---
title: "Load sharing: 900 MW and 450 MW generators"
subject: "Power Plant Equipment"
chapter: 2
key: "semester_7/power_plant_equipment:load-sharing-900-450"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/load-sharing-900-450
site: BCT Engineering
---

# Load sharing: 900 MW and 450 MW generators

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

- Asked in: 2074 Chaitra, 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-900-450
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-2

## Question

A power system consist of two generators operating in parallel and supplying a load of 1200 MW. Generator G1 is rated as 900 MW with 2% drop regulation and generator G2 is rated as 450MW with 3% drop regulation. G1 supplied 750 MW and G2 supplies 450 MW and frequency is 60 Hz. When the load is increased by 150 MW, Calculate the new operating frequency and additional power generated by each generator.

## Short answer

New frequency is 59.85 Hz. G1 adds 112.5 MW (total 862.5 MW), G2 adds 37.5 MW (total 487.5 MW).

## Answer

Both generators move down their droop lines to a common lower frequency; the 150 MW is shared 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: 900 MW, 2 %, supplying 750 MW; G2: 450 MW, 3 %, supplying 450 MW; $f_0 = 60$ Hz; $\Delta P_L = 150$ MW.

$$
\begin{aligned}
K_1 &= \frac{900}{0.02 \times 60} = 750\ \text{MW/Hz} \\
K_2 &= \frac{450}{0.03 \times 60} = 250\ \text{MW/Hz} \\
\Delta f &= -\frac{150}{750 + 250} = -0.15\ \text{Hz} \\
f_{new} &= 60 - 0.15 = 59.85\ \text{Hz} \\
\Delta P_1 &= 750 \times 0.15 = 112.5\ \text{MW} \\
\Delta P_2 &= 250 \times 0.15 = 37.5\ \text{MW}
\end{aligned}
$$

| Generator | Initial (MW) | Additional (MW) | New (MW) |
|---|---|---|---|
| G1 | 750 | 112.5 | 862.5 |
| G2 | 450 | 37.5 | 487.5 |
| Total | 1200 | 150 | 1350 |

**Answer:** New frequency = **59.85 Hz**; G1 generates **112.5 MW** extra and G2 **37.5 MW** extra.

**Note:** G2 is already at its 450 MW rating, so 487.5 MW is an overload. If G2 is held at its limit (gate fully open), G1 must take all 150 MW: $\Delta f = -150/750 = -0.2$ Hz, giving 59.8 Hz and G1 = 900 MW.

## Check yourself

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

1. For a unit with rating Pr, droop R (pu) and rated frequency f0, the stiffness K (MW/Hz) is:
   A. K = R f0 / Pr
   B. K = Pr / (R f0)
   C. K = Pr R / f0
   D. K = f0 / (Pr R)
   Answer: B. From dP = -(Pr/(R f0)) df, the stiffness is K = Pr/(R f0). (Key point: Stiffness formula)

2. What are K1 and K2 for G1 (900 MW, 2%) and G2 (450 MW, 3%) at 60 Hz?
   A. 750 and 250 MW/Hz
   B. 450 and 150 MW/Hz
   C. 750 and 375 MW/Hz
   D. 1500 and 500 MW/Hz
   Answer: A. K1 = 900/(0.02 x 60) = 750 and K2 = 450/(0.03 x 60) = 250 MW/Hz. (Key point: Stiffness values)

3. What is the new frequency after the 150 MW load increase?
   A. 59.85 Hz
   B. 59.70 Hz
   C. 59.80 Hz
   D. 59.25 Hz
   Answer: A. df = -150/(750+250) = -0.15 Hz, so f = 59.85 Hz. (Key point: New frequency)

4. How much extra power does each generator supply?
   A. G1 75 MW, G2 75 MW
   B. G1 100 MW, G2 50 MW
   C. G1 112.5 MW, G2 37.5 MW
   D. G1 37.5 MW, G2 112.5 MW
   Answer: C. dP1 = 750 x 0.15 = 112.5 MW and dP2 = 250 x 0.15 = 37.5 MW, in proportion to K. (Key point: Load sharing)

## Related in Power Plant Equipment

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