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Particle settling

  1. Dec 3, 2015 #1
    Hello!
    We have a tube with air flowing horizontally and we put some particles of different diameter.
    What are the equations that will describe their motion?
    We have the weight and buoyancy acting vertically, but there will be drag force from the stream of air, with unknown size and dimension, any hint?
    thanks!
     
  2. jcsd
  3. Dec 4, 2015 #2
    It depends on what you can neglect. If your particles are small and have a high density compared to the gas density, the equation of motion for a particle moving with velocity U in a gas that has a local velocity V (at the location of the particle) can be reduced to:
    [itex]m_p \frac{dU_i}{dt} = m_pg_i + m_p \frac{V_i - U_i}{\tau_p}[/itex], with the particle relaxation time
    [itex]\tau_p = \frac{\rho_p d_p^2}{ 18 C \mu_g}[/itex]

    If your particle Reynolds number is low (particles moving with approximately the gas velocity), you are in the Stokes regime and the constant C is simply 1. If your Reynolds number is higher, you need a correction factor. For small particles, the Reynolds number usually does not become very large because they quickly adjust to the gas velocity V (usually Re<1000). A simple correction factor based on the particle Reynolds number is
    [itex]C=1+ 0.15 Re_p^{0.687}[/itex]

    Some side remarks:
    Very close to a wall, you will have some extra forces due to the velocity gradients.
    In a closed tube where the pressure gradient in the gas is not negligible, an extra force due to pressure gradients enters the equation of motion.
    When your flow is turbulent and you are using a turbulence model to determine the gas velocities, you do not have the instantaneous velocity of the gas at the location of the particle available. In that case, the model for [itex]V_i[/itex] is very complicated.

    Hope this helps.
     
    Last edited: Dec 4, 2015
  4. Dec 4, 2015 #3
    I don''t understand your question, you use many symbols without explaining them.

    Could you answer my following questions please?

    1) The drag force is calculated by either Stoke's law or Drag Force equation, depending on the particles and Re. Because we are talking about small particles in low velocities, we only apply Stokes law and NOT Drag Force equation (with the Drag Coefficient in it). Right?

    2) The equation of Stoke's Drag Force shows that Force is proportional to the relative velocity of the fluid to the particle. Is this a vector equation?? Ie. what if the particle moves down and the fluid moves horizontally?

    3) What is the point to calculate terminal velocity? How is terminal velocity and horizontal displacement relate?
     
  5. Dec 4, 2015 #4
    I don't have a question, maybe you mean something else? I'm sorry, I assumed you were already familiar with the usual notation: m=mass,g=gravity,V (and U)=velocity,d=diameter,[itex]\mu[/itex]=viscosity,[itex]\rho[/itex]=density. The subscript 'p' stands for particle, 'g' for gas. The index 'i' means the component in direction 'i'.
    You always use a drag force equation, it is just that the drag force equation is Stokes drag when the Reynolds number is smaller than 1. If your Reynolds number is smaller than 1, then C=1 in the equation above.
    The equation of motion is a vector equation, you need an equation for each of the directions. In 3-D, you need an equation for the x,y, and z-direction. If a particle moves down and the fluid moves horizontally, then it can move in at least 2 directions, so you need at least 2 equations of motion.
    What do you mean with 'what is the point'? Terminal velocity says that the net forces on the particle are zero, so in your example, assuming your particle is much heavier than air and there are no pressure gradients, the gravitational force balances the drag force in the vertical direction. In horizontal direction, the only force (in my equation at least) is the drag force, and the drag force is zero when the particle moves with the local gas velocity. The local gas velocity is therefore the terminal velocity in the horizontal direction.
     
  6. Dec 4, 2015 #5
    sorry I didn't mean question, but answer

    I am familiar with that notation, but I don't understand what τ is, I have never seen that equation.

    From what I gathered, in Stoke's regime the drag force is: https://upload.wikimedia.org/math/0/8/f/08ffd06b321494dbc7f23908dae6e978.png

    In Newton's regime the drag force is: https://upload.wikimedia.org/math/5/2/6/52694ea39410aa410b12afef2746d8a4.png

    Am I right?

    If yes, then how do I find the horizontal travel distance of the particles in my first post example?

    I thought that I know the time of the journey (from the vertical kinematic equations, since I know the height of the fall). Then I some how (integration?) find the horizontal displacement. Any hint?
     
  7. Dec 5, 2015 #6
    Also, can you tell me please, is it ok to consider that the particle achieves terminal velocity almost instantly and thus it travels horizontally with that velocity constant?
     
  8. Dec 6, 2015 #7
    if you write [itex]m=\rho \frac{4}{3}\pi r^3 [/itex], you can rewrite [itex]m\frac{V-U}{\tau}[/itex] to an expression similar to your Stokes drag term. Your second equation is simply the definition of the drag coefficient. You can rewrite the Stokes drag to this form using the definition of the Reynolds number, and you will get the Stokes drag coefficient.

    [itex]\tau[/itex] is introduced because its dimension is time. It is the particle response time and it gives an estimate of the time that the particle needs to adjust to the gas velocity. If it is small, you can assume that the particle quickly reaches its terminal velocity.

    What do you really want to know? Do you release a particle in the middle of a horizontal tube with velocity zero, and you want to know when/where the particle will hit the bottom wall? I think these equations already give you the answer you seek.
     
  9. Dec 6, 2015 #8
    OK you say we can assume that the particle reaches its terminal velocity quickly.

    But the terminal settling velocity given by this equation https://upload.wikimedia.org/math/8/7/3/873c9ce3c729789cf01c2f71103a51f0.png
    is about particles that freely fall in a steady fluid, not for particles that fall in a horizontally moving fluid!

    So what is that velocity exactly? The horizontal or the vertical velocity?
    The horizontal velocity can be assumed to be the velocity of the horizontal air flow.
    The vertical velocity???
     
  10. Dec 7, 2015 #9
  11. Dec 10, 2015 #10
    Sorry for late reply.
    I don't know the equations, that's what I am after.

    Can you tell me please how to calculate the time needed for a particle for its vertical velocity to reach the settling velocity?
     
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