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Question 6.21 The blades of a windmill sweep out a circle of area A. (a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t?(b) What is the kinetic energy of the air? (c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A = 30 m2, v = 36 km/h and the density of air is 1.2 kg m–3. What is the electrical power produced?

Chapter Work, Energy And Power Page 137

Answers

Answered by pankaj12je
52
Hey there !!!!!!

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Area of circle swept by wind mill = A

Velocity of wind = v

Density of air = d

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Volume of wind flowing = Av

Mass of wind flowing per second = Volume*density = Avd

Mass of wind flowing for time 't'= Avdt

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Kinetic energy of air = mv²/2 = Avdt*v²/2 = Adtv³/2

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If wind converts 25% of its energy into mechanical energy when

A= 30m²   v=36kmph= 36*5/18=10m/s   d=1.2kg/m³

Power produced = Mechanical energy/time = Atdv³*25%/2t =Adv³*25%/2
        
                                                            =30*1.2*10³*25/200 = 4500W

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Hope this helped you......................


Answered by abhi178
57
the blades of windmill sweep out a circle of area A . and wind flows at a velocity v perpendicular to circle .
(a) hence, volume of air passing through the windmill per second = Av
so, volume of air passing through in time t = Av × t
then, mass = density × volume
= P × Av × t
mass of air = PAvt

(b) kinetic energy of air = 1/mv²
= 1/2( PAvt)v²
= PAv³t/2

(c) electric energy produced = 25% of Kinetic energy { wind's energy }
= 1/4 × (PAv³t)/2
= (PAv³t)/8

so, Electric power produced = electric energy produced /time taken for producing
= (PAv³t)/8t
= (PAv³)/8

hence, Electric power = (PAv³)/8
here,
P = 1.2 kg/m³
A = 30 m²
v = 36 km/h = 36×5/18 = 10m/s

Electric power = 1.2 × 30× 10³/8
= 4500 watt
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