Question

The Cessna Cardinal, a single-engine light airplane, has a wing with an area of 16.2 m2 and an aspect ratio of 7.31. Assume that the span efficiency factor is 0.62. 1 (a) If the airplane is flying at standard sea-level conditions with a velocity of 251 km/h, what is the induced drag when the total weight is 9800 N? (b) What is the induced drag when the airplane is flying at 85.5 km/h (this is the stall speed at sea level when the aircraft is in landing configuration, with flaps down).

Answer 1

To calculate the induced drag, you can use the following formula:

Induced Drag (D_i) = (2 * L^2) / (ρ * V^2 * S * π * e * AR)

Where:
- L is the lift force, which is equal to the weight of the aircraft (9800 N in this case).
- ρ is the air density at standard sea-level conditions (approximately 1.225 kg/m^3).
- V is the velocity in m/s (convert km/h to m/s by dividing by 3.6).
- S is the wing area (16.2 m^2).
- π is approximately 3.14159.
- e is the span efficiency factor (0.62 in this case).
- AR is the aspect ratio (7.31).

Let's calculate it for both cases:

(a) At 251 km/h:
Convert 251 km/h to m/s: 251 km/h / 3.6 = 69.7222 m/s

D_i = (2 * 9800 N^2) / (1.225 kg/m^3 * (69.7222 m/s)^2 * 16.2 m^2 * 3.14159 * 0.62 * 7.31)

Calculating this gives you the induced drag at 251 km/h.

(b) At 85.5 km/h (landing configuration):
Convert 85.5 km/h to m/s: 85.5 km/h / 3.6 = 23.75 m/s

D_i = (2 * 9800 N^2) / (1.225 kg/m^3 * (23.75 m/s)^2 * 16.2 m^2 * 3.14159 * 0.62 * 7.31)

Calculating this gives you the induced drag at 85.5 km/h.

Now, perform the calculations for both cases to find the induced drag in each scenario.