---
title: "Hydraulic turbine characteristics and water starting time"
subject: "Power Plant Equipment"
chapter: 2
key: "semester_7/power_plant_equipment:hydraulic-turbine-water-starting"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/hydraulic-turbine-water-starting
site: BCT Engineering
---

# Hydraulic turbine characteristics and water starting time

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

- Asked in: 2080 Baisakh, 5 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/hydraulic-turbine-water-starting
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-2

## Question

Explain the special characteristics of a hydraulic turbine and what is water starting time?

## Short answer

Hydraulic turbines exhibit non-minimum phase behaviour: initial power change opposite to gate movement due to water inertia, requiring transient droop and slow gate rates to avoid water hammer. Water starting time Tw = L U0/(g H0) is the time needed by head H0 to accelerate water to velocity U0, typically 0.5-4 s.

## Answer

A hydraulic turbine responds to gate movement in an unusual way because of the inertia of the water column in the penstock.

#### Special characteristics
1. **Non-minimum phase behaviour:** the turbine transfer function
$$
\frac{\Delta P_m(s)}{\Delta G(s)} = \frac{1 - T_w s}{1 + 0.5T_w s}
$$
has a zero in the right half of the s-plane ($s = +1/T_w$). So the power first changes in the **opposite** direction to the gate movement.
2. **Initial reverse power:** when the gate is suddenly opened, the flow cannot rise at once because of water inertia, so the head (pressure) at the turbine falls and power first drops to **−2 times** the gate change. It then rises to +1 times the gate change with time constant $T_w/2$: $\Delta P_m(t) = [1 - 3e^{-2t/T_w}]\Delta G$. Closing the gate first gives a power (and pressure) rise.
3. **Water inertia governs the response:** the response speed is set by the water starting time $T_w$, which is not constant; it rises with load because $U_0$ increases.
4. **Need for transient droop:** a governor with only permanent droop would be unstable. Hydro governors use a **large temporary (transient) droop with a reset time**, so they respond slowly at first and are slower than steam unit governors.
5. **Water hammer and gate rate limits:** gate movements must be slow enough to keep pressure rise in the penstock safe.

```text
 dPm
  1 |                ____________
    |            _.-'
  0 |---.     _-'
    |   |   .'
 -2 |   |_/        (unit step in gate)
    +---+----------------------> t
```

#### Water starting time
**Water starting time** $T_w$ is the time needed for the head $H_0$ to accelerate the water in the penstock from standstill to the velocity $U_0$:
$$
T_w = \frac{L\,U_0}{g\,H_0}
$$
where $L$ = length of penstock (m), $U_0$ = water velocity (m/s), $H_0$ = head at the gate (m), $g = 9.81$ m/s². It varies with load and is typically **0.5–4 s** at full load. A large $T_w$ means a stronger reverse-power effect and harder speed control. A surge tank close to the power house reduces $L$ and so reduces $T_w$.

## Check yourself

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

1. What is the effect of the right-half-plane zero at s = 1/Tw?
   A. Power initially changes opposite to gate movement
   B. Power responds faster than the gate
   C. The system has no steady state
   D. Power oscillates forever
   Answer: A. The zero causes initial reverse power. (Key point: Non-minimum phase)

2. For a sudden unit gate opening, power first goes to...
   A. +1 times the gate change
   B. -2 times the gate change
   C. -1 times the gate change
   D. +3 times the gate change
   Answer: B. Head falls as flow cannot rise at once, giving -2 then settling to +1. (Key point: Initial reverse power)

3. What does Tw represent?
   A. Gate opening time
   B. Governor reset time
   C. Time for the head to accelerate water from rest to velocity U0
   D. Rotor inertia time
   Answer: C. Tw = L U0/(g H0). (Key point: Water starting time)

4. How can Tw be reduced?
   A. Longer penstock
   B. Higher water velocity
   C. Lower head
   D. A surge tank near the power house, shortening L
   Answer: D. Tw is proportional to L, and a surge tank effectively shortens the penstock. (Key point: Reducing Tw)

## Related in Power Plant Equipment

- [Special characteristics of hydraulic turbine](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Ahydraulic-turbine-characteristics) — Power Plant Equipment, chapter 2
- [Water turbine transfer function and characteristics](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Awater-turbine-tf-characteristics) — Power Plant Equipment, chapter 2, asked 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×
- [Control of water delivery to turbine](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Awater-delivery-control) — Power Plant Equipment, chapter 1

## Related in other subjects

- [Hydroelectric schemes by hydraulic characteristics](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Ahydroelectric-scheme-types-hydraulic) — Hydropower Engineering, chapter 2
- [Compressor and turbine performance characteristics](https://www.bctengineer.com/question.md?key=semester_7%2Fturbo_machines%3Acompressor-turbine-performance-characteristics) — Turbo Machines, chapter 3