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Pressure gradient term in Navier-Stokes

  1. Mar 14, 2015 #1
    Hi,

    I've been thinking about the Navier-Stokes equations and trying to build skill in implementing it in various situations.
    In a particular situation, if I have a fluid flowing down an inclined surface such that it forms a film of finite height which is smaller than the length of flow, there is no applied pressure. So am I allowed to cross off the pressure term in the momentum conservation equation along the direction of flow?
    In a detailed solution to this problem, our instructor non-dimensionalizes the equation to show that the pressure term goes away.
    However, I'm wondering, since there is no applied pressure, can't I just cross it off (like Couette flow)?

    Also, in another problem (the situation is that oil is flowing up an inclined surface submerged in water) the instructor replaces the pressure gradient by the buoyancy force . So in the inclined flow case, am I not allowed to simply replace the pressure term by the hysdrostatic pressure?

    I think my question boils down to whether the pressure term in the NS equation is an external force or internal property of the fluid.
     
  2. jcsd
  3. Mar 15, 2015 #2
    It doesn't have to be non-dimensionalized to do this. The force balance in the direction perpendicular to the surface at any location shows that the pressure is varying hydrostatically in that direction. So it's equal to the atmospheric pressure plus ρgcosθy, where y is the distance measured downward normal to the air interface. If the film thickness is constant, this means that the pressure is not varying with distance along the incline.
    I don't follow the description of the oil problem. Also, I don't understand what you mean about external and internal. Pressure is part of the stress tensor, so it has to be internal, except that it has to match the adjacent medium at an interface.

    Chet
     
  4. Mar 15, 2015 #3
    So in the problem the film thickness is actually varying (increasing) along the flow direction.So if I follow your method and draw a force balance on it, it means the pressure near the end of the film where its thickness is highest is greater?

    The confusion relating to the 'internal'/'external'-ness of the pressure arises because of the couette flow with no pressure applied i.e. shear flow example. Here fluid flows from one side to another due to the external force of the plate. So there is no external pressure gradient, but the fluid should develop a pressure gradient within itself so that it does indeed flow from one side to the other. Does this sound correct?
     
  5. Mar 15, 2015 #4
    I think you forgot to include the word "if" somewhere in this text. In your problem, is the film thickness changing as the fluid flows down the incline? Even so, if the rate of change of the thickness with distance along the incline is very small, you can still approximate the pressure gradient along the incline as negligible.
    I'm having trouble understanding your description. Are you talking about shear flow between concentric rotating cylinders, or are you talking about shear flow between parallel plates?

    Chet
     
  6. Mar 17, 2015 #5
    Ok, so if the rate of change of thickness along the plane is small, the film is of practically uniform thickness along the incline...and the only net force on a small slice of fluid is gravitational. If the film were changing thickness (say increasing along the incline), the net force on a slice of fluid is the gravitational force (pointing down the incline) minus some net hydrostatic force (up the incline)...then there is a pressure gradient in the fluid such that pressure is greater as you go down the incline?

    Let's say shear flow between parallel plates. Doing a similar force balance above, gravity acts downward (perpendicular to plates) so it doesn't cause acceleration of fluid. There is no applied pressure gradient. The fluid moves because of momentum transfer from the moving plate.

    If we had shear flow between rotating cylinders, it's a similar situation except that if the cylinders are vertical, gravity will come into play and a pressure gradient will come into play.

    Is this all okay?
     
  7. Mar 17, 2015 #6
    Yes. If the film thickness is varying rapidly, there will be a pressure gradient along the direction of flow. Just think of a horizontal trough (open channel) in which you are adding fluid rapidly at the left end of the trough so that there is a depth gradient along the direction of flow from left to right. This depth gradient causes a pressure gradient along the trough, which results in the flow.

    Yes.
    The pressure gradient will be in the vertical direction (hydrostatic), perpendicular to the shear flow. So, the vertical pressure gradient will not affect the shear flow. There will also be a radial pressure gradient to provide the centripetal force. However, this also will be perpendicular to the velocity gradient, so it won't affect the shear flow either. To affect the shear flow, you would need to have a circumferential pressure gradient, but this is prevented by the axial symmetry.

    Chet
     
  8. Mar 18, 2015 #7
    Ok thanks Chet, I'm feeling much better about this now. So to summarize, the pressure in a fluid can be from two main sources: 1. hydrostatic pressure
    2. applied pressure .
    Also, just because a fluid is moving, it doesn't mean that the fluid 'creates its own pressure gradient within itslef' (what I was previously calling 'internal pressure'), (like in the shear flow between parallel plates).
     
  9. Mar 18, 2015 #8
    Well, if you look at the terms in the Navier Stokes equations, you will see that it can be from three sources:

    1. viscous stresses (e.g., flow in a pipe)
    2. gravitational forces (hydrostatic)
    3. inertial forces (like the radial pressure gradient in Couette flow)

    Applied pressure is what you have to impose on the boundaries of the fluid to overcome the viscous forces, gravitational forces, and inertial forces. There is a close connection between the pressure and the boundary conditions. This is because, for an incompressible fluid, the pressure is determined only up to an arbitrary constant. It is necessary for the boundary conditions to establish the value of that constant. For example, in the case of the flow down an inclined plane, the pressure at the free surface was atmospheric.

