Design of Pelton wheel for given power and head

Given: $P=6000$ kW, $H=350$ m, $N=500$ rpm, $C_v=0.98$, $\phi=0.46$, $\eta_o=0.86$, number of jets $n=2$.

Discharge

$$\eta_o=\frac{P}{\rho g Q H}\ \Rightarrow\ Q=\frac{6\times10^{6}}{1000\times9.81\times350\times0.86}=2.032\ \text{m}^3/\text{s}$$

Jet velocity and bucket speed

$\sqrt{2gH}=\sqrt{2\times9.81\times350}=82.87$ m/s.

$$V_1=C_v\sqrt{2gH}=81.21\ \text{m/s},\qquad u=\phi\sqrt{2gH}=38.12\ \text{m/s}$$

Wheel (pitch circle) diameter

$$D=\frac{60u}{\pi N}=\frac{60\times38.12}{\pi\times500}=1.456\ \text{m}$$

Jet diameter

Discharge per jet $=1.016$ m³/s.

$$a=\frac{Q_j}{V_1}=12.511\times10^{-3}\ \text{m}^2,\qquad d=\sqrt{\frac{4a}{\pi}}=126.2\ \text{mm}$$

Jet ratio $m=D/d=11.5$, which lies in the normal range of 10 to 14. Good.

Number of buckets

$$Z=15+\frac{D}{2d}=15+\frac{11.5}{2}=20.8\ \Rightarrow\ Z=21\ \text{(rounded up) per wheel}$$

Specific speed check

$$N_s=\frac{N\sqrt{P}}{H^{5/4}}=\frac{500\sqrt{6000}}{350^{1.25}}=25.6\ \text{for the whole wheel}$$

Based on the power per jet (3000 kW), $N_s=18.1$, which is within the Pelton range of about 10 to 35 (metric units, kW). The design is suitable.

Answer: $Q=2.032$ m³/s; wheel diameter $=1.456$ m; jet diameter $=126.2$ mm (two jets); buckets $=21$; $N_s=18.1$ per jet.