Robustly Optimal Fixed Pitch HAWT - PowerPoint PPT Presentation

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Robustly Optimal Fixed Pitch HAWT

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Robustly Optimal Fixed Pitch HAWT (with Tip Correction & Drag ) Exact trig solution of BEM optimal rotor. New criterion of robust optimality ... – PowerPoint PPT presentation

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Title: Robustly Optimal Fixed Pitch HAWT


1
  • 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

2
s 2psin? r/B ? advance/B. HI/F, F a function
of (R-r)/s.221s. ½? W CL r x W ? -dL
/dA 2? I (W?z) the (BEM) eqn (1)
3
dP xTq ? (dLdD)2? xT (W?z) I (sin? -
cos? tane) dA IFH HNsin(?-?), W?z
Wsin?, WNcos(?-?) dP/dA? xT N2 F
sin2(?-?)(sin2? -½sin2? tane) 2cote
cot? cot (3?-2?) ?o2?/3e/6 O(e2)
4
Robustly Optimal Blade Elements ?CL 4Fsin?
tan(?-?) CL / 2? e sin?, ?-?o½?o-¼e ?
sin (?o -?r)Fsin? tan(½?o-¼e) ? ½e??
¼eBc/r ?cos(?o -?r)d Fsin? tan(½?o -¼e) /
d?o robust c o t ? r cot ?d
(1-½de/da) csc(?d -½e) dF/ Fd?o cos
(pF/2)e-f , fp(R-r)/sB(R-r)/ 2rsin?o
cot?r ? cot ½?d½ecsc?d cot?d -½de/dacsc(?d)
- 2 f cot (½pF) cot?d/pF ?r?½?d/h,h1¼e/?d-
¼de/da-fcot(½pF)/ pF
?r? ½?d e/8 Tip ?r?2?d /3 e/8
5
robust ?r sin?r Fsin?d tan(½?d-¼e) perturb
?????r cot ?r effect on d?/d? vs robust
d?o/d? Fsin? tan(?-?)?sin(?-?)
Fsin?sec2(?-?) u vd?/d? vFsin?
sec2(?-?)-....?cos? d?/d? u/v ?(d?/d?) ?
-(u/v)2 ?v/ u u/vd?o/d?2/3, ?v
?(?cosa)???/ sina d(?-?o)/d??-4/9??tan(?-?)c
os2(?-?)/sin2?r? 4(ad-ar) h2/ 9tan(½?d)

6
Blade Pitch for highest net mean Annual power
  ddP/dA-½? xTN2F sin?d
6(?-?o)2 ½d2?/d?2sin2?d (ad-ao)2
mean 64(ad-ar) ??
cot½?h2/92 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 gt0 Then with tT/Td , ??? g?t sin 2? .
. peak p(t)31-4(t -1)2/2 between t .5 and1.5
(?t)2 .05 (ad-ar)8h4cot2½?d(ad-ao)
d2?/d?2270sin2?d/g2sin23?d? 0
7
(No Transcript)
8
WIND SHEAR as T0(1 msin ß) where m k x /X
p(ß) (1 msinß)3/a averaging over ß,
a13m2/2. t 0 T/T0 ?t0 msin ß in t0 T/T0
a?t03 m2/23m4/8 and a(?t0)2 ½ m29m4/8.
t 0 1?t0 ?13m2/2-15m4/8 and the true minimum
variance about it is (?t0) 2 - ?t02 ?½
m2-15m4/8 , so ?t2 ?½ m2-27m4/8 r at k1/3
1k2(1-k) at x/X.7, about .017 versus .05
above T
i p , ?t2 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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