutorial #2 1. Consider the non-lifting flow over a circular cylinder of a given radius, where V∞ = 6m/s. If V∞ is doubled, that is, V∞ = 12m/s, does the shape of the streamlines change? Explain. Solution: The stream function of the non-lifting flow over a cylinder is   R2 Ψ = V∞ r sin θ 1 − 2 , r from which the radial and circumferential velocities can be derived as follows:     R2 R2 ∂Ψ 1 ∂Ψ = −V∞ sin θ 1 + 2 . and Vθ = − = V∞ cos θ 1 − 2 Vr = r ∂θ ∂r r r The Cartesian velocities are then:   R2 2 2 u = Vr cos θ − Vθ sin θ = V∞ 1 + 2 (sin θ − cos θ) , r v = Vr sin θ + Vθ cos θ = −2V∞

R2 sin θ cos θ. r2

The shape of the streamlines will remain the same if their slope at any given point remains the same. Recalling the slope of a streamline is R2 sin θ cos θ v dy r2 = =− , R2 dx u 2 1 + 2 (sin θ − cos2 θ ) r 2

it does not depend on the freestream velocity V∞ at all. Therefore, the streamlines do not change when V∞ is doubled.

2. Consider the lifting flow over a circular cylinder of a given radius and with a given circulation. If V∞ is doubled, keeping the circulation the same, does the shape of the streamlines change? Explain. Solution: The stream function of the lifting flow over a cylinder is   R2 r Γ Ψ = V∞ r sin θ 1 − 2 + ln , r 2π R 1

from which the radial and circumferential velocities can be derived as follows:     ∂Ψ Γ 1 ∂Ψ R2 R2 = V∞ cos θ 1 − 2 and Vθ = − Vr = . = −V∞ sin θ 1 + 2 − r r 2πr r ∂θ ∂r The Cartesian velocities are then:   Γ R2 2 2 sin θ, u = Vr cos θ − Vθ sin θ = V∞ 1 + 2 (sin θ − cos θ) + 2πr r v = Vr sin θ + Vθ cos θ = −2V∞

Γ R2 cos θ. sin θ cos θ − 2 2πr r

Using the same approach taken in the previous question: Γ R2 cos θ sin θ cos θ + 2 v dy 2πV∞ r r = =− , dx u Γ R2 2 2 sin θ 1 + 2 (sin θ − cos θ) + 2πV∞ r r 2

the slope now is a function of both V∞ and Γ. Therefore, the streamlines are bound to change when V∞ is doubled with Γ fixed.

3. Consider the lifting flow over a circular cylinder. The lift coefficient is 5. Calculate the peak (negative) pressure coefficient. Solution: In the class we obtained the pressure and lift coefficients for the lifting flow past a circular cylinder: cp = 1 −



Γ 2πRV∞

2

cl =

−

2Γ sin θ − 4 sin2 θ, πRV∞

Γ . RV∞

In this question cl = 5 and therefore: Γ = 5, RV∞  2 10 5 − sin θ − 4 sin2 θ, cp = 1 − 2π π

2

The peak (negative) pressure coefficient is achieved at θ = π/2 (prove this):  2 10 5 − − 4 ≅ −6.8164. cp-peak = 1 − π 2π

3...