# The Physics of Tidal Energy

As my colleague Jess Bidgood reported in Friday’s paper, a tidal energy project is moving ahead in Maine, with high costs but high hopes too. But the 180-kilowatt unit that the Ocean Renewable Power Company hopes to put under water next week is really just a first step. The big question is, how well will it withstand the force of the rushing water?

The region, the Bay of Fundy, is famous for strong tides, but the company has picked a spot called Cobscook Bay, where the current is relatively slow, an average of 5.8 knots, or 6.7 miles an hour. That is the speed at which the hardware will produce 180 kilowatts.

If the equipment performs well there, the next step is for the company to put a similar piece of hardware in water moving at 8 knots, or 9.2 miles an hour. In that water, the same equipment would produce 498 kilowatts, or 2.8 times more.

If the water speed is greater by only about one-third, why does power nearly triple? Because the energy in the passing water rises with the cube of its speed; that is, the ratio of the energy in a 5.8 knot current to an 8 knot current is the ratio of 5.8 x 5.8 x 5.8 versus 8 x 8 x 8.

But “the increase isn’t free,” as a spokeswoman for the company, Susy Kist, put it. The stresses on the structure also increase, so it has to be made stronger, she said.

NYT

### 7 Responses to The Physics of Tidal Energy

1. Gamecock

E=mc2 is now E=mc4 ?

2. smokey

I assume you’re referring to the rate of force increase. They didn’t just make that up, similar calculations go into figuring out the applied force of hurricane windspeeds. A person standing around in 25mph winds has trouble holding onto their hat, maybe. A person standing around in 75mph winds… isn’t standing around because the applied force is nine times (not three times) that of the 25mph wind.

The only difference is the fluid in which we’re attempting to stand around.

• benofhouston

Smokey, the wind is 27 times as powerful, not 9. You have to account for increased mass as well. See my reply to GC.

Also, I would like to send a public angry glare at Barry for making me remember my fluid dynamics class.

• Barry looks sheepishly away from screen, moves off to make coffee ———————————->

3. Gamecock

E=1/2mv2 would have been a better formula for kinetic energy.

E=1/2mv3 is bogus.

• benofhouston

GC, the final V is from the increased mass in the flow rate.
Kinetic energy = 1/2 M v^2
Mass Flow Rate = density* v * Width
Power = 1/2 * (density*v*Width)*v^2
Power = 1/2 * density * Width * v^3

• Gamecock

Thx. They didn’t teach us that in limnology class.
Or I forgot it.