EAC from given power-angle curves

Data from the curves: $P_m = 1.0$ pu, pre-fault $P_{max1} = 1.8$ pu, during fault $P_{max2} = 0.4$ pu, post-fault $P_{max3} = 1.3$ pu.

Step 1: Can the system be stable?

$$\begin{aligned} \delta_0 &= \sin^{-1}\frac{1.0}{1.8} = 33.75^\circ = 0.5890\ \text{rad} \\ \delta_{max} &= 180^\circ - \sin^{-1}\frac{1.0}{1.3} = 180^\circ - 50.28^\circ = 129.72^\circ = 2.2640\ \text{rad} \end{aligned}$$

During the fault, $P_{max2} = 0.4 < P_m$, so the rotor accelerates. After clearing, $P_{max3} = 1.3 > P_m$, so a decelerating area exists. The system is stable if the fault is cleared before the critical clearing angle, where $A_1 = A_2$.

Step 2: Critical clearing angle

Equating the accelerating area $A_1$ (on curve II, from $\delta_0$ to $\delta_{cr}$) with the decelerating area $A_2$ (on curve III, from $\delta_{cr}$ to $\delta_{max}$):

$$\cos\delta_{cr} = \frac{P_m(\delta_{max}-\delta_0) + P_{max3}\cos\delta_{max} - P_{max2}\cos\delta_0}{P_{max3}-P_{max2}}$$ $$\begin{aligned} \cos\delta_{cr} &= \frac{1.0(2.2640-0.5890) + 1.3\cos129.72^\circ - 0.4\cos33.75^\circ}{1.3-0.4} \\ &= \frac{1.6750 - 0.8307 - 0.3326}{0.9} \\ &= 0.5685 \\ \delta_{cr} &= 55.35^\circ \end{aligned}$$

 Pe
 1.8 |      ____ I
 1.3 |    /  __ \__ III
 1.0 |---*--|----\---\---- Pm
 0.4 |  / A1| A2  \   \ II
     +--d0--dcr---dmax---- delta
       33.75 55.35 129.72

Answer: The system is stable if the fault is cleared at an angle below the critical clearing angle $\delta_{cr} \approx 55.35^\circ$; if clearing is later, it loses synchronism.