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anonymous
 5 years ago
A wind powered generator converts wind energy into electric energy. Assume that the generator converts a fixed fraction of the wind energy, intercepted by its blade into electric energy. For wind speed v, the electrical power output will be proportional to
(Choose one among the four)
1. v
2. v^2
3. v^3
4. v^4
anonymous
 5 years ago
A wind powered generator converts wind energy into electric energy. Assume that the generator converts a fixed fraction of the wind energy, intercepted by its blade into electric energy. For wind speed v, the electrical power output will be proportional to (Choose one among the four) 1. v 2. v^2 3. v^3 4. v^4

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In a unit of time, a unit of mass of wind will pass the blades. The wind has speed v, so the mass' kinetic energy is proportional to v^2. This energy is transfered in part to the blades, and it again proportional to v^2. Since energy is conserved, and we're told the generator converts a fixed fraction, that fixed fraction of energy is still proportional to v^2. The answer is v^2.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks However, that is not exactly what I asked for. The question asked for the proportionality with the power, NOT Energy.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In my answer, I stated "In a unit of time"...this means an amount of energy proportional to v^2 has been transferred in this unit of time...that's power.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But the book says its proportional to v^3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And has not provided any explanation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok...I know what's happened...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Kinetic energy is \[\frac{1}{2}mv^2\]For a mass of air flowing in a unit of time, we can write mass as a volume flow rate x time x density (=mass). Then\[E=\frac{1}{2}(Avt{\rho})v^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The power is just rate of change of energy, which is means from the above, \[P=\frac{E}{t}=\frac{1}{2}(A{\rho})v^3 \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So the power is proportional to the cube of velocity (but then again, it's also proportional to the square...technically speaking ;)).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Can you see how Av yields volume flow rate?
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