Pelton turbine design (2 m³/s, 500 m)

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: $Q = 2$ m³/s, $H = 500$ m, $\eta_o = 0.85$, $f = 50$ Hz. Assumed from the standard ranges: $C_v = 0.98$ (0.97 to 0.99) and speed ratio $\phi = 0.46$ (0.43 to 0.47).

1. Power developed

$$P = \eta_o \rho g Q H = 0.85 \times 9.81 \times 2 \times 500 = 8338.5\ \text{kW}$$

2. Jet velocity and bucket (peripheral) velocity

$$\begin{aligned} \sqrt{2gH} &= \sqrt{2 \times 9.81 \times 500} = 99.05\ \text{m/s} \\ V_1 &= C_v\sqrt{2gH} = 0.98 \times 99.05 = 97.06\ \text{m/s} \\ u &= \phi\sqrt{2gH} = 0.46 \times 99.05 = 45.56\ \text{m/s} \end{aligned}$$

3. Jet diameter (single jet)

$$\begin{aligned} Q &= \frac{\pi}{4} d^2 V_1 \\ d &= \sqrt{\frac{4 \times 2}{\pi \times 97.06}} = 0.1620\ \text{m} \approx 162\ \text{mm} \end{aligned}$$

4. Speed of the wheel

Trial with jet ratio $m = 12$: $D = 12d = 1.944$ m, so

$$N = \frac{60u}{\pi D} = \frac{60 \times 45.56}{\pi \times 1.944} = 447.7\ \text{rpm}$$

Nearest synchronous speed for 50 Hz: $N = 120f/p = 6000/14 = 428.57$ rpm (14 poles).

5. Wheel (pitch circle) diameter and jet ratio

$$\begin{aligned} D &= \frac{60u}{\pi N} = \frac{60 \times 45.56}{\pi \times 428.57} = 2.030\ \text{m} \\ m &= \frac{D}{d} = \frac{2.030}{0.1620} = 12.54 \end{aligned}$$

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

6. Specific speed check

$$N_s = \frac{N\sqrt{P}}{H^{5/4}} = \frac{428.57 \times 91.32}{2364.4} = 16.6$$

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)

$$z = 15 + \frac{D}{2d} = 15 + \frac{12.54}{2} = 21.27 \approx 22$$

8. Bucket size (usual proportions)

• Bucket width $B \approx 5d = 0.810$ m
• Bucket depth $T \approx 1.2d = 0.194$ m
• Bucket radial length $L \approx 2.5d = 0.405$ m

Summary of design