---
title: "Pelton turbine design (2 m³/s, 500 m)"
subject: "Hydro Power"
chapter: 7
key: "semester_6/hydro_power:pelton-design-500m"
source: https://www.bctengineer.com/q/semester_6/hydro_power/pelton-design-500m
site: BCT Engineering
---

# Pelton turbine design (2 m³/s, 500 m)

Hydro Power · Chapter 7: Hydro-Electric Machines

- Asked in: 2080 Asoj, 8 marks
- Video: [Pelton wheel problems | Solved problems in Pelton Wheel Turbine | Fluid mechanics and Machinery](https://www.youtube.com/watch?v=-SxzvLuk6mc) — Amizhdhagam
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/pelton-design-500m
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-7

## Question

Design a pelton wheel turbine with design discharge of 2 m³/s, overall efficiency of 85% and a net head of 500 m.

## Short answer

Single-jet Pelton wheel: power 8.34 MW at 428.57 rpm (14 poles). Jet diameter 162 mm, pitch circle diameter 2.03 m, jet ratio 12.54. Specific speed 16.6. 22 buckets; bucket width 0.81 m, depth 0.19 m.

## Answer

A Pelton wheel is designed by finding the jet velocity and jet diameter, choosing a synchronous speed that gives a suitable jet ratio $m = D/d$ (10 to 14 for good efficiency), and then sizing the wheel and buckets.

**Data:** $Q = 2$ m³/s, $H = 500$ m, $\eta_o = 0.85$, $f = 50$ Hz. Assumed from the standard ranges: $C_v = 0.98$ (0.97 to 0.99) and speed ratio $\phi = 0.46$ (0.43 to 0.47).

#### 1. Power developed

$$
P = \eta_o \rho g Q H = 0.85 \times 9.81 \times 2 \times 500 = 8338.5\ \text{kW}
$$

#### 2. Jet velocity and bucket (peripheral) velocity

$$
\begin{aligned}
\sqrt{2gH} &= \sqrt{2 \times 9.81 \times 500} = 99.05\ \text{m/s} \\
V_1 &= C_v\sqrt{2gH} = 0.98 \times 99.05 = 97.06\ \text{m/s} \\
u &= \phi\sqrt{2gH} = 0.46 \times 99.05 = 45.56\ \text{m/s}
\end{aligned}
$$

#### 3. Jet diameter (single jet)

$$
\begin{aligned}
Q &= \frac{\pi}{4} d^2 V_1 \\
d &= \sqrt{\frac{4 \times 2}{\pi \times 97.06}} = 0.1620\ \text{m} \approx 162\ \text{mm}
\end{aligned}
$$

#### 4. Speed of the wheel

Trial with jet ratio $m = 12$: $D = 12d = 1.944$ m, so

$$
N = \frac{60u}{\pi D} = \frac{60 \times 45.56}{\pi \times 1.944} = 447.7\ \text{rpm}
$$

Nearest synchronous speed for 50 Hz: $N = 120f/p = 6000/14 = 428.57$ rpm (14 poles).

#### 5. Wheel (pitch circle) diameter and jet ratio

$$
\begin{aligned}
D &= \frac{60u}{\pi N} = \frac{60 \times 45.56}{\pi \times 428.57} = 2.030\ \text{m} \\
m &= \frac{D}{d} = \frac{2.030}{0.1620} = 12.54
\end{aligned}
$$

$m$ lies within 10 to 14, so a **single-jet** wheel is satisfactory.

#### 6. Specific speed check

$$
N_s = \frac{N\sqrt{P}}{H^{5/4}} = \frac{428.57 \times 91.32}{2364.4} = 16.6
$$

This is within the single-jet Pelton range (about 8.5 to 30), so the Pelton wheel is the correct choice.

