A wind turbine is located at the top of a hill where the wind blows steadily at 12 m/s, and stands 37 m tall. The air then exits the turbine at 9 m/s and the same elevation. Find the power generated by the wind if the mass flow rate is 137 kg/s. Report your answer in kW and to 2 decimal places.
Accepted Solution
A:
Answer:4.3155 kWStep-by-step explanation:Given, speed of wind when enters into the turbine, V = 12 m/sspeed of wind when exits from the turbine, U = 9 m/smass flow rate of the wind = 137 kg/sAccording to the law of conservation of energyEnergy generated = change in kinetic energyHence,energy generated in 1 sec can be given by[tex]E\ =\ \dfrac{1}{2}.m.V^2\ -\ \dfrac{1}{2}.m.U^2[/tex] [tex]=\ \dfrac{1}{2}\times 137\times 12^2\ -\ \dfrac{1}{2}\times 137\times 9^2[/tex] [tex]=\ 9864\ -\ 5548.5[/tex] = 4315.5 JSo, the power generated in 1 sec will be given by[tex]P\ =\ \dfrac{energy\ generated}{time}[/tex] [tex]=\ \dfrac{4315.5}{1}[/tex] = 4315.5 W = 4.13 kWSo, the power generated will be 4.13 kW.