Critical clearing for temporary sending-end fault

Data: $f = 60$ Hz, $H = 5$ MJ/MVA, $X_d' = 0.3$, $X_t = 0.2$, $X_l = 0.3$ (each of two lines), $E' = 1.17\angle26.387^\circ$, $V = 1\angle0^\circ$, $P_m = P_e = 0.8$ pu.

Step 1: Pre-fault power-angle curve

$$\begin{aligned} X &= 0.3 + 0.2 + \frac{0.3}{2} = 0.65\ \text{pu} \\ P_{max} &= \frac{1.17\times1}{0.65} = 1.8\ \text{pu} \end{aligned}$$

Check: $1.8\sin26.387^\circ = 0.8$ pu, so $\delta_0 = 26.387^\circ = 0.4605$ rad.

Step 2: During and after the fault

• The fault is at the sending end (bus 1), whose voltage becomes zero. So $P_e = 0$ during the fault.
• The fault is temporary and both lines stay intact after clearing, so the post-fault curve is the same as before: $P_e = 1.8\sin\delta$.

$$\delta_{max} = 180^\circ - 26.387^\circ = 153.613^\circ = 2.6811\ \text{rad}$$

Step 3: Critical clearing angle

From the equal area criterion, with $P_{max2} = 0$ and $P_{max3} = P_{max1}$:

$$\begin{aligned} \cos\delta_{cr} &= (\pi-2\delta_0)\sin\delta_0 - \cos\delta_0 \\ &= (2.2205)(0.4444) - 0.8958 \\ &= 0.0911 \\ \delta_{cr} &= 84.78^\circ = 1.4796\ \text{rad} \end{aligned}$$

Step 4: Critical clearing time

With $P_e = 0$, $\delta = \delta_0 + \dfrac{\pi f P_m}{2H}t^2$:

$$\begin{aligned} t_{cr} &= \sqrt{\frac{2H(\delta_{cr}-\delta_0)}{\pi f P_m}} \\ &= \sqrt{\frac{2\times5\times(1.4796-0.4605)}{\pi\times60\times0.8}} \\ &= \sqrt{0.06758} = 0.260\ \text{s} \end{aligned}$$

Answer: Critical clearing angle $\delta_{cr} \approx 84.78^\circ$; critical clearing time $t_{cr} \approx 0.26$ s (about 15.6 cycles at 60 Hz).