Wave Energy Density and Flux

Introduction

We are interested in the transport of energy by ocean waves. It is important to realise that under the assumptions of linear theory, there is no net motion of particles, but there is a transport of energy (as would be expected). The energy consists of two parts, one kinetic due to the motion of the fluid and the other potential due to the variation in the fluid height. It is the resonance between these two energies which gives rise to the wave motion. The situation is analogous to Simple Harmonic Motion but more complicated.

Energy per volume

Energy Volume

We begin by defining $\mathcal{E}(t)$ as the energy in control volume $\Omega(t)$ given by

$\mathcal{E}(t) = \rho \iiint_\Omega \left( \frac{1}{2} |\mathbf{v}|^2 + gz \right) \mathrm{d}V$

where $\rho$ is the fluid density, $g$ is the acceleration due to gravity and $\mathbf{v}$ is the vector of fluid velocity. The mean energy over a unit horizontal surface area $S$ is given by

$\overline{\mathcal{E}} = \overline{\frac{\mathcal{E}(t)}{S}} = \rho \overline{ \int_{-h}^{\zeta(t)} \left( \frac{1}{2} |\mathbf{v}|^2 + gz \right) \mathrm{d}z} = \frac{1}{2} \rho \overline{ \int_{-h}^{\zeta(t)} |\mathbf{v}|^2 \mathrm{d}z} + \overline{ \frac{1}{2} \rho g ( \zeta^2 - h^2 ) }$

where $\zeta(t) \,$ is free surface elevation and the overbar denotes average (which will be important when we consider waves). Note that we are considering water of constant Finite Depth. We can ignore the term $-\frac{1}{2} \rho g h^2 \,$ which represents the potential energy of the ocean at rest.

The remaining perturbation component is the sum of the kinetic and potential energy components, that is

$\overline{\mathcal{E}} = \overline{\mathcal{E}_{kin}} + \overline{\mathcal{E}_{pot}}$

where

$\overline{\mathcal{E}_{kin}} = \frac{1}{2} \rho \overline{\int_{-h}^{\zeta(t)} |\mathbf{v}|^2 \mathrm{d}z}, \qquad |\mathbf{v}|^2 = \nabla\Phi \cdot \nabla \Phi = \partial_x^2\Phi + \partial_z^2\Phi$

and

$\overline{\mathcal{E}_{pot}} = \overline{\frac{1}{2} \rho g \zeta^2 (t)}$

Note that we are assuming only two dimensions $x$ and $z$.

Energy in Linear Plane Progressive Regular Waves

Consider now as a special case of Linear Plane Progressive Regular Waves by the velocity potential in Infinite Depth water (for simplicity). The velocity potential throughout the fluid domain is then given by

$\Phi = \mathrm{Re} \{ \frac{igA}{\omega} e^{kz-ikx+i\omega t} \}$

The components is the $x$ and $z$ directions are given by

$\partial_x\Phi = \mathrm{Re} \{ \frac{igA}{\omega} (-ik) e^{kz-ikx+i\omega t} \} = A \mathrm{Re} \{ \omega e^{kz-ikx+i\omega t} \}$

and

$\partial_z\Phi = \mathrm{Re} \{ \frac{igA}{\omega} k e^{kz-ikx+i\omega t} \} = A \mathrm{Re} \{ i \omega e^{kz-ikx+i\omega t} \}$

respectively. We require the following Lemma which is easily proved. If

$\mathrm{Re} \{ A e^{i\omega t} \} = A(t) \,\!$

and

$\mathrm{Re} \{ B e^{i\omega t} \} = B(t) \,\!$

then it follows that

$\overline{A(t)B(t)} = \frac{1}{2} \mathrm{Re} \{ A B^* \}.$

This allows us to write the following expression

\begin{align} \overline{\mathcal{E}_{kin}} &= \frac{1}{2} \rho \overline{ ( \int_{-\infty}^0 + \int_0^\zeta ) \left( (\partial_x\Phi)^2 + (\partial_z\Phi)^2 \right) } \mathrm{d}z \\ &= \frac{1}{2} \rho \int_{-\infty}^0 \overline{ \left( (\partial_x\Phi)^2 + (\partial_z\Phi)^2 \right) } \mathrm{d}z + O (A^3) \\ &= \rho \frac{\omega^2 A^2}{4k} = \frac{1}{4} \rho g A^2 , \qquad \mbox{for} \ k=\omega^2/g \\ \overline{\mathcal{E}_{pot}} &= \frac{1}{2} \rho g {\overline{\zeta(t)}}^2 = \frac{1}{4} \rho g A^2 . \end{align}

