---
title: "Speed response without governor and isochronous governor"
subject: "Power Plant Equipment"
chapter: 2
key: "semester_7/power_plant_equipment:speed-response-isochronous-block"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/speed-response-isochronous-block
site: BCT Engineering
---

# Speed response without governor and isochronous governor

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

- Asked in: 2082 Baisakh, 8 marks; 2072 Chaitra, 6 marks (asked 2 times)
- Video: [Droop and Isochronous mode in gas turbines, Power plant guru, Speed droop, droop control, governor](https://www.youtube.com/watch?v=GKyqS5MUCmA) — Power Plant Guru
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_plant_equipment/speed-response-isochronous-block
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-2

## Question

Derive the transfer function for speed response for load change in turbine-generator set in the absence of governor. Also, explain how isochronous governor can be incorporated together with turbine-generator with block diagram.

## Short answer

Without governor, the speed response transfer function is ΔΩ(s)/ΔPL(s) = -1/(2Hs+D), leading to a steady-state speed drop -ΔPL/D. An isochronous governor uses an integrator (K/s) in the forward path, forcing the steady-state speed error to zero by adjusting gate position until Pm equals PL.

## Answer

#### Transfer function without governor
Without a governor, the turbine gate is fixed, so $\Delta P_m = 0$ and the speed is decided only by rotor inertia and load damping.

**Inertia (rotating mass):** for small changes, in per unit, the swing equation is
$$
2H\frac{d\,\Delta\omega}{dt} = \Delta P_m - \Delta P_e
$$
where $H$ = inertia constant (s), $\Delta\omega$ = speed deviation (pu). Taking the Laplace transform,
$$
\Delta\Omega(s) = \frac{1}{2Hs}\big[\Delta P_m(s) - \Delta P_e(s)\big]
$$

**Load damping:** a composite load has frequency-insensitive parts (lighting, heating) and frequency-sensitive parts (motors). So
$$
\Delta P_e = \Delta P_L + D\,\Delta\omega
$$
where $\Delta P_L$ = non-frequency-sensitive load change and $D$ = load damping constant (% change in load per % change in frequency, typically 1–2).

Substituting,
$$
\begin{aligned}
2Hs\,\Delta\Omega(s) &= \Delta P_m(s) - \Delta P_L(s) - D\,\Delta\Omega(s) \\
\Delta\Omega(s) &= \frac{\Delta P_m(s) - \Delta P_L(s)}{2Hs + D}
\end{aligned}
$$

With $\Delta P_m = 0$:
$$
\frac{\Delta\Omega(s)}{\Delta P_L(s)} = \frac{-1}{2Hs + D}
$$

```text
 dPm=0  +    +--------------+
 ------>(+)->|      1       |----+--> d(omega)
         ^ - |   2Hs + D    |    |
         |   +--------------+    |
       dPL
```

For a step load $\Delta P_L$:
$$
\Delta\omega(t) = -\frac{\Delta P_L}{D}\left(1 - e^{-tD/2H}\right)
$$
The speed falls exponentially with time constant $2H/D$ and settles at $-\Delta P_L/D$, which is a large error because $D$ is small.

#### Incorporating an isochronous governor
An **isochronous** (constant speed) governor uses **integral control**. It measures speed, compares with the reference, and integrates the error to move the gate:
$$
\Delta Y(s) = -\frac{K}{s}\,\Delta\Omega(s)
$$
The gate keeps moving as long as there is any speed error, so speed returns exactly to the set value.

```text
                          dPL
                           | -
 +-------+  dPm            v      +-----------+
 |turbine|--------------->(+)---->| 1/(2Hs+D) |--+--> dw
 | Gt(s) |                 +      +-----------+  |
 +-------+                                       |
     ^ dY                                        |
 +-------+    -                                  |
 |  K/s  |<---(+)<-------------------------------+
 +-------+     ^ +
               | dw_ref = 0
```

Here $G_t(s)$ is the turbine transfer function (for hydro, $\frac{1 - T_w s}{1 + 0.5T_w s}$).

**Working:** a load rise makes speed fall; the integrator output grows, gates open, $P_m$ rises above $P_e$, the rotor accelerates and speed comes back to the reference. In steady state $\Delta\omega = 0$ and $\Delta P_m = \Delta P_L$.

The isochronous governor works well for a single unit supplying an isolated load. It cannot be used on several units in parallel, because each would try to fix the frequency at its own set point and they would fight for the load; for that, speed droop is added.

## Check yourself

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

1. Which equation relates speed deviation to the power imbalance in per unit?
   A. 2H d(dw)/dt = dPm - dPe
   B. H d(dw)/dt = dPm + dPe
   C. 2H dw = dPm - dPe
   D. d(dw)/dt = D(dPm - dPe)
   Answer: A. The small-signal swing equation is 2H d(dw)/dt = dPm - dPe, giving dW(s) = (dPm - dPe)/(2Hs). (Key point: Swing equation)

2. What does the damping constant D represent in dPe = dPL + D dw?
   A. Governor droop
   B. Frequency-sensitive part of the load, such as motors
   C. Water starting time
   D. Turbine time constant
   Answer: B. D accounts for loads (motors) whose power changes with frequency, typically 1 to 2 percent per percent. (Key point: Load damping)

3. With no governor (dPm = 0), what is the speed response to a step load dPL?
   A. Speed returns to zero error with integral action
   B. Rises exponentially to +dPL/D
   C. Falls exponentially with time constant 2H/D to -dPL/D
   D. Falls linearly without settling
   Answer: C. dW/dPL = -1/(2Hs+D) gives dw(t) = -(dPL/D)(1 - e^(-tD/2H)). (Key point: No-governor response)

4. How does an isochronous governor make steady-state speed error zero?
   A. Proportional gain on speed error
   B. Adds speed droop 1/R
   C. Fixes the gate position
   D. Integrates the speed error so the gate keeps moving until error is zero
   Answer: D. Integral control dY = -(K/s) dW moves the gate as long as any speed error exists. (Key point: Isochronous governor)

## Related in Power Plant Equipment

- [Transient response of turbine-generator with governor](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Agovernor-transient-response) — Power Plant Equipment, chapter 2, asked 3×
- [Isochronous governor and speed droop model](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aisochronous-droop-model) — Power Plant Equipment, chapter 2
- [Speed control by inertia and load damping](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Ainertia-damping-speed-control) — Power Plant Equipment, chapter 2, asked 6×
- [Isochronous generator transient response](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aisochronous-generator-transient) — Power Plant Equipment, chapter 2
- [Water turbine response to step gate change](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Awater-turbine-step-gate) — Power Plant Equipment, chapter 2, asked 2×
- [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×

## Related in other subjects

- [Armature-controlled DC motor transfer function and block diagram](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-armature-dc-motor) — Control Systems, chapter 2
- [PID controller block diagram, transfer function, characteristics](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Apid-bd-tf-characteristics) — Control Systems, chapter 7
- [Impulse response, transfer function and distortionless transmission](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adistortionless-transfer-function) — Communication Systems, chapter 1
- [DMA controller block diagram and transfer](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_organization_and_architecture%3Adma-controller) — Computer Organization and Architecture, chapter 7, asked 6×