Q.The blades of a windmill sweep out a circle of area $A$.

The key idea here is the Kinetic Energy Flux of the wind.

(a) The volume of air passing through the area $A$ in time $t$ is the product of the area and the distance the wind travels in time $t$.

1. Distance travelled by wind in time $t$: $d = vt$.
2. Volume of air passing through $A$: $V = A \times d = Avt$.
3. Mass of air: $m = \rho V = \rho Avt$.

(b) The kinetic energy of this mass of air is given by the standard formula.

1. Kinetic energy formula: $KE = \frac{1}{2}mv^2$.
2. Substitute the mass $m$ from part (a): $KE = \frac{1}{2}(\rho Avt)v^2 = \frac{1}{2}\rho Av^3t$.

(c) Electrical power is the rate of electrical energy production, which is $25\%$ of the wind's kinetic energy flux (wind power).

1. Convert wind velocity to SI units: $v = 36\ \text{km/h} = 36 \times \frac{1000\ \text{m}}{3600\ \text{s}} = 10\ \text{m/s}$.
2. Wind power (kinetic energy flux): $P_{\text{wind}} = \frac{KE}{t} = \frac{1}{2}\rho Av^3$.
3. Electrical power produced: $P_{\text{electrical}} = 0.25 \times P_{\text{wind}} = 0.25 \times \frac{1}{2}\rho Av^3$. …

When dealing with continuous flows like wind, we often think about the rate at which mass or energy passes through a certain area. This is known as a flux. For a windmill, the blades are constantly interacting with new air, so we are interested in the kinetic energy delivered by the wind per unit time, which is power.

Let's break down the problem step by step.

(a) Mass of air passing through the circle in time $t$

1. Visualize the volume of air: Imagine a cylinder of air that passes through the circular area $A$ swept by the windmill blades. If the wind moves at a velocity $v$ perpendicular to this area, then in a time $t$, all the air initially contained within a cylinder of length $vt$ and cross-sectional area $A$ will have passed through the circle.

   The volume of this cylinder of air is given by:

2. Calculate the mass: The mass $m$ of this volume of air can be found using the definition of density ($\rho = \frac{m}{V}$), so $m = \rho V$. Substituting the expression for $V$:

So, the mass of air passing through the circle in time $t$ is $\rho Avt$.

(b) Kinetic energy of the air

1. Recall the kinetic energy formula: The kinetic energy ($KE$) of a mass $m$ moving with velocity $v$ is given by:

2. Substitute the mass: We found the mass $m$ of the air passing in time $t$ to be $\rho Avt$. Substituting this into the kinetic energy formula:

This is the total kinetic energy of the air that passes through the windmill's area in time $t$.

(c) Electrical power produced

1. Understand power: Power is the rate at which energy is transferred or converted. In this context, the wind delivers kinetic energy to the windmill, and the windmill converts a fraction of this into electrical energy. The power of the wind is the kinetic energy delivered per unit time.

   From part (b), the kinetic energy delivered in time $t$ is $KE = \frac{1}{2}\rho Av^3t$.

   Therefore, the kinetic power of the wind ($P_{wind}$) is:

> [!FORMULA]
> The kinetic power of wind passing through an area $A$ with velocity $v$ and air density $\rho$ is:
> $$P_{wind} = \frac{1}{2}\rho Av^3$$

2. Apply the conversion efficiency: The problem states that the windmill converts $25\%$ of the wind's energy into electrical energy. This means the electrical power produced ($P_{elec}$) is $25\%$ of the wind's kinetic power.

$P_{elec} = 0.25 \times P_{wind}$ …

Concept: Kinetic Energy Flux (Power) of a Flowing Fluid

Step 1: Mass of air through area $A$ in time $t$

Step 2: Kinetic energy of that mass of air

Step 3: Wind power (energy per unit time)

$P_{\text{wind}} = \frac{K}{t} = \tfrac12 \rho A v^3$ …