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Windmill - mass of air hitting blades each second? watch

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    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
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    2.4 * 14 = Volume in m^3

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

    mass each sec=density x volume each sec....use the volume you got above
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    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.
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    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....
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    (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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