---
title: "Isochronous governor and speed droop model"
subject: "Power Plant Equipment"
chapter: 2
key: "semester_7/power_plant_equipment:isochronous-droop-model"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/isochronous-droop-model
site: BCT Engineering
---

# Isochronous governor and speed droop model

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

- Asked in: 2069 Chaitra, 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/isochronous-droop-model
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-2

## Question

What do you mean by isochronous governor and explain its operation? Why it is not suitable for multiple number of generating units operating in parallel. Derive the mathematical model of governor with speed droop characteristics.

## Short answer

An isochronous governor uses integral control to hold speed constant, but parallel units with different set points fight and load sharing is unstable. With droop feedback, the governor transfer function is ΔY/ΔΩ = -1/R * 1/(1+sT_G), giving steady-state droop R = -Δω/ΔY.

## Answer

An **isochronous governor** keeps the turbine speed constant at its set value for all loads ("isochronous" = same speed). It uses integral control: the gate keeps moving until the speed error is zero.

#### Operation

```text
 w_ref +   error  +------+ dY +-------+ dPm
 ---->(+)-------->| K/s  |--->|turbine|---+
       ^ -        +------+    +-------+   |
       |                                  v
       |                         -dPL -->(+)
       |      +-----------+               |
       +------| 1/(2Hs+D) |<--------------+
        speed +-----------+
```

1. A load increase makes $P_e > P_m$, so speed falls.
2. The speed error is integrated: $\Delta Y = -\dfrac{K}{s}\Delta\Omega$. The gate opens and $P_m$ rises.
3. The rotor accelerates back. The integrator stops only when the error is zero, so speed returns exactly to $\omega_{ref}$, with $\Delta P_m = \Delta P_L$.

#### Why not for several units in parallel
- Parallel units run at one common frequency. Each isochronous governor tries to hold the frequency at its own set point.
- Set points are never exactly equal, so the units **fight**: the one with the higher set point keeps opening and the other keeps closing, until one is at full load and the other at zero or motoring.
- Integral control has no fixed relation between speed and load, so the **load sharing is not defined** and is unstable. Only one unit in a system can be isochronous; others need droop.

#### Mathematical model of a governor with speed droop
Droop is obtained by feeding the gate position back through a gain $R$ around the integrator (steady-state feedback):

```text
                 +------+  dY
 -dw ---->(+)--->| K/s  |---+---> to turbine
           ^ -   +------+   |
           |     +-----+    |
           +-----|  R  |<---+
                 +-----+
```

With no change in load reference:
$$
\Delta Y(s) = \frac{K}{s}\left[-\Delta\Omega(s) - R\,\Delta Y(s)\right]
$$
$$
\begin{aligned}
s\,\Delta Y &= -K\Delta\Omega - KR\,\Delta Y \\
\Delta Y(s)(s + KR) &= -K\,\Delta\Omega(s) \\
\frac{\Delta Y(s)}{\Delta\Omega(s)} &= -\frac{K}{s + KR} = -\frac{1}{R}\cdot\frac{1}{1 + sT_G}
\end{aligned}
$$
where $T_G = \dfrac{1}{KR}$ is the governor time constant. Including the load reference,
$$
\Delta Y(s) = \frac{1}{1 + sT_G}\left[\Delta P_{ref}(s) - \frac{1}{R}\Delta\Omega(s)\right]
$$

**Steady state** ($s \to 0$): $\Delta Y = \Delta P_{ref} - \Delta\omega/R$, i.e. $R = -\Delta\omega/\Delta Y$ (pu). This is the **speed droop characteristic**: speed falls linearly as output rises. A 5 % droop means a 5 % speed drop gives a 100 % change in gate (output). It gives a unique and stable load sharing between parallel units in proportion to $1/R$.

## Check yourself

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

1. Why can't several isochronous units operate in parallel?
   A. Set points are never exactly equal, so units fight and load sharing is undefined
   B. Their droop is too high
   C. They produce too little power
   D. They have no governor
   Answer: A. Each tries to hold its own set frequency; the one with the higher set point opens fully while the other closes, with no defined sharing. (Key point: Why not in parallel)

2. How is speed droop produced in the governor model?
   A. By removing the integrator
   B. By feeding gate position back through gain R around the integrator
   C. By adding a second turbine
   D. By feeding speed back through R^2
   Answer: B. Gate position feedback through R around K/s makes the governor proportional in steady state. (Key point: Droop feedback)

3. What is dY/dOmega with droop feedback?
   A. -K/(s + KR) = -(1/R)/(1 + sTG)
   B. -K/s
   C. -1/(R + sK)
   D. -R/(1 + sTG)
   Answer: A. Solving s dY = -K dOmega - KR dY gives dY/dOmega = -K/(s+KR), with TG = 1/(KR). (Key point: Governor transfer function)

4. What does a 5% droop mean?
   A. A 5% speed drop gives 5% change in gate
   B. A 5% speed drop causes 100% change in gate (output)
   C. A 100% speed drop gives 5% change in output
   D. Output falls 5% at full load
   Answer: B. R = -dw/dY in pu, so R = 0.05 means 5% speed change moves output 100%. (Key point: Droop meaning)

## Related in Power Plant Equipment

- [Isochronous governor and parallel operation problem](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aisochronous-governor-parallel) — Power Plant Equipment, chapter 2, asked 6×
- [Speed response without governor and isochronous governor](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aspeed-response-isochronous-block) — 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
- [Speed droop in sharing common load](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aspeed-droop-load-sharing) — Power Plant Equipment, chapter 2, asked 2×
- [Load sharing: 600 MVA and 500 MVA, 4%/6% droop](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aload-sharing-600-500mva-4-6) — Power Plant Equipment, chapter 2, asked 2×
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