---
title: "Pelton turbine design (312.5 m, 5 m³/s)"
subject: "Hydropower Engineering"
chapter: 7
key: "semester_7/hydropower_engineering:pelton-design-312-5m-5cumec"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/pelton-design-312-5m-5cumec
site: BCT Engineering
---

# Pelton turbine design (312.5 m, 5 m³/s)

Hydropower Engineering · Chapter 7: Hydro-electric Machines

- Asked in: 2080 Bhadra, 6 marks
- Video: [Pelton wheel problems | Solved problems in Pelton Wheel Turbine | Fluid mechanics and Machinery](https://www.youtube.com/watch?v=-SxzvLuk6mc) — Amizhdhagam
- 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-312-5m-5cumec
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-7

## Question

Design a Pelton turbine for a hydropower plant having net head 312.5 m and discharge 5 cumec. Take efficiency of turbine 85%, frequency 50 Hz and velocity coefficient 0.98.

## Short answer

Power P = 0.85×9.81×5×312.5 = 13.03 MW. Jet velocity V = 76.74 m/s, bucket speed u = 0.46×√(2gH) = 36.02 m/s. With 2 jets, synchronous speed N = 333.3 rpm (18 poles) gives wheel diameter D = 2.06 m; jet diameter d = 0.204 m (D/d ≈ 10.1); about 20 buckets; specific speed Ns = 29.0.

## Answer

#### Step 1: Power
$$
P = \eta\gamma Q H = 0.85 \times 9.81 \times 5 \times 312.5 = 13{,}029\ \text{kW}
$$

#### Step 2: Jet velocity and bucket speed
$V_1 = C_v\sqrt{2gH} = 0.98\sqrt{2 \times 9.81 \times 312.5} = 76.74$ m/s

Speed ratio $K_u = u/\sqrt{2gH} = 0.46$ (usual range 0.43 to 0.47) with $\sqrt{2gH} = 78.30$ m/s: $u = 0.46 \times 78.30 = 36.02$ m/s

#### Step 3: Number of jets and speed
For a single jet, the empirical (Siervo and Lugaresi) specific speed is $N_s \approx 85.49/H^{0.243} = 21.2$. For $z$ jets the wheel specific speed is $N_s\sqrt{z}$. Take **$z = 2$ jets** (horizontal shaft):

$$
N = \frac{N_s\sqrt{z}\,H^{5/4}}{\sqrt{P}} = 344.6\ \text{rpm}
$$

Synchronous speed for 50 Hz: $N = 6000/p$. Choose $p = 18$ poles, **$N = 333.3$ rpm**.

#### Step 4: Wheel (pitch circle) diameter

$$
D = \frac{60u}{\pi N} = \frac{60 \times 36.02}{\pi \times 333.3} = 2.064\ \text{m}
$$

#### Step 5: Jet diameter
Discharge per jet $q = 5/2 = 2.5\ \text{m}^3/\text{s}$:

$$
d = \sqrt{\frac{4q}{\pi V_1}} = \sqrt{\frac{4 \times 2.5}{\pi \times 76.74}} = 0.2037\ \text{m}
$$

Jet ratio $m = D/d = 10.13$ (acceptable range 10 to 20).

#### Step 6: Bucket details
- Number of buckets $Z = 15 + \dfrac{D}{2d} = 15 + 5.07 \approx 20$
- Bucket width $B \approx 3.5d = 0.71$ m; depth about $1.0d$; length about $2.5d$.

#### Step 7: Specific speed of the wheel
$$
N_s = \frac{N\sqrt{P}}{H^{5/4}} = \frac{333.3\sqrt{13{,}029}}{312.5^{1.25}} = 29.0
$$

(about 20.5 per jet)

**Answer: $P = 13.03$ MW; 2 jets; $N = 333.3$ rpm; $D = 2.06$ m; $d = 0.204$ m ($D/d = 10.1$); about 20 buckets; $N_s = 29.0$.**

## Check yourself

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

1. How is the design power obtained?
   A. P = gamma Q V1^2
   B. P = gamma Q H
   C. P = eta gamma Q H = 13.03 MW
   D. P = eta Q H
   Answer: C. P = 0.85 x 9.81 x 5 x 312.5 = 13,029 kW. (Key point: Power)

2. How does the wheel specific speed relate to the number of jets z?
   A. Ns(wheel) = Ns(single jet) / z
   B. Ns(wheel) = Ns(single jet) x z
   C. Ns(wheel) = Ns(single jet) / sqrt(z)
   D. Ns(wheel) = Ns(single jet) x sqrt(z)
   Answer: D. Specific speed rises with sqrt of the number of jets, which is used to get N for z = 2. (Key point: Number of jets and Ns)

3. Why is the calculated speed 344.6 rpm changed to 333.3 rpm?
   A. Because the speed must be below the runaway speed
   B. To make the jet ratio larger
   C. The generator must run at synchronous speed N = 6000/p for 50 Hz, with p = 18 poles
   D. To reduce the number of buckets
   Answer: C. Speed must be a synchronous value, 6000/p, and 18 poles gives 333.3 rpm. (Key point: Synchronous speed)

4. What does the jet ratio m = D/d = 10.13 show?
   A. It gives the bucket width
   B. It lies within the acceptable range of 10 to 20
   C. It is below the allowed range
   D. It sets the number of jets
   Answer: B. Pelton wheels are acceptable when D/d is in the range 10 to 20. (Key point: Jet ratio check)

## Related in Hydropower Engineering

- [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 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
- [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
- [Types of hydropower plants by head, turbine, load](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Ahydropower-plant-types-classification) — Hydropower Engineering, chapter 2
- [Pelton unit speed, discharge, power](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apelton-unit-quantities-190rpm) — Hydropower Engineering, chapter 7

## Related in other subjects

- [Pelton turbine design (5 m³/s, 312.5 m)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Apelton-design-312-5m) — Hydro Power, chapter 7
- [Design: 300 m head plant, GCB rating](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2081-300m-gcb) — Power Plant Design, chapter 4, asked 11×
- [Design: 230 m head plant, 55 km line](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2074-230m-55km) — Power Plant Design, chapter 4, asked 12×
- [Design: 245 m head plant, standby generator](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2073-245m-standby) — Power Plant Design, chapter 4, asked 14×