Tidal power output: 3 km × 5.5 km barrage

In a tidal barrage, the basin fills at high tide and the trapped water is released through turbines when the sea falls. The potential energy of water of area $A$ and tidal range $h$ (centre of mass at $h/2$) is:

$$E = \rho g A h \cdot \frac{h}{2} = \frac{1}{2}\rho g A h^2$$

Given

• $A = 3\ \text{km} \times 5.5\ \text{km} = 3000 \times 5500 = 1.65\times 10^7\ \text{m}^2$
• $h = 10$ m, $\rho = 1023$ kg/m³, $g = 9.81$ m/s²
• 2 tides per day, $\eta = 0.60$

Step 1: Energy per tide

$$\begin{aligned} E &= 0.5 \times 1023 \times 9.81 \times 1.65\times 10^7 \times 10^2 \\ &= 8.279 \times 10^{12}\ \text{J} \\ &= \frac{8.279\times 10^{12}}{3.6\times 10^9} = 2299.8\ \text{MWh} \end{aligned}$$

Step 2: Energy per day available

$$E_{day} = 2 \times 8.279\times 10^{12} = 1.656\times 10^{13}\ \text{J}$$

Step 3: Average electrical power

$$\begin{aligned} P &= \frac{\eta E_{day}}{86400} = \frac{0.6 \times 1.656\times 10^{13}}{86400} \\ &= 1.150\times 10^{8}\ \text{W} = 115.0\ \text{MW} \end{aligned}$$

Annual energy (for reference)

$$E_{year} = 115.0\ \text{MW} \times 8760\ \text{h} \approx 1007\ \text{GWh}$$

Answer: Average power output capacity $\approx 115$ MW (energy per tide $\approx 2300$ MWh before losses).

Assumption: one emptying of the basin per tide (single-effect ebb generation), averaged over 24 hours.