Pelton turbine design (2 m³/s, 605 m, Ns 100)

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. The specific speed given in the data is checked first.

Data: $Q = 2$ m³/s, $H = 605$ m, $\eta_o = 0.87$, $f = 50$ 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

$$P = \eta_o \rho g Q H = 0.87 \times 9.81 \times 2 \times 605 = 10327.0\ \text{kW}$$

2. Jet velocity and bucket (peripheral) velocity

$$\begin{aligned} \sqrt{2gH} &= \sqrt{2 \times 9.81 \times 605} = 108.95\ \text{m/s} \\ V_1 &= C_v\sqrt{2gH} = 0.98 \times 108.95 = 106.77\ \text{m/s} \\ u &= \phi\sqrt{2gH} = 0.46 \times 108.95 = 50.12\ \text{m/s} \end{aligned}$$

3. Jet diameter (single jet)

$$\begin{aligned} Q &= \frac{\pi}{4} d^2 V_1 \\ d &= \sqrt{\frac{4 \times 2}{\pi \times 106.77}} = 0.1544\ \text{m} \approx 154\ \text{mm} \end{aligned}$$

4. Speed of the wheel

Trial with jet ratio $m = 12$: $D = 12d = 1.853$ m, so

$$N = \frac{60u}{\pi D} = \frac{60 \times 50.12}{\pi \times 1.853} = 516.5\ \text{rpm}$$

Nearest synchronous speed for 50 Hz: $N = 120f/p = 6000/12 = 500$ rpm (12 poles).

Use of the given $N_s = 100$: if the speed is taken from the given specific speed,

$$N = \frac{N_s H^{5/4}}{\sqrt{P}} = \frac{100 \times 3000.5}{101.62} = 2953\ \text{rpm} \approx 3000\ \text{rpm}$$

which gives $D = 60u/(\pi N) = 0.319$ m and a jet ratio $m = D/d = 2.07$, far below the minimum of about 10. A single jet Pelton has $N_s$ of only about 8.5 to 30; even 6 jets raise this only by $\sqrt{6}$ (to about 70). Reaching 100 would need about $(100/16.9)^2 \approx 35$ jets, which is impossible. So $N_s = 100$ cannot be met by a Pelton wheel at this head; the design below uses the practical speed of 500 rpm and the standard proportions, and the actual $N_s$ is reported.

5. Wheel (pitch circle) diameter and jet ratio

$$\begin{aligned} D &= \frac{60u}{\pi N} = \frac{60 \times 50.12}{\pi \times 500} = 1.914\ \text{m} \\ m &= \frac{D}{d} = \frac{1.914}{0.1544} = 12.40 \end{aligned}$$

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

6. Specific speed check

$$N_s = \frac{N\sqrt{P}}{H^{5/4}} = \frac{500 \times 101.62}{3000.5} = 16.9$$

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)

$$z = 15 + \frac{D}{2d} = 15 + \frac{12.40}{2} = 21.20 \approx 22$$

8. Bucket size (usual proportions)

• Bucket width $B \approx 5d = 0.772$ m
• Bucket depth $T \approx 1.2d = 0.185$ m
• Bucket radial length $L \approx 2.5d = 0.386$ m

Summary of design