--- title: "Critical clearing for temporary sending-end fault" subject: "Power System Analysis II" chapter: 6 key: "semester_5/power_system_analysis_ii:critical-clearing-temporary-fault-new" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/critical-clearing-temporary-fault-new site: BCT Engineering --- # Critical clearing for temporary sending-end fault Power System Analysis II · Chapter 6: Power System Stability - Asked in: 2083 Baishakh (new course), 4+3 marks - 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-temporary-fault-new - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6 ## Question A 60 Hz synchronous generator has inertia constant H = 5 MJ/MVA and a direct axis transient reactance Xd' = 0.3 p.u. connected to an infinite bus V = 1∠0°. The generator has internal emf E' = 1.17∠26.387° p.u. and Pe = 0.8 p.u. The reactance of transformer and both transmission line are in p.u. A temporary 3-phase fault occurs at the sending end of the line at point 'F'. When the fault is cleared, both the lines is intact. Determine the critical clearing angle and the critical clearing time. [Figure: generator E' = 1.17∠26.387°, Xd' = 0.3 - transformer Xt = 0.2 - bus 1; two parallel lines Xl = 0.3 each (breakers at both ends) from bus 1 to bus 2, the infinite bus V = 1∠0°; fault F at bus 1 (sending end)] ## Short answer Critical clearing angle from equal area criterion is 84.78°. Critical clearing time, computed from the swing equation with zero accelerating power, is 0.26 s (about 15.6 cycles at 60 Hz). ## Answer **Data:** $f = 60$ Hz, $H = 5$ MJ/MVA, $X_d' = 0.3$, $X_t = 0.2$, $X_l = 0.3$ (each of two lines), $E' = 1.17\angle26.387^\circ$, $V = 1\angle0^\circ$, $P_m = P_e = 0.8$ pu. #### Step 1: Pre-fault power-angle curve $$ \begin{aligned} X &= 0.3 + 0.2 + \frac{0.3}{2} = 0.65\ \text{pu} \\ P_{max} &= \frac{1.17\times1}{0.65} = 1.8\ \text{pu} \end{aligned} $$ Check: $1.8\sin26.387^\circ = 0.8$ pu, so $\delta_0 = 26.387^\circ = 0.4605$ rad. #### Step 2: During and after the fault - The fault is at the sending end (bus 1), whose voltage becomes zero. So $P_e = 0$ during the fault. - The fault is temporary and both lines stay intact after clearing, so the post-fault curve is the same as before: $P_e = 1.8\sin\delta$. $$ \delta_{max} = 180^\circ - 26.387^\circ = 153.613^\circ = 2.6811\ \text{rad} $$ #### Step 3: Critical clearing angle From the equal area criterion, with $P_{max2} = 0$ and $P_{max3} = P_{max1}$: $$ \begin{aligned} \cos\delta_{cr} &= (\pi-2\delta_0)\sin\delta_0 - \cos\delta_0 \\ &= (2.2205)(0.4444) - 0.8958 \\ &= 0.0911 \\ \delta_{cr} &= 84.78^\circ = 1.4796\ \text{rad} \end{aligned} $$ #### Step 4: Critical clearing time With $P_e = 0$, $\delta = \delta_0 + \dfrac{\pi f P_m}{2H}t^2$: $$ \begin{aligned} t_{cr} &= \sqrt{\frac{2H(\delta_{cr}-\delta_0)}{\pi f P_m}} \\ &= \sqrt{\frac{2\times5\times(1.4796-0.4605)}{\pi\times60\times0.8}} \\ &= \sqrt{0.06758} = 0.260\ \text{s} \end{aligned} $$ **Answer: Critical clearing angle $\delta_{cr} \approx 84.78^\circ$; critical clearing time $t_{cr} \approx 0.26$ s (about 15.6 cycles at 60 Hz).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is Pmax found for the pre-fault curve? A. X = 0.3 + 0.2 + 0.3 = 0.8 B. Pmax = 0.8 pu C. X = 0.3 + 0.2 + 0.15 = 0.65, so Pmax = 1.17/0.65 = 1.8 pu D. Pmax = 1.17 pu Answer: C. Two 0.3 pu lines in parallel give 0.15 pu. (Key point: Pre-fault Pmax) 2. Why is the post-fault curve the same as the pre-fault curve? A. A line is removed after clearing B. The fault is permanent C. Because Pm changes D. The fault is temporary and both lines remain intact after clearing Answer: D. So Pmax3 = Pmax1 = 1.8 and delta_max = 180 - 26.387 = 153.6 degrees. (Key point: Post-fault curve) 3. What is delta_cr? A. About 84.78 degrees B. About 26.4 degrees C. About 153.6 degrees D. About 60 degrees Answer: A. cos dcr = (pi - 2 d0) sin d0 - cos d0 = 0.0911. (Key point: Critical angle) 4. How is critical time computed here? A. t uses f = 50 Hz B. t = sqrt(2H(dcr - d0)/(pi f Pm)) with f = 60 Hz and Pm = 0.8, giving about 0.26 s C. t uses Pm = 1.8 D. t = dcr/(2 pi f) Answer: B. Using H = 5, f = 60, Pm = 0.8 and (1.4796 - 0.4605) rad gives 0.26 s. 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