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Bernoulli and N-S

  1. Oct 31, 2012 #1
    hi all, i'm little confuse about the relation of these two equation.
    is it right to say that Bernoulli's equation is just a case(incompressible,inviscid,steady) of navier stoke equation?
     
  2. jcsd
  3. Oct 31, 2012 #2
    No the Bernoulli equation is an equation of energy conservation.

    Navier Stokes is an equation of motion - momentum or forces depending upon format.
     
  4. Nov 1, 2012 #3

    cjl

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    The Bernoulli equation can be directly derived from the Navier Stokes relations, so the two are definitely closely related. A number of derivations can be found online, depending on your preferred form of the N-S equations (and whether or not you are comfortable with tensor notation).
     
  5. Nov 3, 2012 #4
    Hello,

    If you start with the navier-stokes equation and assume steady state (drop ∂/∂t terms), inviscid flow (drop term with μ), and integrate over a streamline with density constant you will arrive at the bernoulli equation.
     
  6. Nov 3, 2012 #5
    And how do you guys get the navier stokes equations in the first place?
     
  7. Nov 3, 2012 #6

    boneh3ad

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    Using Reynolds' transport theorem would be the easiest way to go about it or just a straight integral control volume analysis, which is intimately related to RTT.

    At any rate, to the OP:

    If you start with the Navier-Stokes equations and assume the flow to be inviscid, you get the Euler equation
    [tex]\dfrac{\partial \rho}{\partial t} + \vec{V}\cdot\nabla\vec{V} = -\dfrac{1}{\rho}\nabla p[/tex]

    If you take the streamwise component of this equation, you get
    [tex]u\dfrac{\partial u}{\partial s} = -\dfrac{1}{\rho}\dfrac{\partial p}{\partial s}[/tex]

    Integrating this along a streamline
    [tex]\int\left(u\dfrac{\partial u}{\partial s} + \dfrac{1}{\rho}\dfrac{\partial p}{\partial s}\right) = 0[/tex]

    [tex]\dfrac{\partial}{\partial s}\int\left(u\;d u + \dfrac{d p}{\rho}\right) = 0[/tex]

    [tex]\dfrac{\partial}{\partial s}\left(\dfrac{u^2}{2} + \int\dfrac{d p}{\rho}\right) = 0[/tex]

    which is a statement of Bernoulli's principle where the integrand is constant over a streamline.
     
    Last edited: Nov 3, 2012
  8. Nov 29, 2012 #7
    I was solving N-S using SIMPLER algorithm and found that there is term u*dv/dy, where u=viscosity, v=vertical component of velocity, is u=0 for this term, since there won't be stress perpendicular to the surface. Need help
     
  9. Nov 29, 2012 #8

    boneh3ad

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    You really ought to start your own thread instead of hijacking one. At any rate, unless you have a good reason for doing so, you can't just call viscosity zero.
     
  10. Nov 29, 2012 #9
    Thanks man! Below is something that I know about Bernoulli and N-S

    Bernoulli and N-S are special cases of transport Equations. When the flow is steady and inviscid, the 'Energy(enthalpy)-Transport Equation' reduces to Bernoulli, i.e. Total energy is conserved. The Momentum-Transport Equation is called the N-S Equation, it is actually not a special, they are well known as Momentum Equation.
     
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