---
title: "Load reduction: 500 MW and 250 MW units"
subject: "Power Plant Equipment"
chapter: 2
key: "semester_7/power_plant_equipment:load-reduction-500-250"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/load-reduction-500-250
site: BCT Engineering
---

# Load reduction: 500 MW and 250 MW units

Power Plant Equipment · Chapter 2: Power/Frequency Control in Hydro Generator System

- Asked in: 2080 Baisakh, 10 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_plant_equipment/load-reduction-500-250
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-2

## Question

Two generators of 500 MW 11kV, 4% and 250 MW 11kV, 2% are supplying to a common load. When each generator is fully loaded, they operate at common frequency of 50 Hz. If the system load of 100 MW is reduced, what will be frequency deviation and find load shared by each unit?

## Short answer

Stiffness K = P_r/(R f0): both 250 MW/Hz. Frequency rise = 100/500 = +0.2 Hz, new freq 50.2 Hz. Each unit sheds 50 MW: G1 goes from 500 MW to 450 MW, G2 from 250 MW to 200 MW.

## Answer

When load falls, both units move up their droop lines; the frequency rises and each unit sheds load in proportion to $P_r/R$.

**Method:** a unit with droop $R$ (pu on its own rating $P_r$) and rated frequency $f_0$ changes its output by
$$
\Delta P = -\frac{P_r}{R\,f_0}\,\Delta f = -K\,\Delta f
$$
where $K = P_r/(R f_0)$ is its stiffness in MW/Hz. All units run at the same frequency, so $\Delta P_L = -\Delta f \sum K$.

**Data:** G1: 500 MW, $R_1 = 0.04$; G2: 250 MW, $R_2 = 0.02$; both fully loaded at 50 Hz (total 750 MW). Load reduced: $\Delta P_L = -100$ MW. Rated frequency 50 Hz.

**Stiffness:**
$$
\begin{aligned}
K_1 &= \frac{500}{0.04 \times 50} = 250\ \text{MW/Hz} \\
K_2 &= \frac{250}{0.02 \times 50} = 250\ \text{MW/Hz}
\end{aligned}
$$

**Frequency deviation:**
$$
\Delta f = -\frac{\Delta P_L}{K_1 + K_2} = -\frac{-100}{500} = +0.2\ \text{Hz}
$$
New frequency $= 50.2$ Hz.

**Load reduction on each unit:**
$$
\Delta P_1 = -250 \times 0.2 = -50\ \text{MW}, \qquad \Delta P_2 = -250 \times 0.2 = -50\ \text{MW}
$$

| Unit | Initial (MW) | Change (MW) | New (MW) |
|---|---|---|---|
| G1 (500 MW, 4 %) | 500 | −50 | 450 |
| G2 (250 MW, 2 %) | 250 | −50 | 200 |
| Total | 750 | −100 | 650 |

**Answer:** Frequency rises by **0.2 Hz** (to 50.2 Hz). G1 now supplies **450 MW** and G2 **200 MW**; both have equal stiffness, so they shed the load equally. The 11 kV rating does not affect the result.

## Check yourself

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

1. What are K1 and K2 for 500 MW at 4% and 250 MW at 2%?
   A. 250 and 250 MW/Hz
   B. 500 and 250 MW/Hz
   C. 250 and 500 MW/Hz
   D. 100 and 50 MW/Hz
   Answer: A. K1 = 500/(0.04x50) = 250; K2 = 250/(0.02x50) = 250. (Key point: Stiffness)

2. When load falls by 100 MW, what happens to frequency?
   A. It falls by 0.2 Hz
   B. It rises by 0.2 Hz
   C. It rises by 0.1 Hz
   D. It stays at 50 Hz
   Answer: B. df = -dPL/sumK = +100/500 = +0.2 Hz, giving 50.2 Hz. (Key point: Direction of frequency)

3. How is the 100 MW reduction shared?
   A. 67 and 33 MW by rating
   B. 33 and 67 MW
   C. 50 MW each since stiffness is equal
   D. 100 MW from G1
   Answer: C. Equal K gives -50 MW each. (Key point: Load shedding)

4. What are the final outputs?
   A. G1 400 MW, G2 250 MW
   B. G1 450 MW, G2 250 MW
   C. G1 500 MW, G2 150 MW
   D. G1 450 MW, G2 200 MW
   Answer: D. 500 - 50 = 450; 250 - 50 = 200. (Key point: Final loads)

## Close variants set in other exams

- [Load sharing: 1000 MW and 500 MW units](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aload-sharing-1000-500) — Power Plant Equipment, chapter 2, asked 4×
- [Load sharing: 600 and 300 MW at 49 Hz](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aload-sharing-600-300-49hz) — Power Plant Equipment, chapter 2, asked 4×

## Related in Power Plant Equipment

- [Load decrease: 250 and 500 MW at 49 Hz](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aload-decrease-250-500-49hz) — Power Plant Equipment, chapter 2
- [Load sharing: 250/500 MW, frequency falls to 49.5 Hz](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aload-sharing-250-500-4-2) — Power Plant Equipment, chapter 2, asked 2×
- [Load sharing: 250 MW and 500 MW, 4%/3% droop](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aload-sharing-250-500-4-3) — Power Plant Equipment, chapter 2
- [Load sharing: 500 MW and 250 MW, 5%/6% droop](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aload-sharing-500-250-5-6) — Power Plant Equipment, chapter 2, asked 2×
- [Droop for capacity sharing: 500 and 250 MW](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Adroop-capacity-sharing-500-250) — Power Plant Equipment, chapter 2
- [Load sharing: 250 MW and 500 MW, 20%/10% droop](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aload-sharing-250-500-20-10) — Power Plant Equipment, chapter 2, asked 2×

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- [Short shunt compound generator, 100 A at 250 V](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Agen-short-shunt-100a-250v) — Electrical Machines, chapter 3
- [Two generators 200∠30° and 210∠50° V](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Atwo-generators-200v-210v) — Power System Analysis I, chapter 3