Fault cleared at 45°: stability and maximum swing

Data: $E = 1.2$ pu, $V = 1$ pu, $X_d' = 0.25$, two lines of $0.5$ pu each, $\delta_0 = 20^\circ$.

Assumption: the bus fault is temporary, so after clearing both lines are in service and the post-fault curve equals the pre-fault curve.

Step 1: Normal operation

$$\begin{aligned} X &= 0.25 + \frac{0.5}{2} = 0.5\ \text{pu}, \qquad P_{max} = \frac{1.2\times1}{0.5} = 2.4\ \text{pu} \\ P_e &= 2.4\sin20^\circ = 0.8208\ \text{pu} = P_m \end{aligned}$$

Power delivered at normal operation: 0.821 pu.

Step 2: During the fault

A 3-phase fault at bus (1) puts the line-side voltage at zero, so $P_e = 0$. The rotor accelerates from $20^\circ$ to $45^\circ$.

$$A_1 = P_m(\delta_c - \delta_0) = 0.8208\times(0.7854 - 0.3491) = 0.3582\ \text{pu·rad}$$

Step 3: Area available after clearing

$$\delta_{max} = 180^\circ - 20^\circ = 160^\circ = 2.7925\ \text{rad}$$ $$\begin{aligned} A_{2,max} &= P_{max}(\cos45^\circ - \cos160^\circ) - P_m(\delta_{max}-\delta_c) \\ &= 2.4(0.7071 + 0.9397) - 0.8208(2.7925 - 0.7854) \\ &= 2.3048\ \text{pu·rad} \end{aligned}$$

Since $A_{2,max} = 2.305 > A_1 = 0.358$, the system is stable.

Step 4: Maximum swing angle $\delta_1$

Set $A_2 = A_1$:

$$2.4(\cos45^\circ - \cos\delta_1) - 0.8208(\delta_1 - 0.7854) = 0.3582$$

Solving by trial (or numerically): $\delta_1 = 1.1060$ rad.

Answer: Power delivered = 0.821 pu; the system is stable; maximum swing $\delta_{max} \approx 63.4^\circ$.

For reference, the critical clearing angle for this case is $\cos^{-1}[(\pi-2\delta_0)\sin\delta_0 - \cos\delta_0] = 95.97^\circ$, so clearing at $45^\circ$ leaves a wide margin.