---
title: "Pelton turbine design (6867 kW, 350 m)"
subject: "Hydropower Engineering"
chapter: 7
key: "semester_7/hydropower_engineering:pelton-design-6867kw-350m"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/pelton-design-6867kw-350m
site: BCT Engineering
---

# Pelton turbine design (6867 kW, 350 m)

Hydropower Engineering · Chapter 7: Hydro-electric Machines

- Asked in: 2075 Chaitra, 6 marks
- Video: [Pelton Turbine Design #peltonturbine](https://www.youtube.com/watch?v=GigXiCBF4BM) — Hydropower
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/pelton-design-6867kw-350m
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-7

## Question

A Pelton turbine has to be designed for the following data: power developed = 6867 kW, net head = 350 m, overall efficiency = 80%, speed = 550 rpm, coefficient of velocity ($K_v$) = 0.98 and speed ratio ($K_u$) = 0.46. Ratio of jet dia to wheel dia $(d/D) = 1:12$. Find discharge, number of jets, diameter of jet and diameter of wheel.

## Short answer

Discharge is 2.50 m3/s; with d/D=1/12 the required jets are 3.22, so use 4 jets; adjusted jet diameter is about 0.099 m and wheel diameter is 1.32 m.

## Answer

#### Data
$P = 6867$ kW, $H = 350$ m, $\eta_o = 0.80$, $N = 550$ rpm, $K_v = 0.98$, $K_u = 0.46$, $d/D = 1/12$.

#### Discharge
$$
Q = \frac{P}{\eta_o \gamma H} = \frac{6867}{0.8 \times 9.81 \times 350} = 2.50\ \text{m}^3/\text{s}
$$

#### Wheel diameter
$V_1 = K_v\sqrt{2gH} = 0.98\sqrt{2 \times 9.81 \times 350} = 81.21$ m/s

$u = K_u\sqrt{2gH} = 0.46 \times 82.87 = 38.12$ m/s

$$
D = \frac{60u}{\pi N} = \frac{60 \times 38.12}{\pi \times 550} = 1.324\ \text{m}
$$

#### Jet diameter and number of jets
From $d/D = 1/12$: $d = 1.324/12 = 0.1103$ m.

Discharge through one jet $= \dfrac{\pi}{4}(0.1103)^2 \times 81.21 = 0.776\ \text{m}^3/\text{s}$.

$$
z = \frac{2.50}{0.776} = 3.22 \Rightarrow 4\ \text{jets}
$$

With 4 jets the jet diameter is adjusted to carry the full discharge:

$$
d = \sqrt{\frac{4 \times 2.5/4}{\pi \times 81.21}} = 0.099\ \text{m} \approx 99\ \text{mm}
$$

so $D/d = 13.4$ (within the usual range 10 to 20). Specific speed: $N_s = \dfrac{550\sqrt{6867}}{350^{1.25}} = 30.1$.

**Answer: $Q = 2.50$ m³/s; 4 jets; jet diameter $\approx 0.099$ m; wheel diameter $= 1.32$ m.**

## Check yourself

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

1. How is the discharge found?
   A. Q = P/(eta_o V1)
   B. Q = P/(gamma H)
   C. Q = P/(eta_o gamma H) = 2.50 m3/s
   D. Q = P eta_o/(gamma H)
   Answer: C. Overall efficiency relates shaft power to water power: Q = 6867/(0.8 x 9.81 x 350). (Key point: Discharge)

2. Which velocity is used for bucket speed u = Ku sqrt(2gH)?
   A. Jet velocity Kv sqrt(2gH) = 81.21 m/s
   B. Penstock velocity
   C. Theoretical velocity sqrt(2gH) = 82.87 m/s
   D. Runaway velocity
   Answer: C. Speed ratio Ku is defined on sqrt(2gH), so u = 0.46 x 82.87 = 38.12 m/s and D = 60u/(pi N) = 1.324 m. (Key point: Wheel diameter)

3. How are the number of jets determined?
   A. Equal to D/d
   B. Always one
   C. Equal to the number of poles
   D. Total Q divided by discharge in one jet (d = D/12 at V1), rounded up to 4
   Answer: D. One jet of 0.1103 m carries 0.776 m3/s, so 2.5/0.776 = 3.22, rounded up to 4 jets. (Key point: Number of jets)

4. With 4 jets, why is the jet diameter recalculated as 99 mm?
   A. Because V1 changes with jets
   B. To reduce the wheel diameter
   C. To make D/d equal to 12
   D. So the four jets together carry the full discharge Q
   Answer: D. Each of 4 jets carries 2.5/4 m3/s, which gives d = sqrt(4q/(pi V1)) = 0.099 m. (Key point: Jet diameter adjustment)

## Related in Hydropower Engineering

- [Pelton unit speed, discharge, power](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apelton-unit-quantities-190rpm) — 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
- [Turbine setting; three single-jet Pelton wheels](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aturbine-setting-pelton-three-jets) — 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
- [Francis runner design (40 m head, 10 m³/s)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Afrancis-turbine-velocity-triangle-40m) — 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

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