    But, please, don't try too hard to categorize it. After you have solved some problems, you well get a much better feel for how all this works. You just need to get some experience and have patience.

    Yes. As I said, you will soon have a feel for all this.

    Chet
     
  10. Mar 19, 2015 #9
    Ok, thanks I think I get it :-)
     
  11. Apr 27, 2015 #10
    So I have a related question to what we were discussing earlier. I usually don't need to non-dimensionalize the momentum balance in the y-direction, but this one time I did and I realized I was a bit confused. Please take a look at the description of my non-dimensionalization procedure in the attached. It turns out that after proper rearrangement, the non-dimensionalization in the y direction is very similar to the x-direction one (when I rearrange to bring in the Re_L i.e. the Re taking characteristic length in x direction), except with (L/h) everywhere.
    Now in the situation L>>h, it looks like the convective term foes away entirely and we're left with the pressure term., the gravity term and the viscous terms. So now without the Re being >>1, we can't eliminate the viscous terms. However in most situations we are only supposed to be left with the pressure and gravitational term so that it is just hydrostatics. My Prof got to that conclusion by not going to the extent of rearranging everything to get the Re_L and just said all terms other than the pressure term and gravity term are multiplied by (h/L)^2 so they're small. However isn't there a way to get to that conclusion from where I ended up?
     

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  12. Apr 27, 2015 #11
    I'm pretty sure I can help you, but I'm unable to read the attachments you sent. It's just too small on my computer screen, and, unlike ordinary web pages, I'm unable to zoom on the attachment. Any chance you can Latex it?

    Chet
     
  13. Apr 28, 2015 #12
    So unfortunately I've never used Latex before :-( ...would probably be a useful thing to learn. I've taken a different set of pictures, more zoomed in, so pleasetake a look if this is better. If not, I'll try to learn Latex :-)

    Thanks a lot! (1).JPG (2).JPG (3).JPG
    (1).JPG
     
  14. Apr 28, 2015 #13
    OK. Here's what I got, assuming y is the vertical direction:

    I set ##P_c=ρu^2##. So, for the x direction:

    $$\left[\frac{\partial v_x^*}{\partial t^*}+v_x^*\frac{\partial v_x^*}{\partial x^*}+v_y^*\frac{\partial v_x^*}{\partial y^*}\right]=-\frac{\partial p^*}{\partial x^*}+\frac{ηL}{ρuh^2}\left[\left(\frac{h}{L}\right)^2\frac{\partial ^2 v_x^*}{\partial x^{*2}}+\frac{\partial ^2 v_x^*}{\partial y^{*2}}\right]$$

    For the y direction, I get:
    $$\left(\frac{h}{L}\right)^2\left[\frac{\partial v_y^*}{\partial t^*}+v_x^*\frac{\partial v_y^*}{\partial x^*}+v_y^*\frac{\partial v_y^*}{\partial y^*}\right]=-\frac{\partial p^*}{\partial y^*}+\frac{ηL}{ρuh^2}\left[\left(\frac{h}{L}\right)^4\frac{\partial ^2 v_y^*}{\partial x^{*2}}+\left(\frac{h}{L}\right)^2\frac{\partial ^2 v_y^*}{\partial y^{*2}}\right]+\frac{gh}{u^2}$$

    Chet
     
  15. Apr 28, 2015 #14
    Yup, that's what I get too,but if you rearrange a little more (y direction), you will get Reynolds number (=rho*u*L/mu) multiplied by (L/h)^2 in the viscous terms...except your rho has gone away. So then how would you prove that the y direction equation boils down to hydrostatics? Currently in your equation, you have (h/L)^2 multiplied into all but the pressure and gravity terms, so they're small....but after the rearrangement to get the Re in there, the situation looks a but different.
     
  16. Apr 28, 2015 #15
    Well, for this particular situation, in which the length scale in the y direction is much smaller than the length scale in the x direction, the more appropriate definition of the Reynolds number would be ##Re = \frac{ρuh^2}{ηL}##. So presumably, with this definition, Re would be held finite while h/L becomes small. This would be accomplished with adjustment to u and/or η.

    Chet
     
  17. Apr 29, 2015 #16
    Ok, so is it okay to set Pc=rho*u^2 without knowing if the flow is viscous dominated? Also, in relation to what you were saying, if the h is not much smaller than L, then it seems that the y direction equation doesn't boil down to only hydrostatics...then we'd have a problem. Anyway, I think I'm getting it!
     
  18. Apr 29, 2015 #17
    Sure. Don't forget that the shear stress is equal to the friction factor times rho*u^2 even in laminar flow. Another possible choice is ηu/h. If you have time, why don't you try that and see if it leads to anything interesting.

    I wouldn't characterize this as having a problem. The problem is just more 2D in character, and you just can't use the same simplifying approximation to solve the equations. That doesn't mean that they can't be solved.

    Chet
     
  19. May 2, 2015 #18
    This is a very interesting point, about how even in laminar flow the shear stress is proportional to rho*u^2! I have tried out the other non-dimensionalization with the viscous dominated flows and I just get a 1/Re with the pressure term.


    Hopefully none of my problems on the exam should end up with that :-)

    Thanks for your help,Chet!
     
  20. May 2, 2015 #19
    Actually, no. Don't forget that the friction factor is inversely proportional to the Reynolds number, so the dimensional pressure in rectilinear laminar flow is directly proportional to u.

    Chet
     
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