Line opened: stability and rotor swing

Data: $E' = 1.2$ pu, $V = 1.0$ pu, $X_d' = 0.2$ pu, $X_1 = 0.4$ pu, $P_m = 1.5$ pu.

Assumption: the lower line's reactance is not marked, so it is taken equal to the upper line ($0.4$ pu). Opening breaker A disconnects the lower line.

Step 1: Before and after opening

$$\begin{aligned} \text{Before: } X_I &= 0.2 + \frac{0.4}{2} = 0.4,\quad P_{max1} = \frac{1.2}{0.4} = 3.0\ \text{pu},\quad \delta_0 = \sin^{-1}\frac{1.5}{3.0} = 30^\circ \\ \text{After: } X_{II} &= 0.2 + 0.4 = 0.6,\quad P_{max2} = \frac{1.2}{0.6} = 2.0\ \text{pu} \end{aligned}$$

New equilibrium: $\delta_1 = \sin^{-1}(1.5/2.0) = 48.59^\circ$. Limit: $\delta_{max} = 180^\circ - 48.59^\circ = 131.41^\circ$.

Step 2: (i) Stability check

Accelerating area (from $30^\circ$ to $48.59^\circ$):

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

Maximum decelerating area (from $48.59^\circ$ to $131.41^\circ$):

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

Since $A_{2,max} > A_1$, the system remains stable.

Step 3: (ii) Maximum rotor swing $\delta_2$

From $A_1 = A_2$, the net area from $\delta_0$ to $\delta_2$ is zero:

$$\begin{aligned} \int_{\delta_0}^{\delta_2}(P_m - 2.0\sin\delta)\,d\delta &= 0 \\ 1.5(\delta_2 - 0.5236) - 2.0(0.8660 - \cos\delta_2) &= 0 \end{aligned}$$

Solving by trial: $\delta_2 = 1.2196$ rad.

Answer: (i) stable; (ii) the rotor swings to about 69.9° and then oscillates about the new equilibrium of 48.6°.