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Navier-Stokes Problem

  1. Apr 25, 2016 #1

    joshmccraney

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    Hi PF!

    Assume we have a rectangular cavity (2D) filled with a liquid of dimensions ##L \times H## and that the top plate of the cavity moves with some velocity ##V_0##. Also assume ##L \gg H##. I'll also assume ##L \gg H## implies flow is roughly 1-dimensional, and thus a pressure gradient drives the flow near the still wall in the opposite direction rather than the ##y## component that would "bounce" off the vertical cavity walls. When computing the velocity profile, from the Navier-Stokes equation I get $$v_x \frac{\partial v_x}{\partial x} = -\frac{\partial P}{\partial x} + \mu \frac{\partial^2 v_x}{\partial y^2} + \mu \frac{\partial^2 v_x}{\partial x^2}$$ subject to ##v_x(0) = v_0## and ##v_x(H) = 0##, where I assume the flow is fully developed and steady so ##v_x = f(y)##. Notice now the equation reduces to $$\frac{d P}{d x} = \mu \frac{d^2 v_x}{d y^2}$$

    Clearly all we now must do is solve for the pressure gradient since the boundary conditions determine the two integration constants. To find this, I imagine the net flow rate must be zero since no flow enters or leaves. Thus $$\frac{1}{HL} \int_0^L \int_0^H \, v_x dy \, dx = 0$$. Then we should be able to solve for ##\frac{d P}{d x}##.

    Does this look right to you? Thanks for your time!
     
  2. jcsd
  3. Apr 25, 2016 #2

    boneh3ad

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    Do you mind drawing a picture? I am not sure if I follow your description of the geometry completely.
     
  4. Apr 25, 2016 #3
    This looks really good. This is the kind of flow that exists within the flights of screw pumps and screw extruders. See the book Processing of Thermoplastic Materials by Ernest Bernhardt
     
    Last edited: Apr 25, 2016
  5. Apr 26, 2016 #4

    joshmccraney

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    I've attached a rough sketch. I can add detail if necessary.

    Thanks! I'll have to check this out. So you agree with the integral constraint?
     

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  6. Apr 26, 2016 #5
    Yes.
     
  7. Apr 26, 2016 #6

    joshmccraney

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    Cool. So if we weren't to assume that ##y## velocity was negligible, would our equations be $$\rho\left( v_x\frac{\partial v_x}{\partial x} + v_y \frac{\partial v_x}{\partial y} \right) = \mu \frac{\partial^2 v_x}{\partial y^2}\\ \rho\left( v_x\frac{\partial v_y}{\partial x} + v_y \frac{\partial v_y}{\partial y} \right) = \mu \frac{\partial^2 v_y}{\partial y^2} + \rho g$$ where the pressure gradient is gone since it no longer drives flow. Boundary conditions would be all velocities equal zero along all walls except the top moving wall, where ##v_x=v_0##. The integral constraint holds as well, but this time for both velocity components.

    But then we have 5 conditions to meet (the four boundaries plus the integral constraint). This now seems over specified. Any ideas?
     
  8. Apr 26, 2016 #7
    The pressure gradients are not zero. They are significant.
     
  9. Apr 26, 2016 #8

    joshmccraney

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    Ok, so then this problem is solvable, since we have the unknown pressure gradient vector and the integral constraint over the velocity vector.

    Pressure to me is not intuitive; how would I know to include it in this analysis? I knew it was relevant in the first (simpler) part since something had to drive flow backwards.
     
  10. Apr 26, 2016 #9
    You don't use the integral constraint on this. The velocity components on all 4 boundaries are zero.
    You are asking how to solve the equations for the 2D version, including the pressure gradients?
     
  11. Apr 26, 2016 #10

    joshmccraney

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    Why wouldn't we use the integral constraint? Also, the velocity is zero everywhere except the moving side, right?

    No, I think I understand how to handle the math once it's there. The difficulty is setting up the equation, and in this instance, why the pressure gradient is even there.
     
