Question

The blades of a windmill generating electrical energy sweep out an area of 20 m$^2$. If the efficiency of the windmill is 25% and wind speed is 36 kmph, then the electrical power generated is (Density of air = 1.2 kg m$^{-3}$):

• A. 120 kW
• B. 1200 W
• C. 300 W
• D. 3 kW

Hint:

Wind power depends on the cube of wind velocity, so doubling wind speed increases power eightfold. Efficiency of real windmills rarely exceeds 40% due to mechanical and aerodynamic losses. Always convert km/h to m/s before substitution to maintain unit consistency. The swept area of the blades directly controls total extractable power.

Answer 1

The Correct Option is D

• Given: area $A = 20$ m$^2$, air density $\rho = 1.2$ kg/m$^3$, velocity $v = 36$ km/h $= 10$ m/s, efficiency $\eta = 25% = 0.25$.
• Power available in wind = kinetic energy per second flowing through area $A$: \[ P_{\text{wind}} = \frac{1}{2}\rho A v^3 \] • Substituting: \[ P_{\text{wind}} = 0.5 \times 1.2 \times 20 \times 10^3 = 1.2 \times 10^4 \text{ W} \] • The actual power generated = $\eta P_{\text{wind}} = 0.25 \times 1.2 \times 10^4 = 3000$ W = 3 kW.
• Hence, the electrical power generated is 3 kW.