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Pressure variation in Navier-Stokes Equation

  1. Oct 3, 2013 #1
    Hello everyone,

    I have a concern regarding the conservation of momentum for an incompressible Newtonian fluid with constant viscosity.

    Say you have a volume of fluid sliding down an inclined plane with a velocity Vx with the perpendicular axis facing upward in the y-direction. When you try to solve the set of differential equations for Navier-Stokes in y, you end up with dp/dy=row*g, hence this illustrates the hydrostatic pressure variation. What doesn't make sense is how can a hydrostatic equation for a fluid at rest be applied to a fluid moving? When you derive the hydrostatic balance equation from a simple force balance on an element of fluid you only have 3 forces: Fdown(weight at top surface), Fup(weight at bottom surface), and Fdown(gravity force)... But in our inclined moving fluid system, we should have all the same weight Forces and the y-component of Fg AND also an extra Fdown(due to viscous transportation from the flow in the x-direction). Thus by virtue of this force balance, shouldn't the pressure variation in y be shear stress dependant and not simply a hydrostatic consideration?

    I attached a diagram of the situation.

    Thanks in advance!
     

    Attached Files:

  2. jcsd
  3. Oct 3, 2013 #2
    If you are doing a differential force balance on the element you have shown, there are y-direction shear stresses on the left and the right faces of the element. One of these shear stresses is upward, and the other shear stress is downward. The two shear stresses are equal in magnitude, so they cancel.

    I might also mention that, in your equation for dp/dy, there should be a negative sign and also a cosine theta.
     
  4. Oct 3, 2013 #3
    Thanks for the reply! Could you please explain as to why there would be two shear stresses going in opposite directions here? The top surface is a liquid-gas interphase which experiences zero shear, only the bottom surface experiences shear stress due to the liquid-solid interphase boundary condition.

    Thanks!
     
  5. Oct 3, 2013 #4
    For a Newtonian fluid, [tex]σ_{xy}=σ_{yx}=η\left(\frac{\partial v_x}{\partial y}+\frac{\partial v_y}{\partial x}\right)[/tex]

    For your problem [itex]v_y=0[/itex] so,

    [tex]σ_{xy}=σ_{yx}=η\left(\frac{\partial v_x}{\partial y}\right)[/tex]

    σxy is the shear stress on planes of constant y in the x direction, and σyx is the shear stress on planes of constant x in the y direction. The directionalities of the shear stresses on the left and right sides of your volume element follow from the Cauchy stress relationship.

    Note that, because of the parabolic velocity profile, the shear stresses on the left and right faces of your volume element vary with y.

    Chet
     
  6. Oct 3, 2013 #5
    Thanks again for the help. How is it that when I perform a momentum shell balance instead, I get different results: taking the +ve y direction to be pointing down instead

    Momentum due to flow
    [tex] @ x=0, (ρWΔyv_{x}v_{x})|_{x=0} [/tex]
    [tex] @ x=L, (ρWΔyv_{x}v_{x})|_{x=L} [/tex]

    Momentum due to viscous transport
    [tex] @ y=y, (ζ_{yx}LW)|_{y=y} [/tex]
    [tex] @ y=y+Δy, (ζ_{yx}LW)|_{y=y+Δy} [/tex]

    Force due to pressure gradient
    [tex] @ y=y, (pLW)|_{y=y} [/tex]
    [tex] @ y=y+Δy, -(pLW)|_{y=y+Δy} [/tex]

    Force due to gravity (generation)
    [tex] (ρgcosθ)LWΔy [/tex]

    Assuming that we are at steady state (sum of forces equal to zero, no time accumulation) and assume fully developped flow ( [itex] v_{x=0}=v_{x=L} [/itex], momentum due to flow is zero), then

    [tex] (ζ_{yx}LW)|_{y=y} - (ζ_{yx}LW)|_{y=y+Δy} + (pLW)|_{y=y} - (pLW)|_{y=y+Δy} + (ρgcosθ)LWΔy = 0 [/tex]

    ---divide by LWΔy---

    [tex] \left(\frac{(ζ_{yx})|_{y=y} - (ζ_{yx})|_{y=y+Δy}}{Δy}\right) +\left(\frac{(p)|_{y=y} - (p)|_{y=y+Δy}}{Δy}\right) + (ρgcosθ) = 0 [/tex]

    ---taking the limit as Δy→0 and rearraging the equation---


    [tex] \left(\frac{dζ_{yx}}{dy}\right) + \left(\frac{dp}{dy}\right) = - ρgcosθ [/tex]

    This shows that [itex] \left(\frac{dp}{dy}\right) [/itex] is shear dependant, does it not?
     
  7. Oct 4, 2013 #6
    ζxyLW is a force pointing in the x direction, so it should not be included in the y momentum balance. You need to look up the cauchy stress relationship, relating the stress vector acting on a surface of arbitrary orientation to the stress tensor and to a normal drawn to that surface. For a surface of constant y with a normal pointing in the +y direction, the stress vector is given by σyyiyxyix.
     
  8. Oct 4, 2013 #7
    ζxy? I used ζyx as the shear stress pointing in the y direction due to flow in the x direction. Since there is no flow in y, shouldn't ζxy=0?
     
  9. Oct 4, 2013 #8
    No, it is not equal to zero because of the equation for the shear stresses I gave in post #4. The stress tensor is symmetric, so the xy component is equal to the yx component.

    As you know, a shear stress, by definition is the component of the total stress vector on a surface which acts in the direction tangent to the surface. In terms of your control volume, it acts on the surfaces of constant y (top and bottom) in the x direction, and on the surfaces of constant x (left and right) in the y direction. If you include the shear stresses on the left and right faces of your control volume in the force balance in the y-direction, you will find that they exactly cancel, since the shear stress does not change with x, and the shear stress on one of the faces is directed upward, while the shear stress on the other face is directed downward. So the only shear stresses that matter are the ones on surfaces of constant y, and these are part of the x direction force balance.

    Chet
     
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