Critical clearing, fault at infinite bus, H = 6

Data: $f = 50$ Hz, $H = 6$ MJ/MVA, $P_m = 1.0$ pu, $P_{max} = 2.0$ pu.

Assumption: the fault is at the infinite bus and is temporary, so $P_e = 0$ during the fault. After clearing, the original curve $P_e = 2\sin\delta$ is restored.

Step 1: Initial and maximum angles

$$\begin{aligned} \delta_0 &= \sin^{-1}\frac{1.0}{2.0} = 30^\circ = 0.5236\ \text{rad} \\ \delta_{max} &= 180^\circ - 30^\circ = 150^\circ = 2.6180\ \text{rad} \end{aligned}$$

Step 2: Critical clearing angle

Accelerating area:

$$A_1 = P_m(\delta_{cr}-\delta_0)$$

Decelerating area:

$$A_2 = \int_{\delta_{cr}}^{\delta_{max}}(P_{max}\sin\delta - P_m)\,d\delta = P_{max}(\cos\delta_{cr}-\cos\delta_{max}) - P_m(\delta_{max}-\delta_{cr})$$

Setting $A_1 = A_2$ with $P_m = P_{max}\sin\delta_0$ and $\delta_{max} = \pi - \delta_0$:

$$\begin{aligned} \cos\delta_{cr} &= (\pi - 2\delta_0)\sin\delta_0 - \cos\delta_0 \\ &= (3.1416 - 1.0472)(0.5) - 0.8660 \\ &= 0.1812 \\ \delta_{cr} &= 79.56^\circ = 1.3886\ \text{rad} \end{aligned}$$

Step 3: Critical clearing time

During the fault the swing equation is $\dfrac{H}{\pi f}\dfrac{d^2\delta}{dt^2} = P_m$. Integrating twice from rest at $\delta_0$:

$$\delta = \delta_0 + \frac{\pi f P_m}{2H}t^2$$ $$\begin{aligned} t_{cr} &= \sqrt{\frac{2H(\delta_{cr}-\delta_0)}{\pi f P_m}} \\ &= \sqrt{\frac{2\times6\times(1.3886-0.5236)}{\pi\times50\times1.0}} \\ &= \sqrt{0.06608} = 0.257\ \text{s} \end{aligned}$$

Answer: Critical clearing angle $\delta_{cr} \approx 79.56^\circ$; critical clearing time $t_{cr} \approx 0.257$ s (about 12.9 cycles).