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Q5.21(b):
The blades of a windmill sweep out a circle of area A. What is the kinetic energy of the air ?

Solution :

Step 1: Identifying Physical Parameters

To determine the kinetic energy ($K$) of a stream of air flowing through a windmill, we define the following variables:

  • $A$: Area of the circle swept by the windmill blades ($m^2$).
  • $v$: Velocity of the air stream ($m/s$).
  • $\rho$: Density of the air ($kg/m^3$).
  • $dm$: Infinitesimal mass of air passing through the area $A$ in time $dt$ ($kg$).

Step 2: Defining the Volume and Mass Flow

Consider a cylindrical column of air moving with velocity $v$. In a time interval $dt$, the length of the air column passing through the area $A$ is given by $dx = v \cdot dt$.

The volume ($dV$) of this cylinder is: $dV = A \cdot dx = A \cdot v \cdot dt$ [Per the definition of volumetric flow rate].

Since mass is the product of density and volume, the mass ($dm$) of this air is: $dm = \rho \cdot dV = \rho \cdot A \cdot v \cdot dt$ [Per the definition of density $\rho = m/V$].

Step 3: Calculating Kinetic Energy

The kinetic energy ($dK$) associated with this mass of air is defined by the classical formula $dK = \frac{1}{2}(dm)v^2$.

Substituting the expression for $dm$ derived in Step 2: $dK = \frac{1}{2} (\rho \cdot A \cdot v \cdot dt) v^2$ $dK = \frac{1}{2} \rho A v^3 dt$

Step 4: Power and Energy Relationship

The rate at which kinetic energy is delivered to the windmill (Power, $P$) is the derivative of energy with respect to time: $P = \frac{dK}{dt} = \frac{1}{2} \rho A v^3$ [Per the definition of Power in fluid dynamics].

Visual Representation of Air Flow

Area A v (velocity)

Conclusion on Kinetic Energy

The kinetic energy of the air passing through the windmill per unit time is proportional to the cube of the wind velocity and the cross-sectional area swept by the blades.

Final Solution: The kinetic energy per unit time (power) of the air is given by $P = \frac{1}{2} \rho A v^3$.


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