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.
