Critical clearing angle, Pm = 1.2 pu

Data: $E = 1.1$, $V = 1$, $X_d' = 0.15$, $X_1 = X_2 = 0.4$ pu, $P_m = 1.2$ pu. The fault is on line 2 close to the generator and is cleared by opening line 2.

Step 1: Pre-fault (both lines in service)

$$\begin{aligned} X_I &= 0.15 + \frac{0.4}{2} = 0.35\ \text{pu} \\ P_{max1} &= \frac{1.1\times 1}{0.35} = 3.143\ \text{pu} \\ \delta_0 &= \sin^{-1}\frac{1.2}{3.143} = 22.45^\circ = 0.3918\ \text{rad} \end{aligned}$$

Step 2: During the fault

The fault is at the generator end of line 2, which is the generator-side bus. That bus is at zero voltage, so no power reaches the infinite bus: $P_{max2} = 0$.

Step 3: Post-fault (line 2 open)

$$\begin{aligned} X_{III} &= 0.15 + 0.4 = 0.55\ \text{pu} \\ P_{max3} &= \frac{1.1}{0.55} = 2.0\ \text{pu} \\ \delta_{max} &= 180^\circ - \sin^{-1}\frac{1.2}{2.0} = 180^\circ - 36.87^\circ = 143.13^\circ = 2.4981\ \text{rad} \end{aligned}$$

Step 4: Critical clearing angle

From $A_1 = A_2$:

$$\cos\delta_{cr} = \frac{P_m(\delta_{max}-\delta_0) + P_{max3}\cos\delta_{max} - P_{max2}\cos\delta_0}{P_{max3}-P_{max2}}$$ $$\begin{aligned} \cos\delta_{cr} &= \frac{1.2(2.4981-0.3918) + 2.0\cos 143.13^\circ - 0}{2.0 - 0} \\ &= \frac{2.5276 - 1.6000}{2.0} = 0.4638 \\ \delta_{cr} &= 62.37^\circ \end{aligned}$$

Answer: Critical clearing angle $\delta_{cr} \approx 62.37^\circ$.