Question 7.2: A DFIG is operating in the generator mode at a supersynchron......

\begin{aligned} \omega_{\text {slip }} & =\omega_{d A}=- \\ s & =\frac{\omega_{s l i p}}{\omega_{s y n}}=- \\ T_{e m} & =- \\ P_s & =- \\ P_s & =\nu _{s q} i_{s q}=\omega_d \lambda_{s d} i_{s q} \\ i_{s q} & =- \\ Q_s & =\nu _{s q} i_{s d}=\omega_d \lambda_{s d} i_{s d}=\underbrace{-}_{\text {given }} \\ i_{s d} & =- \\ Q_s & =Q_{\text {mag }}-\frac{Q_r}{s}=-\therefore Q_r=-\text { such that } \frac{Q_r}{s}>Q_{\text {mag }} \\ \omega_{d A} & =\omega_{ syn }-\omega_m=- \\ i_{r q} & =+\left(\text { since } i_{s q}=-\right) \\ P_r & =\nu _{r q} i_{r q}=\omega_{d A} \lambda_{r d} i_{r q}=- \\ i_{s d} & =i_{m d}-i_{r d}=- \\ i_{r d} & =+\text { such that } i_{r d}>i_{m d} \\ Q_r & =\nu _{r q} i_{r d}=\omega_{d A} \lambda_{r d} i_{r d}=- \\ \nu _{r d} & =0 \\ \nu _{r q} & =\omega_{d A} \lambda_{r d}=- \end{aligned}

\begin{bmatrix} \nu _A(t) \\ \nu _B(t) \\ \nu _C(t) \end{bmatrix}=\sqrt{\frac{2}{3}}\begin{bmatrix} \cos \left(\theta_{d A}\right) & -\sin \left(\theta_{d A}\right) \\ \cos \left(\theta_{d A}+\frac{4 \pi}{3}\right) & -\sin \left(\theta_{d A}+\frac{4 \pi}{3}\right) \\ \cos \left(\theta_{d A}+\frac{2 \pi}{3}\right) & -\sin \left(\theta_{d A}+\frac{2 \pi}{3}\right) \end{bmatrix}\begin{bmatrix} \nu _{r d} \\ \nu _{r q} \end{bmatrix}

Various space vectors are shown in Fig. 7-5.