---
title: "Load decrease: 250 and 500 MW at 49 Hz"
subject: "Power Plant Equipment"
chapter: 2
key: "semester_7/power_plant_equipment:load-decrease-250-500-49hz"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/load-decrease-250-500-49hz
site: BCT Engineering
---

# Load decrease: 250 and 500 MW at 49 Hz

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

- Asked in: 2069 Chaitra, 8 marks
- Video: [Speed Droop in Power Control Explained](https://www.youtube.com/watch?v=sQ-AT1WdT3M) — RealPars
- 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-decrease-250-500-49hz
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-2

## Question

Two generators of 250 MW and 500 MW capacities respectively are operating in parallel and supplying power to a common load. When each generator is fully loaded, they operate at a common frequency of 49 Hz. Their droop characteristics are 2% and 4% respectively based on their respective ratings. When the system load is decreased, the frequency increases to 50Hz. calculate the decrease in system load shared by each generator.

## Short answer

Each unit's stiffness is K = P_r/(R f0). G1: 250/(0.02×50)=250 MW/Hz; G2: 500/(0.04×50)=250 MW/Hz. With Δf=+1 Hz, each sheds 250 MW, so total load decrease is 500 MW; G1 falls to 0 MW, G2 to 250 MW.

## Answer

Both units are fully loaded at 49 Hz. As the load falls, frequency rises to 50 Hz and each unit sheds load equal to its stiffness times the rise ($f_0 = 50$ Hz).

**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 $\Delta P_L = -\Delta f \sum K$.

$$
\begin{aligned}
K_1 &= \frac{250}{0.02 \times 50} = 250\ \text{MW/Hz} \\
K_2 &= \frac{500}{0.04 \times 50} = 250\ \text{MW/Hz} \\
\Delta f &= 50 - 49 = +1\ \text{Hz} \\
\Delta P_1 &= -250 \times 1 = -250\ \text{MW} \\
\Delta P_2 &= -250 \times 1 = -250\ \text{MW}
\end{aligned}
$$

| Generator | At 49 Hz (MW) | Decrease (MW) | At 50 Hz (MW) |
|---|---|---|---|
| G1 (250 MW, 2 %) | 250 | 250 | 0 |
| G2 (500 MW, 4 %) | 500 | 250 | 250 |
| Total | 750 | 500 | 250 |

**Answer:** The system load decreases by **500 MW**, shared equally: **250 MW by G1** (which falls to no load) and **250 MW by G2** (which now supplies 250 MW).

## Check yourself

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

1. What are K1 (250 MW, 2%) and K2 (500 MW, 4%) at f0 = 50 Hz?
   A. 250 and 250 MW/Hz
   B. 500 and 500 MW/Hz
   C. 125 and 125 MW/Hz
   D. 250 and 125 MW/Hz
   Answer: A. K1 = 250/(0.02 x 50) = 250 and K2 = 500/(0.04 x 50) = 250. (Key point: Stiffness values)

2. What frequency change is used?
   A. -1 Hz
   B. +1 Hz
   C. +49 Hz
   D. +0.5 Hz
   Answer: B. Frequency rises from 49 Hz to 50 Hz, so df = +1 Hz and outputs fall. (Key point: Frequency change)

3. How much load does each generator shed?
   A. 125 MW each
   B. 250 MW each
   C. 250 MW by G1 and 500 MW by G2
   D. 500 MW each
   Answer: B. dP = -K df = -250 MW for each. (Key point: Load shed by each)

4. What is the total decrease and final output of G1?
   A. 500 MW decrease; G1 at 0 MW
   B. 250 MW decrease; G1 at 125 MW
   C. 500 MW decrease; G1 at 250 MW
   D. 750 MW decrease; G1 at 0
   Answer: A. Both shed 250, so total is 500 MW; G1 was at 250 MW and falls to zero. (Key point: Total and final loading)

## Related in Power Plant Equipment

- [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×
- [Droop for capacity sharing: 250 and 500 MW](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Adroop-capacity-sharing-250-500) — Power Plant Equipment, chapter 2