### Q.21:- 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:-**Area of the circle swept by the windmill = *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