The blades of a windmill sweep out a circle of area A.
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 $A=30 \mathrm{~m}^{2}, v=36 \mathrm{~km} / \mathrm{h}$ and the density of air is $1.2 \mathrm{~kg} \mathrm{~m}^{-3}$. What is the electrical power produced?

solution:

Area of the circle swept by the windmill = A

Velocity of the wind = v

Density of air $=\rho$

(a) Volume of the wind flowing through the windmill per sec = Av

Mass of the wind flowing through the windmill per sec $=\rho A v$Mass $m$, of the wind

flowing through the windmill in time $t=\rho A v t$

(b) Kinetic energy of air $=\frac{1}{2} m v^{2}$

$=\frac{1}{2}(\rho A v t) v^{2}=\frac{1}{2} \rho A v^{3} t$

(c) Area of the circle swept by the windmill $=A=30 \mathrm{~m}^{2}$

Velocity of the wind $=v=36 \mathrm{~km} / \mathrm{h}$

Density of air, $\rho=1.2 \mathrm{~kg} \mathrm{~m}^{-3}$

Electric energy produced $=25 \%$ of the wind energy

$=\frac{25}{100} \times$ Kinetic energy of air

$=\frac{1}{8} \rho A v^{3} t$

Electrical power $=\frac{\text { Electrical energy }}{\text { Time }}$

$=\frac{1}{8} \frac{\rho A v^{3} t}{t}=\frac{1}{8} \rho A v^{3}$

$=\frac{1}{8} \times 1.2 \times 30 \times(10)^{3}$

$=4.5 \times 10^{3} \mathrm{~W}=4.5 \mathrm{~kW}$
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