---
title: "Load sharing: 600 MVA (6%) and 500 MVA (4%)"
subject: "Power Plant Equipment"
chapter: 2
key: "semester_7/power_plant_equipment:load-sharing-600-500mva-6-4"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/load-sharing-600-500mva-6-4
site: BCT Engineering
---

# Load sharing: 600 MVA (6%) and 500 MVA (4%)

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

- Asked in: 2072 Kartik, 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-6-4
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-2

## Question

Two generating units have following ratings and droop: Unit 1: 600MVA, R1 = 6%; Unit 2: 500MVA, R2 = 4%. The units are operating in parallel sharing a load of 900 MW at the normal frequency of 60Hz. Unit 1 supplies 500 MW and unit 2 supplies 400MW. If the load is increased by 90MW calculate the steady state frequency deviation and new generation on each unit.

## Short answer

Steady-state frequency deviation is –0.24 Hz (59.76 Hz). Unit 1 generates 540 MW and Unit 2 generates 450 MW.

## Answer

Both units drop along their droop lines to a common frequency; the 90 MW is shared in proportion to $P_r/R$. MVA ratings are used as the MW base.

**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, $R_1 = 0.06$, 500 MW; Unit 2: 500 MVA, $R_2 = 0.04$, 400 MW; $f_0 = 60$ Hz; $\Delta P_L = 90$ MW.

$$
\begin{aligned}
K_1 &= \frac{600}{0.06 \times 60} = 166.67\ \text{MW/Hz} \\
K_2 &= \frac{500}{0.04 \times 60} = 208.33\ \text{MW/Hz} \\
\Delta f &= -\frac{90}{166.67 + 208.33} = -\frac{90}{375} = -0.24\ \text{Hz}
\end{aligned}
$$
In per unit: $\Delta f = -0.24/60 = -0.004$ pu. New frequency $= 59.76$ Hz.

$$
\begin{aligned}
\Delta P_1 &= 166.67 \times 0.24 = 40\ \text{MW} \\
\Delta P_2 &= 208.33 \times 0.24 = 50\ \text{MW}
\end{aligned}
$$

**Per-unit cross-check (base 1000 MVA):** $R_1 = 0.06 \times 1000/600 = 0.1$, $R_2 = 0.04 \times 1000/500 = 0.08$; $\Delta f = -0.09/(10 + 12.5) = -0.004$ pu ✓.

| Unit | Initial (MW) | Increase (MW) | New (MW) |
|---|---|---|---|
| 1 | 500 | 40 | 540 |
| 2 | 400 | 50 | 450 |

**Answer:** Steady-state frequency deviation = **−0.24 Hz (−0.004 pu)**, i.e. 59.76 Hz; Unit 1 = **540 MW**, Unit 2 = **450 MW**.

## Check yourself

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

1. With 60 Hz base, what are K1 (600 MVA, 6%) and K2 (500 MVA, 4%)?
   A. 166.67 and 208.33 MW/Hz
   B. 100 and 125 MW/Hz
   C. 208.33 and 166.67 MW/Hz
   D. 600 and 500 MW/Hz
   Answer: A. K1 = 600/(0.06 x 60) = 166.67 and K2 = 500/(0.04 x 60) = 208.33. (Key point: Stiffness values)

2. What is the steady-state frequency deviation for the 90 MW increase?
   A. -0.15 Hz
   B. -0.24 Hz
   C. -0.54 Hz
   D. -0.09 Hz
   Answer: B. df = -90/(166.67 + 208.33) = -0.24 Hz, i.e. -0.004 pu. (Key point: Frequency deviation)

3. How much extra does each unit supply?
   A. Unit 1 50 MW, Unit 2 40 MW
   B. Unit 1 45 MW, Unit 2 45 MW
   C. Unit 1 40 MW, Unit 2 50 MW
   D. Unit 1 30 MW, Unit 2 60 MW
   Answer: C. dP1 = 166.67 x 0.24 = 40 MW and dP2 = 208.33 x 0.24 = 50 MW. (Key point: Load sharing)

4. What are the new outputs?
   A. 540 MW and 450 MW
   B. 550 MW and 440 MW
   C. 500 MW and 490 MW
   D. 590 MW and 400 MW
   Answer: A. 500 + 40 = 540 MW and 400 + 50 = 450 MW. (Key point: New generation)

## Related in Power Plant Equipment

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