--- title: "Line opened: stability and rotor swing" subject: "Power System Analysis II" chapter: 6 key: "semester_5/power_system_analysis_ii:breaker-a-opens-rotor-swing" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/breaker-a-opens-rotor-swing site: BCT Engineering --- # Line opened: stability and rotor swing Power System Analysis II · Chapter 6: Power System Stability - Asked in: 2071 Shrawan, 6 marks - Video: [Power system Stability analysis | Solved Problem -2 | PSA | Critical Clearing Angle | Power system](https://www.youtube.com/watch?v=jXHndqQ4s6E) — 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/breaker-a-opens-rotor-swing - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6 ## Question The power system shown in figure below is operating in steady state with which the generator delivering 1.5 p.u power to the infinite bus. The circuit breaker A suddenly opens and remain open. Find (i) whether the system remains stable or not. (ii) Angle to which the rotor swings. [Figure: generator E' = 1.2 p.u, X'd = 0.2 p.u; two parallel lines, the upper with X1 = 0.4 p.u and the lower with circuit breakers A and B; infinite bus E = 1.0 p.u] ## Short answer The system remains stable. The rotor swings to about 69.9° and then oscillates around the new equilibrium of 48.6°. ## Answer **Data:** $E' = 1.2$ pu, $V = 1.0$ pu, $X_d' = 0.2$ pu, $X_1 = 0.4$ pu, $P_m = 1.5$ pu. **Assumption:** the lower line's reactance is not marked, so it is taken equal to the upper line ($0.4$ pu). Opening breaker A disconnects the lower line. #### Step 1: Before and after opening $$ \begin{aligned} \text{Before: } X_I &= 0.2 + \frac{0.4}{2} = 0.4,\quad P_{max1} = \frac{1.2}{0.4} = 3.0\ \text{pu},\quad \delta_0 = \sin^{-1}\frac{1.5}{3.0} = 30^\circ \\ \text{After: } X_{II} &= 0.2 + 0.4 = 0.6,\quad P_{max2} = \frac{1.2}{0.6} = 2.0\ \text{pu} \end{aligned} $$ New equilibrium: $\delta_1 = \sin^{-1}(1.5/2.0) = 48.59^\circ$. Limit: $\delta_{max} = 180^\circ - 48.59^\circ = 131.41^\circ$. #### Step 2: (i) Stability check Accelerating area (from $30^\circ$ to $48.59^\circ$): $$ \begin{aligned} A_1 &= P_m(\delta_1-\delta_0) - P_{max2}(\cos\delta_0-\cos\delta_1) \\ &= 1.5(0.8481 - 0.5236) - 2.0(0.8660 - 0.6614) = 0.0775\ \text{pu·rad} \end{aligned} $$ Maximum decelerating area (from $48.59^\circ$ to $131.41^\circ$): $$ \begin{aligned} A_{2,max} &= P_{max2}(\cos\delta_1-\cos\delta_{max}) - P_m(\delta_{max}-\delta_1) \\ &= 2.0(0.6614+0.6614) - 1.5(2.2935-0.8481) = 0.4775\ \text{pu·rad} \end{aligned} $$ Since $A_{2,max} > A_1$, **the system remains stable**. #### Step 3: (ii) Maximum rotor swing $\delta_2$ From $A_1 = A_2$, the net area from $\delta_0$ to $\delta_2$ is zero: $$ \begin{aligned} \int_{\delta_0}^{\delta_2}(P_m - 2.0\sin\delta)\,d\delta &= 0 \\ 1.5(\delta_2 - 0.5236) - 2.0(0.8660 - \cos\delta_2) &= 0 \end{aligned} $$ Solving by trial: $\delta_2 = 1.2196$ rad. **Answer: (i) stable; (ii) the rotor swings to about 69.9° and then oscillates about the new equilibrium of 48.6°.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are Pmax before and after breaker A opens? A. 2.0 pu before, 3.0 pu after B. 3.0 pu both C. 6.0 pu before, 3.0 pu after D. 3.0 pu before, 2.0 pu after Answer: D. Before: X = 0.2 + 0.2 = 0.4, Pmax = 3.0. After the lower line is out: X = 0.6, Pmax = 2.0. (Key point: Curves before and after) 2. What are the new equilibrium angle and delta_max? A. 48.59 degrees and 131.41 degrees B. 30 degrees and 150 degrees C. 48.59 degrees and 90 degrees D. 30 degrees and 131.41 degrees Answer: A. On the new curve sin d1 = 1.5/2.0, and delta_max = 180 - d1. (Key point: New equilibrium and limit) 3. Why does the system remain stable? A. A1 is greater than A2,max B. A2,max (0.4775) is greater than A1 (0.0775) C. Pm is less than Pmax2 so it never moves D. The line opening does not change Pe Answer: B. Little accelerating energy is gained between 30 and 48.59 degrees compared with the available decelerating area. (Key point: Stability condition) 4. What happens to the rotor after the maximum swing of about 69.9 degrees? A. It returns to 30 degrees B. It goes out of step C. It oscillates about the new equilibrium at 48.6 degrees D. It stays at 69.9 degrees Answer: C. Swinging equally beyond equilibrium, it oscillates around 48.6 degrees (and damps out). 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