
Yuxuan Z. answered 09/12/20
PhD in Physics
I don't know what the exact meaning of "... turbine that has rotors of 5 m in length." I guess it gives you the size of the swept area by the turbine blades. But is the 5-m the diameter or the radius? I am not sure. So let's just assume that the 5-m be the radius. Therefore the swept area A = π·r2 = 3.14×5×5 = 78.5 (m2). With the wind speed of 4 m/s, the volume of the air moving across the turbine is 78.5×4 = 314 m3/s. Under 25°C condifition, the density of the air is 1.184 kg/m3 (from wiki), so the mass of the moving air is 314×1.184 = 371.8 kg/s. The kinetic energy of the air in one second is 1/2 × m × v2 = 0.5×371.8×42 = 2974 kg·(m/s)2. Consider the overall efficiency of 25%, the power generated from the wind will be 2974×0.25 = 743 w = 0.743 kw.