--- title: "Critical clearing time and angle, fault at P" subject: "Power System Analysis II" chapter: 6 key: "semester_5/power_system_analysis_ii:critical-clearing-time-point-p-h5" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/critical-clearing-time-point-p-h5 site: BCT Engineering --- # Critical clearing time and angle, fault at P Power System Analysis II · Chapter 6: Power System Stability - Asked in: 2076 Asoj, 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-time-point-p-h5 - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6 ## Question Calculate the critical clearing time and critical angle if the fault is at point P for the network having inertia 5 MJ/MVA shown below. The machine is delivering 1.0 pu and both the terminal voltage and the infinite bus voltage are 1.0 pu. [Figure: generator G (j0.2) - transformer j0.1 - bus with V1 = 1.0; two parallel lines each j0.4 with breakers at both ends to the bus V2 and the infinite bus; fault P at the sending end of one line, which is opened to clear the fault] ## Short answer Using equal area criterion with zero electrical power during the fault, the critical clearing angle is 59.23° and the critical clearing time is 0.187 s (about 9.3cycles at 50 Hz). ## Answer **Data:** $X_d' = 0.2$, $X_T = 0.1$, two lines of $0.4$ pu each, $H = 5$ MJ/MVA, $f = 50$ Hz (assumed), $P_m = 1.0$ pu. $V_1 = 1.0$ pu at the bus marked in the figure (the HV bus after the transformer), and $V_2 = 1.0$ pu at the infinite bus. **Assumptions:** P is at the sending end of one line, and the fault is cleared by opening that line. #### Step 1: Pre-fault operating point Reactance from $V_1$ to the infinite bus: $0.4/2 = 0.2$ pu. $$ \begin{aligned} \sin\theta &= \frac{P X}{V_1V_2} = \frac{1\times 0.2}{1} = 0.2 \;\Rightarrow\; \theta = 11.54^\circ \\ I &= \frac{1\angle11.54^\circ - 1\angle0^\circ}{j0.2} = 1.0 + j0.1010\ \text{pu} \\ E' &= V_1 + j(0.2+0.1)I = (0.9798 + j0.2) + j0.3(1.0 + j0.1010) \\ &= 0.9495 + j0.5 = 1.0731\angle 27.77^\circ \end{aligned} $$ So $\delta_0 = 27.77^\circ = 0.4847$ rad. #### Step 2: Power-angle curves | Condition | $X$ (pu) | $P_{max}$ (pu) | |---|---|---| | Pre-fault | $0.2+0.1+0.2 = 0.5$ | $1.0731/0.5 = 2.146$ | | During fault (P at bus) | — | $0$ | | Post-fault (one line) | $0.2+0.1+0.4 = 0.7$ | $1.0731/0.7 = 1.533$ | $$ \delta_{max} = 180^\circ - \sin^{-1}\frac{1}{1.533} = 139.28^\circ = 2.4310\ \text{rad} $$ #### Step 3: Critical clearing angle $$ \begin{aligned} \cos\delta_{cr} &= \frac{P_m(\delta_{max}-\delta_0) + P_{max3}\cos\delta_{max}}{P_{max3}} \\ &= \frac{1(2.4310-0.4847) + 1.533\cos139.28^\circ}{1.533} \\ &= 0.5116 \\ \delta_{cr} &= 59.23^\circ = 1.0337\ \text{rad} \end{aligned} $$ #### Step 4: Critical clearing time With $P_e = 0$ during the fault: $$ \begin{aligned} t_{cr} &= \sqrt{\frac{2H(\delta_{cr}-\delta_0)}{\pi f P_m}} \\ &= \sqrt{\frac{2\times5\times(1.0337-0.4847)}{\pi\times50\times1}} \\ &= 0.187\ \text{s} \end{aligned} $$ **Answer: $\delta_{cr} \approx 59.2^\circ$, $t_{cr} \approx 0.187$ s (about 9.3 cycles).** If "terminal voltage" is instead taken at the generator terminals (before the transformer), then $E' = 1.05\angle28.44^\circ$, $P_{max1} = 2.1$, $P_{max3} = 1.5$, which gives $\delta_{cr} = 57.9^\circ$ and $t_{cr} = 0.181$ s. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With the fault at point P (sending end bus), what is Pe during the fault? A. Pe = 0, because the bus voltage collapses to zero B. Pe = Pm, since the machine still delivers 1.0 pu C. Pe = 1.533 sin(delta), the post-fault curve D. Pe = 2.146 sin(delta), the pre-fault curve Answer: A. A 3-phase fault at the sending end bus makes the bus voltage zero, so no power is transferred and Pe = 0. (Key point: Power-angle curve during fault) 2. Which reactances give the pre-fault and post-fault Pmax values? A. Pre-fault 0.7 pu, post-fault 0.5 pu B. Pre-fault 0.5 pu (two lines), post-fault 0.7 pu (one line) C. Both 0.5 pu because the transformer and generator do not change D. Pre-fault 0.3 pu, post-fault 0.7 pu Answer: B. Pre-fault: 0.2 + 0.1 + 0.4/2 = 0.5. After opening the faulted line: 0.2 + 0.1 + 0.4 = 0.7, so Pmax falls from 2.146 to 1.533. (Key point: Pre-fault and post-fault reactance) 3. With Pe = 0 during the fault, which expression gives cos(delta_cr)? A. [Pm(dmax - d0) - Pmax3 cos dmax] / Pmax1 B. Pm / Pmax3 C. [Pm(dmax - d0) + Pmax3 cos dmax] / Pmax3 D. [Pm(dmax + d0)] / Pmax3 Answer: C. Equating accelerating area Pm(dcr - d0) to decelerating area on the post-fault curve with Pmax2 = 0 gives this expression, using Pmax3. (Key point: Critical clearing angle formula) 4. How is the critical clearing time found once delta_cr is known? A. t = (dcr - d0)/(2 pi f) B. t = H/(pi f) x Pm C. t = sqrt(2H(dmax - d0)/(pi f Pmax3)) D. t = sqrt(2H(dcr - d0)/(pi f Pm)), since Pe = 0 during the fault Answer: D. With Pe = 0 the swing equation integrates to delta = d0 + (pi f Pm/2H)t^2, giving this t. 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