--- title: "Critical clearing angle, Pmax 1.5, 0.5, 1.2" subject: "Power System Analysis II" chapter: 6 key: "semester_5/power_system_analysis_ii:critical-clearing-angle-reactive-line" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/critical-clearing-angle-reactive-line site: BCT Engineering --- # Critical clearing angle, Pmax 1.5, 0.5, 1.2 Power System Analysis II · Chapter 6: Power System Stability - Asked in: 2074 Asoj, 8 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-angle-reactive-line - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6 ## Question A generator is delivering 0.9 pu to an infinite bus through a purely reactive transmission line. Maximum power that could be delivered is 1.5 pu. A fault occurs such that maximum electrical power output reduces to 0.5 pu. When the fault is cleared, the maximum power that can be delivered is 1.2 pu. Determine the critical clearing angle using equal area criterion. ## Short answer Given pre-fault, fault, and post-fault power maxima (1.5, 0.5, 1.2 pu) and mechanical power 0.9 pu, the equal area criterion yields a critical clearing angle of approximately 65.4°. ## Answer **Data:** $P_m = 0.9$ pu, $P_{max1} = 1.5$ pu (pre-fault), $P_{max2} = 0.5$ pu (during fault), $P_{max3} = 1.2$ pu (post-fault). The line is purely reactive, so there are no losses. #### Step 1: Angles $$ \begin{aligned} \delta_0 &= \sin^{-1}\frac{0.9}{1.5} = 36.87^\circ = 0.6435\ \text{rad} \\ \delta_{max} &= 180^\circ - \sin^{-1}\frac{0.9}{1.2} = 180^\circ - 48.59^\circ = 131.41^\circ = 2.2935\ \text{rad} \end{aligned} $$ #### Step 2: Equal area criterion $$ \int_{\delta_0}^{\delta_{cr}}(P_m - P_{max2}\sin\delta)\,d\delta = \int_{\delta_{cr}}^{\delta_{max}}(P_{max3}\sin\delta - P_m)\,d\delta $$ $$ \Rightarrow\ \cos\delta_{cr} = \frac{P_m(\delta_{max}-\delta_0) + P_{max3}\cos\delta_{max} - P_{max2}\cos\delta_0}{P_{max3}-P_{max2}} $$ #### Step 3: Substitute $$ \begin{aligned} \cos\delta_{cr} &= \frac{0.9(2.2935-0.6435) + 1.2\cos131.41^\circ - 0.5\cos36.87^\circ}{1.2-0.5} \\ &= \frac{1.4850 - 0.7937 - 0.4000}{0.7} \\ &= 0.4161 \\ \delta_{cr} &= 65.41^\circ \end{aligned} $$ ```text Pe 1.5 | ___ I (pre) 1.2 | / _ \__ III (post) 0.9 |--*--|----\--------- Pm 0.5 | / A1| A2 \ II (fault) +--d0-dcr----dmax---- delta ``` **Answer: Critical clearing angle $\delta_{cr} \approx 65.4^\circ$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the initial angle delta_0 found? A. sin^-1(Pm/Pmax1) = sin^-1(0.9/1.5) = 36.87 degrees B. sin^-1(Pm/Pmax2) = sin^-1(0.9/0.5) C. sin^-1(Pm/Pmax3) = 48.59 degrees D. cos^-1(0.9/1.5) Answer: A. Before the fault the machine operates on the pre-fault curve, where Pm = Pmax1 sin(delta_0). (Key point: Initial angle) 2. How is delta_max found? A. 180 degrees - sin^-1(Pm/Pmax1) B. 180 degrees - sin^-1(Pm/Pmax3) = 131.41 degrees C. sin^-1(Pm/Pmax3) = 48.59 degrees D. 90 degrees Answer: B. delta_max is the second crossing of Pm with the post-fault curve, so the post-fault Pmax3 = 1.2 is used. (Key point: Maximum angle) 3. Which formula gives cos(delta_cr) for three different curves? A. [Pm(dmax - d0) + Pmax3 cos dmax]/Pmax3 B. [Pm(dmax - d0)]/(Pmax1 - Pmax2) C. [Pm(dmax - d0) + Pmax3 cos dmax - Pmax2 cos d0]/(Pmax3 - Pmax2) D. [Pmax3 cos dmax - Pmax2 cos d0]/Pm Answer: C. Fault Pmax2 is not zero here, so the accelerating area has a -Pmax2(cos d0 - cos dcr) term, which gives this formula. (Key point: Equal-area formula) 4. What is the critical clearing angle? A. About 36.9 degrees B. About 131.4 degrees C. About 48.6 degrees D. About 65.4 degrees Answer: D. cos(dcr) = 0.2913/0.7 = 0.4161, so dcr = 65.41 degrees. 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