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 electricalenergy , and that A = 30 m^(2) , v = 36 km/h and the density of air is 1.2 m^(2) v = 36 km/h and the density of air is 1.2 kg m^(-3) .What is the electrical power produced ?

The blades of a windmill sweep out a circle of area A . (a) If the win

1.	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 electricalenergy , and that A = 30 m^(2) , v = 36 km/h and the density of air is 1.2 m^(2) v = 36 km/h and the density of air is 1.2 kg m^(-3) .What is the electrical power produced ?
Answer» Solution :(a) Suppose area of the circle swept by the windmill is A, velocity of the WIND is v and density of air = `rho ` Distance covered of air passing through circular area A , V = Ad ` :. V = Avt [ :. d = vt] ` ` :. ` MASS of flowing air = volume `xx` density m =` Avtrho` (b) Kinetic energy of air pasing through area of A in time t , `K = 1/2 mv^(2)` `=1/2 xxAvt rho xx v^(2)` ` K = 1/2 A rho vt^(3)` (c ) Input kinetic energy is energy of wind and output energy GENERATED , ` :. "EFFICIENCY " eta = (" Output energy")/("Input energy")` 25 % = `("ELECTRICAL energy ")/(1/2Av^(3))` ` :. ` Electric energy produced ` =1/4 xx1/2 xx 0 xx 1.2 xx(10)^(3) xxt` = 4500 t where `v = (36xx1000)/(3600) = 10 ms^(-1)` ` :. "Electric power " = (" Electric energy")/("Time")` = 4500 W `=4.5 W`