GATE Papers >> EEE >> 2017 >> Question No 55

With one transmission line removal, the effective rectance increases so P_max decreases, the power angle by equal area criterion

$\int\limits_{{\mathrm\delta}_0}^{{\mathrm\delta}_1}$ (1 – P_max sin $\delta$)d$\delta$ = $\int\limits_{{\mathrm\delta}_1}^{{\mathrm\delta}_2}$ (P_max sin$\delta$ – 1) d$\delta$

$\begin{array}{l}\left({\mathrm\delta}_1-{\mathrm\delta}_0\right)-{\mathrm P}_\max\left({\mathrm{cosδ}}_0-{\mathrm{cosδ}}_1\right)={\mathrm P}_\max({\mathrm{cosδ}}_1-{\mathrm{cosδ}}_2)-\left({\mathrm\delta}_2-{\mathrm\delta}_1\right)\\\left({\mathrm\delta}_2-{\mathrm\delta}_0\right)={\mathrm P}_\max\left({\mathrm{cosδ}}_0-{\mathrm{cosδ}}_2\right)\\{\mathrm P}_\max=\frac{{\mathrm\delta}_2-{\mathrm\delta}_0}{\left({\mathrm{cosδ}}_0-{\mathrm{cosδ}}_2\right)}\\{\mathrm\delta}_0=\sin^{-1}\frac1{1.5}=0.72972\\{\mathrm P}_\max=\frac{\left(1.221-0.72972\right)}{\cos\left(0.7297\right)-\cos\left(1.221\right)}=1.22\mathrm{pu}\end{array}$