Why Are Gravity Effects Neglected in Stokes Flow Analysis?

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SUMMARY

In Stokes flow analysis, gravity effects are often neglected due to the assumption of low Reynolds number (Re→0), allowing the material derivative of velocity (Dv/Dt) to equal zero. The gravitational force per unit mass, represented as \mathbf{g} = -\nabla\left(-\int\mathbf{g} \cdot d\mathbf{x}\right), can be incorporated into the pressure term, leading to the concept of "modified pressure." This approach simplifies the mathematical treatment of fluid dynamics without altering the fundamental equations.

PREREQUISITES
  • Understanding of Stokes flow and low Reynolds number fluid dynamics.
  • Familiarity with the concept of material derivatives in fluid mechanics.
  • Knowledge of pressure terms in fluid equations, specifically modified pressure.
  • Basic grasp of vector calculus and gradient operations.
NEXT STEPS
  • Study the implications of low Reynolds number on fluid behavior in Stokes flow.
  • Research the derivation and applications of modified pressure in fluid dynamics.
  • Explore the mathematical treatment of gravitational effects in incompressible flow.
  • Learn about the role of density assumptions in fluid mechanics, particularly in Stokes flow.
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Fluid dynamics researchers, mechanical engineers, and students studying advanced fluid mechanics concepts, particularly those focusing on Stokes flow and pressure dynamics.

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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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The density is being treated as constant, so the gravational force per unit mass is \mathbf{g} = -\nabla\left(-\int\mathbf{g} \cdot d\mathbf{x}\right).

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

You can therefore absorb it into the pressure term. It makes no difference to the mathematics if you work with p or p' = p - \int\mathbf{g} \cdot d\mathbf{x}.
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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