# Energy in an ocean current vs energy in an air current

1. Nov 19, 2005

### R0man

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!

2. Nov 19, 2005

### Staff: Mentor

Considering mass flow rate as $\rho$VA and the specific kinetic energy as V2/2, and the same turbine area, the numbers seem about right.

What number is considered low?

3. Nov 19, 2005

### R0man

I think it's low because of info found http://www.bluenergy.com/technology.html".
To quote them: "Sea water is 832 times denser than air and a non-compressible medium, an 8 knot tidal current is the equivalent of a 390 km/hr wind."

Last edited by a moderator: Apr 21, 2017
4. Nov 19, 2005

### Bystander

Don't confuse energy with power.

5. Nov 19, 2005

### Staff: Mentor

well if one uses the relationship - (ryo_water)(v_water^2)=(ryo_air)(v_air^2), then one would get closer to 390 km/h.

However, using 8 knots = 14.8 km/h and the ratio of densities of seawater to air = 832, I get more like 426 km/h for the wind speed.

I'd have to give this more thought.