Critical clearing angle, Pmax 1.5, 0.5, 1.2

Data: $P_m = 0.9$ pu, $P_{max1} = 1.5$ pu (pre-fault), $P_{max2} = 0.5$ pu (during fault), $P_{max3} = 1.2$ pu (post-fault). The line is purely reactive, so there are no losses.

Step 1: Angles

$$\begin{aligned} \delta_0 &= \sin^{-1}\frac{0.9}{1.5} = 36.87^\circ = 0.6435\ \text{rad} \\ \delta_{max} &= 180^\circ - \sin^{-1}\frac{0.9}{1.2} = 180^\circ - 48.59^\circ = 131.41^\circ = 2.2935\ \text{rad} \end{aligned}$$

Step 2: Equal area criterion

$$\int_{\delta_0}^{\delta_{cr}}(P_m - P_{max2}\sin\delta)\,d\delta = \int_{\delta_{cr}}^{\delta_{max}}(P_{max3}\sin\delta - P_m)\,d\delta$$ $$\Rightarrow\ \cos\delta_{cr} = \frac{P_m(\delta_{max}-\delta_0) + P_{max3}\cos\delta_{max} - P_{max2}\cos\delta_0}{P_{max3}-P_{max2}}$$

Step 3: Substitute

$$\begin{aligned} \cos\delta_{cr} &= \frac{0.9(2.2935-0.6435) + 1.2\cos131.41^\circ - 0.5\cos36.87^\circ}{1.2-0.5} \\ &= \frac{1.4850 - 0.7937 - 0.4000}{0.7} \\ &= 0.4161 \\ \delta_{cr} &= 65.41^\circ \end{aligned}$$

 Pe
 1.5 |     ___ I (pre)
 1.2 |   /  _ \__ III (post)
 0.9 |--*--|----\--------- Pm
 0.5 | / A1| A2  \  II (fault)
     +--d0-dcr----dmax---- delta

Answer: Critical clearing angle $\delta_{cr} \approx 65.4^\circ$.