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Navier-Stokes why qg=0?

  1. Apr 5, 2017 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    Navier-Stokes equations of motion

    3. The attempt at a solution

    I did everything well but, my question is, why we assume last term rho*gz=0? in the N-S equation?

    Also why do we use Navier Stokes equations in terms of velocity gradients for newtonian fluid with constants rho and mu, and not Navier Stokes in terms of Tau?

    When to use Navier Stokes in terms of Tau, and when to use Navier Stokes in terms of Rho and Mu?
    Last edited: Apr 5, 2017
  2. jcsd
  3. Apr 5, 2017 #2
    They are defining P as ##P=p-\rho g_zz## to simplify the equation.

    The equations of motion in terms of tau's won't allow you to easily solve for the velocities in complicated problems. And they don't directly take into account the viscous behavior of the fluid.
  4. Apr 6, 2017 #3
    Thank you for your reply, I have 2 questions if you don't mind

    1)why ρgz=0? on the right hand size, what is the phenomena behind, so they make that assumption

    2) On the left hand side on the Navier Stokes equation for z component in terms of velocity gradients with constant ρ and μ, I don't get why they make all the left hand side terms=0?.
    Like everything on the left side=0. I thought δVz / δx different from 0. I knew that Vz(x), so it should not be 0. Also what happens with ρ term on left hand side in Navier Stokes, is it also =0?
  5. Apr 6, 2017 #4
    Who says it was taken to be zero? What does the equation I wrote mean to you.
    The flow is assumed to be occurring at a constant downward velocity with a balance between viscous drag, gravitiational forces, and pressure forces.
    If is different from zero. What is zero are the velocities in the x and y directions, the derivative of Vz with respect to z, and the derivative of Vz with respect to time.
  6. Apr 6, 2017 #5
    1)P=p-ρgz=p+ρgh, I saw that in the solution they cut out that term.

    2)So how exactly does this affect the left hand side term, that it makes it =0? Means no accumulation?
    Why instead of 0=right hand side, why is it not ρ*δVz / δx= right hand side

    3)I understood that δVz / δx ≠0 and the rest are 0 but , why does it not stay there then, on the left side if it is ≠0?
    Last edited: Apr 6, 2017
  7. Apr 6, 2017 #6
    What do you mean that they cut out that term? The equation they were solving was: $$0=-\frac{dP}{dz}+\frac{d^2 v_z}{dx^2}$$ where $$P=p-\rho g z$$In this way the analysis automatically includes a pressure gradient in the z direction, a downward gravitational force on the fluid, and/or a combination of both.
    How can there be accumulation if the fluid is incompressible?
    The term you are asking about is $$\rho v_x \frac{\partial v_z}{\partial x}$$This term is zero because ##v_x=0##
    As I said, it is multiplied by ##v_x##, which is zero.

    Please take the time to learn LaTex. I expect you to have learned it by the end of tomorrow. Otherwise, I'm not sure I can continue to answer your questions.
  8. Apr 7, 2017 #7
    Alright, understood.

    Last questions, if you don't mind

    1)What is the purpose of these Navier-Stokes equations, what exactly are they used for, what can we get by using them?

    2)Could you please give me an example, when to use Navier-Stokes equations?

    3)When to know when to use the Navier-Stokes equations?What exactly is the criteria that determines we need to use Navier-Stokes equations?
  9. Apr 7, 2017 #8
    The Navier Stokes equations are extremely useful and are employed abundantly in practice. They apply to laminar flow. Suppose you have a fluid flowing in laminar flow through a pipe, or a blood vessel, or an IV capillary, and you wanted to know the relationship between the volumetric flow rate and the pressure drop. Suppose you had molten polymer flowing through a complicated piece of processing equipment like a film casting die, and you needed to design the die so that the flow out the die lips produces a uniform film, and the flow within the die does not have any regions where polymer can thermally degrade (thus compromising product quality). Suppose you are doing atmospheric science, and you need to know the velocity at which water droplets are falling through the air. Suppose you have a piece of equipment that is transferring heat to a fluid. Before you quantify the heat transfer at the wall and the temperature distribution within the fluid, you first need to know the flow velocity distribution.

    This is just the tip of the iceberg.
  10. Apr 8, 2017 #9
    Are the Navier Stokes equations applicable to both momentum and mass(mole) transfer? or just to only 1 of them?
  11. Apr 8, 2017 #10
    What are you getting at here? Can you elaborate?
  12. Apr 8, 2017 #11
    How to know when to use Navier-Stokes equations? Could you explain this please? Could you give me some examples of tasks that involve using Navier-Stokes, or when they are required to get involved?
  13. Apr 8, 2017 #12
    I feel that I adequately addressed this in post #8. All I'm saying is that, in any process that involves fluid flow, it is likely that the Navier Stokes equations are going to be employed in design and improvement. And Computational Fluid Dynamics (CFD) packages are commercially available to solve the Navier Stokes equations for systems involving complicated flow geometry.
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