Critical clearing angle and time, H = 2

Data: $E = 1$ pu, $V = 1$ pu, $X_g = 0.05$ pu, two lines of $0.2$ pu each, $P_m = 1.2$ pu, $H = 2$ MJ/MVA, $f = 50$ Hz.

Assumptions: the fault is at the infinite-bus end of one line, so the generator cannot deliver power during the fault ($P_e = 0$). The fault is cleared by opening the faulted line.

Step 1: Pre-fault

$$\begin{aligned} X_I &= 0.05 + \frac{0.2}{2} = 0.15\ \text{pu}, \qquad P_{max1} = \frac{1\times1}{0.15} = 6.667\ \text{pu} \\ \delta_0 &= \sin^{-1}\frac{1.2}{6.667} = 10.37^\circ = 0.1810\ \text{rad} \end{aligned}$$

Step 2: During and after the fault

• During the fault: $P_{max2} = 0$.
• After the fault (one line): $X_{III} = 0.05 + 0.2 = 0.25$ pu, so $P_{max3} = 4.0$ pu.

$$\delta_{max} = 180^\circ - \sin^{-1}\frac{1.2}{4.0} = 180^\circ - 17.46^\circ = 162.54^\circ = 2.8369\ \text{rad}$$

Step 3: Critical clearing angle

$$\begin{aligned} \cos\delta_{cr} &= \frac{P_m(\delta_{max}-\delta_0) + P_{max3}\cos\delta_{max}}{P_{max3}} \\ &= \frac{1.2(2.8369-0.1810) + 4.0\cos162.54^\circ}{4.0} \\ &= \frac{3.1871 - 3.8158}{4.0} = -0.1572 \\ \delta_{cr} &= 99.04^\circ = 1.7286\ \text{rad} \end{aligned}$$

Step 4: Critical clearing time

With $P_e = 0$ during the fault, $\delta = \delta_0 + \dfrac{\pi f P_m}{2H}t^2$. So

$$\begin{aligned} t_{cr} &= \sqrt{\frac{2H(\delta_{cr}-\delta_0)}{\pi f P_m}} \\ &= \sqrt{\frac{2\times2\times(1.7286-0.1810)}{\pi\times50\times1.2}} \\ &= \sqrt{0.03284} = 0.181\ \text{s} \end{aligned}$$

Answer: Critical clearing angle $\delta_{cr} \approx 99.04^\circ$; critical clearing time $t_{cr} \approx 0.181$ s (about 9 cycles).