# 8.2: Wavepower

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Ocean waves carry very much power – many things convince us of this, for instance: the erosion of the coastal rocks, the noise the waves attacking the Devil’s Churn at the Oregon Coast make, and last but not least, how a wave can accelerate a wave surfer. The power of the waves on oceans, seas, lakes and rivers is all converted solar power. This is a two-step conversion: first, solar power gives rise to winds – and the the winds pass their energy to waves. What is the exact physical mechanism of generating waves by wind is explained, e.g., in this Web article by Ahmed Sawan, or in this Wikipedia article. In Fig. $$\PageIndex{1}$$ there is a plot by S. J. Yang form this latter source that is used to explain the role of pressure fluctuations and the so-called “shear instability” in the formation of waves.

We will not try to get very deep into the physics of waves of water. We will only limit our attention to the power a wave carries. The power from a wave is conventionally expressed in W/m units, i.e., the power in Watts over one meter of crest length (distance along an individual crest).

Please accept without derivation the formulae for the wave power and energy density. The power, per unit length of the wave's crest, is:

$P=\frac{\rho g^{2} T H^{2}}{64 \pi} \mathrm{W} / \mathrm{m}$

where:

$$\rho=$$ the density of sea water $$=1025 \mathrm{~kg} / \mathrm{m}^{3}$$;

$$g=$$ acceleration due to gravity $$=9.81 \mathrm{~m} / \mathrm{s}^{2}$$;

$$T=$$ period of wave (in seconds; the averages are: winter $$8 \mathrm{~s}$$, summer $$6 \mathrm{~s}$$ );

$$H=$$ wave height (in meters; the averages are: winter $$3.5 \mathrm{~m}$$, summer $$1.5 \mathrm{~m}$$ ).Here $$H$$ is the significant height of the wave, i.e., its height measured from the "trough to the crest".

The energy density per a unit area is:

$E=\frac{\rho g H^{2}}{16} \mathrm{~J} / \mathrm{m}^{2}$

The dependence of the power of waves reaching Oregon Coast on the time of year is shown in Fig. $$\PageIndex{2}$$.

If we use the Eq. $$8.1$$ to calculate the power of waves at Oregon Coast in wintertime $$(H=3.5 \mathrm{~m}, T=8 \mathrm{~s})$$, we get:

$P=\frac{1025 \times 9.81^{2} \times 3.5^{2} \times 8}{64 \pi}=48079 \mathrm{~W} / \mathrm{m}=48.1 \mathrm{~kW} / \mathrm{m}$
quite consistent with the graph in Fig. $$\PageIndex{2}$$.

Oregon has about 300 miles of coastal line, or 480 kilometers. If we multiply it by $$50 \mathrm{~kW} / \mathrm{m}$$, we get $$24 \mathrm{GW}$$. The electric energy consumption per year in Oregon, according to US Energy Information Administration, is $$60.2 \mathrm{TWh}$$, which corresponds to average power consumption of $$6.9 \mathrm{GW}-$$ which means that in wintertime the poor Oregon Coast is beaten mercilessly by waves with a power three and a half times greater than the electric power consumed by the entire state!

In another US EIA Web page we find that annual energy potential of wave off the coasts of the United States is estimated to be as much as $$2.64$$ trillion kilowatt-hours, or the equivalent of about 65% of U.S. electricity generation in 2015.

The data in Fig. $$\PageIndex{3}$$, taken from on OSU NMMREC Web page, show the distribution of wavepower not only at the coastal regions, but all over the global oceans. In order to find the total energy of all ocean waves worldwide, one has to integrate such data. It’s difficult to find such information in the available sources, though. If fact, the total energy of all waves at all oceans is not particularly relevant. Why? Well, nobody will ever try to harness wave energy from waters located far away from the coasts. Obviously, only the coastal zones are really important, for two reasons – first, devices capturing wave power have to be anchored, which is not possible too far from the shore. And second, the electric power power generated has to be sent back to the coast by cables that cannot be too long.

But even making estimates of how much wave power can be potentially retrieve from coastal regions is not a straightforward task – the best illustration of this is that in a 2011 report of the International Energy Agency, a highly respected autonomous Paris-based intergovernmental organization, this estimate is given with a very high uncertainly margin: namely, from 8,000 to 80,000 TWh per year. In order to set these numbers in a right perspective, let’s say that the total global electric energy consumption in 2015 was 20,900 TWh – in other words, the realistic assessment of the available wavepower resources is that it can satisfy between 0.4 to nearly 4 times the global demand for electric power. Even if only the former of the two numbers were realistic, it’s definitely worth going for it!

8.2: Wavepower is shared under a CC BY 1.3 license and was authored, remixed, and/or curated by Tom Giebultowicz.