---
title: "Number of jets of Pelton wheel (300 m, 1.5 m³/s)"
subject: "Hydropower Engineering"
chapter: 7
key: "semester_7/hydropower_engineering:pelton-jets-300m-1-5cumec"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/pelton-jets-300m-1-5cumec
site: BCT Engineering
---

# Number of jets of Pelton wheel (300 m, 1.5 m³/s)

Hydropower Engineering · Chapter 7: Hydro-electric Machines

- Asked in: 2072 Kartik, 4 marks
- Video: [Numerical problem|| Pelton wheel turbine|| hydraulic efficiency of turbine|| Technical classes](https://www.youtube.com/watch?v=yMXMgqgu1Jk) — TECHNICAL CLASSES
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/pelton-jets-300m-1-5cumec
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-7

## Question

Water is being supplied to a Pelton wheel under a head of 300 m through 100 mm diameter pipes. If the quantity of water supplied to the wheel is 1.50 m³/s, find the number of jets in the wheel. Assume coefficient of velocity is 0.96.

## Short answer

Jet velocity is 73.65 m/s and each 100 mm jet carries 0.5785 m3/s; 1.5 m3/s divided by this gives 2.59, so use 3 jets.

## Answer

#### Interpretation
The 100 mm diameter is taken as the diameter of each jet (nozzle outlet). Head $H = 300$ m, $C_v = 0.96$, total $Q = 1.5\ \text{m}^3/\text{s}$.

#### Velocity of jet
$$
V_1 = C_v\sqrt{2gH} = 0.96\sqrt{2 \times 9.81 \times 300} = 73.65\ \text{m/s}
$$

#### Discharge through one jet
$$
q = \frac{\pi}{4}d^2V_1 = \frac{\pi}{4}(0.1)^2 \times 73.65 = 0.5785\ \text{m}^3/\text{s}
$$

#### Number of jets
$$
n = \frac{Q}{q} = \frac{1.5}{0.5785} = 2.59 \approx 3
$$

**Answer: 3 jets (the next whole number above 2.59).**

## Check yourself

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

1. What is the 100 mm diameter taken to be in the solution?
   A. The wheel diameter
   B. The diameter of each jet at the nozzle outlet
   C. The nozzle inlet diameter
   D. The penstock diameter
   Answer: B. The answer takes 100 mm as the jet diameter, so each jet has area pi/4 x 0.1^2. (Key point: Interpretation of 100 mm)

2. What is the jet velocity formula and result?
   A. sqrt(2gH) = 76.7 m/s
   B. Cv sqrt(2gH) = 73.65 m/s
   C. Cv sqrt(gH) = 52 m/s
   D. Cv H = 288 m/s
   Answer: B. V1 = 0.96 sqrt(2 x 9.81 x 300) = 73.65 m/s. (Key point: Jet velocity)

3. How is 2.59 treated to get the number of jets?
   A. Doubled to 5
   B. Rounded down to 2
   C. Rounded up to the next whole number, 3
   D. Left as 2.59
   Answer: C. Jets must be whole and must carry at least Q, so n = Q/q = 2.59 becomes 3. (Key point: Number of jets)

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