An electical question

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?

Reply to
Student
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First of all, do not take all those significant figures seriously.

It looks like your arithmetic is OK. Nevertheless, you did not answer the question. It asked for the energy obtainable during a whole year

Of course the problem is greatly oversimplified. I know of no place that will give uniform wind.

Bill

-- Ferme le Bush

Reply to
Salmon Egg

The wind is not given as uniform. 600 watts per square meter is given as the "average" windpower density.

Chuck

Reply to
chuck

You need to understand the difference between power and energy.

600 watts per square meter is a power

Energy is the product of power times time.

Look up the defination of kWh, then you should be able to calculate it from your power calculations. Do not expect someone to do your homework.

Bill Kaszeta Photovoltaic Resources Int'l Tempe Arizona USA snipped-for-privacy@pvri-removethis.biz

Reply to
Bill Kaszeta / Photovoltaic Resources

Oh, it looks like you are correct. I answered in kilowatts, not KWH. How do i interconvert the two?

Reply to
Student

So I need to multiply the number of watts generated by the number of hours in a year?

so the final answer should be 24 x 365 x 73.632 = 645016.32 KWh

Is this right?

btw, I'm not looking to have someone else complete my homework. I just don't have anyone else to learn from, and it seems like the people in this group are quite smart.

Bill Kaszeta / Photovoltaic Resources wrote:

Reply to
Student

No where in your calculation do you take time into account. The question asks about kwh - which needs time to be factored into the calculation. (h = hour)

Ed

Reply to
ehsjr

Holy cats!

You read the instructions!

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job.

The rest of you, take a lesson from this student.

Reply to
thrugoodmarshall

Good.

One hint: carry the units through the equation with you. That way, problems will jump out if the units don't match up. Like this:

PI * (12.5 m)^2 * 0.25 * (600 W / m^2) * (24 h/day) * (365 day/year) * (

1/1000 k ) = 645016* kWh/year

Cancel out the units:

(m)^2 * ( W / m^2 ) * (h/day) * (day/year) * ( k )= kWh / year

If they don't come out the way they're supposed to, you forgot something.

Yes. Strange, isn't it? ;-)

Reply to
Paul Hovnanian P.E.

Did you notice the "PE" behind some of their names? It stands for "pretty expensive".

Dean

Reply to
Dean Hoffman

Pizza Eater

Reply to
Paul Hovnanian P.E.

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