--- title: "Critical clearing angle and time, H = 2" subject: "Power System Analysis II" chapter: 6 key: "semester_5/power_system_analysis_ii:critical-clearing-h2-parallel-lines" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/critical-clearing-h2-parallel-lines site: BCT Engineering --- # Critical clearing angle and time, H = 2 Power System Analysis II · Chapter 6: Power System Stability - Asked in: 2070 Chaitra, 10 marks - Video: [KTU_EET304_PS II_Module 3_problems related to critical clearing angle](https://www.youtube.com/watch?v=_RvcCOCNF2k) — anu sunny - 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-h2-parallel-lines - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6 ## Question The figure below shows the generator connected to infinite bus using parallel transmission line. The generator is rated for 50 Hz and H constant of 2 pu. In steady state, the generator delivers a power of 1.2 pu to the infinite bus. A sudden 3 phase to ground fault occurs in one of the line near infinite bus as shown in figure. Determine the critical clearing angle and critical clearing time before which fault must be cleared so that system remains in stable state. [Figure: generator E = 1 p.u with j0.05 pu; two parallel lines j0.2 pu each to the infinite bus V = 1 p.u; fault on one line near the infinite bus] ## Short answer Critical clearing angle is 99.04°. With zero power during fault and H = 2 pu, the critical clearing time is 0.181 s. ## Answer **Data:** $E = 1$ pu, $V = 1$ pu, $X_g = 0.05$ pu, two lines of $0.2$ pu each, $P_m = 1.2$ pu, $H = 2$ MJ/MVA, $f = 50$ Hz. **Assumptions:** the fault is at the infinite-bus end of one line, so the generator cannot deliver power during the fault ($P_e = 0$). The fault is cleared by opening the faulted line. #### Step 1: Pre-fault $$ \begin{aligned} X_I &= 0.05 + \frac{0.2}{2} = 0.15\ \text{pu}, \qquad P_{max1} = \frac{1\times1}{0.15} = 6.667\ \text{pu} \\ \delta_0 &= \sin^{-1}\frac{1.2}{6.667} = 10.37^\circ = 0.1810\ \text{rad} \end{aligned} $$ #### Step 2: During and after the fault - During the fault: $P_{max2} = 0$. - After the fault (one line): $X_{III} = 0.05 + 0.2 = 0.25$ pu, so $P_{max3} = 4.0$ pu. $$ \delta_{max} = 180^\circ - \sin^{-1}\frac{1.2}{4.0} = 180^\circ - 17.46^\circ = 162.54^\circ = 2.8369\ \text{rad} $$ #### Step 3: Critical clearing angle $$ \begin{aligned} \cos\delta_{cr} &= \frac{P_m(\delta_{max}-\delta_0) + P_{max3}\cos\delta_{max}}{P_{max3}} \\ &= \frac{1.2(2.8369-0.1810) + 4.0\cos162.54^\circ}{4.0} \\ &= \frac{3.1871 - 3.8158}{4.0} = -0.1572 \\ \delta_{cr} &= 99.04^\circ = 1.7286\ \text{rad} \end{aligned} $$ #### Step 4: Critical clearing time With $P_e = 0$ during the fault, $\delta = \delta_0 + \dfrac{\pi f P_m}{2H}t^2$. So $$ \begin{aligned} t_{cr} &= \sqrt{\frac{2H(\delta_{cr}-\delta_0)}{\pi f P_m}} \\ &= \sqrt{\frac{2\times2\times(1.7286-0.1810)}{\pi\times50\times1.2}} \\ &= \sqrt{0.03284} = 0.181\ \text{s} \end{aligned} $$ **Answer: Critical clearing angle $\delta_{cr} \approx 99.04^\circ$; critical clearing time $t_{cr} \approx 0.181$ s (about 9 cycles).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is Pe during the fault near the infinite bus end of one line? A. Equal to Pm B. Pre-fault value C. Half the pre-fault value D. Zero (assumed), so Pmax2 = 0 Answer: D. The solution assumes the fault is at the infinite-bus end so the generator cannot deliver power during the fault. (Key point: Fault near infinite bus) 2. Which Pmax applies after clearing? A. 4.0 pu, from X = 0.05 + 0.2 = 0.25 B. 6.667 pu from both lines C. 0 pu D. 2.0 pu Answer: A. The faulted line is switched out, leaving a single 0.2 pu line, so Pmax3 = 1/0.25. (Key point: Post-fault curve) 3. What is the critical clearing angle? A. About 10.37 degrees B. About 99.04 degrees C. About 162.5 degrees D. About 17.5 degrees Answer: B. cos dcr = [1.2(2.8369 - 0.1810) + 4cos162.54]/4 = -0.1572. (Key point: Critical clearing angle) 4. What does t_cr depend on? A. Only dcr B. Only Pmax3 C. H, f, Pm and (dcr - d0): t = sqrt(2H(dcr - d0)/(pi f Pm)) D. Pmax1 and f Answer: C. During the fault the swing equation has constant acceleration, giving t = 0.181 s. 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