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?

#### Solution 1

*A*

Velocity of the wind = *v*

Density of air = *ρ* (a) Volume of the wind flowing through the windmill per sec =

*Av*

Mass of the wind flowing through the windmill per sec = *ρ**Av*

Mass *m*, of the wind flowing through the windmill in time *t* = *ρ**Avt*(b) Kinetic energy of air = (1/2)

*mv*

^{2}

= (1/2) (*ρ*A*vt*)*v*^{2} = (1/2)*ρAv*^{3}t

(c) 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

= (25/100) × Kinetic energy of air

= (1/8) *ρ* *A* v^{3}t

Electrical power = Electrical energy / Time

= (1/8)* ρ* *A* *v*^{3}t / t

= (1/8) *ρ* *A* *v*^{3}

= (1/8) × 1.2 × 30 × (10)^{3}

= 4.5 kW

#### Solution 2

(a) Volume of wind flowing per second = Av Mass of wind flowing per second = Avρ

Mass of air passing in second = Avρt

(b) Kineic energy of air = `1/2mv^2 = 1/2 (Avrhot)v^2 = 1/2 Av^3rhot`

(c) Electrical energy produced = `25/100 xx 1/2 Av^3 rhot = (Av^3rhot)/8`

Electrical power = `(Av^3rhot)/"8t" = (Av^3rho)/8`

Now `A = 30m^2, v = 36 kmh^(-1) = 36 xx 5/18 ms^(-1) 10 ms^(-1), rho = 1.2 kg ms^(-1)`

:. Electrical power = `((30xx10xx10xx10xx1.2)/8)W = 4500 W = 4.5 KW`