---
title: "Multi-jet Pelton wheel design (200 m, 4 m³/s)"
subject: "Hydropower Engineering"
chapter: 7
key: "semester_7/hydropower_engineering:pelton-multijet-200m-4cumec"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/pelton-multijet-200m-4cumec
site: BCT Engineering
---

# Multi-jet Pelton wheel design (200 m, 4 m³/s)

Hydropower Engineering · Chapter 7: Hydro-electric Machines

- Asked in: 2078 Bhadra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/pelton-multijet-200m-4cumec
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-7

## Question

Consider the design of a multi-jet Pelton wheel with parameters and operating conditions as given below: head = 200 m; flow rate = 4 m³/s; nozzle velocity coefficient = 0.98; wheel dia. = 1.47 m; mechanical efficiency = 86%; blade speed to jet speed ratio = 0.47; jet dia. to wheel dia. ratio = 0.113. (i) Calculate the wheel rotational speed (rev/min). (ii) Calculate the power output (MW). (iii) Determine the number of nozzles required. (iv) Calculate the specific speed of the machine.

## Short answer

Jet velocity V = 61.39 m/s, bucket speed u = 28.85 m/s gives N = 374.9 rpm. Power output = 0.86×9.81×4×200 = 6.75 MW. Jet diameter from ratio 0.113×1.47 = 0.166 m, flow per jet = 1.33 m³/s, so 3 nozzles. Specific speed Ns = N√P/H^1.25 = 40.9.

## Answer

#### Data
$H = 200$ m, $Q = 4\ \text{m}^3/\text{s}$, $C_v = 0.98$, $D = 1.47$ m, $\eta_m = 0.86$, $\phi = 0.47$, $d/D = 0.113$.

Jet velocity: $V_1 = C_v\sqrt{2gH} = 0.98\sqrt{2 \times 9.81 \times 200} = 61.39$ m/s.

#### (i) Rotational speed
$u = \phi V_1 = 0.47 \times 61.39 = 28.85$ m/s

$$
N = \frac{60u}{\pi D} = \frac{60 \times 28.85}{\pi \times 1.47} = 374.9\ \text{rpm}
$$

#### (ii) Power output
Taking the 86 % efficiency as the efficiency from the water power to the shaft:

$$
P = \eta\,\gamma Q H = 0.86 \times 9.81 \times 4 \times 200 = 6749\ \text{kW} = 6.75\ \text{MW}
$$

#### (iii) Number of nozzles
Jet diameter $d = 0.113 \times 1.47 = 0.1661$ m; jet area $= 0.02167\ \text{m}^2$.

Discharge per jet $q = 0.02167 \times 61.39 = 1.330\ \text{m}^3/\text{s}$.

$$
n = \frac{Q}{q} = \frac{4}{1.330} = 3.01 \approx 3\ \text{nozzles}
$$

#### (iv) Specific speed
$$
N_s = \frac{N\sqrt{P}}{H^{5/4}} = \frac{374.9\sqrt{6749}}{200^{1.25}} = 40.9
$$

**Answer: $N = 374.9$ rpm; $P = 6.75$ MW; 3 nozzles; $N_s = 40.9$ (kW, m, rpm).**

## Check yourself

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

1. What is the jet velocity used in the design?
   A. Cv H = 196 m/s
   B. Q/A of the wheel
   C. sqrt(2gH) = 62.6 m/s
   D. Cv sqrt(2gH) = 61.39 m/s
   Answer: D. V1 = 0.98 sqrt(2 x 9.81 x 200) = 61.39 m/s. (Key point: Jet velocity)

2. How is the rotational speed found?
   A. N = 60 V1/(pi D)
   B. N = 120 f/p with p unknown
   C. u = phi V1, then N = 60u/(pi D) = 374.9 rpm
   D. N = Ns H^(5/4)/sqrt(P) with Ns unknown
   Answer: C. Bucket speed u = 0.47 x 61.39 = 28.85 m/s, so N = 60u/(pi D). (Key point: Wheel speed)

3. How is the power output computed with the given 86% efficiency?
   A. P = 0.86 gamma Q H / 2
   B. P = gamma Q H / 0.86
   C. P = 0.86 x Q V1^2
   D. P = 0.86 gamma Q H = 6.75 MW
   Answer: D. Output = efficiency x water power = 0.86 x 9.81 x 4 x 200 = 6749 kW. (Key point: Power output)

4. How is the number of nozzles determined?
   A. Wheel diameter divided by jet diameter
   B. Total Q divided by discharge of one jet (area from d = 0.113D times V1), about 3
   C. Total power divided by 1 MW
   D. Specific speed divided by 10
   Answer: B. One jet carries 0.02167 x 61.39 = 1.33 m3/s, so n = 4/1.33 = 3. (Key point: Number of nozzles)

## Related in Hydropower Engineering

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