---
title: "Inertia constant on 500 MVA base"
subject: "Power System Analysis II"
chapter: 6
key: "semester_5/power_system_analysis_ii:inertia-constant-500mva-base"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/inertia-constant-500mva-base
site: BCT Engineering
---

# Inertia constant on 500 MVA base

Power System Analysis II · Chapter 6: Power System Stability

- Asked in: 2070 Asar, 4 marks
- Video: [Example Calculating H, H refered to a new base and acelerating time -Example_H_Calculation_v1.mlx](https://www.youtube.com/watch?v=1T0BzAinzAs) — Francisco Gonzalez-Longatt
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/inertia-constant-500mva-base
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6

## Question

A 2 pole, 50 Hz, 11.5 kV turbo generator has a rating of 100 MW, power factor 0.85 lagging. Calculate the inertia constant in MJ/MVA on a base of 500 MVA and its momentum in MJ-sec/elec-degree.

## Short answer

On a 500 MVA base, the inertia constant is H = 0.987 MJ/MVA. The angular momentum is M = 0.0548 MJ·s/elec deg.

## Answer

**Assumption:** the paper does not give the rotor inertia. The standard version of this problem uses a moment of inertia of $J = 10{,}000$ kg·m², and that value is assumed here.

#### Step 1: Rating and speed

$$
\begin{aligned}
S &= \frac{P}{\text{pf}} = \frac{100}{0.85} = 117.65\ \text{MVA} \\
N_s &= \frac{120f}{P} = \frac{120\times50}{2} = 3000\ \text{rpm} \\
\omega_{sm} &= \frac{2\pi\times3000}{60} = 314.16\ \text{rad/s}
\end{aligned}
$$

#### Step 2: Stored kinetic energy

$$
KE = \tfrac{1}{2}J\omega_{sm}^2 = \tfrac{1}{2}\times10^4\times(314.16)^2 = 493.48\times10^6\ \text{J} = 493.48\ \text{MJ}
$$

#### Step 3: Inertia constant

On the machine's own rating: $H = 493.48/117.65 = 4.19$ MJ/MVA.

On a 500 MVA base:

$$
H_{500} = \frac{493.48}{500} = 0.987\ \text{MJ/MVA}
$$

#### Step 4: Angular momentum

$$
M = \frac{GH}{180f} = \frac{KE}{180f} = \frac{493.48}{180\times50} = 0.0548\ \text{MJ·s/elec deg}
$$

In per unit on the 500 MVA base, this is $0.0548/500 = 1.097\times10^{-4}$ pu·s/elec deg.

**Answer: $H = 0.987$ MJ/MVA on 500 MVA (4.19 MJ/MVA on its own 117.65 MVA rating); $M = 0.0548$ MJ·s/elec deg.**

## Check yourself

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

1. How is the machine MVA rating obtained?
   A. S = P x pf
   B. S = 100 MVA
   C. S = 500 MVA
   D. S = P/pf = 100/0.85 = 117.65 MVA
   Answer: D. MVA is MW divided by power factor. (Key point: Machine MVA rating)

2. How is the stored kinetic energy computed?
   A. KE = (1/2) J omega^2, with omega = 2 pi x 3000/60 = 314.16 rad/s
   B. KE = J omega
   C. KE = (1/2) J omega with omega in rpm
   D. KE = J omega^2 x pf
   Answer: A. A 2-pole 50 Hz machine runs at 3000 rpm; KE = 0.5 x 10^4 x 314.16^2 = 493.48 MJ. (Key point: Stored kinetic energy)

3. How does H change when moving to a 500 MVA base?
   A. H stays 4.19
   B. H = KE/500 = 0.987 MJ/MVA, the stored energy is divided by the new base
   C. H = KE x 500
   D. H = 4.19 x 500
   Answer: B. H is stored energy per MVA of the chosen base, so a bigger base gives a smaller H. (Key point: Changing the MVA base)

4. How is M in MJ-s/elec-degree computed?
   A. M = KE x 180 f
   B. M = KE/f
   C. M = KE/(180 f) = 493.48/9000 = 0.0548
   D. M = H/(pi f) x 360
   Answer: C. M = GH/(180 f) and GH equals the stored energy. (Key point: Momentum M)

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