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Robustly Optimal Fixed Pitch HAWT Blade Elements Design & Optimization

Explore exact trigonometric solutions for optimal rotor design, with robust criteria for maximizing annual power. The blade pitch, chord, and aerodynamics are considered, leading to efficient wind turbine performance.

erin-fletcher
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Robustly Optimal Fixed Pitch HAWT Blade Elements Design & Optimization

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  1. RobustlyOptimal Fixed Pitch HAWT • (with Tip Correction & Drag ) • Exact trig solution of BEM optimal rotor • New criterion of robust optimality • Max annual power pitch& chord

  2. s= 2psinr/B advance/B. H=I/F, F a function of (R-r)/s.221s. ½WCL rx W= -dL /dA= 2I(Wz)the (BEM) eqn(1)

  3. dP= xTq (dL+dD)=2 xT(Wz) I (sin- cos tane) dA I=FH H=Nsin(-), Wz =Wsin, W=Ncos(-) dP/dA=xT N2F sin2(-)(sin2-½sin2 tane) 2cote= cot +cot (3-2) o=2/3+e/6 +O(e2)

  4. Robustly Optimal Blade Elements CL =4Fsintan(-) CL / 2= esin, -o=½o-¼e sin (o -r)=Fsintan(½o-¼e)  =½e =¼eBc/r cos(o -r)=d { Fsintan(½o-¼e)} / dorobust c o t  r = cot d + (1-½de/da) csc(d-½e)+ dF/ Fdo cos (pF/2)=e-f , f=p(R-r)/s=B(R-r)/ 2rsino cotrcot½d+½ecscdcotd-½de/dacsc(d) - 2 f cot (½pF) cotd/pF r½d/h,h=1+¼e/d-¼de/da-fcot(½pF)/ pF r ½d –e/8 Tip r2d /3 –e/8

  5. robust r sinr =Fsindtan(½d-¼e) perturb=rcot r effect on d/dvs robust do/d Fsintan(-)=sin(-) Fsinsec2(-)= u= vd/d v=Fsinsec2(-)-....+cos d/d =u/v(d/d) -(u/v)2 v/ u u/v=do/d=2/3, v =(cosa)=/ sina d(-o)/d-4/9tan(-)cos2(-)/sin2r4(ad-ar) h2/ 9tan(½d)

  6. Blade Pitch for highest net mean Annual power ddP/dA=-½ xTN2Fsind {6(-o)2+ ½d2/d2sin2d(ad-ao)2} mean 6{4(ad-ar)cot½h2/9}2+ ½d2/d2sin2d(ad-ao)2 Relate , x to T: synchronous generator X as T-1. p.d. rotary pump fixed torque has T2CT =T2Cp/X so XT2. Say xTg, g2 >0 Then with τ=T/Td , gτ sin 2. . peak p(τ)=3{1-4(τ-1)2}/2 betweenτ=.5 and1.5 (τ)2= .05 (ad-ar)8h4cot2½d+(ad-ao) d2/d2270sin2d/g2sin23d0

  7. WIND SHEAR as T0(1+ msin β) where m= k x /X p(β)= (1+ msinβ)3/a averaging over β, a=1+3m2/2. t0= T/T0 t0= msin β in τ0= T/T0aτ0=3 m2/2+3m4/8 and a(τ0)2 = ½ m2+9m4/8. τ 0= 1+τ01+3m2/2-15m4/8 and the true minimum variance about itis (τ0)2 - { τ0}2 ½ m2-15m4/8 , so {τ}2 ½ m2-27m4/8 ; rat k=1/3 1+k=2(1-k) at x/X=.7, about .017 versus .05 above; T i p , {τ}2 is exactly .029 . Such independent variances ADD. YAW the  variance is proportional to the mean 4th power of yaw angle. Need wind data and yawcontrol algorithm.

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