Question

If the diameter (d) of horizontal axis rotor is doubled and wind speed (V) is halved, the available wind power (Pa ) will

• A. Remain same
• B. Increase by two times
• C. Reduced to half
• D. Increase by four times

Hint:

Wind power is extremely sensitive to velocity changes because of the cubic relationship (\(V^3\)).
Even though doubling the diameter quadruples the area (\(\times 4\)), halving the wind speed reduces the kinetic energy density by an eighth (\(\times 1/8\)), resulting in a net reduction to half (\(4/8 = 1/2\)).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The kinetic energy of wind passing through a swept area of a rotor determines the maximum theoretical power available for harvest.

Step 2: Key Formula or Approach:
The power \(P_a\) available in the wind is given by:
\[ P_a = \frac{1}{2} \rho A V^3 \] where:
\(\rho\) = density of air
\(A\) = swept area of the rotor, \(A = \frac{\pi d^2}{4}\)
\(V\) = wind velocity
Thus, the power relation is:
\[ P_a \propto d^2 V^3 \]

Step 3: Detailed Explanation:
Let the initial rotor diameter be \(d_1\) and initial wind velocity be \(V_1\).
The new parameters are:
\[ d_2 = 2 d_1 \] \[ V_2 = \frac{1}{2} V_1 \] Comparing the new available power \(P_{a2}\) to the initial power \(P_{a1}\):
\[ \frac{P_{a2}}{P_{a1}} = \left( \frac{d_2}{d_1} \right)^2 \times \left( \frac{V_2}{V_1} \right)^3 \] Substitute the new values:
\[ \frac{P_{a2}}{P_{a1}} = (2)^2 \times \left( \frac{1}{2} \right)^3 \] \[ \frac{P_{a2}}{P_{a1}} = 4 \times \frac{1}{8} = \frac{4}{8} = \frac{1}{2} \] Thus, the power is reduced to half.

Step 4: Final Answer:
The correct option is 3, which corresponds to "Reduced to half".