Question

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$\sf\huge{\underline{\mathbb{Question}}}$

$<b><u>The blades of windmill sweep out a circle of area A.(a) If the wind blows at a velocity of 'v' perpendicular to the circle , what is the mass of air passing through it in time 't'?.(b)What is the Kinetic energy of air?.(c)Assume that the windmill converts 25% of the wind's energy into electrical enegy, and given that , A= 30[tex]m^{2}$, v = 36km\h and density of air is 1.2$kg\times m^{-3}$.What is the electrical power produced?![/tex]

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Answer 1

$\tt\huge{\underline{\mathbb{Answer}}}$

$\huge{\underline{Question}}$

▶The blades of windmill sweep out a circle of area A.(a) If the wind blows at a velocity of 'v' perpendicular to the circle , what is the mass of air passing through it in time 't'?.(b)What is the Kinetic energy of air?.(c)Assume that the windmill converts 25% of the wind's energy into electrical enegy, and given that , A= 30$m^{2}$, v = 36km\h and density of air is 1.2$kg\times m^{-3}$.What is the electrical power produced?!

$\sf{\underline{Solution}}$>>

(a). Volume of wind blowing per second= $A\times vm^{3}$

since, v =$\frac{s}{t}$

if t = 1

,then v = s..

SIMILARLY, mass of wind blowing per second = $V\times p$

where, V is volume ,and p is density

m = $A\times v\times p$....(mass blowing in 1 second)

for 't' second..

m = $A\times v\times p\times t$

(b). KINETIC ENERGY of air=>

K.E= $\frac{mv^{2}}{2}$

K.E = $\frac{A\times v\times p\times t\times v^{2}}{2}$

K.E = $\frac{A\times p\times t\times v^{3}}{2}$

3. Electric enegy produced = $efficiency\times KINETIC energy$

K.E(n) = $\frac{A\times p\times t\times v^{3}}{2}\times n$

where n is the efficiency..

Electric power(P) = $\frac{[tex]\frac{A\times p\times t\times v^{3}}{2}\times n$}{t}[/tex]

P = $\frac{A\times p\times n\times v^{3}}{2}$

__Given: A = 30$m^{2}$

v =36km\h = $\frac{36\times 5}{18}$ =10m/s

n = 0.25

p =1.2$kg\times m^{-3}$

P = $\frac{30\times 1.2\times 0.25\times 10^{3}}{2}$

P = $\frac{9000}{2}$

P = 4.5kw = 4500 watt..

Answer 2

Answer:

The blades of windmill sweep out a circle of area A.(a) If the wind blows at a velocity of 'v' perpendicular to the circle , what is the mass of air passing through it in time 't'?.(b)What is the Kinetic energy of air?.(c)Assume that the windmill converts 25% of the wind's energy into electrical enegy, and given that , A= 30m^{2}m

2

, v = 36km\h and density of air is 1.2kg\times m^{-3}kg×m

−3

.What is the electrical power produced?!

\sf{\underline{Solution}}

Solution

>>

(a). Volume of wind blowing per second= A\times vm^{3}A×vm

3

since, v =\frac{s}{t}

t

s

if t = 1

,then v = s..

SIMILARLY, mass of wind blowing per second = V\times pV×p

where, V is volume ,and p is density

m = A\times v\times pA×v×p ....(mass blowing in 1 second)

for 't' second..

m = A\times v\times p\times tA×v×p×t

(b). KINETIC ENERGY of air=>

K.E= \frac{mv^{2}}{2}

2

mv

2

K.E = \frac{A\times v\times p\times t\times v^{2}}{2}

2

A×v×p×t×v

2

K.E = \frac{A\times p\times t\times v^{3}}{2}

2

A×p×t×v

3

3. Electric enegy produced = efficiency\times KINETIC energyefficiency×KINETICenergy

K.E(n) = \frac{A\times p\times t\times v^{3}}{2}\times n

2

A×p×t×v

3

×n

where n is the efficiency..

Electric power(P) = \frac{$\frac{A\times p\times t\times v^{3}}{2}\times n }{t}$

P = \frac{A\times p\times n\times v^{3}}{2}

2

A×p×n×v

3

__Given: A = 30m^{2}m

2

v =36km\h = \frac{36\times 5}{18}

18

36×5

=10m/s

n = 0.25

p =1.2kg\times m^{-3}kg×m

−3

P = \frac{30\times 1.2\times 0.25\times 10^{3}}{2}

2

30×1.2×0.25×10

3

P = \frac{9000}{2}

2

9000

P = 4.5kw = 4500 watt!!