External forces on Stokes flow

  • Thread starter Est120
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  • #1
Est120
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Homework Statement:
find the appropiate Navier-Stokes equation in the low Re number limit
Relevant Equations:
Navier Stokes equations
in the limit as Re→0 , we can neglect the material derivate of v ( Dv/Dt =0 ) but why in books they always make the gravity effects equal to 0?
i can't understand and no one really explains this stuff
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  • #2
pasmith
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The density is being treated as constant, so the gravational force per unit mass is [itex]\mathbf{g} = -\nabla\left(-\int\mathbf{g} \cdot d\mathbf{x}\right)[/itex].

You can therefore absorb it into the pressure term. It makes no difference to the mathematics if you work with [itex]p[/itex] or [itex]p' = p - \int\mathbf{g} \cdot d\mathbf{x}[/itex].
 
  • #3
Est120
50
3
The density is being treated as constant, so the gravational force per unit mass is [itex]\mathbf{g} = -\nabla\left(-\int\mathbf{g} \cdot d\mathbf{x}\right)[/itex].

You can therefore absorb it into the pressure term. It makes no difference to the mathematics if you work with [itex]p[/itex] or [itex]p' = p - \int\mathbf{g} \cdot d\mathbf{x}[/itex].
yeah, that's the so called "modified pressure" right? i hardly managed to get that but still almost no source of information talks about that...
 

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