The single line diagram of a lossless system is shown in the figure. The system is operating in steady-state at a stable equilibrium point with the power output of the generator being P_maxsin⁡δ, where δ is the load angle and the mechanical power input is 0.5P_max. A fault occurs on line 2 such that the power output of the generator is less than 0.5P_max during the fault. After the fault is cleared by opening line 2, the power output of the generator is $\left\{P_{max}/\sqrt{2}\right\}\sin \delta$. If the critical fault clearing angle is $\pi /2$ radians, the accelerating area on the power angle curve is __________ times P_max (rounded off to decimal places).

The correct answer is 0.10.

Step-by-step derivation:
1. Identify the System Parameters:
From the single line diagram and the problem statement, we have:
- Pre-fault electrical power output: $P_{e1}=P_{\text{max}}\sin \delta$
- Mechanical power input: $P_m=0.5P_{\text{max}}$
- Post-fault electrical power output (after line 2 is opened to clear the fault): $P_{e3}=\frac{P_{\text{max}}}{\sqrt{2}}\sin \delta$
- Critical clearing angle: ${\delta }_{cr}=\frac{\pi }{2}\text{radians}$

2. Determine the Initial Load Angle (${\delta }_0$):
In steady-state pre-fault condition, the mechanical input power equals the electrical power output:
$P_m=P_{e1}\left({\delta }_0\right)$
$0.5P_{\text{max}}=P_{\text{max}}\sin {\delta }_0$
$\sin {\delta }_0=0.5\Rightarrow {\delta }_0=\frac{\pi }{6}\text{radians}$

3. Determine the Maximum Rotor Angle (${\delta }_{\text{max}}$):
The maximum limit of the rotor angle for transient stability occurs where the mechanical power line intersects the post-fault power curve on the right-hand side:
$P_m=P_{e3}\left({\delta }_{\text{max}}\right)$
$0.5P_{\text{max}}=\frac{P_{\text{max}}}{\sqrt{2}}\sin {\delta }_{\text{max}}$
$\sin {\delta }_{\text{max}}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$
Since ${\delta }_{\text{max}}$ is in the second quadrant:
${\delta }_{\text{max}}=\pi -\frac{\pi }{4}=\frac{3\pi }{4}\text{radians}$

4. Find the Accelerating Area ($A_1$) Using the Equal Area Criterion:
According to the Equal Area Criterion, at the critical fault clearing angle, the accelerating area ($A_1$) is equal to the decelerating area ($A_2$):
$A_1=A_2$
The decelerating area ($A_2$) lies between the critical clearing angle ${\delta }_{cr}$ and the maximum angle ${\delta }_{\text{max}}$ under the post-fault curve:
$A_2={\int }_{{\delta }_{cr}}^{{\delta }_{\text{max}}}(P_{e3}-P_m)d\delta$
Substitute the values into the integration:
$A_2={\int }_{\pi /2}^{3\pi /4}(\frac{P_{\text{max}}}{\sqrt{2}}\sin \delta -0.5P_{\text{max}})d\delta$
$A_2=P_{\text{max}}{[-\frac{1}{\sqrt{2}}\cos \delta -0.5\delta ]}_{\pi /2}^{3\pi /4}$
Evaluate at the upper limit ($\delta =\frac{3\pi }{4}$):
$-\frac{1}{\sqrt{2}}\cos \left(\frac{3\pi }{4}\right)-0.5\left(\frac{3\pi }{4}\right)=-\frac{1}{\sqrt{2}}(-\frac{1}{\sqrt{2}})-\frac{3\pi }{8}=0.5-\frac{3\pi }{8}$
Evaluate at the lower limit ($\delta =\frac{\pi }{2}$):
$-\frac{1}{\sqrt{2}}\cos \left(\frac{\pi }{2}\right)-0.5\left(\frac{\pi }{2}\right)=0-\frac{\pi }{4}$
Subtract the lower limit evaluation from the upper limit evaluation:
$A_2=P_{\text{max}}\left[\right(0.5-\frac{3\pi }{8})-(-\frac{\pi }{4}\left)\right]$
$A_2=P_{\text{max}}[0.5-\frac{3\pi }{8}+\frac{2\pi }{8}]=P_{\text{max}}[0.5-\frac{\pi }{8}]$
Using $\pi \approx 3.14159$:
$A_2\approx P_{\text{max}}[0.5-0.3927]\approx 0.1073P_{\text{max}}$
Therefore, the accelerating area $A_1=A_2\approx 0.10$ times $P_{\text{max}}$ (rounded off to two decimal places).