Deep-Water Wave Power per Metre of Crest

Energy density says how much is there. Power says how fast it arrives, and it arrives at the group velocity, because that is what carries energy in a dispersive wave field. Multiply $E = \rho g H^{2}/8$ by $c_g = gT/4\pi$ and you get $P = \rho g^{2}H^{2}T/64\pi$, watts per metre of crest.

Per metre of what, exactly, and why? Because a wave front has no width of its own. A wind turbine intercepts a definite disc and a solar panel a definite rectangle, so both have a natural area to normalise by. A wave train stretches along its crest indefinitely, and the only sensible way to quote the resource is as a flux across a line of unit length drawn along that crest. Hence kilowatts per metre, the unit every wave resource map on earth is drawn in, and hence the $\mathrm{W/m}$ dimension this page uses.

The numbers are worth having a feel for. A modest 1 m, 8-second sea carries about 4 kW/m. A 2 m, 8-second sea carries 16 — the square again. The best annual-average resources in the world, off western Ireland, the Hebrides, southern Chile, Tasmania and the exposed western edge of Vancouver Island, run 50 to 70 kW/m; a winter storm at those sites can exceed 1000 kW/m for a few hours. The convenient shortcut, with $H_s$ in metres and $T_e$ in seconds and seawater density assumed, is $P \approx 0.5\,H_s^{2}\,T_e$ kW/m, and it is accurate enough to check any published figure against.

This is the resource, not the harvest, and the distinction is the most abused number in marine energy. What a device produces is the incident flux multiplied by a capture width ratio, a power take-off efficiency, an availability figure and a transmission efficiency — and the product of those is nothing like one. Even a well-performing prototype converts a modest fraction of the flux crossing its own width, over a year, and the honest question to ask of any claimed output is: how many metres of wave front is this machine standing in? A press release that quotes the resource as though it were the output is quoting a number several times too large, and it happens constantly.

Three technical cautions sit behind the formula itself. First, it is the DEEP-WATER form, built on $c_g = c/2$; nearshore the group-velocity ratio is different, and the flux has already been cut by refraction, bottom friction and breaking. A 60 kW/m offshore resource is commonly 20 to 30 kW/m by the 10 m contour, which is awkward, because that is where devices are easiest to install and maintain. Second, the period wanted is the ENERGY period $T_e = m_{-1}/m_0$ from the spectral moments, not the peak period $T_p$ read off the top of the spectrum; $T_e$ is typically 0.85 to 0.90 of $T_p$, so substituting $T_p$ overstates the resource by about a tenth. Third, this treats the sea as a single regular wave, whereas a real sea state is a spectrum, and the proper calculation integrates the flux across it.

The economics fall out of the same asymmetry. Height enters squared and period only linearly, so a 40% rise in height doubles the power while a 40% rise in period only adds 40%. The best resources are therefore where the storms are, which is exactly where the survival problem is hardest. A wave machine's production case and its destruction case are the same physics differing only in degree, and that — more than any conversion efficiency — is why the industry has been slower than wind.