27. 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 proportional to (A) v (B) v2 (C) v3 (D) v4

MATHEMATICAL MODEL
The following table shows the definition of variousvariables used in this model:
E = Kinetic Energy(J)
ρ = Density(kg/m3)
m = Mass (kg) A = Swept Area(m²)
v = Wind Speed(m/s)
C_p = PowerCoefficient
P = Power (W)
r = Radius (m)
dm/dt = Mass flow rate(kg/s)
x = distance (m)
dE/dt = Energy FlowRate (J/s)
t = time (s)

Under constant acceleration, the kinetic energy of
an object having mass m and velocity v is equal
to the work done W in displacing that object from
rest to a distance s under a force F , i.e.:
E =W = Fs
According to Newton’s Law, we have:
F = ma
Hence,
E = mas … (1)
Using the third equation of motion:
v²-u²=2as
we get:
a=(v²-u²)/2s
Since the initial velocity of the object is zero, i.e.
u = 0 , we get:
a=v²/2s
Substituting it in equation (1), we get that the
kinetic energy of a mass in motions is:

E = mv²/2 … (2)
The power in the wind is given by the rate of
change of energy:
P(Power)= dE/dt= (v²/2)dm/dt… (3)
As mass flow rate is given by:
dm/dt = ρAdx/dt
and the rate of change of distance is given by:
dx/dt = v

we get:
dm/dt = ρAv

Hence, from equation (3), the power can be
defined as:
P = (v²/2)ρAv= ρAv³/2… (4)
So, P(Power) is proportional to v³