Wind Power Energy Sources Fall 2014

Wind Potential Wind energy is the most abundant renewable energy source after solar 120 GW of peak world capacity installed by 2008

Drag Wind Power Earliest type of wind power Velocity limited to wind speed

Drag Power Mechanics Force on surface equal to drag D = C D [½ρ(u – v) 2 cℓ] Power = Drag · Velocity P = Dv = C D [½ρ(u – v) 2 cℓ]v = ½ρu 3 C D (1 – v/u) 2 cℓ(v/u) = ½ρu 3 A P C D (1 – v/u) 2 (v/u)  where A P = cℓ is the cross-sectional area u v D c ℓ C D = 1.4 for this shape

Drag Power Limitations ½ρu 3 A P C D (1 – v/u) 2 (v/u)  At v = 0, power = 0 (from v/u)  At v = u, power = 0 (from 1 – v/u)  v cannot exceed u Maximum power at v/u = 1/3  Greatest coefficient of power C P = 4 / 27 C D

Lift-type Wind Turbines Wind is parallel to axis of rotation and perpendicular to motion of blades v cross section u c

Velocity Triangle v u v r = u - v β θ α v/u = tanβ

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v β Force Triangle L = c·ℓ·½ρv r 2 ·C L D = c·ℓ·½ρv r 2 ·C D  ℓ is a length in the radial direction F = L·sinβ – D·cosβ  net force in direction of blade motion lift, L drag, D net force β useful force, F u r vrvr n = 2

v β Power P= F·v = (L·sinβ – D·cosβ)v = c·ℓ·½ρ · v r 2 · v · (C L ·sinβ – C D ·cosβ) = A P ·½ρ(v r 2 /u 2 )(v/u)·u 3 ·sinβ(C L – C D /tanβ) = ½ρA P ·(1/sin 2 β)(v/u)u 3 sinβ(C L – C D /tanβ)  Since sinβ = u/v r and tanβ=u/v lift, L drag, D net force β useful force, F u multiply by u 3 ·u -3 v r = u - v only a function of v and u

At any radius, with area dAp

If (v/u) 2 » 1, the “1” in the square root is insignificant and the equation becomes  C P, the dimensionless Coefficient of Power, is the LHS of the equation Coefficient of Power

Max C P occurs at The maximum value is

Calculating Power (approx.) A P, the “plan form” area, is the total solid blade area, n·R·C ave where C ave is computed to ensure that actual blade area seen from the axial direction is equal to R·C ave A is the swept area R C ave r dℓdℓ n = 3 r min

Torque Recall: F = L·sinβ – D·cosβ, u/v = tanβ  As radius ↑, v ↑ Torque = Force × radius r Torque = Force × (r/R) × R Both Force and Radius would increase with r  Two reasons why T ↑ as r ↑ We want torque to stay the same over entire blade to minimize internal bending moment

Keeping Torque Uniform We can change  θ, blade tilt (but it cannot be negative)  c, chord (blade width) only a function of r (and u but wind speed is uniform over blade) We are assuming that blade shape drag and lift behavior is same. Rpm is given

Example A 20-m radius blade has a coefficient of lift of 1.2 when α = 12º and decreases linearly to C L = 0.4 when the angle of attack is 0. The coefficient of drag is 0.1 over the whole blade, and the stall angle is 12º. The blades begin at radius = 4 m. The wind speed is 7 m/s.  What is the rotor speed if the “speed ratio” (blade speed over wind speed) remains under the max C P value?  What is the optimum pitch angle at various radii for the greatest extraction of power from the wind?  Determine the chord at the tip for uniform torque on the blade (chord at base is 1 m)

Solution (speed) Blade velocity is greatest at tip Max C P occurs at  The blade tip moves (relative to ground) at 56 m/s  This is 56/20 = 2.8 rad/sec or 26.7 RPM

Solution (table) radiusv/uβαθCLCL C L /C D XKT/c mººº N·m/m First, pick a bunch of radii from the base to the tip Then, calculate u/v for each; u is constant and v is angular velocity × radius u/v = tanβ β v u Now find β, the angle between blade velocity relative and blade velocity The angle of attack is the smallest of β or the stall angle The pitch, the angle from the blade profile to blade velocity, is α – β The lift coefficient scales from 0.4 to 1.2 as α goes from 0º to 12º; C D remains constant The specific torque is proportional to (r/R)·X, so torque (per chord length) times an arbitrary constant K, will equal that expression

Solution (chord length) The specific torque at the blade tip is 1.43/.60 = 4.2 times that at the base Therefore, the chord at the base must be 4.2 times that at the tip Since the base chord is 1 m, the tip chord is 1 ÷ 4.2 =.24 m

Useful approx for Cp max Function of B=no of blades, D/L, and tip- speed ratio λ at design wind speed 24

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Wind Characteristics The relative wind speed (u/c) at a given time follows the Rayleigh distribution   c is scale parameter Shape factor (k) describes variance  k = 2 common for wind Φ(u)

Rayleigh Averages Mean wind speed ū = cΓ(1 + 1/k)  Γ is “gamma function”  Γ(1.5) = 0.89 But power is proportional to the cube of wind speed   Average of the cubes ≠ the cube of the averages

Power Averages Power Density [W/m²],  ū = c·0.89 → c = ū/0.89 So if the average wind speed ↑ by 15%, the annual energy ↑ by 50%  ≈ 1.5

Meteorological Data

Performance Data 39 m Diameter, 500 kW Nameplate Capacity Turbine

Electrical Output