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

# Swing equation and inertia constant

Power System Analysis II · Chapter 6: Power System Stability

- Asked in: 2076 Asoj, 6+2 marks
- Video: [The Swing Equation - Power System Stability - Power System 3](https://www.youtube.com/watch?v=mxrM2FMpUqo) — Ekeeda
- 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-inertia-constant
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6

## Question

Derive an expression for the swing equation of a synchronous machine. Signify the importance of inertia constant in the machine.

## Short answer

From Newton's law, J d2δm/dt2 = Tm - Te; multiplying by speed and using H = ½Jωs²/S gives (2H/ωs) d2δ/dt2 = Pm - Pe in per unit, or (H/πf)d2δ/dt2 = Pm - Pe. H is stored kinetic energy per MVA; larger H slows rotor acceleration and increases critical clearing time.

## Answer

The **swing equation** describes how the rotor angle $\delta$ of a synchronous machine changes when the mechanical input and electrical output 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.

#### Importance of the inertia constant $H$
$$
H = \frac{\text{stored kinetic energy at synchronous speed (MJ)}}{\text{machine rating (MVA)}}
$$
1. **Rate of rotor swing:** from $\frac{d^2\delta}{dt^2} = \frac{\pi f}{H}(P_m - P_e)$, a larger $H$ gives a smaller acceleration for the same power imbalance, so $\delta$ rises more slowly after a fault.
2. **Critical clearing time:** for a terminal fault $t_{cr} = \sqrt{\frac{4H(\delta_{cr}-\delta_0)}{\omega_s P_m}}$, so $t_{cr} \propto \sqrt H$. Higher $H$ allows more time for breakers to clear the fault.
3. **Narrow range:** $H$ lies in a narrow range for each machine type (about 2-9 s), e.g. turbo-alternators 4-9 s, hydro generators 2-4 s, so typical values can be assumed when data are missing.
4. **Equivalent machine:** machines swinging together can be combined: $H_{eq} = \sum H_i S_i / S_{base}$.
5. **Frequency response:** $H$ also decides how fast the system frequency falls after loss of generation.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How does the derivation get from torques to powers?
   A. Divide by omega
   B. Multiply the torque equation by rotor speed (P = T omega), with omega_m ~ omega_sm so M = J omega is constant
   C. Square both sides
   D. Integrate over time
   Answer: B. Speed changes little in a swing, so M = J omega_sm is treated as constant. (Key point: Power step in derivation)

2. From d2(delta)/dt2 = (pi f/H)(Pm - Pe), a larger H gives:
   A. Larger acceleration
   B. No change
   C. Smaller acceleration, so delta rises more slowly
   D. Negative acceleration
   Answer: C. Acceleration is inversely proportional to H for the same power imbalance. (Key point: Effect of H on acceleration)

3. How does critical clearing time vary with H for a terminal fault?
   A. Inversely with H
   B. Independent of H
   C. Proportional to H
   D. Proportional to sqrt(H)
   Answer: D. t_cr = sqrt(4H(delta_cr - delta_0)/(omega_s Pm)), so it goes as sqrt(H). (Key point: Critical clearing time)

4. Machines swinging together combine as:
   A. H_eq = sum(Hi Si)/S_base
   B. H_eq = product of Hi
   C. H_eq = average of Hi
   D. H_eq = max Hi
   Answer: A. Inertia constants are weighted by MVA ratings onto a common base. (Key point: Equivalent inertia)

## Related in Power System Analysis II

- [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
- [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
- [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
- [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
- [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
- [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

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