---
title: "Load sharing: 600 and 500 MVA, 900 MVA load"
subject: "Power Plant Equipment"
chapter: 2
key: "semester_7/power_plant_equipment:load-sharing-600-500mva-to-100"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/load-sharing-600-500mva-to-100
site: BCT Engineering
---

# Load sharing: 600 and 500 MVA, 900 MVA load

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

- Asked in: 2074 Asoj, 8 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-sharing-600-500mva-to-100
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-2

## Question

Two generating units of capacity 600MVA and 500MVA having droop of 4% and 6% are supplying 500MVA and 400MVA respectively to a common load of 900MVA at 60Hz. Calculate the new frequency and the new generation on each unit if load is increased to 100MVA.

## Short answer

New frequency is approximately 59.74 Hz. Unit 1 generates 564.29 MVA and Unit 2 generates 435.71 MVA.

## Answer

Taking "increased to 100 MVA" as **increased by 100 MVA** (total 1000 MVA), and treating MVA as MW (unity power factor), both units share the increase 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:** Unit 1: 600 MVA, 4 %, supplying 500; Unit 2: 500 MVA, 6 %, supplying 400; $f_0 = 60$ Hz; $\Delta P_L = 100$.

$$
\begin{aligned}
K_1 &= \frac{600}{0.04 \times 60} = 250\ \text{MVA/Hz} \\
K_2 &= \frac{500}{0.06 \times 60} = 138.89\ \text{MVA/Hz} \\
\Delta f &= -\frac{100}{388.89} = -0.2571\ \text{Hz} \\
f_{new} &= 60 - 0.2571 = 59.74\ \text{Hz} \\
\Delta P_1 &= 250 \times 0.2571 = 64.29 \\
\Delta P_2 &= 138.89 \times 0.2571 = 35.71
\end{aligned}
$$

| Unit | Initial | Increase | New |
|---|---|---|---|
| 1 (600 MVA, 4 %) | 500 | 64.29 | 564.29 |
| 2 (500 MVA, 6 %) | 400 | 35.71 | 435.71 |
| Total | 900 | 100 | 1000 |

**Answer:** New frequency ≈ **59.74 Hz**; Unit 1 generates **564.29 MVA (MW)** and Unit 2 **435.71 MVA (MW)**.

## Check yourself

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

1. What are K1 (600 MVA, 4%) and K2 (500 MVA, 6%) at 60 Hz?
   A. 250 and 138.89 per Hz
   B. 150 and 83.3 per Hz
   C. 250 and 83.3 per Hz
   D. 600 and 500 per Hz
   Answer: A. K1 = 600/(0.04 x 60) = 250; K2 = 500/(0.06 x 60) = 138.89. (Key point: Stiffness values)

2. What is the new frequency?
   A. 59.74 Hz
   B. 59.90 Hz
   C. 59.50 Hz
   D. 60.26 Hz
   Answer: A. df = -100/(250+138.89) = -0.2571 Hz, so f = 59.74 Hz. (Key point: New frequency)

3. What are the new outputs of unit 1 and unit 2?
   A. 550 and 450
   B. 564.29 and 435.71
   C. 535.71 and 464.29
   D. 600 and 400
   Answer: B. dP1 = 250 x 0.2571 = 64.29 and dP2 = 35.71, so 564.29 and 435.71. (Key point: Load sharing)

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