Tidal power generation and output: 3 km × 3 km

Generation of electric power from a tidal project

Tides are the periodic rise and fall of sea level caused by the moon and sun's gravity; most coasts have two high and two low tides a day. A tidal power plant uses the head between the sea and a basin.

        SEA                BARRAGE            BASIN
  high tide  ~~~~~      +-----------+
                        | sluice    |     ~~~~~ stored
  low tide   ~~~~       | turbine   |           water
                        +-----------+
   <--- flow through turbine on ebb (basin -> sea)

1. A barrage (dam) is built across a bay or estuary, with sluice gates and low-head bulb (Kaplan type) turbines.
2. Filling: at rising tide the gates open and the basin fills.
3. Holding: at high tide the gates close and water is held.
4. Generation (ebb): when the sea falls, a head forms; water flows from basin to sea through the turbines, driving generators.
5. In double-effect schemes, power is also generated on the flood tide (sea to basin).

Examples: La Rance (France, 240 MW), Sihwa (South Korea, 254 MW).

Power estimation

Energy of water in the basin per tide (centre of gravity at $h/2$):

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

Given: $A = 3000 \times 3000 = 9\times 10^6$ m², $h = 10$ m, $\rho = 1025$ kg/m³, $g = 9.81$ m/s², 2 tides per day, $\eta = 0.30$.

$$\begin{aligned} E &= 0.5 \times 1025 \times 9.81 \times 9\times 10^6 \times 10^2 \\ &= 4.525\times 10^{12}\ \text{J per tide} \\ &= 1256.9\ \text{MWh per tide} \\ E_{day} &= 2 \times 4.525\times 10^{12} = 9.050\times 10^{12}\ \text{J} \\ P_{avg} &= \frac{\eta E_{day}}{86400} = \frac{0.30 \times 9.050\times 10^{12}}{86400} \\ &= 3.142\times 10^{7}\ \text{W} \end{aligned}$$

Answer: Average power output $\approx 31.4$ MW (about 275 GWh per year).

Assumption: one basin emptying per high tide, averaged over 24 hours.