Critical clearing time and angle, fault at P

Data: $X_d' = 0.2$, $X_T = 0.1$, two lines of $0.4$ pu each, $H = 5$ MJ/MVA, $f = 50$ Hz (assumed), $P_m = 1.0$ pu. $V_1 = 1.0$ pu at the bus marked in the figure (the HV bus after the transformer), and $V_2 = 1.0$ pu at the infinite bus.

Assumptions: P is at the sending end of one line, and the fault is cleared by opening that line.

Step 1: Pre-fault operating point

Reactance from $V_1$ to the infinite bus: $0.4/2 = 0.2$ pu.

$$\begin{aligned} \sin\theta &= \frac{P X}{V_1V_2} = \frac{1\times 0.2}{1} = 0.2 \;\Rightarrow\; \theta = 11.54^\circ \\ I &= \frac{1\angle11.54^\circ - 1\angle0^\circ}{j0.2} = 1.0 + j0.1010\ \text{pu} \\ E' &= V_1 + j(0.2+0.1)I = (0.9798 + j0.2) + j0.3(1.0 + j0.1010) \\ &= 0.9495 + j0.5 = 1.0731\angle 27.77^\circ \end{aligned}$$

So $\delta_0 = 27.77^\circ = 0.4847$ rad.

Step 2: Power-angle curves

$$\delta_{max} = 180^\circ - \sin^{-1}\frac{1}{1.533} = 139.28^\circ = 2.4310\ \text{rad}$$

Step 3: Critical clearing angle

$$\begin{aligned} \cos\delta_{cr} &= \frac{P_m(\delta_{max}-\delta_0) + P_{max3}\cos\delta_{max}}{P_{max3}} \\ &= \frac{1(2.4310-0.4847) + 1.533\cos139.28^\circ}{1.533} \\ &= 0.5116 \\ \delta_{cr} &= 59.23^\circ = 1.0337\ \text{rad} \end{aligned}$$

Step 4: Critical clearing time

With $P_e = 0$ during the fault:

$$\begin{aligned} t_{cr} &= \sqrt{\frac{2H(\delta_{cr}-\delta_0)}{\pi f P_m}} \\ &= \sqrt{\frac{2\times5\times(1.0337-0.4847)}{\pi\times50\times1}} \\ &= 0.187\ \text{s} \end{aligned}$$

Answer: $\delta_{cr} \approx 59.2^\circ$, $t_{cr} \approx 0.187$ s (about 9.3 cycles).

If "terminal voltage" is instead taken at the generator terminals (before the transformer), then $E' = 1.05\angle28.44^\circ$, $P_{max1} = 2.1$, $P_{max3} = 1.5$, which gives $\delta_{cr} = 57.9^\circ$ and $t_{cr} = 0.181$ s.