Wind Energy

156 A. Rauh et al.
ways. First the 10-min speeds Ūi are inserted into the empirical power curve
function:
PC= 1
N1 �
LPC( Ūi); N1 = N/600. (27.2)
N1
i=1
Pictorially, this amounts to shifting the points of Fig. 27.3 vertically onto the
power curve. This average is compared with the true average of the measured
1-s powers L(ti), or equivalently the average of the 10-min values ¯ Li (N1 =
N/600 be an integer):
exp= 1
N
N�
L(ti) = 1
N1 �
¯Li. (27.3)
N1
i=1
i=1
An inspection of Fig. 27.3 indicates that LPC significantly overestimates the
power output exp, in particular since in the region of the plateau most
data points have to be shifted by a relatively large distance from below onto
the power curve. As a matter of fact, due to safety reasons, power output
is kept limited near the rated power. Also in the large time interval of 24 h
our power curve average overestimates exp, i.e., by 17%. In comparison
with this, the Tjareborg power curve, available in the world wide Web [4],
overestimates the same 24-h data by about 8%.
One reason for this difference may lie in the fact that our 24 h data base for
establishing the power curve is rather small. However, in both cases neglection
of the finite response time causes systematic errors.
27.3 Relaxation Model
In order to include the delayed reponse of the WEC to power prediction, we
recently proposed the following relaxation model [1]:
d
dt L(t) =r(t)[LPC(U(t)) − L(t)] ; r(t) > 0, (27.4)
where LPC denotes the power curve and U(t), L(t) the instantaneous wind
speed and power, respectively. Because the relaxation function r(t) is positive,
the above model exhibits the attraction property of the power curve. In
principle, the model could be nonlinearly extended by adding uneven powers
of LPC(U(t)) − L(t) with positive coefficients to preserve attraction.
In the simplest case, we may choose r(t) =r0 = const. Defining the mean
power as usual by the time average one finds that
=< LPC(U(t)) >
� � ��
1
1+O . (27.5)
r0T
Thus, the average based on the power curve predicts the true mean-power
output, provided the averaging time T is much larger than the relaxation
time τ := 1/r0. To see this, one integrates (27.4) from time t =0toT :