Question

The blades of a windmill sweep out a circle of area A. (a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t ? (b) What is the kinetic energy of the air ? (c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A = 30 \(m^2\) , v = 36 km/h and the density of air is 1.2 kg \(m^{-3}\). What is the electrical power produced ?

Answer 1

Area of the circle swept by the windmill = A 
Velocity of the wind = v
Density of air = ρ
Volume of the wind flowing through the windmill per sec = Av
Mass of the wind flowing through the windmill per sec = ρAv
Mass m, of the wind flowing through the windmill in time t =ρAvt
Kinetic energy of air = \(\frac{1}{2}mv^2\)
= \(\frac{1}{2}\) (ρAvt)\(v^2\)= 1 / 2 ρA\(v^3\)t
Area of the circle swept by the windmill = A = 30 \(m^2\)
Velocity of the wind = v = 36 km/h
Density of air, ρ = 1.2 kg \(m^{-3}\)
Electric energy produced = 25% of the wind energy
= \(\frac{25}{100}\) × kinetic energy of air
= \(\frac{1}{8}\) ρAv3t
Electrical power = \(\frac{Electrical \space energy}{Time}\)
= \(\frac{1}{8}\) \(\frac{pAv^3t}{t}\) = \(\frac{1}{8}\)ρA\(v^3\)
 = \(\frac{1}{8}\)× 1.2 × 30 × \((10)^3\)
= 4.5 × \(10^3\) W = 4.5 kW