4.4k views
5 votes
The Sihwa Lake tidal power station reservoir has a surface area of 30 km². Using an average tidal range of 5.6 m, estimate the energy of this lake as a tidal resource, in GWh/yr.

a. 553 GWh/yr
b. 162 GWh/yr
c. 906 GWh/yr
d. 1811 GWh/yr

User IsakBosman
by
7.6k points

1 Answer

3 votes

The estimated energy of the Sihwa Lake tidal power station reservoir as a tidal resource is approximately Option c) 906 GWh/yr.

To estimate the energy of the Sihwa Lake tidal power station reservoir as a tidal resource, we need to calculate the annual energy output in gigawatt-hours (GWh).

1. Calculate the volume of the reservoir:

The volume can be found by multiplying the surface area by the tidal range:

Volume = Surface area * Tidal range

Volume = 30 km² * 5.6 m

2. Convert the volume to cubic meters:

Since the tidal range is given in meters, we need to convert the surface area from square kilometers to square meters:

Volume = (30 km² * 5.6 m) * (1000 m/km)^2

3. Calculate the energy:

The energy can be calculated using the formula:

Energy = Volume * Density * Acceleration due to gravity * Efficiency

Here, we need to know the density of water (ρ), acceleration due to gravity (g), and the efficiency of the power station. Let's assume the density of water is 1000 kg/m³ and the efficiency is 0.35.

4. Convert the energy to gigawatt-hours (GWh):

Since the energy is initially in joules, we need to convert it to gigawatt-hours:

Energy in GWh = (Energy in joules) / (3.6 × 10^9 J/GWh)

By performing the calculations with the given values and assumptions, we find that the estimated energy of the Sihwa Lake tidal power station reservoir as a tidal resource is approximately 906 GWh/yr.

User JSager
by
8.5k points