Page 300 - Wind Energy Handbook

274 DESIGN LOADS FOR HORIZONTAL-AXIS WIND TURBINES

If the wind speed variation due to wind shear is assumed to be linear with height,
i.e., u ¼ U(kr=R), and the teeter moment is calculated from the expression on the
right-hand side of Equation (5.94), which assumes a frozen wake instead of the
equilibrium wake resulting from momentum theory, a very simple expression for
the teeter angle results in the case of zero ä 3 angle. The teeter moment becomes
ð
dC l Uk R
3
M T ¼ M TO cos Ùt ¼ rÙ c(r)r dr cos Ùt (5:98)
1
2 dÆ R R
Substitution of Equation (5.98) in Equation (5.97), with ø n set equal to Ù for the case
of a zero ä 3 angle, results in the following expression for the teeter angle

Uk
æ ¼ cos(Ùt ð=2)) (5:99)
ÙR
Thus the teeter response lags the excitation by 908 and the magnitude of the teeter
excursion is simply equal to the velocity gradient divided by the rotational speed.
For a hub height of 35 m, the equivalent uniform velocity gradient over the rotor
disc for the case above is 0:125U=R ¼ 0:075 m=s per m, giving a teeter excursion of
0.024 radians or 1:48. This differs from the earlier value of 1:058 because of the
frozen wake assumption.

Teeter response to stochastic loads

As usual, it is convenient to analyse the response to the stochastic loads in the
frequency domain. The teeter moment providing excitation is given by the right-
hand side of Equation (5.94). By following a similar method to that used for the
generalized load in Section 5.8.6, the following expression for the power spectrum
of the teeter moment can be derived:
ð ð
2 R R
o
S MT (n) ¼ 1 rÙ dC l S (r 1 , r 2 , n)c(r 1 )c(r 2 )r 1 r 2 jr 1 jjr 2 j dr 1 dr 2 (5:100)
2 dÆ R R u
o
where S (r 1 , r 2 , n) is the rotationally sampled cross spectrum. In practice,
u
o
S (r 1 , r 2 , n) is evaluated for a few discrete radius values, and the integrals replaced
u
by summations.
The power spectrum of the teeter angle response is related to the teeter moment
power spectrum by a formula analagous to Equation (5.90), as follows:
S MT (n) 1
S æ (n) ¼
(5:101)
2 2
2 2
2
(Iø ) [(1 2ðn=ø n ) ) þ (2î:2ðn=ø n ) ]
n
2
2 2
This can be written S æ (n) ¼ (S MT (n)=(Iø ) )[DMR] where DMR stands for the
n
dynamic magnification ratio.
Figure 5.30 shows the teeter angle power spectrum, S æ (n), for a two-bladed rotor