---
title: "Pelton wheel 70 kW: flow rate and diameter"
subject: "Hydropower Engineering"
chapter: 7
key: "semester_7/hydropower_engineering:pelton-wheel-70kw-100m-400rpm"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/pelton-wheel-70kw-100m-400rpm
site: BCT Engineering
---

# Pelton wheel 70 kW: flow rate and diameter

Hydropower Engineering · Chapter 7: Hydro-electric Machines

- Asked in: 2079 Baishakh, 2+2 marks
- Video: [Design of Pelton wheel Turbine || Hydropower Engineering](https://www.youtube.com/watch?v=eSXfoRBETT4) — OPengineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/pelton-wheel-70kw-100m-400rpm
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-7

## Question

A Pelton wheel develops 70 kW under a head of 100 m of water, it rotates at 400 rev/min. The diameter of the penstock is 200 mm. The ratio of bucket speed to jet velocity is 0.46 and overall efficiency of the installation is 85%. Calculate (i) volumetric flow rate, (ii) wheel diameter.

## Short answer

Volumetric flow rate Q = P/(ηγH) = 0.0839 m³/s. Jet velocity V ≈ 44.29 m/s, bucket speed u = 0.46V = 20.37 m/s; wheel diameter D = 60u/(πN) ≈ 0.97 m.

## Answer

#### Data
$P = 70$ kW (shaft), $H = 100$ m, $N = 400$ rpm, $\eta_o = 0.85$, speed ratio $\phi = u/V_1 = 0.46$, penstock 200 mm.

#### (i) Volumetric flow rate
$$
Q = \frac{P}{\eta_o \gamma H} = \frac{70}{0.85 \times 9.81 \times 100} = 0.0839\ \text{m}^3/\text{s}
$$

($= 83.9$ litres per second; velocity in the 200 mm penstock $= 2.67$ m/s.)

#### (ii) Wheel diameter
Jet velocity (taking $C_v \approx 1$, as no nozzle coefficient is given): $V_1 = \sqrt{2gH} = \sqrt{2 \times 9.81 \times 100} = 44.29$ m/s.

Bucket speed $u = 0.46 \times 44.29 = 20.37$ m/s.

$$
D = \frac{60u}{\pi N} = \frac{60 \times 20.37}{\pi \times 400} = 0.972\ \text{m}
$$

(If $C_v = 0.98$ were used, $V_1 = 43.40$ m/s, $u = 19.97$ m/s and $D = 0.953$ m.)

**Answer: $Q = 0.0839$ m³/s; wheel diameter $\approx 0.97$ m.**

## Check yourself

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

1. How is the flow rate found from the shaft power?
   A. Q = P/(gamma H) ignoring efficiency
   B. Q = P/(eta_o sqrt(2gH))
   C. Q = P eta_o/(gamma H)
   D. Q = P/(eta_o gamma H)
   Answer: D. Shaft power = eta_o gamma Q H, so Q = 70/(0.85 x 9.81 x 100) = 0.0839 m3/s. (Key point: Discharge from power)

2. How is the bucket speed u obtained?
   A. u = Q/A of the penstock
   B. u = V1 / 0.46
   C. u = 0.46 x V1 where V1 = sqrt(2gH) (Cv about 1)
   D. u = pi D N only
   Answer: C. Speed ratio u/V1 = 0.46 with V1 = 44.29 m/s gives u = 20.37 m/s. (Key point: Bucket speed)

3. Which formula gives the wheel diameter from bucket speed?
   A. D = 60 u/(pi N)
   B. D = pi N/(60 u)
   C. D = 2u/N
   D. D = u/(N)
   Answer: A. u = pi D N/60, so D = 60u/(pi N) = 0.972 m. (Key point: Wheel diameter)

## Related in Hydropower Engineering

- [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
- [Number of jets of Pelton wheel (300 m, 1.5 m³/s)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apelton-jets-300m-1-5cumec) — 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
- [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 unit speed, discharge, power](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apelton-unit-quantities-190rpm) — Hydropower Engineering, chapter 7
- [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

## Related in other subjects

- [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
- [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
- [Water jet on moving blade, 2 in diameter](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Awater-jet-moving-blade-2in) — Applied Mechanics (Dynamics), chapter 3
- [Water hammer in 200 mm pipe](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Awater-hammer-200mm-pipe) — Hydro Power, chapter 1