tag:blogger.com,1999:blog-7800828841873738105.post5317988017840496219..comments2023-07-02T06:58:33.737-05:00Comments on WebberEnergyBlog: "Anaconda" a new way to harness wave energyMichael E. Webber, Ph.D.http://www.blogger.com/profile/12416546342365493633noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-7800828841873738105.post-91017364641358296832009-05-10T20:53:00.000-05:002009-05-10T20:53:00.000-05:00When I read that there was only one commercial wav...When I read that there was only one commercial wave farm I thought why aren't there more? Is the reason that there are so fewer wave farms than wind farms that there isn't one established technology for harnessing the energy or that the ocean is far away from where most people are or that they are too difficult to maintain? Maybe it's a combination of all of the above and more. <br /><br />I found some other challenges to wind power:<br />-wave farms can displace fisherman<br />-wave farms in the US are hindered by limited R&D funding<br />-total cost of generation is too high to be competitive<br />-the devices have to be able to withstand salt water corrosion and storms so producing such robust devices may be cost prohibitive<br />-hard to efficiently convert wave motion to electricity<br /><br />http://www.ocsenergy.anl.gov/documents/docs/OCS_EIS_WhitePaper_Wave.pdfLiane Millerhttps://www.blogger.com/profile/03379700018435088854noreply@blogger.comtag:blogger.com,1999:blog-7800828841873738105.post-61864729979662511952009-05-10T16:09:00.000-05:002009-05-10T16:09:00.000-05:00Waves have a very intense amount of energy. You c...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. <br /><br />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. <br /><br />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 (http://www.worldwildweather.com/forecast/Global/Wave%20period?t0=0&tz=0), earths surface area of 361,132,000 Km^2 (http://en.wikipedia.org/wiki/Earth) , 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.<br /><br />P=0.5*1025kg/m^3*361,132,000,000,000m*(2*2meters/8seconds))^3 = 2.3*10^16 watts<br /><br />This seems like a lot. Maybe one of the engineers in the class knows if I am way off on this.contangohttps://www.blogger.com/profile/03281610372906976026noreply@blogger.com