Francis size and setting (150 m, 160 m³/s)

Step 1: Power and units

$$P = \eta\gamma Q H = 0.85 \times 9.81 \times 160 \times 150 = 200{,}124\ \text{kW}$$

Number of units: adopt 4 units, each $Q = 40$ m³/s and $P_u = 50{,}031$ kW.

Step 2: Specific speed and rated speed

For Francis turbines (metric units, kW, m, rpm) the empirical relation of Schweiger and Gregory is

$$N_s = \frac{3763}{H^{0.854}} + 49.4 = 101.5$$ $$N = \frac{N_s H^{5/4}}{\sqrt{P_u}} = 238.3\ \text{rpm}$$

Synchronous speed at 50 Hz: $N = 6000/p$. Adopt $p = 26$ poles, $N = 230.8$ rpm. Actual specific speed:

$$N_s = \frac{N\sqrt{P_u}}{H^{5/4}} = 98.3$$

Step 3: Runner diameter

Peripheral speed $u = K_u\sqrt{2gH}$ with $K_u = 0.7$ (range 0.6 to 0.9):

$$u = 0.7\sqrt{2 \times 9.81 \times 150} = 37.97\ \text{m/s}, \qquad D = \frac{60u}{\pi N} = \frac{60 \times 37.97}{\pi \times 230.8} = 3.14\ \text{m}$$

Step 4: Setting (Thoma cavitation coefficient)

$$\sigma_c = 0.0432\left(\frac{N_s}{100}\right)^2 = 0.0418$$

Sea-level site assumed: $H_{atm} = 10.3$ m, $H_v = 0.24$ m.

$$H_s = H_{atm} - H_v - \sigma_c H = 10.3 - 0.24 - 0.0418 \times 150 = 3.79\ \text{m}$$

This is the highest permissible level of the runner outlet above tailwater; in practice set lower (safety margin 1 to 2 m, and reduce $H_{atm}$ by about 1 m per 900 m of site elevation).

Answer: 4 units of 50 MW; $N = 230.8$ rpm; runner diameter $D \approx 3.14$ m; setting height $H_s \le 3.8$ m above tailwater level.