---
title: "Aeroplane angle of attack and take-off power"
subject: "Fluid Mechanics"
chapter: 9
key: "semester_3/fluid_mechanics:aeroplane-angle-of-attack-power"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/aeroplane-angle-of-attack-power
site: BCT Engineering
---

# Aeroplane angle of attack and take-off power

Fluid Mechanics · Chapter 9: Flow past through submerged bodies

- Asked in: 2078 Bhadra, 6+2 marks
- Video: [Understanding Aerodynamic Lift](https://www.youtube.com/watch?v=E3i_XHlVCeU) — The Efficient Engineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/aeroplane-angle-of-attack-power
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-9

## Question

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: $C_D = 0.008\,(1 + \alpha)$, $C_L = 0.35\,(1 + 0.2\alpha)$. For small $\alpha$, where $\alpha$ 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.

## 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.

## Answer

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

#### Given

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

#### Required lift coefficient

$$
L = W = C_L\cdot\tfrac12\rho S V^2
\;\Rightarrow\;
C_L = \frac{2W}{\rho S V^2} = \frac{2(13\,500)}{1.29(30)(30)^2} = 0.7752
$$

#### Angle of attack

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

#### Drag and power

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

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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What condition fixes CL at take-off?
   A. Drag equals weight
   B. Thrust equals lift
   C. Lift is zero
   D. Lift equals weight at the take-off speed
   Answer: D. CL = 2W/(rho S V^2) = 0.775. (Key point: Take-off condition)

2. What required CL results?
   A. About 0.35
   B. About 0.056
   C. About 0.775
   D. About 1.55
   Answer: C. 2(13500)/(1.29 x 30 x 900) = 0.7752. (Key point: CL value)

3. From CL = 0.35(1 + 0.2 alpha) = 0.775 the angle is
   A. About 2.2 degrees
   B. About 6.07 degrees
   C. About 11 degrees
   D. About 0.78 degrees
   Answer: B. 1 + 0.2 alpha = 2.215 gives alpha = 6.07. (Key point: Angle of attack)

4. How is the power required computed?
   A. Power = drag x speed, with F_D from CD at alpha: about 29.6 kW
   B. Power = lift x speed
   C. Power = weight x speed
   D. Power = CD x V
   Answer: A. F_D = 985.6 N, P = 985.6 x 30 = 29.6 kW. (Key point: Power)

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