Question

For a smooth sphere with a 11 cm diameter thrown through air (air density of 1.25 kg/m3, air viscosity 1.83*10-5 Pa-s), what is the speed (in cm/s) when the Reynolds number is 2,511

Answer 1

Answer:

u = 0.334 m/s

Explanation:

The formula for Reynold's number is given as follows:

Re = ρuL/μ

where,

Re = Reynold's Number = 2511

ρ = Density of air = 1.25 kg/m³

μ = Dynamic Viscosity of Air = 1.83 x 10⁻⁵ Pa.s

L = Characteristic Length = Diameter of Sphere = 11 cm = 0.11 m

u = Speed = ?

Therefore,

2511 = (1.25 kg/m³)(u)(0.11 m)/(1.83 x 10⁻⁵ Pa.s)

u = (2511)/(7513.66 s/m)

u = 0.334 m/s