Note that it is a standard feature of linear oscillations that the average potential and kinetic energies are equal. Hence

$\overline{\mathcal{E}} = \overline{\mathcal{E}_{kin}} + \overline{\mathcal{E}_{pot}} = \frac{1}{2} \rho g A^2$

Energy Flux

Moving Volume

Energy flux $\mathcal{P}(t)$ is the rate of change of energy density $\epsilon(t)$. It is the flux of energy which is critical to ocean waves. While the individual fluid particles do not move the waves carry energy. We begin by deriving the energy flux in general conditions.

$\mathcal{P}(t) \equiv \frac{\mathrm{d}\mathcal{E}(t)}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t} \iiint_{\Omega(t)} \epsilon(t) \mathrm{d}V = \iint_{\partial\Omega(t)} \frac{\partial \epsilon(t)}{\partial t} \mathrm{d}V + \iint_{\partial\Omega(t)} \epsilon(t) U_n \mathrm{d} S$

(the last result following from the transport theorem) where $U_n$ is the normal velocity of surface $\partial\Omega(t)$ outwards of the enclosed volume $\Omega(t)$. We know that

\begin{align} \frac{\partial \epsilon}{\partial t} &= \frac{\partial}{\partial t} \{ \frac{1}{2} \rho |\mathbf{v}|^2 + \rho g z \} \\ &= \frac{1}{2} \rho \frac{\partial}{\partial t} ( \nabla\Phi \cdot \nabla\Phi) \\ &= \rho \nabla \cdot \left( \frac{\partial\Phi}{\partial t} \nabla\Phi \right) - \rho \frac{\partial\Phi}{\partial t} \nabla^2 \Phi \end{align}

so that

$\mathcal{P}(t) = \rho \iiint_{\Omega(t)} \nabla \cdot \left( \frac{\partial \Phi}{\partial t} \nabla \Phi \right) \mathrm{d}V + \rho \iint_{\partial\Omega(t)} \left( \frac{1}{2} |\mathbf{v}|^2 + gz \right) U_n \mathrm{d}S$

Invoking the scalar form of Gauss's theorem in the first term, we obtain:

$\mathcal{P}(t) = \rho \iint_{\partial\Omega(t)} \frac{\partial\Phi}{\partial t} \nabla \Phi \cdot \mathbf{n} \mathrm{d}S + \rho \iint_{\partial\Omega(t)} \left( \frac{1}{2} |\mathbf{v}|^2 + gz \right) U_n \mathrm{d}S$

where $\mathbf{n}$ is the unit normal.

An alternative form for the energy flux $\mathcal{P}(t) \,$ crossing the closed control surface $\partial\Omega(t) \,$ is obtained by invoking Bernoulli's equation in the second term. Recall that:

$\frac{P-P_a}{\rho} + \frac{\partial\Phi}{\partial t} + \frac{1}{2} \nabla\Phi \dot \nabla\Phi + gz = 0$

at any point in the fluid domain and on the boundary. Here we allowed $\ P_a$ the atmospheric pressure to be non-zero for the sake of physical clarity. Upon substitution in the equation above for $\mathcal{P}(t)$ we obtain the alternate form:

$\mathcal{P}(t) = \rho \iint_{\partial\Omega(t)} \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial n} \mathrm{d}s - \rho \iint_{\partial\Omega(t)} \left( \frac{P-P_a}{\rho} + \frac{\partial\Phi}{\partial t} \right) U_n \mathrm{d}s$

where $\frac{\partial\phi}{\partial n}$ is $\nabla\phi\cdot\mathbf{n}$ So the energy flux across $\partial\Omega(t)\,$ is given by the terms under the integral sign. They can be collected in the more compact form:

$\mathcal{P}(t) = \iint \left\{ \rho \frac{\partial\Phi}{\partial t} \left( \frac{\partial\Phi}{\partial n} - U_n \right) - ( P - P_a) U_n \right\} \mathrm{d}s$

Note that $\mathcal{P}(t) \,$ measures the energy flux into the volume $\Omega(t) \,$ or the rate of growth of the energy density $\mathcal{E}(t)\,$.

