Forecasting the energy produced by a windmill on a yearly basis

Abstract

The objective of this article is to study as extensively as possible the uncertainties affecting the annual energy produced by a windmill. In the literature, the general approach is to estimate the mean annual energy from a transformation of a Weibull distribution law. Then the issue is reduced to estimating the coefficients of this distribution. This is obtained by classical statistical methods. Therefore, the uncertainties are mostly limited to those resulting from the statistical procedures. But in fact, the real uncertainty of the random variable which represents the annual energy cannot been reduced to the uncertainty on its mean and to the uncertainties induced from the estimation procedure. We propose here a model, which takes advantage of the fact that the annual energy production is the sum of many random variables representing the 10 min energy production during the year. Under some assumptions, we make use of the central limit theorem and show that an intrinsic uncertainties of wind power, usually not considered, carries an important risk. We also explain an observation coming from practice that the forecasted annual production is always overestimated, which creates a risk of reducing the profitability of the operation.

Appendices

Proof of CLT for annual wind power production (Proposition 3.1)

Let us refer to (Doukhan 1994, Dedecker 2007) for a proof of point ii). Stationarity implies that ${\mathbb{E} \left(P_t \right)=\mathcal{P}}$ for all index t, next Formula (3.3) after having checked that

$$ \hbox{var} \left(\sum_{i=1}^T P_t\right) = T\times {\mathcal{V}}\times \Upgamma_T^2. $$

This is provided by the following calculation

$$ \begin{aligned} \hbox{var} \left(\sum_{i=1}^T P_t\right)& ={\mathbb{E}}\left(\sum_{t,s=1}^T cov(P_t, P_s) \right) \quad = \quad \sum_{i,j=1}^T\left[\rho_P(i-j)\times {\mathcal{V}}\right] \\ &={\mathcal{V}}\times \left\{T\rho_P(0) + 2\sum_{i=1}^T \rho_P(k) (T-k) \right\} \\ &=T\, {\mathcal{V}}\times \left\{1 + 2\sum_{i=1}^{T-1} \rho_P(k) (1-k/T) \right\} \end{aligned} $$

since ρ _P (0) = 1. This completes the proof of point i) and then the proof of Proposition 3.1.

Proof of error bound on mean 10 min power (Proposition 5.1)

By Taylor formula

$$ \begin{aligned} f (\widetilde{V}_t)&= &f\left(V_t + \sigma(V_t)\cdot\eta_t\right)\\ &= & f(V_t)+ \sum_{p=1}^{3} \left(\frac {1}{p!}\,f^{(p)} (V_t )\cdot \sigma(V_t)^p \cdot\eta_t^p\right) + \frac {1}{4!} M_4 \cdot \sigma(V_t)^4 \cdot\eta_t^4 \end{aligned} $$

where M ₄ is a r.v. bounded by $\|f^{(4)}\|_{L^\infty}.$ Next, by using successively the independency of η _t and V _t and that η _t is a zero mean Gaussian r.v. with variance 1, we can deduce that

$$ \begin{aligned}{\mathbb{E}}\left[f\left(\widetilde{V}_t\right)\right] &={\mathbb{E}}\left[f\left(V_{t} \right)\right]+ \sum_{p=1}^{3} \left(\frac {1} {p!}\,{\mathbb{E}}\left[f^{(p)}\left(V_{t}\right)\cdot \sigma(V_{t})^p \cdot\eta_{t}^p\right]\right) + \frac {1} {4!} {\mathbb{E}}\left[ M_{4} \cdot \sigma(V_{t})^4 \cdot\eta_{t}^4\right]\\ &={\mathbb{E}}\left[f\left(V_{t}\right)\right] + \sum_{p=1}^{3} \left(\frac {1} {p!}\,{\mathbb{E}}\left[f^{(p)}\left(V_{t}\right)\cdot \sigma(V_{t})^p\right] \cdot{\mathbb{E}}\left[ \eta_{t}^p\right]\right) + \frac {1} {4!} {\mathbb{E}}\left[ M_{4} \cdot \sigma(V_{t})^4 \cdot\eta_{t}^4\right]\\ &={\mathbb{E}}\left[f\left(V_{t}\right)\right] + \left(\frac {1} {2!}\,{\mathbb{E}}\left[f^{(2)}\left(V_{t}\right)\cdot \sigma(V_{t})^2\right] \cdot{\mathbb{E}}\left[ \eta_{t}^2\right]\right) + \frac {1}{4!} {\mathbb{E}}\left[ M_{4} \cdot \sigma(V_{t})^4 \cdot\eta_{t}^4\right]\\ &={\mathbb{E}}\left[f\left(V_{t}\right)\right] +\, \frac{1}{2}\,{\mathbb{E}}\left[f^{\prime\prime}\left(V_{t}\right)\cdot \sigma(V_{t})^2 \right] + \frac{1}{24}\, {\mathbb{E}}\left[M_{4} \cdot \sigma(V_{t})^4 \cdot\eta_{t}^4\right]. \end{aligned} $$

