Critical clearing time, H = 8 MJ/MVA

Assumption: the internal emf magnitude is not given in the figure, so $|E| = 1.0$ pu and $f = 50$ Hz are assumed. The method is the same for any other $E$.

Data: $X_d' = 0.3$, $X_T = 0.2$, $L_1 = L_2 = 0.3$ pu, $P_m = 1$ pu, $H = 8$ MJ/MVA, $V = 1\angle0^\circ$.

Step 1: Pre-fault

$$\begin{aligned} X_I &= 0.3 + 0.2 + \frac{0.3}{2} = 0.65\ \text{pu} \\ P_{max1} &= \frac{1\times1}{0.65} = 1.5385\ \text{pu} \\ \delta_0 &= \sin^{-1}\frac{1}{1.5385} = 40.54^\circ = 0.7076\ \text{rad} \end{aligned}$$

Step 2: During the fault

The fault is at the end of line 1, either at bus 1 or at the infinite bus 2. In both cases one of the buses on the power path is at zero voltage, so $P_{max2} = 0$ and the power transferred is zero.

Step 3: Post-fault (line 1 removed)

$$\begin{aligned} X_{III} &= 0.3 + 0.2 + 0.3 = 0.8\ \text{pu} \\ P_{max3} &= \frac{1}{0.8} = 1.25\ \text{pu} \\ \delta_{max} &= 180^\circ - \sin^{-1}\frac{1}{1.25} = 126.87^\circ = 2.2143\ \text{rad} \end{aligned}$$

Since $P_{max3} > P_m$, the system can be stable if the fault is cleared in time.

Step 4: Critical clearing angle

$$\begin{aligned} \cos\delta_{cr} &= \frac{P_m(\delta_{max}-\delta_0) + P_{max3}\cos\delta_{max}}{P_{max3}} \\ &= \frac{1(2.2143-0.7076) + 1.25\cos 126.87^\circ}{1.25} \\ &= \frac{1.5067 - 0.75}{1.25} = 0.6054 \\ \delta_{cr} &= 52.74^\circ = 0.9206\ \text{rad} \end{aligned}$$

Step 5: Critical clearing time

With $P_e = 0$ during the fault, $\delta = \delta_0 + \dfrac{\pi f P_m}{2H}t^2$. So

$$\begin{aligned} t_{cr} &= \sqrt{\frac{2H(\delta_{cr}-\delta_0)}{\pi f P_m}} \\ &= \sqrt{\frac{2\times 8\times(0.9206-0.7076)}{\pi\times 50\times 1}} \\ &= 0.147\ \text{s} \end{aligned}$$

Answer (with $E = 1.0$ pu): $\delta_{cr} = 52.74^\circ$, critical clearing time $t_{cr} \approx 0.147$ s (about 7.4 cycles).