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There is as much energy in an ocean current moving at "X" knots as there is in an air current moving at "Y" kph.

I tried using P=1/2(ryo)(v^3)(pi)(r^2), where P is the power produced by the turbine, ryo is the density, v is the velocity, and r is the radius of the turbine.

Assuming P is equal in either case, I set the equations (one of water, one for air) equal to each other. I also assumed r was equal in either case, so r, pi, and 1/2 cancel, leaving:

(ryo_water)(v_water^3)=(ryo_air)(v_air^3)

Using:

ryo_air = 1.275kg/m^3

ryo_water = 1025kg/m^3 (seawater)

I end up with v_water = 0.1075(v_air)

So if I have my conversions right an 8 knot current would have as much energy as a 137.8kph wind. This seems reasonable to me, but it is very low compared to data I can find on the net.

Do anyone have any insight on this? Thanks in advance!