robustly optimal fixed pitch hawt ( with tip correction & drag )
DESCRIPTION
Robustly Optimal Fixed Pitch HAWT ( with Tip Correction & Drag ) Exact trig solution of BEM optimal rotor New criterion of robust optimality Max annual power pitch& chord. s= 2 p sin r/B advance/ B . H = I /F, F a function of (R-r)/s .221 s . - PowerPoint PPT PresentationTRANSCRIPT
Robustly Optimal Fixed Pitch HAWT
(with Tip Correction & Drag )
•Exact trig solution of BEM optimal rotor
•New criterion of robust optimality
•Max annual power pitch& chord
s= 2psin r/B advance/B. H=I/F, F a function of (R-r)/s.221s. ½ W CL r x W = -dL /dA= 2 I (Wz) the (BEM) eqn (1)
dP= xT (dL+dD)=2 xT (Wz) I (sin - cos tane) dA I=FH H=Nsin(-), Wz =Wsin, W=Ncos(-)
dP/dA= xT N2 F sin2(-)(sin2 -½sin2 tane)
2cote= cot +cot (3-2) =2/3+e/6 +
Robustly Optimal Blade Elements
CL =4Fsin tan(-) CL / 2= e sin, -=½¼e sin ( -r)=Fsin tan(½¼e) =½e =¼eBc/r
cos( -r)=d { Fsin tan(½ ¼e)} / d robust
c o t r = cot d + (1-½de/d) csc(d ½e)+ dF/ Fd
cos (pF/2)=e-f , f=p(R-r)/s=B(R-r)/ 2rsino
cotr cot ½d+½ecscd cotd -½de/dcsc(d) - 2 f cot (½pF) cotd/pF
r½d/h,h=1+¼/d-¼d/d-fcot(½F)/ F
r ½d –/8 Tip r2d /3 –/8
robust r sinr =Fsind tan(½d¼e)
perturb =r cot r effect on d/d vs robust d/d
Fsin tan(-)=sin(-)
Fsinsec2(-)= u= vd/d v=Fsin sec2(-)-....+cos
d/d =u/v (d/d) -(u/v)2 v/ u
u/v=d/d=2/3, v =(cos)=/ sin
d(-)/d-4/9tan(-)cos2(-)/sin2r 4(d-r) h2/ 9tan(½d)
Blade Pitch for highest net mean Annual power dP/dA=-½ xTN2F sind
{6(-o)2 + ½d2/d2sin2d (ad-ao)2} mean 6{4(ad-ar) cot½h2/9}2 + ½d2/d2sin2d(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 xTg, g2 >0 Then with τ=T/Td , gτ sin 2 . .
peak p(τ)=3{1-4(τ -1)2}/2 between τ =.5 and1.5 (τ)2= .05 (ad-ar)8h4cot2½d+(ad-ao) d2/d2270sin2d/g2sin23d 0
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. t 0= T/T0 t0= msin β
in τ0= T/T0 aτ0=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 it is
(τ0) 2 - { τ0}2 ½ m2-15m4/8 , so {τ}2 ½ m2-27m4/8 ;
r at 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.