--- title: "Rotor dynamics for incremental angle change" subject: "Power System Analysis II" chapter: 6 key: "semester_5/power_system_analysis_ii:incremental-rotor-dynamics" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/incremental-rotor-dynamics site: BCT Engineering --- # Rotor dynamics for incremental angle change Power System Analysis II · Chapter 6: Power System Stability - Asked in: 2078 Kartik, 6 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/incremental-rotor-dynamics - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6 ## Question Starting from the swing equation of the rotor of a synchronous machine connected to an infinite bus, obtain the mathematical expression representing the rotor dynamics for an incremental change in rotor angle of a synchronous machine. ## Short answer Linearizing the swing equation about delta0 gives M d2DeltaDelta/dt2 + Ps DeltaDelta = 0, where Ps = Pmax cos delta0. With damping, M d2DeltaDelta/dt2 + D dDeltaDelta/dt + Ps DeltaDelta = 0. Stability requires Ps > 0, i.e. delta0 < 90 degrees, and natural frequency is omega_n = sqrt(Ps/M). ## Answer For a small change in rotor angle, the swing equation can be **linearised** about the operating point. The result is a second-order linear differential equation that shows the rotor oscillates about $\delta_0$ with a natural frequency set by the synchronising power coefficient and the inertia. #### Swing equation For a machine connected to an infinite bus through a total reactance $X$: $$ M\frac{d^2\delta}{dt^2} = P_m - P_e = P_m - P_{max}\sin\delta, \qquad M = \frac{H}{\pi f}\ \text{(pu)} $$ #### Linearising for a small change Let the rotor angle change by a small amount $\Delta\delta$ from the steady-state angle $\delta_0$, with $P_m$ constant: $$ \begin{aligned} \delta &= \delta_0 + \Delta\delta \\ M\frac{d^2(\delta_0+\Delta\delta)}{dt^2} &= P_m - P_{max}\sin(\delta_0+\Delta\delta) \end{aligned} $$ Expand the sine. For small $\Delta\delta$: $\cos\Delta\delta \approx 1$, $\sin\Delta\delta \approx \Delta\delta$, so $$ \sin(\delta_0+\Delta\delta) \approx \sin\delta_0 + \Delta\delta\cos\delta_0 $$ At steady state $P_m = P_{max}\sin\delta_0$ and $d^2\delta_0/dt^2 = 0$. These terms cancel, which leaves $$ M\frac{d^2\Delta\delta}{dt^2} + \left(P_{max}\cos\delta_0\right)\Delta\delta = 0 $$ The term $P_s = \left.\dfrac{dP_e}{d\delta}\right|_{\delta_0} = P_{max}\cos\delta_0$ is the **synchronising power coefficient**. So: $$ M\frac{d^2\Delta\delta}{dt^2} + P_s\,\Delta\delta = 0 $$ #### Including damping If there is damping power $D\,\dfrac{d\Delta\delta}{dt}$ (from damper windings), the equation becomes $$ M\frac{d^2\Delta\delta}{dt^2} + D\frac{d\Delta\delta}{dt} + P_s\,\Delta\delta = 0 $$ Its characteristic equation is $Ms^2 + Ds + P_s = 0$, which gives $$ \omega_n = \sqrt{\frac{P_s}{M}} = \sqrt{\frac{\pi f P_s}{H}}\ \text{rad/s}, \qquad \zeta = \frac{D}{2\sqrt{M P_s}} $$ The damped frequency of oscillation is $\omega_d = \omega_n\sqrt{1-\zeta^2}$. #### Interpretation - If $P_s > 0$ (that is, $0 < \delta_0 < 90^\circ$), the roots have negative real parts or lie on the imaginary axis. $\Delta\delta$ then oscillates about $\delta_0$, and the system is **steady-state stable**. - If $P_s < 0$ ($\delta_0 > 90^\circ$), one root is real and positive. $\Delta\delta$ then grows exponentially and the machine loses synchronism. - The machine is most stable at small $\delta_0$, where $P_s$ is large. The limit is reached at $\delta_0 = 90^\circ$, where $P_s = 0$. - A larger $H$ lowers the frequency of the rotor oscillation; it does not change the stability limit. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What approximation is used to linearise sin(delta0 + Ddelta)? A. sin(delta0) + Ddelta cos(delta0) B. sin(delta0) - Ddelta cos(delta0) C. sin(delta0) cos(delta0) D. Ddelta sin(delta0) Answer: A. For small Ddelta, cos(Ddelta) ~ 1 and sin(Ddelta) ~ Ddelta, giving sin(delta0) + Ddelta cos(delta0). (Key point: Linearisation) 2. The synchronising power coefficient Ps is: A. Pmax sin(delta0) B. Pm/Pmax C. Pmax cos(delta0) = dPe/d(delta) at delta0 D. H/(pi f) Answer: C. Ps is the slope of the power-angle curve at the operating point. (Key point: Synchronising power coefficient) 3. When is the machine steady-state stable according to the linearised model? A. When delta0 > 90 degrees B. When Ps < 0 C. When H is very small D. When Ps > 0, i.e. 0 < delta0 < 90 degrees Answer: D. With Ps > 0 the roots have non-positive real parts and the angle oscillates about delta0; for Ps < 0 one root is positive real. (Key point: Stability condition) 4. How does a larger H affect the natural frequency wn = sqrt(pi f Ps / H)? A. It lowers it B. It raises it C. It leaves it unchanged D. It makes it zero Answer: A. wn is inversely proportional to sqrt(H); a larger H gives slower oscillation, without changing the stability limit. (Key point: Effect of H on oscillation) ## Related in Power System Analysis II - [Rotor angle and swing equation, infinite bus](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Arotor-angle-swing-equation) — Power System Analysis II, chapter 6 - [Swing equation of synchronous machine](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aswing-equation-derivation) — Power System Analysis II, chapter 6 - [Stability definitions and swing equation from dynamics](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Astability-define-swing-from-dynamics) — Power System Analysis II, chapter 6 - [Critical clearing angle/time, fault at infinite bus](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Acritical-clearing-infinite-bus-h5) — Power System Analysis II, chapter 6 - [Stability classification and swing equation](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Astability-classification-swing) — Power System Analysis II, chapter 6 - [Initial rotor angle and critical clearing angle](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Arotor-angle-critical-clearing-2076) — Power System Analysis II, chapter 6 ## Related in other subjects - [Construction of synchronous machine and rotor types](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Asynchronous-machine-construction-rotors) — Electric Machines II, chapter 1, asked 2× - [Power angle characteristics of cylindrical rotor machine](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Apower-angle-cylindrical-rotor) — Electric Machines II, chapter 1, asked 2×