Applicability of Error Limit in Forecasting and Scheduling of Wind and Solar Power in India

Abstract

Forecasting of power generation is an essential requirement for high penetration of variable renewable energy in existing grid system as the major purpose of forecasting is to reduce the uncertainty of renewable generation, so that its variability can be more precisely accommodated. This paper focuses on the statistical behaviour of error in solar and wind power forecasting considering Indian regulations and analyses the applicability of the error limit in calculating the energy accuracy of forecasting and the stability of the grid.

Appendix

$$e(i) = \frac{1}{Avc}\left| {x_{A} (i) - x_{S} (i)} \right| = \frac{1}{Avc}\left| {\left( {x_{A} (i) - \overline{{x_{A} }} } \right) - \left( {x_{S} (i) - \overline{{x_{S} }} } \right) + \left( {\overline{{x_{A} }} - \overline{{x_{S} }} } \right)} \right|$$

(A.1)

For a good forecasting system, we can consider $\overline{{x_{S} }} \to \overline{{x_{A} }}$ and $\sigma_{A} \to \sigma_{S}$. Hence, using the no-penalty band in Table 1, for some n we can state that

$$\begin{aligned} m^{2} & \ge \frac{1}{n}\sum\limits_{i = 1}^{n} {e^{2} (i) = \left( {\frac{1}{AvC}} \right)^{2} \frac{1}{n}} \sum\limits_{i = 1}^{n} {\left( {(x_{A} (i) - \overline{{x_{A} }} ) - (x_{S} (i) - \overline{{x_{S} }} )} \right)^{2} } \\ & = \left( {\frac{1}{AvC}} \right)^{2} \frac{1}{n}\sum\limits_{i = 1}^{n} {\left[ {\left( {x_{A} (i) - \overline{{x_{A} }} } \right)^{2} + \left( {x_{S} (i) - \overline{{x_{S} }} } \right)^{2} - 2\left( {x_{A} (i) - \overline{{x_{A} }} } \right)\left( {x_{S} (i) - \overline{{x_{S} }} } \right)} \right]} \\ \end{aligned}$$

(A.2)

$= \sigma_{A}^{2} + \sigma_{S}^{2} - 2r\sigma_{A} \sigma_{S} = 2(1 - r)\sigma_{A}^{2} ,$ which implies (3).

Using (4) and (6), we get

$$C \approx k\int\limits_{m}^{\infty } {\lambda e^{2} \exp ( - \lambda e){\text{d}}e}$$

Integrating, we get

$$C \approx k\left[ {\left( {m + \frac{1}{\lambda }} \right)^{2} + \left( {\frac{1}{\lambda }} \right)^{2} } \right]\exp ( - \lambda m)$$

Using (8), $P(m) = 1 - \exp ( - \lambda m)$ and for exponential distribution $\sigma_{e} = 1/\lambda .$ Hence, replacing $\exp ( - \lambda m)$ and λ, we get (7).