This page describes the models used for the transmission lines and constant impedance loads in VFlexP. Transmission lines are modelled with the pi-model and the inductive and capacitive parts are separated, including the capacitive part in the parallel RC model of each bus. The different models used for dynamic and static consideration of inductive and capacitive elements in d-q reference frame in VFlexP are described in the following lines:

Recalling the equations of an inductive element in d-q reference frame, in per unit variables: \begin{equation} \left\{ \begin{array}{l} v_{d} = \displaystyle \frac{1}{\omega_{base}} L \displaystyle \frac{di_{d}}{dt} - \omega L i_{q} + Ri_{d} \\ \ \\ v_{q} = \displaystyle \frac{1}{\omega_{base}} L \displaystyle \frac{di_{q}}{dt} + \omega L i_{d} + Ri_{q} \end{array} \right. \end{equation} Considering static behavior of the inductive element in a reduced model: \begin{equation} \left\{ \begin{array}{l} v_{d} = - \omega L i_{q} + Ri_{d} \\ v_{q} = \omega L i_{d} + Ri_{q} \end{array} \right. \rightarrow \begin{bmatrix} v_d \\ v_q \end{bmatrix} = \begin{bmatrix} R & -\omega L \\ \omega L & R \end{bmatrix} \begin{bmatrix} i_d \\ i_q \end{bmatrix} \end{equation} Solving the system for the current: \begin{equation} \begin{bmatrix} i_d \\ i_q \end{bmatrix} = \begin{bmatrix} R & -\omega L \\ \omega L & R \end{bmatrix} ^{-1} \begin{bmatrix} v_d \\ v_q \end{bmatrix} \end{equation} The matrix of a static inductive element remains: \begin{equation} \begin{bmatrix} R & -\omega L \\ \omega L & R \end{bmatrix} ^{-1} = \frac{1}{R^2+(\omega L)^2} \begin{bmatrix} R & \omega L \\ -\omega L & R \end{bmatrix} \end{equation} Recalling the equations of a capacitive element in d-q reference frame, in per unit variables: \begin{equation} \left\{ \begin{array}{l} i_{d} = \displaystyle \frac{1}{\omega_{base}} C \displaystyle \frac{dv_{d}}{dt} - \omega C v_{q} + \frac{1}{R}v_{d} \\ \ \\ i_{q} = \displaystyle \frac{1}{\omega_{base}} C \displaystyle \frac{dv_{q}}{dt} + \omega C v_{d} + \frac{1}{R}v_{q} \end{array} \right. \end{equation} Considering static behavior of the capacitive element in a reduced model: \begin{equation} \left\{ \begin{array}{l} i_{d} = - \omega C v_{q} + \displaystyle \frac{1}{R}v_{d} \\ \ \\ i_{q} = \omega C v_{d} + \displaystyle \frac{1}{R}v_{q} \end{array} \right. \rightarrow \begin{bmatrix} i_d \\ i_q \end{bmatrix} = \begin{bmatrix} 1/R & -\omega C \\ \omega C & 1/R \end{bmatrix} \begin{bmatrix} v_d \\ v_q \end{bmatrix} \end{equation} Solving the system for the voltage: \begin{equation} \begin{bmatrix} v_d \\ v_q \end{bmatrix} = \begin{bmatrix} 1/R & -\omega C \\ \omega C & 1/R \end{bmatrix} ^{-1} \begin{bmatrix} i_d \\ i_q \end{bmatrix} \end{equation} The matrix of a static capacitive element remains: \begin{equation} \begin{bmatrix} 1/R & -\omega C \\ \omega C & 1/R \end{bmatrix} ^{-1} = \frac{1}{\left(1/R\right)^2+(\omega C)^2} \begin{bmatrix} 1/R & \omega C \\ -\omega C & 1/R \end{bmatrix} \end{equation}