Energy transfer for each boundary

Break $\partial\Omega(t) \,$ into its components and derive specialized forms of $\mathcal{P}(t) \,$ pertinent to each.

• $\partial\Omega_F$ nonlinear position of the free surface. On this $\partial_n\Phi= U_n;$ and $P = P_a$, so the fluid pressure is equal to the atmospheric pressure. Therefore over $\partial\Omega_F$ $\mathcal{P}(t)=0$, as expected, i.e. there is no energy flow into the atmosphere.
• $\partial\Omega_B$ is the non-moving solid boundary, $U_n =$ and $\frac{\partial\Phi}{\partial n} = U_n=0$ which is the no-normal flux condition.
• $\partial\Omega^\pm$ which are the fluid boundaries fixed in space relative to an earth frame $U_n = 0$ and $, \frac{\partial\Phi}{\partial n} \ne 0$
• $\partial\Omega_U$ the fluid boundaries moving with velocity $\mathbf{u}$ relative to an earth frame.
• $U_n = \mathbf{u} \cdot \mathbf{n}, \quad \frac{\partial\Phi}{\partial n} \ne 0$. This case will be of interest for ships moving with constant velocity $U$.

The formula derived above are very general for potential flows with a free surface and solid boundaries. We are now ready to apply them to plane progressive waves.

Surface Wave Problem

We are ready now to apply the above formulas to the surface wave problem.

Energy flux across a vertical fluid boundary fixed in space

\begin{align} \mathcal{P}(t) &= - \rho \int_{-\infty}^{\zeta(t)} \frac{\partial\Phi}{\partial t} \Phi_n \mathrm{d}z \\ &= - \rho \left( \int_{-\infty}^0 + \int_0^\zeta \right) \frac{\partial\Phi}{\partial t} \Phi_n \mathrm{d}z \\ &= - \rho \int_{-\infty}^0 \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial n} \mathrm{d}z + O(A^3) \end{align}

Mean energy flux for a Linear Plane Progressive Regular Waves follows upon substitution of the regular wave velocity potential and taking mean values:

$\overline{\mathcal{P}} = - \rho \int_{-\infty}^0 \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial x} \mathrm{d}z = \frac{1}{2} \rho g A^2 \left( \frac{1}{2} \frac{g}{\omega} \right)$

where $A$ is the wave amplitude. or

$\overline{\mathcal{P}} = \overline{\mathcal{E}} c_g, \quad c_g = \mbox{group velocity} = \frac{1}{2} c_p$
.

It follows from this that the mean energy flux of a plane progressive wave is the product of its mean energy density times the group velocity of deep water waves.

A more formal proof that this is the velocity with which the energy flux of plane progressive waves propagates is to consider what needs to be the horizontal velocity $U_n \equiv U$ of a fluid boundary so that the mean energy flux across it vanishes?

This can be found from the solution of the following equation:

$\overline{\mathcal{P}(t)} = \rho \ {\overline{\int_{-\infty}^0 \partial_t\Phi \partial_x\Phi \mathrm{d}z}} - U \ {\overline{\int_{-\infty}^0 \left(\frac{P}{\rho} + \frac{\partial\Phi}{\partial t} \right) \mathrm{d}z}} = 0$

Where terms of $O(A^3)$ have been neglected. Note that within linear theory, energy density and energy flux are quantities of $O(A^2)$. If higher-order terms are kept then we need to consider the treatment of second-order surface wave theory, at least. Solving the above equation for $U$ we obtain:

$U = \dfrac{\rho \ {\overline{\int_{-\infty}^0 \dfrac{\partial\Phi}{\partial t} \dfrac{\partial\Phi}{\partial x} \mathrm{d}z}}} {{\overline{\int_{-\infty}^0 \left( \dfrac{P}{\rho} + \dfrac{\partial\Phi}{\partial t} \right) \mathrm{d}z}}}$

