Answer:
The correct answer for (a) D = 1.5 KN (b) r₀ = 55.8 m²/s
Step-by-step 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