  12. May 3, 2016 #11

    joshmccraney

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    Hey Chet, did you forget about me? :frown:
     
  13. May 12, 2016 #12
    The qualitative nature of the solution does not change when you just make the channel deeper.
     
  14. May 12, 2016 #13
    It is automatically satisfied when you satisfy the boundary conditions on the edge faces.
    yes.
    The thing to do on this problem is to first consider the case where the inertial forces are very low, so that they can be neglected. That is, at very low Reynolds number. If you can't solve it for this (linear) case, you'll never be able to solve for the case where inertia is included (non-linear case). What are your equations in the case where inertia is neglected?
     
  15. May 12, 2016 #14

    joshmccraney

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    I believe we have two, one for each velocity component. Let's say ##u## is velocity in ##x-##direction and ##v## is velocity in ##y##-direction. Then we have $$\frac{\partial P}{\partial x} = \mu \left( \frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 u}{\partial x^2} \right)\\
    \frac{\partial P}{\partial y} = \mu \left( \frac{\partial^2 v}{\partial y^2}+\frac{\partial^2 v}{\partial x^2} \right)$$
    I believe both derivatives on the right hand side are relevant since we are not assuming ##y \ll x##, which invalidates the scaling argument. I am neglecting gravity though.

    Does this look right to you?
     
  16. May 12, 2016 #15
    Yes. Now, in addition to these, you have the continuity equation. Please include that.
     
  17. May 12, 2016 #16

    joshmccraney

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    Of course! That's $$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$$Boundary conditions are no slip on the four walls, which implies velocity is zero along the cavity except the top wall, where ##u=U_0## which is the speed of the top shelf.

    So we have three equations and three unknowns (##P,u,v##). We have 8 boundary conditions (4 for ##u## and 4 for ##v##, each evaluated along the wall.
     
  18. May 12, 2016 #17
    Here's a way of proceeding further with this problem: Define a stream function ##\psi## for the flow as follows:
    $$u=\frac{\partial \psi}{\partial y}$$
    $$v=-\frac{\partial \psi}{\partial x}$$
    These equations satisfy the continuity equation exactly. What do you get if you substitute them into your two differential equations?

    What do you think the streamline pattern looks like for this flow? The streamlines are the lines of constant ##\psi##.
     
  19. May 13, 2016 #18

    joshmccraney

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    Oh yea, the stream function ##\psi \equiv \nabla \times \vec{v}##. Plugging these in yields third derivatives and mixed partials of ##\psi##.

    $$\frac{\partial P}{\partial x} = \mu \left( \frac{\partial^3 \psi}{\partial y^3}+\frac{\partial^3 \psi}{\partial x^2 \partial y} \right)\\
    \frac{\partial P}{\partial y} = -\mu \left( \frac{\partial^3 \psi}{\partial y^2 \partial x}+\frac{\partial^3 \psi}{\partial x^3} \right)$$

    I have no idea what they look like? Maybe a circle roughly?
     
  20. May 13, 2016 #19
    Good. Now take the partial derivative of the first equation with respect to y and subtract the partial derivative of the second equation with respect to x. What do you get?
    Very good. So you envision that there will be a recirculating flow in the cavity, and the streamlines will each be closed.
     
  21. May 13, 2016 #20

    joshmccraney

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    $$\frac{\partial^4 \psi}{\partial x^4} + \frac{\partial^4 \psi}{\partial y^4}+2\frac{\partial^4 \psi}{\partial x^2 \partial y^2}=0$$

    which I believe is separable. I've worked with similar equations before, where separability exists but was unusual. The mixed partial evokes such equations. Letting ##\psi = X(x)Y(y)## implies $$\frac{d_x^4X}{X}+ \frac{d_y^4 Y}{Y}+2 \frac{X''Y''}{XY}=0$$
    Hmmmmm tricks that have worked in the past fail now. Any ideas? By inspection sines and cosines would all differentiate back to original functions. Perhaps a good ansatz is to let ##X=\sin(ax)## or ##X=\cos(ax)## and ##Y=\sin(cy)## or ##Y=\cos(cy)##. Then we have an algebraic expression that behaves like ##a^4+c^4+2c^2c^2=0##. Quadratic formula could be used to represent one in terms of the other, but before continuing does this idea look right to you?

    Another option is to let ##Y## and ##X## be linear functions. And rethinking the above, exponentials would be easier to work with than trig, but trig stood out first.
     
    Last edited: May 14, 2016
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