GATE 2024 EE Q.56: solution and answer

Official answer: 0.10 to 0.12Source: Official answer key, GATE 2024 EE (IISc Bengaluru)

1. Work in units of $P_{max}$. Mechanical input $P_m=0.5$. Before the fault $\sin {\delta }_0=0.5$, so ${\delta }_0=\pi /6$.

2. After the fault is cleared (line 2 open) $P_e=\frac{1}{\sqrt{2}}\sin \delta$. The rotor swings forward until $P_e=P_m$ again on the falling side of the curve: $\frac{1}{\sqrt{2}}\sin {\delta }_{max}=0.5\Rightarrow \sin {\delta }_{max}=\frac{1}{\sqrt{2}}$, so ${\delta }_{max}=\pi -\frac{\pi }{4}=\frac{3\pi }{4}$.

3. At the critical clearing angle ${\delta }_c=\pi /2$ the rotor just reaches ${\delta }_{max}$, so the equal area criterion gives $A_{acc}=A_{dec}$. The during-fault curve is not given, so $A_{acc}$ is found from $A_{dec}$.

4. Decelerating area: $$A_{dec}={\int }_{\pi /2}^{3\pi /4}\left(\frac{1}{\sqrt{2}}\sin \delta -0.5\right)d\delta$$

5. Evaluate: $\frac{1}{\sqrt{2}}{\left[-\cos \delta \right]}_{\pi /2}^{3\pi /4}=\frac{1}{\sqrt{2}}\left(0+\frac{1}{\sqrt{2}}\right)=0.5$, and $0.5\left(\frac{3\pi }{4}-\frac{\pi }{2}\right)=\frac{\pi }{8}=0.3927$.

6. $A_{dec}=0.5-0.3927=0.1073$, hence $A_{acc}=A_{dec}\approx 0.11P_{max}$.