--- title: "Critical clearing angle, breakers 1 and 2" subject: "Power System Analysis II" chapter: 6 key: "semester_5/power_system_analysis_ii:critical-clearing-angle-breakers-1-2" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/critical-clearing-angle-breakers-1-2 site: BCT Engineering --- # Critical clearing angle, breakers 1 and 2 Power System Analysis II · Chapter 6: Power System Stability - Asked in: 2080 Bhadra, 8 marks; 2074 Chaitra, 10 marks (asked 2 times) - Video: [POWER SYSTEM - Critical clearing angle (using equal area criterion)](https://www.youtube.com/watch?v=-bJ_X-5Lyfk) — Tulsiram shinde - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/critical-clearing-angle-breakers-1-2 - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6 ## Question A three phase fault is applied at the point P as shown in figure below. Find the critical clearing angle for clearing the fault with simultaneous opening of the breakers 1 and 2. The reactance values of various components are indicated in the diagram. The generator is delivering 1.0 pu power at the instant preceding the fault. [Figure: generator |E| = 1.2 pu, reactance j0.25 to the sending bus; two parallel lines j0.5 and j0.4 between the sending and receiving buses; the j0.4 line has breakers 1 (sending end) and 2 (receiving end) and the fault P is on it near breaker 2; reactance j0.05 from the receiving bus to the infinite bus |V| = 1∠0] ## Short answer Pre-fault: X1=0.522 pu, Pmax1=2.298 pu, δ0=25.8°. During fault at receiving bus: Pmax2=0. Post-fault (j0.4 line removed): X3=0.8 pu, Pmax3=1.5 pu, δmax=138.2°. Using equal area criterion with Pmax2=0: cosδcr = [Pm(δmax-δ0) + Pmax3cosδmax]/Pmax3 = 0.5624. Critical clearing angle δcr = 55.8°. ## Answer **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 ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the pre-fault Pmax1 found in this problem? A. EV divided by 0.25 + 0.5 + 0.4 + 0.05 B. EV divided by 0.25 + (0.5 parallel 0.4) + 0.05 = 0.5222 pu, giving 2.298 pu C. EV divided by 0.5 parallel 0.4 only D. EV divided by 0.25 + 0.05 Answer: B. Before the fault both lines are in, so their parallel combination (0.2222) is added to the generator and 0.05 reactances. (Key point: Pre-fault Pmax and initial angle) 2. Why is Pmax2 = 0 during the fault? A. The generator is disconnected from the system B. The fault at P is practically at the receiving bus, the only path to the infinite bus, so that bus is at zero voltage C. The j0.5 line is open during the fault D. The generator voltage E falls to zero Answer: B. All power to the infinite bus passes through the receiving bus; a 3-phase fault there makes its voltage zero so no power is delivered. (Key point: Power during the fault) 3. After breakers 1 and 2 open the faulted j0.4 line, what is Pmax3? A. 2.298 pu B. 1.2/0.5 = 2.4 pu C. 1.5 pu, from X = 0.25 + 0.5 + 0.05 = 0.8 D. 1.0 pu Answer: C. Only the j0.5 line remains, so X3 = 0.8 pu and Pmax3 = 1.2/0.8 = 1.5 pu. (Key point: Post-fault reactance) 4. With Pmax2 = 0, equal areas A1 = A2 up to delta_max = 138.19° give which critical clearing angle? A. About 25.8° B. About 138.2° C. About 41.8° D. About 55.8° Answer: D. cos(delta_cr) = [1.0(2.4119 - 0.4502) + 1.5 cos138.19°]/1.5 = 0.5624, so delta_cr is about 55.8°. 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