Subtopic: Sources of power on the farm – human, animal, mechanical, electrical, wind, solar and biomass
Question Statement:
A small horizontal-axis wind turbine (HAWT) is being designed to deliver 2 kW shaft power at a steady wind speed of 7 m·s⁻¹. The HAWT operates with an aerodynamic power coefficient of 0.36, drivetrain efficiency of 0.90, and air density of 1.225 kg·m⁻³. Neglecting other losses, the required rotor diameter is X m. When the wind speed drops to 6 m·s⁻¹, the shaft power is Y kW. Identify the nearest correct option for the X and Y combination. (Take π = 3.14)
Options:
(A) X = 6.12, Y = 1.28
(B) X = 3.06, Y = 0.63
(C) X = 6.12, Y = 2.00
(D) X = 6.75, Y = 1.50
Official Answer Key:
A
Step-by-Step Solution & Explanation
Part X — rotor diameter. Shaft power extracted from the wind:
P = ½ ρ A v³ · C_p · η_drivetrain
2000 = 0.5 × 1.225 × A × 7³ × 0.36 × 0.90
2000 = 0.5 × 1.225 × A × 343 × 0.324
2000 = 68.07 A
A = 29.38 m²
D = √(4A/π) = √(4 × 29.38/3.14) = √37.43 = 6.12 m ✔
Part Y — power at reduced wind speed. With rotor area, C_p, η and ρ unchanged, P ∝ v³:
Y = 2 × (6/7)³ = 2 × (216/343) = 2 × 0.6297 = 1.26 kW
The nearest listed value is 1.28 kW, confirming option (A). Note (C) is wrong because power cannot remain at 2 kW when wind speed falls, and (B) halves the diameter incorrectly.
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