--- title: "Critical clearing, fault at infinite bus, H = 6" subject: "Power System Analysis II" chapter: 6 key: "semester_5/power_system_analysis_ii:critical-clearing-fault-infinite-bus-h6" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/critical-clearing-fault-infinite-bus-h6 site: BCT Engineering --- # Critical clearing, fault at infinite bus, H = 6 Power System Analysis II · Chapter 6: Power System Stability - Asked in: 2068 Chaitra, 8 marks - Video: [Power system Stability analysis | Solved Problem -1 | Critical Clearing Angle | PSA](https://www.youtube.com/watch?v=ne0so8Xe52k) — Simple n Easy Learning - 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-fault-infinite-bus-h6 - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6 ## Question A 3-phase, 50 Hz generator is connected to an infinite bus via transmission line. The inertia constant of the generator is 6 MJ/MVA and mechanical power input is 1.0 pu. The maximum power that can be delivered by generator is 2.0 pu. A 3-phase fault occurs at the infinite bus. Determine the critical clearing angle and critical clearing time. ## Short answer Using the equal area criterion with zero electrical power during the fault yields a critical clearing angle of 79.56°. The swing equation with constant acceleration gives a critical clearing time of 0.257 s (12.9 cycles). ## Answer **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).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With Pm = 1.0 and Pmax = 2.0, what are delta_0 and delta_max? A. 30 degrees and 90 degrees B. 60 degrees and 120 degrees C. 30 degrees and 150 degrees D. 45 degrees and 135 degrees Answer: C. delta_0 = sin^-1(0.5) = 30 degrees and delta_max = 180 - 30. (Key point: Initial and maximum angle) 2. What is Pe during the fault and after clearing? A. 2 sin(delta) throughout B. Zero throughout C. 1 pu during the fault D. Zero during the fault, then back to 2 sin(delta) Answer: D. The fault is temporary at the infinite bus; the original curve returns after clearing. (Key point: Fault at infinite bus) 3. Which result is correct for the critical clearing angle? A. 79.56 degrees from cos dcr = (pi - 2 d0) sin d0 - cos d0 B. 30 degrees C. 150 degrees D. 60 degrees Answer: A. Evaluating (3.1416 - 1.0472)(0.5) - 0.866 = 0.1812 gives 79.56 degrees. (Key point: Critical clearing angle) 4. How does the swing equation give time during the fault? A. delta = d0 + Pm t B. delta = d0 + (pi f Pm/2H) t^2 since acceleration is constant C. delta = d0 + 2H t D. delta grows linearly in t Answer: B. With Pe = 0, acceleration is constant, giving t = 0.257 s. (Key point: Critical clearing time) ## Related in Power System Analysis II - [Critical clearing time, H = 8 MJ/MVA](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Acritical-clearing-time-h8) — Power System Analysis II, chapter 6 - [Critical clearing angle/time, fault at infinite bus](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Acritical-clearing-infinite-bus-h5) — Power System Analysis II, chapter 6 - [Critical clearing for temporary sending-end fault](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Acritical-clearing-temporary-fault-new) — Power System Analysis II, chapter 6 - [Critical clearing time and angle, fault at P](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Acritical-clearing-time-point-p-h5) — Power System Analysis II, chapter 6 - [Critical clearing angle, Pmax 2, 0.5, 1.5](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Acritical-clearing-angle-pmax-two) — Power System Analysis II, chapter 6 - [Critical clearing angle, Pm = 1.2 pu](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Acritical-clearing-line2-fault-pm1-2) — Power System Analysis II, chapter 6 ## Related in other subjects - [Generator on infinite bus with 15% excitation rise](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Agenerator-infinite-bus-excitation-15) — Electric Machines II, chapter 1 - [Power transfer and pf of motor on infinite bus](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Amotor-infinite-bus-power-transfer) — Electric Machines II, chapter 2 - [Line with 50+j50 load, first maximum, power](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Aline-50-j50-load-first-maximum) — Electromagnetism, chapter 5 - [Bus-bar reactor scheme: 30 MVA and 50 MVA](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Abus-bar-reactor-30-50-mva) — Power Plant Equipment, chapter 4, asked 2×