Multi-jet Pelton wheel design (200 m, 4 m³/s)

Data

$H = 200$ m, $Q = 4\ \text{m}^3/\text{s}$, $C_v = 0.98$, $D = 1.47$ m, $\eta_m = 0.86$, $\phi = 0.47$, $d/D = 0.113$.

Jet velocity: $V_1 = C_v\sqrt{2gH} = 0.98\sqrt{2 \times 9.81 \times 200} = 61.39$ m/s.

(i) Rotational speed

$u = \phi V_1 = 0.47 \times 61.39 = 28.85$ m/s

$$N = \frac{60u}{\pi D} = \frac{60 \times 28.85}{\pi \times 1.47} = 374.9\ \text{rpm}$$

(ii) Power output

Taking the 86 % efficiency as the efficiency from the water power to the shaft:

$$P = \eta\,\gamma Q H = 0.86 \times 9.81 \times 4 \times 200 = 6749\ \text{kW} = 6.75\ \text{MW}$$

(iii) Number of nozzles

Jet diameter $d = 0.113 \times 1.47 = 0.1661$ m; jet area $= 0.02167\ \text{m}^2$.

Discharge per jet $q = 0.02167 \times 61.39 = 1.330\ \text{m}^3/\text{s}$.

$$n = \frac{Q}{q} = \frac{4}{1.330} = 3.01 \approx 3\ \text{nozzles}$$

(iv) Specific speed

$$N_s = \frac{N\sqrt{P}}{H^{5/4}} = \frac{374.9\sqrt{6749}}{200^{1.25}} = 40.9$$

Answer: $N = 374.9$ rpm; $P = 6.75$ MW; 3 nozzles; $N_s = 40.9$ (kW, m, rpm).