# The Blades of a Windmill Sweep Out a Circle of Area 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 And What is the Kinetic Energy of the Air and What is the Electrical Power Produced - Physics

The blades of a windmill sweep out a circle of area A. (a) If the wind flows at a velocity 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 = 30 m2= 36 km/h and the density of air is 1.2 kg m–3. What is the electrical power produced?

#### Solution 1

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 = ρAv

Mass m, of the wind flowing through the windmill in time t = ρAvt

(b) Kinetic energy of air = (1/2) mv2

= (1/2) (ρAvt)v2 = (1/2)ρAv3t

(c) Area of the circle swept by the windmill = = 30 m2

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 v3t

Electrical power = Electrical energy / Time

= (1/8) ρ A v3t / t

= (1/8) ρ A v3

= (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

Concept: Power
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#### APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 6 Work, Energy and Power
Q 21 | Page 137