---
title: "Load sharing: 500 MW and 250 MW, 5%/6% droop"
subject: "Power Plant Equipment"
chapter: 2
key: "semester_7/power_plant_equipment:load-sharing-500-250-5-6"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/load-sharing-500-250-5-6
site: BCT Engineering
---

# Load sharing: 500 MW and 250 MW, 5%/6% droop

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

- Asked in: 2076 Asoj, 8 marks
- Video: [What is Droop setting in Governor of Generators? How Load of Generators in parallel is controlled?](https://www.youtube.com/watch?v=YXeVxP5-utk) — Electro Globe
- 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-500-250-5-6
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-2

## Question

Two generating units of 500MW and 250MW capacities respectively are operating in parallel and supplying power to a common load. When each generator is half loaded, they operate at a common frequency of 50Hz. The droop regulations are 5% and 6% respectively based on their respective ratings. If the load is increased by 100 MW, Calculate: a) New frequency at which they operates b) Power supplied by each generator

## Short answer

Stiffness: G1 200, G2 83.33 MW/Hz. Frequency drop = 100/(sum) = 0.3529 Hz, new freq = 49.65 Hz. G1 picks up 70.59 MW to 320.59 MW; G2 picks up 29.41 MW to 154.41 MW.

## Answer

The two units move down their droop lines to a common lower frequency, sharing the 100 MW 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, 5 %, initial 250 MW; G2: 250 MW, 6 %, initial 125 MW; $f_0 = 50$ Hz; $\Delta P_L = 100$ MW.

$$
\begin{aligned}
K_1 &= \frac{500}{0.05 \times 50} = 200\ \text{MW/Hz} \\
K_2 &= \frac{250}{0.06 \times 50} = 83.33\ \text{MW/Hz} \\
K_1 + K_2 &= 283.33\ \text{MW/Hz}
\end{aligned}
$$

#### a) New frequency
$$
\Delta f = -\frac{100}{283.33} = -0.3529\ \text{Hz}, \qquad f = 50 - 0.3529 = 49.65\ \text{Hz}
$$

#### b) Power supplied by each generator
$$
\begin{aligned}
\Delta P_1 &= 200 \times 0.3529 = 70.59\ \text{MW} \\
\Delta P_2 &= 83.33 \times 0.3529 = 29.41\ \text{MW}
\end{aligned}
$$

| Generator | Initial (MW) | Increase (MW) | New (MW) |
|---|---|---|---|
| G1 | 250 | 70.59 | 320.59 |
| G2 | 125 | 29.41 | 154.41 |
| Total | 375 | 100 | 475 |

**Answer:** New frequency ≈ **49.65 Hz**; G1 supplies **320.59 MW**, G2 supplies **154.41 MW**.

## 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 5% and 250 MW at 6%?
   A. 200 and 83.33 MW/Hz
   B. 100 and 41.7 MW/Hz
   C. 200 and 125 MW/Hz
   D. 500 and 250 MW/Hz
   Answer: A. K1 = 500/(0.05x50) = 200; K2 = 250/(0.06x50) = 83.33. (Key point: Stiffness)

2. What is the new frequency for 100 MW rise?
   A. 49.50 Hz
   B. 49.65 Hz
   C. 50.35 Hz
   D. 49.88 Hz
   Answer: B. df = -100/283.33 = -0.3529 Hz. (Key point: New frequency)

3. How much extra does G1 take?
   A. 50 MW
   B. 29.41 MW
   C. 70.59 MW
   D. 100 MW
   Answer: C. 200 x 0.3529 = 70.59 MW. (Key point: G1 share)

4. What are the final G1 and G2 outputs?
   A. 300 MW and 175 MW
   B. 350 MW and 125 MW
   C. 320.59 MW and 125 MW
   D. 320.59 MW and 154.41 MW
   Answer: D. 250 + 70.59 and 125 + 29.41. (Key point: Final outputs)

## Close variants set in other exams

- [Load sharing: 600 MW and 300 MW, 7%/8% droop](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aload-sharing-600-300-7-8) — Power Plant Equipment, chapter 2, asked 2×

## Related in Power Plant Equipment

- [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, 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: 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: 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×
- [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
- [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

## Related in other subjects

- [Load sharing between two stations in parallel](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Aload-sharing-two-stations) — Electric Machines II, chapter 1
- [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