---
title: "Design of Pelton wheel for given power and head"
subject: "Fluid Machines"
chapter: 3
key: "semester_5/fluid_machines:pelton-wheel-design-numerical"
source: https://www.bctengineer.com/q/semester_5/fluid_machines/pelton-wheel-design-numerical
site: BCT Engineering
---

# Design of Pelton wheel for given power and head

Fluid Machines · Chapter 3: Water Turbine

- Asked in: year not given, 8 marks
- Video: [Design of Pelton wheel Turbine || Hydropower Engineering](https://www.youtube.com/watch?v=eSXfoRBETT4) — OPengineer
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/fluid_machines/pelton-wheel-design-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_5/fluid_machines/chapter-3

## Question

Design a Pelton wheel for a power station that must deliver 6000 kW shaft power under a net head of 350 m at 500 rpm. Take the nozzle coefficient 0.98, speed ratio 0.46 and overall efficiency 86 %. Use two jets. Find the diameter of the wheel, the diameter of each jet and the number of buckets, and check the specific speed.

## Short answer

Discharge Q = P/(ρgH η_o) = 2.032 m³/s. Jet velocity Vⱼ = 81.21 m/s, bucket speed u = 38.12 m/s. Wheel diameter D = 1.456 m. With two jets, d = 126.2 mm. Number of buckets Z = 15 + D/(2d) ≈ 21. Specific speed per jet N_s = 18.1, within Pelton range.

## Answer

**Given:** $P=6000$ kW, $H=350$ m, $N=500$ rpm, $C_v=0.98$, $\phi=0.46$, $\eta_o=0.86$, number of jets $n=2$.

#### Discharge

$$
\eta_o=\frac{P}{\rho g Q H}\ \Rightarrow\ Q=\frac{6\times10^{6}}{1000\times9.81\times350\times0.86}=2.032\ \text{m}^3/\text{s}
$$

#### Jet velocity and bucket speed

$\sqrt{2gH}=\sqrt{2\times9.81\times350}=82.87$ m/s.

$$
V_1=C_v\sqrt{2gH}=81.21\ \text{m/s},\qquad u=\phi\sqrt{2gH}=38.12\ \text{m/s}
$$

#### Wheel (pitch circle) diameter

$$
D=\frac{60u}{\pi N}=\frac{60\times38.12}{\pi\times500}=1.456\ \text{m}
$$

#### Jet diameter

Discharge per jet $=1.016$ m³/s.

$$
a=\frac{Q_j}{V_1}=12.511\times10^{-3}\ \text{m}^2,\qquad d=\sqrt{\frac{4a}{\pi}}=126.2\ \text{mm}
$$

Jet ratio $m=D/d=11.5$, which lies in the normal range of 10 to 14. Good.

#### Number of buckets

$$
Z=15+\frac{D}{2d}=15+\frac{11.5}{2}=20.8\ \Rightarrow\ Z=21\ \text{(rounded up) per wheel}
$$

#### Specific speed check

$$
N_s=\frac{N\sqrt{P}}{H^{5/4}}=\frac{500\sqrt{6000}}{350^{1.25}}=25.6\ \text{for the whole wheel}
$$

Based on the power per jet (3000 kW), $N_s=18.1$, which is within the Pelton range of about 10 to 35 (metric units, kW). The design is suitable.

**Answer:** $Q=2.032$ m³/s; wheel diameter $=1.456$ m; jet diameter $=126.2$ mm (two jets); buckets $=21$; $N_s=18.1$ per jet.

## Check yourself

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

1. Discharge is found from overall efficiency:
   A. Q = V1 a
   B. Q = P/(g H)
   C. Q = P/(rho H)
   D. Q = P/(rho g H eta_o) = 2.032 m3/s
   Answer: D. eta_o = P/(rho g Q H), so Q = 2.032 m3/s. (Key point: Discharge)

2. The wheel diameter is from:
   A. D = 60 V1/(pi N)
   B. D = 60u/(pi N) = 1.456 m
   C. D = V1/N
   D. D = 2u
   Answer: B. u = 0.46 x 82.87 = 38.12 m/s, so D = 1.456 m. (Key point: Wheel diameter)

3. With two jets, jet diameter is computed using:
   A. Q x 2
   B. Q/2 per jet over V1, giving 126.2 mm
   C. Total Q
   D. Q/u
   Answer: B. Q per jet = 1.016 m3/s; a = Q_j/V1, d = 126.2 mm. (Key point: Jet diameter)

4. Number of buckets Z uses:
   A. 15 + D/(2d)
   B. 2D/d
   C. 15 x d
   D. D/d
   Answer: A. Z = 15 + 11.5/2 = 20.8, rounded up to 21. (Key point: Number of buckets)

5. The jet ratio m = D/d = 11.5 is acceptable because:
   A. It equals 1
   B. It is above 50
   C. It is below 5
   D. It is within the normal range 10 to 14
   Answer: D. The normal range is 10 to 14. (Key point: Jet ratio)

## Related in Fluid Machines

- [Pelton wheel power with penstock losses](https://www.bctengineer.com/question.md?key=semester_5%2Ffluid_machines%3Apelton-wheel-numerical-penstock) — Fluid Machines, chapter 3
- [Pelton wheel: components and working](https://www.bctengineer.com/question.md?key=semester_5%2Ffluid_machines%3Apelton-wheel-working-components) — Fluid Machines, chapter 3

## Related in other subjects

- [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 wheel 70 kW: flow rate and diameter](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apelton-wheel-70kw-100m-400rpm) — Hydropower Engineering, chapter 7
- [Multi-jet Pelton wheel design (200 m, 4 m³/s)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apelton-multijet-200m-4cumec) — Hydropower Engineering, chapter 7
- [Pelton unit speed, discharge, power](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apelton-unit-quantities-190rpm) — Hydropower Engineering, chapter 7