--- title: "Critical clearing angle/time, fault at infinite bus" subject: "Power System Analysis II" chapter: 6 key: "semester_5/power_system_analysis_ii:critical-clearing-infinite-bus-h5" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/critical-clearing-infinite-bus-h5 site: BCT Engineering --- # Critical clearing angle/time, fault at infinite bus Power System Analysis II · Chapter 6: Power System Stability - Asked in: 2082 Baishakh, 8 marks - Video: [Critical Clearing Angle | Power system Stability analysis | Transient Stability](https://www.youtube.com/watch?v=9NS0Bv_nyOk) — 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-infinite-bus-h5 - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6 ## Question A 50 Hz synchronous generator connected to infinite bus as shown in the figure below has a H constant of 5 p.u. Under normal condition (just before disturbance), the generator injects 1 p.u. real power to the infinite bus with generator excitation voltage of 1.1 p.u. and infinite bus voltage 1 p.u. Suddenly a 3-phase bolted fault occurs at the infinite bus. Compute the critical fault clearing angle and critical fault clearing time for system to be transient stable. [Figure: generator X = 10%, transformer X = 5%, two parallel lines each X = 0.8 pu, to the infinite bus] ## Short answer Pre-fault Pmax=2 pu, delta0=30°. During fault Pe=0. Equal area gives delta_cr=79.56°. Then t_cr = sqrt(2H(delta_cr-delta0)/(pi f Pm)) = 0.235 s. ## Answer **Data:** $H = 5$ MJ/MVA, $f = 50$ Hz, $P_m = 1$ pu, $E = 1.1$ pu, $V = 1$ pu, $X_g = 0.10$, $X_T = 0.05$, two parallel lines of $0.8$ pu each. **Assumption:** the fault is a temporary fault at the infinite bus. After it is cleared, the network is the same as before the fault. #### Step 1: Pre-fault power-angle curve $$ \begin{aligned} X_1 &= 0.10 + 0.05 + \frac{0.8}{2} = 0.55\ \text{pu} \\ P_{max1} &= \frac{EV}{X_1} = \frac{1.1\times 1}{0.55} = 2.0\ \text{pu} \\ \delta_0 &= \sin^{-1}\frac{1}{2.0} = 30^\circ = 0.5236\ \text{rad} \end{aligned} $$ #### Step 2: During and after the fault A bolted fault at the infinite bus makes $V = 0$, so $P_{max2} = 0$ and $P_e = 0$ during the fault. After clearing, $P_{max3} = P_{max1} = 2.0$ pu. Then $$ \delta_{max} = 180^\circ - \delta_0 = 150^\circ = 2.618\ \text{rad} $$ #### Step 3: Critical clearing angle (equal areas) $$ \begin{aligned} A_1 &= P_m(\delta_{cr}-\delta_0) \\ A_2 &= \int_{\delta_{cr}}^{\delta_{max}}(P_{max}\sin\delta - P_m)\,d\delta \end{aligned} $$ Setting $A_1 = A_2$ gives $$ \begin{aligned} \cos\delta_{cr} &= (\pi - 2\delta_0)\sin\delta_0 - \cos\delta_0 \\ &= (3.1416 - 1.0472)(0.5) - 0.8660 \\ &= 0.1812 \\ \delta_{cr} &= 79.56^\circ = 1.3886\ \text{rad} \end{aligned} $$ #### Step 4: Critical clearing time With $P_e = 0$ during the fault, the swing equation is $\dfrac{H}{\pi f}\dfrac{d^2\delta}{dt^2} = P_m$. Integrating twice: $$ \delta = \delta_0 + \frac{\pi f P_m}{2H}t^2 \;\Rightarrow\; t_{cr} = \sqrt{\frac{2H(\delta_{cr}-\delta_0)}{\pi f P_m}} $$ $$ t_{cr} = \sqrt{\frac{2\times 5\times(1.3886-0.5236)}{\pi\times 50\times 1}} = \sqrt{0.05507} = 0.2347\ \text{s} $$ **Answer: Critical clearing angle $\delta_{cr} = 79.56^\circ$; critical clearing time $t_{cr} = 0.235$ s (about 11.7 cycles).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are Pmax1 and delta0 for E = 1.1, V = 1, X = 0.55 and Pm = 1 pu? A. Pmax1 = 2.0 pu, delta0 = 30 degrees B. Pmax1 = 1.0 pu, delta0 = 90 degrees C. Pmax1 = 2.0 pu, delta0 = 60 degrees D. Pmax1 = 0.55 pu, delta0 = 30 degrees Answer: A. Pmax1 = 1.1/0.55 = 2.0 pu and delta0 = asin(1/2) = 30 degrees. (Key point: Pre-fault power angle) 2. Why is Pe = 0 during a bolted fault at the infinite bus? A. The generator trips B. V becomes 0, so Pmax2 = EV/X = 0 C. The reactance becomes infinite D. Pm becomes zero Answer: B. The bus voltage V is zero, so no power is transferred during the fault. (Key point: Fault at infinite bus) 3. With Pe = 0 during the fault, the critical clearing time is: A. sqrt(pi f Pm / (2H(dcr - d0))) B. 2H(dcr - d0)/(pi f Pm) C. sqrt(2H(dcr - d0)/(pi f Pm)) D. (dcr - d0) / (pi f) Answer: C. Integrating (H/pi f) d2delta/dt2 = Pm twice gives delta = d0 + (pi f Pm/2H) t^2. 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