---
title: "Pelton turbine design (5 m³/s, 312.5 m)"
subject: "Hydro Power"
chapter: 7
key: "semester_6/hydro_power:pelton-design-312-5m"
source: https://www.bctengineer.com/q/semester_6/hydro_power/pelton-design-312-5m
site: BCT Engineering
---

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

Hydro Power · Chapter 7: Hydro-Electric Machines

- Asked in: 2080 Chaitra, 8 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_6/hydro_power/pelton-design-312-5m
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-7

## Question

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

## Short answer

Single-jet Pelton wheel: power 13.03 MW at 200 rpm (30 poles). Jet diameter 288 mm, wheel diameter 3.44 m, jet ratio 11.94. Specific speed 17.4. 21 buckets; bucket width 1.44 m, depth 0.35 m.

## Answer

A Pelton wheel is designed by finding the jet velocity and jet diameter, choosing a synchronous speed that gives a suitable jet ratio $m = D/d$ (10 to 14 for good efficiency), and then sizing the wheel and buckets.

**Data:** $Q = 5$ m³/s, $H = 312.5$ m, $\eta_o = 0.85$, $f = 50$ Hz, coefficient of velocity of the jet $C_v = 0.98$ (given). Assumed from the standard range: speed ratio $\phi = 0.46$ (0.43 to 0.47).

#### 1. Power developed

$$
P = \eta_o \rho g Q H = 0.85 \times 9.81 \times 5 \times 312.5 = 13028.9\ \text{kW}
$$

#### 2. Jet velocity and bucket (peripheral) velocity

$$
\begin{aligned}
\sqrt{2gH} &= \sqrt{2 \times 9.81 \times 312.5} = 78.30\ \text{m/s} \\
V_1 &= C_v\sqrt{2gH} = 0.98 \times 78.30 = 76.74\ \text{m/s} \\
u &= \phi\sqrt{2gH} = 0.46 \times 78.30 = 36.02\ \text{m/s}
\end{aligned}
$$

#### 3. Jet diameter (single jet)

$$
\begin{aligned}
Q &= \frac{\pi}{4} d^2 V_1 \\
d &= \sqrt{\frac{4 \times 5}{\pi \times 76.74}} = 0.2880\ \text{m} \approx 288\ \text{mm}
\end{aligned}
$$

#### 4. Speed of the wheel

Trial with jet ratio $m = 12$: $D = 12d = 3.456$ m, so

$$
N = \frac{60u}{\pi D} = \frac{60 \times 36.02}{\pi \times 3.456} = 199.0\ \text{rpm}
$$

Nearest synchronous speed for 50 Hz: $N = 120f/p = 6000/30 = 200$ rpm (30 poles).

#### 5. Wheel (pitch circle) diameter and jet ratio

$$
\begin{aligned}
D &= \frac{60u}{\pi N} = \frac{60 \times 36.02}{\pi \times 200} = 3.440\ \text{m} \\
m &= \frac{D}{d} = \frac{3.440}{0.2880} = 11.94
\end{aligned}
$$

$m$ lies within 10 to 14, so a **single-jet** wheel is satisfactory.

#### 6. Specific speed check

$$
N_s = \frac{N\sqrt{P}}{H^{5/4}} = \frac{200 \times 114.14}{1313.9} = 17.4
$$

This is within the single-jet Pelton range (about 8.5 to 30), so the Pelton wheel is the correct choice.

#### 7. Number of buckets (Tygun's formula)

$$
z = 15 + \frac{D}{2d} = 15 + \frac{11.94}{2} = 20.97 \approx 21
$$

#### 8. Bucket size (usual proportions)

- Bucket width $B \approx 5d = 1.440$ m
- Bucket depth $T \approx 1.2d = 0.346$ m
- Bucket radial length $L \approx 2.5d = 0.720$ m

#### Summary of design

| Item | Value |
|---|---|
| Power | 13.03 MW |
| Speed | 200 rpm |
| Number of jets | 1 |
| Jet diameter $d$ | 288 mm |
| Wheel diameter $D$ | 3.44 m |
| Jet ratio $m$ | 11.94 |
| Specific speed $N_s$ | 17.4 |
| Number of buckets | 21 |
| Bucket width × depth | 1.44 m × 0.35 m |

## Check yourself

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

1. How is the power developed by the Pelton turbine calculated?
   A. P = overall efficiency x rho g Q H = 0.85 x 9.81 x 5 x 312.5 = 13028.9 kW
   B. P = rho g Q H with no efficiency
   C. P = eta x Q x H / g
   D. P = eta x 0.5 rho Q V1^2 with V1 = sqrt(2gH) only, ignoring head
   Answer: A. Output power is the hydraulic power times the overall efficiency. (Key point: Power developed)

2. How are the jet velocity V1 and bucket velocity u found?
   A. V1 = sqrt(2gH) with no coefficient, and u = V1
   B. V1 = Cv sqrt(2gH) = 76.74 m/s and u = phi sqrt(2gH) = 36.02 m/s
   C. V1 = phi sqrt(2gH) and u = Cv sqrt(2gH)
   D. V1 = sqrt(gH) and u = 2 V1
   Answer: B. The jet velocity uses the velocity coefficient Cv (0.98), the bucket velocity uses the speed ratio phi (0.46), both times sqrt(2gH). (Key point: Jet and bucket velocities)

3. How is the jet diameter d found for a single jet?
   A. From Q = d V1, so d = Q/V1
   B. From Q = pi d^2 u, using the bucket velocity
   C. From Q = (pi/4) d^2 V1, so d = sqrt(4Q/(pi V1)) = 0.2880 m (288 mm)
   D. From P = pi d^2 V1, using the power
   Answer: C. The discharge equals jet area times jet velocity. (Key point: Jet diameter)

4. How are the wheel speed and diameter fixed?
   A. Use any speed and accept any m, because only d matters
   B. Take D = 12d always and N follows with no synchronous restriction
   C. Choose N from Ns = 100 and ignore m
   D. Pick a synchronous speed N = 120f/p (200 rpm), then D = 60u/(pi N) = 3.440 m, giving m = D/d = 11.94, within 10 to 14
   Answer: D. The speed must suit the generator, and the resulting jet ratio D/d must lie in 10 to 14 for good efficiency. (Key point: Speed and jet ratio)

5. What do the specific speed check and bucket count show?
   A. Ns = N sqrt(P)/H^(5/4) = 17.4, inside the single-jet range 8.5 to 30; z = 15 + m/2 = 20.97, so 21 buckets
   B. Ns = 17.4 is above 30, so several jets are needed; z = 15 + m = 25 buckets
   C. Ns must equal 100 exactly; z = D/d buckets
   D. Ns is irrelevant to Pelton choice; z = 2m = 23 buckets
   Answer: A. The Pelton wheel suits low specific speed (about 8.5 to 30 for a single jet); bucket number follows Tygun's formula. (Key point: Specific speed and buckets)

## Related in Hydro Power

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