Pelton turbine design (5 m³/s, 310 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: $Q = 5$ m³/s, $H = 310$ m, $\eta_o = 0.86$, $f = 50$ 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

$$P = \eta_o \rho g Q H = 0.86 \times 9.81 \times 5 \times 310 = 13076.7\ \text{kW}$$

2. Jet velocity and bucket (peripheral) velocity

$$\begin{aligned} \sqrt{2gH} &= \sqrt{2 \times 9.81 \times 310} = 77.99\ \text{m/s} \\ V_1 &= C_v\sqrt{2gH} = 0.98 \times 77.99 = 76.43\ \text{m/s} \\ u &= \phi\sqrt{2gH} = 0.46 \times 77.99 = 35.87\ \text{m/s} \end{aligned}$$

3. Jet diameter (single jet)

$$\begin{aligned} Q &= \frac{\pi}{4} d^2 V_1 \\ d &= \sqrt{\frac{4 \times 5}{\pi \times 76.43}} = 0.2886\ \text{m} \approx 289\ \text{mm} \end{aligned}$$

4. Speed of the wheel

Trial with jet ratio $m = 12$: $D = 12d = 3.463$ m, so

$$N = \frac{60u}{\pi D} = \frac{60 \times 35.87}{\pi \times 3.463} = 197.8\ \text{rpm}$$

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

5. Wheel (pitch circle) diameter and jet ratio

$$\begin{aligned} D &= \frac{60u}{\pi N} = \frac{60 \times 35.87}{\pi \times 200} = 3.426\ \text{m} \\ m &= \frac{D}{d} = \frac{3.426}{0.2886} = 11.87 \end{aligned}$$

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

6. Specific speed check

$$N_s = \frac{N\sqrt{P}}{H^{5/4}} = \frac{200 \times 114.35}{1300.8} = 17.6$$

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)

$$z = 15 + \frac{D}{2d} = 15 + \frac{11.87}{2} = 20.93 \approx 21$$

8. Bucket size (usual proportions)

• Bucket width $B \approx 5d = 1.443$ m
• Bucket depth $T \approx 1.2d = 0.346$ m
• Bucket radial length $L \approx 2.5d = 0.722$ m

Summary of design