Initial rotor angle and critical clearing angle

Data: $|V_t| = 1.08$ pu, $X_d' = 0.2$, $X_T = 0.1$, lines $j0.5 \parallel j0.4$ between bus 1 and bus 3, bus 3 tied directly to the infinite bus $1\angle0^\circ$, $P_m = 1$ pu.

Assumption: the fault at bus 3 is a temporary fault. After it is cleared, the network is unchanged.

Step 1: Reactances

$$\begin{aligned} X_{lines} &= \frac{0.5\times 0.4}{0.5+0.4} = 0.2222\ \text{pu} \\ X_{t\text{-}\infty} &= 0.1 + 0.2222 = 0.3222\ \text{pu}\ \ \text{(terminal to infinite bus)} \end{aligned}$$

Step 2: Terminal voltage angle

$$\begin{aligned} P &= \frac{|V_t||V|}{X}\sin\theta \;\Rightarrow\; \sin\theta = \frac{1\times 0.3222}{1.08\times 1} = 0.2984 \\ \theta &= 17.36^\circ, \quad V_t = 1.08\angle17.36^\circ = 1.0308 + j0.3222 \end{aligned}$$

Step 3: Current and internal emf

$$\begin{aligned} I &= \frac{V_t - V}{jX} = \frac{0.0308 + j0.3222}{j0.3222} = 1.0 - j0.0956 = 1.0044\angle{-5.46^\circ}\ \text{pu} \\ E' &= V_t + jX_d' I = (1.0308 + j0.3222) + j0.2(1.0 - j0.0956) \\ &= 1.0499 + j0.5222 = 1.1726\angle 26.45^\circ\ \text{pu} \end{aligned}$$

Rotor angle before the fault: $\delta_0 = 26.45^\circ$.

Step 4: Power-angle curves

$$\begin{aligned} X_{total} &= 0.2 + 0.3222 = 0.5222\ \text{pu} \\ P_{max} &= \frac{1.1726\times 1}{0.5222} = 2.2455\ \text{pu} \end{aligned}$$

Check: $2.2455\sin 26.45^\circ = 1.0$ pu.

During the fault, bus 3 is shorted, and it is tied directly to the infinite bus, so $P_e = 0$. After clearing, $P_{max3} = P_{max1} = 2.2455$ pu. Then $\delta_{max} = 180^\circ - 26.45^\circ = 153.55^\circ$.

Step 5: Critical clearing angle

$$\begin{aligned} \cos\delta_{cr} &= (\pi - 2\delta_0)\sin\delta_0 - \cos\delta_0 \\ &= (3.1416 - 0.9231)(0.4453) - 0.8954 \\ &= 0.0926 \\ \delta_{cr} &= 84.69^\circ \end{aligned}$$

Answer: Initial rotor angle $\delta_0 = 26.45^\circ$ ($E' = 1.173$ pu); critical clearing angle $\delta_{cr} \approx 84.69^\circ$.