View source for Wave Energy Density and Flux

{{Ocean Wave Interaction with Ships and Offshore Structures
 | chapter title = [[Wave Energy Density and Flux]]
 | next chapter = [[Wave Momentum Flux]]
 | previous chapter =   [[Linear Plane Progressive Regular Waves]]
}}


= 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 not net motion of particles, but there is a transport of energy (as would be expected). The energy consists of two part, one kinetic due to the motion of the fluid and the other potential due to the the variation in the fluid height. It is the resonance between these two energies which gives rise to the wave motion. The situation is analagous to Simple Harmonic Motion but more complicated. 

= Energy per volume =

[[Image:Energy_volume.png|thumb|thumb|right|600px|Energy Volume]]

We begin by defining 
<math> \varepsilon(t) </math> as the energy in control volume <math> \gamma(t) </math> given by
<center><math> \varepsilon (t) = \rho \iiint_\Omega \left( \frac{1}{2} |\mathbf{v}|^2 + gz \right) dV </math></center>
where <math>\rho</math> is the fluid density, <math>g</math> is the acceleration due to gravity and <math>\mathbf{v}</math> is the vector
of fluid velocity. 

The mean energy over a unit horizontal surface area <math> S </math> is given by 
<center><math> \overline{\varepsilon} = \overline{\frac{\varepsilon(t)}{S}} = \rho \overline{ \int_{-H}^{\zeta(t)} \left( \frac{1}{2} V^2 + gZ \right) dZ} = \frac{1}{2} \rho \overline{ \int_{-H}^{\zeta(t)} V^2 dZ} + \overline{ \frac{1}{2} \rho g ( \zeta^2 - H^2 ) } </math></center>
where <math> \zeta(t) \, </math> is free surface elevation and the overbar denotes average (which will be important when we consider waves). Not that we are considering water of constant [[Finite Depth]].
We can ignore the term <math> -\frac{1}{2} \rho g H^2 \, </math> 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
<center><math> \overline{\varepsilon} = \overline{\varepsilon_{kin}} + \overline{\varepsilon_{pot}} </math></center>
where
<center><math> \overline{\varepsilon_{kin}} = \frac{1}{2} \rho \overline{\int_{-H}^{\zeta(t)} V^2 dZ}, \qquad V^2 = \nabla\Phi \cdot \nabla \Phi = \Phi_X^2 + \Phi_Z^2 </math></center>
and
<center><math> \overline{\varepsilon_{pot}} = \overline{\frac{1}{2} \rho g \zeta^2 (t)} </math></center>

= Energy in Linear 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 
<center><math> \Phi = \mathbf{Re} \{ \frac{igA}{\omega} e^{KZ-iKX+i\omega t} \} </math></center>
The components is the x and z directions are given by 
<center><math> \Phi_X = \mathbf{Re} \{ \frac{igA}{\omega} (-iK) e^{KZ-iKX+i\omega t} \} = A \mathbf{Re} \{ \omega e^{KZ-iKX+i\omega t} \} </math></center>
and
<center><math> \Phi_Z = \mathbf{Re} \{ \frac{igA}{\omega} K e^{KZ-iKX+i\omega t} \} 
 = A \mathbf{Re} \{ i \omega e^{KZ-iKX+i\omega t} \} </math></center>
respectively. 

We require the following <u>Lemma</u> which is easily proved. If
<center><math> \mathbf{Re} \{ A e^{i\omega t} \} = A(t) </math></center>
and
<center><math> \mathbf{Re} \{ B e^{i\omega t} \} = B(t) </math></center>
then it follows that
<center><math> \overline{A(t)B(t)} = \frac{1}{2} \mathbf{Re} \{ A B^* \} </math></center>.

This allows us to write the following expression 
<center><math> \overline{\epsilon_{kin}} = \frac{1}{2} \rho \overline{ ( \int_{-\infty}^0 + \int_0^\zeta ) \left( \Phi_X^2 + \Phi_Z^2 \right) } dZ </math></center>

<center><math> = \frac{1}{2} \rho \int_{-\infty}^0 \left( \Phi_X^2 + \Phi_Z^2 \right) dZ + O (A^3) </math></center>

<center><math> = \rho \frac{\omega^2 A^2}{4K} = \frac{1}{4} \rho g A^2 , \qquad \mbox{for} \ K=\omega^2/g </math></center>
<center>
<math> \overline{\varepsilon_{pot}} = \frac{1}{2} \rho g {\overline{\zeta(t)}}^2 = \frac{1}{4} \rho g A^2 </math>
</center>
Hence:
<center><math> \overline{\varepsilon} = \overline{\varepsilon_{kin}} + \overline{\varepsilon_{pot}} = \frac{1}{2} \rho g A^2 </math></center>

= Energy Flux =

[[Image:Moving_volume.jpg|thumb|right|600px|Moving Volume]]

<u> Energy flux </u> is rate of change of energy density <math> \varepsilon(t) </math>. 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.

