---
title: "Turbine selection method and governing factors"
subject: "Power Plant Design"
chapter: 3
key: "semester_8/power_plant_design:turbine-selection-method-factors"
source: https://www.bctengineer.com/q/semester_8/power_plant_design/turbine-selection-method-factors
site: BCT Engineering
---

# Turbine selection method and governing factors

Power Plant Design · Chapter 3: Hydro Power Plant Design

- Asked in: 2078 Kartik, 4+4 marks
- Video: [Lecture Video6_18EE42_, Module-1_Water Turbines-Chacs, Governing & Selection,Units Size_R.Gunasekari](https://www.youtube.com/watch?v=1p8opVrJuXs) — Sairam EE Video
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/power_plant_design/turbine-selection-method-factors
- Whole chapter: https://www.bctengineer.com/md/semester_8/power_plant_design/chapter-3

## Question

What is scientific method of turbine selection? Mention major factors governing turbine during selection of appropriate turbine type.

## Short answer

Select turbine by specific speed Ns=N√P/H^1.25 and head-discharge data; choose type from Ns ranges (Pelton 10-35, Francis 60-300, Kaplan 300-1000). Governing factors: net head, discharge variation, part-load efficiency, cavitation, sediment, cost.

## Answer

#### Scientific method of turbine selection
Turbine selection by the scientific method is based on **specific speed** and the site's head–discharge data, checked against standard application charts.

1. Find **net head** $H$ and **design discharge** per unit $Q$ (e.g. $Q_{45}$ from the FDC less riparian release).
2. Find unit output $P = \eta_t \rho g Q H$.
3. Choose a synchronous speed $N = 120f/p$.
4. Calculate
$$
N_s = \frac{N\sqrt{P}}{H^{5/4}}
$$
(N in rpm, P in kW, H in m).
5. Select the type from the $N_s$ range: Pelton 10–35 (single jet), Francis 60–300, Kaplan 300–1000.
6. Cross-check with the head range (Pelton 50–1300 m, Francis 10–350 m, Kaplan 2–40 m) and the manufacturer's $H$–$Q$ chart; check cavitation (Thoma's $\sigma$) and part-load efficiency.

#### Major factors governing turbine selection
1. **Net head:** primary factor; fixes the broad type.
2. **Discharge and its variation:** low head–high flow suits Kaplan; high head–low flow suits Pelton.
3. **Specific speed and rotational speed:** higher speed gives smaller generator but more cavitation risk.
4. **Part-load efficiency:** Pelton and Kaplan have flat efficiency curves; Francis drops at low load.
5. **Cavitation and setting:** reaction turbines need submergence, adding excavation cost.
6. **Sediment content:** high silt erodes Francis runners; Pelton runners are easier to repair.
7. **Number of units and unit size**, transport limits.
8. **Cost, maintenance and runaway speed.**

#### Example
Net head 300 m, design flow 4 m³/s for one unit: $P \approx 9.81\times 4\times 300\times 0.9 \approx 10.6$ MW. At 600 rpm, $N_s = 600\sqrt{10595}/300^{5/4} \approx 49$, which is in the multi-jet Pelton range; a 2-jet or 4-jet Pelton (or a slow Francis) would be compared on cost and part-load efficiency.

## Check yourself

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

1. Which is the specific speed expression used in selection?
   A. Ns = N H^(5/4) / sqrt(P)
   B. Ns = N sqrt(P) / H^(5/4)
   C. Ns = N P / H
   D. Ns = sqrt(N) P / H^2
   Answer: B. Ns combines speed, power (kW) and head (m) into one number. (Key point: Specific speed formula)

2. Which Ns range corresponds to Francis?
   A. 10-35
   B. 300-1000
   C. 60-300
   D. 1000-2000
   Answer: C. Pelton is 10-35, Francis 60-300, Kaplan 300-1000. (Key point: Ns ranges)

3. Which statement on part-load efficiency is correct?
   A. Francis is flat; Pelton drops sharply
   B. All turbines are equally flat
   C. Kaplan drops below 80% load
   D. Pelton and Kaplan stay flat; Francis drops at low load
   Answer: D. Francis efficiency falls at low load, with vibration. (Key point: Part-load)

4. For H = 300 m and Q = 4 m3/s, what is the unit power (eta 0.9)?
   A. About 10.6 MW
   B. About 1.06 MW
   C. About 11.8 MW
   D. About 106 MW
   Answer: A. P = 9.81 x 4 x 300 x 0.9 = 10,595 kW. (Key point: Example)

## Close variants set in other exams

- [Criteria for turbine selection](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Aturbine-selection-criteria) — Power Plant Design, chapter 3, asked 3×

## Related in Power Plant Design

- [Scientific method of turbine selection](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Aturbine-selection-scientific) — Power Plant Design, chapter 3, asked 2×
- [Hydropower site selection factors](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Asite-selection-discharge-factors) — Power Plant Design, chapter 3

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