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Help Deriving the Navier Stoke's Eq

  1. Mar 3, 2014 #1

    joshmccraney

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    hey pf!

    i was hoping one of you could help me with deriving the navier stokes equations. i know the equations state something like this [tex]\sum \vec{F}=m\vec{a}[/tex] and for fluids we have something like this for the forces: [tex]\underbrace{-\iint Pd\vec{s}}_{\text{pressure}}+\underbrace{\vec{F}}_{\text{viscous forces}}-\underbrace{\iint \rho\vec{V}\vec{V}\cdot d\vec{s}}_{\text{momentum leaving flux}}[/tex] and for the acceleration terms we have [tex]\underbrace{\frac{\partial}{\partial t}\iiint \rho\vec{V} dv}_{\text{time rate of change of momentum}}[/tex]

    now ultimately we can use the divergence theorem, some vector/tensor identities, and arrive at the navier stokes equation: [tex]\rho \frac {D \vec{V}}{Dt} = - \nabla P + \mu \nabla^2 \vec{V}[/tex]

    my question is, can someone please explain to me how to arrive from my [itex]\vec{F}[/itex] viscous force to the [itex]\mu \nabla^2 \vec{V}[/itex]

    i'm good with the rest, but i'm just not sure how to start from an integral calculus perspective.

    thanks!
     
  2. jcsd
  3. Mar 3, 2014 #2

    maajdl

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  4. Mar 3, 2014 #3
    The viscous force per unit volume is [itex]∇\centerdot \vec{σ}_v[/itex], where [itex]\vec{σ}_v[/itex] is the viscous portion of the stress tensor, and is related to the kinematics of the deformation by
    [itex]\vec{σ}_v=2μ\vec{E}[/itex]
    where [itex]\vec{E}[/itex] is the rate of deformation tensor:

    [tex]\vec{E}=\frac{(∇\vec{v})+(∇\vec{v})^T}{2}[/tex]

    In this equation, [itex]\vec{v}[/itex] is the velocity vector.
     
  5. Mar 4, 2014 #4

    joshmccraney

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    can you explain where [itex]\vec{E}[/itex] comes from and how it relates to the rate of deformation (this is not intuitive to me)? looking at wikipedia it appears we have two matrices; namely the [itex]\vec{E}[/itex] as you have listed and another, they name [itex]\vec{R}[/itex] defined as [itex]\vec{R}=\frac{(∇\vec{v})-(∇\vec{v})^T}{2}[/itex]. i'm just not too sure what is going on or where these two are coming from ([itex]\vec{E}[/itex] and [itex]\vec{E}[/itex] that is)

    thanks!
     
  6. Mar 4, 2014 #5
    The rate of deformation tensor relates to the rate at which the various material points in the fluid separate from one another as the fluid deforms. It's kind of a rate of stretching. The R tensor is the vorticity tensor, which relates to how rapidly fluid elements are rotating in the flow. This rotation does not contribute to the rate of separation of the material points. The relationship between the stress tensor and the rate of deformation tensor (in terms of the viscosity) is the 3D generalization of Newton's law of viscosity, and reduces to Newton's law of viscosity for the case of shear between parallel plates.
     
  7. Mar 16, 2014 #6
    I have a similar question, but here I'm just interested in the derivation of the Stokes equation (Material Equation, etc):
    Where a=dV/dt and V is a vector of d/dt(u,v,w) (i,j,k are respective here). But my fundamental question is why u, v, and w would be functions of (x,y,z,t)? I understand t and x being functions of, say u, because the u is defined as the velocity in the i direction.

    Any help is appreciated.
     
  8. Mar 16, 2014 #7
    The x component of velocity u can also be a function of y and z. For example, suppose you have a "shear flow" where [itex]u=γy[/itex], where y is the distance from a horizontal wall, and γ is the "shear rate."
     
  9. Mar 17, 2014 #8
    That was easy! Thanks much.
     
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