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Aeroplane angle of attack and take-off power

Short answer

Equating lift to weight gives C_L = 2W/(ρSV²) = 0.775; solving 0.35(1+0.2α) = 0.775 yields angle of attack α ≈ 6.07°. Drag coefficient C_D = 0.008(1+α) = 0.0566, drag force ≈ 986 N, so required power is 29.6 kW.

An aeroplane is designed according to the following specifications: Weight = 13.5 kN, wing Area = 30 m², Take off speed = 30 m/s. Model tests show that the lift and drag coefficient vary with the angle of attack of the wing according to following approximate relations: CD=0.008 (1+α)C_D = 0.008\,(1 + \alpha), CL=0.35 (1+0.2α)C_L = 0.35\,(1 + 0.2\alpha). For small α\alpha, where α\alpha is the angle of attack measured in degree. The atmospheric density is 1.29 kg/m³. Find the angle of attack that ensures take-off at the design speed and power required for take off.

For take-off, the lift must equal the weight at the take-off speed. This gives the required CLC_L, then the angle of attack from the given relation; the drag and power follow.

Given

W=13.5W = 13.5 kN =13 500= 13\,500 N, S=30 m2S = 30\ \text{m}^2, V=30V = 30 m/s, ρ=1.29 kg/m3\rho = 1.29\ \text{kg/m}^3.

Required lift coefficient

L=W=CL⋅12ρSV2  ⇒  CL=2WρSV2=2(13 500)1.29(30)(30)2=0.7752L = W = C_L\cdot\tfrac12\rho S V^2 \;\Rightarrow\; C_L = \frac{2W}{\rho S V^2} = \frac{2(13\,500)}{1.29(30)(30)^2} = 0.7752

Angle of attack

0.35 (1+0.2α)=0.7752  ⇒  1+0.2α=2.215  ⇒  α=6.07∘0.35\,(1 + 0.2\alpha) = 0.7752 \;\Rightarrow\; 1 + 0.2\alpha = 2.215 \;\Rightarrow\; \alpha = 6.07^\circ

Drag and power

CD=0.008(1+6.07)=0.05659FD=CD⋅12ρSV2=0.05659(0.5)(1.29)(30)(30)2=985.6 NP=FD V=985.6×30=29 567 W≈29.6 kW\begin{aligned} C_D &= 0.008(1 + 6.07) = 0.05659 \\ F_D &= C_D\cdot\tfrac12\rho S V^2 = 0.05659(0.5)(1.29)(30)(30)^2 = 985.6\ \text{N} \\ P &= F_D\,V = 985.6 \times 30 = 29\,567\ \text{W} \approx 29.6\ \text{kW} \end{aligned}

Answer: angle of attack α≈6.07∘\alpha \approx 6.07^\circ; power required for take-off ≈29.6\approx 29.6 kW.

✓ Answer checked by Claude Opus 5.5 (high effort)

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