---
title: "Swing equation of synchronous machine"
subject: "Power System Analysis II"
chapter: 6
key: "semester_5/power_system_analysis_ii:swing-equation-derivation"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/swing-equation-derivation
site: BCT Engineering
---

# Swing equation of synchronous machine

Power System Analysis II · Chapter 6: Power System Stability

- Asked in: 2076 Chaitra, 6 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/swing-equation-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6

## Question

Derive the swing equation of rotor of a synchronous machine.

## Short answer

The swing equation is (2H/omega_s) d^2 delta/dt^2 = P_m - P_e. Start from J d^2 theta_m/dt^2 = T_m - T_e, convert torque to power, introduce H = (1/2)J omega_s^2/S, and change to electrical angle. With P_e = EV/X sin delta, it becomes (H/(pi f)) d^2 delta/dt^2 = P_m - P_max sin delta.

## Answer

The **swing equation** describes the relative motion of the rotor (load angle $\delta$) of a synchronous machine with respect to the synchronously rotating air-gap field when the mechanical and electrical powers are not balanced.

#### 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. The derivation starts from Newton's law of rotation:
   A. J d2(theta_m)/dt2 = Tm - Te (accelerating torque)
   B. J dtheta/dt = Tm
   C. T = J omega
   D. Pm = Pe
   Answer: A. Moment of inertia times angular acceleration equals net accelerating torque Tm - Te. (Key point: Starting law)

2. Why is the angle measured from a synchronously rotating frame (theta = omega_s t + delta)?
   A. To make J constant
   B. So that the second derivative of theta equals that of delta, i.e. only the departure from synchronism matters
   C. To remove Te
   D. To convert to MVA
   Answer: B. Since omega_s t has zero second derivative, d2(theta)/dt2 = d2(delta)/dt2. (Key point: Synchronous reference)

3. H is defined as:
   A. Rotor mass in kg
   B. Torque per MVA
   C. Stored kinetic energy at synchronous speed per MVA rating
   D. Power per radian
   Answer: C. H = (1/2 J omega_sm^2)/S, giving M = 2HS/omega_sm. (Key point: Inertia constant H)

4. The final per-unit swing equation is:
   A. H/(pi f) d2(delta)/dt2 = Pe - Pm
   B. H d2(delta)/dt2 = pi f Pm
   C. (2H/omega_s) d2(delta)/dt2 = Pm - Pe, i.e. H/(pi f) d2(delta)/dt2 = Pm - Pe
   D. d(delta)/dt = Pm - Pe
   Answer: C. Dividing M d2(delta)/dt2 = Pm - Pe by S and using electrical angle gives this form. (Key point: Final form)

## 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
- [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
- [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
- [Swing equation and inertia constant](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aswing-equation-inertia-constant) — Power System Analysis II, chapter 6
- [Swing equation and swing curve](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aswing-equation-swing-curve) — Power System Analysis II, chapter 6
- [Equivalent swing equation, coherent machines](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aequivalent-swing-equation-coherent) — Power System Analysis II, chapter 6

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