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Work, Power and Energy

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Question : 21 of 30
Marks: +1, -0
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 thewindmill converts 25% of the wind’s energy into electrical energy, and that A=30 m2,  v=36 km/hA = 30\,\mathrm{m}^2,\; v = 36\,\mathrm{km/h} and the density of air is 1.2 kg m−3\mathrm{kg}\,\mathrm{m}^{-3}. What is the electrical power produced?
Solution:  
(a) Volume of wind flowing/second =Av= Av
Mass of air passing t  s=Avρtt\; s = Av\rho t
(b) K.E. of air =12mv2=12(Avρt)v2=12Av3ρt= \frac{1}{2} m v^{2} = \frac{1}{2}(Av\rho t) v^{2} = \frac{1}{2} A v^{3} \rho t
(c) Electrical energy produced =(25100)×K.E.= \left(\frac{25}{100}\right) \times \mathrm{K.E.} of air =14×12Av3ρt= \frac{1}{4} \times \frac{1}{2} A v^{3} \rho t
Electrical power =18Av3ρtt=18Av3ρ= \frac{1}{8} \frac{A v^{3} \rho t}{t} = \frac{1}{8} A v^{3} \rho
P=18×30×(10)3×1.2=4500P = \frac{1}{8} \times 30 \times (10)^{3} \times 1.2 = 4500 watt =4.5 kW= 4.5\,\mathrm{kW}
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