Question

A payload is to be delivered by airdrop. The vertical velocity at landing has to be less than 7m/s. The total weight of the payload and parachute system is 400kg. Determine the minimum diameter of the open parachute, assuming a drag coefficient Cd 1.20.

Answer 1

Answer:

The minimum diameter of the open parachute = 11.77 m

Explanation:

For the open parachute + payload setup to drop, its weight must at least match the drag force so that the load+parachute can drop with its constant terminal velocity without accelerating.

Drag force = CρAv²/2

C = drag coefficient = 1.2

ρ = density of fluid = density of air = 1.225 kg/m³

A = Area of the body (falling through the fluid) facing the fluid = ?

v = velocity of Body falling through a fluid = 7 m/s

Weight = mg = 400 × 9.8 = 3920 N

(CρAv²/2) - 3920 = 0

(CρAv²/2) = 3920

A = (2×3920)/(Cρv²) = 7840/(1.2 × 1.225 × 7²)

A = 108.84 m²

The open parachute is assumed to be circular

A = πD²/4

108.84 = πD²/4

D² = (4 × 108.84)/π

D = 11.77 m