Pelton turbine design (6867 kW, 350 m)

Data

$P = 6867$ kW, $H = 350$ m, $\eta_o = 0.80$, $N = 550$ rpm, $K_v = 0.98$, $K_u = 0.46$, $d/D = 1/12$.

Discharge

$$Q = \frac{P}{\eta_o \gamma H} = \frac{6867}{0.8 \times 9.81 \times 350} = 2.50\ \text{m}^3/\text{s}$$

Wheel diameter

$V_1 = K_v\sqrt{2gH} = 0.98\sqrt{2 \times 9.81 \times 350} = 81.21$ m/s

$u = K_u\sqrt{2gH} = 0.46 \times 82.87 = 38.12$ m/s

$$D = \frac{60u}{\pi N} = \frac{60 \times 38.12}{\pi \times 550} = 1.324\ \text{m}$$

Jet diameter and number of jets

From $d/D = 1/12$: $d = 1.324/12 = 0.1103$ m.

Discharge through one jet $= \dfrac{\pi}{4}(0.1103)^2 \times 81.21 = 0.776\ \text{m}^3/\text{s}$.

$$z = \frac{2.50}{0.776} = 3.22 \Rightarrow 4\ \text{jets}$$

With 4 jets the jet diameter is adjusted to carry the full discharge:

$$d = \sqrt{\frac{4 \times 2.5/4}{\pi \times 81.21}} = 0.099\ \text{m} \approx 99\ \text{mm}$$

so $D/d = 13.4$ (within the usual range 10 to 20). Specific speed: $N_s = \dfrac{550\sqrt{6867}}{350^{1.25}} = 30.1$.

Answer: $Q = 2.50$ m³/s; 4 jets; jet diameter $\approx 0.099$ m; wheel diameter $= 1.32$ m.