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