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Navier-stokes derivation question

  1. Apr 2, 2014 #1

    joshmccraney

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    hey pf!

    so i have a small question when deriving the navier-stokes equations from newton's 2nd law. specifically, newton states that $$\Sigma \vec{F} = m \vec{a} = m \frac{d \vec{v}}{dt}$$

    when setting a control volume of fluid and dealing with the time rate-of-change of momentum we write $$m \frac{d \vec{v}}{dt} = \frac{\partial}{\partial t} \iiint_V \rho \vec{v} dV$$ but isnt it true that $$\frac{\partial}{\partial t} \iiint_V \rho \vec{v} dV = \frac{d (m \vec{v})}{dt}$$

    can someone please help me out here?

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

    Andy Resnick

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    A few points:
    $$\Sigma \vec{F} = \frac{d(m\vec{v})}{dt}$$

    $$\frac{d}{dt} \iiint_V A dV = \iiint_V \frac{\partial A}{\partial t} dV + \iint_{\partial V} A \vec{v} \bullet \vec{n} dS$$ (Reynolds Transport Theorem)
     
  4. Apr 2, 2014 #3

    joshmccraney

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    thanks andy. are you saying newtons second law is not force = mass * acceleration but rather rate of change of mass * acceleration?

    also, i have not invoked reynolds transport theorem and was hoping i could go through the derivation without it (or at least this step). do we need to use reynolds theorem here?
     
  5. Apr 2, 2014 #4

    Andy Resnick

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    Force = the time rate of change of momentum (momentum = mass*velocity)

    There are probably ways to derive the Navier-Stokes momentum equation from Newton's second law without the Reynolds transport theorem, but I can't think of one right now.
     
  6. Apr 2, 2014 #5

    boneh3ad

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    The only way I have ever seen it done without the Reynolds Transport Theorem involves using a control volume and volume integrals, which simply reduces to the Reynolds Transport Theorem anyway.
     
  7. Apr 2, 2014 #6

    joshmccraney

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    thanks to you both! would it be okay with you two if i asked a few more questions about the derivation for navier stokes? i'm writing a .pdf so i can be sure i understand "most" of what is going on. i do, however, still have a few questions, if you guys dont mind?
     
  8. Apr 2, 2014 #7

    joshmccraney

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    actually, to save time, if you both (or either) are okay with assisting me in going through the derivation, perhaps i'll start by asking if the following is correct starting from newton's second law: ##\frac{d(m\vec v)}{dt}=\Sigma \vec{F}##

    [tex]
    \underbrace{\frac{\partial}{\partial t} \iiint_V \rho \vec{v} dV}_{\text{Momentum Rate of Change}} = \underbrace{-\iint_{\partial V} \rho \vec{v}\vec{v} \cdot d\vec{S}}_{\text{Momentum Flux}}\:\:\: + \underbrace{- \iint_{\partial V} P d\vec{S}}_{\text{Pressure Force}} \:\:\:+ \underbrace{\iiint_V \rho \vec{g} dV}_{\text{Body Force (gravity)}} +\underbrace{-\iint_{\partial V} \vec{{\tau}} \cdot d\vec{S}}_{\text{Shearing Forces}}
    [/tex]
    where all notation is basic, although i will say for clarity that ##\vec\tau## is the stress tensor (sorry, im not sure how to bold within physics forums, as i had to use a package in latex). i think everything else is obvious. also, ##\vec v \vec v## is a 2nd rank tensor, using the dyadic product. i assume only gravity as a body force for this derivation, although im not too concerned here.
     
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