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!