Wave Power – A Simple Analysis

How much power can we extract from ocean waves?

Energy required to disturb the surface of water:

Before we ask how much power we can extract from waves, we have to estimate how much energy there is in a wave. Consider a moving disturbance in the surface of a body of water that looks something like Fig.1, which is not a realistic shape but a good starting point for analysis:

Fig.1. A simple square wave.


The energy required to create this disturbance is just the work done in moving the water that was in the trough of the wave up to the crest of the wave (Fig.2).

Fig. 2. How water has to be displaced to make one wavelength of a wave.

Thus a body of water height h, width w and length L has been moved vertically up a distance h (centre-of-mass to centre-of-mass) as shown in Fig.3.

Fig.3. Finding the potential energy in one wavelength of a square wave.


If the density of water is ρ then the mass m involved in the move is ρwhL. As the work done W to raise this mass a height h is mgh (Eqn.1):

$W = \rho g w h^2 L \tag{1}$

Power in a wave-train

In the open ocean such disturbances usually occur repetitively, with a frequency f and a spacing λ, i.e. the speed of the waves, v = f λ. If the disturbances are packed together, λ = 2w.

We won’t worry about the unnatural shape of the waves for now. Rectangles are easier to deal with than real wave shapes.

If we have some kind of energy absorber at the end of this wave train capable of using waver power to generate, say, electricity, the rate at which waver energy crosses this absorber is (Eqn.2):

$P = \rho g w h^2 L f = \rho g (\dfrac{1}{2} \lambda ) h^2 L f = \dfrac{1}{2} \rho g h^2 L v  \tag{2}$

Consider a typical ocean wave train: h = 1 m, λ = 10 m, f = 0.1 Hz (period = 10 s), i.e. v = 1 m/s.

Assume the energy absorber is L = 10 m long, typical for such an installation (Eqn.3):

$P = \dfrac{1}{2} (1000 \textnormal{ kg/m}^3)(10 \textnormal{m/s}^2)(1 \textnormal{ m})^2 (10 \textnormal{ m})(1 \textnormal{ m/s}) = 50,000 \textnormal{ W , i.e. 50 kW}   \tag{3}$

One big omission we have made here is that of kinetic energy. The wave form we show is not static. The water is moving. Fortunately there is a very powerful result known as the Virial Theorem[note] Virial Theorem, https://en.wikipedia.org/wiki/Virial_theorem [2019-10-15].[/note] that says that in this instance of a gravitational restoring force, the magnitude of the kinetic energy will be the same as that of the potential energy. Thus $P$ will be doubled. However, our crude square will overestimate $P$ compared to a somewhat more realistic sine wave[note] Stokes Wave, https://en.wikipedia.org/wiki/Stokes_wave [2019-10-16].[/note], again by a factor of two. Thus, these two considerations will cancel each other out.

Although still a rough approximation, this result is close to that obtained with a much more sophisticated water-wave model. It shows that a significant amounts of power are potentially available in water waves. However, one must remember that 50 kW from this 10-metre size device would only service the energy needs of a few single-family dwelling. And we haven’t said anything yet about how this power can be extracted[note]Wave Power,  http://en.wikipedia.org/wiki/Wave_power [2019-10-15].[/note].

For real current wave data, suitable for use in student projects, see the U.S. NOAA National Data Buoy Center. Zoom in on a buoy and click to see current conditions[note] National Oceanic and Atmospheric Administration, National Data Buoy Center, http://www.ndbc.noaa.gov/ [2019-10-15]. [/note].


Updated (CEW) 2019-10-15