Stress that the r.v. M ₄ is not independent of η _t . However, since M ₄ is bounded, we can use the Cauchy-Schwarz inequality and deduce that

$$ \begin{aligned} {\mathbb{E}}\left[M_4 \cdot \sigma(V_t)^4 \cdot\eta_t^4\right] &\le& \|f^{(4)}\|_{L^\infty}\cdot {\mathbb{E}}\left[\sigma(V_t)^8\right]^{1/2} \cdot{\mathbb{E}}\left[\eta_t^8\right]^{1/2}\\ &=& \sqrt{105}\cdot \|f^{(4)}\|_{L^\infty}\cdot {\mathbb{E}}\left[\sigma(V_t)^8\right]^{1/2}. \end{aligned}$$

This completes the proof of Proposition 5.1.$\square$

Estimation of the parameters of Weibull pdf

First, we estimate the probability of zero wind p ₀ by the empirical frequency of zero wind $\widehat{p}_0,$ that is

$$ \widehat{p}_0= Card\{i, V_i=0\}/N, $$

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where Card  I denote the cardinal of the index set I and N denote the size of dataset. In the example of Fig. 1, we find $\widehat{p}_0=0.0285$.

Secondly, we estimate the two parameters λ and k of the Weibull pdf for positive wind speed. As pointed out by (Ramírez 2005, Ramírez 2006), the MLE is the more efficient wether the moment method is the more simple to describe.

3.1 Moment method for estimating Weibull pdf

Indeed, if V follows a Weibull law, its moments are given by

$$ M_i= {\mathbb{E}}( V^i)= \lambda^i \,\Upgamma(1+i/k). $$

Then by considering the first and third moment we get the equation satisfied by the moment estimators

$$ \begin{aligned} \widehat{\lambda}\,\Upgamma(1+1/\widehat{k})&= & \overline{M}_1:= \frac 1 n \sum_{i=1}^n V_i,\\ \widehat{\lambda}^3\,\Upgamma(1+3/\widehat{k})&= &\overline{M}_3:= \frac 1 n \sum_{i=1}^n V_i^3 \end{aligned} $$

which implies $\frac{\Upgamma(1+3/\widehat{k})}{\Upgamma(1+1/\widehat{k}) ^3}= \frac{\overline{M}_3}{\overline{M}_1^3}$ and after $\widehat{\lambda}=\frac{\Upgamma(1+1/\widehat{k})}{\overline{M}_1}.$ The numerical solution of the first equation is easy since the function $x\mapsto \Upgamma(1+3/x)/\Upgamma(1+1/x)^3$ is decreasing.

3.2 MLE for estimating Weibull pdf

On the other hand, following Seguro and Lambert (2000), Ramírez and Carta (2006), Carta et al. (2009), MLE of the parameter $\widehat{k}$ is furnished by the implicit solution of the equation

$$ \widehat{k}= \left[ \frac{\sum_{i=1}^n V_i^k \ln V_i }{\sum_{i=1}^n V_i^k }- \frac{1}{n} \sum_{i=1}^n \ln V_i \right]^{-1} $$

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and after $\widehat{\lambda}$ is given by

$$ \widehat{\lambda}= \left[\frac{1}{n} \sum_{i=1}^n V_i^k \right]^{1/k}. $$

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Moreover, we can use the value of k provided by the moment method as the initial value for the iterative Newton algorithm.