A wind-powered generator converts wind energy | vedclass

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(D) The kinetic energy of a mass $m$ of air moving with speed $v$ is $K = \frac{1}{2}mv^2$.The mass of air passing through an area $A$ in time $t$ is

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$v^3$

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(D) The kinetic energy of a mass $m$ of air moving with speed $v$ is $K = \frac{1}{2}mv^2$.The mass of air passing through an area $A$ in time $t$ is $m = \rho A v t$,where $\rho$ is the density of air.The rate of flow of kinetic energy (power) is $P = \frac{dK}{dt} = \frac{1}{2} \left(\frac{dm}{dt}\right) v^2$.Since $\frac{dm}{dt} = \rho A v$,we substitute this into the power equation:$P = \frac{1}{2} (\rho A v) v^2 = \frac{1}{2} \rho A v^3$.Since the generator converts a fixed fraction of this energy,the electrical power output is proportional to $v^3$.

$A$ wind-powered generator converts wind energy into electrical energy. Assume that the generator converts a fixed fraction of the wind energy intercepted by its blades into electrical energy. For wind speed $v$,the electrical power output will be most likely proportional to

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