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Wind Turbine Blade Aerodynamic Work Solution

Formulas & Definitions

In renewable energy systems, wind turbine blades convert kinetic airflow energy into mechanical shaft power. Aerospace and energy engineers analyze the dot product between aerodynamic lift/drag force vectors and the local blade velocity/displacement vectors to evaluate instantaneous power generation.

A wind turbine rotor blade rotates in a vertical plane. Over an infinitesimal time interval Δt\Delta tΔt, a specific blade airfoil section undergoes a displacement vector of d⃗=2i^+5j^ meters\vec{d} = 2\hat{i} + 5\hat{j}\text{ meters}d=2i^+5j^​ meters. The combined aerodynamic lift and drag force vector exerted by the passing wind on this section is F⃗=120i^−30j^ Newtons\vec{F} = 120\hat{i} - 30\hat{j}\text{ Newtons}F=120i^−30j^​ Newtons.

  1. Compute the mechanical work done by the wind on this blade section using the dot product.
  2. Calculate the angle θ\thetaθ between the aerodynamic force vector F⃗\vec{F}F and the displacement vector d⃗\vec{d}d.
  3. Explain physically why the work is positive despite the negative yyy-component of force.

Step-by-Step Solution:

  1. Calculate Mechanical Work (WWW): Using the algebraic dot product definition:

    W=F⃗⋅d⃗=(120)(2)+(−30)(5)=240−150=90 JoulesW = \vec{F} \cdot \vec{d} = (120)(2) + (-30)(5) = 240 - 150 = 90\text{ Joules}W=F⋅d=(120)(2)+(−30)(5)=240−150=90 Joules

  2. Calculate the Angle (θ\thetaθ): Compute vector magnitudes:

    ∣F⃗∣=1202+(−30)2=14400+900=15300≈123.69 N|\vec{F}| = \sqrt{120^2 + (-30)^2} = \sqrt{14400 + 900} = \sqrt{15300} \approx 123.69\text{ N}∣F∣=1202+(−30)2​=14400+900​=15300​≈123.69 N
    ∣d⃗∣=22+52=4+25=29≈5.385 m|\vec{d}| = \sqrt{2^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29} \approx 5.385\text{ m}∣d∣=22+52​=4+25​=29​≈5.385 m
    Apply geometric dot product formula:
    cos⁡θ=F⃗⋅d⃗∣F⃗∣∣d⃗∣=90(123.69)(5.385)=90666.07≈0.1351\cos\theta = \frac{\vec{F} \cdot \vec{d}}{|\vec{F}| |\vec{d}|} = \frac{90}{(123.69)(5.385)} = \frac{90}{666.07} \approx 0.1351cosθ=∣F∣∣d∣F⋅d​=(123.69)(5.385)90​=666.0790​≈0.1351
    θ=cos⁡−1(0.1351)≈82.24∘\theta = \cos^{-1}(0.1351) \approx 82.24^\circθ=cos−1(0.1351)≈82.24∘

  3. Physical Interpretation: Because θ≈82.24∘<90∘\theta \approx 82.24^\circ < 90^\circθ≈82.24∘<90∘, the net force vector forms an acute angle with the displacement vector (cos⁡θ>0\cos\theta > 0cosθ>0). The positive forward thrust along xxx (240 J240\text{ J}240 J) exceeds the negative resisting work along yyy (−150 J-150\text{ J}−150 J), resulting in net positive energy transfer (+90 J+90\text{ J}+90 J) driving rotor rotation.

Meaning of the Answer & Real-Life Application:

Aerospace and wind energy engineers optimize blade pitch angles to ensure the aerodynamic resultant force maintains an acute angle (θ<90∘\theta < 90^\circθ<90∘) relative to blade motion across varying wind speeds, maximizing positive work extraction and power output (P=W/ΔtP = W / \Delta tP=W/Δt).