Number of Kaplan units (380 m³/s, 30 m)

Given: $Q = 380$ m³/s, $H = 30$ m, $N = 150$ rpm, overall efficiency $\eta_o = 0.90$, specific speed of Kaplan unit $N_s = 620$ (SI: rpm, kW, m).

Formulas:

$$P = \eta_o\,\rho g Q H, \qquad N_s = \frac{N\sqrt{P}}{H^{5/4}}$$

1. Total power available

$$\begin{aligned} P_{total} &= 0.90 \times 9.81 \times 380 \times 30 \\ &= 100{,}650.6\ \text{kW} \end{aligned}$$

2. Power of one unit from specific speed

$$\begin{aligned} H^{5/4} &= 30^{1.25} = 70.21 \\ \sqrt{P} &= \frac{N_s H^{5/4}}{N} = \frac{620 \times 70.21}{150} = 290.20 \\ P_{unit} &= 290.20^2 = 84{,}218\ \text{kW} \end{aligned}$$

3. Number of units

$$n = \frac{P_{total}}{P_{unit}} = \frac{100{,}650.6}{84{,}218} = 1.195$$

Adopt 2 units (the next whole number, so that the specific speed of each unit does not exceed 620).

4. Output of each unit

$$P_{each} = \frac{100{,}650.6}{2} = 50{,}325.3\ \text{kW} \approx 50.3\ \text{MW}$$

Check: actual specific speed $= \dfrac{150\sqrt{50{,}325.3}}{70.21} = 479$, within the Kaplan range.

Answer: number of Kaplan units $= 2$; output of each unit $\approx 50{,}325$ kW (50.3 MW).