---
title: "Load sharing: 600 MW and 300 MW, 7%/8% droop"
subject: "Power Plant Equipment"
chapter: 2
key: "semester_7/power_plant_equipment:load-sharing-600-300-7-8"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/load-sharing-600-300-7-8
site: BCT Engineering
---

# Load sharing: 600 MW and 300 MW, 7%/8% droop

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

- Asked in: 2081 Bhadra, 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-sharing-600-300-7-8
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-2

## Question

Two generating units of 600 MW and 300 MW capacities respectively are operating in parallel and supplying power to common load. When each generator is half loaded, they operate at a common frequency of 50 Hz. The droop regulations are 7% and 8% respectively based on their respective ratings. If the load is increased by 150 MW, Calculate: a) New frequency at which they operate. b) Power supplied by each generator.

## Short answer

Compute stiffness K = P_r/(R f0): G1 171.43 MW/Hz, G2 75 MW/Hz. Frequency drop = 150/(sum K) = 0.6087 Hz, new freq = 49.39 Hz. G1 adds 104.35 MW to 404.35 MW, G2 adds 45.65 MW to 195.65 MW.

## Answer

Both units drop along their droop lines to a common new frequency, each taking a share of the extra 150 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: 600 MW, $R_1 = 0.07$, initial 300 MW; G2: 300 MW, $R_2 = 0.08$, initial 150 MW; $f_0 = 50$ Hz; $\Delta P_L = 150$ MW.

**Stiffness:**
$$
\begin{aligned}
K_1 &= \frac{600}{0.07 \times 50} = 171.43\ \text{MW/Hz} \\
K_2 &= \frac{300}{0.08 \times 50} = 75\ \text{MW/Hz} \\
K_1 + K_2 &= 246.43\ \text{MW/Hz}
\end{aligned}
$$

#### a) New frequency
$$
\begin{aligned}
\Delta f &= -\frac{150}{246.43} = -0.6087\ \text{Hz} \\
f_{new} &= 50 - 0.6087 = 49.39\ \text{Hz}
\end{aligned}
$$

#### b) Power supplied by each generator
$$
\begin{aligned}
\Delta P_1 &= 171.43 \times 0.6087 = 104.35\ \text{MW} \\
\Delta P_2 &= 75 \times 0.6087 = 45.65\ \text{MW}
\end{aligned}
$$

| Generator | Initial (MW) | Increase (MW) | New (MW) |
|---|---|---|---|
| G1 (600 MW, 7 %) | 300 | 104.35 | 404.35 |
| G2 (300 MW, 8 %) | 150 | 45.65 | 195.65 |
| Total | 450 | 150 | 600 |

**Answer:** New frequency ≈ **49.39 Hz**; G1 supplies **404.35 MW** and G2 supplies **195.65 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 600 MW at 7% and 300 MW at 8%?
   A. 171.43 and 75 MW/Hz
   B. 85.7 and 37.5 MW/Hz
   C. 171.43 and 150 MW/Hz
   D. 120 and 75 MW/Hz
   Answer: A. K1 = 600/(0.07x50) = 171.43; K2 = 300/(0.08x50) = 75. (Key point: Stiffness)

2. What is the new frequency for a 150 MW rise?
   A. 49.65 Hz
   B. 49.39 Hz
   C. 50.61 Hz
   D. 49.0 Hz
   Answer: B. df = -150/246.43 = -0.6087 Hz. (Key point: New frequency)

3. How much extra does each unit take?
   A. 75 MW each
   B. 100 MW and 50 MW
   C. 104.35 MW and 45.65 MW
   D. 45.65 MW and 104.35 MW
   Answer: C. dP1 = 171.43x0.6087 = 104.35; dP2 = 75x0.6087 = 45.65. (Key point: Load increase shares)

4. What are the final outputs of G1 and G2?
   A. 350 MW and 250 MW
   B. 300 MW and 150 MW
   C. 404.35 MW and 45.65 MW
   D. 404.35 MW and 195.65 MW
   Answer: D. Initial 300 + 104.35 and 150 + 45.65. (Key point: New generation)

## Close variants set in other exams

- [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×

## Related in Power Plant Equipment

- [Load sharing: 600 and 300 MW, 51 Hz no-load](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aload-sharing-600-300-51hz) — Power Plant Equipment, chapter 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
- [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 calculation: 800 MW and 300 MW units](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Adroop-calculation-800-300) — 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

## 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 50 kVA and 25 kVA transformers in parallel](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aparallel-50kva-25kva-60kva-load) — Electrical Machines I, chapter 2
- [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