Q. 214.1( 16 Votes )

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 m², v = 36 km/h and the density of air is 1.2 kg m^–3. What is the electrical power produced?

Class 11th NCERT - Physics Part-I 6. Work, Energy and Power

Answer :

Given,

Area swept by the blades = A

Velocity of wind flow = v

Density of air = ρ

(a) Volume of the wind flowing through the windmill per second, V = Av

Mass of the wind flowing through the mill per second,

m = ρAv

Mass of the wind flowing through the mill in time t,

M = ρAvt

(b) Kinetic energy of air, K = (1/2)Mv²

⇒ K = (1/2)(ρAvt)v²

⇒ K = (1/2)ρAv³t

Area swept by the blades, A = 30 m²

Velocity of wind flow, v = 36 km/h

⇒ v = (36×1000 m)/(60×60 sec)

⇒ v = 10 m/s

Density of air, ρ = 1.2 kg m^-3

Electric energy produced, E = 25% of wind energy

⇒ E = (25/100)(1/2) ρAv³t

⇒ E = (25/100)(1/2)× 1.2 kg m^-3 × 30 m² × (10 m s^-1)³ × t

⇒ E = 4500t J

Electric power, P = Electric energy (E)/ Time (t)

⇒ P = 4500t/t

⇒ P = 4500 W = 4.5 kW

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