Stability after tripping one circuit

Data: $E = 1.2$ pu, $V = 1$ pu, $X_s = 0.2$ pu, each line $0.4$ pu, $P_m = 1.5$ pu.

Step 1: Power-angle curves

$$\begin{aligned} \text{Both lines: } X_I &= 0.2 + \frac{0.4}{2} = 0.4\ \text{pu}, \quad P_{max1} = \frac{1.2\times1}{0.4} = 3.0\ \text{pu} \\ \text{One line: } X_{II} &= 0.2 + 0.4 = 0.6\ \text{pu}, \quad P_{max2} = \frac{1.2}{0.6} = 2.0\ \text{pu} \end{aligned}$$

Step 2: Angles

$$\begin{aligned} \delta_0 &= \sin^{-1}\frac{1.5}{3.0} = 30^\circ = 0.5236\ \text{rad} \\ \delta_1 &= \sin^{-1}\frac{1.5}{2.0} = 48.59^\circ = 0.8481\ \text{rad}\ \ \text{(new equilibrium)} \\ \delta_{max} &= 180^\circ - 48.59^\circ = 131.41^\circ = 2.2935\ \text{rad} \end{aligned}$$

When the line trips, $P_e$ drops to $2.0\sin30^\circ = 1.0 < P_m$, so the rotor accelerates.

Step 3: Equal areas

Accelerating area, from $\delta_0$ to $\delta_1$:

$$\begin{aligned} A_1 &= P_m(\delta_1-\delta_0) - P_{max2}(\cos\delta_0-\cos\delta_1) \\ &= 1.5(0.8481-0.5236) - 2.0(0.8660-0.6614) \\ &= 0.4868 - 0.4092 = 0.0775\ \text{pu·rad} \end{aligned}$$

Maximum decelerating area, from $\delta_1$ to $\delta_{max}$:

$$\begin{aligned} A_{2,max} &= P_{max2}(\cos\delta_1-\cos\delta_{max}) - P_m(\delta_{max}-\delta_1) \\ &= 2.0(0.6614+0.6614) - 1.5(2.2935-0.8481) \\ &= 2.6458 - 2.1681 = 0.4775\ \text{pu·rad} \end{aligned}$$

 Pe
 3.0 |      ___ before (3 sin d)
 2.0 |    /  _ \__ after (2 sin d)
 1.5 |---/-*-----\----\---- Pm
     |  /A1| A2   \    \
     +-30--48.6--69.9--131.4 delta

Step 4: Result

$A_{2,max} = 0.4775 > A_1 = 0.0775$, so the system is stable and keeps synchronism.

Equating $A_1 = A_2$, that is $1.5(\delta_2-0.5236) = 2.0(0.8660-\cos\delta_2)$, gives a maximum swing of $\delta_2 = 69.88^\circ$. The rotor then settles at $48.59^\circ$.

Answer: Stable; the rotor swings to about 69.9° and settles at 48.6°.