Sunday, May 10, 2009

Wave power potential

I had originally posted this as a comment to another post below, but was hoping to see if anyone who is an expert on wave power might be able to tell me if my logic is close or just way off.

Waves have a very intense amount of energy. You can feel it when you go swimming at the beach and see it when you look at the aftermath of hurricanes on the shoreline.

I am not an expert on wind or wave power but I would bet that it is related to the density of the water itself. For wind, Pwind=.5*Pair*A*V^3 and I would imagine that the equation for water energy is similar. However, when waves move through water, they do not actually move the water in the direction of the wave. Rather they displace the water(medium) in a direction that is perpendicular to the direction of the wave. Therefore, the speed part of the equation would be 2 times the height of the wave (h) divided by the period of the wave (t). Therefore, the power in the wave would be, Pwave=0.5*Pwater*A*(2*h/t)^3.

If air density (Pair) is 1.275 Kg/m^3 and salt water density (Pwater) is 1,025 Kg/m^3 than the energy density of water should be ~800 times more than that of air for a given cross sectional area moving at the same V. If 75% of our planet is covered by water and you assume an average wave height of 2 meters and an average period of 8 seconds (, earths water surface area of 361,132,000 Km^2 ( , then you get total wave power of 2.3*10^16 watts or 23,130,000 Giga-Watts of total wave power availability on earth.

P=0.5*1025kg/m^3*361,132,000,000,000m*(2*2meters/8seconds))^3 = 2.3*10^16 watts

This seems like a lot. Maybe someone can tell me if I am way off on this.

1 comment:

Windy said...

Check out the below websites for interesting research on the potential of wave power (specifically focused on ocean energy programs).