An airplane weighing 73.6 kN has elliptic wings 15.23 m in span. For a speed of 90 ms^-1 in straight and level flight at low altitude, find (a) the induced drag and (b) the circulation around sections halfway along the wings.

The correct answer for (a) D = 1.5 KN (b) r₀ = 55.8 m²/s

Explanation:

Solution

The first step is to solve for the induced drag

(a) Given that,

The Aeroplane weight L = 73.6 kN, which is called the lift force

The wings in span = 15.23 m

The speed = 90 ms^-1

At a lower altitude, the air density p = 1.225 * kg/m³

(a) Now,

The induced drag = D = L²/1/2 p V²π b² e

Where e = efficiency

= (73.6 * 10³) ²/1/2 * 1.225 * 90² * π * 15.23²

Thus,

D = 1.5 KN

(b) The circulations around sections halfway along the wings

The overall wings is denoted by

L = π / 4 p V r₀ b

where r₀ is the wings center of circulation

now,

73.6 * 10³ = π / 4 * 1.225 * 90 * r₀ * 15.23

r₀ = 4 * 73.6 * 10³ / π * 1.225 * 90 * 15.23

Therefore r₀ = 55.8 m²/s