<center><math> P(t) \equiv \frac{d\varepsilon(t)}{dt} , \ \varepsilon = \iiint_{V(t)} (\frac{1}{2} \rho V^2 +gZ ) dV </math></center>

<center><math> P(t) = \frac{d \varepsilon(t)}{dt} = \frac{d}{dt} \iiint_{V(t)} \epsilon(t) dV = \iint_{S(t)} \frac{\partial \epsilon(t)}{\partial t} dV + \iint_{S(t)} \epsilon(t) U_n dS </math></center> 

Transport theorem where <math> U_n </math> is normal velocity of surface <math> S(t) </math> outwards of the enclosed volume <math> V </math>.

<center><math> \frac{\partial \epsilon}{\partial t} = \frac{\partial}{\partial t} \{ \frac{1}{2} \rho V^2 + \rho g Z \} = \frac{1}{2} \rho \frac{\partial}{\partial t} ( \nabla\Phi \cdot \nabla\Phi) </math></center>

<center><math> = \rho \nabla \cdot \left( \frac{\partial\Phi}{\partial t} \nabla\Phi \right) - \rho \frac{\partial\Phi}{\partial t} \nabla^2 \Phi </math></center>

<center><math> P(t) = \frac{d \varepsilon(t)}{dt} = \rho \iiint_{V(t)} \nabla \cdot \left( \frac{\partial \Phi}{\partial t} \nabla \Phi \right) dV + \rho \iint_{S(t)} \left( \frac{1}{2} V^2 + gZ \right) U_n dS </math></center>

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

<center><math> P(t) = \rho \iint \frac{\partial\Phi}{\partial t} \nabla \Phi \dot \vec n ds + \rho \iint \left( \frac{1}{2} V^2 + gZ \right) U_n ds </math></center>

An alternative form for the energy flux <math> P(t) \, </math> crossing the closed control surface <math> S(t) \, </math> is obtained by invoking Bernoulli's equation in the second term. Recall that:

<center><math> \frac{P-P_a}{\rho} + \frac{\partial\Phi}{\partial t} + \frac{1}{2} \nabla\Phi \dot \nabla\Phi + gZ = 0 </math></center>
at any point in the fluid domain and on the boundary.

Here we allowed <math> \ P_a \equiv \mbox{Atmospheric pressure} \ </math> to be non-zero for the sake of physical clarity. Upon substitution in <math> P(t) </math> we obtain the alternate form:

<center><math> P(t) = \rho \iint \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial n} ds - \rho \iint \left( \frac{P-P_a}{\rho} + \frac{\partial\Phi}{\partial t} \right) U_n ds </math></center>

So the energy flux across <math> S(t)\, </math> is given by the terms under the interral sign. They can be collected in the more compact form:

<center><math> 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\} ds </math></center>

Note that <math> P(t) \, </math> measures the energy flux into the volume <math> V(t) \, </math> or the rate of growth of the energy density <math> \varepsilon(t)\, </math>.

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

Break <math> S(t) \, </math> into tis components and derive specialized forms of <math> P(t) \, </math> pertinent to each.

<center><math> S_F : \ \mbox{nonlinear position of the free surface} </math></center>

<center><math> \frac{\partial\Phi}{\partial n} = U_n; \ \mbox{normal flow velocity} \equiv \mbox{normal velocity of free surface boundary; kinematic condition}. </math></center>

<center><math> P = P_a; \ \mbox{fluid pressure} \equiv \mbox{atmospheric} </math></center>

Therefore over <math> S_F; \ P(t) \equiv 0 </math> as expected. No energy can flow into the atmosphere!

<center><math> S_B: \ \mbox{non-moving solid boundary} </math></center>

<center><math> U_n = 0, \ \frac{\partial\Phi}{\partial n} = U_n; \ \mbox{no-normal flux condition} </math></center>

<center><math> S^\pm: \ \mbox{fluid boundaries fixed in space relative to an earth frame} </math></center>

<center><math> U_n = 0, \ \frac{\partial\Phi}{\partial n} \ne 0 </math></center>

<center><math> S_U: \ \mbox{fluid boundaries moving w. velocity} \ \vec{U} \ \mbox{relative to an earth frame} </math></center>

<center><math> U_n = \vec U \cdot \vec n, \quad \frac{\partial\Phi}{\partial n} \ne 0 </math></center>

This case will be of interest for ships moving with constant velocity <math> U </math>.

The formulae 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.

= Energy flux across a vertical fluid boundary fixed in space. =

<center><math> P(t) = - \rho \int_{-\infty}^{\zeta(t)} \frac{\partial\Phi}{\partial t} \Phi_n dZ = - \rho \left( \int_{-\infty}^0 + \int_0^\zeta \right) \frac{\partial\Phi}{\partial t} \Phi_n dZ = - \rho \int_{-\infty}^0 \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial n} dZ + O(A^3) </math></center>

Mean energy flux for a plane progressive wave follows upon substitution of the regular wave velocity potential and taking mean values:

<center><math> \bar P = - \rho \int_{-\infty}^0 \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial x} dZ = \frac{1}{2} \rho g A^2 \left( \frac{1}{2} \frac{g}{\omega} \right) </math></center>

or

<center><math> \bar P = \bar \varepsilon V_g, \quad V_g = \mbox{group velocity} = \frac{1}{2} V_p \equiv \frac{1}{2} C </math></center>

It follows from this that the mean energy flux of a plane progressive wave is the product of its mean energy density times a velocity which equals <math> \frac{1}{2} </math> the phase velocity in deep water. We call this the [http://en.wikipedia.org/wiki/Group_velocity group velocity] of deep water waves and it is defined as

<center><math> V_g = \frac{1}{2} V_P = \frac{1}{2} \frac{g}{\omega} </math></center>

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 <math>U_n \equiv U </math> of a fluid boundary so that the mean energy flux across it vanishes?

