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R0man
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This isn't actually a homework question, but it does relate to a research project I'm doing relating to wind/ocean turbines. What I'd like to be able to do is state that:
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!
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!