Question

A wind mill converts a fixed fraction of the wind energy intercepted by its blades into electrical energy. The electrical power output \( P \) is related to the velocity of wind \( v \) as

• A. \( v^3 \)
• B. \( v^2 \)
• C. \( v^{3/2} \)
• D. \( v^{5/2} \)

Hint:

Wind power depends strongly on velocity. Doubling wind speed increases power by eight times.

Answer 1

The Correct Option is A

Step 1: Expression for kinetic energy of wind.
Kinetic energy of air mass is:
\[ KE = \frac{1}{2}mv^2 \]

Step 2: Mass flow rate of air.
Mass of air passing per second:
\[ m = \rho A v \]

Step 3: Substitute in energy expression.
\[ KE = \frac{1}{2} (\rho A v) v^2 \]
\[ KE = \frac{1}{2} \rho A v^3 \]

Step 4: Power relation.
Power is energy per unit time:
\[ P \propto v^3 \]

Step 5: Fraction conversion.
Since windmill converts a fixed fraction, proportionality remains unchanged.

Step 6: Final relation.
\[ P \propto v^3 \]

Step 7: Final Answer.
\[ \boxed{v^3} \]