70 Ocean surface wave energy

When an ocean surface wave passes overhead at a fixed location, the surface moves up and down with the frequency of the wave,

    \[ \eta = A \hspace{2pt} cos(\omega t). \]

Let’s find the wave energy (per m2 of sea surface) for the entire water column below by vertically integrating the sum of kinetic (KE) and potential (PE) energy per unit volume.

    \[ \frac{Energy}{Area} =\int_{-H}^{\eta}\frac{PE}{volume}dz+\int_{-H}^{\eta}\frac{KE}{volume}dz. \]

Without loss of generality, we will derive this for a shallow water wave.

(1) The potential energy associated with the wave is the difference in potential energy with the wave present and with the water column sitting stationary with is sea surface at z = 0.

    \[  \int_{-H}^{\eta}\frac{PE}{volume}dz=\int_{-H}^{\eta}\frac{mgz}{volume}dz-\int_{-H}^{0}\frac{mgz}{volume}dz \]

Next, we simplify the difference of the two (definite) integrals and use the definition of density

    \[  \int_{-H}^{\eta}\frac{PE}{volume}dz=\int_{0}^{\eta}\rho gz \hspace{2pt} dz = \rho g\int_{0}^{\eta}z \hspace{2pt} dz = \frac{1}{2}\rho g \eta^2.  \]

When averaged over a full wave period (where the integral of the cosine = 1/2), we get

    \[ \int_{-H}^{\eta}\frac{PE}{volume}dz= \frac{1}{4}\rho g A^2 \]

(2) The kinetic energy associated with a shallow water wave, where w<<u is

    \[ \int_{-H}^{\eta}\frac{KE}{volume}dz=\int_{-H}^{\eta}\frac{1}{2} \rho u^2 dz \]

Substituting in our expression for the velocity

    \[ u=g \frac{k}{\omega}Acos(\omega t) = \frac{g}{C}\eta \]

where we have used the definition of wave speed C=\omega/k. Now using the relationship between shallow water wave speed and water depth, C=\sqrt{gH}, we get

    \[ u^2=\frac{g^2}{gH} \eta^2 = \frac{g}{H} \eta^2. \]

The kinetic energy becomes

    \[ \int_{-H}^{\eta}\frac{1}{2} \rho u^2 dz =\frac{1}{2}\frac{\rho g}{H}\eta^2 (\eta+H) \approx \frac{1}{2} \rho g \eta^2 \]

for small values of sea surface height displacement relative to water depth (\eta<<H).

When averaged over a full wave period, we get

    \[ \int_{-H}^{\eta}\frac{KE}{volume}dz=\frac{1}{4}\rho g A^2 \]

Note that the kinetic and potential energy terms are the same.  The wave energy is said to be equipartitioned between these energy forms (in fact, energy converts back and forth between the two forms over the wave cycle).

Key Takeaways

The total (vertically integrated) wave energy for the water column per unit m2 sea surface disturbed by a sinusoid of amplitude, A, is

    \[ E=\frac{1}{2}\rho g A^2 \quad J m^{-2} \]