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

Question

A body of mass $4$ kg is placed at a point $P$ having coordinates $ (3,4) $ m. Under the action of force $ \mathbf{F} = (2\hat{i} + 3\hat{j}) $ N, it moves to a new point $Q$ having coordinates $ (6,10) $ m in $4$ sec. The average power and instantaneous power at the end of $4$ sec are in the ratio:

Answer 1

Given Data (Part c)

Parameter	Value
Blade area ($A$)	30 m²
Wind speed ($v$)	36 km/h = 10 m/s
Air density ($\rho$)	1.2 kg/m³
Efficiency ($\eta$)	25% = 0.25

(a) Mass of Air Passing Through

Volume swept per time $t$: $A \times v \times t$

$$m = \rho \times (\text{volume}) = \rho A v t$$

Mass = $\rho A v t$

(b) Kinetic Energy of Air

KE of moving air mass:

$$KE = \frac{1}{2} m v^2 = \frac{1}{2} (\rho A v t) v^2 = \frac{1}{2} \rho A v^3 t$$

KE = $\frac{1}{2} \rho A v^3 t$

(c) Electrical Power Produced

Power = energy per unit time. Wind power available:

$$P_\text{wind} = \frac{KE}{t} = \frac{1}{2} \rho A v^3$$

Electrical power (25% efficiency):

$$P_\text{elec} = \eta P_\text{wind} = 0.25 \times \frac{1}{2} \rho A v^3$$

Calculate:

$$P_\text{wind} = \frac{1}{2} \times 1.2 \times 30 \times 10^3 = 0.6 \times 30 \times 1000 = 18{,}000 \, \text{W}$$ $$P_\text{elec} = 0.25 \times 18{,}000 = 4{,}500 \, \text{W} = 4.5 \, \text{kW}$$

Electrical Power

$P = \textbf{4.5 kW}$

Summary

Part	Result
(a)	$m = \rho A v t$
(b)	$KE = \frac{1}{2} \rho A v^3 t$
(c)	$P = 4.5 \, \text{kW}$

Solution and Explanation