#### 7. Number of buckets (Tygun's formula)

$$
z = 15 + \frac{D}{2d} = 15 + \frac{12.54}{2} = 21.27 \approx 22
$$

#### 8. Bucket size (usual proportions)

- Bucket width $B \approx 5d = 0.810$ m
- Bucket depth $T \approx 1.2d = 0.194$ m
- Bucket radial length $L \approx 2.5d = 0.405$ m

#### Summary of design

| Item | Value |
|---|---|
| Power | 8.34 MW |
| Speed | 428.57 rpm |
| Number of jets | 1 |
| Jet diameter $d$ | 162 mm |
| Wheel diameter $D$ | 2.03 m |
| Jet ratio $m$ | 12.54 |
| Specific speed $N_s$ | 16.6 |
| Number of buckets | 22 |
| Bucket width × depth | 0.81 m × 0.19 m |

## Check yourself

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

1. How is the power developed by the Pelton turbine calculated?
   A. P = rho g Q H with no efficiency
   B. P = eta x Q x H / g
   C. P = eta x 0.5 rho Q V1^2 with V1 = sqrt(2gH) only, ignoring head
   D. P = overall efficiency x rho g Q H = 0.85 x 9.81 x 2 x 500 = 8338.5 kW
   Answer: D. Output power is the hydraulic power times the overall efficiency. (Key point: Power developed)

2. How are the jet velocity V1 and bucket velocity u found?
   A. V1 = Cv sqrt(2gH) = 97.06 m/s and u = phi sqrt(2gH) = 45.56 m/s
   B. V1 = sqrt(2gH) with no coefficient, and u = V1
   C. V1 = phi sqrt(2gH) and u = Cv sqrt(2gH)
   D. V1 = sqrt(gH) and u = 2 V1
   Answer: A. The jet velocity uses the velocity coefficient Cv (0.98), the bucket velocity uses the speed ratio phi (0.46), both times sqrt(2gH). (Key point: Jet and bucket velocities)

3. How is the jet diameter d found for a single jet?
   A. From Q = d V1, so d = Q/V1
   B. From Q = (pi/4) d^2 V1, so d = sqrt(4Q/(pi V1)) = 0.1620 m (162 mm)
   C. From Q = pi d^2 u, using the bucket velocity
   D. From P = pi d^2 V1, using the power
   Answer: B. The discharge equals jet area times jet velocity. (Key point: Jet diameter)

4. How are the wheel speed and diameter fixed?
   A. Use any speed and accept any m, because only d matters
   B. Take D = 12d always and N follows with no synchronous restriction
   C. Pick a synchronous speed N = 120f/p (428.57 rpm), then D = 60u/(pi N) = 2.030 m, giving m = D/d = 12.54, within 10 to 14
   D. Choose N from Ns = 100 and ignore m
   Answer: C. The speed must suit the generator, and the resulting jet ratio D/d must lie in 10 to 14 for good efficiency. (Key point: Speed and jet ratio)

5. What do the specific speed check and bucket count show?
   A. Ns = 16.6 is above 30, so several jets are needed; z = 15 + m = 26 buckets
   B. Ns must equal 100 exactly; z = D/d buckets
   C. Ns is irrelevant to Pelton choice; z = 2m = 25 buckets
   D. Ns = N sqrt(P)/H^(5/4) = 16.6, inside the single-jet range 8.5 to 30; z = 15 + m/2 = 21.27, so 22 buckets
   Answer: D. The Pelton wheel suits low specific speed (about 8.5 to 30 for a single jet); bucket number follows Tygun's formula. (Key point: Specific speed and buckets)

## Close variants set in other exams

- [Pelton turbine design (1.8 m³/s, 605 m)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Apelton-design-605m-1-8) — Hydro Power, chapter 7, asked 2×

## Related in Hydro Power

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- [Pelton turbine design (5 m³/s, 310 m)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Apelton-design-310m) — Hydro Power, chapter 7
- [Pelton turbine design (2 m³/s, 605 m, Ns 100)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Apelton-design-605m-ns100) — Hydro Power, chapter 7
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## Related in other subjects

- [Design of Pelton wheel for given power and head](https://www.bctengineer.com/question.md?key=semester_5%2Ffluid_machines%3Apelton-wheel-design-numerical) — Fluid Machines, chapter 3
- [Pelton wheel design (310 m, 5 m³/s)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apelton-design-310m-5cumec-specific-speed) — Hydropower Engineering, chapter 7
- [Pelton turbine design (6867 kW, 350 m)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apelton-design-6867kw-350m) — Hydropower Engineering, chapter 7
- [Pelton turbine design (312.5 m, 5 m³/s)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apelton-design-312-5m-5cumec) — Hydropower Engineering, chapter 7