Pelton turbine design (3.75 m³/s, 380 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: $Q = 3.75$ m³/s, $H = 380$ m, $\eta_o = 0.86$, $f = 50$ Hz. Given: coefficient of velocity of the jet $C_v = 0.97$ and speed ratio $\phi = 0.46$.

1. Power developed

$$P = \eta_o \rho g Q H = 0.86 \times 9.81 \times 3.75 \times 380 = 12022.2\ \text{kW}$$

2. Jet velocity and bucket (peripheral) velocity

$$\begin{aligned} \sqrt{2gH} &= \sqrt{2 \times 9.81 \times 380} = 86.35\ \text{m/s} \\ V_1 &= C_v\sqrt{2gH} = 0.97 \times 86.35 = 83.76\ \text{m/s} \\ u &= \phi\sqrt{2gH} = 0.46 \times 86.35 = 39.72\ \text{m/s} \end{aligned}$$

3. Jet diameter (single jet)

$$\begin{aligned} Q &= \frac{\pi}{4} d^2 V_1 \\ d &= \sqrt{\frac{4 \times 3.75}{\pi \times 83.76}} = 0.2388\ \text{m} \approx 239\ \text{mm} \end{aligned}$$

4. Speed of the wheel

Trial with jet ratio $m = 12$: $D = 12d = 2.865$ m, so

$$N = \frac{60u}{\pi D} = \frac{60 \times 39.72}{\pi \times 2.865} = 264.8\ \text{rpm}$$

Nearest synchronous speed for 50 Hz: $N = 120f/p = 6000/22 = 272.73$ rpm (22 poles).

5. Wheel (pitch circle) diameter and jet ratio

$$\begin{aligned} D &= \frac{60u}{\pi N} = \frac{60 \times 39.72}{\pi \times 272.73} = 2.781\ \text{m} \\ m &= \frac{D}{d} = \frac{2.781}{0.2388} = 11.65 \end{aligned}$$

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

6. Specific speed check

$$N_s = \frac{N\sqrt{P}}{H^{5/4}} = \frac{272.73 \times 109.65}{1677.8} = 17.8$$

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)

$$z = 15 + \frac{D}{2d} = 15 + \frac{11.65}{2} = 20.82 \approx 21$$

8. Bucket size (usual proportions)

• Bucket width $B \approx 5d = 1.194$ m
• Bucket depth $T \approx 1.2d = 0.287$ m
• Bucket radial length $L \approx 2.5d = 0.597$ m

Summary of design