--- title: "Stability classification and swing equation" subject: "Power System Analysis II" chapter: 6 key: "semester_5/power_system_analysis_ii:stability-classification-swing" source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/stability-classification-swing site: BCT Engineering --- # Stability classification and swing equation Power System Analysis II · Chapter 6: Power System Stability - Asked in: 2075 Chaitra, 8 marks - Video: [Swing Equation derivation | Power system Stability | Power system Analysis](https://www.youtube.com/watch?v=ZpfAQ3G_8DE) — 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/stability-classification-swing - Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6 ## Question How power system stability is classified? Derive the swing equation for the rotor angle of the synchronous machine. ## Short answer Stability is classified as rotor angle stability (steady-state/small-signal, dynamic, transient), voltage stability, and frequency stability. The swing equation, derived from Newton's law via torque-power conversion and inertia constant H, is (2H/ωs)d2δ/dt2 = Pm - Pe in per unit. ## Answer #### Classification of power system stability ```text Power system stability +----------------+-----------------+ Rotor angle Frequency Voltage stability stability stability +------+ +--------+ Small- Transient Large-dist. Small-dist. signal (large-dist.) (steady state / dynamic) ``` 1. **Rotor angle stability:** ability of synchronous machines to remain in synchronism. - **Steady state (small-signal) stability:** small, gradual changes; limit $EV/X$. - **Dynamic stability:** small disturbances with controllers (AVR, governor, PSS) acting; concerns damping of oscillations over several seconds. - **Transient stability:** large sudden disturbances such as faults; first swing, about 1 s. 2. **Voltage stability:** ability to keep acceptable voltages at all buses; lost through lack of reactive power, leading to voltage collapse. 3. **Frequency stability:** ability to maintain frequency after a large imbalance between generation and load. #### Derivation Consider the rotor of a synchronous generator. Let - $J$ = moment of inertia of rotor and turbine (kg·m²) - $\theta_m$ = angular position of the rotor with respect to a stationary axis (mech. rad) - $T_m$ = mechanical (shaft) torque, $T_e$ = electromagnetic torque (N·m) **Step 1: Newton's law of rotation.** Accelerating torque = moment of inertia × angular acceleration: $$ J\frac{d^2\theta_m}{dt^2} = T_a = T_m - T_e $$ **Step 2: Measure angle from a synchronously rotating frame.** The rotor angle $\delta_m$ is the angle of the rotor relative to a reference axis rotating at synchronous speed $\omega_{sm}$: $$ \theta_m = \omega_{sm}t + \delta_m \;\Rightarrow\; \frac{d\theta_m}{dt} = \omega_{sm} + \frac{d\delta_m}{dt}, \quad \frac{d^2\theta_m}{dt^2} = \frac{d^2\delta_m}{dt^2} $$ so $$ J\frac{d^2\delta_m}{dt^2} = T_m - T_e $$ **Step 3: Convert torque to power.** Multiply by the rotor speed $\omega_m$ (power = torque × speed): $$ J\omega_m\frac{d^2\delta_m}{dt^2} = P_m - P_e $$ $M = J\omega_m$ is the **angular momentum** (inertia constant). Since the speed changes very little during a swing, $\omega_m \approx \omega_{sm}$ and $M$ is taken as constant: $$ M\frac{d^2\delta_m}{dt^2} = P_m - P_e $$ **Step 4: Introduce the inertia constant $H$.** $H$ is the stored kinetic energy at synchronous speed per MVA rating $S$: $$ H = \frac{\tfrac12 J\omega_{sm}^2}{S}\ \ \text{(MJ/MVA or s)} \;\Rightarrow\; M = J\omega_{sm} = \frac{2HS}{\omega_{sm}} $$ Substituting and dividing by $S$: $$ \frac{2H}{\omega_{sm}}\frac{d^2\delta_m}{dt^2} = \frac{P_m - P_e}{S} = P_m - P_e\ \ \text{(pu)} $$ **Step 5: Use electrical angle.** For a machine with $p$ poles, $\delta = \frac{p}{2}\delta_m$ and $\omega_s = \frac{p}{2}\omega_{sm}$, so $\delta/\omega_s = \delta_m/\omega_{sm}$: $$ \boxed{\frac{2H}{\omega_s}\frac{d^2\delta}{dt^2} = P_m - P_e} \qquad \text{or} \qquad \frac{H}{\pi f}\frac{d^2\delta}{dt^2} = P_m - P_e $$ with $\delta$ in electrical radians and $\omega_s = 2\pi f$. If $\delta$ is in electrical degrees, $\dfrac{H}{180f}\dfrac{d^2\delta}{dt^2} = P_m - P_e$. For a machine connected to an infinite bus through reactance $X$, $P_e = \dfrac{EV}{X}\sin\delta = P_{max}\sin\delta$, so $$ \frac{H}{\pi f}\frac{d^2\delta}{dt^2} = P_m - P_{max}\sin\delta $$ This is the **swing equation**. It is a second-order non-linear differential equation; its solution $\delta(t)$ describes how the rotor swings after a disturbance. If damping is included, a term $D\frac{d\delta}{dt}$ is added on the left. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Power system stability is classified into: A. Rotor angle, frequency and voltage stability B. Voltage, current and power stability C. Steady and dynamic only D. Fault and load stability Answer: A. The three main classes are rotor angle, frequency and voltage stability; rotor angle includes small-signal and transient. (Key point: Classification) 2. Rotor angle stability is the ability of: A. Buses to hold voltage B. Frequency to stay constant C. Lines to carry current D. Synchronous machines to remain in synchronism Answer: D. Rotor angle stability concerns keeping synchronous machines in step. (Key point: Rotor angle stability) 3. In the derivation, the net torque causing rotor acceleration is: A. Te - Tm B. Tm - Te C. Tm + Te D. Tm only Answer: B. Accelerating torque Ta = Tm - Te. (Key point: Accelerating torque) 4. Why does delta/omega_s = delta_m/omega_sm allow the swing equation in electrical units? A. Because poles cancel: delta = (p/2) delta_m and omega_s = (p/2) omega_sm B. Because H is dimensionless C. Because Pe is constant D. Because f is 50 Hz Answer: A. Both electrical quantities scale by p/2, so the ratio is unchanged and the equation holds in electrical radians. 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