--- title: "Stability after tripping one circuit" subject: "Power System Analysis II" chapter: 6 key: "semester_5/power_system_analysis_ii:one-line-tripped-stability" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/one-line-tripped-stability site: BCT Engineering --- # Stability after tripping one circuit Power System Analysis II · Chapter 6: Power System Stability - Asked in: 2070 Asar, 8 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/one-line-tripped-stability - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6 ## Question For the system shown in figure below, the numerical values for different quantities are: E = 1.2 pu, V = 1 pu, Xs = 0.2 pu and reactance of each line is 0.4 pu. Initially the generator is delivering a power of 1.5 pu. If one of the double circuit lines is now tripped out, using equal area criteria determine whether the system would be able to maintain its stability or not. [Figure: generator E∠δ - double circuit line, each circuit X = j0.4 p.u with circuit breakers at both ends - infinite bus V∠0°] ## Short answer The system remains stable. The rotor swings to approximately 69.9° and settles at the new equilibrium angle of 48.6°. ## Answer **Data:** $E = 1.2$ pu, $V = 1$ pu, $X_s = 0.2$ pu, each line $0.4$ pu, $P_m = 1.5$ pu. #### Step 1: Power-angle curves $$ \begin{aligned} \text{Both lines: } X_I &= 0.2 + \frac{0.4}{2} = 0.4\ \text{pu}, \quad P_{max1} = \frac{1.2\times1}{0.4} = 3.0\ \text{pu} \\ \text{One line: } X_{II} &= 0.2 + 0.4 = 0.6\ \text{pu}, \quad P_{max2} = \frac{1.2}{0.6} = 2.0\ \text{pu} \end{aligned} $$ #### Step 2: Angles $$ \begin{aligned} \delta_0 &= \sin^{-1}\frac{1.5}{3.0} = 30^\circ = 0.5236\ \text{rad} \\ \delta_1 &= \sin^{-1}\frac{1.5}{2.0} = 48.59^\circ = 0.8481\ \text{rad}\ \ \text{(new equilibrium)} \\ \delta_{max} &= 180^\circ - 48.59^\circ = 131.41^\circ = 2.2935\ \text{rad} \end{aligned} $$ When the line trips, $P_e$ drops to $2.0\sin30^\circ = 1.0 < P_m$, so the rotor accelerates. #### Step 3: Equal areas Accelerating area, from $\delta_0$ to $\delta_1$: $$ \begin{aligned} A_1 &= P_m(\delta_1-\delta_0) - P_{max2}(\cos\delta_0-\cos\delta_1) \\ &= 1.5(0.8481-0.5236) - 2.0(0.8660-0.6614) \\ &= 0.4868 - 0.4092 = 0.0775\ \text{pu·rad} \end{aligned} $$ Maximum decelerating area, from $\delta_1$ to $\delta_{max}$: $$ \begin{aligned} A_{2,max} &= P_{max2}(\cos\delta_1-\cos\delta_{max}) - P_m(\delta_{max}-\delta_1) \\ &= 2.0(0.6614+0.6614) - 1.5(2.2935-0.8481) \\ &= 2.6458 - 2.1681 = 0.4775\ \text{pu·rad} \end{aligned} $$ ```text Pe 3.0 | ___ before (3 sin d) 2.0 | / _ \__ after (2 sin d) 1.5 |---/-*-----\----\---- Pm | /A1| A2 \ \ +-30--48.6--69.9--131.4 delta ``` #### Step 4: Result $A_{2,max} = 0.4775 > A_1 = 0.0775$, so **the system is stable** and keeps synchronism. Equating $A_1 = A_2$, that is $1.5(\delta_2-0.5236) = 2.0(0.8660-\cos\delta_2)$, gives a maximum swing of $\delta_2 = 69.88^\circ$. The rotor then settles at $48.59^\circ$. **Answer: Stable; the rotor swings to about 69.9° and settles at 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 one line trips? A. 2.0 pu and 3.0 pu B. 3.0 pu and 1.5 pu C. 1.5 pu and 3.0 pu D. 3.0 pu and 2.0 pu Answer: D. Both lines: X = 0.4, Pmax = 3.0. One line: X = 0.6, Pmax = 2.0. (Key point: Curves before and after tripping) 2. Right after the trip at 30 degrees, why does the rotor accelerate? A. Pe drops to 2.0 sin30 = 1.0 pu, below Pm = 1.5 pu B. Pe rises above Pm C. Pm falls to zero D. Pe becomes zero Answer: A. Mechanical input is greater than electrical output, so there is net accelerating power. (Key point: Why rotor accelerates) 3. Between which angles is A1 calculated? A. From 0 to 30 degrees B. From delta_0 = 30 degrees to the new equilibrium 48.59 degrees C. From 48.59 to 131.41 degrees D. From 30 to 150 degrees Answer: B. Pe is below Pm until the new curve crosses Pm at 48.59 degrees. (Key point: Accelerating area limits) 4. What shows stability? A. A1 greater than A2,max B. delta_1 is less than delta_0 C. A2,max = 0.4775 is greater than A1 = 0.0775 D. Pmax2 is less than Pmax1 Answer: C. The rotor swings to about 69.9 degrees and settles at 48.59 degrees. 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