electrical question from a HS student.

[Student]

I'm a HS sophomore, and I am hoping to become an engineer of some sort
someday. I'm working on a homework problem which I can't seem to solve.
I'll post the question and how I went about trying to solve it. Please
let me know if I've done anything wrong.


"Assume a wind turbine with a hub 50 m above ground, a rotor diameter
of 25 m and a wind-conversion efficientcy of 25%. The turbine will
operate in an area w/ an average windpower density of 600 watts per
square meter at 50m altitude. How many kwh can the turbine generate per
year?"

I started out by using
pi x r^2

pi x (12.5)^2=490.8738521 square meters

then I multiplied by .25 (the efficiency) and multiplied by 600.

That gives me 73631.07782 watts, which equals 73.632 Kilwatts

I've been told that 73.632 is incorrect... why?

[Jack]

You're doing fine but you just stopped a little too soon.
73.632kw is correct as the POWER that the turbine generates under average
conditions but the question posed was how much ENERGY does the turbine
generate in a year.

Energy = Power x Time
KilowattHours = Kilowatts x Hours
Answer = 73.632 x 24 hours/day x 365.25 days per year = 645,458 kwh/year

Post by Student
I'm a HS sophomore, and I am hoping to become an engineer of some sort
someday. I'm working on a homework problem which I can't seem to solve.
I'll post the question and how I went about trying to solve it. Please
let me know if I've done anything wrong.
"Assume a wind turbine with a hub 50 m above ground, a rotor diameter
of 25 m and a wind-conversion efficientcy of 25%. The turbine will
operate in an area w/ an average windpower density of 600 watts per
square meter at 50m altitude. How many kwh can the turbine generate per
year?"
I started out by using
pi x r^2
pi x (12.5)^2=490.8738521 square meters
then I multiplied by .25 (the efficiency) and multiplied by 600.
That gives me 73631.07782 watts, which equals 73.632 Kilwatts
I've been told that 73.632 is incorrect... why?