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Windmill - mass of air hitting blades each second? watch

1. This is an exam question from an A2 edexcel specimin paper.

Information given:
area swept out by blades in one revolution: 2.4m^2
power output: 1kW at wind speed of 12.5
typical operating speed of blades: 600 revolutions per minute
density of air: 1.3 kgm^-3

So how do you work out the mass of air hitting the blades each second when the ind speed is 14ms^-1

Some of this information will be unnecessary for this part of the question, they're relevent to other bits, i just put it all in in case.

thanks
2. 2.4 * 14 = Volume in m^3

Then use dentisty to find mass.
3. volume of air each second= 2.4 x 14

mass each sec=density x volume each sec....use the volume you got above
4. The thing is that it's a fundamentally flawed question - only about 1/10th of the wind of the calculated cylinder will actually touch the windmill at all - a windmill's rotors don't form a full circle, do they?

That makes the above question completely retarded.
5. Well I think its an mass flow = density x Area x Velocity question.

Where it has 600 revs/min = 100 rev/sec.

Therefore Area/rev x 100. Others are given.

Have been known to be wrong though....
6. (Original post by yodude888)
The thing is that it's a fundamentally flawed question - only about 1/10th of the wind of the calculated cylinder will actually touch the windmill at all - a windmill's rotors don't form a full circle, do they?

That makes the above question completely retarded.
at 600 rpm the difference is negligible, but if this were to be done for accuracy a lot more factors would have to be taken into account.

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