Critical clearing angle, breakers 1 and 2

Data: $|E| = 1.2$ pu, $|V| = 1.0$ pu, $P_m = 1.0$ pu. Generator reactance j0.25, parallel lines j0.5 and j0.4, j0.05 to the infinite bus. The fault P is on the j0.4 line next to breaker 2, i.e. practically at the receiving-end bus.

1. Pre-fault (both lines in)

$$\begin{aligned} X_1 &= 0.25 + (0.5\|0.4) + 0.05 = 0.25 + 0.2222 + 0.05 = 0.5222\ \text{pu} \\ P_{max1} &= \frac{EV}{X_1} = \frac{1.2 \times 1}{0.5222} = 2.298\ \text{pu} \\ \delta_0 &= \sin^{-1}\frac{P_m}{P_{max1}} = \sin^{-1}\frac{1}{2.298} = 25.80^\circ = 0.4502\ \text{rad} \end{aligned}$$

2. During fault

The fault at P is at the receiving bus, which is the only path from the generator to the infinite bus. That bus is at zero voltage, so no power reaches the infinite bus:

$$P_{max2} = 0$$

3. Post-fault (j0.4 line removed by breakers 1 and 2)

$$\begin{aligned} X_3 &= 0.25 + 0.5 + 0.05 = 0.80\ \text{pu} \\ P_{max3} &= \frac{1.2 \times 1}{0.8} = 1.5\ \text{pu} \\ \delta_{max} &= 180^\circ - \sin^{-1}\frac{1}{1.5} = 180^\circ - 41.81^\circ = 138.19^\circ = 2.4119\ \text{rad} \end{aligned}$$

4. Equal area criterion

 Pe
  ^           post-fault 1.5 sin(d)
  |            .--''''--.
  |         .'    A2      '.
 1|....+---+---------------+.... Pm = 1.0
  |    |A1 |                \
  |    |   |                 \
  +----+---+------------------+--> d
      d0   dc               dmax
 During fault Pe = 0, so A1 = Pm(dc - d0)

Accelerating area $A_1 = P_m(\delta_c - \delta_0)$ must equal decelerating area $A_2 = \int_{\delta_c}^{\delta_{max}}(P_{max3}\sin\delta - P_m)\,d\delta$. This gives

$$\cos\delta_{cr} = \frac{P_m(\delta_{max} - \delta_0) + P_{max3}\cos\delta_{max} - P_{max2}\cos\delta_0}{P_{max3} - P_{max2}}$$

Substituting ($P_{max2} = 0$):

$$\begin{aligned} \cos\delta_{cr} &= \frac{1.0\,(2.4119 - 0.4502) + 1.5\cos138.19^\circ}{1.5} \\ &= \frac{1.9617 + 1.5(-0.7454)}{1.5} = \frac{1.9617 - 1.1180}{1.5} = 0.5624 \\ \delta_{cr} &= \cos^{-1}(0.5624) = 55.78^\circ \end{aligned}$$

Answer: critical clearing angle $\delta_{cr} \approx 55.8^\circ$ (initial angle 25.8°). Breakers 1 and 2 must open before the rotor swings beyond 55.8°, otherwise the machine loses synchronism.