---
title: "Load sharing: 250/500 MW, frequency falls to 49.5 Hz"
subject: "Power Plant Equipment"
chapter: 2
key: "semester_7/power_plant_equipment:load-sharing-250-500-4-2"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/load-sharing-250-500-4-2
site: BCT Engineering
---

# Load sharing: 250/500 MW, frequency falls to 49.5 Hz

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

- Asked in: 2070 Chaitra, 8 marks; 2071 Chaitra, 5 marks (asked 2 times)
- Video: [Speed Load characteristics | Two Generators rated 200MW and 400MW | Solved Problem -1 | PSOC | droop](https://www.youtube.com/watch?v=MEGzoyxTT6w) — Simple n Easy Learning
- 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-sharing-250-500-4-2
- 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 connected to a common bus. When Each generator is half loaded, they operates at a common frequency of 50 Hz. Their droops characteristics are 4% and 2% respectively based on their respective capacities. When the system load is increased, the frequency decreases to 49.5 Hz. Calculate the increase in system load and load shared by each generator.

## Short answer

Stiffness: K1=125 MW/Hz, K2=500 MW/Hz. With 0.5 Hz drop, system load increases by 312.5 MW. G1 takes 62.5 MW (now 187.5 MW), G2 takes 250 MW (now 500 MW, fully loaded).

## Answer

Both generators move down their droop lines to the common frequency of 49.5 Hz; the load increase is the sum of what each picks up.

**Method:** with droop $R_i$ (pu, on its own rating $P_{ri}$) and rated frequency $f_0$, the droop line of each unit gives
$$
\Delta P_i = -\frac{P_{ri}}{R_i\, f_0}\,\Delta f = -K_i\,\Delta f
$$
where $K_i = P_{ri}/(R_i f_0)$ is the stiffness of the unit in MW/Hz. For the units together, $\Delta P_L = -\Delta f\,\sum K_i$.

**Data:** $P_{r1} = 250$ MW, $R_1 = 0.04$; $P_{r2} = 500$ MW, $R_2 = 0.02$; $f_0 = 50$ Hz. Initial loads: 125 MW and 250 MW (half load). $\Delta f = 49.5 - 50 = -0.5$ Hz.

**Stiffness:**
$$
\begin{aligned}
K_1 &= \frac{250}{0.04 \times 50} = 125\ \text{MW/Hz} \\
K_2 &= \frac{500}{0.02 \times 50} = 500\ \text{MW/Hz}
\end{aligned}
$$

**Increase in load on each generator:**
$$
\begin{aligned}
\Delta P_1 &= 125 \times 0.5 = 62.5\ \text{MW} \\
\Delta P_2 &= 500 \times 0.5 = 250\ \text{MW}
\end{aligned}
$$

**Increase in system load:**
$$
\Delta P_L = 62.5 + 250 = 312.5\ \text{MW}
$$

| Generator | Rating (MW) | Droop | Initial (MW) | Increase (MW) | New (MW) |
|---|---|---|---|---|---|
| G1 | 250 | 4 % | 125 | 62.5 | 187.5 |
| G2 | 500 | 2 % | 250 | 250 | 500 |
| Total | 750 | – | 375 | 312.5 | 687.5 |

**Answer:** The system load increased by **312.5 MW**. G1 shares **62.5 MW** (now 187.5 MW) and G2 shares **250 MW** (now 500 MW, i.e. fully loaded). Any further load increase would overload G2.

## Check yourself

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

1. What are K1 and K2 for 250 MW at 4% and 500 MW at 2% on 50 Hz?
   A. 125 and 500 MW/Hz
   B. 50 and 100 MW/Hz
   C. 62.5 and 250 MW/Hz
   D. 500 and 125 MW/Hz
   Answer: A. K1 = 250/(0.04x50) = 125; K2 = 500/(0.02x50) = 500. (Key point: Stiffness)

2. What df is used for the final frequency of 49.5 Hz?
   A. -1 Hz
   B. -0.5 Hz
   C. +0.5 Hz
   D. -49.5 Hz
   Answer: B. df = 49.5 - 50 = -0.5 Hz. (Key point: Frequency deviation)

3. What is the total load increase?
   A. 625 MW
   B. 250 MW
   C. 312.5 MW
   D. 375 MW
   Answer: C. 62.5 + 250 = 312.5 MW. (Key point: System load increase)

4. What is notable about G2's final loading?
   A. It falls to 250 MW
   B. It exceeds 600 MW
   C. It is unloaded
   D. It is 500 MW, fully loaded, so further load would overload it
   Answer: D. G2 goes from 250 to 500 MW, its rating. (Key point: Overload check)

## 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 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×
- [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×
- [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×
- [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

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- [Load sharing of two transformers in parallel](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aparallel-transformers-load-sharing) — Electrical Machines, chapter 2