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Cylindrical rod velocity profile

  1. Apr 5, 2017 #1
    1. The problem statement, all variables and given/known data
    cylindrical rod.JPG
    2. Relevant equations
    0=viscous forces
    0=in-out+sum of forces(0)
    2*pi*r*L*τrz evaluated at r-2*pi*r*L*τrz evaluated at r+ΔR (eq1)

    Dividing by volume and taking limit as δr->0 in eq1 I get :

    (δτrz)/(δr) =0

    My question is, how did they in solution got
    (rδτrz)/(δr) =0, how do they have that additional r in the numerator?

    3. The attempt at a solution

    cylindrical rod 2.JPG
  2. jcsd
  3. Apr 5, 2017 #2
    My question is: how did they in solution got (rδτrz)/(δr) =0, how do they have that additional r in the numerator?

    Because when dividing by volume and taking limit as Δr->0 I get (δτrz)/(δr) =0, so my solution is lacking that r in the numerator, compared with theirs, but why, what is wrong?
  4. Apr 5, 2017 #3
    Suppose they wrote the force balance equation the way it should have been written, as
    $$2\pi rL [\tau_{rz}]_r-2\pi (r+\Delta r)L [\tau_{rz}]_{r+\Delta r}=0$$
    Would this make any more sense to you?
  5. Apr 7, 2017 #4
    Thank you.

    I know that in order to determine the surface, the axis needs to be perpendicular to the plane.

    Can you also tell me how to determine the flow direction in a cylinder or sphere?What is the criteria?

    And also how to determine the pressure direction in cylinder or sphere?What is the criteria?

    I do not want to go by intuition, I want to understand how to identify the flow direction and pressure direction.
  6. Apr 7, 2017 #5
    I don't understand these questions. Can you discuss in terms of a specific situation?

    By the way, with regard to our other thread on a film sliding down an inclined plane, I see that you are using BSL. I was trying to teach you a little very useful and simple tensor analysis in that thread (starting with the little trick I showed you), but the material I was going to cover is all done very nicely in Appendix A3 of BSL. Please read over this appendix, and get back to me with questions.
  7. Apr 7, 2017 #6
    Thank you for your reply.

    In the below picture(Fig1) it is pretty much intuitive that the flow will be on y surface in x direction, and that P(y). Capture.JPG
    fig1-illustrates falling film

    However in the following pictures, how does he know that the velocity and pressure is like that(please see the highlighted text)

    fig2-illustrates The space between two coaxial cylinders is filled with an incompressible fluid at constant temperature. The radii of the inner and outer wetted surfaces are κR and R, respectively. The angular velocities of rotation of the inner and outer cylinders are Ωi and Ωo. Determine the velocity distribution in the fluid and the torques on the two cylinders needed to maintain the motion

    fig3-illustrating A very viscous Newtonian fluid flows in the space between two concentric spheres (as shown in Figure 2). It is desired to find the rate of flow in the system as a function of the imposed pressure difference. Neglect end effects and postulate that vθ depends only on r and θ with the other velocity components zero.

    fig4-is the solution for fig3

    For example in fig 4(above) how did they know the velocity is Vθ and that it depends on r and θ?

    I know that by using continuity equation we can show that velocity does not depend on a specific direction.

    I also know how to define and find the surfaces because the axis needs to be perpendicular on the plane.Take the cube in fig1 for example, the x surface corresponds to the left hand side square, y surface corresponds to the top square of the cube, and z surface goes into the board.

    I want to understand and deduce how the velocity and pressure are acting on different surfaces and directions, without guessing, or without intuition.

    How did they knew that is like that(the yellow highlighted information)?What are the criterias?
    Last edited: Apr 7, 2017
  8. Apr 7, 2017 #7
    The pressure is a function of y by applying the NS equation (or the equation of motion) in the y direction.
    The first equations follow from the no-slip boundary condition. The condition that V is a function only of r follows from symmetry.
    It follows from symmetry.
    The latitudinal velocity being a function independent of longitude follows from symmetry.
    Step 1. Consider the symmetry
    Step 2. Write down the equation of motion for a direction perpendicular to what you think is the main direction of flow. This can give you information about the velocity in that direction and how the pressure varies.
    Step 3. Consider that at the interface between two media, the components of the stress vector (or, equivalently, the momentum flux) are continuous. If the free surface is air, then the stress vector is zero and the momentum flux is zero.
    Step 4. Gain experience doing lots of problems. You'll get the idea.
  9. Apr 7, 2017 #8
    Thank you fir your reply!

    I know that the no slip boundary condition implies that the velocity at the wall=0.

    I will follow your recipe, but I do not understand what you mean by symmetry?Can you explain it in detail please? Or give some examples on this term(symmetry)?
  10. Apr 7, 2017 #9
    Furthermore, could you please tell me how you know(what is the criteria) that is on y direction and not on x or z ?
  11. Apr 7, 2017 #10
    In the concentric rotating cylinders example, the geometry is the same at all values of ##\theta##. So symmetry means, "how does the fluid know what value of ##\theta## it is at? The flow has to be the same at all values of ##\theta##.

    In the concentric spheres example, the geometry is the same at all values of longitude. So symmetry means, "how doe the fluid know what longitude it is at." The flow has to be the same at all values of longitude.
  12. Apr 7, 2017 #11
    The geometry is exactly the same at all values of x and z. so how does the fluid know which way to vary with x and z? The velocity in the y direction must be zero, or the fluid would flow through the wall. The flow in the z direction is zero, because the fluid would not know which way to go. If the velocity is zero in the y direction, from the NS equation in the y direction, ##\frac{dp}{dy}=\rho g_y ##, subject to the boundary condition ##p=p_{air}## at y = 0.
  13. Apr 8, 2017 #12
    I still did not fully understood it.

    Does this mean that we can identify the velocity profile direction just by observing where the geometry is the same?
  14. Apr 8, 2017 #13
    I don't know how to explain it any better. Just put yourself in the place of the fluid, and ask yourself which way you would flow.
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