
A wind powered generator converts wind energy into electrical energy. Assume that the generator converts a fixed fraction of the wind energy intercepted by its blades into electrical energy. For wind speed \[v\], the electrical power output will be most likely proportional to:
A) ${v^4}$
B) ${v^9}$
C) $v$
D) ${v^3}$
Answer
173.7k+ views
Hint: This question is totally based upon the concepts of force and power. We need to relate the power with force and velocity. Also, we need to relate the force exerted with the mass of the wind and its density and then we need to solve the equation further in order to get the required answer of the given question.
Complete step by step answer:
As we know, that power of a body is the work done by the body per unit time. Mathematically we can represent it by,
$Power,P = \dfrac{W}{t}$……………….. (i)
Now, work done is the force required to displace the body. Mathematically it can be represented as,
$W = F.S$
And, the distance is the product of velocity and time. Mathematically, it can be represented as
$S = vt$…………………(ii)
Again, we know that force is the product of mass and acceleration, and acceleration is the change in velocity. Mathematically, we can represent it as
$F = ma = m\dfrac{{\Delta v}}{t}$………….(iii)
Also, we know that mass per time of a body can be represented by the product of its area, density and velocity. Mathematically, it can be represented by
$\dfrac{m}{t} = \rho Av$………………(iv)
Now, putting the values from equation (ii), (iii) and (iv), we get,
$P = \dfrac{{F.S}}{t}$
$ \Rightarrow P = \dfrac{{\rho A{v^2}}}{t} \times vt$
$\therefore P = \rho A{v^3}$
Clearly, we can see that the electrical power is proportional to ${v^3}$.
Hence, option (D), i.e. ${v^3}$ is the correct option for the given question.
Note: We should know this fact that density, $\rho = \dfrac{{mass}}{{volume}} = \dfrac{m}{V}$.
Now, volume $V = Area \times length = AS$
And $S = vt$
When we relate all these we get,
$\rho = \dfrac{m}{{AS}} = \dfrac{m}{{Avt}}$
Now, $m = \rho Avt$
Or, $\dfrac{m}{t} = \rho Av$
We need to remember all these relations in order to conclude the correct answer.
Complete step by step answer:
As we know, that power of a body is the work done by the body per unit time. Mathematically we can represent it by,
$Power,P = \dfrac{W}{t}$……………….. (i)
Now, work done is the force required to displace the body. Mathematically it can be represented as,
$W = F.S$
And, the distance is the product of velocity and time. Mathematically, it can be represented as
$S = vt$…………………(ii)
Again, we know that force is the product of mass and acceleration, and acceleration is the change in velocity. Mathematically, we can represent it as
$F = ma = m\dfrac{{\Delta v}}{t}$………….(iii)
Also, we know that mass per time of a body can be represented by the product of its area, density and velocity. Mathematically, it can be represented by
$\dfrac{m}{t} = \rho Av$………………(iv)
Now, putting the values from equation (ii), (iii) and (iv), we get,
$P = \dfrac{{F.S}}{t}$
$ \Rightarrow P = \dfrac{{\rho A{v^2}}}{t} \times vt$
$\therefore P = \rho A{v^3}$
Clearly, we can see that the electrical power is proportional to ${v^3}$.
Hence, option (D), i.e. ${v^3}$ is the correct option for the given question.
Note: We should know this fact that density, $\rho = \dfrac{{mass}}{{volume}} = \dfrac{m}{V}$.
Now, volume $V = Area \times length = AS$
And $S = vt$
When we relate all these we get,
$\rho = \dfrac{m}{{AS}} = \dfrac{m}{{Avt}}$
Now, $m = \rho Avt$
Or, $\dfrac{m}{t} = \rho Av$
We need to remember all these relations in order to conclude the correct answer.
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