---
title: "Stored energy and rotor acceleration"
subject: "Power System Analysis II"
chapter: 6
key: "semester_5/power_system_analysis_ii:turbo-generator-energy-acceleration"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/turbo-generator-energy-acceleration
site: BCT Engineering
---

# Stored energy and rotor acceleration

Power System Analysis II · Chapter 6: Power System Stability

- Asked in: 2081 Bhadra, 2+3+3 marks
- Video: [Example Q2.1.  Inertia Constant and Swing Equation Applications](https://www.youtube.com/watch?v=XkUNV8GSCRg) — 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/turbo-generator-energy-acceleration
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6

## Question

A 50 Hz, 4 pole turbo generator 100 MVA, 11 kV has an inertia constant 8 MJ/MVA. (i) find the stored energy in the rotor at synchronous speed. (ii) If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find rotor acceleration, neglecting mechanical and electrical losses. (iii) if acceleration calculated in part (ii) is maintained for 10 cycles, find the change in torque and rotor speed in revolutions/minute at end of this period.

## Short answer

Stored energy = GH = 800 MJ. Acceleration alpha = 337.5 elec deg/s^2 (28.125 rpm/s). After 10 cycles, rotor angle increases by 6.75 elec deg and speed reaches 1505.625 rpm.

## Answer

**Data:** $G = 100$ MVA, $H = 8$ MJ/MVA, $f = 50$ Hz, 4 poles. Synchronous speed $N_s = 120f/P = 1500$ rpm.

#### (i) Stored energy at synchronous speed

$$
KE = G H = 100 \times 8 = 800\ \text{MJ}
$$

#### (ii) Rotor acceleration

Accelerating power: $P_a = P_m - P_e = 80 - 50 = 30$ MW.

The inertia constant in MJ·s/elec-degree is

$$
M = \frac{GH}{180 f} = \frac{800}{180\times 50} = 0.08889\ \text{MJ·s/elec deg}
$$

Swing equation: $M\,\dfrac{d^2\delta}{dt^2} = P_a$, so

$$
\begin{aligned}
\alpha &= \frac{P_a}{M} = \frac{30}{0.08889} = 337.5\ \text{elec deg/s}^2 \\
&= \frac{337.5}{2} = 168.75\ \text{mech deg/s}^2 \quad (\text{4 poles} = 2\ \text{pole pairs}) \\
&= \frac{168.75\times 60}{360} = 28.125\ \text{rpm/s}
\end{aligned}
$$

#### (iii) After 10 cycles

Time: $t = 10/50 = 0.2$ s. Starting from synchronous speed with constant $\alpha$:

The change in **torque (load) angle** is

$$
\Delta\delta = \tfrac{1}{2}\alpha t^2 = \tfrac{1}{2}\times 337.5\times 0.2^2 = 6.75\ \text{elec deg}\ (= 3.375\ \text{mech deg})
$$

The rotor speed at the end of 10 cycles is

$$
\begin{aligned}
\Delta N &= 28.125\times 0.2 = 5.625\ \text{rpm} \\
N &= 1500 + 5.625 = 1505.625\ \text{rpm}
\end{aligned}
$$

"Change in torque" is read here as the change in torque (load) angle, which is the usual form of this problem. For reference, the accelerating torque is $T_a = P_a/\omega_{sm} = 30\times10^6/(2\pi\times1500/60) = 1.91\times10^5$ N·m.

**Answer: (i) 800 MJ; (ii) $\alpha = 337.5$ elec deg/s² (28.125 rpm/s); (iii) $\Delta\delta = 6.75$ elec deg, speed $= 1505.625$ rpm.**

## Check yourself

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

1. Stored kinetic energy of a 100 MVA machine with H = 8 MJ/MVA is:
   A. 8 MJ
   B. 80 MJ
   C. 12.5 MJ
   D. 800 MJ
   Answer: D. KE = G x H = 100 x 8 = 800 MJ. (Key point: Stored energy)

2. With mechanical input 80 MW and electrical load 50 MW, the accelerating power is:
   A. 130 MW
   B. 30 MW
   C. 50 MW
   D. 80 MW
   Answer: B. Pa = Pm - Pe = 80 - 50 = 30 MW (losses neglected). (Key point: Accelerating power)

3. With M = GH/(180 f) = 0.08889 MJ s/elec deg, the acceleration is:
   A. 168.75 elec deg/s2
   B. 30 elec deg/s2
   C. 337.5 elec deg/s2
   D. 0.08889 elec deg/s2
   Answer: C. alpha = Pa/M = 30/0.08889 = 337.5 elec deg/s2, i.e. 28.125 rpm/s for a 4-pole machine. (Key point: Acceleration)

## Related in Power System Analysis II

- [Inertia constant on 500 MVA base](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Ainertia-constant-500mva-base) — Power System Analysis II, chapter 6
- [Critical clearing time, H = 8 MJ/MVA](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Acritical-clearing-time-h8) — Power System Analysis II, chapter 6
- [L-G fault and 3-phase fault, 30 MVA generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Alg-fault-30mva-voltages-3phase) — Power System Analysis II, chapter 5
- [Generator-reactor system fault MVA](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Agenerator-reactor-fault-mva) — Power System Analysis II, chapter 3
- [L-L fault on 40 MVA generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3All-fault-40mva-generator) — Power System Analysis II, chapter 5
- [L-L fault currents and voltages, 30 MVA generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3All-fault-30mva-currents-voltages) — Power System Analysis II, chapter 5

## Related in other subjects

- [Differential protection: 11 kV, 100 MVA generator](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Agenerator-differential-11kv-6-ohm) — Switchgear and Protection, chapter 7
- [440 V IM with 50 kW rotor input](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Aim-440v-rotor-input-50kw) — Electrical Machine, chapter 5
- [Nominal-T 100 km line, 50 MVA at 132 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Anominal-t-100km-50mva) — Power System Analysis I, chapter 6
- [Speed of 4 and 6 pole alternators for 50 Hz](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aalternator-rpm-4-6-pole-50hz) — Electrical Machines, chapter 6