Wave Momentum Flux

Momentum flux in potential flow

[math]\displaystyle{ \frac{d\overrightarrow{M(t)}}{dt} = \rho \frac{d}{dt} \iiint_V(t) \overrightarrow V dV = \rho \iiint_V(t) \frac{\partial\overrightarrow{V}}{\partial t} dV + \rho \oint_{S(t)} \overrightarrow{V} U_n dS, }[/math]

by virtue of the transport theorem

Invoking Euler's equations in inviscid flow

[math]\displaystyle{ \frac{\partial\overrightarrow{V}}{\partial t} + (\overrightarrow{V} \cdot \nabla ) \overrightarrow V = - \frac{1}{e} \nabla P + \overrightarrow g }[/math]

We may recast the rate of change of the momentum ([math]\displaystyle{ \equiv \, }[/math] momentum flux) in the form

[math]\displaystyle{ \frac{d\overrightarrow{M(t)}}{dt} = - \rho \iiint_V(t) [ \nabla ( \frac{P}{\rho} + g Z ) + ( \overrightarrow{V} \cdot \nabla ) \overrightarrow{V} ] dV + \rho \oint_{s(t)} \overrightarrow{V} U_n dS }[/math]

So far [math]\displaystyle{ V(t)\, }[/math] is and arbitrary closed time dependent volume bounded by the time dependent surface [math]\displaystyle{ S(t)\, }[/math]. Here we need to invoke an important and complex vector theorem.

Recall from the proof of Bernoulli's equation that:

[math]\displaystyle{ (\overrightarrow{V} \cdot \nabla ) \overrightarrow{V} = \nabla ( \frac{1}{2} \overrightarrow{V} \cdot {V} ) - \overrightarrow{V} \times (\nabla \times \overrightarrow{V} ) }[/math]

By virtue of Gauss's vector theorem:

[math]\displaystyle{ \iiint_{V(t)} \nabla ( \frac{1}{2} \overrightarrow{V} \cdot \overrightarrow{V} ) dV = \frac{1}{2} \oint_{S(t)} \overrightarrow{V} \cdot \overrightarrow{V} \overrightarrow{n} dS }[/math]

where in potential flow: [math]\displaystyle{ \overrightarrow{V} = \nabla \Phi \, }[/math].

In potential flow it can be shown that:

[math]\displaystyle{ \oint_{S(t)} \frac{1}{2} ( \overrightarrow{V} \cdot \overrightarrow{V} ) \overrightarrow{n} dS = \oint_{S(t)} \frac{\partial\Phi}{\partial n} \nabla \Phi dS = \oint_{S(t)} V_n \overrightarrow{V} dS }[/math]

Proof left as an exercise! Just prove that for [math]\displaystyle{ \nabla^2 \Phi = 0 \, }[/math];

[math]\displaystyle{ \oint_S \frac{1}{2} ( \nabla\Phi \cdot \nabla\Phi) \overrightarrow{n} dS \equiv \oint_S \frac{\partial\Phi}{\partial n} \nabla\Phi dS. }[/math]

Upon substitution inthe momentum flux formula, we obtain:

[math]\displaystyle{ \frac{d\overrightarrow{M}}{dt} = - \rho \oint_S(t) [ ( \frac{P}{\rho} + gZ ) \overrightarrow{n} + \overrightarrow{V} (V_n - U_n )] dS }[/math]

Following an application of the vector theorem of Gauss. In the expression above [math]\displaystyle{ \vec{n} \, }[/math] is the unit vector pointing out of the volume [math]\displaystyle{ V(t), V_n = \vec{n} \cdot \nabla \Phi }[/math] and [math]\displaystyle{ U_n \, }[/math] is the outward normal velocity of the surface [math]\displaystyle{ S(t)\, }[/math].

This formula is of central importance in potential flow marine hydrodynamics because the rate of change of the linear momentum defined above is just [math]\displaystyle{ \pm \, }[/math] the force acting on the fluid volume. When its mean value can be shown to vanish, important force expressions on solid boundaries follow and will be derived in what follows.

Consider separately the term in the momentum flux expression involving the hydrostatic pressure:

[math]\displaystyle{ \frac{d\overrightarrow{M_H}}{dt} = - \rho \oint_S gZ \vec{n}, \ \mbox{where} \ S=S_F + S^{\pm} + S_B + S_{-\infty} }[/math]

The integral over the body surface, assuming a fully submerged body is:

[math]\displaystyle{ \frac{d\overrightarrow{M_{H,B}}}{dt} = - \rho \oint_{S_B} gZ \vec{n} = \rho g V \vec{k} = \overrightarrow{B} }[/math]

By virtue of the vector theorem of Gauss this is none other than the principle of archimedes!

