Question

I have a cylindrical buoy with the flat parts on top and bottom respectively. The radius of the buoy is 3 mts. It's hit by a wave with an amplitude of 1.5 mts and a period of 7.3 sec. I want to know how much energy I can harvest with the buoy. I have read any number of papers but I can't find anything with an example, so I can't plug in values and understand what's going on, it's just the formulas and in most cases they are different or have the units wrong. How can I find out the force, work, power and energy I can get from the buoy?

Answer 1

To estimate the energy harvested from a wave hitting a cylindrical buoy, we calculate the gravitational potential energy of the wave using its amplitude and density of water. The mass, gravitational acceleration, and height displaced are used in the formula E = mgh, and then apply the buoy's efficiency to compute the potential power output.

Step-by-step explanation:

To estimate the energy that can be harvested from a wave hitting a cylindrical buoy, we need to understand the wave's gravitational potential energy (GPE) and how this energy may be converted by the buoy. First, we must determine the mass of water displaced by the wave and the height it is raised to calculate GPE using the formula E = mgh, where E is the energy, m is the mass, g is the acceleration due to gravity (approximately 9.81 m/s²), and h is the height.

Considering a wave with an amplitude (A) of 1.5 meters, we can assume the height (h) of the displacement is 7A/8. The area of the wave crest is represented as Aλ/27 where λ is the wavelength, and hence not given here, we assume a length along the wave (l). The mass of the displaced water will be the product of the density of water (ρ), which is approximately 1000 kg/m³, the area Aλ/27, and the length l. Therefore, the mass (m) will be ρAλl/27. The gravitational potential energy is then calculated as mgh = ρAλlgh/27.

Once GPE is known, the power output of the buoy can be calculated, considering its efficiency in converting the wave's mechanical energy to useable energy. Usually, a certain percentage, say η (efficiency), of GPE can be converted to electric power. Power P is often represented as P = η ⋅ ρ ⋅ A ⋅ g ⋅ h, where in this case, A refers to the cross-sectional area of the buoy that is facing the wave and η includes all conversion losses.

In reality, additional variables and complex integrations over the wave's phase would be considered for a precise calculation of power and total energy harvested, but to get an elementary estimate, simplified models and assumptions as depicted can be used.