Critical clearing angle/time, fault at infinite bus

Data: $H = 5$ MJ/MVA, $f = 50$ Hz, $P_m = 1$ pu, $E = 1.1$ pu, $V = 1$ pu, $X_g = 0.10$, $X_T = 0.05$, two parallel lines of $0.8$ pu each.

Assumption: the fault is a temporary fault at the infinite bus. After it is cleared, the network is the same as before the fault.

Step 1: Pre-fault power-angle curve

$$\begin{aligned} X_1 &= 0.10 + 0.05 + \frac{0.8}{2} = 0.55\ \text{pu} \\ P_{max1} &= \frac{EV}{X_1} = \frac{1.1\times 1}{0.55} = 2.0\ \text{pu} \\ \delta_0 &= \sin^{-1}\frac{1}{2.0} = 30^\circ = 0.5236\ \text{rad} \end{aligned}$$

Step 2: During and after the fault

A bolted fault at the infinite bus makes $V = 0$, so $P_{max2} = 0$ and $P_e = 0$ during the fault.

After clearing, $P_{max3} = P_{max1} = 2.0$ pu. Then

$$\delta_{max} = 180^\circ - \delta_0 = 150^\circ = 2.618\ \text{rad}$$

Step 3: Critical clearing angle (equal areas)

$$\begin{aligned} A_1 &= P_m(\delta_{cr}-\delta_0) \\ A_2 &= \int_{\delta_{cr}}^{\delta_{max}}(P_{max}\sin\delta - P_m)\,d\delta \end{aligned}$$

Setting $A_1 = A_2$ gives

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

Step 4: Critical clearing time

With $P_e = 0$ during the fault, the swing equation is $\dfrac{H}{\pi f}\dfrac{d^2\delta}{dt^2} = P_m$. Integrating twice:

$$\delta = \delta_0 + \frac{\pi f P_m}{2H}t^2 \;\Rightarrow\; t_{cr} = \sqrt{\frac{2H(\delta_{cr}-\delta_0)}{\pi f P_m}}$$ $$t_{cr} = \sqrt{\frac{2\times 5\times(1.3886-0.5236)}{\pi\times 50\times 1}} = \sqrt{0.05507} = 0.2347\ \text{s}$$

Answer: Critical clearing angle $\delta_{cr} = 79.56^\circ$; critical clearing time $t_{cr} = 0.235$ s (about 11.7 cycles).