Pelton unit speed, discharge, power

Data

$N = 190$ rpm, $P = 5150.25$ kW, $H = 220$ m, $\eta_o = 0.80$.

Discharge from power:

$$Q = \frac{P}{\eta_o \gamma H} = \frac{5150.25}{0.8 \times 9.81 \times 220} = 2.983\ \text{m}^3/\text{s}$$

i) Unit quantities (head 1 m)

$$N_u = \frac{N}{\sqrt{H}} = \frac{190}{\sqrt{220}} = 12.81\ \text{rpm}$$ $$Q_u = \frac{Q}{\sqrt{H}} = \frac{2.983}{\sqrt{220}} = 0.2011\ \text{m}^3/\text{s}$$ $$P_u = \frac{P}{H^{3/2}} = \frac{5150.25}{220^{1.5}} = 1.578\ \text{kW}$$

ii) At $H = 140$ m (same efficiency, similar operation)

$$N = N_u\sqrt{140} = 12.81 \times 11.832 = 151.6\ \text{rpm}$$ $$Q = Q_u\sqrt{140} = 0.2011 \times 11.832 = 2.380\ \text{m}^3/\text{s}$$ $$P = P_u \times 140^{1.5} = 1.578 \times 1656.5 = 2614\ \text{kW}$$

Answer: $N_u = 12.81$ rpm, $Q_u = 0.2011$ m³/s, $P_u = 1.578$ kW; at 140 m: $N = 151.6$ rpm, $Q = 2.38$ m³/s, $P = 2614$ kW.