---
title: "Specific speed of a turbine"
subject: "Power Plant Design"
chapter: 3
key: "semester_8/power_plant_design:specific-speed-purpose"
source: https://www.bctengineer.com/q/semester_8/power_plant_design/specific-speed-purpose
site: BCT Engineering
---

# Specific speed of a turbine

Power Plant Design · Chapter 3: Hydro Power Plant Design

- Asked in: 2073 Bhadra, 6 marks; 2071 Magh, 6 marks (asked 2 times)
- Video: [Specific Speed of Turbine in hindi || Fluid mechanics in Hindi](https://www.youtube.com/watch?v=5NvcyX6z8BQ) — Gear Institute Mechanical Engineering Videos
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/power_plant_design/specific-speed-purpose
- Whole chapter: https://www.bctengineer.com/md/semester_8/power_plant_design/chapter-3

## Question

How the 'specific speed' is defined for a turbine? Discuss its purpose.

## Short answer

Specific speed Ns is the rpm of a geometrically similar turbine producing 1 kW under 1 m head. It classifies turbines (Pelton 10-35, Francis 60-300, Kaplan 300-1000), enables model-to-prototype scaling, and guides speed selection and cavitation check.

## Answer

**Specific speed** ($N_s$) of a turbine is the speed at which a geometrically similar model turbine would run if it produced **1 kW of power under a head of 1 m**. It is a type number that describes the shape of the runner.

#### Derivation (outline)
For similar turbines, the jet/flow velocity varies as $V \propto \sqrt{H}$ and the peripheral speed $u = \pi D N/60 \propto \sqrt{H}$, so $D \propto \sqrt{H}/N$. Discharge $Q \propto D^2\sqrt{H}$ and power $P \propto QH \propto D^2 H^{3/2}$. Substituting $D$:
$$
\begin{aligned}
P &\propto \frac{H}{N^2}\,H^{3/2} = \frac{H^{5/2}}{N^2} \\
\Rightarrow N &= K\,\frac{H^{5/4}}{\sqrt{P}}
\end{aligned}
$$
When $P = 1$ kW and $H = 1$ m, $N = K = N_s$. Hence
$$
N_s = \frac{N\sqrt{P}}{H^{5/4}}
$$
with $N$ in rpm, $P$ in kW and $H$ in m.

#### Purpose
1. **Selection of turbine type:** each type works best in a band of $N_s$:

| Turbine | $N_s$ range (kW, m) |
|---|---|
| Pelton (single jet) | 10–35 |
| Pelton (multi jet) | 35–60 |
| Francis | 60–300 |
| Kaplan / propeller | 300–1000 |

2. **Comparison of turbines:** turbines of different size are compared on one scale.
3. **Model testing:** a model and prototype with the same $N_s$ are geometrically and dynamically similar, so model results can be scaled up.
4. **Fixing speed and unit size:** for a given head and power, choosing $N_s$ fixes the rotational speed, and hence the generator poles and cost.
5. **Cavitation check:** the Thoma critical $\sigma$ rises with $N_s$, so $N_s$ fixes how low the runner must be set.

**Example:** $N = 600$ rpm, $P = 5000$ kW, $H = 200$ m gives $N_s = 600\sqrt{5000}/200^{1.25} \approx 56.4$, which lies at the boundary of multi-jet Pelton and slow Francis, so both would be compared.

## Check yourself

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

1. Specific speed is the speed of a similar model that produces:
   A. 1 MW under 1 m head
   B. 1 kW under 1 m head
   C. 1 kW under 100 m head
   D. 1 hp under 10 m head
   Answer: B. Ns is the speed at 1 kW power and 1 m head. (Key point: Definition)

2. From similarity, power P is proportional to:
   A. H^(5/2) / N^2
   B. H^2 / N
   C. N^2 / H^(5/2)
   D. H^(3/2) N
   Answer: A. With D proportional to sqrt(H)/N, P proportional to D^2 H^(3/2) gives H^(5/2)/N^2. (Key point: Derivation)

3. A turbine with Ns of 400 is most likely:
   A. Pelton
   B. Francis
   C. Kaplan / propeller
   D. Multi-jet Pelton
   Answer: C. Kaplan/propeller range is 300-1000. (Key point: Selection)

4. Why is Ns useful for model testing?
   A. Same Ns means similar shape so model results scale to the prototype
   B. It removes the need for models
   C. It gives the cost
   D. It measures sediment
   Answer: A. A model and prototype with equal Ns are geometrically and dynamically similar. (Key point: Model testing)

5. For N = 600 rpm, P = 5000 kW, H = 200 m, Ns is about:
   A. 5.6
   B. 56
   C. 560
   D. 5600
   Answer: B. Ns = 600 x 70.7 / 752 = 56.4, at the multi-jet Pelton and slow Francis boundary. (Key point: Calculation)

## Related in Power Plant Design

- [Head, specific speed and part load in turbine choice](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Aturbine-head-speed-part-load) — Power Plant Design, chapter 3
- [Low specific speed reaction turbines](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Alow-specific-speed-reaction) — Power Plant Design, chapter 3
- [Wind turbine output: 12 m blade, 18 m/s](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Awind-numerical-12m-18ms) — Power Plant Design, chapter 1, asked 3×
- [Wind turbine output: 6.5 m blade, 9 and 18 m/s](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Awind-numerical-6-5m-9ms) — Power Plant Design, chapter 1, asked 3×
- [Wind turbine output: 6 m blade, 8 and 16 m/s](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Awind-numerical-6m-8ms) — Power Plant Design, chapter 1, asked 3×

## Related in other subjects

- [Specific speed of a turbine](https://www.bctengineer.com/question.md?key=semester_5%2Ffluid_machines%3Aspecific-speed-turbine-derivation) — Fluid Machines, chapter 3
- [Model testing and specific speed of turbine](https://www.bctengineer.com/question.md?key=semester_5%2Ffluid_machines%3Aturbine-model-similarity-numerical) — Fluid Machines, chapter 3
- [Purpose of governor and governing of Francis turbine](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Agoverning-francis-turbine) — Hydro Power, chapter 7
- [Pump specific speed and affinity laws](https://www.bctengineer.com/question.md?key=semester_5%2Ffluid_machines%3Apump-specific-speed-affinity-numerical) — Fluid Machines, chapter 4