--- title: "Fault cleared at 45°: stability and maximum swing" subject: "Power System Analysis II" chapter: 6 key: "semester_5/power_system_analysis_ii:fault-cleared-45-max-swing" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/fault-cleared-45-max-swing site: BCT Engineering --- # Fault cleared at 45°: stability and maximum swing Power System Analysis II · Chapter 6: Power System Stability - Asked in: 2075 Asoj, 10 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/fault-cleared-45-max-swing - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6 ## Question The power system shown below is operating initially at power angle δ = 20°. Calculate the power delivered to the infinite bus at normal operation. If a 3-phase to ground fault occurs at bus (1) and fault is cleared when the power angle δ reaches 45°, determine whether system come back to stable or not? If yes, calculate the maximum δ-swing angle. [Figure: generator E = 1.2 pu∠δ, j0.25 pu to bus (1); two parallel lines each j0.5 pu (breakers at both ends) from bus (1) to bus (2), the infinite bus V = 1 pu∠0°] ## Short answer Initial power is 0.821 pu. During the three-phase fault at bus 1, electrical power is zero; clearing at 45° leaves the decelerating area larger than the accelerating area, so the system is stable and the maximum rotor swing is about 63.4°. ## Answer **Data:** $E = 1.2$ pu, $V = 1$ pu, $X_d' = 0.25$, two lines of $0.5$ pu each, $\delta_0 = 20^\circ$. **Assumption:** the bus fault is temporary, so after clearing both lines are in service and the post-fault curve equals the pre-fault curve. #### Step 1: Normal operation $$ \begin{aligned} X &= 0.25 + \frac{0.5}{2} = 0.5\ \text{pu}, \qquad P_{max} = \frac{1.2\times1}{0.5} = 2.4\ \text{pu} \\ P_e &= 2.4\sin20^\circ = 0.8208\ \text{pu} = P_m \end{aligned} $$ **Power delivered at normal operation: 0.821 pu.** #### Step 2: During the fault A 3-phase fault at bus (1) puts the line-side voltage at zero, so $P_e = 0$. The rotor accelerates from $20^\circ$ to $45^\circ$. $$ A_1 = P_m(\delta_c - \delta_0) = 0.8208\times(0.7854 - 0.3491) = 0.3582\ \text{pu·rad} $$ #### Step 3: Area available after clearing $$ \delta_{max} = 180^\circ - 20^\circ = 160^\circ = 2.7925\ \text{rad} $$ $$ \begin{aligned} A_{2,max} &= P_{max}(\cos45^\circ - \cos160^\circ) - P_m(\delta_{max}-\delta_c) \\ &= 2.4(0.7071 + 0.9397) - 0.8208(2.7925 - 0.7854) \\ &= 2.3048\ \text{pu·rad} \end{aligned} $$ Since $A_{2,max} = 2.305 > A_1 = 0.358$, **the system is stable**. #### Step 4: Maximum swing angle $\delta_1$ Set $A_2 = A_1$: $$ 2.4(\cos45^\circ - \cos\delta_1) - 0.8208(\delta_1 - 0.7854) = 0.3582 $$ Solving by trial (or numerically): $\delta_1 = 1.1060$ rad. **Answer: Power delivered = 0.821 pu; the system is stable; maximum swing $\delta_{max} \approx 63.4^\circ$.** For reference, the critical clearing angle for this case is $\cos^{-1}[(\pi-2\delta_0)\sin\delta_0 - \cos\delta_0] = 95.97^\circ$, so clearing at $45^\circ$ leaves a wide margin. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the power delivered at delta = 20 degrees found? A. Pe = (E V / X) sin(delta) with X = 0.25 + 0.5/2 = 0.5 pu, giving 0.821 pu B. Pe = 1.2 sin(20) with X = 0.25 pu only C. Pe = 2.4 cos(20 degrees) D. Pe = 1.2 x 1 x 0.5 Answer: A. Two parallel lines give 0.25 reactance, so total X = 0.5, Pmax = 2.4 pu and Pe = 2.4 sin20 = 0.821 pu. (Key point: Normal operation power) 2. During the 3-phase fault at bus (1), the accelerating area is A1 = Pm(dc - d0). Why? A. Pe is half of Pm during the fault B. Pe = 0 during the fault, so the full Pm accelerates the rotor C. The rotor decelerates during the fault D. The fault doubles Pmax Answer: B. Bus (1) voltage is zero, so no electrical power is transferred and the whole mechanical input accelerates the rotor. (Key point: Fault at bus (1)) 3. How is stability decided after clearing at 45 degrees? A. Stable if A1 exceeds A2,max B. Stable if dc is greater than d0 C. Stable if the maximum available decelerating area A2,max exceeds A1 D. Stable if Pm is greater than Pmax Answer: C. Here A2,max = 2.305 exceeds A1 = 0.358, so the rotor is brought back before reaching delta_max = 160 degrees. (Key point: Stability decision) 4. What condition gives the maximum swing angle delta_1? A. delta_1 = 160 degrees, delta_max B. Pe = Pm at delta_1 = 45 degrees C. A2 = 0 D. Decelerating area A2 equals accelerating area A1 Answer: D. The rotor swings until the decelerating area equals the accelerating area, giving about 63.4 degrees, well below 160 degrees. 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