Francis design (90 m, 60 m³/s, 94%)

Step 1: Power and units

$$P = \eta\gamma Q H = 0.94 \times 9.81 \times 60 \times 90 = 49{,}796\ \text{kW}$$

Number of units: a single 50 MW unit would need a large runner (3.6 m) and shaft. Adopt 2 units, each $Q = 30$ m³/s and $P_u = 24{,}898$ kW (gives easier transport, maintenance and part-load operation).

Step 2: Specific speed and rated speed

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

$$N_s = \frac{3763}{H^{0.854}} + 49.4 = 130.1$$ $$N = \frac{N_s H^{5/4}}{\sqrt{P_u}} = 228.5\ \text{rpm}$$

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

$$N_s = \frac{N\sqrt{P_u}}{H^{5/4}} = 131.4$$

Step 3: Runner diameter

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

$$u = 0.7\sqrt{2 \times 9.81 \times 90} = 29.41\ \text{m/s}, \qquad D = \frac{60u}{\pi N} = \frac{60 \times 29.41}{\pi \times 230.8} = 2.43\ \text{m}$$

Step 4: Setting (Thoma cavitation coefficient)

$$\sigma_c = 0.0432\left(\frac{N_s}{100}\right)^2 = 0.0745$$

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

$$H_s = H_{atm} - H_v - \sigma_c H = 10.3 - 0.24 - 0.0745 \times 90 = 3.35\ \text{m}$$

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: 2 units of 24.9 MW; $N_s \approx 131$; $N = 230.8$ rpm; $D \approx 2.43$ m; setting $H_s \le 3.35$ m above tailwater level.