Wave Momentum Flux: Difference between revisions

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<u> Momentum flux in potnetial flow </u>
<u> Momentum flux in potential flow </u>


<center><math> \frac{d\overline{M(t)}}{dt} = \rho \frac{d}{dt} \iiint_V(t) \bar V dV = \rho \iiint_V(t) \frac{\partial\bar{V}}{\partial t} dV + \rho \oiint_{S(t)} \bar{V} U_n dS, \ \mbox{by virtue of the transport theorem} </math></center>
<center><math> \frac{d\overline{M(t)}}{dt} = \rho \frac{d}{dt} \iiint_V(t) \bar V dV = \rho \iiint_V(t) \frac{\partial\bar{V}}{\partial t} dV + \rho \oint_{S(t)} \bar{V} U_n dS, </math></center>
by virtue of the transport theorem


Invoking Euler's equations in inviscid flow
Invoking Euler's equations in inviscid flow

Revision as of 22:36, 16 February 2007

Momentum flux in potential flow

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

by virtue of the transport theorem

Invoking Euler's equations in inviscid flow

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

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