We may therefore consider the second part of the integral involving wave effects independently and in the absence of the body, assumed fully submerged. In the case of a surface piercing body and in the fully nonlinear case matters are more complex but tractable!

Consider the application of the momentum conservation theorem in the case of a submerged or floating body in steep waves:

Here we consider the two-dimensional case in order to present the concepts. Extensions to three dimensions are then trivial.

Note that unlike the energy conservation principle, the momentum conservation theorem derived above is a vector identity with a horizontal and a vertical component.

The integral of the hydrostatic term over the remaining surfaces leads to:

[math]\displaystyle{ \frac{d \overrightarrow{M}_{H,S}}{dt} = - \rho \iint_{S_F+S^+ + S^- +S_{\infty}} gZ \bar{n} dS = - \rho g \forall_{\mbox{Fluid}} \vec{K} }[/math]

This is simply the static weight of the volume of fluid bounded by [math]\displaystyle{ S_F, S^+, S^- \, }[/math] and [math]\displaystyle{ S_{\infty}\, }[/math]. With no waves present, this is simply the weight of the ocean water "column" bounded by [math]\displaystyle{ S\, }[/math] which does not concern us here. This weight does not change in principle when waves are present at least when [math]\displaystyle{ S^+, S^- \, }[/math] are placed sufficiently far away that the wave amplitude has decreased to zero. So "in principle" this term being of hydrostatic origin may be ignored. However, it is in principle more "retional" to apply the momentum conservation theorem over the "linearized" volume [math]\displaystyle{ V_L(t) \, }[/math] which is perfectly possible within the framework derived above. In this case [math]\displaystyle{ \frac{d\overrightarrow{M}_H}{dt} \, }[/math] is exactly equal to the staic weight of the water column and can be ignored in the wave-body interaction problem. On [math]\displaystyle{ S_F; P=P_a=0 \, }[/math] and hence all terms within the free-surface integral and over [math]\displaystyle{ S_{\infty} \, }[/math] (seafloor) can be neglected. It follows that:

[math]\displaystyle{ \frac{d \overrightarrow{M}}{dt} = - \rho \iint_{S^\pm +S_B} \left[ \frac{P}{\rho} \vec{n} + \overrightarrow{V} (V_n -U_n) \right] dS }[/math]

Note that the free surface integrals also vanish for the horizontal component since the hydrostatic force is always vertical!

This momentum flux formula is of central importance in wave-body interactions and has many important applications, some of which are discussed bellow.

Note that the mathematical derivations involved in its proof apply equally when the volume [math]\displaystyle{ V\, }[/math] and its enclosed surface are selected to be at their linearized positions. In such a case it is essential to set [math]\displaystyle{ U_n=0\, }[/math] and [math]\displaystyle{ V_n \ne 0 }[/math]. Let the math take over and suggest the proper expression for the force. In the fully nonlinear case, [math]\displaystyle{ U_n \ne 0 \, }[/math] on [math]\displaystyle{ S_F\, }[/math] and [math]\displaystyle{ P=0 \, }[/math] on [math]\displaystyle{ S_F \, }[/math]!

On a solid boundary:

[math]\displaystyle{ U_n =V_n \, }[/math]

and

[math]\displaystyle{ \overrightarrow{F}_B = \iint_{S_B} P \vec{n} dS }[/math]

With [math]\displaystyle{ \vec{n} \, }[/math] pointing inside the body.

We may therefore recast the momentum conservation theorem in the form:

[math]\displaystyle{ \overrightarrow{F}_B (t) = - \frac{d \overrightarrow{M}}{dt} - \rho \iint{S^\pm} \left[ \frac{P}{\rho} \vec{n} + \overrightarrow{V} (V_n - U_n) \right] dS }[/math]

Where [math]\displaystyle{ S^\pm \, }[/math] are fluid boundaries at some distance from the body.

If the volume of fluid surrounded by the body, free surface and the furfaces [math]\displaystyle{ S^\pm \, }[/math] does not grow in time, then the momentum of the enclosed fluid cannot grow either, so [math]\displaystyle{ \overrightarrow{M}(t) \, }[/math] is a stationary physical quantity.

It is a well known result that the mean value in time of the time derivative of a stationary quantity is zero. So:

[math]\displaystyle{ {\overline{\frac{d\overrightarrow{M}}{dt}}}^t = 0 }[/math]

Proof :