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

<center><math> \overline{P(t)} = 0 = \rho \ {\overline{\int_{-\infty}^0 \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial x} dZ}}^t - U \ {\overline{\int_{-\infty}^0 \left(\frac{P}{\rho} + \frac{\partial\Phi}{\partial t} \right) dZ}}^t = 0 </math></center>

Where terms of <math> O(A^3) </math> have been neglected. Note that within linear theory, energy density and energy flux are quantities of <math>O(A^2)</math>. 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 <math> U </math> we obtain:

<center><math> U = \frac{\rho \ {\overline{\int_{-\infty}^0 \frac{\partial\Phi}{\partial t} \frac{\partial\Phi}{\partial x} dZ}}^t}{{\overline{\int_{-\infty}^0 \left( \frac{P}{\rho} + \frac{\partial\Phi}{\partial t} \right) dZ}}^t}  </math></center>

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

<center><math> U \equiv V_g = \frac{1}{2} \frac{g}{\omega} = \frac{1}{2} V_P </math></center>

Note that <math> U \equiv V_g </math> by definition. If the above exercise is repeated in water of finite depth the solution for <math> U </math> after some algebra is:

<center><math> U = V_g = \left( \frac{1}{2} + \frac{KH}{\sinh 2KH} \right) V_P </math></center>

with

<center><math> \omega^2 = gK \tanh KH \,</math></center>

It may be shown that the group velocity <math> V_g </math> is given in terms of <math> \omega \ne k \, </math> by the relation

<center><math> V_g = \frac{d\omega}{d K} </math></center>

= Rayleigh's proof of the group velocity formula =

This relation follows from the very elegant "device" due to [http://en.wikipedia.org/wiki/John_Strutt%2C_3rd_Baron_Rayleigh 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

<center><math> \zeta(x,t) = A cos ( \omega_1 t - K_1 x) + A cos ( \omega_2 t - K_2 x) \, </math></center>

where the amplitude is assumed to be common and:

<center><math> \omega_2 = \omega_1 + \Delta \omega, \quad | \Delta\omega | \ll \omega_1 , \omega_2 </math></center>

<center><math> K_2 = K_1 + \Delta K, \quad | \Delta K | \ll K_1 , K_2 </math></center>

Converting into complex notation:

<center><math> \zeta(x,t) = A \mathbf{Re} \{ e^{i\omega_1 t - i K_1 x} + e^{i\omega_2 t - i K_2 x} \} = A \mathbf{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} \} </math></center> <br>
<center><math> = A \mathbf{Re} \{ e^{i\omega_1 t - i K_1 x} \left( 1 + e^{i\Delta\omega t - i \Delta K x} \right) \} </math></center>

The combined wave elevation <math> \zeta \,</math> vanishes identically where <math> F \equiv \left( 1 + e^{i\Delta\omega t - i \Delta K x} \right) = 0 \, </math>.

<math> F = 0 \, \ </math> when:

<center><math> e^{i(\Delta t - \Delta K x)} = -1 \,</math></center>

or when:

<center><math> \Delta \omega t - \Delta K x = ( 2 n + 1 ) \pi, \qquad n = 0, 1, 2, \cdots </math></center>

Solving for <math> x \, </math> we obtain:

<center><math> x = \frac{1}{\Delta K} \{ (2n+1)\pi + t \Delta\omega \} \equiv X(t) </math></center>

For values of <math> X(t)\, </math> given above, <math> \zeta \equiv 0 \, </math>. These are the nodes of the bi-chromatic wave train where at all times the elevation vanishes and hence the evergy density <math> \equiv 0 </math>. The wave group has the form of consequtive packets separated by nodes. 

The speed of the nodes is <math> \frac{dx}{dt} = \frac{\Delta\omega}{\Delta K} \to \frac{d\omega}{dK} \,  </math> and the energy trapped within two consecutive nodes cannot escape so it must travel at the group velocity: <math> V_g = \frac{dx}{dt} = \frac{d\omega}{dK} \, </math> !

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

<center><math> V_g = \frac{d\omega}{dK} = \left( \frac{1}{2} + \frac{KH}{\tanh KH} \right) \frac{\omega}{K} </math></center>

= 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 "retreived" at some distance away. Deriving expressions of the energy flux retreived 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 Flux| Wave Momentum]].

-----
This article is based on the MIT open course notes and the original article can be found
[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/C2653597-0443-4705-A4E5-07A93AFCA3F5/0/lecture4.pdf here]