Question:hard

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

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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 (\(4/8 = 1/2\)).
  • Remain same
  • Increase by two times
  • Reduced to half
  • Increase by four times
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The Correct Option is C

Solution and Explanation

Step 1: Write the wind power formula and identify the variables.
The power available in wind passing through a rotor is \[ P_a = \frac{1}{2} \rho A V^3 \] where the swept area is \( A = \frac{\pi d^2}{4} \), so the power is proportional to \( d^2 V^3 \).
Step 2: Apply the given changes to each variable separately.
Doubling the diameter multiplies the swept area term by \( 2^2 = 4 \). Halving the wind speed multiplies the velocity term by \( (0.5)^3 = 0.125 \), that is, one eighth.
Step 3: Combine the two factors to get the net change.
\[ \frac{P_{new}}{P_{old}} = 4 \times \frac{1}{8} = \frac{1}{2} \] So the new available wind power is exactly half of the original power, since the cubic dependence on wind speed dominates the squared dependence on diameter.
\[ \boxed{\text{Reduced to half}} \]
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