Upon substitution of the plane progressive wave velocity potential and definition of pressure from Bernoulli's equation we obtain:

$U \equiv c_g = \frac{1}{2} \frac{g}{\omega} = \frac{1}{2} c_p$

Note that $U \equiv c_g$ by definition. If the above exercise is repeated in water of finite depth the solution for $U$ after some algebra is:

$U = c_g = \left( \frac{1}{2} + \frac{kh}{\sinh 2kh} \right) c_p$

with

$\omega^2 = gk \tanh kh\,$

It may be shown that the group velocity $c_g$ is given by the relation

$c_g = \frac{\mathrm{d}\omega}{\mathrm{d} k}$

Rayleigh's proof of the group velocity formula

This relation follows from the very elegant "device" due to Rayleigh which applies to any wave form: Consider two plane progressive waves of nearly equal frequencies and hence wavenumbers. Their joint wave elevation is given by

$\zeta(x,t) = A \cos ( \omega_1 t - k_1 x) + A \cos ( \omega_2 t - k_2 x) \,$

where the amplitude is assumed to be common and

\begin{align} \omega_2 &= \omega_1 + \Delta \omega, \quad | \Delta\omega | \ll \omega_1 , \omega_2 \\ k_2 &= k_1 + \Delta k, \quad | \Delta k | \ll k_1 , k_2 \end{align}

Converting into complex notation:

\begin{align} \zeta(x,t) &= A \mathrm{Re} \{ e^{i\omega_1 t - i k_1 x} + e^{i\omega_2 t - i k_2 x} \} \\ &= A \mathrm{Re} \{ e^{i\omega_1 t - i k_1 x} + e^{i\omega_1 t - i k_1 x + i \Delta\omega t - i \Delta k x} \} \\ &= A \mathrm{Re} \{ e^{i\omega_1 t - i k_1 x} \left( 1 + e^{i\Delta\omega t - i \Delta k x} \right) \} \end{align}

The combined wave elevation $\zeta \,$ vanishes identically where $\left( 1 + e^{i\Delta\omega t - i \Delta k x} \right) = 0 \,$, i.e.

$e^{i(\Delta\omega t - \Delta k x)} = -1 \,$

or when

$\Delta \omega t - \Delta k x = ( 2 n + 1 ) \pi, \qquad n = 0, 1, 2, \cdots$

Solving for $x(t) \,$ we obtain:

$x(t) = \frac{1}{\Delta k} \{ (2n+1)\pi + t \Delta\omega \}$

For values of $x(t)\,$ given above, $\zeta = 0 \,$. These are the nodes of the bi-chromatic wave train where at all times the elevation vanishes and hence the evergy density is zero. The wave group has the form of consecutive packets separated by nodes. The speed of the nodes is $\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\Delta\omega}{\Delta k} \to \frac{\mathrm{d}\omega}{\mathrm{d}k} \,$ and the energy trapped within two consecutive nodes cannot escape so it must travel at the group velocity: $c_g = \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\mathrm{d}\omega}{\mathrm{d}k} \,$. Note that Rayleigh's proof applies equally to waves in finite depth or deep water and in principle to any propagating wave form. In finite depth it can be shown after some algebra that

$c_g = \frac{\mathrm{d}\omega}{\mathrm{d}k} = \left( \frac{1}{2} + \frac{kh}{\sinh 2kh} \right) \frac{\omega}{k}$

Summary

The formulae for the energy flux derived above are very general and for potential flow nonlinear surface waves that are not breaking constitute the energy conservation principle. Energy flux (power) input into the fluid domain by any mechanism, wavemaker, wind (in a conservative manner), a ship or any floating body must be "retrieved" at some distance away. Deriving expressions of the energy flux retrieved at "infinity" is a powerful method for estimating the wave resistance of ships (more on this later), the wave damping of floating bodies, etc. Yet, the only general way of evaluating wave forces on floating bodies (moving or not) or solid boundaries is by applying the Wave Momentum.